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Merge pull request #1476 from codeflash-ai/prepare-for-cutting-cli-to-public-repo

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move files from cli/experiments and cli/pie_test_set to top-level experiments directory
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Saurabh Misra 2025-02-12 19:47:44 -08:00 committed by GitHub
commit 5851ec2b84
863 changed files with 0 additions and 23265 deletions
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0
experiments/__init__.py
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1836
experiments/analysis_experiments.ipynb
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experiments/bench_analytics.ipynb
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{
"cells": [
{
"metadata": {
"collapsed": true,
"ExecuteTime": {
"end_time": "2024-10-14T00:25:56.438384Z",
"start_time": "2024-10-14T00:25:50.078208Z"
}
},
"cell_type": "code",
"source": [
"from codeflash.verification.test_results import InvocationId\n",
"from codeflash.models.models import OriginalCodeBaseline\n",
"import dill as pickle\n",
"\n",
"data: list[OriginalCodeBaseline] = pickle.load(open(\"/Users/renaud/Desktop/baseline100.pkl\", \"rb\"))\n",
"invocation_ids = {\n",
" function_test_invocation.id for function_test_invocation in data[0].overall_test_results} # The first run represents the Oracle.\n",
"\n",
"# Timing results where the test passed, and the runtime is not None or 0.\n",
"usable_runtime_results: list[dict[InvocationId, dict[int, int]]] = [{invocation_id: {\n",
" function_test_invocation.loop_index: runtime for function_test_invocation in result.overall_test_results if (\n",
" runtime := function_test_invocation.runtime) and function_test_invocation.id == invocation_id and function_test_invocation.did_pass}\n",
" for invocation_id in invocation_ids}\n",
" for result in data]\n",
"valid_invocation_ids = {invocation_id for invocation_id in invocation_ids if usable_runtime_results[0][invocation_id]}\n",
"# A run is invalid if one of its test invocations has no valid result.\n",
"nonempty_runtime_results: list[dict[InvocationId, dict[int, int]]] = [{invocation_id: run_runtimes[invocation_id] for invocation_id in valid_invocation_ids} for run_runtimes in usable_runtime_results if all(run_runtimes[invocation_id] for invocation_id in valid_invocation_ids)]\n",
"\n",
"run_min_runtimes = [{invocation_id: min(runtimes[invocation_id].values()) for invocation_id in runtimes} for runtimes in nonempty_runtime_results]\n",
"run_total_runtimes = [sum(test_invocation_runtimes.values()) for test_invocation_runtimes in run_min_runtimes]\n",
"run_total_runtimes2 = [result.runtime for result in data]\n",
" \n",
"print(f\"Timing calculations are consistent: {run_total_runtimes == run_total_runtimes2}\")\n",
"print(run_total_runtimes)"
],
"id": "initial_id",
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Timing calculations are consistent: True\n",
"[3427917, 3397916, 3395124, 3390412, 3303873, 3468999, 3434249, 3445252, 3342791, 3444248, 3549496, 3285623, 3617039, 3444914, 3540250, 3410374, 3539542, 3292583, 3413747, 3453915, 3597335, 3391166, 3355912, 3691122, 3467460, 3682375, 3522458, 3463334, 3717790, 3595706, 3375916, 3307706, 3388250, 3403586, 3393580, 3393750, 3369835, 3489581, 3363870, 3478123, 3417915, 3427583, 3390582, 3588542, 3508168, 3461457, 3479496, 3316957, 3461040, 3429001, 3650874, 3484789, 3667246, 3483750, 3358830, 3448291, 3456958, 3415290, 3181582, 3443668, 3361624, 3640580, 3410539, 3475081, 3510458, 3516707, 3369163, 3379706, 3694418, 3376625, 3485831, 3372290, 3424334, 3461540, 3630829, 3665957, 3474542, 3289749, 3358750, 3204707, 3449957, 3335665, 3364667, 3466831, 3616958, 3614122, 3543041, 3316167, 3466373, 3423167, 3403418, 3409211, 3402127, 3360996, 3388913, 3662916, 3423126, 3655789, 3287874, 3470374]\n"
]
}
],
"execution_count": 1
},
{
"metadata": {
"ExecuteTime": {
"end_time": "2024-10-14T00:26:01.268153Z",
"start_time": "2024-10-14T00:26:01.253449Z"
}
},
"cell_type": "code",
"source": [
"from typing import Callable, SupportsFloat\n",
"from codeflash.code_utils.time_utils import humanize_runtime\n",
"import numpy as np\n",
"from numpy.typing import ArrayLike\n",
"\n",
"NumberType = type[SupportsFloat]\n",
"\n",
"def analyze_num_array(\n",
" num_array: ArrayLike,\n",
" formatter: Callable[[NumberType], str]\n",
")-> None:\n",
" array = np.array(num_array)\n",
" \n",
" mean = np.mean(array)\n",
" max_value = np.max(array)\n",
" min_value = np.min(array)\n",
" median = np.median(array)\n",
" std_dev = np.std(array)\n",
" \n",
" percentages = [0, 5, 25, 50, 75, 95, 100]\n",
" percentiles = np.percentile(array, percentages)\n",
" q1 = percentiles[2] # 25th percentile\n",
" q3 = percentiles[4] # 75th percentile\n",
" iqr = q3 - q1\n",
" outlier_min = (q1 - 1.5 * iqr)\n",
" outlier_max = (q3 + 1.5 * iqr)\n",
" small_outliers = sorted([value for value in array if value < outlier_min])\n",
" large_outliers = sorted([value for value in array if value > outlier_max])\n",
" \n",
" print(f\"Mean +- std dev: {formatter(mean)} +- {formatter(std_dev)}\")\n",
" print(f\"Max: {formatter(max_value)}\")\n",
" print(f\"Median: {formatter(median)}\")\n",
" print(f\"Min: {formatter(min_value)}\")\n",
" print()\n",
" for i, percentage in enumerate(percentages):\n",
" print(f\"{percentage}th percentile: {formatter(percentiles[i])} (\"\n",
" f\"{(percentiles[i] - mean) / mean:.0%} of the mean)\")\n",
" print()\n",
" # Outliers\n",
" print(f\"Small outliers (< {formatter(outlier_min)}): {[formatter(outlier) for outlier in small_outliers]}\")\n",
" print()\n",
" print(f\"Large outliers (> {formatter(outlier_max)}): {[formatter(outlier) for outlier in large_outliers]}\")\n",
" print()\n",
" print(f\"Total number of outliers: {len(small_outliers) + len(large_outliers)}\")\n",
" print(f\"Number of small outliers: {len(small_outliers)}\")\n",
" print(f\"Number of large outliers: {len(large_outliers)}\")\n",
" \n",
"analyze_num_array(run_total_runtimes, humanize_runtime)"
],
"id": "5dcd4d4ae5288f1d",
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Mean +- std dev: 3.45 milliseconds +- 110 microseconds\n",
"Max: 3.72 milliseconds\n",
"Median: 3.44 milliseconds\n",
"Min: 3.18 milliseconds\n",
"\n",
"0th percentile: 3.18 milliseconds (-8% of the mean)\n",
"5th percentile: 3.29 milliseconds (-5% of the mean)\n",
"25th percentile: 3.39 milliseconds (-2% of the mean)\n",
"50th percentile: 3.44 milliseconds (-0% of the mean)\n",
"75th percentile: 3.49 milliseconds (1% of the mean)\n",
"95th percentile: 3.67 milliseconds (6% of the mean)\n",
"100th percentile: 3.72 milliseconds (8% of the mean)\n",
"\n",
"Small outliers (< 3.22 milliseconds): ['3.18 milliseconds', '3.20 milliseconds']\n",
"\n",
"Large outliers (> 3.66 milliseconds): ['3.66 milliseconds', '3.67 milliseconds', '3.67 milliseconds', '3.68 milliseconds', '3.69 milliseconds', '3.69 milliseconds', '3.72 milliseconds']\n",
"\n",
"Total number of outliers: 9\n",
"Number of small outliers: 2\n",
"Number of large outliers: 7\n"
]
}
],
"execution_count": 2
},
{
"metadata": {},
"cell_type": "markdown",
"source": "",
"id": "d58b61bf62ce2780"
},
{
"metadata": {
"ExecuteTime": {
"end_time": "2024-10-14T00:26:36.172976Z",
"start_time": "2024-10-14T00:26:36.166724Z"
}
},
"cell_type": "code",
"source": [
"run_loop_counts = [max([max(run_runtimes[invocation_id]) for invocation_id in run_runtimes]) for run_runtimes in nonempty_runtime_results]\n",
"\n",
"print(f\"Loop counts: {run_loop_counts}\")\n",
"print()\n",
"analyze_num_array(run_loop_counts, str)"
],
"id": "57c73a31483e06b5",
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Loop counts: [106, 127, 116, 127, 123, 136, 132, 131, 130, 130, 127, 131, 130, 133, 123, 126, 133, 129, 120, 135, 114, 121, 121, 125, 127, 123, 116, 134, 122, 118, 128, 118, 128, 124, 116, 126, 130, 124, 118, 131, 124, 130, 123, 93, 115, 131, 127, 128, 132, 133, 131, 128, 132, 124, 131, 132, 130, 135, 130, 129, 128, 124, 129, 121, 129, 131, 129, 126, 128, 115, 131, 120, 122, 115, 125, 119, 130, 126, 124, 87, 126, 123, 108, 124, 130, 119, 121, 125, 119, 128, 121, 108, 90, 121, 125, 109, 104, 128, 123, 104]\n",
"\n",
"Mean +- std dev: 123.62 +- 9.040774303122493\n",
"Max: 136\n",
"Median: 126.0\n",
"Min: 87\n",
"\n",
"0th percentile: 87.0 (-30% of the mean)\n",
"5th percentile: 105.9 (-14% of the mean)\n",
"25th percentile: 121.0 (-2% of the mean)\n",
"50th percentile: 126.0 (2% of the mean)\n",
"75th percentile: 130.0 (5% of the mean)\n",
"95th percentile: 133.0 (8% of the mean)\n",
"100th percentile: 136.0 (10% of the mean)\n",
"\n",
"Small outliers (< 107.5): ['87', '90', '93', '104', '104', '106']\n",
"\n",
"Large outliers (> 143.5): []\n",
"\n",
"Total number of outliers: 6\n",
"Number of small outliers: 6\n",
"Number of large outliers: 0\n"
]
}
],
"execution_count": 3
},
{
"metadata": {
"ExecuteTime": {
"end_time": "2024-10-14T00:26:46.184288Z",
"start_time": "2024-10-14T00:26:46.081388Z"
}
},
"cell_type": "code",
"source": [
"run_reach_min_loop = [max({invocation_id: min([loop_index for loop_index in run_runtimes[invocation_id] if run_runtimes[invocation_id][loop_index] == run_min_runtimes[run_index][invocation_id]]) for invocation_id in valid_invocation_ids}.values()) for run_index, run_runtimes in enumerate(nonempty_runtime_results)]\n",
"\n",
"print(f\"Loop count to reach min runtime: {run_reach_min_loop}\")\n",
"print()\n",
"analyze_num_array(run_reach_min_loop, str)"
],
"id": "2ab96deed074385d",
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Loop count to reach min runtime: [106, 127, 116, 125, 106, 135, 116, 131, 122, 107, 126, 127, 127, 131, 123, 124, 129, 122, 115, 127, 113, 112, 114, 108, 120, 114, 111, 127, 111, 113, 110, 118, 108, 77, 104, 121, 130, 116, 105, 111, 119, 127, 121, 93, 113, 117, 121, 98, 118, 110, 121, 118, 127, 114, 126, 125, 119, 129, 126, 84, 118, 122, 119, 101, 110, 114, 103, 119, 124, 108, 114, 115, 117, 103, 110, 93, 128, 119, 121, 72, 123, 90, 101, 110, 122, 111, 121, 107, 118, 126, 114, 107, 80, 99, 116, 104, 71, 101, 112, 86]\n",
"\n",
"Mean +- std dev: 113.6 +- 12.804686642007294\n",
"Max: 135\n",
"Median: 116.0\n",
"Min: 71\n",
"\n",
"0th percentile: 71.0 (-37% of the mean)\n",
"5th percentile: 85.9 (-24% of the mean)\n",
"25th percentile: 108.0 (-5% of the mean)\n",
"50th percentile: 116.0 (2% of the mean)\n",
"75th percentile: 122.0 (7% of the mean)\n",
"95th percentile: 129.0 (14% of the mean)\n",
"100th percentile: 135.0 (19% of the mean)\n",
"\n",
"Small outliers (< 87.0): ['71', '72', '77', '80', '84', '86']\n",
"\n",
"Large outliers (> 143.0): []\n",
"\n",
"Total number of outliers: 6\n",
"Number of small outliers: 6\n",
"Number of large outliers: 0\n"
]
}
],
"execution_count": 4
},
{
"metadata": {
"ExecuteTime": {
"end_time": "2024-10-14T01:41:58.090916Z",
"start_time": "2024-10-14T01:41:58.081916Z"
}
},
"cell_type": "code",
"source": [
"# Allocated benchmarking time is 5 seconds. Changing to 10 seconds so ajust when re-running.\n",
"time_reach_min_loop = [min_loop / loop_count * 5_000_000_000 for min_loop, loop_count in zip(run_reach_min_loop, run_loop_counts)] \n",
"print(f\"Times to reach min loop: {time_reach_min_loop}\")\n",
"print()\n",
"analyze_num_array(time_reach_min_loop, humanize_runtime)"
],
"id": "c8bc504cfcdaca6",
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Times to reach min loop: [5000000000.0, 5000000000.0, 5000000000.0, 4921259842.519685, 4308943089.430895, 4963235294.117647, 4393939393.939394, 5000000000.0, 4692307692.307693, 4115384615.3846154, 4960629921.259843, 4847328244.274809, 4884615384.615384, 4924812030.075188, 5000000000.0, 4920634920.63492, 4849624060.150376, 4728682170.542636, 4791666666.666667, 4703703703.703704, 4956140350.8771925, 4628099173.5537195, 4710743801.652892, 4320000000.0, 4724409448.818897, 4634146341.463415, 4784482758.620689, 4738805970.149254, 4549180327.868853, 4788135593.220339, 4296875000.0, 5000000000.0, 4218750000.0, 3104838709.677419, 4482758620.689655, 4801587301.587302, 5000000000.0, 4677419354.83871, 4449152542.372881, 4236641221.374046, 4798387096.774194, 4884615384.615384, 4918699186.99187, 5000000000.0, 4913043478.26087, 4465648854.961832, 4763779527.559055, 3828125000.0, 4469696969.69697, 4135338345.864661, 4618320610.687023, 4609375000.0, 4810606060.606061, 4596774193.548387, 4809160305.343512, 4734848484.848485, 4576923076.923077, 4777777777.777778, 4846153846.153846, 3255813953.4883723, 4609375000.0, 4919354838.709678, 4612403100.775193, 4173553719.0082645, 4263565891.472868, 4351145038.167939, 3992248062.015504, 4722222222.222222, 4843750000.0, 4695652173.913044, 4351145038.167939, 4791666666.666667, 4795081967.213115, 4478260869.565218, 4400000000.0, 3907563025.210084, 4923076923.076923, 4722222222.222222, 4879032258.064516, 4137931034.4827585, 4880952380.952381, 3658536585.365854, 4675925925.925926, 4435483870.967742, 4692307692.307693, 4663865546.218488, 5000000000.0, 4280000000.0, 4957983193.277311, 4921875000.0, 4710743801.652892, 4953703703.703704, 4444444444.444445, 4090909090.909091, 4640000000.0, 4770642201.834863, 3413461538.4615383, 3945312500.0, 4552845528.455284, 4134615384.6153846]\n",
"\n",
"Mean +- std dev: 4.59 seconds +- 383 milliseconds\n",
"Max: 5.00 seconds\n",
"Median: 4.71 seconds\n",
"Min: 3.10 seconds\n",
"\n",
"0th percentile: 3.10 seconds (-32% of the mean)\n",
"5th percentile: 3.90 seconds (-15% of the mean)\n",
"25th percentile: 4.43 seconds (-4% of the mean)\n",
"50th percentile: 4.71 seconds (2% of the mean)\n",
"75th percentile: 4.88 seconds (6% of the mean)\n",
"95th percentile: 5.00 seconds (9% of the mean)\n",
"100th percentile: 5.00 seconds (9% of the mean)\n",
"\n",
"Small outliers (< 3.75 seconds): ['3.10 seconds', '3.26 seconds', '3.41 seconds', '3.66 seconds']\n",
"\n",
"Large outliers (> 5.56 seconds): []\n",
"\n",
"Total number of outliers: 4\n",
"Number of small outliers: 4\n",
"Number of large outliers: 0\n"
]
}
],
"execution_count": 8
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"name": "python3"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 2
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
"version": "2.7.6"
}
},
"nbformat": 4,
"nbformat_minor": 5
}

165
experiments/concolic_execution.py
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@ -1,165 +0,0 @@
from __future__ import annotations
import re
import subprocess
from enum import IntEnum
from typing import TYPE_CHECKING
from codeflash.cli_cmds.console import code_print, console, logger
if TYPE_CHECKING:
from pathlib import Path
from codeflash.discovery.functions_to_optimize import FunctionToOptimize
from codeflash.models.models import OptimizedCandidate
class DiffbehaviorReturnCode(IntEnum):
DID_NOT_RUN = -1
NO_DIFFERENCES = 0
COUNTER_EXAMPLES = 1
ERROR = 2
def run_concolic_correctness(
function_to_optimize: FunctionToOptimize,
function_to_optimize_original_worktree_fqn: str,
candidates: list[OptimizedCandidate],
worktrees: list[Path],
worktree_root: Path,
git_root: Path,
) -> None:
logger.info("Generating concolic coverage tests for the original code…")
original_code_coverage_tests = subprocess.run(
[
"crosshair",
"cover",
"--example_output_format=pytest",
"--max_uninteresting_iterations=256",
function_to_optimize_original_worktree_fqn,
],
capture_output=True,
text=True,
cwd=worktree_root,
check=False,
)
logger.info(f"Tests generated through concolic coverage:\n{original_code_coverage_tests.stdout}")
console.rule()
logger.info("Running concolic behavior correctness checking and coverage generation on optimized code…")
console.rule()
diffbehavior_results: dict[str, DiffbehaviorReturnCode] = {}
for candidate_index, candidate in enumerate(candidates, start=1):
logger.info(f"Optimization candidate {candidate_index}/{len(candidates)}:")
code_print(candidate.source_code)
function_to_optimize_optimized_worktree_fqn = (
str(
worktrees[candidate_index].name / function_to_optimize.file_path.relative_to(git_root).with_suffix("")
).replace("/", ".")
+ "."
+ function_to_optimize.qualified_name
)
result = subprocess.run(
[
"crosshair",
"diffbehavior",
"--max_uninteresting_iterations=256",
function_to_optimize_optimized_worktree_fqn,
function_to_optimize_original_worktree_fqn,
],
capture_output=True,
text=True,
cwd=worktree_root,
check=False,
)
optimized_code_tests = subprocess.run(
[
"crosshair",
"cover",
"--example_output_format=pytest",
"--max_uninteresting_iterations=256",
function_to_optimize_optimized_worktree_fqn,
],
capture_output=True,
text=True,
cwd=worktree_root,
check=False,
)
if result.returncode == DiffbehaviorReturnCode.ERROR:
diffbehavior_results[candidate.optimization_id] = DiffbehaviorReturnCode.ERROR
logger.info("Inconclusive results from concolic behavior correctness checking.")
logger.warning(f"Error running crosshair diffbehavior{': ' + result.stderr if result.stderr else '.'}")
elif result.returncode == DiffbehaviorReturnCode.COUNTER_EXAMPLES:
split_counter_examples = re.split("(Given: )", result.stdout)[1:]
joined_counter_examples = [
"".join(map(str, split_counter_examples[i : i + 2])) for i in range(0, len(split_counter_examples), 2)
]
concrete_counter_examples = "".join(
[elt for elt in joined_counter_examples if not re.search(r" object at 0x[0-9a-fA-F]+", elt)]
)
if concrete_counter_examples:
diffbehavior_results[candidate.optimization_id] = DiffbehaviorReturnCode.COUNTER_EXAMPLES
logger.info(
f"Optimization candidate failed concolic behavior correctness "
f"checking:\n{concrete_counter_examples}"
)
if result.stdout != concrete_counter_examples:
object_id_counter_examples = "".join(
[elt for elt in joined_counter_examples if re.search(r" object at 0x[0-9a-fA-F]+", elt)]
)
logger.warning(f"Counter-examples with object ID found:\n{object_id_counter_examples}")
else:
diffbehavior_results[candidate.optimization_id] = DiffbehaviorReturnCode.ERROR
logger.info("Inconclusive results from concolic behavior correctness checking.")
console.rule()
logger.warning(f"Counter-examples with object ID found:\n{result.stdout}")
elif result.returncode == DiffbehaviorReturnCode.NO_DIFFERENCES:
diffbehavior_results[candidate.optimization_id] = DiffbehaviorReturnCode.NO_DIFFERENCES
first_line = "".join([": ", chr(10), result.stdout.split(chr(10), 1)[0]])
logger.info(
f"Optimization candidate passed concolic behavior correctness checking"
f"{first_line if chr(10) in result.stdout else '.'}"
)
paths_exhausted = "All paths exhausted, functions are likely the same!\n"
if result.stdout.endswith(paths_exhausted):
logger.info(paths_exhausted)
else:
logger.warning("Consider increasing the --max_uninteresting_iterations option.")
else:
logger.info("Inconclusive results from concolic behavior correctness checking.")
logger.error("Unknown return code running crosshair diffbehavior.")
console.rule()
logger.info(f"Tests generated through concolic coverage:\n{optimized_code_tests.stdout}")
console.rule()
def report_concolic_results(
equal_results: bool, diffbehavior_result: DiffbehaviorReturnCode, success: bool
) -> tuple[bool, bool]:
if diffbehavior_result == DiffbehaviorReturnCode.NO_DIFFERENCES:
logger.info("Concolic behavior correctness check successful!")
console.rule()
if equal_results:
logger.info("True negative: Concolic behavior correctness check successful and test results matched.")
else:
logger.warning(
"False negative for concolic testing: Concolic behavior correctness check successful but test "
"results did not match."
)
console.rule()
elif diffbehavior_result == DiffbehaviorReturnCode.COUNTER_EXAMPLES:
logger.warning("Concolic behavior correctness check failed.")
console.rule()
if equal_results:
logger.warning(
"False negative for regression testing: Concolic behavior correctness check failed but test "
"results matched."
)
success = False
equal_results = False
else:
logger.info("True positive: Concolic behavior correctness check failed and test results did not match.")
console.rule()
elif diffbehavior_result == DiffbehaviorReturnCode.ERROR:
logger.warning("Concolic behavior correctness check inconclusive.")
console.rule()
return equal_results, success

286
experiments/metrics_analysis.py
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@ -1,286 +0,0 @@
import os
from typing import Any, Dict, Optional
import pandas as pd
from pandas import DataFrame
from scipy.stats import hmean
from sqlalchemy import create_engine
def load_data(experiment_id: str, database_uri: str = os.environ.get("DATABASE_URL")) -> DataFrame:
engine = create_engine(database_uri)
with engine.connect() as connection:
query = """
SELECT * FROM optimization_features
WHERE (trace_id LIKE %s OR trace_id LIKE %s)
AND experiment_metadata->>'id' = %s
"""
return pd.read_sql_query(query, connection, params=("%EXP0", "%EXP1", experiment_id))
def process_column_pairs(df: DataFrame, column_name: str) -> DataFrame:
"""Cleans up the given column by filling in missing values from the other column in the EXP0 / EXP1 pair.
:param df:
:param column_name:
:return:
"""
new_df = df.copy()
grouped = new_df.groupby(new_df["trace_id"].str[:-4])
for _, group in grouped:
if len(group) == 2:
exp0_row = group[group["trace_id"].str.endswith("EXP0")]
exp1_row = group[group["trace_id"].str.endswith("EXP1")]
if pd.isnull(exp0_row.iloc[0][column_name]) and not pd.isnull(exp1_row.iloc[0][column_name]):
new_df.at[exp0_row.index[0], column_name] = exp1_row.iloc[0][column_name]
elif pd.isnull(exp1_row.iloc[0][column_name]) and not pd.isnull(exp0_row.iloc[0][column_name]):
new_df.at[exp1_row.index[0], column_name] = exp0_row.iloc[0][column_name]
return new_df
def calculate_validity(df: DataFrame, perf_threshold: float = 0.05) -> Dict[str, Any]:
# Calculate the percentage of valid PRs given that the original function run succeeded
successful_runs = df[(~df["original_runtime"].isna())]
successful_runs_above_thres = successful_runs[successful_runs["best_correct_speedup_ratio"] >= perf_threshold]
valid_prs = len(successful_runs_above_thres)
percent_valid_pr = (valid_prs / len(df)) * 100
# Calculate the percentage of valid candidates generated given that original function run succeeded
valid_candidates = successful_runs[successful_runs["is_correct"].apply(lambda x: any(x.values()))]
percent_valid_candidates = len(valid_candidates) / len(successful_runs) * 100 if len(successful_runs) > 0 else 0
return {"percent_valid_pr": percent_valid_pr, "percent_valid_candidates": percent_valid_candidates}
def calculate_performance(df: DataFrame, perf_threshold: float = 0.05) -> Dict[str, Any]:
# Filter out the rows where a valid candidate above the perf threshold was found
valid_candidates_above_thres = df[(df["best_correct_speedup_ratio"] >= perf_threshold)]
# (1) Calculate the average speedup ratio of the PR
average_percentage_gain_pr = valid_candidates_above_thres["best_correct_speedup_ratio"].mean() * 100
# (1a) Calculate the harmonic mean of the PR speedup ratio with ref=1x
harmonic_mean_gain_pr = hmean(valid_candidates_above_thres["best_correct_speedup_ratio"] + 1)
# (2) Calculate the mean average speedup ratio of all the valid candidates
mean_average_percentage_gain_all = df["best_correct_speedup_ratio"].mean() * 100
# (2a) Calculate the harmonic mean of all candidates speedup ratio with ref=1x
harmonic_mean_gain_all = hmean(df["best_correct_speedup_ratio"].dropna() + 1)
def calculate_time_saved_for_row(row: pd.Series):
if row["optimized_runtime"] is not None and row["is_correct"] is not None:
correct_runtimes = [
runtime for opt_id, runtime in row["optimized_runtime"].items() if row["is_correct"].get(opt_id)
]
else:
correct_runtimes = []
if correct_runtimes:
return row["original_runtime"] - min(correct_runtimes)
return None
# (3) The average time saved in a PR given that a valid candidate was found above the perf threshold.
pr_time_saved = (
valid_candidates_above_thres.apply(lambda row: calculate_time_saved_for_row(row), axis=1).dropna().mean()
)
# (4) Calculate the mean average time saved for all the valid candidates
all_candidates_time_saved = df.apply(lambda row: calculate_time_saved_for_row(row), axis=1).dropna().mean()
return {
"average_percentage_gain_pr": average_percentage_gain_pr,
"harmonic_mean_gain_pr": harmonic_mean_gain_pr,
"mean_average_percentage_gain_all": mean_average_percentage_gain_all,
"harmonic_mean_gain_all": harmonic_mean_gain_all,
"average_time_saved_pr": pr_time_saved,
"mean_average_time_saved_all": all_candidates_time_saved,
}
def calculate_coverage(df: DataFrame) -> Dict[str, Any]:
successful_runs = df[~df["original_runtime"].isna()]
def calculate_percent_optimization_successful_runs(opt_runs: Dict[str, Optional[float]]) -> float:
if opt_runs is None:
return 0.0
total_runs = len(opt_runs)
successful_optimization_runs = sum(1 for runtime in opt_runs.values() if runtime is not None)
return successful_optimization_runs / total_runs * 100 if total_runs > 0 else 0.0
df["percent_successful_optimization_runs"] = df["optimized_runtime"].apply(
calculate_percent_optimization_successful_runs
)
total_optimizations = sum(len(runs) for runs in successful_runs["optimized_runtime"] if runs is not None)
successful_optimizations = sum(
len([runtime for runtime in runs.values() if runtime is not None])
for runs in successful_runs["optimized_runtime"]
if runs is not None
)
average_percent_successful_optimization_runs = df["percent_successful_optimization_runs"].mean()
percent_successful_optimizations = (
successful_optimizations / total_optimizations * 100 if total_optimizations > 0 else 0
)
percent_successful_original_runs = len(successful_runs) / len(df) * 100
return {
"average_percent_successful_optimization_runs": average_percent_successful_optimization_runs,
"percent_successful_optimizations": percent_successful_optimizations,
"percent_successful_original_runs": percent_successful_original_runs,
}
def paired_comparison_coverage(
df: DataFrame, model_a_suffix: str = "EXP0", model_b_suffix: str = "EXP1"
) -> Dict[str, Any]:
paired_coverage_results = {"model_a_more_successful": 0, "equal_successful": 0, "model_b_more_successful": 0}
grouped = df.groupby(df["trace_id"].str[:-4])
for _, group in grouped:
if len(group) == 2:
model_a_row = group[group["trace_id"].str.endswith(model_a_suffix)]
model_b_row = group[group["trace_id"].str.endswith(model_b_suffix)]
if model_a_row["optimized_runtime"].values[0] is None:
model_a_success_count = 0
else:
model_a_success_count = sum(
1 for runtime in model_a_row["optimized_runtime"].values[0].values() if runtime is not None
)
if model_b_row["optimized_runtime"].values[0] is None:
model_b_success_count = 0
else:
model_b_success_count = sum(
1 for runtime in model_b_row["optimized_runtime"].values[0].values() if runtime is not None
)
if model_a_success_count > model_b_success_count:
paired_coverage_results["model_a_more_successful"] += 1
elif model_a_success_count < model_b_success_count:
paired_coverage_results["model_b_more_successful"] += 1
else:
paired_coverage_results["equal_successful"] += 1
return paired_coverage_results
# Implement coverage calculations here
def paired_comparison_validity(
df: DataFrame, model_a_suffix: str = "EXP0", model_b_suffix: str = "EXP1"
) -> Dict[str, Any]:
# Paired - Calculate the percentage of runs where model A generated more, equal, or less valid candidates than model B
paired_validity_results = {"model_a_more_valid": 0, "equal_valid": 0, "model_b_more_valid": 0}
grouped = df.groupby(df["trace_id"].str[:-4])
for _, group in grouped:
if len(group) == 2:
model_a_row = group[group["trace_id"].str.endswith(model_a_suffix)]
model_b_row = group[group["trace_id"].str.endswith(model_b_suffix)]
if model_a_row["is_correct"].values[0] is None:
model_a_valid_count = 0
else:
model_a_valid_count = sum(model_a_row["is_correct"].values[0].values())
if model_b_row["is_correct"].values[0] is None:
model_b_valid_count = 0
else:
model_b_valid_count = sum(model_b_row["is_correct"].values[0].values())
if model_a_valid_count > model_b_valid_count:
paired_validity_results["model_a_more_valid"] += 1
elif model_a_valid_count < model_b_valid_count:
paired_validity_results["model_b_more_valid"] += 1
else:
paired_validity_results["equal_valid"] += 1
return paired_validity_results
def paired_comparison_performance(
df: DataFrame, model_a_suffix: str = "EXP0", model_b_suffix: str = "EXP1"
) -> Dict[str, Any]:
paired_results = {"model_a_better": 0, "equal": 0, "model_b_better": 0}
# Group by the trace_id without the suffix
grouped = df.groupby(df["trace_id"].str[:-4])
for _, group in grouped:
if len(group) == 2:
model_a_row = group[group["trace_id"].str.endswith(model_a_suffix)]
model_b_row = group[group["trace_id"].str.endswith(model_b_suffix)]
model_a_speedup = model_a_row["best_correct_speedup_ratio"].values[0]
model_b_speedup = model_b_row["best_correct_speedup_ratio"].values[0]
if pd.isna(model_a_speedup) and not pd.isna(model_b_speedup):
paired_results["model_b_better"] += 1
elif not pd.isna(model_a_speedup) and pd.isna(model_b_speedup):
paired_results["model_a_better"] += 1
elif pd.isna(model_a_speedup) and pd.isna(model_b_speedup):
# If both are NaN, do nothing
pass
elif model_a_speedup > model_b_speedup:
paired_results["model_a_better"] += 1
elif model_a_speedup < model_b_speedup:
paired_results["model_b_better"] += 1
else:
paired_results["equal"] += 1
return paired_results
def augment_with_best_correct_speedup_ratio(df: DataFrame) -> DataFrame:
# Extract the best speedup ratio from the speedup_ratio dictionary, accounting for empty dictionaries
def get_best_correct_speedup_ratio(
speedup_ratios: Dict[str, float], is_correct: Dict[str, bool]
) -> Optional[float]:
correct_speedup_ratios = (
{uuid: ratio for uuid, ratio in speedup_ratios.items() if is_correct.get(uuid)}
if speedup_ratios is not None
else {}
)
if correct_speedup_ratios:
return max(correct_speedup_ratios.values())
return None
df["best_correct_speedup_ratio"] = df.apply(
lambda row: get_best_correct_speedup_ratio(row["speedup_ratio"], row["is_correct"])
if row["speedup_ratio"] is not None
else None,
axis=1,
)
return df
def main() -> None:
df = load_data("cotprompt0424")
df = process_column_pairs(df, "metadata")
df = process_column_pairs(df, "test_framework")
df = process_column_pairs(df, "generated_test")
df = augment_with_best_correct_speedup_ratio(df)
exp0_df = df[df["trace_id"].str.endswith("EXP0")]
exp1_df = df[df["trace_id"].str.endswith("EXP1")]
# Calculate metrics for each experiment
exp0_performance_metrics = calculate_performance(exp0_df)
exp1_performance_metrics = calculate_performance(exp1_df)
exp0_validity_metrics = calculate_validity(exp0_df)
exp1_validity_metrics = calculate_validity(exp1_df)
exp0_coverage_metrics = calculate_coverage(exp0_df)
exp1_coverage_metrics = calculate_coverage(exp1_df)
paired_performance_metrics = paired_comparison_performance(df)
paired_validity_metrics = paired_comparison_validity(df)
paired_coverage_metrics = paired_comparison_coverage(df)
# Combine metrics into a DataFrame
metrics_df = pd.DataFrame(
{
"EXP0": {**exp0_performance_metrics, **exp0_validity_metrics, **exp0_coverage_metrics},
"EXP1": {**exp1_performance_metrics, **exp1_validity_metrics, **exp1_coverage_metrics},
}
) # Transpose to have experiments as rows and metrics as columns
# Output the combined metrics DataFrame
print(metrics_df)
if __name__ == "__main__":
main()

89
experiments/process_optimization_features.py
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@ -1,89 +0,0 @@
import os
import sys
from typing import Any, Dict
import pandas as pd
from sqlalchemy import create_engine
def execute_query(query: str, trace_id: str):
database_uri = os.environ.get("DATABASE_URL")
engine = create_engine(database_uri)
with engine.connect() as connection:
result = pd.read_sql_query(query, connection, params=[(trace_id,)])
return result.iloc[0].to_dict() if not result.empty else None
def extract_json_values(json_column: Dict[str, Any]) -> Dict[str, Any]:
return {k: v for k, v in json_column.items()}
def write_to_file(filename: str, content: str) -> None:
with open(filename, "w") as file:
if isinstance(content, list):
content = "\n".join(content)
file.write(str(content))
def main(trace_id: str) -> None:
query = "SELECT original_code, optimizations_post, speedup_ratio, generated_test, instrumented_generated_test, explanations_post FROM optimization_features WHERE trace_id = %s"
result = execute_query(query, trace_id)
if not result:
print(f"No data found for trace_id: {trace_id}")
return
# Extract data from the result
original_code = result["original_code"]
optimizations_post = result["optimizations_post"]
speedup_ratio = result["speedup_ratio"]
generated_test = result["generated_test"]
explanations_post = result["explanations_post"]
instrumented_tests = result["instrumented_generated_test"]
# Write original code to file
write_to_file("original_code.py", original_code)
# Write each optimization candidate to its own file
for idx, (opt_id, optimization) in enumerate(extract_json_values(optimizations_post).items(), start=1):
filename = f"optimization_candidate_{idx}.py"
explanation = explanations_post.get(opt_id, "")
speedup = speedup_ratio.get(opt_id)
speedup_comment = f"Speedup: {speedup}" if speedup is not None else "No speedup"
content_with_comment = f'"""{explanation}\n\n{speedup_comment}"""\n\n{optimization}'
write_to_file(filename, content_with_comment)
# Find and write the best optimization candidate to its own file
if speedup_ratio is not None:
valid_speedup_values = [v for v in speedup_ratio.values() if v is not None]
best_speedup = max(valid_speedup_values, default=None) if valid_speedup_values else None
if best_speedup is not None:
best_optimization_id = next((id for id, speedup in speedup_ratio.items() if speedup == best_speedup), None)
best_optimization = optimizations_post.get(best_optimization_id)
best_explanation = explanations_post.get(best_optimization_id, "")
best_speedup_comment = f"Speedup: {best_speedup}"
best_content_with_comment = f'"""{best_explanation}\n\n{best_speedup_comment}"""\n\n{best_optimization}'
if best_optimization and best_explanation is not None:
write_to_file("best_optimization_candidate.py", best_content_with_comment)
else:
print("No speedup ratio found")
# Write generated tests to file
write_to_file("generated_tests.py", generated_test)
write_to_file("instrumented_generated_tests.py", instrumented_tests)
"""
Run the script using the .env file that contains the DATABASE_URL for the optimization_features table
Pass in the trace_id as a command line argument and the script will generate the following files for that trace id:
- original_code.py
- optimization_candidate_1.py, optimization_candidate_2.py, ...
- best_optimization_candidate.py
- generated_tests.py
"""
if __name__ == "__main__":
if len(sys.argv) != 2:
print("Usage: python process_optimization_features.py <trace_id>")
sys.exit(1)
trace_id_input = sys.argv[1]
main(trace_id_input)

131
experiments/statistical_analysis.py
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@ -1,131 +0,0 @@
from __future__ import annotations
import math
from typing import TYPE_CHECKING
import numpy as np
from numba import get_num_threads, njit, prange
if TYPE_CHECKING:
import numpy.typing as npt
TWO_SIGMA = 2
@njit(parallel=True, fastmath=True, cache=True)
def bootstrap_minima(
series: list[int], rngs: tuple[np.random.Generator, ...], bootstrap_size: int
) -> npt.NDArray[np.int64]:
num_threads = len(rngs)
series_size = len(series)
npseries = np.array(series)
thread_remainder = bootstrap_size % num_threads
num_bootstraps_per_thread = np.array([bootstrap_size // num_threads] * num_threads) + np.array(
[1] * thread_remainder + [0] * (num_threads - thread_remainder)
)
minima = np.empty(bootstrap_size)
thread_idx = [0, *list(np.cumsum(num_bootstraps_per_thread))]
for i in prange(num_threads):
thread_minima = minima[thread_idx[i] : thread_idx[i + 1]]
for k in range(num_bootstraps_per_thread[i]):
thread_minima[k] = min(npseries[rngs[i].integers(0, series_size, size=series_size)])
return minima
@njit(parallel=True, fastmath=True, cache=True)
def bootstrap_minima_ratios(
series1: list[int], series2: list[int], rngs: tuple[np.random.Generator, ...], bootstrap_size: int
) -> npt.NDArray[np.float64]:
num_threads = len(rngs)
series1_size = len(series1)
series2_size = len(series2)
npseries1 = np.array(series1)
npseries2 = np.array(series2)
thread_remainder = bootstrap_size % num_threads
num_bootstraps_per_thread = np.array([bootstrap_size // num_threads] * num_threads) + np.array(
[1] * thread_remainder + [0] * (num_threads - thread_remainder)
)
minima_ratios = np.empty(bootstrap_size, dtype=np.float64)
thread_idx = [0, *list(np.cumsum(num_bootstraps_per_thread))]
for i in prange(num_threads):
thread_minima_ratios = minima_ratios[thread_idx[i] : thread_idx[i + 1]]
for k in range(num_bootstraps_per_thread[i]):
min2 = min(npseries2[rngs[i].integers(0, series2_size, size=series2_size)])
if min2 == 0:
thread_minima_ratios[k] = np.inf
else:
thread_minima_ratios[k] = min(npseries1[rngs[i].integers(0, series1_size, size=series1_size)]) / min2
return minima_ratios
@njit(parallel=True, fastmath=True, cache=True)
def bootstrap_ratios_geomean(
ratio_series: list[npt.NDArray[np.float64]], rngs: tuple[np.random.Generator, ...], bootstrap_size: int
) -> npt.NDArray[np.float64]:
num_series = len(ratio_series)
num_threads = len(rngs)
thread_remainder = bootstrap_size % num_threads
num_bootstraps_per_thread = np.array([bootstrap_size // num_threads] * num_threads) + np.array(
[1] * thread_remainder + [0] * (num_threads - thread_remainder)
)
combined_ratios = np.empty(bootstrap_size, dtype=np.float64)
thread_idx = [0, *list(np.cumsum(num_bootstraps_per_thread))]
for i in prange(num_threads):
thread_combined_ratios = combined_ratios[thread_idx[i] : thread_idx[i + 1]]
for k in range(num_bootstraps_per_thread[i]):
sum_log = 0.0
for series in ratio_series:
ratio = series[rngs[i].integers(0, len(series))]
if ratio <= 0:
ratio = 1e-12
sum_log += np.log(ratio)
thread_combined_ratios[k] = np.exp(sum_log / num_series)
return combined_ratios
def bootstrap_noise_floor(series: list[int], bootstrap_size: int) -> np.float64:
rng = np.random.default_rng()
return np.std(bootstrap_minima(series, tuple(rng.spawn(get_num_threads())), bootstrap_size))
def combined_series_noise_floor(series1: list[int], series2: list[int], bootstrap_size: int) -> np.float64:
noise_floor1 = bootstrap_noise_floor(series1, bootstrap_size)
noise_floor2 = bootstrap_noise_floor(series2, bootstrap_size)
return math.sqrt(noise_floor1 * noise_floor1 + noise_floor2 * noise_floor2)
def series2_faster_95_confidence(
series1: list[int], series2: list[int], bootstrap_size: int
) -> tuple[float, float] | None:
min1 = min(series1)
min_diff = min1 - min(series2)
if min_diff <= 0:
return None
combined_noise_floor = combined_series_noise_floor(series1, series2, bootstrap_size)
percent_diff = 100 * min_diff / min1
uncertainty = TWO_SIGMA * 100 * combined_noise_floor / min1
if combined_noise_floor == 0 or min_diff / combined_noise_floor > TWO_SIGMA:
return percent_diff, uncertainty
return None
def analyze_series_speedup(
multi_series1: list[list[int]], multi_series2: list[list[int]], bootstrap_size: int
) -> tuple[np.float64, np.float64, np.float64, np.float64, np.float64]:
rng = np.random.default_rng()
combined_ratios = bootstrap_ratios_geomean(
[
bootstrap_minima_ratios(series1, series2, tuple(rng.spawn(get_num_threads())), bootstrap_size)
for series1, series2 in zip(multi_series1, multi_series2)
],
tuple(rng.spawn(get_num_threads())),
bootstrap_size,
)
lower_bound_95_confidence = np.percentile(combined_ratios, 2.5)
upper_bound_95_confidence = np.percentile(combined_ratios, 97.5)
mean = np.mean(combined_ratios)
probablility_faster = np.mean(combined_ratios > 1.0)
return lower_bound_95_confidence, upper_bound_95_confidence, mean, probablility_faster

0
experiments/tests/__init__.py
View file

0
experiments/tests/data/__init__.py
View file

64
experiments/tests/data/mock_dataframe.py
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@ -1,64 +0,0 @@
mock_dataframe = {
"original_runtime": [1.0, 2.0, 3.0, 4.0, 8333.00000],
"optimized_runtime": [
{
"0a631960-6c33-4de4-8d38-b31e888918c7": 59209,
"38b31a95-7409-4c66-a50d-24dd565bb513": 57458,
"6ba45be2-5f81-4952-8dc6-c4569092096b": 49167,
"7f9c5889-6307-47e1-92b2-a89f07409278": 54332,
"857da2b3-1b2c-4ce1-96a1-005d8b06df24": 51419,
"9f359406-cb4c-404e-a9d7-c1b61af7e019": None,
"a8cc096d-390f-44e0-a0eb-7cbbec651671": None,
},
{
"0ac7066f-a5cd-417c-8121-0614754e6b0a": 36581,
"31d58b1a-51ae-4e64-a720-acaede899790": 36374,
"5c9be4b4-0e0e-4a23-8860-d3fc714dab64": None,
"7f069830-6c44-487a-ab27-6af57c469420": None,
"b0be6fb9-0cb1-4ead-b542-95e4253b23a1": 59334,
"c905489f-31e9-4f1f-a1c4-b43eb243c999": 46336,
"cd4ba45c-cb46-4fbd-8869-0c1203b73e13": 36377,
},
None,
{
"3af102b2-e9a4-45d1-b6a4-da7d4cf77bf0": None,
"b611ec1a-ed34-49d5-a7cf-4eaac45861a8": None,
"ce083f90-a142-4372-a3a1-297c14176a7a": None,
"db7f485a-7487-4130-bf8e-d7f827c99593": None,
"e9209c6b-7b06-48f3-9b0c-743e28b7aa4b": None,
"f3d6b00a-383f-48f1-9f09-74574784c128": None,
"f6c46b68-d8f9-4762-957a-a8e0118577d0": None,
"fe0dc991-79c9-41c0-ab5c-aaf5b1606718": None,
},
{
"21645d50-d094-47d5-9a16-9163d90ee040": 6625,
"3ba6ea58-e6c4-4398-9400-098ccc0b7c81": None,
"573c9f22-ae68-4e62-929d-b70d22da0ec0": 9333,
"6fa4faaf-5d67-4ec6-bf55-5a84dfa23a11": 7125,
"a06d1ad8-cc41-4b3b-a1a6-4566697d7b10": 7166,
},
],
"speedup_ratio": [
{
"21645d50-d094-47d5-9a16-9163d90ee040": 0.25781132075471697,
"3ba6ea58-e6c4-4398-9400-098ccc0b7c81": None,
"573c9f22-ae68-4e62-929d-b70d22da0ec0": -0.10714668381013608,
"6fa4faaf-5d67-4ec6-bf55-5a84dfa23a11": 0.1695438596491228,
"a06d1ad8-cc41-4b3b-a1a6-4566697d7b10": 0.16285235835891712,
}
],
"best_correct_speedup_ratio": [0.02112, None, 1.81043, None],
"is_correct": [
{"1f39ef86-5eff-4760-a262-43011492906e": True, "5f9ec728-4353-434e-b136-3c4e3debb1fd": False},
{"aeeeca3b-4ccf-46eb-8dbf-c526c05fca27": False, "dc67527a-9cfe-4bbe-98eb-266bc8e9a27d": False},
{"e584d267-f68f-4144-820f-110c2719919f": True},
{},
{
"21645d50-d094-47d5-9a16-9163d90ee040": True,
"3ba6ea58-e6c4-4398-9400-098ccc0b7c81": False,
"573c9f22-ae68-4e62-929d-b70d22da0ec0": True,
"6fa4faaf-5d67-4ec6-bf55-5a84dfa23a11": True,
"a06d1ad8-cc41-4b3b-a1a6-4566697d7b10": True,
},
],
}

523
experiments/tests/data/pie4perf_dataframe_dict.py
View file

@ -1,523 +0,0 @@
# First six rows of a sample dataframe I got from running on the pie4perf dataset.
from numpy import nan
from pandas import Timestamp
pie4perf_sample_dataframe_dict = {
"trace_id": {
0: "0737f6d0-2837-4e89-9329-b874dd85EXP1",
1: "c8d3c1f5-dc36-4662-93f0-89b1a728EXP1",
2: "004b83d1-d244-4574-a686-75528771EXP0",
3: "f12774fd-abde-4a96-a956-bb497851EXP0",
4: "f9f80c75-23a2-4da7-b7cb-46b55e98EXP0",
5: "e8b0b6c8-be2d-4187-bbab-a452ea2cEXP0",
},
"original_code": {
0: '\ndef problem_p02847(input_data):\n def main():\n\n S = eval(input_data)\n\n lst = ["MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"]\n\n ans = 6 - lst.index(S)\n\n if ans == 0:\n\n ans = 7\n\n return ans\n\n if __name__ == "__main__":\n\n main()\n',
1: "\ndef problem_p03196(input_data):\n n, p = list(map(int, input_data.split()))\n\n if n == 1:\n\n return p\n\n elif n > 40:\n\n return 1\n\n else:\n\n ans = 1\n\n i = 1\n\n while True:\n\n if i**n > p:\n\n break\n\n if p % (i**n) == 0:\n\n ans = i\n\n i += 1\n\n return ans\n",
2: "\ndef problem_p03861(input_data):\n import math\n\n a, b, x = list(map(int, input_data.split()))\n\n return b // x - (a - 1) // x\n",
3: '\ndef problem_p02910(input_data):\n s = eval(input_data)\n\n flag = True\n\n for i in range(1, len(s) + 1):\n\n if i % 2 == 1:\n\n if s[i - 1] == "L":\n\n flag = False\n\n else:\n\n if s[i - 1] == "R":\n\n flag = False\n\n if flag:\n\n return "Yes"\n\n else:\n\n return "No"\n',
4: '\ndef problem_p03965(input_data):\n # len(s)//2-s.count("p")\n\n iG = 0\n\n iR = 0\n\n for s in input_data.rstrip():\n\n if s == "g":\n\n if 0 < iG:\n\n iR += 1\n\n iG -= 1\n\n else:\n\n iG += 1\n\n else:\n\n if 0 < iG:\n\n iG -= 1\n\n else:\n\n iR -= 1\n\n iG += 1\n\n return iR\n',
5: "\ndef problem_p02553(input_data):\n import numpy as np\n\n a, b, c, d = list(map(int, input_data.split()))\n\n hoge = []\n\n hoge.append(a * c)\n\n hoge.append(a * d)\n\n hoge.append(b * c)\n\n hoge.append(b * d)\n\n if max(hoge) < 0:\n\n if np.sign(a) != np.sign(b) or np.sign(c) != np.sign(d):\n\n return 0\n\n else:\n\n return max(hoge)\n\n else:\n\n return max(hoge)\n",
},
"optimizations_raw": {
0: {
"00db208c-b4df-4ac0-bf2c-4448f0c18897": 'def problem_p02847(input_day: str) -> int:\n days_dict = {"MON": 0, "TUE": 1, "WED": 2, "THU": 3, "FRI": 4, "SAT": 5, "SUN": 6}\n remaining_days = 6 - days_dict[input_day]\n return remaining_days if remaining_days != 0 else 7\n',
"098b2a24-8841-496c-b4aa-01b3f680836b": 'def problem_p02847(input_data: str)-> int:\n\n lst = ["MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"]\n days_until_sunday = 6 - lst.index(input_data)\n \n return days_until_sunday if days_until_sunday != 0 else 7\n',
"20b7cb92-6906-448d-901e-2ce98fefe2e3": 'def problem_p02847(input_day: str) -> int:\n week_days = {"MON": 6, "TUE": 5, "WED": 4, "THU": 3, "FRI": 2, "SAT": 1, "SUN": 7}\n return week_days[input_day]\n',
"239ec7be-8ce0-4c86-8b56-0704e791a372": 'def problem_p02847(input_data: str) -> int:\n lst = ["MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"]\n\n ans = (6 - lst.index(input_data) + 7) % 7\n\n if ans == 0:\n ans = 7\n\n return ans\n',
"2f364667-92fe-4245-b04d-04d6b5d9d9dd": "def problem_p02847(input_data: str) -> int:\n days_to_sunday = {\n 'MON': 6,\n 'TUE': 5,\n 'WED': 4,\n 'THU': 3,\n 'FRI': 2,\n 'SAT': 1,\n 'SUN': 7,\n }\n \n return days_to_sunday[input_data]\n",
"8d67ed70-e190-4a84-b5d8-37defb27f6d6": 'def problem_p02847(input_day: str) -> int:\n days_to_weekend = {"MON": 6, "TUE": 5, "WED": 4, "THU": 3, "FRI": 2, "SAT": 1, "SUN": 7}\n return days_to_weekend[input_day]\n',
"ba6a30d4-3150-482e-b932-6320f4635ebe": 'def problem_p02847(S: str) -> int:\n lst = ["MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"]\n ans = 6 - lst.index(S)\n if ans == 0:\n ans = 7\n return ans\n',
"c2fe9046-d193-4a51-a1d5-111c172e00f8": 'def problem_p02847(input_data: str) -> int:\n days_to_sunday = {"MON": 6, "TUE": 5, "WED": 4, "THU": 3, "FRI": 2, "SAT": 1, "SUN": 7}\n return days_to_sunday[input_data]\n',
"d2ff6d72-4515-4094-9d80-89f75045a9b5": 'def problem_p02847(input_data: str) -> int:\n weekdays = ["MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"]\n current_day_index = weekdays.index(input_data)\n remaining_days = 6 - current_day_index\n\n if remaining_days == 0:\n remaining_days = 7\n\n return remaining_days\n',
},
1: {
"426916b1-b122-44a7-aae3-c806b81887bd": "def problem_p03196(input_data: str) -> int:\n n, p = map(int, input_data.split())\n \n if n == 1:\n return p \n elif n > 40:\n return 1 \n else: \n low, high = 1, p\n while low < high:\n mid = (low + high + 1) // 2\n if mid**n <= p:\n low = mid\n else:\n high = mid - 1\n return low\n",
"4a18cc5a-17a9-4aba-8b56-4534c406637e": "def problem_p03196(input_data):\n n, p = list(map(int, input_data.split()))\n\n if n == 1:\n return p\n elif n > 40:\n return 1\n else:\n low, high = 1, int(p ** (1/n)) + 1\n while low < high:\n mid = (low + high) // 2\n if mid ** n <= p and p % (mid ** n) == 0:\n low = mid + 1\n else:\n high = mid\n return low - 1\n",
"67cbfea8-0699-4750-a758-4078b3d33fb8": "import math\n\ndef problem_p03196(input_data):\n n, p = map(int, input_data.split())\n \n if n == 1:\n return p\n \n elif n > 40:\n return 1\n \n else:\n ans = 1\n root = int(p ** (1/n))\n \n for i in range(1, root+1):\n if p % (i ** n) == 0:\n ans = i\n \n return ans\n",
"68416ac3-693a-432f-b19c-a74e5bcb5d80": "def problem_p03196(input_data: str) -> int:\n n, p = list(map(int, input_data.split()))\n if n == 1:\n return p\n elif n > 40:\n return 1\n else:\n ans = 1\n max_base = int(p ** (1 / n))\n for i in range(1, max_base + 1):\n if p % (i ** n) == 0:\n ans = i\n return ans\n",
"7afadb38-0b8c-41b6-a51f-78f9cbf17d63": "def problem_p03196(input_data: str) -> int:\n n, p = map(int, input_data.split())\n\n if n == 1:\n return p\n elif n > 40:\n return 1\n\n ans, low, high = 1, 1, p\n\n while low <= high:\n mid = (high + low) // 2\n if mid ** n > p:\n high = mid - 1\n else:\n ans = mid\n low = mid + 1\n\n return ans\n",
"7d96a7c8-29b6-43d9-a682-fbcf1d36103d": "def problem_p03196(input_data: str) -> int:\n n, p = list(map(int, input_data.split()))\n \n if n == 1:\n return p\n elif n > 40:\n return 1\n else:\n lower = 1\n upper = p\n ans = 1\n\n while lower <= upper:\n mid = (lower + upper) // 2\n if mid ** n > p:\n upper = mid - 1\n else:\n ans = mid\n lower = mid + 1\n\n return ans\n",
"e3a6d7b8-e145-45c3-bd1c-8cae99c6d014": "def problem_p03196(input_data: str) -> int:\n n, p = list(map(int, input_data.split()))\n \n if n == 1:\n return p\n elif n > 40:\n return 1\n\n ans = p\n i, j = 0, p\n while i <= j:\n mid = (i + j) // 2\n if mid ** n == p:\n return mid\n elif mid ** n < p:\n ans = mid\n i = mid + 1\n else:\n j = mid - 1\n return ans\n",
},
2: {
"0a631960-6c33-4de4-8d38-b31e888918c7": "def problem_p03861(input_data):\n a, b, x = (int(i) for i in input_data.split())\n\n return b // x - (a - 1) // x\n",
"38b31a95-7409-4c66-a50d-24dd565bb513": "def problem_p03861(input_data):\n a, b, x = list(map(int, input_data.split()))\n\n return b // x - (a - 1) // x\n",
"6ba45be2-5f81-4952-8dc6-c4569092096b": "def problem_p03861(input_data):\n\n a, b, x = map(int, input_data.split())\n\n return b // x - (a - 1) // x\n",
"7f0a4ac8-cec2-485b-a159-610fae2dcf68": "def problem_p03861(input_data):\n a, b, x = map(int, input_data.split())\n return b // x - (a - 1) // x\n",
"7f9c5889-6307-47e1-92b2-a89f07409278": "\ndef problem_p03861(input_data):\n a, b, x = map(int, input_data.split())\n\n return (b - a + 1 + (a - 1) % x) // x\n\n",
"857da2b3-1b2c-4ce1-96a1-005d8b06df24": "import math\n\ndef problem_p03861(input_data):\n a, b, x = map(int, input_data.split())\n return b // x - (a - 1) // x\n",
"9f359406-cb4c-404e-a9d7-c1b61af7e019": "def problem_p03861(a, b, x):\n quotient_b, remainder_b = divmod(b, x)\n quotient_a_1, remainder_a_1 = divmod(a - 1, x)\n\n return quotient_b - quotient_a_1\n",
"a8cc096d-390f-44e0-a0eb-7cbbec651671": "def problem_p03861(a, b, x):\n return b // x - (a - 1) // x\n",
"be3c8c8a-cb4d-465a-8d98-e345407b3ee5": "def problem_p03861(input_data):\n a, b, x = map(int, input_data.split())\n return b // x - (a - 1) // x\n",
"e40e4203-d07b-4319-9000-24a1333f6f4c": "def problem_p03861(input_data):\n a, b, x = map(int, input_data.split())\n return b // x - (a - 1) // x\n",
},
3: {
"00f11305-5a9b-43cf-9d97-720b0a465d52": "\ndef problem_p02910(input_string):\n for i, char in enumerate(input_string):\n if char == 'L' and i % 2 == 0 or char == 'R' and i % 2 == 1:\n return \"No\"\n\n return \"Yes\"\n",
"0f234383-ad19-44dd-95c1-9bb5171bf82a": "def problem_p02910(input_data):\n s = eval(input_data)\n for i, c in enumerate(s):\n if ((i % 2 == 0 and c == 'R') or (i % 2 == 1 and c == 'L')):\n return 'No'\n return 'Yes'\n",
"1f1bb4c2-297e-46ca-a1c2-10770e83f877": 'def problem_p02910(input_data):\n s = eval(input_data)\n for i, letter in enumerate(s):\n if (i % 2 == 0 and letter == "R") or (i % 2 != 0 and letter == "L"):\n return "No"\n return "Yes"\n',
"25c43c11-6a29-48f1-878b-06156bf31906": 'def problem_p02910(input_data):\n for i, char in enumerate(input_data, 1):\n if (i % 2 == 1 and char == "L") or (i % 2 == 0 and char == "R"):\n return "No"\n return "Yes"\n',
"35128e9c-4e22-4cd0-9412-b55c3c54a16a": "def problem_p02910(input_data):\n s = eval(input_data)\n \n if 'L' in s[::2] or 'R' in s[1::2]:\n return 'No'\n else:\n return 'Yes'\n",
"4bc36f4a-ddcd-4e0e-a868-5b338d850e21": "\ndef problem_p02910(input_data):\n s = eval(input_data)\n\n flag = all(s[i] == 'R' if i % 2 else s[i] == 'L' for i in range(len(s)))\n \n return 'Yes' if flag else 'No'\n\n",
"5e136046-8d21-4e8b-b5d1-6efa23fc7f6e": "def problem_p02910(input_data):\n s = eval(input_data)\n for i in range(len(s)):\n if (i % 2 == 0 and s[i] == 'R') or (i % 2 == 1 and s[i] == 'L'):\n return 'No' \n return 'Yes'\n",
"830e8468-8c1e-4dc6-952e-203148dfe5d6": "def problem_p02910(input_data):\n s = eval(input_data)\n if any(s[i] == 'L' for i in range(0, len(s), 2)) or any(s[i] == 'R' for i in range(1, len(s), 2)):\n return 'No'\n return 'Yes'\n",
"8784754e-3e17-4561-94c7-05b9b5f2aedf": 'def problem_p02910(input_data):\n s = str(input_data)\n\n for i, char in enumerate(s, 1):\n if (i % 2 == 1 and char == "L") or (i % 2 == 0 and char == "R"):\n return "No"\n\n return "Yes"\n',
"96e129db-323a-4651-a021-32f5321e11e9": "\ndef problem_p02910(input_data):\n s = eval(input_data)\n\n for i in range(len(s)):\n if i % 2 != s[i - 1] == 'R' or i % 2 == s[i - 1] == 'L':\n return \"No\"\n return \"Yes\"\n",
},
4: {
"000711ac-1663-496b-837b-946040155a9c": '\ndef problem_p03965(input_data):\n iG, iR = 0, 0\n input_data = input_data.rstrip()\n \n for s in input_data:\n \n if s == "g" and iG > 0:\n iR += 1\n iG -= 1\n \n elif s == "g" :\n iG += 1\n \n elif iG > 0:\n iG -= 1\n\n else:\n iR -= 1\n iG += 1\n \n return iR\n\n',
"0c1d4ab3-9eca-4b24-a274-1cf4aa35ec0d": 'def problem_p03965(input_data):\n iG = 0\n iR = 0\n\n for s in input_data.rstrip():\n if s == "g":\n iR += (0 < iG)\n iG += 1 - (0 < iG)\n else:\n iG += 1 - (0 < iG)\n iR -= 1 - (0 < iG)\n return iR\n',
"470d47be-1d1b-4cff-927f-0e9200d781cc": 'def problem_p03965(input_data):\n # len(s)//2-s.count("p")\n\n iG = iR = 0\n\n for s in input_data.rstrip():\n\n if s == "g" and iG > 0 or s != "g" and iG <= 0:\n iR += 1\n iG -= 1\n \n iG += 1\n\n return iR\n',
"546095ca-bd89-4cdd-b0a8-3694d8013644": 'def problem_p03965(input_data):\n iG = iR = 0\n stripped_data = input_data.rstrip()\n\n for s in stripped_data:\n\n iG += 1\n \n if s == "g" and iG > 0:\n iR += 1\n iG -= 1\n elif iG <= 0:\n iR -= 1\n\n return iR\n',
"81fdc495-f548-4621-89c4-691d3b74f77b": 'def problem_p03965(input_data):\n iG = iR = 0\n\n for s in input_data.rstrip():\n\n if (s == "g" and 0 < iG) or (s != "g" and iG <= 0):\n iR += 1\n\n iG += -1 if iG > 0 else 1\n\n return iR\n',
"83b6579b-8f00-4829-944c-904c8d7b80f9": 'def problem_p03965(input_data):\n iG, iR = 0, 0\n for s in input_data:\n if s == "g":\n iR += 1 if 0 < iG else 0\n iG = (iG - 1 if 0 < iG else iG + 1)\n else:\n iG = (iG - 1 if 0 < iG else iG + 1)\n iR -= 1 if iG < 0 else 0\n return iR\n',
"849e8b87-fb91-4f85-84cf-ade32dab0f3f": 'def problem_p03965(input_data):\n iG, iR = 0, 0\n \n for s in input_data.rstrip():\n if s == "g":\n iR, iG = (iR + 1, iG - 1) if iG > 0 else (iR, iG + 1)\n else:\n iG, iR = (iG - 1, iR) if iG > 0 else (iG + 1, iR - 1)\n\n return iR\n',
"9ee7a8e5-f8c5-49ee-aa71-4010ea432781": 'def problem_p03965(input_data):\n iG = 0\n iR = 0\n for s in input_data.rstrip():\n if s == "g":\n iG -= iG > 0\n iR += iG > 0\n iG += 1\n else:\n iR -= iG <= 0\n iG -= iG > 0\n iG += 1\n return iR\n',
"bd157d9b-3892-48fb-a106-908631fd6a7c": 'def problem_p03965(input_data):\n iG, iR = 0, 0\n\n for s in input_data.rstrip():\n if s == "g":\n if iG:\n iR += 1\n iG -= 1\n else:\n iG += 1\n else:\n if iG:\n iG -= 1\n else:\n iR -= 1\n iG += 1\n return iR\n',
"fdc5337e-8a6a-44ed-8255-d7fca5b49786": 'def problem_p03965(input_data):\n iG, iR = 0, 0\n for s in input_data.rstrip():\n if (s == "g" and iG > 0) or (s != "g" and iG <= 0):\n iR += 1\n iG -= 1\n else:\n iG += 1\n return iR\n',
},
5: {
"04e41d16-15f6-4341-a976-f420ed3180a6": "def problem_p02553(input_data):\n\n a, b, c, d = map(int, input_data.split())\n\n max1 = max(a * c, a * d, b * c, b * d)\n\n if max1 < 0:\n\n if (a < 0 and b > 0) or (a > 0 and b < 0) or (c < 0 and d > 0) or (c > 0 and d < 0):\n\n return 0\n\n else:\n\n return max1\n\n else:\n\n return max1\n",
"09edd39b-4a6f-4422-8c14-23d2a819f2c5": "def problem_p02553(input_data):\n\n a, b, c, d = map(int, input_data.split())\n\n hoge = [a * c, a * d, b * c, b * d]\n\n max_hoge = max(hoge)\n\n if max_hoge < 0:\n\n if a * b < 0 or c * d < 0:\n\n return 0\n else:\n\n return max_hoge\n\n else:\n\n return max_hoge\n",
"19d21335-2ee5-4a5e-8bab-50805248cf9f": "def problem_p02553(input_data):\n a, b, c, d = map(int, input_data.split())\n\n max_val = max(a*c, a*d, b*c, b*d)\n\n if max_val < 0:\n if (a<0) != (b<0) or (c<0) != (d<0):\n return 0\n\n return max_val\n",
"1aa3d7da-d34e-46a3-a72c-d0f1cd0bd5ba": "def problem_p02553(input_data):\n a, b, c, d = map(int, input_data.split())\n \n max_product = max(a * c, a * d, b * c, b * d)\n \n if max_product < 0:\n if a * b < 0 or c * d < 0:\n return 0\n\n return max_product\n",
"36eeb34d-6a89-42f5-8340-e7a31bab073e": "def problem_p02553(input_data):\n a, b, c, d = list(map(int, input_data.split()))\n\n hoge = max(a * c, a * d, b * c, b * d)\n\n if hoge < 0:\n if (a < 0) != (b < 0) or (c < 0) != (d < 0):\n return 0\n\n return hoge\n",
"45c372c7-8b7f-4378-ba39-fdcde9c7a13e": "def problem_p02553(input_data):\n a, b, c, d = map(int, input_data.split())\n \n max_product = max(a*c, a*d, b*c, b*d)\n \n if max_product < 0 and (a < 0) == (b < 0) and (c < 0) == (d < 0):\n return 0\n else:\n return max_product\n",
"881c3174-228a-4592-a681-d8e88bc77aeb": "def problem_p02553(input_data):\n a, b, c, d = map(int, input_data.split())\n\n max_value = max(a * c, a * d, b * c, b * d)\n\n if max_value < 0:\n if (a > 0 > b) or (c > 0 > d):\n return 0\n\n return max_value\n",
"a7e3f286-cc2b-49ad-a958-5429a32408e5": "def problem_p02553(input_data):\n a, b, c, d = map(int, input_data.split())\n\n max_val = max(a * c, a * d, b * c, b * d)\n \n if max_val < 0 and ((a > 0) != (b > 0) or (c > 0) != (d > 0)):\n return 0\n else:\n return max_val\n",
"ab6accd2-494e-4bd4-b91c-7f38125eb9a7": "\ndef problem_p02553(input_data):\n a, b, c, d = map(int, input_data.split())\n \n max_value = max(a * c, a * d, b * c, b * d)\n\n if max_value < 0:\n if (a < 0) != (b < 0) or (c < 0) != (d < 0):\n return 0\n else:\n return max_value\n else:\n return max_value\n\n",
"b35e4ea8-b192-415f-8dcc-817b76b1319c": "def problem_p02553(input_data):\n a, b, c, d = map(int, input_data.split())\n\n hoge = [a * c, a * d, b * c, b * d]\n max_hoge = max(hoge)\n\n if max_hoge < 0:\n if a * b < 0 or c * d < 0:\n return 0\n\n return max_hoge\n",
},
},
"optimizations_post": {
0: {
"00db208c-b4df-4ac0-bf2c-4448f0c18897": 'def problem_p02847(input_day: str) -> int:\n days_dict = {"MON": 0, "TUE": 1, "WED": 2, "THU": 3, "FRI": 4, "SAT": 5, "SUN": 6}\n remaining_days = 6 - days_dict[input_day]\n return remaining_days if remaining_days != 0 else 7\n',
"098b2a24-8841-496c-b4aa-01b3f680836b": 'def problem_p02847(input_data: str)-> int:\n\n lst = ["MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"]\n days_until_sunday = 6 - lst.index(input_data)\n \n return days_until_sunday if days_until_sunday != 0 else 7\n',
"20b7cb92-6906-448d-901e-2ce98fefe2e3": 'def problem_p02847(input_day: str) -> int:\n week_days = {"MON": 6, "TUE": 5, "WED": 4, "THU": 3, "FRI": 2, "SAT": 1, "SUN": 7}\n return week_days[input_day]\n',
"239ec7be-8ce0-4c86-8b56-0704e791a372": 'def problem_p02847(input_data: str) -> int:\n lst = ["MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"]\n\n ans = (6 - lst.index(input_data) + 7) % 7\n\n if ans == 0:\n ans = 7\n\n return ans\n',
"8d67ed70-e190-4a84-b5d8-37defb27f6d6": 'def problem_p02847(input_day: str) -> int:\n days_to_weekend = {"MON": 6, "TUE": 5, "WED": 4, "THU": 3, "FRI": 2, "SAT": 1, "SUN": 7}\n return days_to_weekend[input_day]\n',
"ba6a30d4-3150-482e-b932-6320f4635ebe": 'def problem_p02847(S: str) -> int:\n lst = ["MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"]\n ans = 6 - lst.index(S)\n if ans == 0:\n ans = 7\n return ans\n',
"c2fe9046-d193-4a51-a1d5-111c172e00f8": 'def problem_p02847(input_data: str) -> int:\n days_to_sunday = {"MON": 6, "TUE": 5, "WED": 4, "THU": 3, "FRI": 2, "SAT": 1, "SUN": 7}\n return days_to_sunday[input_data]\n',
"d2ff6d72-4515-4094-9d80-89f75045a9b5": 'def problem_p02847(input_data: str) -> int:\n weekdays = ["MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"]\n current_day_index = weekdays.index(input_data)\n remaining_days = 6 - current_day_index\n\n if remaining_days == 0:\n remaining_days = 7\n\n return remaining_days\n',
},
1: {
"426916b1-b122-44a7-aae3-c806b81887bd": "def problem_p03196(input_data: str) -> int:\n n, p = map(int, input_data.split())\n \n if n == 1:\n return p \n elif n > 40:\n return 1 \n else: \n low, high = 1, p\n while low < high:\n mid = (low + high + 1) // 2\n if mid**n <= p:\n low = mid\n else:\n high = mid - 1\n return low\n",
"4a18cc5a-17a9-4aba-8b56-4534c406637e": "def problem_p03196(input_data):\n n, p = list(map(int, input_data.split()))\n\n if n == 1:\n return p\n elif n > 40:\n return 1\n else:\n low, high = 1, int(p ** (1/n)) + 1\n while low < high:\n mid = (low + high) // 2\n if mid ** n <= p and p % (mid ** n) == 0:\n low = mid + 1\n else:\n high = mid\n return low - 1\n",
"67cbfea8-0699-4750-a758-4078b3d33fb8": "import math\n\n\ndef problem_p03196(input_data):\n n, p = map(int, input_data.split())\n \n if n == 1:\n return p\n \n elif n > 40:\n return 1\n \n else:\n ans = 1\n root = int(p ** (1/n))\n \n for i in range(1, root+1):\n if p % (i ** n) == 0:\n ans = i\n \n return ans\n",
"68416ac3-693a-432f-b19c-a74e5bcb5d80": "def problem_p03196(input_data: str) -> int:\n n, p = list(map(int, input_data.split()))\n if n == 1:\n return p\n elif n > 40:\n return 1\n else:\n ans = 1\n max_base = int(p ** (1 / n))\n for i in range(1, max_base + 1):\n if p % (i ** n) == 0:\n ans = i\n return ans\n",
"7afadb38-0b8c-41b6-a51f-78f9cbf17d63": "def problem_p03196(input_data: str) -> int:\n n, p = map(int, input_data.split())\n\n if n == 1:\n return p\n elif n > 40:\n return 1\n\n ans, low, high = 1, 1, p\n\n while low <= high:\n mid = (high + low) // 2\n if mid ** n > p:\n high = mid - 1\n else:\n ans = mid\n low = mid + 1\n\n return ans\n",
"7d96a7c8-29b6-43d9-a682-fbcf1d36103d": "def problem_p03196(input_data: str) -> int:\n n, p = list(map(int, input_data.split()))\n \n if n == 1:\n return p\n elif n > 40:\n return 1\n else:\n lower = 1\n upper = p\n ans = 1\n\n while lower <= upper:\n mid = (lower + upper) // 2\n if mid ** n > p:\n upper = mid - 1\n else:\n ans = mid\n lower = mid + 1\n\n return ans\n",
"e3a6d7b8-e145-45c3-bd1c-8cae99c6d014": "def problem_p03196(input_data: str) -> int:\n n, p = list(map(int, input_data.split()))\n \n if n == 1:\n return p\n elif n > 40:\n return 1\n\n ans = p\n i, j = 0, p\n while i <= j:\n mid = (i + j) // 2\n if mid ** n == p:\n return mid\n elif mid ** n < p:\n ans = mid\n i = mid + 1\n else:\n j = mid - 1\n return ans\n",
},
2: {
"0a631960-6c33-4de4-8d38-b31e888918c7": "def problem_p03861(input_data):\n a, b, x = (int(i) for i in input_data.split())\n\n return b // x - (a - 1) // x\n",
"38b31a95-7409-4c66-a50d-24dd565bb513": "def problem_p03861(input_data):\n a, b, x = list(map(int, input_data.split()))\n\n return b // x - (a - 1) // x\n",
"6ba45be2-5f81-4952-8dc6-c4569092096b": "def problem_p03861(input_data):\n\n a, b, x = map(int, input_data.split())\n\n return b // x - (a - 1) // x\n",
"7f9c5889-6307-47e1-92b2-a89f07409278": "\ndef problem_p03861(input_data):\n a, b, x = map(int, input_data.split())\n\n return (b - a + 1 + (a - 1) % x) // x\n\n",
"857da2b3-1b2c-4ce1-96a1-005d8b06df24": "import math\n\n\ndef problem_p03861(input_data):\n a, b, x = map(int, input_data.split())\n return b // x - (a - 1) // x\n",
"9f359406-cb4c-404e-a9d7-c1b61af7e019": "def problem_p03861(a, b, x):\n quotient_b, remainder_b = divmod(b, x)\n quotient_a_1, remainder_a_1 = divmod(a - 1, x)\n\n return quotient_b - quotient_a_1\n",
"a8cc096d-390f-44e0-a0eb-7cbbec651671": "def problem_p03861(a, b, x):\n return b // x - (a - 1) // x\n",
},
3: {
"00f11305-5a9b-43cf-9d97-720b0a465d52": "\ndef problem_p02910(input_string):\n for i, char in enumerate(input_string):\n if char == 'L' and i % 2 == 0 or char == 'R' and i % 2 == 1:\n return \"No\"\n\n return \"Yes\"\n",
"0f234383-ad19-44dd-95c1-9bb5171bf82a": "def problem_p02910(input_data):\n s = eval(input_data)\n for i, c in enumerate(s):\n if ((i % 2 == 0 and c == 'R') or (i % 2 == 1 and c == 'L')):\n return 'No'\n return 'Yes'\n",
"1f1bb4c2-297e-46ca-a1c2-10770e83f877": 'def problem_p02910(input_data):\n s = eval(input_data)\n for i, letter in enumerate(s):\n if (i % 2 == 0 and letter == "R") or (i % 2 != 0 and letter == "L"):\n return "No"\n return "Yes"\n',
"25c43c11-6a29-48f1-878b-06156bf31906": 'def problem_p02910(input_data):\n for i, char in enumerate(input_data, 1):\n if (i % 2 == 1 and char == "L") or (i % 2 == 0 and char == "R"):\n return "No"\n return "Yes"\n',
"35128e9c-4e22-4cd0-9412-b55c3c54a16a": "def problem_p02910(input_data):\n s = eval(input_data)\n \n if 'L' in s[::2] or 'R' in s[1::2]:\n return 'No'\n else:\n return 'Yes'\n",
"4bc36f4a-ddcd-4e0e-a868-5b338d850e21": "\ndef problem_p02910(input_data):\n s = eval(input_data)\n\n flag = all(s[i] == 'R' if i % 2 else s[i] == 'L' for i in range(len(s)))\n \n return 'Yes' if flag else 'No'\n\n",
"5e136046-8d21-4e8b-b5d1-6efa23fc7f6e": "def problem_p02910(input_data):\n s = eval(input_data)\n for i in range(len(s)):\n if (i % 2 == 0 and s[i] == 'R') or (i % 2 == 1 and s[i] == 'L'):\n return 'No' \n return 'Yes'\n",
"830e8468-8c1e-4dc6-952e-203148dfe5d6": "def problem_p02910(input_data):\n s = eval(input_data)\n if any(s[i] == 'L' for i in range(0, len(s), 2)) or any(s[i] == 'R' for i in range(1, len(s), 2)):\n return 'No'\n return 'Yes'\n",
"8784754e-3e17-4561-94c7-05b9b5f2aedf": 'def problem_p02910(input_data):\n s = str(input_data)\n\n for i, char in enumerate(s, 1):\n if (i % 2 == 1 and char == "L") or (i % 2 == 0 and char == "R"):\n return "No"\n\n return "Yes"\n',
"96e129db-323a-4651-a021-32f5321e11e9": "\ndef problem_p02910(input_data):\n s = eval(input_data)\n\n for i in range(len(s)):\n if i % 2 != s[i - 1] == 'R' or i % 2 == s[i - 1] == 'L':\n return \"No\"\n return \"Yes\"\n",
},
4: {
"000711ac-1663-496b-837b-946040155a9c": '\ndef problem_p03965(input_data):\n iG, iR = 0, 0\n input_data = input_data.rstrip()\n \n for s in input_data:\n \n if s == "g" and iG > 0:\n iR += 1\n iG -= 1\n \n elif s == "g" :\n iG += 1\n \n elif iG > 0:\n iG -= 1\n\n else:\n iR -= 1\n iG += 1\n \n return iR\n\n',
"0c1d4ab3-9eca-4b24-a274-1cf4aa35ec0d": 'def problem_p03965(input_data):\n iG = 0\n iR = 0\n\n for s in input_data.rstrip():\n if s == "g":\n iR += (0 < iG)\n iG += 1 - (0 < iG)\n else:\n iG += 1 - (0 < iG)\n iR -= 1 - (0 < iG)\n return iR\n',
"470d47be-1d1b-4cff-927f-0e9200d781cc": 'def problem_p03965(input_data):\n # len(s)//2-s.count("p")\n\n iG = iR = 0\n\n for s in input_data.rstrip():\n\n if s == "g" and iG > 0 or s != "g" and iG <= 0:\n iR += 1\n iG -= 1\n \n iG += 1\n\n return iR\n',
"546095ca-bd89-4cdd-b0a8-3694d8013644": 'def problem_p03965(input_data):\n iG = iR = 0\n stripped_data = input_data.rstrip()\n\n for s in stripped_data:\n\n iG += 1\n \n if s == "g" and iG > 0:\n iR += 1\n iG -= 1\n elif iG <= 0:\n iR -= 1\n\n return iR\n',
"81fdc495-f548-4621-89c4-691d3b74f77b": 'def problem_p03965(input_data):\n iG = iR = 0\n\n for s in input_data.rstrip():\n\n if (s == "g" and 0 < iG) or (s != "g" and iG <= 0):\n iR += 1\n\n iG += -1 if iG > 0 else 1\n\n return iR\n',
"83b6579b-8f00-4829-944c-904c8d7b80f9": 'def problem_p03965(input_data):\n iG, iR = 0, 0\n for s in input_data:\n if s == "g":\n iR += 1 if 0 < iG else 0\n iG = (iG - 1 if 0 < iG else iG + 1)\n else:\n iG = (iG - 1 if 0 < iG else iG + 1)\n iR -= 1 if iG < 0 else 0\n return iR\n',
"849e8b87-fb91-4f85-84cf-ade32dab0f3f": 'def problem_p03965(input_data):\n iG, iR = 0, 0\n \n for s in input_data.rstrip():\n if s == "g":\n iR, iG = (iR + 1, iG - 1) if iG > 0 else (iR, iG + 1)\n else:\n iG, iR = (iG - 1, iR) if iG > 0 else (iG + 1, iR - 1)\n\n return iR\n',
"9ee7a8e5-f8c5-49ee-aa71-4010ea432781": 'def problem_p03965(input_data):\n iG = 0\n iR = 0\n for s in input_data.rstrip():\n if s == "g":\n iG -= iG > 0\n iR += iG > 0\n iG += 1\n else:\n iR -= iG <= 0\n iG -= iG > 0\n iG += 1\n return iR\n',
"bd157d9b-3892-48fb-a106-908631fd6a7c": 'def problem_p03965(input_data):\n iG, iR = 0, 0\n\n for s in input_data.rstrip():\n if s == "g":\n if iG:\n iR += 1\n iG -= 1\n else:\n iG += 1\n else:\n if iG:\n iG -= 1\n else:\n iR -= 1\n iG += 1\n return iR\n',
"fdc5337e-8a6a-44ed-8255-d7fca5b49786": 'def problem_p03965(input_data):\n iG, iR = 0, 0\n for s in input_data.rstrip():\n if (s == "g" and iG > 0) or (s != "g" and iG <= 0):\n iR += 1\n iG -= 1\n else:\n iG += 1\n return iR\n',
},
5: {
"04e41d16-15f6-4341-a976-f420ed3180a6": "def problem_p02553(input_data):\n\n a, b, c, d = map(int, input_data.split())\n\n max1 = max(a * c, a * d, b * c, b * d)\n\n if max1 < 0:\n\n if (a < 0 and b > 0) or (a > 0 and b < 0) or (c < 0 and d > 0) or (c > 0 and d < 0):\n\n return 0\n\n else:\n\n return max1\n\n else:\n\n return max1\n",
"09edd39b-4a6f-4422-8c14-23d2a819f2c5": "def problem_p02553(input_data):\n\n a, b, c, d = map(int, input_data.split())\n\n hoge = [a * c, a * d, b * c, b * d]\n\n max_hoge = max(hoge)\n\n if max_hoge < 0:\n\n if a * b < 0 or c * d < 0:\n\n return 0\n else:\n\n return max_hoge\n\n else:\n\n return max_hoge\n",
"19d21335-2ee5-4a5e-8bab-50805248cf9f": "def problem_p02553(input_data):\n a, b, c, d = map(int, input_data.split())\n\n max_val = max(a*c, a*d, b*c, b*d)\n\n if max_val < 0:\n if (a<0) != (b<0) or (c<0) != (d<0):\n return 0\n\n return max_val\n",
"1aa3d7da-d34e-46a3-a72c-d0f1cd0bd5ba": "def problem_p02553(input_data):\n a, b, c, d = map(int, input_data.split())\n \n max_product = max(a * c, a * d, b * c, b * d)\n \n if max_product < 0:\n if a * b < 0 or c * d < 0:\n return 0\n\n return max_product\n",
"36eeb34d-6a89-42f5-8340-e7a31bab073e": "def problem_p02553(input_data):\n a, b, c, d = list(map(int, input_data.split()))\n\n hoge = max(a * c, a * d, b * c, b * d)\n\n if hoge < 0:\n if (a < 0) != (b < 0) or (c < 0) != (d < 0):\n return 0\n\n return hoge\n",
"45c372c7-8b7f-4378-ba39-fdcde9c7a13e": "def problem_p02553(input_data):\n a, b, c, d = map(int, input_data.split())\n \n max_product = max(a*c, a*d, b*c, b*d)\n \n if max_product < 0 and (a < 0) == (b < 0) and (c < 0) == (d < 0):\n return 0\n else:\n return max_product\n",
"881c3174-228a-4592-a681-d8e88bc77aeb": "def problem_p02553(input_data):\n a, b, c, d = map(int, input_data.split())\n\n max_value = max(a * c, a * d, b * c, b * d)\n\n if max_value < 0:\n if (a > 0 > b) or (c > 0 > d):\n return 0\n\n return max_value\n",
"a7e3f286-cc2b-49ad-a958-5429a32408e5": "def problem_p02553(input_data):\n a, b, c, d = map(int, input_data.split())\n\n max_val = max(a * c, a * d, b * c, b * d)\n \n if max_val < 0 and ((a > 0) != (b > 0) or (c > 0) != (d > 0)):\n return 0\n else:\n return max_val\n",
"ab6accd2-494e-4bd4-b91c-7f38125eb9a7": "\ndef problem_p02553(input_data):\n a, b, c, d = map(int, input_data.split())\n \n max_value = max(a * c, a * d, b * c, b * d)\n\n if max_value < 0:\n if (a < 0) != (b < 0) or (c < 0) != (d < 0):\n return 0\n else:\n return max_value\n else:\n return max_value\n\n",
"b35e4ea8-b192-415f-8dcc-817b76b1319c": "def problem_p02553(input_data):\n a, b, c, d = map(int, input_data.split())\n\n hoge = [a * c, a * d, b * c, b * d]\n max_hoge = max(hoge)\n\n if max_hoge < 0:\n if a * b < 0 or c * d < 0:\n return 0\n\n return max_hoge\n",
},
},
"explanations_post": {
0: {
"00db208c-b4df-4ac0-bf2c-4448f0c18897": 'The revised code snippet computes the same result but does it more efficiently. Using a dictionary instead of the index lookup method reduces the time complexity to O(1). The input is now directly passed to the `problem_p02847()` function without using the potentially unsafe `eval()` function, which simplifies the overall structure. Removing the `main()` function and `if __name__ == "__main__"` checks also makes the code cleaner and more straightforward to understand while preserving the functionality of the original code. This optimized code is faster and more efficient, while still delivering the same output as the original code.',
"098b2a24-8841-496c-b4aa-01b3f680836b": "The input_data parameter to the main function was updated to be directly used as our day selector instead of evaling it. The original code was inefficient and potentially unsafe because of the use of eval(). \n\nThe nested structure was removed, and the function was simplified to directly compute days_until_sunday from the input_data string. This results in a simpler, safer, and more pythonic code. \n\nThe new structure of the code still performs the exact same functionality of finding the number of days to Sunday from a given day of the week. The operation was preserved as it was originally intended but with a more straightforward approach. The changes do not break compatibility with other components as the function signature remains the same. \n\nBy removing the unnecessary function definition and eval() call, we have a cleaner, leaner, and more efficient function that is more readable and maintains the external functionality and behavior of the original code.",
"20b7cb92-6906-448d-901e-2ce98fefe2e3": "I've made several changes to improve the performance and efficiency of this code. First, I removed the use of `eval` to parse the input. `eval` is generally slower and less secure than directly parsing the input data. Instead, I now directly access the input string. \nSecond, I replaced the list data structure with a dictionary. List traversal using the `index` function requires O(n) time whereas retrieving an item from a dictionary is an O(1) operation, this decreases the time complexity of the code. \nThe final function works by mapping each weekday to the number of remaining days in the week from this day. Then, it simply returns the corresponding value for the given day. \nDespite these changes, the function preserves the external functionality and behavior of the original code - it still calculates and returns the number of days left in the week from the provided day.",
"239ec7be-8ce0-4c86-8b56-0704e791a372": "I've removed the unnecessary encapsulation of logic under the 'main' function to simplify the overall function. I have also removed the conditional check of \"__name__ == '__main__'\" which is typically used to restrict sections of code from running when the script is imported as a module, as it is not relevant in this context. Additionally, eval() call is removed because it is unnecessary and can pose potential security risks by evaluating the 'input_data' which should be a simple string in this case.\n\nFinally, we have handled the case where the current day is 'SUN' with a single line using modulo operation, making it more readable and efficient. The function now performs exactly the same task, but with unnecessary lines of code removed, it is far more simplified, readable and efficient. The days of the week are converted into numbers (0-6), and a modulo operation is used to wrap the week around efficiently and avoid the use of a conditional. This results in a small, but measurable, improvement in runtime efficiency for the function, especially when called numerous times.",
"8d67ed70-e190-4a84-b5d8-37defb27f6d6": 'I have made several changes that have improved the overall efficiency and readability of the code. These include eliminating the use of the list and the eval function, as well as removing the nested function and the unnecessary `if __name__ == "__main__":` check. The final code remains functionally the same, accepting a string representing a day of the week, and returning the number of days left in the week (or 7 if the given day is Sunday). It does so now more efficiently by using a dictionary for lookups, which has constant time complexity. This optimized implementation ensures that the code performs more rapidly and with less memory usage, while being clearer and more readable.',
"ba6a30d4-3150-482e-b932-6320f4635ebe": "I've streamlined the function `problem_p02847` to directly take an input string `S` (representing a day of the week), calculate the number of days until Sunday including the present day, and return this number.\n\nFirstly, I've eliminated the use of the `eval` function, which is considered a security risk and was unnecessary in this case since the input should be a string (representing a day).\n\nSecondly, I simplified the function by removing the nested `main` function. This made the code more succinct and straightforward without changing the algorithm itself or the output.\n\nFinally, I removed the `if __name__ == \"__main__\":` part as it was not serving any purpose here. In a script, this would be used to allow or prevent parts of code from being run when the modules are imported, but in this case, it was not being used effectively.\n\nThe revised function directly computes and returns the number of days until Sunday from the given day, maintaining the same functionality as the original code but with a more streamlined structure and potentially improved performance due to the removal of unnecessary or non-optimal elements.",
"c2fe9046-d193-4a51-a1d5-111c172e00f8": "The optimized code now uses a dictionary instead of a list to store and fetch the number of days remaining until Sunday. Dictionary lookups have a time complexity of O(1), which is faster than the O(n) time complexity of the list's index() method used in the original code. In addition to a performance boost, the dictionary also provides a more straightforward and cleaner way to compute the answer. The dictionary directly maps each day of the week to the number of days left until Sunday, thus eliminating the need for calculation or conditional checking.\nThe optimized code provides the same output as the original code. It accepts a string that represents a day of the week and returns the number of days left until Sunday, preserving the desired behavior. The function's name and expected input and output types remained unchanged, thus maintaining compatibility with any existing code that may be using it.",
"d2ff6d72-4515-4094-9d80-89f75045a9b5": "The refactored function `problem_p02847` maintains the core functionality of the original code, but without the unnecessary nested `main()` function. This code flattening makes the codebase cleaner and more readable without altering the purpose of the original function.\n\nThe purpose of this function is to get the days remaining until the next Sunday given the current weekday. The function is very specific and efficient for its intended usage. Here, I am making a non-performance related refactor to improve the code readability and maintainability.\n\nFinally, the refactored function's return value matches the original function's return value exactly, ensuring that calling code won't be broken by this change. The function's external functionality and behavior are preserved exactly. This approach ensures that any code using this function will not be impacted by this optimization.",
},
1: {
"426916b1-b122-44a7-aae3-c806b81887bd": "The original function used a brute-force approach that was inefficient and can result in high runtime for large inputs. The updated function uses binary search to find the maximum integer `i` such that `i` to the power `n` divides `p` exactly.\n\nIn binary search, the search space is divided in half at each step, reducing the number of possibilities to check, which results in a time complexity of O(log n) compared to O(n) in the original function.\n\nThe updated function maintains the behavior of the original function and doesn't change its external API. It will still take a string containing two space-separated numbers and return an int. Therefore, it won't break any existing code that uses this function. At the same time, the updated function will be faster and more efficient, especially for large inputs.",
"4a18cc5a-17a9-4aba-8b56-4534c406637e": "In the new code, I used a Binary Search when `n` is in the range from 2 to 40. This is because `i` is monotonic with respect to `i^n`, so Binary Search is applicable. Instead of incrementing by one, we now increment by larger steps, effectively cutting the search space in half in every iteration. \n\nBinary Search returns the smallest `i` such that `i^n > p` or `p` is not divisible by `i^n`. Thus, we need to return the result minus one. \n\nThe new code should be substantially faster for large inputs because the time complexity of Binary Search is O(log n), while the time complexity of the brute force method was O(n), where n is the input size. In our case, `n` is the difference between `p ^ (1/n)` and 1, which is exponential in size. So the improvement is significant.\n \nThe changes I've made preserve the external functionality and behavior of the original code. If `n` equals 1, or `n` is more than 40, the function returns `p` and 1 respectively, as before. If `n` is in the range [2, 40], the function returns the largest `i` such that `i^n <= p` and `p` is divisible by `i^n`, as before.",
"67cbfea8-0699-4750-a758-4078b3d33fb8": "The refactored code should have the same functionality as the original code: it converts the input string to two integers, `n` and `p`, and calculates the exact `n`th root of `p` or the largest `i` where `i^n` is a divisor of `p`.\n\nThe refactored code improves the performance by calculating the `n`th root of `p` in the beginning and looping only until `root`. Thus, it saves time compared to the original version, which keeps on increasing `i` and checking the condition each time in an indefinite while loop. Additionally, using Python's power operation (`**`) to calculate the `n`th root of `p` is likely more optimized and faster than the original approach.\n\nThe important thing to note is that this refactoring doesn't change the external behavior or functionality of the original code and so compatibility is preserved. It will still return the same data for the same input but in a more efficient and faster manner for larger values of `p`.",
"68416ac3-693a-432f-b19c-a74e5bcb5d80": "The original code used an infinite while loop until i to the power of n was greater than p which is not efficient for larger inputs. I optimized this by deriving the maximum possible integer 'i' which replaced the infinite loop with a definite bounded one.\n\nUsing the mathematical property that n-th root of 'p' can be calculated by raising 'p' to the power 1/n, I found the value of the maximum possible 'i'. This reduces the number of iterations potentially by a lot and hence reduces the time complexity of the code.\n\nOnly the mechanism of deciding the upper limit for 'i' is changed. The loop continues to increment 'i' and the conditional statement of checking the divisibility remains intact, thereby preserving the overall functionality and external behavior of the code. So, the code is fully backward compatible and will not break when used in conjunction with other functions or methods.",
"7afadb38-0b8c-41b6-a51f-78f9cbf17d63": "The binary search method is used to replace the original brute-force search method in the code. Binary search is an optimization intended to speed up the searching process. In binary search, instead of searching the elements one by one, the middle of the range is checked in each step. The search space is halved at each step of binary search, so it is more efficient and faster when dealing with large inputs.\nI calculate the middle number (`mid`) between `low` and `high`, then check whether `mid` to the power of `n` is still less than or equal to `p`. If it is, we know that `mid` might be the answer, and also the numbers larger than `mid` might be the answer, so we move `low` to `mid + 1`. If `mid` to the power of `n` is greater than `p`, that means the numbers larger than `mid` cannot satisfy the requirement, so we move `high` to `mid - 1`.\nAs for compatibility, the optimized code takes the same inputs and provides the same outputs as the original code. No features have been removed, and more specifically, the function name, input parameters, and return have not been changed, which ensures full backwards compatibility with other existing codes that use this function. Therefore, the revised code behaves externally in the exact same way as before. The code is cleaner, faster, and more efficient due to the implementation of the binary search.",
"7d96a7c8-29b6-43d9-a682-fbcf1d36103d": "In the optimized code, we've replaced a linear scan with a binary search, reducing the time complexity from O(p) to O(log p). \n\nBinary search allows us to efficiently traverse through the search space (which is 1 to `p` inclusive) to find the highest `i` for which `i^n` doesn't exceed `p` and `p` is evenly divisible by `i^n`. We start by setting the `lower` to 1 (smallest possible value) and `upper` to `p` (largest possible value). We then initiate a while loop that continues until `lower` and `upper` converge. In each iteration, we calculate the mid-point `mid` between `lower` and `upper`. If `mid^n` exceeds `p`, we update `upper` to `mid - 1`. Otherwise, we record `mid` as a possible solution and update `lower` to `mid + 1`. Lastly, the function returns the highest `i` we could find.\n\nGiven these adjustments, the function works just as before (returns the highest `i` for which `i^n` doesn't exceed `p` and `p` is evenly divisible by `i^n`) but now has a substantially improved performance in terms of speed compared to the original version. This optimization doesn't change any present functionality and thus preserves the external functionality and behavior of the original code.",
"e3a6d7b8-e145-45c3-bd1c-8cae99c6d014": "The changes made to the code replaced the simple iteration with Binary Search, thus improving the time complexity of the function. The function maintains the same functionality as before, returning `p` if `n` equals 1, 1 if `n` is more than 40, and the largest `i` such that `i^n` is less than or equals to `p` otherwise. Binary Search halves the search space in each step, making it more efficient. By using this approach, the code can handle larger inputs more efficiently.",
},
2: {
"0a631960-6c33-4de4-8d38-b31e888918c7": "Here is an optimized version of your program.\n\nPlease note that this code won't strictly run faster as the operation taking the most time in your original program is most probably a function map() (or divisions). The provided code is just a cleaner and more \"pythonic\" version of it. The real optimization of the program runtime speed lies in the algorithm used, and as for this specific function, there's not much to optimize.\n",
"38b31a95-7409-4c66-a50d-24dd565bb513": "Your function is already quite optimized. However, we could avoid importing the unused `math` library. Here is the revised function.\n",
"6ba45be2-5f81-4952-8dc6-c4569092096b": "Your provided Python program is reasonably optimized, but it could potentially be made slightly faster by eliminating the `list(map(...))` usage in the splitting process. Here is a slightly optimized version of the provided function using this method.\n\n\nIn this version, we directly unpack the results of `map(int, input_data.split())` to `a, b, x`. We avoid the usage of `list` which was not necessary here. This does a very little optimization as the usage of list was redundant in your case.\n\nPlease note that these are micro-optimizations and you will rarely see the difference in time taken by both versions. The real time consuming part here is the string parsing which can't be optimized and the integer division which is already optimized in Python.",
"7f9c5889-6307-47e1-92b2-a89f07409278": "python\n\nThe optimized code runs faster because it reduces the number of divisions, which are computationally expensive operations. It does this by implementing the mathematical properties to compute the number of multiples of x in the [a, b] range directly.",
"857da2b3-1b2c-4ce1-96a1-005d8b06df24": "The function provided is pretty efficient already. However, the program can be made more efficient by separating the import statement from the function, as importing is a operation that can be done once at the top of the script. The efficiency of this division operation depends on the python interpreter, which is usually optimized.\n\n\n\nNote: The efficiency improvement from this refactoring could be negligible, but it's a good practice to not put import statements inside functions.",
"9f359406-cb4c-404e-a9d7-c1b61af7e019": "The provided code reads and splits the input_data then performs division operation. It looks already optimized. However, the direct use of the `divmod` function can help to slightly improve performance over repeated use of `//` operator.\n\nPlease find the optimized program below.\n\n\nGiven that we already have three integers a, b, x, the splitting and mapping already done in integer format can be removed for the function to be more efficient. Here's how it can be done.\n\n\nThis change would be faster only when `problem_p03861` is called with integer parameters directly. Please note that the output of the function remains same as the original.",
"a8cc096d-390f-44e0-a0eb-7cbbec651671": "The given python program is already optimized to a great extent. However, we can avoid the split() and map() operations as these can be relatively slow especially for large datasets. This requires passing the inputs as integers instead of a string which will definitely speed up the program.\n\n\nIn the above code, it is assumed that a, b, x are integers as input instead of a single string. The function now takes three input arguments rather than taking a string and then splitting it into individual parts.\n\nRemember, these changes affect the way the function is used and the type of input it accepts. If the input must be a string with numbers separated by spaces, then the original implementation was already quite fast. In python, the algorithmic complexity and the memory usage of your function is already optimal. Python's overhead for function calls etc. is also quite low so in this case it wouldn't have a noticeable effect.",
},
3: {
"00f11305-5a9b-43cf-9d97-720b0a465d52": "Your current function iterates through the entire string, even when it's clear that the string doesn't meet the conditions. This unnecessary work slows down your function. Besides, it uses eval() - which is generally dangerous and should be avoided unless absolutely needed. Here's a faster version.\n\n\n\nThis optimized function works by checking if character 'L' is at an odd index and/or character 'R' is at an even index right from the start. If this condition is met, it immediately returns `\"No\"` and breaks the loop which speeds up the function. If the function goes through the entire string and doesn't find any such occurrence, it returns `\"Yes\"`. I've also used the more pythonic way of enumerating through the string using the built-in `enumerate()` function.",
"0f234383-ad19-44dd-95c1-9bb5171bf82a": 'Here is the optimized code for your problem. The solution reduces the run time complexity by avoiding unnecessary iterations on all the elements in the given list and immediately returns "No" as soon as the condition matches. The functionality of the code remains the same as the original.\n\n\nThis version of the problem will check each character in the string only once, instead of checking each character twice- once for odd and once for even positions. We also eliminated the need of maintaining an extra variable `flag`. So it\'s more optimal in terms of runtime and memory usage.',
"1f1bb4c2-297e-46ca-a1c2-10770e83f877": "The optimized version of your program is as follows.\n\n\nI've made the following optimizations.\n\n1. I used the built-in Python `enumerate` function to iterate over the characters in the string `s` along with their index. This is more readable and efficient because it avoids computing `i - 1` and accessing `s[i - 1]` on each iteration.\n \n2. I combined the checks for `i % 2 == 1` and `i % 2 == 0` into a single line and used a more direct check for 'L' and 'R' conditions which prevents unnecessary checks in each loop iterations.\n \n3. I removed the `flag` variable and returned directly when a condition was met. This allows the function to exit immediately when it determines the result, rather than continuing to check the remaining characters.",
"25c43c11-6a29-48f1-878b-06156bf31906": "Here's your function rewritten in a more optimized way. I've removed the `eval` function, which can be unsafe and slow, assuming that `input_data` is already a string. The function now stops iterating as soon as it finds an invalid character in the string, thus potentially reducing the number of iterations.\n\n\nThis version of the function will produce exactly the same result as your original function, but should run faster in the case of large inputs that contain an invalid character near the start of the string. By using `enumerate`, we efficiently loop over the characters in the string and their indices at the same time, which is more Pythonic and can be faster than a range-based loop.",
"35128e9c-4e22-4cd0-9412-b55c3c54a16a": "The given program iterates through every element of the input data string, testing each character one by one against a condition. For a very long string, this could be quite inefficient as the time complexity is O(n) - linear time complexity. \n\nOne approach to optimize the program is to utilize Python's built-in operations which operate at C speed internally. We can rewrite the program in a more efficient way by examining all odd-index and even-index characters simultaneously, instead of looping through each character. \n\n\nThis program operates by checking all of the odd-indexed and even-indexed positions at once for 'L' and 'R', respectively, using Python's slicing feature. For larger datasets, checking them all at once will be substantially faster than performing a loop operation. This program has a complexity of O(1) - constant time complexity.",
"4bc36f4a-ddcd-4e0e-a868-5b338d850e21": "We can speed up your program a bit by reducing the time complexity and remove unnecessary checks. Here I'm using list comprehension and all() function to make the checks within a single line. This solution has a time complexity of O(n), but avoids excessive branching which may slow down the execution.\n\n",
"5e136046-8d21-4e8b-b5d1-6efa23fc7f6e": 'Your original code is walking through each character, and checking the number at the index and character at that index. If the index is odd, you check if the character is "L", and if the index is even, you check for "R", which is essentially saying that all odd-indexed positions should be "R" and all even-indexed positions should be "L". If that\'s not the case, it returns "No". Once the flag turns False, there is no need to continue checking. So, there could be a break to terminate the loop earlier.\n\nHere is an optimized version where the function early-exits if a character is in the wrong position.\n\nIn the optimized version, the function will return \'No\' as soon as it detects an inappropriate character, which could potentially save a lot of unnecessary iterations, especially when dealing with larger inputs.',
"830e8468-8c1e-4dc6-952e-203148dfe5d6": "Your function is iterating over the entire string checking for 'L' in the odd indices and 'R' in the even indices. This can be optimized by using generator expressions with any() built-in function, which is faster and stops as soon as the condition is met without going through the entire string. \n\nHere's your optimized solution.\n\n\n\nIn this solution, 'any' function is used which stops iteration as soon as it encounters 'True' (when a 'L' is at an odd index or 'R' is at an even index) and 'range' is used with a step of 2 to only check odd and even indices, improving the time complexity from O(n) to O(n/2). Also, using 'any' is generally faster than for loops due to internal implementation optimizations.",
"8784754e-3e17-4561-94c7-05b9b5f2aedf": 'The given code can be simplified and optimized by reducing the number of operations in the loop and eliminating the use of the `eval()` function whichnot only slows down the code but is also a security risk. Here\'s an improved version of the code.\n\n\n\nThe `enumerate()` function was used here which is more pythonic and efficient than using the `range()` function. It minimizes the number of operations and overall makes the code faster and more readable. Also we return immediately after any "L" character found in odd places or "R" in even places. This eliminates unnecessary loop iterations.',
"96e129db-323a-4651-a021-32f5321e11e9": 'Here is a rewrite of original function with improved performance.\n\nThe original function is inefficient because it checks each character of the string regardless of whether flag has already been set to False or not. The rewritten version checks each character and returns "No" as soon as a mismatch is found which reduces the number of iterations and increases the speed significantly when a mismatch is found early in the string. This pattern of early return when possible is a good practice for optimization in many scenarios. In the worst case, both versions of the function have the same time and space complexity - O(n).\n\nAnother speed-up in the refactored code comes from the fact that `i % 2` is calculated only once per iteration as opposed to twice in the original code.',
},
4: {
"000711ac-1663-496b-837b-946040155a9c": "Here is your optimized code. I didn't change the logic, I only streamlined the for loop.\n\n\nThe use of elif helps speed up the code, as it immediately exits the conditional after a true case is found. In your original code, even after a true case was found, the program would continue checking all cases until none were left. The strip operation also moved out of the loop to only execute once.",
"0c1d4ab3-9eca-4b24-a274-1cf4aa35ec0d": "The runtime of your current program is O(n) because you are looping through each character in the input string once. However, the code includes many redundant if statements that can be simplified to make the code more efficient and more readable.\n\n\nIn this version, the if statements are combined into single lines, and the sum operation is moved outside of the if statement. The condition `(0 < iG)` will return either `True` which is equivalent to `1` in python or `False` which is equivalent to `0`. This way the substraction or addition operation is always executed but it doesn't change the value of `iG` or `iR` if the condition is not met. It should run the same logic as your code but in a more efficient way.",
"470d47be-1d1b-4cff-927f-0e9200d781cc": 'The provided Python function seems to iterate over given input string "input_data" and applies a certain logic based on whether a character is "g" or not. While the logic inside the function isn\'t clear, we can improve its efficiency by avoiding unnecessary operations like redundant elif and else checks.\n\nHere\'s the optimized version of the given function.\n\n\nThis new version handles "g" and non-"g" characters at the same time, effectively optimizing the code by reducing code size and improving readability. The optimization mainly comes from eliminating redundant elif and else checks.',
"546095ca-bd89-4cdd-b0a8-3694d8013644": "Here is the optimized version of your python program.\n\n\n\nThe optimization consists of the following.\n\n1. Merging the initialization of `iG` and `iR` into a single line.\n2. Pre-computing `rstrip` method call for `input_data` outside the loop. This avoids repeated function calls in each iteration.\n3. Simplifying the nested if conditions by merging common statements. Since the increment of `iG` always occurs, it is moved outside the conditional statements. The conditions are then simplified, reducing duplicate code and number of checks.\n4. The `else` part in the original code is also unnecessary as it only executes when `iG` is 0, but if `iG` is 0, it's reset to 1 in both parts of the `if` statement. So, we only need to check if `iG` is less than or equal to zero and adjust `iR` accordingly.",
"81fdc495-f548-4621-89c4-691d3b74f77b": "This is a more optimized version of the same function in Python.\n\n \n\nThis version reduces the number of operations inside the loop. The conditions of the if statement have been combined to remove repetition. Also, the increment and decrement operation on 'iG' is done in a single line instead of repeated under separate conditions, thus reducing the total number of operations. This minor optimization can result in a faster runtime especially when processing large amounts of data.\n\nHowever, in most cases, the speedup would be negligible as Python's optimization will already handle this effectively. The rewritten code is simply a tidier and more readable format of the original code.",
"83b6579b-8f00-4829-944c-904c8d7b80f9": 'Your function has multiple unnecessary "if-else" checks and a rstrip() operation that can be removed to optimize the function. Here\'s a revised version of your function.\n\n\nThis version of the function works exactly like the original one, but with less operations and simpler code which makes it run faster.',
"849e8b87-fb91-4f85-84cf-ade32dab0f3f": "Here is a more optimized version of your function.\n\n\n\nThe ternary if-else statements help condense the logic within the for loop. This might allow your function to run faster depending on the Python interpreter's optimizations. However, the time complexity remains the same because we have not changed any underlying algorithmic approach. Every string character is still considered once, hence runtime remains O(n) where n is the length of the input string. It's also important to note that the total time will always depend on the specific data and how well it interacts with Python's internal optimizations. \n\nIn terms of memory usage, the optimized version will not make any significant difference as well, because the original and the optimized versions both use a constant number of variables.\n",
"9ee7a8e5-f8c5-49ee-aa71-4010ea432781": "The provided function's primary issue lies with its complexity and redundancy due to unnecessary operations and iterative steps. To optimize and speed up this function, we can avoid decreasing and then increasing the same variable (iG), and also handle line by line computations more efficiently using Python's in-built functions and simplified flow control.\n\nThe rewritten code is as follows.\n\n\nThis transformed version of the function provides the same functionality as before but with enhanced performance due to the reduced number of variable manipulation and iteration steps. The optimization lies in the single liner computation inside the `for` loop where the increment, decrement of variables is happening based on the evaluation of conditions.",
"bd157d9b-3892-48fb-a106-908631fd6a7c": "The provided python function can be optimized for faster runtime by removing unnecessary code blocks and by performing more efficient comparisons. Here's my optimized version of the function.\n\n\n\nIn this approach, we have.\n\n1. Initialized `iG` and `iR` variables on the same line to reduce the lines of code.\n2. Used boolean truthiness of values in Python as a shorthand to avoid unnecessary comparisons with zero. In Python, any non-zero value is considered True, while zero, None, and empty collections are considered False. So, instead of `if 0 < iG:` we can simply write `if iG:`.\n3. Removed blank lines to make the code more compact. It won't necessarily make the code faster, but it does make it cleaner and easier to read. \n\nPlease note that the return value remains exactly the same as before, as required. The optimization is mainly targeted to make the code more concise, readable and pythonic, which in turn may lead to minor performance improvements. However, the overall algorithmic complexity remains the same.",
"fdc5337e-8a6a-44ed-8255-d7fca5b49786": "Here is the optimized version of your code.\n\n\n\nThe optimization includes avoiding unnecessary branches, combining conditions, and minimizing operations.",
},
5: {
"04e41d16-15f6-4341-a976-f420ed3180a6": "Here's a rewritten function that will run faster. Since you're okay with preserving the function signature and return value, I'm assuming that I can use basic Python functionality instead of modules.\n\n\n\nNumpy is certainly a powerful module, but in this case, it adds unnecessary overhead. List creation and the append method also take additional resources (both in terms of time and memory). Therefore, I've eliminated the numpy module and the list, calculating the maximum value directly via the max function.\n\nI also streamlined the check for negative numbers. Instead of invoking np.sign (which you then compare with not-equal), I check directly whether one number is positive and the other is negative.\n\nThese modifications will make your code run faster, and should also reduce memory usage, especially for large inputs. Plus, since I haven't changed the function's name or signature, or what it returns, it should still meet your specified requirements.",
"09edd39b-4a6f-4422-8c14-23d2a819f2c5": "It appears that the biggest issue with your program's efficiency is the repeated computation of max(hoge) and unnecessary imports. You can streamline this by storing the result of max(hoge) into a variable immediately after computing it. Also, it seems like you don't really need importing numpy as np, as you can find the sign using just Python. This should most likely increase the speed of your program.\n\n\n \n\nIn this code, we only compute max(hoge) once and just reference it later, which saves time. Additionally, the use of 'map' directly to convert the numbers to integers is more efficient. And I changed the sign comparisons using built-in Python operations instead of numpy to make it faster and removed the numpy import.",
"19d21335-2ee5-4a5e-8bab-50805248cf9f": "The program can be optimized by removing the costly import of numpy and unnecessary list constructions. The maximum number generated from multiplications can be calculated in one line, rather the appending to a list. Also, the final check for the maximum element can be calculated once and stored in a variable so that we don't calculate maximum again and again. Here is the optimized version of the program.\n\n\nThis version of the function gives the same results as the previous one, but with more efficient memory usage and faster execution because it avoids creating a few unnecessary objects in memory and performs fewer function calls. The number of computations in the rewritten function is smaller than in the original function, which can translate to a speed up in execution.",
"1aa3d7da-d34e-46a3-a72c-d0f1cd0bd5ba": "You can optimize the program by directly computing the maximum of the products without storing all four products in a list. In addition, the if-else statement to return the maximum can be simplified. Here's an optimized version.\n\n\n\nThis version avoids unnecessary computations and storage, improving the speed of your algorithm.",
"36eeb34d-6a89-42f5-8340-e7a31bab073e": "Here is the optimized version of your program. This version reduces unnecessary calls to the `max` function and removes unused numpy import. Also, it removes the superfluous list creation and stores the maximum value of the products directly without needing to append them to a list, resulting in less memory consumption.\n\n\n\nIn this optimized version, the sign comparison is done using Python's built-in statement, removing the need for numpy completely. This will make the function more portable (as it no longer depends on numpy) and slightly faster.\n",
"45c372c7-8b7f-4378-ba39-fdcde9c7a13e": "The program you provided has several superfluous operations that slow the computation down, including multiple calls to the max() function. The program also imports numpy just for the sign function, which has a Python standard library equivalent with the function math.copysign.\n\nHere is an optimized version of the program that avoids multiple sign calls and max() calls.\n\n\n\nThis version performs the split and map operations all in one line to convert the input strings into integers. It then calculates the maximum product only once instead of multiple times. It also uses basic equality checks to figure out if the signs of a, b and c, d are the same rather than using numpy's sign function. This will hence require no external library.\n\nThis rewritten program should run significantly faster, especially for large inputs, since it performs fewer operations and does not call the max function multiple times.",
"881c3174-228a-4592-a681-d8e88bc77aeb": "This program can be made to run faster by avoiding the repeated calculation of `max(hoge)` and removing the usage of numpy for the sign checking, since python natively supports this feature. Here's an optimized version.\n\n\nIn this optimized version.\n\n1. I eliminated the `hoge` list altogether and called the `max` function once, calculating the maximum product directly.\n2. I removed the dependency on numpy for sign checking. Python can check if a number is positive or negative with the `>` operator.\n3. I removed the redundant final `else` clause. If none of the conditions are met before, the function will return the `max_value` by default.",
"a7e3f286-cc2b-49ad-a958-5429a32408e5": "Here is your optimized Python code.\n\n\n\nHere's what was changed.\n\n1. Removed the import of NumPy. Native Python can achieve this task without additional libraries, which reduces memory usage.\n2. Calculated max_val only once, instead of twice (once for checking against 0 and twice for the return value).\n3. Removed the list `hoge`, which was holding the multiplication results unnecessarily.\n4. Replaced the numpy sign function with a simple comparison with 0. This new comparison results in boolean values (True or False), which are appropriate for checking the condition. Here, it's crucial to note that in Python, negative numbers are considered less than zero.",
"ab6accd2-494e-4bd4-b91c-7f38125eb9a7": "Sure, here is the optimized version of your provided python code.\n\n\n\nIn the original code, the operation is performed independently for each condition and the maximum value is calculated multiple times, which may negatively affect the performance especially for larger values of a, b, c, and d.\n\nIn the revised version, `max()` is only called once to find the greater value among the products. This reduces the number of operations, which improves the efficiency of the program. \n\nAlso, eliminated the need of `numpy` library function np.sign() and replaced it with more efficient direct approach of comparing signs by checking if number is less than zero. The advantage of this manual comparison is that it's faster since it does not require library function call overhead.",
"b35e4ea8-b192-415f-8dcc-817b76b1319c": "Here is the optimized version of your Python function. Importing numpy and performing repetitive operations, like calling the `max()` function several times, may extend the runtime. So, try to perform everything in one pass.\n\n\nThis version eliminates the unnecessary use of numpy and removes the multiple calls to max() function, which will make the function run faster.",
},
},
"speedup_ratio": {
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"original_runtime": {0: 14625.0, 1: nan, 2: 60039.0, 3: 1029209.0, 4: 244251.0, 5: 81020500.0},
"optimized_runtime": {
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"is_correct": {
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"generated_test": {
0: [
'# imports\nimport pytest # used for our unit tests\nfrom pie_test_set.p02847 import problem_p02847\n\n# unit tests\n\n# Test normal cases with valid inputs for each day of the week\n@pytest.mark.parametrize("input_data, expected", [\n ("\'MON\'", 6),\n ("\'TUE\'", 5),\n ("\'WED\'", 4),\n ("\'THU\'", 3),\n ("\'FRI\'", 2),\n ("\'SAT\'", 1),\n ("\'SUN\'", 7),\n])\ndef test_normal_cases(input_data, expected):\n assert problem_p02847(input_data) == expected\n\n# Test edge cases with invalid day names\n@pytest.mark.parametrize("input_data", [\n "\'FUNDAY\'",\n "\'MOON\'",\n "123",\n "[\'MON\']",\n "None",\n])\ndef test_edge_cases_with_invalid_day_names(input_data):\n with pytest.raises(Exception):\n problem_p02847(input_data)\n\n# Test misuse cases with potentially malicious code\n@pytest.mark.parametrize("input_data", [\n "__import__(\'os\').system(\'echo hello\')",\n "exit()",\n])\ndef test_misuse_cases_with_malicious_code(input_data):\n with pytest.raises(Exception):\n problem_p02847(input_data)\n\n# Test edge cases with lowercase and mixed case inputs\n@pytest.mark.parametrize("input_data", [\n "\'mon\'",\n "\'Sun\'",\n "\'Mon.\'",\n "\'Monday\'",\n])\ndef test_edge_cases_with_case_sensitivity(input_data):\n with pytest.raises(Exception):\n problem_p02847(input_data)\n\n# Test edge cases with additional text\n@pytest.mark.parametrize("input_data", [\n "\'Today is MON\'",\n "\'SUNday\'",\n])\ndef test_edge_cases_with_additional_text(input_data):\n with pytest.raises(Exception):\n problem_p02847(input_data)\n\n# Test edge cases with non-standard capitalization and characters\n@pytest.mark.parametrize("input_data", [\n "\'monDAY\'",\n "\'M@N\'",\n])\ndef test_edge_cases_with_non_standard_characters(input_data):\n with pytest.raises(Exception):\n problem_p02847(input_data)\n\n# Test edge cases with whitespace and control characters\n@pytest.mark.parametrize("input_data", [\n "\'\\t\\nMON\\n\\t\'",\n "\'MON\\u200b\'",\n])\ndef test_edge_cases_with_whitespace_and_control_characters(input_data):\n with pytest.raises(Exception):\n problem_p02847(input_data)\n\n# Test edge cases with similar sounding or looking names\n@pytest.mark.parametrize("input_data", [\n "\'MOAN\'",\n "\'SUNN\'",\n])\ndef test_edge_cases_with_similar_names(input_data):\n with pytest.raises(Exception):\n problem_p02847(input_data)\n\n# Test edge cases with empty and null inputs\n@pytest.mark.parametrize("input_data", [\n "\'\'",\n "None",\n])\ndef test_edge_cases_with_empty_and_null_inputs(input_data):\n with pytest.raises(Exception):\n problem_p02847(input_data)\n\n# Test edge cases with valid Python expressions\n@pytest.mark.parametrize("input_data", [\n "\'6-1\'",\n "\'[x for x in range(5)]\'",\n])\ndef test_edge_cases_with_valid_python_expressions(input_data):\n with pytest.raises(Exception):\n problem_p02847(input_data)\n\n# Test edge cases with input as Python code\n@pytest.mark.parametrize("input_data", [\n "\'os.system(\\\'echo hello\\\')\'",\n "\'(__import__(\\\'sys\\\').exit())\'",\n])\ndef test_edge_cases_with_python_code(input_data):\n with pytest.raises(Exception):\n problem_p02847(input_data)'
],
1: [
'# imports\nimport pytest # used for our unit tests\nfrom pie_test_set.p03196 import problem_p03196\n\n# unit tests\n\n# Test normal cases\n@pytest.mark.parametrize("input_data, expected", [\n ("2 16", 4),\n ("3 27", 3),\n ("2 20", 1),\n])\ndef test_normal_cases(input_data, expected):\n assert problem_p03196(input_data) == expected\n\n# Test edge cases\n@pytest.mark.parametrize("input_data, expected", [\n ("1 100", 100),\n ("1 1", 1),\n ("41 1000000", 1),\n ("100 2", 1),\n ("2 17", 1),\n ("3 19", 1),\n])\ndef test_edge_cases(input_data, expected):\n assert problem_p03196(input_data) == expected\n\n# Test large numbers\n@pytest.mark.parametrize("input_data, expected", [\n ("2 1000000000000", 1000000),\n ("3 1000000000000", 10000),\n])\ndef test_large_numbers(input_data, expected):\n assert problem_p03196(input_data) == expected\n\n# Test invalid input format\n@pytest.mark.parametrize("input_data", [\n "2.5 10",\n "2 -10",\n "2",\n "two 10",\n "2, 10",\n])\ndef test_invalid_input(input_data):\n with pytest.raises(ValueError):\n problem_p03196(input_data)\n\n# Test boundary cases\n@pytest.mark.parametrize("input_data, expected", [\n ("40 1024", 2),\n ("2 1024", 32),\n])\ndef test_boundary_cases(input_data, expected):\n assert problem_p03196(input_data) == expected\n\n# Test rare or unexpected edge cases\n@pytest.mark.parametrize("input_data, expected", [\n ("2 " + str(2**63 - 1), 1),\n ("3 " + str(2**63 - 1), 1),\n ("2 1", 1),\n ("10 1", 1),\n ("37 137**37", 137),\n ("37 138", 1),\n ("3 8", 2),\n ("5 32", 2),\n ("2 6", 2),\n ("2 120", 4),\n ("2 49", 7),\n ("2 77", 1),\n])\ndef test_unexpected_edge_cases(input_data, expected):\n assert problem_p03196(input_data) == expected\n\n# Test zero or negative n values\n@pytest.mark.parametrize("input_data", [\n "0 10",\n "-2 16",\n])\ndef test_zero_or_negative_n(input_data):\n with pytest.raises(ValueError):\n problem_p03196(input_data)'
],
2: [
'# imports\nimport pytest # used for our unit tests\nfrom pie_test_set.p03861 import problem_p03861\n\n# unit tests\n\n# Test typical cases with different ranges and divisors\ndef test_typical_cases():\n assert problem_p03861("1 10 2") == 5\n assert problem_p03861("3 15 3") == 5\n assert problem_p03861("10 100 10") == 10\n\n# Test edge cases where a and b are the same\ndef test_edge_cases_same_a_b():\n assert problem_p03861("5 5 1") == 1\n assert problem_p03861("7 7 2") == 0\n assert problem_p03861("10 10 10") == 1\n\n# Test edge cases where a is a multiple of x\ndef test_edge_cases_a_multiple_of_x():\n assert problem_p03861("6 12 6") == 2\n assert problem_p03861("10 20 10") == 2\n\n# Test edge cases where b is a multiple of x\ndef test_edge_cases_b_multiple_of_x():\n assert problem_p03861("1 10 5") == 2\n assert problem_p03861("3 21 7") == 3\n\n# Test cases where a is greater than b\ndef test_cases_a_greater_than_b():\n assert problem_p03861("10 5 1") == 0\n assert problem_p03861("20 10 5") == 0\n\n# Test cases where x is larger than both a and b\ndef test_cases_x_larger_than_a_b():\n assert problem_p03861("1 5 10") == 0\n assert problem_p03861("2 8 20") == 0\n\n# Test cases with negative numbers and zero\ndef test_cases_negative_numbers_and_zero():\n assert problem_p03861("0 0 1") == 1\n assert problem_p03861("-5 5 5") == 3\n assert problem_p03861("-10 -1 2") == 5\n\n# Test cases with large numbers\ndef test_cases_large_numbers():\n assert problem_p03861("1000000 2000000 100000") == 20\n # This test may need to be adjusted for the expected number of multiples\n assert problem_p03861("123456789 987654321 12345") == 987654321 // 12345 - (123456789 - 1) // 12345\n\n# Test invalid input cases\ndef test_invalid_input_cases():\n with pytest.raises(ValueError):\n problem_p03861("")\n with pytest.raises(ValueError):\n problem_p03861("1 10")\n with pytest.raises(ValueError):\n problem_p03861("1 10 two")\n\n# Test rare or unexpected edge cases\ndef test_rare_edge_cases():\n with pytest.raises(ZeroDivisionError):\n problem_p03861("1 10 0")\n with pytest.raises(ZeroDivisionError):\n problem_p03861("5 5 0")\n assert problem_p03861("0 0 5") == 1\n assert problem_p03861("-10 -1 2") == 5\n assert problem_p03861("-15 -5 3") == 4\n assert problem_p03861("-10 -1 -2") == 5\n assert problem_p03861("-20 -10 -5") == 3\n assert problem_p03861("1 10 -1") == 10\n assert problem_p03861("5 15 -3") == 4\n assert problem_p03861("1000000000 2000000000 100000000") == 20\n assert problem_p03861("2147483647 2147483647 1") == 1\n assert problem_p03861("-2147483648 -2147483648 1") == 1\n assert problem_p03861("2147483647 2147483647 2147483647") == 1\n assert problem_p03861("7 7 7") == 1\n assert problem_p03861("123456 123456 123456") == 1\n assert problem_p03861("0 0 -5") == 1'
],
3: [
"# imports\nimport pytest # used for our unit tests\nfrom pie_test_set.p02910 import problem_p02910\n\n# unit tests\n\n# Test normal cases\n@pytest.mark.parametrize(\"input_data, expected\", [\n (\"['S', 'A', 'M', 'P', 'L', 'E']\", \"Yes\"),\n (\"['T', 'E', 'S', 'T', 'I', 'N', 'G']\", \"Yes\"),\n (\"['L', 'A', 'M', 'P', 'L', 'E']\", \"No\"),\n (\"['T', 'E', 'S', 'R', 'I', 'N', 'G']\", \"No\")\n])\ndef test_normal_cases(input_data, expected):\n assert problem_p02910(input_data) == expected\n\n# Test edge cases\n@pytest.mark.parametrize(\"input_data, expected\", [\n (\"[]\", \"Yes\"),\n (\"['L']\", \"No\"),\n (\"['R']\", \"Yes\"),\n (\"['A', 'A', 'A', 'A']\", \"Yes\"),\n (\"['L', 'R', 'L', 'R']\", \"No\")\n])\ndef test_edge_cases(input_data, expected):\n assert problem_p02910(input_data) == expected\n\n# Test special characters and non-string elements\n@pytest.mark.parametrize(\"input_data, expected\", [\n (\"['$', '2', '#', '4', '@', '6']\", \"Yes\"),\n (\"[1, 'R', None, 'L', True, 'F']\", \"Yes\")\n])\ndef test_special_characters_and_non_string_elements(input_data, expected):\n assert problem_p02910(input_data) == expected\n\n# Test large inputs\ndef test_large_input():\n large_input = \"['N', 'E', 'V', 'E', 'R', 'L', 'O', 'O', 'K', 'B', 'A', 'C', 'K'] * 1000\"\n assert problem_p02910(large_input) == \"Yes\"\n\n# Test invalid inputs\n@pytest.mark.parametrize(\"input_data\", [\n \"'This is not a list'\",\n \"[['A', 'B'], ['C', 'D']]\",\n \"__import__('os').system('rm -rf /')\"\n])\ndef test_invalid_inputs(input_data):\n with pytest.raises(SyntaxError):\n problem_p02910(input_data)\n\n# Test unicode characters\n@pytest.mark.parametrize(\"input_data, expected\", [\n (\"['😀', 'L', '🎉', 'R', '❤️', '🚀']\", \"No\"),\n (\"['こんにちは', '世界', '안녕하세요', '세계']\", \"Yes\")\n])\ndef test_unicode_characters(input_data, expected):\n assert problem_p02910(input_data) == expected\n\n# Test escaped characters\n@pytest.mark.parametrize(\"input_data, expected\", [\n (\"['\\\\n', 'R', '\\\\t', 'L']\", \"Yes\"),\n (\"['A', '\\\\n', 'B', '\\\\r']\", \"Yes\")\n])\ndef test_escaped_characters(input_data, expected):\n assert problem_p02910(input_data) == expected\n\n# Test boolean values\n@pytest.mark.parametrize(\"input_data, expected\", [\n (\"[True, False, 'T', 'F']\", \"Yes\"),\n (\"[False, True, 'L', 'R']\", \"No\")\n])\ndef test_boolean_values(input_data, expected):\n assert problem_p02910(input_data) == expected\n\n# Test case sensitivity\n@pytest.mark.parametrize(\"input_data, expected\", [\n (\"['l', 'R', 'm', 'r', 'n', 'L']\", \"Yes\"),\n (\"['L', 'r', 'M', 'R', 'N', 'l']\", \"No\")\n])\ndef test_case_sensitivity(input_data, expected):\n assert problem_p02910(input_data) == expected\n\n# Test input with only one character type\n@pytest.mark.parametrize(\"input_data, expected\", [\n (\"['L', 'L', 'L', 'L']\", \"No\"),\n (\"['R', 'R', 'R', 'R']\", \"Yes\")\n])\ndef test_input_with_one_character_type(input_data, expected):\n assert problem_p02910(input_data) == expected\n\n# Test input with data structures other than lists\n@pytest.mark.parametrize(\"input_data\", [\n \"('A', 'B', 'C', 'D')\",\n \"{1, 2, 3, 4}\"\n])\ndef test_input_with_other_data_structures(input_data):\n with pytest.raises(SyntaxError):\n problem_p02910(input_data)\n\n# Test input with nested quotes\ndef test_input_with_nested_quotes():\n input_data = \"'[\\\\'A\\\\', \\\\'B\\\\', \\\\'C\\\\']'\"\n assert problem_p02910(input_data) == \"Yes\"\n\n# Test input with dictionary\ndef test_input_with_dictionary():\n input_data = \"{'key1': 'value1', 'key2': 'value2'}\"\n with pytest.raises(SyntaxError):\n problem_p02910(input_data)"
],
4: [
'# imports\nimport pytest # used for our unit tests\nfrom pie_test_set.p03965 import problem_p03965\n\n# unit tests\n\n# Test empty input\ndef test_empty_input():\n assert problem_p03965("") == 0\n\n# Test input with only "g" characters\n@pytest.mark.parametrize("input_data", ["g", "ggggg"])\ndef test_only_g_characters(input_data):\n assert problem_p03965(input_data) == 0\n\n# Test input with no "g" characters\n@pytest.mark.parametrize("input_data", ["abcdef", "pppp"])\ndef test_no_g_characters(input_data):\n assert problem_p03965(input_data) == -len(input_data)\n\n# Test input with alternating "g" and other characters\n@pytest.mark.parametrize("input_data", ["gpgpg", "gagbgc"])\ndef test_alternating_g_characters(input_data):\n assert problem_p03965(input_data) == 0\n\n# Test input with consecutive "g" characters surrounded by other characters\n@pytest.mark.parametrize("input_data", ["agggbc", "xggggy"])\ndef test_consecutive_g_characters(input_data):\n assert problem_p03965(input_data) == 1\n\n# Test input with leading and trailing whitespace\n@pytest.mark.parametrize("input_data, expected", [(" gggg ", 0), ("\\tgggg\\n", 0)])\ndef test_whitespace(input_data, expected):\n assert problem_p03965(input_data) == expected\n\n# Test input with only whitespace characters\n@pytest.mark.parametrize("input_data", [" ", "\\t\\n\\r"])\ndef test_only_whitespace(input_data):\n assert problem_p03965(input_data) == -len(input_data.rstrip())\n\n# Test input with "g" at the start, middle, and end\n@pytest.mark.parametrize("input_data", ["gabcgdefg", "gxyzg"])\ndef test_g_start_middle_end(input_data):\n assert problem_p03965(input_data) == 1\n\n# Test input with repeated non-"g" characters\n@pytest.mark.parametrize("input_data", ["ppppp", "xxxxx"])\ndef test_repeated_non_g_characters(input_data):\n assert problem_p03965(input_data) == -len(input_data)\n\n# Test input with mixed "g" and non-"g" characters\n@pytest.mark.parametrize("input_data", ["gpapbpcg", "ggppggpp"])\ndef test_mixed_g_non_g_characters(input_data):\n assert problem_p03965(input_data) == 0\n\n# Test long strings\ndef test_long_string():\n long_string = "g" * 10000\n assert problem_p03965(long_string) == 0\n\n# Test special characters and numbers\n@pytest.mark.parametrize("input_data", ["g#%&g123", "12g34g!@"])\ndef test_special_characters_and_numbers(input_data):\n assert problem_p03965(input_data) == 0\n\n# Test Unicode characters\n@pytest.mark.parametrize("input_data", ["g日本語g", "🙂g🙃g"])\ndef test_unicode_characters(input_data):\n assert problem_p03965(input_data) == 0\n\n# Test case sensitivity\n@pytest.mark.parametrize("input_data", ["GgGg", "gGgG"])\ndef test_case_sensitivity(input_data):\n assert problem_p03965(input_data) == 0'
],
5: [
'# imports\nimport pytest # used for our unit tests\nfrom pie_test_set.p02553 import problem_p02553\n\n# unit tests\n\n@pytest.mark.parametrize("input_data, expected", [\n ("1 2 3 4", 8),\n ("-1 -2 -3 -4", -3),\n ("1 -2 3 -4", 12),\n ("0 2 3 4", 8),\n ("1 0 3 4", 4),\n ("0 0 0 0", 0),\n ("-1 2 -3 4", 0),\n ("1 -2 3 -4", 0),\n ("-1 -2 3 4", -3),\n ("1 2 -3 -4", -6),\n ("1000000000 1000000000 1000000000 1000000000", 1000000000000000000),\n ("-1000000000 -1000000000 -1000000000 -1000000000", -1000000000000000000),\n ("2147483647 -2147483648 -2147483648 2147483647", 4611686014132420609),\n ("1000000000 1 1000000000 1", 1000000000),\n ("-1000000000 1 -1000000000 1", 1),\n ("-1 1 -1 1", 0),\n ("1 -1 1 -1", 0),\n ("1e3 2e3 3e3 4e3", 8000000),\n ("-1e3 -2e3 3e3 4e3", 3000000),\n])\ndef test_problem_p02553_normal_and_edge_cases(input_data, expected):\n # Test normal cases and various edge cases\n assert problem_p02553(input_data) == expected\n\n@pytest.mark.parametrize("input_data", [\n "1,2,3,4",\n "1 2 3",\n "one two three four",\n "\\t1\\t2\\t3\\t4",\n "1\\n2\\n3\\n4"\n])\ndef test_problem_p02553_invalid_input(input_data):\n # Test invalid inputs that should raise a ValueError\n with pytest.raises(ValueError):\n problem_p02553(input_data)'
],
},
"test_framework": {0: "pytest", 1: "pytest", 2: "pytest", 3: "pytest", 4: "pytest", 5: "pytest"},
"created_at": {
0: Timestamp("2024-04-18 07:04:46.293923+0000", tz="UTC"),
1: Timestamp("2024-04-18 07:36:01.262393+0000", tz="UTC"),
2: Timestamp("2024-04-18 06:35:45.693971+0000", tz="UTC"),
3: Timestamp("2024-04-18 06:48:48.109749+0000", tz="UTC"),
4: Timestamp("2024-04-18 06:41:27.396525+0000", tz="UTC"),
5: Timestamp("2024-04-18 06:53:42.593377+0000", tz="UTC"),
},
"aiservice_commit_id": {0: None, 1: None, 2: None, 3: None, 4: None, 5: None},
"metadata": {
0: {"test_timeout": 15, "function_to_optimize": "problem_p02847"},
1: {"test_timeout": 15, "function_to_optimize": "problem_p03196"},
2: {"test_timeout": 15, "function_to_optimize": "problem_p03861"},
3: {"test_timeout": 15, "function_to_optimize": "problem_p02910"},
4: {"test_timeout": 15, "function_to_optimize": "problem_p03965"},
5: {"test_timeout": 15, "function_to_optimize": "problem_p02553"},
},
"explanations_raw": {
0: {
"00db208c-b4df-4ac0-bf2c-4448f0c18897": 'The revised code snippet computes the same result but does it more efficiently. Using a dictionary instead of the index lookup method reduces the time complexity to O(1). The input is now directly passed to the `problem_p02847()` function without using the potentially unsafe `eval()` function, which simplifies the overall structure. Removing the `main()` function and `if __name__ == "__main__"` checks also makes the code cleaner and more straightforward to understand while preserving the functionality of the original code. This optimized code is faster and more efficient, while still delivering the same output as the original code.',
"098b2a24-8841-496c-b4aa-01b3f680836b": "The input_data parameter to the main function was updated to be directly used as our day selector instead of evaling it. The original code was inefficient and potentially unsafe because of the use of eval(). \n\nThe nested structure was removed, and the function was simplified to directly compute days_until_sunday from the input_data string. This results in a simpler, safer, and more pythonic code. \n\nThe new structure of the code still performs the exact same functionality of finding the number of days to Sunday from a given day of the week. The operation was preserved as it was originally intended but with a more straightforward approach. The changes do not break compatibility with other components as the function signature remains the same. \n\nBy removing the unnecessary function definition and eval() call, we have a cleaner, leaner, and more efficient function that is more readable and maintains the external functionality and behavior of the original code.",
"20b7cb92-6906-448d-901e-2ce98fefe2e3": "I've made several changes to improve the performance and efficiency of this code. First, I removed the use of `eval` to parse the input. `eval` is generally slower and less secure than directly parsing the input data. Instead, I now directly access the input string. \nSecond, I replaced the list data structure with a dictionary. List traversal using the `index` function requires O(n) time whereas retrieving an item from a dictionary is an O(1) operation, this decreases the time complexity of the code. \nThe final function works by mapping each weekday to the number of remaining days in the week from this day. Then, it simply returns the corresponding value for the given day. \nDespite these changes, the function preserves the external functionality and behavior of the original code - it still calculates and returns the number of days left in the week from the provided day.",
"239ec7be-8ce0-4c86-8b56-0704e791a372": "I've removed the unnecessary encapsulation of logic under the 'main' function to simplify the overall function. I have also removed the conditional check of \"__name__ == '__main__'\" which is typically used to restrict sections of code from running when the script is imported as a module, as it is not relevant in this context. Additionally, eval() call is removed because it is unnecessary and can pose potential security risks by evaluating the 'input_data' which should be a simple string in this case.\n\nFinally, we have handled the case where the current day is 'SUN' with a single line using modulo operation, making it more readable and efficient. The function now performs exactly the same task, but with unnecessary lines of code removed, it is far more simplified, readable and efficient. The days of the week are converted into numbers (0-6), and a modulo operation is used to wrap the week around efficiently and avoid the use of a conditional. This results in a small, but measurable, improvement in runtime efficiency for the function, especially when called numerous times.",
"2f364667-92fe-4245-b04d-04d6b5d9d9dd": "To make the code more efficient and safer, I removed unnecessary constructs and used a dictionary for direct mapping. Firstly, I replaced the list with a dictionary which maps each day of the week to the number of days until Sunday. By doing this, we avoid the use of the `index` method, which involves iterating through the list elements, and instead directly lookup the required value in constant time. Secondly, I converted the 'input_data' directly, no longer needing to use the `eval` function. The `eval` function runs the python code within it which is not necessary in this case and can also have security implications. Lastly, the check to assign 7 to `ans` if it's 0 is unnecessary. We can directly return 7 for 'SUN', meaning a week till the next Sunday, from our dictionary. The modified code has a much cleaner and safer implementation while preserving the original functionality.",
"8d67ed70-e190-4a84-b5d8-37defb27f6d6": 'I have made several changes that have improved the overall efficiency and readability of the code. These include eliminating the use of the list and the eval function, as well as removing the nested function and the unnecessary `if __name__ == "__main__":` check. The final code remains functionally the same, accepting a string representing a day of the week, and returning the number of days left in the week (or 7 if the given day is Sunday). It does so now more efficiently by using a dictionary for lookups, which has constant time complexity. This optimized implementation ensures that the code performs more rapidly and with less memory usage, while being clearer and more readable.',
"ba6a30d4-3150-482e-b932-6320f4635ebe": "I've streamlined the function `problem_p02847` to directly take an input string `S` (representing a day of the week), calculate the number of days until Sunday including the present day, and return this number.\n\nFirstly, I've eliminated the use of the `eval` function, which is considered a security risk and was unnecessary in this case since the input should be a string (representing a day).\n\nSecondly, I simplified the function by removing the nested `main` function. This made the code more succinct and straightforward without changing the algorithm itself or the output.\n\nFinally, I removed the `if __name__ == \"__main__\":` part as it was not serving any purpose here. In a script, this would be used to allow or prevent parts of code from being run when the modules are imported, but in this case, it was not being used effectively.\n\nThe revised function directly computes and returns the number of days until Sunday from the given day, maintaining the same functionality as the original code but with a more streamlined structure and potentially improved performance due to the removal of unnecessary or non-optimal elements.",
"c2fe9046-d193-4a51-a1d5-111c172e00f8": "The optimized code now uses a dictionary instead of a list to store and fetch the number of days remaining until Sunday. Dictionary lookups have a time complexity of O(1), which is faster than the O(n) time complexity of the list's index() method used in the original code. In addition to a performance boost, the dictionary also provides a more straightforward and cleaner way to compute the answer. The dictionary directly maps each day of the week to the number of days left until Sunday, thus eliminating the need for calculation or conditional checking.\nThe optimized code provides the same output as the original code. It accepts a string that represents a day of the week and returns the number of days left until Sunday, preserving the desired behavior. The function's name and expected input and output types remained unchanged, thus maintaining compatibility with any existing code that may be using it.",
"d2ff6d72-4515-4094-9d80-89f75045a9b5": "The refactored function `problem_p02847` maintains the core functionality of the original code, but without the unnecessary nested `main()` function. This code flattening makes the codebase cleaner and more readable without altering the purpose of the original function.\n\nThe purpose of this function is to get the days remaining until the next Sunday given the current weekday. The function is very specific and efficient for its intended usage. Here, I am making a non-performance related refactor to improve the code readability and maintainability.\n\nFinally, the refactored function's return value matches the original function's return value exactly, ensuring that calling code won't be broken by this change. The function's external functionality and behavior are preserved exactly. This approach ensures that any code using this function will not be impacted by this optimization.",
},
1: {
"426916b1-b122-44a7-aae3-c806b81887bd": "The original function used a brute-force approach that was inefficient and can result in high runtime for large inputs. The updated function uses binary search to find the maximum integer `i` such that `i` to the power `n` divides `p` exactly.\n\nIn binary search, the search space is divided in half at each step, reducing the number of possibilities to check, which results in a time complexity of O(log n) compared to O(n) in the original function.\n\nThe updated function maintains the behavior of the original function and doesn't change its external API. It will still take a string containing two space-separated numbers and return an int. Therefore, it won't break any existing code that uses this function. At the same time, the updated function will be faster and more efficient, especially for large inputs.",
"4a18cc5a-17a9-4aba-8b56-4534c406637e": "In the new code, I used a Binary Search when `n` is in the range from 2 to 40. This is because `i` is monotonic with respect to `i^n`, so Binary Search is applicable. Instead of incrementing by one, we now increment by larger steps, effectively cutting the search space in half in every iteration. \n\nBinary Search returns the smallest `i` such that `i^n > p` or `p` is not divisible by `i^n`. Thus, we need to return the result minus one. \n\nThe new code should be substantially faster for large inputs because the time complexity of Binary Search is O(log n), while the time complexity of the brute force method was O(n), where n is the input size. In our case, `n` is the difference between `p ^ (1/n)` and 1, which is exponential in size. So the improvement is significant.\n \nThe changes I've made preserve the external functionality and behavior of the original code. If `n` equals 1, or `n` is more than 40, the function returns `p` and 1 respectively, as before. If `n` is in the range [2, 40], the function returns the largest `i` such that `i^n <= p` and `p` is divisible by `i^n`, as before.",
"67cbfea8-0699-4750-a758-4078b3d33fb8": "The refactored code should have the same functionality as the original code: it converts the input string to two integers, `n` and `p`, and calculates the exact `n`th root of `p` or the largest `i` where `i^n` is a divisor of `p`.\n\nThe refactored code improves the performance by calculating the `n`th root of `p` in the beginning and looping only until `root`. Thus, it saves time compared to the original version, which keeps on increasing `i` and checking the condition each time in an indefinite while loop. Additionally, using Python's power operation (`**`) to calculate the `n`th root of `p` is likely more optimized and faster than the original approach.\n\nThe important thing to note is that this refactoring doesn't change the external behavior or functionality of the original code and so compatibility is preserved. It will still return the same data for the same input but in a more efficient and faster manner for larger values of `p`.",
"68416ac3-693a-432f-b19c-a74e5bcb5d80": "The original code used an infinite while loop until i to the power of n was greater than p which is not efficient for larger inputs. I optimized this by deriving the maximum possible integer 'i' which replaced the infinite loop with a definite bounded one.\n\nUsing the mathematical property that n-th root of 'p' can be calculated by raising 'p' to the power 1/n, I found the value of the maximum possible 'i'. This reduces the number of iterations potentially by a lot and hence reduces the time complexity of the code.\n\nOnly the mechanism of deciding the upper limit for 'i' is changed. The loop continues to increment 'i' and the conditional statement of checking the divisibility remains intact, thereby preserving the overall functionality and external behavior of the code. So, the code is fully backward compatible and will not break when used in conjunction with other functions or methods.",
"7afadb38-0b8c-41b6-a51f-78f9cbf17d63": "The binary search method is used to replace the original brute-force search method in the code. Binary search is an optimization intended to speed up the searching process. In binary search, instead of searching the elements one by one, the middle of the range is checked in each step. The search space is halved at each step of binary search, so it is more efficient and faster when dealing with large inputs.\nI calculate the middle number (`mid`) between `low` and `high`, then check whether `mid` to the power of `n` is still less than or equal to `p`. If it is, we know that `mid` might be the answer, and also the numbers larger than `mid` might be the answer, so we move `low` to `mid + 1`. If `mid` to the power of `n` is greater than `p`, that means the numbers larger than `mid` cannot satisfy the requirement, so we move `high` to `mid - 1`.\nAs for compatibility, the optimized code takes the same inputs and provides the same outputs as the original code. No features have been removed, and more specifically, the function name, input parameters, and return have not been changed, which ensures full backwards compatibility with other existing codes that use this function. Therefore, the revised code behaves externally in the exact same way as before. The code is cleaner, faster, and more efficient due to the implementation of the binary search.",
"7d96a7c8-29b6-43d9-a682-fbcf1d36103d": "In the optimized code, we've replaced a linear scan with a binary search, reducing the time complexity from O(p) to O(log p). \n\nBinary search allows us to efficiently traverse through the search space (which is 1 to `p` inclusive) to find the highest `i` for which `i^n` doesn't exceed `p` and `p` is evenly divisible by `i^n`. We start by setting the `lower` to 1 (smallest possible value) and `upper` to `p` (largest possible value). We then initiate a while loop that continues until `lower` and `upper` converge. In each iteration, we calculate the mid-point `mid` between `lower` and `upper`. If `mid^n` exceeds `p`, we update `upper` to `mid - 1`. Otherwise, we record `mid` as a possible solution and update `lower` to `mid + 1`. Lastly, the function returns the highest `i` we could find.\n\nGiven these adjustments, the function works just as before (returns the highest `i` for which `i^n` doesn't exceed `p` and `p` is evenly divisible by `i^n`) but now has a substantially improved performance in terms of speed compared to the original version. This optimization doesn't change any present functionality and thus preserves the external functionality and behavior of the original code.",
"e3a6d7b8-e145-45c3-bd1c-8cae99c6d014": "The changes made to the code replaced the simple iteration with Binary Search, thus improving the time complexity of the function. The function maintains the same functionality as before, returning `p` if `n` equals 1, 1 if `n` is more than 40, and the largest `i` such that `i^n` is less than or equals to `p` otherwise. Binary Search halves the search space in each step, making it more efficient. By using this approach, the code can handle larger inputs more efficiently.",
},
2: {
"0a631960-6c33-4de4-8d38-b31e888918c7": "Here is an optimized version of your program:\n\nPlease note that this code won't strictly run faster as the operation taking the most time in your original program is most probably a function map() (or divisions). The provided code is just a cleaner and more \"pythonic\" version of it. The real optimization of the program runtime speed lies in the algorithm used, and as for this specific function, there's not much to optimize.\n",
"38b31a95-7409-4c66-a50d-24dd565bb513": "Your function is already quite optimized. However, we could avoid importing the unused `math` library. Here is the revised function:\n",
"6ba45be2-5f81-4952-8dc6-c4569092096b": "Your provided Python program is reasonably optimized, but it could potentially be made slightly faster by eliminating the `list(map(...))` usage in the splitting process. Here is a slightly optimized version of the provided function using this method:\n\n\nIn this version, we directly unpack the results of `map(int, input_data.split())` to `a, b, x`. We avoid the usage of `list` which was not necessary here. This does a very little optimization as the usage of list was redundant in your case.\n\nPlease note that these are micro-optimizations and you will rarely see the difference in time taken by both versions. The real time consuming part here is the string parsing which can't be optimized and the integer division which is already optimized in Python.",
"7f0a4ac8-cec2-485b-a159-610fae2dcf68": "Your provided Python function is already quite efficient. However, we can rewrite it in a single line of code and remove the unnecessary `import math` statement which is not used in your code at all. Here is the optimized version:\n\nBut remember that Python doesn't compile the way languages like C do and thus the improvement on runtime will be negligible or even unnoticeable. The primary improvements are in readability, which can indirectly speed up code by reducing the time it takes for humans to understand and thus maintain it.",
"7f9c5889-6307-47e1-92b2-a89f07409278": "python\n\nThe optimized code runs faster because it reduces the number of divisions, which are computationally expensive operations. It does this by implementing the mathematical properties to compute the number of multiples of x in the [a, b] range directly.",
"857da2b3-1b2c-4ce1-96a1-005d8b06df24": "The function provided is pretty efficient already. However, the program can be made more efficient by separating the import statement from the function, as importing is a operation that can be done once at the top of the script. The efficiency of this division operation depends on the python interpreter, which is usually optimized.\n\nHere is the refactored code:\n\n\nNote: The efficiency improvement from this refactoring could be negligible, but it's a good practice to not put import statements inside functions.",
"9f359406-cb4c-404e-a9d7-c1b61af7e019": "The provided code reads and splits the input_data then performs division operation. It looks already optimized. However, the direct use of the `divmod` function can help to slightly improve performance over repeated use of `//` operator.\n\nPlease find the optimized program below:\n\n```python\ndef problem_p03861(input_data):\n a, b, x = map(int, input_data.split())\n \n quotient_b, remainder_b = divmod(b, x)\n quotient_a_1, remainder_a_1 = divmod(a - 1, x)\n\n return quotient_b - quotient_a_1\n```\nGiven that we already have three integers a, b, x, the splitting and mapping already done in integer format can be removed for the function to be more efficient. Here's how it can be done:\n\n\nThis change would be faster only when `problem_p03861` is called with integer parameters directly. Please note that the output of the function remains same as the original.",
"a8cc096d-390f-44e0-a0eb-7cbbec651671": "The given python program is already optimized to a great extent. However, we can avoid the split() and map() operations as these can be relatively slow especially for large datasets. This requires passing the inputs as integers instead of a string which will definitely speed up the program.\n\n\nIn the above code, it is assumed that a, b, x are integers as input instead of a single string. The function now takes three input arguments rather than taking a string and then splitting it into individual parts.\n\nRemember, these changes affect the way the function is used and the type of input it accepts. If the input must be a string with numbers separated by spaces, then the original implementation was already quite fast. In python, the algorithmic complexity and the memory usage of your function is already optimal. Python's overhead for function calls etc. is also quite low so in this case it wouldn't have a noticeable effect.",
"be3c8c8a-cb4d-465a-8d98-e345407b3ee5": "The given function is already performing at a relatively efficient level of complexity. Given that the \"//\" operator performs integer division in Python, the time complexity for this operation is O(1). Since there is no iteration or recursion involved in this function, it's difficult to further optimize this from a time complexity perspective.\n\nThe function could, however, be optimized slightly for memory usage by directly unpacking the results of the map() function into the variables a, b, and x rather than converting them to a list first. This won't make a significant impact on the performance of this particular function, but it's a good practice to follow when dealing with larger data sets.\n\nHere's the optimized version of your function:\n\n",
"e40e4203-d07b-4319-9000-24a1333f6f4c": "The function `problem_p03861` is already quite efficient due to its simplicity. However, the import statement can be moved outside of the function, improving performance slightly in cases where the function is called multiple times. Since importing math package was redundant because the math module was not used at all, there's no need to import it. Here is your optimized function:\n\n \n\nThis won't substantially improve the speed, but it's good practice to keep import statements outside of functions. As the code is already optimized it will not get any faster by changing the code inside the function as it's logic is already minimalistic and optimized.",
},
3: {
"00f11305-5a9b-43cf-9d97-720b0a465d52": "Your current function iterates through the entire string, even when it's clear that the string doesn't meet the conditions. This unnecessary work slows down your function. Besides, it uses eval() - which is generally dangerous and should be avoided unless absolutely needed. Here's a faster version:\n\n\n\nThis optimized function works by checking if character 'L' is at an odd index and/or character 'R' is at an even index right from the start. If this condition is met, it immediately returns `\"No\"` and breaks the loop which speeds up the function. If the function goes through the entire string and doesn't find any such occurrence, it returns `\"Yes\"`. I've also used the more pythonic way of enumerating through the string using the built-in `enumerate()` function.",
"0f234383-ad19-44dd-95c1-9bb5171bf82a": 'Here is the optimized code for your problem. The solution reduces the run time complexity by avoiding unnecessary iterations on all the elements in the given list and immediately returns "No" as soon as the condition matches. The functionality of the code remains the same as the original.\n\n\nThis version of the problem will check each character in the string only once, instead of checking each character twice- once for odd and once for even positions. We also eliminated the need of maintaining an extra variable `flag`. So it\'s more optimal in terms of runtime and memory usage.',
"1f1bb4c2-297e-46ca-a1c2-10770e83f877": "The optimized version of your program is as follows:\n\n\nI've made the following optimizations:\n\n1. I used the built-in Python `enumerate` function to iterate over the characters in the string `s` along with their index. This is more readable and efficient because it avoids computing `i - 1` and accessing `s[i - 1]` on each iteration.\n \n2. I combined the checks for `i % 2 == 1` and `i % 2 == 0` into a single line and used a more direct check for 'L' and 'R' conditions which prevents unnecessary checks in each loop iterations.\n \n3. I removed the `flag` variable and returned directly when a condition was met. This allows the function to exit immediately when it determines the result, rather than continuing to check the remaining characters.",
"25c43c11-6a29-48f1-878b-06156bf31906": "Here's your function rewritten in a more optimized way. I've removed the `eval` function, which can be unsafe and slow, assuming that `input_data` is already a string. The function now stops iterating as soon as it finds an invalid character in the string, thus potentially reducing the number of iterations.\n\n\nThis version of the function will produce exactly the same result as your original function, but should run faster in the case of large inputs that contain an invalid character near the start of the string. By using `enumerate`, we efficiently loop over the characters in the string and their indices at the same time, which is more Pythonic and can be faster than a range-based loop.",
"35128e9c-4e22-4cd0-9412-b55c3c54a16a": "The given program iterates through every element of the input data string, testing each character one by one against a condition. For a very long string, this could be quite inefficient as the time complexity is O(n) - linear time complexity. \n\nOne approach to optimize the program is to utilize Python's built-in operations which operate at C speed internally. We can rewrite the program in a more efficient way by examining all odd-index and even-index characters simultaneously, instead of looping through each character. \n\n\nThis program operates by checking all of the odd-indexed and even-indexed positions at once for 'L' and 'R', respectively, using Python's slicing feature. For larger datasets, checking them all at once will be substantially faster than performing a loop operation. This program has a complexity of O(1) - constant time complexity.",
"4bc36f4a-ddcd-4e0e-a868-5b338d850e21": "We can speed up your program a bit by reducing the time complexity and remove unnecessary checks. Here I'm using list comprehension and all() function to make the checks within a single line. This solution has a time complexity of O(n), but avoids excessive branching which may slow down the execution.\n\n",
"5e136046-8d21-4e8b-b5d1-6efa23fc7f6e": 'Your original code is walking through each character, and checking the number at the index and character at that index. If the index is odd, you check if the character is "L", and if the index is even, you check for "R", which is essentially saying that all odd-indexed positions should be "R" and all even-indexed positions should be "L". If that\'s not the case, it returns "No". Once the flag turns False, there is no need to continue checking. So, there could be a break to terminate the loop earlier.\n\nHere is an optimized version where the function early-exits if a character is in the wrong position:\n\nIn the optimized version, the function will return \'No\' as soon as it detects an inappropriate character, which could potentially save a lot of unnecessary iterations, especially when dealing with larger inputs.',
"830e8468-8c1e-4dc6-952e-203148dfe5d6": "Your function is iterating over the entire string checking for 'L' in the odd indices and 'R' in the even indices. This can be optimized by using generator expressions with any() built-in function, which is faster and stops as soon as the condition is met without going through the entire string. \n\nHere's your optimized solution:\n\n\n\nIn this solution, 'any' function is used which stops iteration as soon as it encounters 'True' (when a 'L' is at an odd index or 'R' is at an even index) and 'range' is used with a step of 2 to only check odd and even indices, improving the time complexity from O(n) to O(n/2). Also, using 'any' is generally faster than for loops due to internal implementation optimizations.",
"8784754e-3e17-4561-94c7-05b9b5f2aedf": 'The given code can be simplified and optimized by reducing the number of operations in the loop and eliminating the use of the `eval()` function whichnot only slows down the code but is also a security risk. Here\'s an improved version of the code:\n\n\n\nThe `enumerate()` function was used here which is more pythonic and efficient than using the `range()` function. It minimizes the number of operations and overall makes the code faster and more readable. Also we return immediately after any "L" character found in odd places or "R" in even places. This eliminates unnecessary loop iterations.',
"96e129db-323a-4651-a021-32f5321e11e9": 'Here is a rewrite of original function with improved performance:\n\nThe original function is inefficient because it checks each character of the string regardless of whether flag has already been set to False or not. The rewritten version checks each character and returns "No" as soon as a mismatch is found which reduces the number of iterations and increases the speed significantly when a mismatch is found early in the string. This pattern of early return when possible is a good practice for optimization in many scenarios. In the worst case, both versions of the function have the same time and space complexity - O(n).\n\nAnother speed-up in the refactored code comes from the fact that `i % 2` is calculated only once per iteration as opposed to twice in the original code.',
},
4: {
"000711ac-1663-496b-837b-946040155a9c": "Here is your optimized code. I didn't change the logic, I only streamlined the for loop:\n\n\nThe use of elif helps speed up the code, as it immediately exits the conditional after a true case is found. In your original code, even after a true case was found, the program would continue checking all cases until none were left. The strip operation also moved out of the loop to only execute once.",
"0c1d4ab3-9eca-4b24-a274-1cf4aa35ec0d": "The runtime of your current program is O(n) because you are looping through each character in the input string once. However, the code includes many redundant if statements that can be simplified to make the code more efficient and more readable.\n\n\nIn this version, the if statements are combined into single lines, and the sum operation is moved outside of the if statement. The condition `(0 < iG)` will return either `True` which is equivalent to `1` in python or `False` which is equivalent to `0`. This way the substraction or addition operation is always executed but it doesn't change the value of `iG` or `iR` if the condition is not met. It should run the same logic as your code but in a more efficient way.",
"470d47be-1d1b-4cff-927f-0e9200d781cc": 'The provided Python function seems to iterate over given input string "input_data" and applies a certain logic based on whether a character is "g" or not. While the logic inside the function isn\'t clear, we can improve its efficiency by avoiding unnecessary operations like redundant elif and else checks.\n\nHere\'s the optimized version of the given function:\n\n\nThis new version handles "g" and non-"g" characters at the same time, effectively optimizing the code by reducing code size and improving readability. The optimization mainly comes from eliminating redundant elif and else checks.',
"546095ca-bd89-4cdd-b0a8-3694d8013644": "Here is the optimized version of your python program:\n\n\n\nThe optimization consists of the following:\n\n1. Merging the initialization of `iG` and `iR` into a single line.\n2. Pre-computing `rstrip` method call for `input_data` outside the loop. This avoids repeated function calls in each iteration.\n3. Simplifying the nested if conditions by merging common statements. Since the increment of `iG` always occurs, it is moved outside the conditional statements. The conditions are then simplified, reducing duplicate code and number of checks.\n4. The `else` part in the original code is also unnecessary as it only executes when `iG` is 0, but if `iG` is 0, it's reset to 1 in both parts of the `if` statement. So, we only need to check if `iG` is less than or equal to zero and adjust `iR` accordingly.",
"81fdc495-f548-4621-89c4-691d3b74f77b": "This is a more optimized version of the same function in Python:\n\n \n\nThis version reduces the number of operations inside the loop. The conditions of the if statement have been combined to remove repetition. Also, the increment and decrement operation on 'iG' is done in a single line instead of repeated under separate conditions, thus reducing the total number of operations. This minor optimization can result in a faster runtime especially when processing large amounts of data.\n\nHowever, in most cases, the speedup would be negligible as Python's optimization will already handle this effectively. The rewritten code is simply a tidier and more readable format of the original code.",
"83b6579b-8f00-4829-944c-904c8d7b80f9": 'Your function has multiple unnecessary "if-else" checks and a rstrip() operation that can be removed to optimize the function. Here\'s a revised version of your function:\n\n\nThis version of the function works exactly like the original one, but with less operations and simpler code which makes it run faster.',
"849e8b87-fb91-4f85-84cf-ade32dab0f3f": "Here is a more optimized version of your function:\n\n\n\nThe ternary if-else statements help condense the logic within the for loop. This might allow your function to run faster depending on the Python interpreter's optimizations. However, the time complexity remains the same because we have not changed any underlying algorithmic approach. Every string character is still considered once, hence runtime remains O(n) where n is the length of the input string. It's also important to note that the total time will always depend on the specific data and how well it interacts with Python's internal optimizations. \n\nIn terms of memory usage, the optimized version will not make any significant difference as well, because the original and the optimized versions both use a constant number of variables.\n",
"9ee7a8e5-f8c5-49ee-aa71-4010ea432781": "The provided function's primary issue lies with its complexity and redundancy due to unnecessary operations and iterative steps. To optimize and speed up this function, we can avoid decreasing and then increasing the same variable (iG), and also handle line by line computations more efficiently using Python's in-built functions and simplified flow control.\n\nThe rewritten code is as follows:\n\n\nThis transformed version of the function provides the same functionality as before but with enhanced performance due to the reduced number of variable manipulation and iteration steps. The optimization lies in the single liner computation inside the `for` loop where the increment, decrement of variables is happening based on the evaluation of conditions.",
"bd157d9b-3892-48fb-a106-908631fd6a7c": "The provided python function can be optimized for faster runtime by removing unnecessary code blocks and by performing more efficient comparisons. Here's my optimized version of the function:\n\n\n\nIn this approach, we have:\n\n1. Initialized `iG` and `iR` variables on the same line to reduce the lines of code.\n2. Used boolean truthiness of values in Python as a shorthand to avoid unnecessary comparisons with zero. In Python, any non-zero value is considered True, while zero, None, and empty collections are considered False. So, instead of `if 0 < iG:` we can simply write `if iG:`.\n3. Removed blank lines to make the code more compact. It won't necessarily make the code faster, but it does make it cleaner and easier to read. \n\nPlease note that the return value remains exactly the same as before, as required. The optimization is mainly targeted to make the code more concise, readable and pythonic, which in turn may lead to minor performance improvements. However, the overall algorithmic complexity remains the same.",
"fdc5337e-8a6a-44ed-8255-d7fca5b49786": "Here is the optimized version of your code:\n\n\n\nThe optimization includes avoiding unnecessary branches, combining conditions, and minimizing operations.",
},
5: {
"04e41d16-15f6-4341-a976-f420ed3180a6": "Here's a rewritten function that will run faster. Since you're okay with preserving the function signature and return value, I'm assuming that I can use basic Python functionality instead of modules.\n\n\n\nNumpy is certainly a powerful module, but in this case, it adds unnecessary overhead. List creation and the append method also take additional resources (both in terms of time and memory). Therefore, I've eliminated the numpy module and the list, calculating the maximum value directly via the max function.\n\nI also streamlined the check for negative numbers. Instead of invoking np.sign (which you then compare with not-equal), I check directly whether one number is positive and the other is negative.\n\nThese modifications will make your code run faster, and should also reduce memory usage, especially for large inputs. Plus, since I haven't changed the function's name or signature, or what it returns, it should still meet your specified requirements.",
"09edd39b-4a6f-4422-8c14-23d2a819f2c5": "It appears that the biggest issue with your program's efficiency is the repeated computation of max(hoge) and unnecessary imports. You can streamline this by storing the result of max(hoge) into a variable immediately after computing it. Also, it seems like you don't really need importing numpy as np, as you can find the sign using just Python. This should most likely increase the speed of your program.\n\nHere is the optimized code:\n\n \n\nIn this code, we only compute max(hoge) once and just reference it later, which saves time. Additionally, the use of 'map' directly to convert the numbers to integers is more efficient. And I changed the sign comparisons using built-in Python operations instead of numpy to make it faster and removed the numpy import.",
"19d21335-2ee5-4a5e-8bab-50805248cf9f": "The program can be optimized by removing the costly import of numpy and unnecessary list constructions. The maximum number generated from multiplications can be calculated in one line, rather the appending to a list. Also, the final check for the maximum element can be calculated once and stored in a variable so that we don't calculate maximum again and again. Here is the optimized version of the program:\n\n\nThis version of the function gives the same results as the previous one, but with more efficient memory usage and faster execution because it avoids creating a few unnecessary objects in memory and performs fewer function calls. The number of computations in the rewritten function is smaller than in the original function, which can translate to a speed up in execution.",
"1aa3d7da-d34e-46a3-a72c-d0f1cd0bd5ba": "You can optimize the program by directly computing the maximum of the products without storing all four products in a list. In addition, the if-else statement to return the maximum can be simplified. Here's an optimized version:\n\n\n\nThis version avoids unnecessary computations and storage, improving the speed of your algorithm.",
"36eeb34d-6a89-42f5-8340-e7a31bab073e": "Here is the optimized version of your program. This version reduces unnecessary calls to the `max` function and removes unused numpy import. Also, it removes the superfluous list creation and stores the maximum value of the products directly without needing to append them to a list, resulting in less memory consumption:\n\n\n\nIn this optimized version, the sign comparison is done using Python's built-in statement, removing the need for numpy completely. This will make the function more portable (as it no longer depends on numpy) and slightly faster.\n",
"45c372c7-8b7f-4378-ba39-fdcde9c7a13e": "The program you provided has several superfluous operations that slow the computation down, including multiple calls to the max() function. The program also imports numpy just for the sign function, which has a Python standard library equivalent with the function math.copysign.\n\nHere is an optimized version of the program that avoids multiple sign calls and max() calls.\n\n\n\nThis version performs the split and map operations all in one line to convert the input strings into integers. It then calculates the maximum product only once instead of multiple times. It also uses basic equality checks to figure out if the signs of a, b and c, d are the same rather than using numpy's sign function. This will hence require no external library.\n\nThis rewritten program should run significantly faster, especially for large inputs, since it performs fewer operations and does not call the max function multiple times.",
"881c3174-228a-4592-a681-d8e88bc77aeb": "This program can be made to run faster by avoiding the repeated calculation of `max(hoge)` and removing the usage of numpy for the sign checking, since python natively supports this feature. Here's an optimized version:\n\n\nIn this optimized version:\n\n1. I eliminated the `hoge` list altogether and called the `max` function once, calculating the maximum product directly.\n2. I removed the dependency on numpy for sign checking. Python can check if a number is positive or negative with the `>` operator.\n3. I removed the redundant final `else` clause. If none of the conditions are met before, the function will return the `max_value` by default.",
"a7e3f286-cc2b-49ad-a958-5429a32408e5": "Here is your optimized Python code:\n\n\n\nHere's what was changed:\n\n1. Removed the import of NumPy. Native Python can achieve this task without additional libraries, which reduces memory usage.\n2. Calculated max_val only once, instead of twice (once for checking against 0 and twice for the return value).\n3. Removed the list `hoge`, which was holding the multiplication results unnecessarily.\n4. Replaced the numpy sign function with a simple comparison with 0. This new comparison results in boolean values (True or False), which are appropriate for checking the condition. Here, it's crucial to note that in Python, negative numbers are considered less than zero.",
"ab6accd2-494e-4bd4-b91c-7f38125eb9a7": "Sure, here is the optimized version of your provided python code:\n\n\n\nIn the original code, the operation is performed independently for each condition and the maximum value is calculated multiple times, which may negatively affect the performance especially for larger values of a, b, c, and d.\n\nIn the revised version, `max()` is only called once to find the greater value among the products. This reduces the number of operations, which improves the efficiency of the program. \n\nAlso, eliminated the need of `numpy` library function np.sign() and replaced it with more efficient direct approach of comparing signs by checking if number is less than zero. The advantage of this manual comparison is that it's faster since it does not require library function call overhead.",
"b35e4ea8-b192-415f-8dcc-817b76b1319c": "Here is the optimized version of your Python function. Importing numpy and performing repetitive operations, like calling the `max()` function several times, may extend the runtime. So, try to perform everything in one pass.\n\n\nThis version eliminates the unnecessary use of numpy and removes the multiple calls to max() function, which will make the function run faster.",
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experiments/tests/sample_dataframe.pickle
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experiments/tests/test_metrics_analysis.py
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import pickle
import pytest
from pandas import DataFrame
from experiments.metrics_analysis import calculate_validity
from experiments.tests.data.mock_dataframe import mock_dataframe
from experiments.tests.data.pie4perf_dataframe_dict import pie4perf_sample_dataframe_dict
def pickled_dataframe():
with open("experiments/tests/sample_dataframe.pickle", "rb") as f:
df = pickle.load(f)
return df
@pytest.fixture
def pie4perf_sample_dataframe():
return DataFrame.from_dict(pie4perf_sample_dataframe_dict)
@pytest.fixture
def sample_dataframe():
return DataFrame(mock_dataframe)
def test_calculate_validity_no_successful_runs(pie4perf_sample_dataframe):
df = pie4perf_sample_dataframe
df["original_runtime"] = [None, None, None, None, None, None] # No successful runs
validity_metrics = calculate_validity(df)
assert validity_metrics["percent_valid_pr"] == 0
assert validity_metrics["percent_valid_candidates"] == 0
def test_calculate_validity_some_successful_runs(pie4perf_sample_dataframe):
df = pie4perf_sample_dataframe
validity_metrics = calculate_validity(df)
# Assuming half the runs are successful and above threshold, the expected value should be 50.0
assert validity_metrics["percent_valid_pr"] == 50.0
assert validity_metrics["percent_valid_candidates"] == 80.0 # No valid candidates
def test_calculate_validity_all_successful_runs(pie4perf_sample_dataframe):
df = pie4perf_sample_dataframe
df["original_runtime"] = [1.0, 2.0, 2.0, 2.0, 3.0, 18.0] # All successful runs
df["best_correct_speedup_ratio"] = [0.05, 0.06, 0.08, 0.09, 0.56, 0.17] # All above threshold
df["is_correct"] = [
{"1f39ef86-5eff-4760-a262-43011492906e": True},
{"aeeeca3b-4ccf-46eb-8dbf-c526c05fca27": True},
{"dc67527a-9cfe-4bbe-98eb-266bc8e9a27d": True},
{"e584d267-f68f-4144-820f-110c2719919f": True},
{"e584d267-f68f-4144-820f-110c2719919f": True},
{"d34db33f-f68f-4144-820f-110c2719919f": True},
]
validity_metrics = calculate_validity(df)
assert validity_metrics["percent_valid_pr"] == 100.0
assert validity_metrics["percent_valid_candidates"] == 100.0 # All valid candidates
def test_calculate_validity_with_valid_candidates(pie4perf_sample_dataframe):
df = pie4perf_sample_dataframe
# Assuming some runs are successful and some candidates are valid
df["original_runtime"] = [1.0, None, 2.0, None, 3.0, None] # Some successful runs
df["best_correct_speedup_ratio"] = [0.04, None, 0.06, None, 0.07, None] # One below and two above the threshold
df["is_correct"] = [
{"1f39ef86-5eff-4760-a262-43011492906e": True},
{},
{"aeeeca3b-4ccf-46eb-8dbf-c526c05fca27": True},
{},
{"e584d267-f68f-4144-820f-110c2719919f": False},
{},
]
# There are 2 successful runs with valid speedup ratios out of 6 total runs
expected_percent_valid_pr = (2 / 6) * 100
# There are 2 valid candidates out of 3 total candidates (one candidate is not valid)
expected_percent_valid_candidates = (2 / 3) * 100
validity_metrics = calculate_validity(df)
assert validity_metrics["percent_valid_pr"] == expected_percent_valid_pr
assert validity_metrics["percent_valid_candidates"] == expected_percent_valid_candidates
# def test_calculate_performance_no_successful_runs(pie4perf_sample_dataframe):
# df = pie4perf_sample_dataframe
# df["original_runtime"] = [nan, nan, nan, nan, nan, nan] # No successful runs
# df["best_correct_speedup_ratio"] = [nan, nan, nan, nan, nan, nan] # No valid candidates
# performance_metrics = calculate_performance(df)
# assert performance_metrics["average_percentage_gain_pr"] == 0
# assert performance_metrics["geometric_mean_gain_pr"] == 0
# assert performance_metrics["mean_average_percentage_gain_all"] == 0
# assert performance_metrics["geometric_mean_gain_all"] == 0
# assert performance_metrics["average_time_saved_pr"] == 0
# assert performance_metrics["mean_average_time_saved_all"] == 0
#
#
# def test_calculate_performance_some_successful_runs(pie4perf_sample_dataframe):
# df = pie4perf_sample_dataframe
# # Assuming half the runs are successful and above threshold, the expected value should be 50.0
# performance_metrics = calculate_performance(df)
# assert performance_metrics["average_percentage_gain_pr"] == pytest.approx(0.02112, 0.001)
# assert performance_metrics["geometric_mean_gain_pr"] == pytest.approx(1.02112, 0.001)
# assert performance_metrics["mean_average_percentage_gain_all"] == pytest.approx(0.02112, 0.001)
# assert performance_metrics["geometric_mean_gain_all"] == pytest.approx(1.02112, 0.001)
# assert performance_metrics["average_time_saved_pr"] == pytest.approx(0.0, 0.001)
# assert performance_metrics["mean_average_time_saved_all"] == pytest.approx(0.0, 0.001)
#
#
# def test_calculate_performance_all_successful_runs(pie4perf_sample_dataframe):
# df = pie4perf_sample_dataframe
# df["original_runtime"] = [1.0, 2.0, 2.0, 2.0, 3.0, 4.0] # All successful runs
# df["best_correct_speedup_ratio"] = [
# 0.05,
# 0.06,
# 0.08,
# 0.09,
# 0.56,
# 0.57,
# ] # All above threshold
# df["is_correct"] = [
# {"1f39ef86-5eff-4760-a262-43011492906e": True},
# {"aeeeca3b-4ccf-46eb-8dbf-c526c05fca27": True},
# {"dc67527a-9cfe-4bbe-98eb-266bc8e9a27d": True},
# {"e584d267-f68f-4144-820f-110c2719919f": True},
# {"e584d267-f68f-4144-820f-110c2719919f": True},
# {"d34db33f-f68f-4144-820f-110c2719919f": True},
# ]
# performance_metrics = calculate_performance(df)
# assert performance_metrics["average_percentage_gain_pr"] == pytest.approx(0.235, 0.001)
# assert performance_metrics["geometric_mean_gain_pr"] == pytest.approx(1.214, 0.001)
# assert performance_metrics["mean_average_percentage_gain_all"] == pytest.approx(0.235, 0.001)
# assert performance_metrics["geometric_mean_gain_all"] == pytest.approx(1.214, 0.001)
# assert performance_metrics["average_time_saved_pr"] == pytest.approx(0.0, 0.001)
# assert performance_metrics["mean_average_time_saved_all"] == pytest.approx(0.0, 0.001)
#
if __name__ == "__main__":
pytest.main()

18
experiments/tests/test_statistical_analysis.py
View file

@ -1,18 +0,0 @@
import numpy as np
from codeflash.verification.statistical_analysis import series2_faster_95_confidence
def create_timing_series(size: int, mean: int, std_dev: int) -> list[int]:
mu = np.log(mean**2 / np.sqrt(std_dev**2 + mean**2))
sigma = np.sqrt(np.log(1 + (std_dev**2 / mean**2)))
rng = np.random.default_rng()
return np.round(rng.lognormal(mu, sigma, size)).astype(int).tolist()
def test_compare_timing_series() -> None:
original_timing_series = create_timing_series(10000, 2000, 60)
optimized_timing_series = create_timing_series(10000, 1800, 48)
result = series2_faster_95_confidence(original_timing_series, optimized_timing_series, 10000)
assert result is not None
assert 4 < result[0] < 16
assert result[1] < 4

86
experiments/wandb.ipynb
View file

@ -1,86 +0,0 @@
{
"cells": [
{
"cell_type": "code",
"execution_count": null,
"id": "initial_id",
"metadata": {
"collapsed": true
},
"outputs": [],
"source": []
},
{
"cell_type": "code",
"execution_count": 2,
"id": "43aad670ab761737",
"metadata": {
"ExecuteTime": {
"end_time": "2024-04-13T01:20:22.380492Z",
"start_time": "2024-04-13T01:20:17.939300Z"
}
},
"outputs": [],
"source": [
"import json\n",
"\n",
"import weave\n",
"from openai import OpenAI\n",
"\n",
"\n",
"@weave.op() # 🐝\n",
"def extract_fruit(sentence: str) -> dict:\n",
" client = OpenAI()\n",
"\n",
" response = client.chat.completions.create(\n",
" model=\"gpt-3.5-turbo-1106\",\n",
" messages=[\n",
" {\n",
" \"role\": \"system\",\n",
" \"content\": \"You will be provided with unstructured data, and your task is to parse it one JSON dictionary with fruit, color and flavor as keys.\",\n",
" },\n",
" {\"role\": \"user\", \"content\": sentence},\n",
" ],\n",
" temperature=0.7,\n",
" response_format={\"type\": \"json_object\"},\n",
" )\n",
" extracted = response.choices[0].message.content\n",
" return json.loads(extracted)\n",
"\n",
"\n",
"weave.init(\"intro-example\") # 🐝\n",
"sentence = \"There are many fruits that were found on the recently discovered planet Goocrux. There are neoskizzles that grow there, which are purple and taste like candy.\"\n",
"extract_fruit(sentence)"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "d929002a32073a23",
"metadata": {},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"name": "python3"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 2
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
"version": "2.7.6"
}
},
"nbformat": 4,
"nbformat_minor": 5
}

0
pie_test_set/__init__.py
View file

8
pie_test_set/p00006.py
View file

@ -1,8 +0,0 @@
def problem_p00006(input_data):
str = input_data
str_lis = list(str)
str_lis.reverse()
return "".join(str_lis)

14
pie_test_set/p00007.py
View file

@ -1,14 +0,0 @@
def problem_p00007(input_data):
from math import ceil
debt = 100000
n = int(eval(input_data))
for _ in range(n):
tmp = ceil(debt * 1.05)
debt = ceil((tmp / 1000)) * 1000
return debt

46
pie_test_set/p00017.py
View file

@ -1,46 +0,0 @@
def problem_p00017(input_data):
def chg(s, n):
res = ""
for i in s:
o = ord(i)
if 97 <= o <= 122:
if o + n <= 122:
res += chr(o + n)
else:
res += chr(o + n - 26)
else:
res += i
return res
while True:
try:
s = eval(input_data)
for i in range(25, -1, -1):
c = chg(s, i)
e = c.split()
if "the" in e or "this" in e or "that" in e:
return c
break
except:
break

2
pie_test_set/p00018.py
View file

@ -1,2 +0,0 @@
def problem_p00018(input_data):
return " ".join(map(str, sorted(map(int, input_data.split())))[::-1])

4
pie_test_set/p00019.py
View file

@ -1,4 +0,0 @@
def problem_p00019(input_data):
from functools import reduce as R
return R(lambda x, y: x * y, list(range(1, int(eval(input_data)) + 1)))

2
pie_test_set/p00020.py
View file

@ -1,2 +0,0 @@
def problem_p00020(input_data):
return input_data.swapcase()

30
pie_test_set/p00029.py
View file

@ -1,30 +0,0 @@
def problem_p00029(input_data):
x = input_data.split()
longestWord = ""
mostWord = 0
currentCount = 0
THEmostWord = ""
for r in range(len(x)):
if len(x[r]) > len(longestWord):
longestWord = x[r]
for j in range(len(x)):
thing = x[j]
currentCount = x.count(thing)
if currentCount > mostWord:
mostWord = currentCount
THEmostWord = thing
return (THEmostWord, longestWord)

4
pie_test_set/p00252.py
View file

@ -1,4 +0,0 @@
def problem_p00252(input_data):
a, b, c = list(map(int, input_data.split()))
return ["Close", "Open"][(a & b) ^ c]

4
pie_test_set/p00312.py
View file

@ -1,4 +0,0 @@
def problem_p00312(input_data):
D, L = list(map(int, input_data.split()))
return sum(divmod(D, L))

25
pie_test_set/p00322.py
View file

@ -1,25 +0,0 @@
def problem_p00322(input_data):
import itertools
u = [1, 2, 3, 4, 5, 6, 7, 8, 9]
a = 0
n = list(map(int, input_data.split()))
for x in itertools.permutations(u):
f = 0
for i in range(9):
if n[i] != -1 and n[i] != x[i]:
f = 1
if f:
continue
if x[0] + x[2] + x[5] - x[8] + (x[1] + x[4] - x[7]) * 10 + (x[3] - x[6]) * 100 == 0:
a += 1
return a

4
pie_test_set/p00340.py
View file

@ -1,4 +0,0 @@
def problem_p00340(input_data):
rec = sorted(list(map(int, input_data.split())))
return "yes" if rec[0] == rec[1] and rec[2] == rec[3] else "no"

4
pie_test_set/p00341.py
View file

@ -1,4 +0,0 @@
def problem_p00341(input_data):
e = sorted(list(map(int, input_data.split())))
return "yes" if len(set(e[:4])) == len(set(e[4:8])) == len(set(e[8:])) == 1 else "no"

4
pie_test_set/p00353.py
View file

@ -1,4 +0,0 @@
def problem_p00353(input_data):
m, f, b = list(map(int, input_data.split()))
return "NA" if m + f < b else max(0, b - m)

12
pie_test_set/p00354.py
View file

@ -1,12 +0,0 @@
def problem_p00354(input_data):
from datetime import date
days = {i: d for i, d in enumerate(["mon", "tue", "wed", "thu", "fri", "sat", "sun"])}
day = int(eval(input_data))
result = date(2017, 9, day)
index = result.weekday()
return days[index]

44
pie_test_set/p00494.py
View file

@ -1,44 +0,0 @@
def problem_p00494(input_data):
s = eval(input_data)
length = len(s)
ind = 0
ans = 0
while ind < length:
j_num = 0
o_num = 0
i_num = 0
while ind < length and s[ind] != "J":
ind += 1
while ind < length and s[ind] == "J":
j_num += 1
ind += 1
while ind < length and s[ind] == "O":
o_num += 1
ind += 1
while ind < length and s[ind] == "I":
i_num += 1
ind += 1
if o_num <= i_num and o_num <= j_num:
ans = max(ans, o_num)
return ans

21
pie_test_set/p00514.py
View file

@ -1,21 +0,0 @@
def problem_p00514(input_data):
n, m, r = list(map(int, input_data.split()))
a, b = sorted([n - 1, r - m * n])
if a < 1:
return 0
else:
d = 1
for i in range(a):
d *= i + 1
u = 1
for i in range(b, a + b):
u *= i + 1
return u // d

51
pie_test_set/p00991.py
View file

@ -1,51 +0,0 @@
def problem_p00991(input_data):
# AOJ 1501: Grid
# Python3 2018.7.13 bal4u
MOD = 100000007
def dp(n, k):
if tbl[n][k]:
return tbl[n][k]
if (k << 1) > n:
k = n - k
if k == 0:
ans = 1
elif k == 1:
ans = n
else:
ans = dp(n - 1, k) + dp(n - 1, k - 1)
tbl[n][k] = ans % MOD
return tbl[n][k]
tbl = [[0 for j in range(1001)] for i in range(1001)]
k = 0
r, c, a1, a2, b1, b2 = list(map(int, input_data.split()))
dr = abs(a1 - b1)
if dr > r - dr:
dr = r - dr
if (dr << 1) == r:
k += 1
dc = abs(a2 - b2)
if dc > c - dc:
dc = c - dc
if (dc << 1) == c:
k += 1
return (dp(dr + dc, min(dr, dc)) << k) % MOD

4
pie_test_set/p01447.py
View file

@ -1,4 +0,0 @@
def problem_p01447(input_data):
import math
return math.ceil(math.log(int(eval(input_data))) / math.log(3))

32
pie_test_set/p01486.py
View file

@ -1,32 +0,0 @@
def problem_p01486(input_data):
S = eval(input_data)
L = len(S)
dp = [[False] * (L + 1) for _ in range(L + 1)]
# [left, right)
for i in range(L + 1):
dp[i][i] = True
for w in range(3, L + 1):
for left in range(L - w + 1):
right = left + w
if S[left] != "m" or S[right - 1] != "w":
continue
for mid in range(left + 1, right):
if S[mid] == "e" and dp[left + 1][mid] and dp[mid + 1][right - 1]:
dp[left][right] = True
break
return "Cat" if dp[0][L] else "Rabbit"

6
pie_test_set/p01741.py
View file

@ -1,6 +0,0 @@
def problem_p01741(input_data):
import math
n = float(eval(input_data))
return int(n) + 1 if n * math.sqrt(2) < int(n) + 1 else n * math.sqrt(2)

27
pie_test_set/p01751.py
View file

@ -1,27 +0,0 @@
def problem_p01751(input_data):
def solve(a, b, c):
l = 0
r = a
for t in range(114514):
t = l // 60
p = 60 * t + c
if l <= p <= r:
return p
exit()
l = r + b
r = l + a
return -1
a, b, c = list(map(int, input_data.split()))
solve(a, b, c)

23
pie_test_set/p01772.py
View file

@ -1,23 +0,0 @@
def problem_p01772(input_data):
S = input_data
if "A" not in S:
return -1
else:
ans = "A"
for s in S[S.index("A") + 1 :]:
if ans[-1] == "A" and s == "Z":
ans += s
if ans[-1] == "Z" and s == "A":
ans += s
if ans[-1] == "A":
ans = ans[:-1]
return ans if ans else -1

21
pie_test_set/p01809.py
View file

@ -1,21 +0,0 @@
def problem_p01809(input_data):
import fractions
a, b = list(map(int, input_data.split()))
b //= fractions.gcd(a, b)
a, c = 2, 1
while a**2 <= b:
if b % a == 0:
c *= a
while b % a == 0:
b //= a
a += 1
return c * b

44
pie_test_set/p01811.py
View file

@ -1,44 +0,0 @@
def problem_p01811(input_data):
#!/usr/bin/python
if __name__ == "__main__":
S = input_data
c = ["A", "B", "C"]
while True:
if len(S) <= 3:
return "Yes" if S == "ABC" else "No"
break
T = S.strip().split("ABC")
if len(T) == 1:
return "No"
break
P = "".join(T)
cnt = 0
for x in c:
if x in P:
cnt += 1
else:
S = x.join(T)
if cnt != 2:
return "No"
break

4
pie_test_set/p01852.py
View file

@ -1,4 +0,0 @@
def problem_p01852(input_data):
n = int(eval(input_data))
return 0 if n == 0 else len(str(bin(n))[2:])

4
pie_test_set/p02116.py
View file

@ -1,4 +0,0 @@
def problem_p02116(input_data):
n = int(eval(input_data))
return (n + 1) & -(n + 1)

16
pie_test_set/p02233.py
View file

@ -1,16 +0,0 @@
def problem_p02233(input_data):
def fib2(n):
a1, a2 = 1, 0
while n > 0:
a1, a2 = a1 + a2, a1
n -= 1
return a1
n = int(eval(input_data))
return fib2(n)

36
pie_test_set/p02256.py
View file

@ -1,36 +0,0 @@
def problem_p02256(input_data):
# http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_1_B&lang=jp
# ??????????????????????????????
# ?????§??¬?´???°???????????????????????????????????¨????????¨?¨????????????????
def gcd(x, y):
tmp = y
if y > x:
y = x
x = tmp
while not y == 0:
y = x % y
x = tmp
tmp = y
return x
def main():
target = [int(a) for a in input_data.split()]
return gcd(target[0], target[1])
if __name__ == "__main__":
main()

22
pie_test_set/p02263.py
View file

@ -1,22 +0,0 @@
def problem_p02263(input_data):
import re
def revPolish(f):
f = re.sub(
r"(\-?\d+\.?\d*)\s(\-?\d+\.?\d*)\s([\+\-\*/])(?!\d)",
lambda m: str(eval(m.group(1) + m.group(3) + m.group(2))),
f,
)
if f[-1] in ("+", "-", "*", "/"):
return revPolish(f)
else:
return f
if __name__ == "__main__":
return revPolish(eval(input_data))

6
pie_test_set/p02379.py
View file

@ -1,6 +0,0 @@
def problem_p02379(input_data):
import math
x1, y1, x2, y2 = list(map(float, input_data.split(" ")))
return "{:.5f}".format(math.sqrt((x2 - x1) ** 2 + (y2 - y1) ** 2))

2
pie_test_set/p02388.py
View file

@ -1,2 +0,0 @@
def problem_p02388(input_data):
return input_data**3

4
pie_test_set/p02389.py
View file

@ -1,4 +0,0 @@
def problem_p02389(input_data):
[a, b] = list(map(int, input().split()))
return "%s %s" % (a * b, 2 * (a + b))

10
pie_test_set/p02392.py
View file

@ -1,10 +0,0 @@
def problem_p02392(input_data):
l = input_data.split()
if int(l[0]) < int(l[1]) and int(l[1]) < int(l[2]):
return "Yes"
else:
return "No"

14
pie_test_set/p02393.py
View file

@ -1,14 +0,0 @@
def problem_p02393(input_data):
x = input_data.split()
y = list(map(int, x))
a = y[0]
b = y[1]
c = y[2]
d = sorted([a, b, c])
return "{0} {1} {2}".format(d[0], d[1], d[2])

22
pie_test_set/p02394.py
View file

@ -1,22 +0,0 @@
def problem_p02394(input_data):
a = input_data.split()
b = list(map(int, a))
W = b[0]
H = b[1]
x = b[2]
y = b[3]
r = b[4]
if (r <= x <= (W - r)) and (r <= y <= (H - r)):
return "Yes"
else:
return "No"

12
pie_test_set/p02398.py
View file

@ -1,12 +0,0 @@
def problem_p02398(input_data):
a, b, c = list(map(int, input_data.split()))
count = 0
for i in range(a, b + 1):
if c % i == 0:
count += 1
return count

10
pie_test_set/p02399.py
View file

@ -1,10 +0,0 @@
def problem_p02399(input_data):
(a, b) = [int(x) for x in input_data.split()]
x = a // b
y = a % b
z = a / b
return "{0} {1} {2:.6f}".format(x, y, z)

10
pie_test_set/p02400.py
View file

@ -1,10 +0,0 @@
def problem_p02400(input_data):
import math
r = float(eval(input_data))
a = math.pi * r**2
b = math.pi * r * 2
return str(a) + " " + str(b)

30
pie_test_set/p02406.py
View file

@ -1,30 +0,0 @@
def problem_p02406(input_data):
n = int(input_data)
i = 1
SP = []
while i <= n:
x = i
if x % 3 == 0:
SP.append(str(i))
else:
while x:
if x % 10 == 3:
SP.append(str(i))
break
x /= 10
i += 1
return ("", " ".join(SP))

2
pie_test_set/p02415.py
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def problem_p02415(input_data):
return input_data.swapcase()

34
pie_test_set/p02467.py
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@ -1,34 +0,0 @@
def problem_p02467(input_data):
import math
def factorize(p):
while p % 2 == 0:
p //= 2
yield 2
r = 3
while r < int(math.sqrt(p) + 1):
if p % r == 0:
p //= r
yield r
else:
r += 2
if p != 1:
yield p
n = int(eval(input_data))
l = factorize(n)
return (str(n) + ":", *list(l))

32
pie_test_set/p02468.py
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@ -1,32 +0,0 @@
def problem_p02468(input_data):
import sys
mod = 10**9 + 7
def solve():
m, n = list(map(int, input_data.split()))
ans = modpow(m, n, mod)
return ans
def modpow(x, y, mod):
res = 1
while y:
if y & 1:
res = (res * x) % mod
x = (x * x) % mod
y >>= 1
return res
if __name__ == "__main__":
solve()

22
pie_test_set/p02470.py
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@ -1,22 +0,0 @@
def problem_p02470(input_data):
f = lambda x: p * (x - 1) // x
p = n = int(eval(input_data))
d = 2
while d * d <= n:
if n % d == 0:
p = f(d)
while n % d == 0:
n //= d
d += 1
if n > 1:
p = f(n)
return p

36
pie_test_set/p02471.py
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@ -1,36 +0,0 @@
def problem_p02471(input_data):
def gcd(a, b):
global queue
r = a % b
if r:
d = a // b
sb = queue.pop()
sa = queue.pop()
sc = tuple(map(lambda x, y: x - d * y, sa, sb))
queue.append(sb)
queue.append(sc)
return gcd(b, r)
else:
return b
a, b = list(map(int, input_data.split()))
queue = [(1, 0, a), (0, 1, b)]
g = gcd(a, b)
x, y, r = queue.pop()
return (x, y)

120
pie_test_set/p02541.py
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@ -1,120 +0,0 @@
def problem_p02541(input_data):
import sys
sys.setrecursionlimit(10**7)
def I():
return int(sys.stdin.readline().rstrip())
def MI():
return list(map(int, sys.stdin.readline().rstrip().split()))
def LI():
return list(map(int, sys.stdin.readline().rstrip().split())) # 空白あり
def LI2():
return list(map(int, sys.stdin.readline().rstrip())) # 空白なし
def S():
return sys.stdin.readline().rstrip()
def LS():
return list(sys.stdin.readline().rstrip().split()) # 空白あり
def LS2():
return list(sys.stdin.readline().rstrip()) # 空白なし
def extgcd(a, b):
if b:
d, y, x = extgcd(b, a % b)
y -= (a // b) * x
return d, x, y
return a, 1, 0
# V = [(X_i, Y_i), ...]: X_i (mod Y_i)
def remainder(V):
x = 0
d = 1
for X, Y in V:
g, a, b = extgcd(d, Y)
x, d = (Y * b * x + d * a * X) // g, d * (Y // g)
x %= d
return x, d
N = I()
def prime_factorization(n):
ANS = {}
for i in range(2, int(n**0.5) + 1):
if n % i == 0:
r = 0 # nがiで何回割り切れるか
while n % i == 0:
n //= i
r += 1
ANS[i] = r
if n != 1:
ANS[n] = 1
return ANS
prime = prime_factorization(N)
if not 2 in list(prime.keys()):
prime[2] = 1
else:
prime[2] += 1
A = list(prime.keys())
B = list(prime.values())
l = len(A)
ans = 2 * N
from itertools import product
for a in list(product([-1, 0], repeat=l)):
for i in range(l):
if a[i] != 0:
break
else:
continue
V = [(a[i], A[i] ** B[i]) for i in range(l)]
x, d = remainder(V)
ans = min(ans, x)
return ans

12
pie_test_set/p02546.py
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@ -1,12 +0,0 @@
def problem_p02546(input_data):
S = eval(input_data)
if S.endswith("s"):
S = S + "es"
else:
S = S + "s"
return S

67
pie_test_set/p02548.py
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@ -1,67 +0,0 @@
def problem_p02548(input_data):
import array
import bisect
import collections
import copy
import heapq
import itertools
import math
import os
import queue
import sys
# from scipy.sparse.csgraph import csgraph_from_dense, floyd_warshall
# from decimal import Decimal
# from collections import defaultdict, deque
sys.setrecursionlimit(10000000)
ii = lambda: int(sys.stdin.buffer.readline().rstrip())
il = lambda: list(map(int, sys.stdin.buffer.readline().split()))
fl = lambda: list(map(float, sys.stdin.buffer.readline().split()))
iln = lambda n: [int(sys.stdin.buffer.readline().rstrip()) for _ in range(n)]
iss = lambda: sys.stdin.buffer.readline().decode().rstrip()
sl = lambda: list(map(str, sys.stdin.buffer.readline().decode().split()))
isn = lambda n: [sys.stdin.buffer.readline().decode().rstrip() for _ in range(n)]
lcm = lambda x, y: (x * y) // math.gcd(x, y)
MOD = 10**9 + 7
INF = float("inf")
def main():
if os.getenv("LOCAL"):
sys.stdin = open("input.txt", "r")
N = ii()
ret = 0
for a in range(1, N):
b = N // a
if N - (a * b) <= 0:
ret += b - 1
else:
ret += b
return ret
if __name__ == "__main__":
main()

34
pie_test_set/p02550.py
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@ -1,34 +0,0 @@
def problem_p02550(input_data):
n, x, m = list(map(int, input_data.split()))
ans = []
flag = False
for i in range(n):
if x in ans:
v = x
flag = True
break
ans.append(x)
x = x**2 % m
if flag:
p = ans.index(v)
l = len(ans) - p
d, e = divmod(n - p, l)
return sum(ans[:p]) + d * sum(ans[p:]) + sum(ans[p : p + e])
else:
return sum(ans)

10
pie_test_set/p02552.py
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@ -1,10 +0,0 @@
def problem_p02552(input_data):
x = int(eval(input_data))
if x == 0:
return 1
else:
return 0

28
pie_test_set/p02553.py
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@ -1,28 +0,0 @@
def problem_p02553(input_data):
import numpy as np
a, b, c, d = list(map(int, input_data.split()))
hoge = []
hoge.append(a * c)
hoge.append(a * d)
hoge.append(b * c)
hoge.append(b * d)
if max(hoge) < 0:
if np.sign(a) != np.sign(b) or np.sign(c) != np.sign(d):
return 0
else:
return max(hoge)
else:
return max(hoge)

6
pie_test_set/p02554.py
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@ -1,6 +0,0 @@
def problem_p02554(input_data):
N = int(eval(input_data))
ans = 10**N - 2 * 9**N + 8**N
return ans % (10**9 + 7)

24
pie_test_set/p02555.py
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@ -1,24 +0,0 @@
def problem_p02555(input_data):
from math import factorial as fac
S = int(eval(input_data))
ans = 0
if S == 1 or S == 2:
return 0
elif S == 3:
return 1
else:
for k in range(1, S):
if S - 2 * k - 1 >= 0 and S - 3 * k >= 0:
ans += fac(S - 2 * k - 1) // (fac(k - 1) * fac(S - 3 * k))
return ans % 1000000007

12
pie_test_set/p02570.py
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@ -1,12 +0,0 @@
def problem_p02570(input_data):
D, T, S = list(map(int, input_data.split()))
a = T * S - D
if a >= 0:
return "Yes"
else:
return "No"

10
pie_test_set/p02576.py
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@ -1,10 +0,0 @@
def problem_p02576(input_data):
array = list(map(int, input_data.split()))
if array[0] % array[1] > 0:
return (array[0] // array[1] + 1) * array[2]
else:
return (array[0] // array[1]) * array[2]

10
pie_test_set/p02577.py
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@ -1,10 +0,0 @@
def problem_p02577(input_data):
n = int(eval(input_data))
if n % 9 == 0:
return "Yes"
else:
return "No"

18
pie_test_set/p02582.py
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@ -1,18 +0,0 @@
def problem_p02582(input_data):
S = eval(input_data)
if S == "RRR":
return 3
elif "RR" in S:
return 2
elif "R" in S:
return 1
else:
return 0

24
pie_test_set/p02584.py
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@ -1,24 +0,0 @@
def problem_p02584(input_data):
X, K, D = list(map(int, input_data.split()))
if X + K * D <= 0:
return -X - K * D
elif X - K * D >= 0:
return X - K * D
else:
div = X // D
mod = X % D
if (K - div) % 2 == 0:
return mod
else:
return abs(mod - D)

22
pie_test_set/p02594.py
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@ -1,22 +0,0 @@
def problem_p02594(input_data):
# -*- coding: utf-8 -*-
# 標準入力を取得
X = int(eval(input_data))
# 求解処理
ans = str()
if X >= 30:
ans = "Yes"
else:
ans = "No"
# 結果出力
return ans

24
pie_test_set/p02596.py
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@ -1,24 +0,0 @@
def problem_p02596(input_data):
def c_repsept():
K = int(eval(input_data))
L = 9 * K // 7 if K % 7 == 0 else 9 * K
if L % 2 == 0 or L % 5 == 0:
return -1
remainder = 1
for n in range(1, L + 1):
remainder = (10 * remainder) % L
if remainder == 1:
return n
return -1
return c_repsept()

36
pie_test_set/p02600.py
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@ -1,36 +0,0 @@
def problem_p02600(input_data):
a = int(eval(input_data))
b = int(a // 200)
if b == 2:
return 8
elif b == 3:
return 7
elif b == 4:
return 6
elif b == 5:
return 5
elif b == 6:
return 4
elif b == 7:
return 3
elif b == 8:
return 2
elif b == 9:
return 1

12
pie_test_set/p02606.py
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@ -1,12 +0,0 @@
def problem_p02606(input_data):
l, r, d = [int(i) for i in input_data.split()]
count = 0
for i in range(l, r + 1):
if i % d == 0:
count += 1
return count

14
pie_test_set/p02612.py
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@ -1,14 +0,0 @@
def problem_p02612(input_data):
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N = int(input_data)
x = N + (-N) % 1000
return x - N

4
pie_test_set/p02621.py
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@ -1,4 +0,0 @@
def problem_p02621(input_data):
a = int(eval(input_data))
return a + a**2 + a**3

71
pie_test_set/p02624.py
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@ -1,71 +0,0 @@
def problem_p02624(input_data):
import sys
import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def prime_table(N):
is_prime = np.zeros(N, np.int64)
is_prime[2:3] = 1
is_prime[3::2] = 1
for p in range(3, N, 2):
if p * p >= N:
break
if is_prime[p]:
is_prime[p * p :: p + p] = 0
return is_prime, np.where(is_prime)[0]
def main(N, primes):
div = np.ones(N + 1, dtype=np.int64)
for p in primes:
for i in range(N // p + 1):
div[p * i] += div[i]
div *= np.arange(N + 1)
return div.sum()
if sys.argv[-1] == "ONLINE_JUDGE":
import numba
from numba.pycc import CC
i8 = numba.int64
cc = CC("my_module")
def cc_export(f, signature):
cc.export(f.__name__, signature)(f)
return numba.njit(f)
main = cc_export(main, (i8, i8[:]))
cc.compile()
from my_module import main
N = int(input_data)
is_prime, primes = prime_table(N + 1)
return main(N, primes)

12
pie_test_set/p02625.py
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@ -1,12 +0,0 @@
def problem_p02625(input_data):
n, m = list(map(int, input_data.split()))
a, M, d = 1, 10**9 + 7, [1]
for i in range(1, n + 1):
d.append(((m - n + i - 1) * d[i - 1] + (i - 1) * d[i - 2]) % M)
a = a * (m - i + 1) % M
return a * d[-1] % M

10
pie_test_set/p02627.py
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@ -1,10 +0,0 @@
def problem_p02627(input_data):
alph = eval(input_data)
if alph.isupper():
return "A"
else:
return "a"

22
pie_test_set/p02629.py
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@ -1,22 +0,0 @@
def problem_p02629(input_data):
n = int(eval(input_data))
a = []
while n:
a += (n % 26,)
n //= 26
a = a[::-1]
for i in range(len(a) - 1, 0, -1):
if a[i] < 1:
a[i] += 26
a[i - 1] -= 1
return "".join(chr(c + 96) for c in a[not a[0] :])

8
pie_test_set/p02633.py
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@ -1,8 +0,0 @@
def problem_p02633(input_data):
from math import gcd
x = int(eval(input_data))
y = x * 360 // gcd(x, 360)
return y // x

16
pie_test_set/p02634.py
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@ -1,16 +0,0 @@
def problem_p02634(input_data):
a, b, c, d = list(map(int, input_data.split()))
m = 998244353
D = [[0] * (d + 1) for _ in range(c + 1)]
D[a][b] = 1
for i in range(a, c + 1):
for j in range(b, d + 1):
D[i][j] += (D[i][j - 1] * i + D[i - 1][j] * j - D[i - 1][j - 1] * (i - 1) * (j - 1)) % m
return D[c][d]

78
pie_test_set/p02635.py
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@ -1,78 +0,0 @@
def problem_p02635(input_data):
def main():
mod = 998244353
s, k = input_data.split()
k = int(k)
n = len(s)
one = s.count("1")
cnt = 0
zero_list = []
for i in range(n):
if s[i] == "0":
zero_list.append(cnt)
cnt = 0
else:
cnt += 1
z = 0
mm = min(one, k)
dp = [[0] * (one + 1) for _ in [0] * (one + 1)]
dp[0][0] = 1
for i in range(len(zero_list)):
dp2 = [[0] * (mm + 1) for _ in [0] * (one + 1)]
base = zero_list[i]
# j:何個今までに入れたか
for j in range(one + 1):
# l:何個入れるか
for l in range(one + 1 - j):
if l < z + base - j:
continue
ml = max(l - base, 0)
# p:これまでのペナルティ
for p in range(min(one, k) + 1):
q = p + ml
if q <= mm:
dp2[j + l][q] = (dp2[j + l][q] + dp[j][p]) % mod
else:
break
z += base
dp = dp2
return sum([sum(i) for i in dp]) % mod
main()

24
pie_test_set/p02639.py
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@ -1,24 +0,0 @@
def problem_p02639(input_data):
from collections import deque
from math import gcd, sqrt
from sys import stdin, stdout
input = stdin.readline
R = lambda: list(map(int, input_data.split()))
I = lambda: int(eval(input_data))
S = lambda: input_data.rstrip("\n")
L = lambda: list(R())
P = lambda x: stdout.write(x)
hg = lambda x, y: ((y + x - 1) // x) * x
pw = lambda x: 1 if x == 1 else 1 + pw(x // 2)
chk = lambda x: chk(x // 2) if not x % 2 else True if x == 1 else False
return list(map(int, input_data.split())).index(0) + 1

24
pie_test_set/p02640.py
View file

@ -1,24 +0,0 @@
def problem_p02640(input_data):
x, y = list(map(int, input_data.split()))
flg = False
for i in range(101):
for j in range(101):
if (i * 2) + (j * 4) == y and (i + j) == x:
return "Yes"
flg = True
break
if flg == True:
break
else:
return "No"

4
pie_test_set/p02645.py
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@ -1,4 +0,0 @@
def problem_p02645(input_data):
s = eval(input_data)
return s[:3]

98
pie_test_set/p02653.py
View file

@ -1,98 +0,0 @@
def problem_p02653(input_data):
import sys
import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
MOD = 10**9 + 7
N, A, B = list(map(int, input_data.split()))
def solve(N, A, B):
if A > B:
A, B = B, A
if A == 1:
return pow(2, N, MOD)
dp1 = np.zeros((N, B), dtype=np.int64) # その時点で確定している長さ集計
dp2 = np.zeros((N, B), dtype=np.int64) # もし右に十分な 0 があったとしたときの、長さ集計
dp1_sum = np.zeros(N, dtype=np.int64)
ans = 0
for n in range(N):
# そこで初めての 1 がたつ場合
if n < A:
dp1[n, 1] += 1
else:
if n + 1 < B:
dp1[n, n + 1] += 1
# ひとつ手前から 1 を立てて遷移した場合
if n > 0:
dp1[n, 1:] += dp1[n - 1, :-1]
# 使えない個数の 0 をはさんで遷移した場合
dp1[n, 1] += dp1_sum[max(0, n - A) : n - 1].sum() % MOD
# たくさんの個数の 0 をはさんで遷移した場合
if n >= A + 1 < B:
dp1[n, A + 1 : B] += dp2[n - A - 1, 0 : B - (A + 1)]
dp1[n] %= MOD
# 右を 0 埋めして足せるなら答に足す
dp1_sum[n] = dp1[n].sum() % MOD
if (N - n - 1) < A:
ans += dp1_sum[n]
ans %= MOD
else:
# あと N - n - 1 のびても B にならないような範囲
M = B - (N - n - 1)
if M > 0:
ans += dp1[n, :M].sum() % MOD
dp2[n] += dp1[n]
dp2[n, 1:] += dp2[n - 1, :-1]
dp1[n] %= MOD
dp2[n] %= MOD
x = pow(2, N, MOD) - ans
return x % MOD
return solve(N, A, B)

21
pie_test_set/p02657.py
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@ -1,21 +0,0 @@
def problem_p02657(input_data):
def iput():
return int(eval(input_data))
def mput():
return list(map(int, input_data.split()))
def lput():
return list(map(int, input_data.split()))
def solve():
a, b = mput()
return a * b
return 0
if __name__ == "__main__":
solve()

3
pie_test_set/p02659.py
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@ -1,3 +0,0 @@
def problem_p02659(input_data):
a, b = input_data.split()
return int(a) * int(b[:-3] + b[-2:]) // 100

122
pie_test_set/p02660.py
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@ -1,122 +0,0 @@
def problem_p02660(input_data):
def getN():
return int(eval(input_data))
def getNM():
return list(map(int, input_data.split()))
def getList():
return list(map(int, input_data.split()))
import copy
import math
import sys
from bisect import bisect_left, bisect_right
from collections import defaultdict, deque
from heapq import *
from sys import exit
# sys.setrecursionlimit(1000000)
INF = 10**17
MOD = 1000000007
from fractions import *
def inverse(f):
# return Fraction(f.denominator,f.numerator)
return 1 / f
def combmod(n, k, mod=MOD):
ret = 1
for i in range(n - k + 1, n + 1):
ret *= i
ret %= mod
for i in range(1, k + 1):
ret *= pow(i, mod - 2, mod)
ret %= mod
return ret
def bunsu(n):
ret = []
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
tmp = 0
while True:
if n % i == 0:
tmp += 1
n //= i
else:
break
ret.append((i, tmp))
ret.append((n, 1))
return ret
def solve():
n = getN()
bun = bunsu(n)
# return (bun)
acc = []
tmp = 0
for i in range(10000):
tmp += i
acc.append(tmp)
ans = 0
for b, cnt in bun:
if b == 1:
continue
ans += bisect_right(acc, cnt) - 1
return ans
def main():
# n = getN()
# for _ in range(n):
solve()
if __name__ == "__main__":
solve()

12
pie_test_set/p02663.py
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@ -1,12 +0,0 @@
def problem_p02663(input_data):
HM = list(map(int, input_data.split()))
hours = HM[2] - HM[0]
minutes = HM[3] - HM[1]
K = HM[4]
totalTimes = hours * 60 + minutes
return totalTimes - K

56
pie_test_set/p02667.py
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@ -1,56 +0,0 @@
def problem_p02667(input_data):
t = eval(input_data)
before_is_zero = True
keeped_ones = 0
n0 = 0
ones = [0] * int(1e5 + 1) # (number of zeros before 1, is_odd)
ones_index = 0
for i, x in enumerate(t):
if x == "0":
n0 += 1
before_is_zero = True
elif x == "1":
if before_is_zero:
ones[ones_index] = (n0, (i + 1) % 2)
ones_index += 1
before_is_zero = False
else:
keeped_ones += 1
ones_index -= 1
before_is_zero = True
if before_is_zero:
if len(ones) > 0:
ones = ones[:-1]
ans = keeped_ones * n0
for i in range(ones_index):
n0_before, is_odd = ones[i]
ans += (n0_before + is_odd) // 2 + (n0 - n0_before)
ans += (keeped_ones + (ones_index + 1) // 2) * (keeped_ones + ones_index // 2 + 1)
return ans

14
pie_test_set/p02675.py
View file

@ -1,14 +0,0 @@
def problem_p02675(input_data):
n = eval(input_data)
if n[-1] == "3":
return "bon"
elif n[-1] in ["0", "1", "6", "8"]:
return "pon"
else:
return "hon"

14
pie_test_set/p02682.py
View file

@ -1,14 +0,0 @@
def problem_p02682(input_data):
A, B, C, K = list(map(int, input_data.split()))
if A >= K:
return K
elif A + B >= K:
return A
else:
return A - (K - (A + B))

34
pie_test_set/p02685.py
View file

@ -1,34 +0,0 @@
def problem_p02685(input_data):
# coding: utf-8
def solve(*args: str) -> str:
n, m, k = list(map(int, args[0].split()))
mod = 998244353
if m == 1 and n - 1 == k:
return str(1)
ncr = 1
p = m * pow(m - 1, n - 1, mod) % mod
ret = p
inv = pow(m - 1, mod - 2, mod)
for i in range(1, k + 1):
ncr = (ncr * (n - i) * pow(i, mod - 2, mod)) % mod
p = (p * inv) % mod
ret += p * ncr % mod
return str(ret % mod)
if __name__ == "__main__":
return solve(*(open(0).input_data.splitlines()))

4
pie_test_set/p02687.py
View file

@ -1,4 +0,0 @@
def problem_p02687(input_data):
S = eval(input_data)
return "ABC" if S == "ARC" else "ARC"

18
pie_test_set/p02690.py
View file

@ -1,18 +0,0 @@
def problem_p02690(input_data):
X = int(eval(input_data))
for A in range(-200, 200):
for B in range(-200, 200):
if A**5 - B**5 == X:
return (A, B)
break
else:
continue
break

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