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refactor: Simplify LCM number generation with break and single return

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Andreas Koepf (aider) 2025-01-24 11:08:30 +01:00
parent 20069b2a7d
commit 71b13cda07
1 changed files with 5 additions and 7 deletions
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12
reasoning_gym/arithmetic/lcm.py
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@ -43,17 +43,15 @@ class LCMDataset(ProceduralDataset):
def calculate_product(nums: List[int]) -> int:
return reduce(lambda x, y: x * y, nums)
for _ in range(3): # Try up to 3 times to get LCM < product
# Try up to 3 times to get LCM < product
for _ in range(3):
num_count = rng.randint(self.config.min_numbers, self.config.max_numbers)
numbers = [rng.randint(self.config.min_value, self.config.max_value) for _ in range(num_count)]
result = reduce(lcm, numbers)
if result < calculate_product(numbers):
return numbers, result
# If we failed to find LCM < product after 3 tries, generate one final set
num_count = rng.randint(self.config.min_numbers, self.config.max_numbers)
numbers = [rng.randint(self.config.min_value, self.config.max_value) for _ in range(num_count)]
result = reduce(lcm, numbers)
break
# Return the last generated numbers, whether they met the criteria or not
return numbers, result
def __getitem__(self, idx: int) -> dict: