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24fb140672
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| 24fb140672 | |||
| 2df02a8204 | |||
| 8580160d67 | |||
| 5cee1f7998 |
@@ -9,8 +9,9 @@ Solution for problem 008 of Project Euler
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https://projecteuler.net/problem=8
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https://projecteuler.net/problem=8
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"""
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"""
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import collections
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from collections import deque
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from itertools import islice
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from itertools import islice
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from math import prod
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from project_euler_python.utils import timeit
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from project_euler_python.utils import timeit
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@@ -19,7 +20,7 @@ from project_euler_python.utils import timeit
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def sliding_window(iterable, n):
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def sliding_window(iterable, n):
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# sliding_window('ABCDEFG', 4) -> ABCD BCDE CDEF DEFG
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# sliding_window('ABCDEFG', 4) -> ABCD BCDE CDEF DEFG
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it = iter(iterable)
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it = iter(iterable)
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window = collections.deque(islice(it, n), maxlen=n)
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window = deque(islice(it, n), maxlen=n)
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if len(window) == n:
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if len(window) == n:
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yield tuple(window)
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yield tuple(window)
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for x in it:
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for x in it:
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@@ -65,12 +66,10 @@ def compute():
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num = NUM.replace("\n", "").replace(" ", "")
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num = NUM.replace("\n", "").replace(" ", "")
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adjacent_digits = 13
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adjacent_digits = 13
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ans = 0
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ans = 0
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for nums in sliding_window(num, adjacent_digits):
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prod = 1
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for digits in sliding_window(num, adjacent_digits):
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for num in nums:
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window_product = prod(int(digit) for digit in digits)
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prod *= int(num)
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ans = max(ans, window_product)
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if prod > ans:
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ans = prod
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return ans
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return ans
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@@ -24,9 +24,9 @@ def compute():
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Find the product abc.
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Find the product abc.
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"""
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"""
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upper_limit = 1000
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upper_limit = 1000 + 1
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for a in range(1, upper_limit + 1):
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for a in range(1, upper_limit):
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for b in range(a + 1, upper_limit + 1):
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for b in range(a + 1, upper_limit):
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c = upper_limit - a - b
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c = upper_limit - a - b
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if a * a + b * b == c * c:
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if a * a + b * b == c * c:
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# It is now implied that b < c, because we have a > 0
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# It is now implied that b < c, because we have a > 0
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@@ -9,7 +9,7 @@ Solution for problem 012 of Project Euler
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https://projecteuler.net/problem=12
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https://projecteuler.net/problem=12
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"""
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"""
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from itertools import count
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from itertools import accumulate, count
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from math import floor, sqrt
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from math import floor, sqrt
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from project_euler_python.utils import timeit
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from project_euler_python.utils import timeit
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@@ -53,12 +53,10 @@ def compute():
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divisors?
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divisors?
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"""
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"""
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triangle = 0
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limit = 500
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for i in count(1):
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for triangle in accumulate(count(1)):
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# This is the ith triangle number, i.e. num = 1 + 2 + ... + i =
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# This is the ith triangle number, i.e. num = 1 + 2 + ... + i = i*(i+1)/2
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# = i*(i+1)/2
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if num_divisors(triangle) > limit:
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triangle += i
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if num_divisors(triangle) > 500:
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return str(triangle)
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return str(triangle)
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@@ -12,14 +12,14 @@ https://projecteuler.net/problem=14
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from project_euler_python.utils import timeit
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from project_euler_python.utils import timeit
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def chain_length(n, terms):
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def chain_length(n: int, terms: dict[int, int]) -> int:
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length = 0
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length = 0
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while n != 1:
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while n != 1:
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if n in terms:
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if n in terms:
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length += terms[n]
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length += terms[n]
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break
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break
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if n % 2 == 0:
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if n % 2 == 0:
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n = n / 2
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n //= 2
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else:
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else:
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n = 3 * n + 1
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n = 3 * n + 1
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length += 1
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length += 1
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@@ -27,7 +27,7 @@ def chain_length(n, terms):
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@timeit("Problem 014")
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@timeit("Problem 014")
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def compute():
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def compute() -> int:
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"""
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"""
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The following iterative sequence is defined for the set of positive
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The following iterative sequence is defined for the set of positive
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integers:
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integers:
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@@ -49,15 +49,17 @@ def compute():
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NOTE: Once the chain starts the terms are allowed to go above one million.
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NOTE: Once the chain starts the terms are allowed to go above one million.
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"""
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"""
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ans = 0
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limit = 1_000_000
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limit = 1_000_000
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score = 0
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ans = 0
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terms = dict()
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longest_length = 0
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for i in range(1, limit):
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chain_lengths = {1: 0}
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terms[i] = chain_length(i, terms)
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if terms[i] > score:
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for start in range(2, limit):
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score = terms[i]
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current_length = chain_length(start, chain_lengths)
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ans = i
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chain_lengths[start] = current_length
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if current_length > longest_length:
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ans = start
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longest_length = current_length
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return ans
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return ans
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