95 lines
2.3 KiB
Python
95 lines
2.3 KiB
Python
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def odd(n):
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return n % 2 == 1
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def is_square(n):
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""" From the internet. I should implement my own version of this. """
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## Trivial checks
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if type(n) != int: ## integer
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return False
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if n < 0: ## positivity
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return False
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if n == 0: ## 0 pass
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return True
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## Reduction by powers of 4 with bit-logic
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while n&3 == 0:
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n=n>>2
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## Simple bit-logic test. All perfect squares, in binary,
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## end in 001, when powers of 4 are factored out.
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if n&7 != 1:
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return False
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if n==1:
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return True ## is power of 4, or even power of 2
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## Simple modulo equivalency test
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c = n%10
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if c in {3, 7}:
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return False ## Not 1,4,5,6,9 in mod 10
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if n % 7 in {3, 5, 6}:
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return False ## Not 1,2,4 mod 7
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if n % 9 in {2,3,5,6,8}:
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return False
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if n % 13 in {2,5,6,7,8,11}:
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return False
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## Other patterns
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if c == 5: ## if it ends in a 5
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if (n//10)%10 != 2:
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return False ## then it must end in 25
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if (n//100)%10 not in {0,2,6}:
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return False ## and in 025, 225, or 625
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if (n//100)%10 == 6:
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if (n//1000)%10 not in {0,5}:
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return False ## that is, 0625 or 5625
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else:
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if (n//10)%4 != 0:
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return False ## (4k)*10 + (1,9)
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## Babylonian Algorithm. Finding the integer square root.
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## Root extraction.
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s = (len(str(n))-1) // 2
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x = (10**s) * 4
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A = {x, n}
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while x * x != n:
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x = (x + (n // x)) >> 1
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if x in A:
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return False
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A.add(x)
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return True
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def euler_094():
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n_max = 10 ** 9
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n, total_peri = 2, 0
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while n < n_max:
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n += 2
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for b2, c in [(n, n + 1), (n, n - 1)]:
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b = b2 // 2
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a_squared = c * c - b * b
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if not is_square(a_squared):
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continue
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peri = 2 * c + b2
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# print("b2={} c={} peri={}".format(b2, c, peri))
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if peri > n_max:
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return total_peri
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total_peri += peri
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if b2 > 10000:
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n = int(n * 3.725) - 5
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if odd(n):
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n -= 1
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if __name__ == "__main__":
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solution = euler_094()
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print("e094.py: " + str(solution))
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assert(solution == 518408346)
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