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89 lines
2.3 KiB
Python
89 lines
2.3 KiB
Python
"""
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Project Euler Problem 58:https://projecteuler.net/problem=58
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Starting with 1 and spiralling anticlockwise in the following way,
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a square spiral with side length 7 is formed.
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37 36 35 34 33 32 31
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38 17 16 15 14 13 30
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39 18 5 4 3 12 29
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40 19 6 1 2 11 28
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41 20 7 8 9 10 27
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42 21 22 23 24 25 26
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43 44 45 46 47 48 49
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It is interesting to note that the odd squares lie along the bottom right
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diagonal ,but what is more interesting is that 8 out of the 13 numbers
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lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.
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If one complete new layer is wrapped around the spiral above,
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a square spiral with side length 9 will be formed.
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If this process is continued,
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what is the side length of the square spiral for which
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the ratio of primes along both diagonals first falls below 10%?
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Solution: We have to find an odd length side for which square falls below
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10%. With every layer we add 4 elements are being added to the diagonals
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,lets say we have a square spiral of odd length with side length j,
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then if we move from j to j+2, we are adding j*j+j+1,j*j+2*(j+1),j*j+3*(j+1)
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j*j+4*(j+1). Out of these 4 only the first three can become prime
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because last one reduces to (j+2)*(j+2).
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So we check individually each one of these before incrementing our
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count of current primes.
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"""
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from math import isqrt
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def isprime(number: int) -> int:
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"""
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returns whether the given number is prime or not
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>>> isprime(1)
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0
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>>> isprime(17)
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1
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>>> isprime(10000)
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0
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"""
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if number == 1:
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return 0
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if number % 2 == 0 and number > 2:
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return 0
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for i in range(3, isqrt(number) + 1, 2):
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if number % i == 0:
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return 0
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return 1
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def solution(ratio: float = 0.1) -> int:
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"""
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returns the side length of the square spiral of odd length greater
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than 1 for which the ratio of primes along both diagonals
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first falls below the given ratio.
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>>> solution(.5)
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11
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>>> solution(.2)
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309
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>>> solution(.111)
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11317
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"""
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j = 3
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primes = 3
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while primes / (2 * j - 1) >= ratio:
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for i in range(j * j + j + 1, (j + 2) * (j + 2), j + 1):
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primes = primes + isprime(i)
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j = j + 2
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return j
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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