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Simple problem of abs values of primes. . . Working on problem 36 Signed-off-by: Julien Lengrand-Lambert <julien@lengrand.fr>
76 lines
2.1 KiB
Python
76 lines
2.1 KiB
Python
#!/usr/bin/env python
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"""
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##---
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# Julien Lengrand-Lambert
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#Created on : Thu Jan 19 10:12:06 CET 2012
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#
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# DESCRIPTION : Solves problem 27 of Project Euler
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A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
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Euler published the remarkable quadratic formula:
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n^2 + n + 41
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It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39.
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Considering quadratics of the form:
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n^2 + an + b, where |a| = 1000 and |b| = 1000
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where |n| is the modulus/absolute value of n
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e.g. |11| = 11 and |4| = 4
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Find the product of the coefficients, a and b,
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for the quadratic expression that produces the maximum number of
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primes for consecutive values of n, starting with n = 0.
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##---
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# TODO : enhance with memoization?
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"""
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def is_prime(value):
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"""
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Returns True or False depending whether value is prime or not.
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"""
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start = 2
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while (start < value / 2):
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if value % start == 0 :
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return False
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else :
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start += 1
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return True
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def quad_fun(a, b, n):
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"""
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Returns the result of the Euler quadratic function:
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n^2 + a*n + b
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"""
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return (pow(n, 2) + a * n + b)
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def prime_series(a_range, b_range):
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"""
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Returns the product of a and b for the quad_fun that produces the max number of primes;
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a_range, b_range being the ranges for a and b
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"""
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max = 0
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max_prod = 0
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a_curr = 0
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for a in a_range:
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a_curr += 1
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print "%d/%d" %(a_curr, len(a_range))
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for b in b_range:
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n = 0
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value = quad_fun(a, b, n)
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while is_prime(abs(value)):
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n += 1
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value = quad_fun(a, b, n)
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# checking if current serie if the best we got :
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if n > max:
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max = n
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max_prod = a * b
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return max_prod
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if __name__ == '__main__' :
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a_range = range(-999, 1000)
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b_range = range(-999, 1000)
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print "Answer is : %d" % (prime_series(a_range, b_range))
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raw_input() |