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44_pentagonal_numbers.py
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44_pentagonal_numbers.py
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# https://projecteuler.net/problem=44
#
# Pentagonal numbers are generated by the formula, Pn=n(3n-1)/2. The first ten
# pentagonal numbers are:
#
# 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
#
# It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their difference,
# 70 - 22 = 48, is not pentagonal.
#
# Find the pair of pentagonal numbers, Pj and Pk, for which their sum and
# difference are pentagonal and D = |Pk - Pj| is minimised; what is the
# value of D?
import utils
limit = 10000 # limit in our search for pentagonal nums to prevent hangs
def main():
p = [utils.pentagonal_number(n) for n in xrange(1, limit)]
p_set = set(p) # for fast lookups
# p_k > p_j as p_k - p_j must be pentagonal ie. positive
return min([p[k] - p[j] for k in xrange(0, len(p)) for j in xrange(0, k)
if p[k] + p[j] in p_set and p[k] - p[j] in p_set])
# the long way:
# d = 100000000 # our smallest pentagonal difference
# for k in xrange(0, len(p_nums)):
# # progress print
# if not k % 1000:
# print k
# p_k = p_nums[k]
# # p_k > p_j as p_k - p_j must be pentagonal ie. positive
# for j in xrange(0, k):
# p_j = p_nums[j]
# diff = p_k - p_j
# if p_j + p_k in p_nums_set and diff in p_nums_set:
# print "found valid pentagonal pair %d and %d" % (p_j, p_k)
# if diff < d:
# d = diff
# print("p_j %d and p_k %d produce smallest diff so far: %d"
# % (p_j, p_k, d))
if __name__ == "__main__":
print main()