Project Euler 51: Ruby
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by Ben Griswold
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Published on Tue, 14 Sep 2010 03:33:46 +0000
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2010/12/06
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In my attempt to learn Ruby out in the open, here’s my solution for Project Euler Problem 51.
I know I started back up with Python this week, but I have three more Ruby solutions in the hopper and I wanted to share. For the record, Project Euler 51 was the second hardest Euler problem for me thus far. Yeah.
As always, any feedback is welcome.
# Euler 51 # http://projecteuler.net/index.php?section=problems&id=51 # By replacing the 1st digit of *3, it turns out that six # of the nine possible values: 13, 23, 43, 53, 73, and 83, # are all prime. # # By replacing the 3rd and 4th digits of 56**3 with the # same digit, this 5-digit number is the first example # having seven primes among the ten generated numbers, # yielding the family: 56003, 56113, 56333, 56443, # 56663, 56773, and 56993. Consequently 56003, being the # first member of this family, is the smallest prime with # this property. # # Find the smallest prime which, by replacing part of the # number (not necessarily adjacent digits) with the same # digit, is part of an eight prime value family. timer_start = Time.now require 'mathn' def eight_prime_family(prime) 0.upto(9) do |repeating_number| # Assume mask of 3 or more repeating numbers if prime.count(repeating_number.to_s) >= 3 ctr = 1 (repeating_number + 1).upto(9) do |replacement_number| family_candidate = prime.gsub(repeating_number.to_s, replacement_number.to_s) ctr += 1 if (family_candidate.to_i).prime? end return true if ctr >= 8 end end false end # Wanted to loop through primes using Prime.each # but it took too long to get to the starting value. n = 9999 while n += 2 next if !n.prime? break if eight_prime_family(n.to_s) end puts n puts "Elapsed Time: #{(Time.now - timer_start)*1000} milliseconds"
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