Project Euler 51: Ruby

Posted by Ben Griswold on Johnny Coder See other posts from Johnny Coder or by Ben Griswold
Published on Tue, 14 Sep 2010 03:33:46 +0000 Indexed on 2010/12/06 16:59 UTC
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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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