31 hitting memory limit, want to try on machine.

SamGerber · SamGerber · commit a818bb7d9624 · 2015-10-03T22:15:07.000Z
diff --git a/project-euler/31-pandigital_products.rb b/project-euler/31-pandigital_products.rb
@@ -0,0 +1,50 @@
+# Pandigital products
+# Problem 32
+# We shall say that an n-digit number is pandigital if it makes use of all the 
+# digits 1 to n exactly once; for example, 
+# the 5-digit number, 15234, is 1 through 5 pandigital.
+
+# The product 7254 is unusual, as the identity, 39 × 186 = 7254, 
+# containing multiplicand, multiplier, and product is 1 through 9 pandigital.
+
+# Find the sum of all products whose multiplicand/multiplier/product identity
+# can be written as a 1 through 9 pandigital.
+
+# HINT: Some products can be obtained in more than one way so be sure to only
+# include it once in your sum.
+
+# Solution by Sam Gerber
+
+# This method blah
+def pandigital_products
+  characters = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 0]
+  equations = characters.permutation.sort.uniq
+  results = []
+  
+  equations.each do |equation|
+    star_index = equation.index(0)
+    break unless star_index > 0 && star_index < 5
+    number1 = equation[0...star_index].join.to_i
+    equation = equation[star_index + 1..-1]
+    equal_index = equation.index(0)
+    break unless equal_index > 0 && equal_index < equation.length - 1
+    number2 = equation[0...equal_index].join.to_i
+    product = equation[equal_index + 1..-1].join.to_i
+    results << [number1, number2, product] if number1 * number2 == product
+  end
+  
+  results
+end
+
+
+
+
+
+# This method returns true if the numbers contain all the
+# digits from 1 to n
+def is_pandigital?(*numbers, n)
+  numbers = numbers.join.split("").sort
+  numbers.count == numbers.uniq.count && numbers.first == "1" && numbers.last == n.to_s
+end
+
+print pandigital_products
