# Fwd: Please Forward: Ruby Quiz Submission

**URL:** <https://www.ruby-forum.com/t/fwd-please-forward-ruby-quiz-submission/229091>\
**Category:** Ruby\
**Created:** [December 17, 2006, 5:46pm UTC](https://www.ruby-forum.com/t/fwd-please-forward-ruby-quiz-submission/229091 "2006-12-17T17:46:31Z")\
**Posts on this page:** 2\
**Page:** 1

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**Author:** ![bbazzarrakk](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/b/47e85d/32.png) [@bbazzarrakk](https://www.ruby-forum.com/u/bbazzarrakk)\
**Post date:** [December 17, 2006, 5:46pm UTC](https://www.ruby-forum.com/t/fwd-please-forward-ruby-quiz-submission/229091/1 "2006-12-17T17:46:31Z")

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Begin forwarded message:

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<div class="post-metadata">

**Author:** ![bbazzarrakk](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/b/47e85d/32.png) [@bbazzarrakk](https://www.ruby-forum.com/u/bbazzarrakk)\
**Post date:** [December 18, 2006, 2:11am UTC](https://www.ruby-forum.com/t/fwd-please-forward-ruby-quiz-submission/229091/2 "2006-12-18T02:11:48Z")

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On Mon, 18 Dec 2006 01:44:16 +0900, James Edward G. II wrote:

> > #960 it will have to iterate about 20,000 times, however it still  
> > while h.size\<i.to\_i  
> > p=p.sort\_by{rand}  
> > $_=p\*’ ’  
> > h[$_]=1if~/R.\*K.\*R/&&~/B(…)\*B/  
> > end  
> > puts"Position ##{i}:",$\_

My solution borrows heavily from Darren’s, but I don’t like the idea of  
creating permutations by bogosorting, since there’s no guarantee you’ll  
ever see every necessary permutation. (Here, he does see every  
permutation, quickly, but that’s not guaranteed for different random  
seeds.)

#!/usr/bin/env ruby  
require ‘rubygems’  
require ‘facets/core/enumerable/permutation’  
require ‘set’  
i=ARGV[0] || rand(960)+1  
i=i.to\_i  
fail “A number between 1 and 960 is required” if not (1…960).include? i  
pieces=%w’K R N B B Q N R’  
generated=Set.new  
pieces.each\_permutation do |p|  
arrangement=p.join  
if arrangement=~/R.\*K.\*R/ and arrangement=~/B(…)\*B/ and  
not generated.include?(arrangement)

```
#the "generated" set is needed because identical objects will
#nevertheless appear in both possible permutations, like so:
# [1,1].each_permutation {|pe| p pe}
# gives the output
# [1, 1]
# [1, 1]
generated.add arrangement

if generated.size==i
  print "Position ##{generated.size}:", arrangement, "\n"
  break
end

```

end  
end