This was a popular problem and we saw many different ways to go about solving it. The discussion on the mailing list was also very good, with many ideas shared and explained. I'm going to go through a few solutions today, because they are not overly long and I think we can learn something from each of them. Let's begin with some code by Dennis Frommknecht. Here's the entire solution: #!/usr/bin/env ruby -W # This solution uses 3 nested loops to divide the # numbers into 4 groups (using regular expressions). # Then the 3 allowed combinations of plus and minus # are inserted between the groups. # Finally the result is calculated using eval NUMBERS = "123456789" CORRECT_RES = 100 OPS = [['+', '-', '-'], ['-', '+', '-'], ['-', '-', '+']] num_of_ops = OPS[0].length equ_counter = 0 1.upto(NUMBERS.length - num_of_ops) do |i| 1.upto(NUMBERS.length - num_of_ops - i + 1) do |j| 1.upto(NUMBERS.length - num_of_ops - i + 1 - j + 1) do |k| if NUMBERS.match(/(\d{#{i}})(\d{#{j}})(\d{#{k}})(\d+)/) then OPS.each do |o| command = "#{$1} #{o[0]} #{$2} #{o[1]} #{$3} #{o[2]} #{$4}" res = eval command equ_counter += 1 puts "*" * 15 if res == CORRECT_RES puts "#{command} = #{res}" puts "*" * 15 if res == CORRECT_RES end end end end end puts "#{equ_counter} possible equations tested" There are really two aspects to solving the base quiz. First, you need to find the possible permutation of the operators. There are really only three orders they can come up in and Dennis chose to just hardcode those in the OPS constant. The other problem is about partitioning. You need to divide the digits into four groupings that you can insert the three operators between. Again, Dennis pretty much hardcoded the logic for this using three nested iterators. The interesting side of this solution though is how they work. Basically, Dennis just counts off some number of digits from the front of the line. Then he counts of some number of digits after that. A third count is made for some number of digits after that, ensuring we leave at least one at the end. This gives us three lengths, which one Regexp turns into four groups. From there, the expression is constructed, eval()ed to get an answer, and finally printed. This solution does a good job of showing the process, but there are some shortcuts to be found. For example, if you move from considering just the ordering of the three operators to the actual positioning of the three operators, you can drop the partitioning step altogether. Let me use Paul's code to show you what I mean: class String def unique_permutations # modified to get unique permutations from # http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/... # which says it was inspired by a RubyQuiz! :) return [self] if self.length < 2 perms = Hash.new 0.upto(self.length - 1) do |n| rest = self.split(//u) # for UTF-8 encoded strings picked = rest.delete_at(n) rest.join.unique_permutations.each { |x| perms[picked + x] = nil } end perms.keys end end digits = ARGV[0] ops = ARGV[1] target = ARGV[2].to_i # pad ops list with spaces to match the number of slots between the digits ops = ops + " " * (digits.size - ops.size - 1) # build a format string with slots between the digits digits = digits.split("").join("%s") operator_perms = ops.unique_permutations operator_perms.each do |p| # build expression by inserting the ops into the format string, # after converting spaces to empty strings exp = digits % p.split("").map{|x|x.chomp(" ")} val = eval(exp) puts "*******************" if val==target puts exp + " = " + val.to_s puts "*******************" if val==target end puts puts "%d possible equations tested" % operator_perms.size Paul's permutation code is the first new element that jumps out at us here. It works by recursively combining smaller and smaller permutations until it has generated all possible combinations. The permutations are stored in Hash keys so duplicates will automatically be eliminated. Next we see that Paul nails pretty much all of the extra credit by importing the digits, operators, and target from command-line parameters. This means that he can't hardcode any logic, because he can't count on what will be passed to his script. The next two lines are the interesting part. Paul counts the number of spaces between all the digits. These are the positions in which operators need to be placed. Paul pads the operator list to include extra spaces, so it will match the position count. Then the digit list is transformed into a printf() style pattern that will insert an operator, or extra space, between each number. From there it just takes one call to the unique_permutations() method to walk the choices. Paul's output code is very similar to the code we saw from Dennis earlier. Going one step further, let's examine how Christian Neukirchen walked the same position list: (00000000..'22222222'.to_i(3)).map { |x| x.to_s(3).rjust(8, "0"). tr('012', '-+ ') }. find_all { |x| x.count("-") == 2 and x.count("+") == 1 }. map { |x| t = "1" + x.split(//).zip((2..9).to_a).join.delete(" ") [eval(t), t] }.sort.each { |s, x| puts "*****************" if s == 100 puts "#{x}: #{s}" puts "*****************" if s == 100 } This code does the same permutation of positions Paul's code did, using simple counting. You see, counting from 00000000 (a pretty way to show 0 with eight positions) to whatever number '22222222'.to_i(3) is, in base 3, will yield all possible combinations of 0s, 1s, and 2s. Christian then culls the list for the proper mixes of operators and transliterates those digits to operators or spaces. The rest of the code we have seen before. Together these solutions show quite a few options for handling permutations in Ruby. Another common option from the submissions I didn't show was to load an external library that handles permutations. This is a great option in practice, because you get road-tested code, but I wanted to show how you might handle such a thing for this summary. My thanks to all the people who sent in permutations of Ruby code to solve this permutations problem. They are great stuff. Tomorrow's problem is something you can handle on your fingers...

on 2007-04-12 15:01