The three rules of Ruby Q.: 1. Please do not post any solutions or spoiler discussion for this quiz until 48 hours have passed from the time on this message. 2. Support Ruby Q. by submitting ideas as often as you can: http://www.rubyquiz.com/ 3. Enjoy! Suggestion: A [QUIZ] in the subject of emails about the problem helps everyone on Ruby T. follow the discussion. Please reply to the original quiz message, if you can. -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= A famous set of computer problems involve verbal arithmetic. In these problems, you are given equations of words like: send + more ------ money or: forty ten + ten ------- sixty The goal is to find a single digit for each letter that makes the equation true. Normal rules of number construction apply, so the first digit of a multi-digit number should be nonzero and each letter represents a different digit. This week's quiz is to build a program that reads in equations and outputs solutions. You can decide how complex of an equation you want to support, with the examples above being the minimum implementation. Here's a solution you can test against: $ ruby verbal_arithmetic.rb 'send+more=money' s: 9 e: 5 n: 6 d: 7 m: 1 o: 0 r: 8 y: 2

on 2007-06-15 16:24

on 2007-06-15 16:55

I really don't get it :) How did you figured out that 2 = y in your example? > y: 2 > -- greets one must still have chaos in oneself to be able to give birth to a dancing star

on 2007-06-15 16:58

On Jun 15, 2007, at 7:55 AM, anansi wrote: > I really don't get it :) How did you figured out that 2 = y in your > example? Well, a non-clever way is to try an exhaustive search of each digit for each letter. James Edward G. II

on 2007-06-15 17:10

ohh I totally misunderstood the task here.. I thought the input would be send+more and my app should give money as answer trough calculation.. James Edward G. II wrote: > On Jun 15, 2007, at 7:55 AM, anansi wrote: > >> I really don't get it :) How did you figured out that 2 = y in your >> example? > > Well, a non-clever way is to try an exhaustive search of each digit for > each letter. > > James Edward G. II > -- greets one must still have chaos in oneself to be able to give birth to a dancing star

on 2007-06-15 17:17

anansi wrote, On 6/15/2007 8:10 AM: > ohh I totally misunderstood the task here.. I thought the input would > be send+more and my app should give money as answer trough calculation.. > For perhaps a reason to change the wording, that's what I thought as well until I read the explanation to anansi's question. Sam

on 2007-06-15 17:26

On Jun 15, 2007, at 8:17 AM, Sammy L. wrote: > anansi wrote, On 6/15/2007 8:10 AM: >> ohh I totally misunderstood the task here.. I thought the input >> would be send+more and my app should give money as answer trough >> calculation.. >> > > For perhaps a reason to change the wording, that's what I thought > as well until I read the explanation to anansi's question. Sorry for the confusion folks. You get both sides of the equation and are expected to fine the digit to represent each letter. James Edward G. II

on 2007-06-15 17:27

Hmm maybe this helps to clarify (for emacs users) the problem is like mpuzz :) (mpuz?) Robert

on 2007-06-15 19:20

I hate these puzzles, but to clarify what you want to do: send +more ---------- money using the key becomes: s: 9 e: 5 n: 6 d: 7 m: 1 o: 0 r: 8 y: 2 9567 + 1085 ---------- 10652 So your task is to find the key. Jason

on 2007-06-15 21:36

```
Jason,
> So your task is to find the key.
Just think of these as a cryptogram with numbers instead of letters.
- Warren B.
```

on 2007-06-17 03:43

Mi implementation is "working". For instance: ./quiz_128.rb "a+b=c" a: 5 b: 1 c: 6 But ./quiz_128.rb "send+more=money" is taking ages! time ./quiz_128.rb "send+more=money" m: 1 y: 2 n: 6 o: 0 d: 7 e: 5 r: 8 s: 9 real 7m17.065s user 2m46.806s sys 0m6.676s Is there some trick to cut the solution space (I'm just scanning the solution space) ? I know I have a crappy PC but I have a feeling that this is not the real problem! When can I send my solution to be reviewed? > > + more > > $ ruby verbal_arithmetic.rb 'send+more=money' > s: 9 > e: 5 > n: 6 > d: 7 > m: 1 > o: 0 > r: 8 > y: 2 > > -- "Es tambiÃ©n nuestra intenciÃ³n erradicar la corrupciÃ³n, ofreciendo como norma la honestidad, la idoneidad y la eficiencia. Con madurez y sentido de unidad es fÃ¡cil pensar en la recomposiciÃ³n del ser argentino. Ese ser argentino, basado en madurez y en sentido de unidad, permitirÃ¡ inspirar para elevarnos por encima de la miseria que la antinomia nos ha planteado, para dejar, de una vez por todas, ese ser "anti" y ser, de una vez por todas, "pro": "Pro argentinos"" Jorge Rafael Videla para el 25 de mayo de 1976

on 2007-06-17 04:10

quoth the Aureliano C.: > When can I send my solution to be > reviewed? I think about 12 hours from now... -d

on 2007-06-17 18:04

Hi, Here's my solution: http://pastie.caboo.se/71188 It's of the dumb-brute-force-slow-as-hell variety: $ time ./rq128_verbalarithmetic_rafc.rb 'send + more = money' {"m"=>1, "y"=>2, "n"=>6, "o"=>0, "d"=>7, "e"=>5, "r"=>8, "s"=>9} real 2m51.197s user 2m39.722s sys 0m11.397s $ time ./rq128_verbalarithmetic_rafc.rb 'forty + ten + ten = sixty' {"x"=>4, "n"=>0, "y"=>6, "o"=>9, "e"=>5, "f"=>2, "r"=>7, "s"=>3, "i"=>1, "t"=>8} real 7m28.151s user 6m55.114s sys 0m32.938s But on the other hand, it can handle any operator as well as bases other than 10. Did you know for instance that in base 9, ruby is worth up to 4 perls with some extra fun thrown in? $ ./rq128_verbalarithmetic_rafc.rb 'n * perl + fun = ruby' 9 {"l"=>8, "b"=>7, "y"=>0, "n"=>1, "e"=>5, "p"=>3, "f"=>6, "r"=>4, "u"=>2} {"l"=>8, "b"=>7, "y"=>0, "n"=>1, "e"=>6, "p"=>3, "f"=>5, "r"=>4, "u"=>2} {"l"=>1, "b"=>7, "y"=>4, "n"=>2, "e"=>6, "p"=>3, "f"=>5, "r"=>8, "u"=>0} {"l"=>2, "b"=>5, "y"=>0, "n"=>3, "e"=>7, "p"=>1, "f"=>8, "r"=>6, "u"=>4} {"l"=>4, "b"=>5, "y"=>6, "n"=>3, "e"=>0, "p"=>2, "f"=>8, "r"=>7, "u"=>1} {"l"=>4, "b"=>0, "y"=>6, "n"=>3, "e"=>1, "p"=>2, "f"=>8, "r"=>7, "u"=>5} {"l"=>4, "b"=>7, "y"=>6, "n"=>3, "e"=>5, "p"=>2, "f"=>1, "r"=>8, "u"=>0} {"l"=>2, "b"=>6, "y"=>3, "n"=>4, "e"=>7, "p"=>1, "f"=>5, "r"=>8, "u"=>0} Regards, Raf

on 2007-06-17 19:12

Well, I think I can send it now. My implementation is on http://pastie.caboo.se/71198. On 6/16/07, Aureliano C. <removed_email_address@domain.invalid> wrote: > time ./quiz_128.rb "send+more=money" > user 2m46.806s > > 1. Please do not post any solutions or spoiler discussion for this quiz until > > if you can. > > > > number should be nonzero and each letter represents a different digit. > > n: 6 > "Es tambiÃ©n nuestra intenciÃ³n erradicar la corrupciÃ³n, ofreciendo como > norma la honestidad, la idoneidad y la eficiencia. Con madurez y > sentido de unidad es fÃ¡cil pensar en la recomposiciÃ³n del ser > argentino. Ese ser argentino, basado en madurez y en sentido de > unidad, permitirÃ¡ inspirar para elevarnos por encima de la miseria que > la antinomia nos ha planteado, para dejar, de una vez por todas, ese > ser "anti" y ser, de una vez por todas, "pro": "Pro argentinos"" > > Jorge Rafael Videla para el 25 de mayo de 1976 > -- "Es tambiÃ©n nuestra intenciÃ³n erradicar la corrupciÃ³n, ofreciendo como norma la honestidad, la idoneidad y la eficiencia. Con madurez y sentido de unidad es fÃ¡cil pensar en la recomposiciÃ³n del ser argentino. Ese ser argentino, basado en madurez y en sentido de unidad, permitirÃ¡ inspirar para elevarnos por encima de la miseria que la antinomia nos ha planteado, para dejar, de una vez por todas, ese ser "anti" y ser, de una vez por todas, "pro": "Pro argentinos"" Jorge Rafael Videla para el 25 de mayo de 1976

on 2007-06-17 21:19

Hello, here is my solution for the arithmetic quiz. I am a ruby beginner and this is my first ruby program (ok, my second one, the first was "puts 'hello world'"). So please forgive me the dirty trick using eval (why build an own parser if ruby provides such a fine solution?). I build in a brute force search algorithm via an iterative method for permutations. Therefore, any valid ruby expression is allowed, e.g. verbal_arithmetic.rb 'a+b==c && a+c==d-b && a*c==d' solving a*b*c*d!=0 && a+b==c && a+c==d-b && a*c==d -- Solution --- a: 2 b: 1 c: 3 d: 6 Please feel free to send any comments. Regards Holger #!/usr/bin/ruby -w # # Solution to ruby quiz #128 # http://www.rubyquiz.com/quiz128.html # by Holger # # Usage: # verbal_arithmetic.rb <equation> # # Examples: # verbal_arithmetic.rb 'send+more=money' # verbal_arithmetic.rb 'a+b==c && a+c==d-b && a*c==d' # #********************************************************************* # Permutator which gives all combinations of <m> elements out of # array <n> # # usage: # perms(m, n) { |x| ... } # #********************************************************************* def perms(m, n) p = [nil] * m t = [-1] * m k = 0 while k >= 0 if k==m yield p k = k-1 end n[t[k]] = p[k] if t[k]>=0 while(t[k]<n.length()) t[k] = t[k]+1 if n[t[k]] p[k] = n[t[k]] n[t[k]] = nil k = k+1 t[k] = -1 break end end k = k-1 if t[k]==n.length() end end # Read from command line and make valid ruby expression (= -> == if not already present) puzzle = ARGV[0].gsub(/=+/,"==") # Extract all letters and all first letters digits = puzzle.gsub(/\W/,"").split(//).uniq starts = puzzle.gsub(/(\w)\w*|\W/,"\\1").split(//).uniq if digits.length()>= 10 puts "oops, too much letters" else # String containing all digits digitss = digits.join # Build "first digit must not be zero" condition cond0 = starts.join("*") + "!=0" # And now perform an exhaustive search puts "solving #{cond0} && #{puzzle}" perms(digits.length(), (0...10).to_a) { |v| p0 = cond0.tr(digitss, v.join) p1 = puzzle.tr(digitss, v.join) if eval(p0) && eval(p1) # Hint: first evaluate p0 as p1 may not be a valid expression puts '-- Solution ---' [digits, v].transpose.each do |x,y| puts "#{x}: #{y}" end end } end

on 2007-06-17 21:26

My solution works with addition and multiplication of any sized list of words, and subtraction of a list of exactly two words. Its also a fair bit faster than brute force. Unfortunately it got pretty messy, and after some rough debugging I'm not in the mood to clean it up just now. I came up with the basic idea by noticing that the equations can be built up and checked from simpler equations taken from the right-hand side. Er.. lemme give an example to explain better: 9567 + 1085 ------ 10652 Start off by looking for ways to satisfy: 7 + 5 --- 2 Then move further to the left: 67 + 85 ---- 152 The 52 is right, but not the 1. For addition and multiplication, we can just take it mod 100, and if that works then its a possible partial solution. For subtraction of two numbers, though, it doesn't work when it goes negative. The trick in that case is to just mod the left-most digit: 9567 - 1085 ------ 8482 7 - 5 --- 2 OK 67 - 85 ---- -22 => 82 (-2 mod 10 = 8) verbal_arithmetic.rb contains (old, ugly) code for generating permutations and combinations. So the program works from right-to-left, finding partial solutions that work by going through ways of mapping letters to numbers, and for each one trying to move further left. Basically a depth-first search. There are a number of other subteties, but like I said, I'm tired of messing with this now, so I'll leave it there. Ask if you must. Examples: $ time ./verbal_arithmetic.rb 'send more' + money Found mapping: m: 1 y: 2 n: 6 o: 0 d: 7 e: 5 r: 8 s: 9 real 0m1.074s user 0m0.993s sys 0m0.019s $ ./verbal_arithmetic.rb 'forty ten ten' + sixty Found mapping: x: 4 y: 6 n: 0 o: 9 e: 5 f: 2 r: 7 s: 3 t: 8 i: 1 $ ./verbal_arithmetic.rb 'foo bar' - 'zag' Found mapping: a: 0 b: 3 z: 4 o: 1 f: 7 r: 9 g: 2 $ ./verbal_arithmetic.rb 'fo ba' \* 'wag' Found mapping: w: 4 a: 2 b: 1 o: 5 f: 3 g: 0

on 2007-06-17 21:27

[Note: this didn't seem to get through the first time. Apologies if it did.] My solution works with addition and multiplication of any sized list of words, and subtraction of a list of exactly two words. Its also a fair bit faster than brute force. Unfortunately it got pretty messy, and after some rough debugging I'm not in the mood to clean it up just now. I came up with the basic idea by noticing that the equations can be built up and checked from simpler equations taken from the right-hand side. Er.. lemme give an example to explain better: 9567 + 1085 ------ 10652 Start off by looking for ways to satisfy: 7 + 5 --- 2 Then move further to the left: 67 + 85 ---- 152 The 52 is right, but not the 1. For addition and multiplication, we can just take it mod 100, and if that works then its a possible partial solution. For subtraction of two numbers, though, it doesn't work when it goes negative. The trick in that case is to just mod the left-most digit: 9567 - 1085 ------ 8482 7 - 5 --- 2 OK 67 - 85 ---- -22 => 82 (-2 mod 10 = 8) verbal_arithmetic.rb contains (old, ugly) code for generating permutations and combinations. So the program works from right-to-left, finding partial solutions that work by going through ways of mapping letters to numbers, and for each one trying to move further left. Basically a depth-first search. There are a number of other subteties, but like I said, I'm tired of messing with this now, so I'll leave it there. Ask if you must. Examples: $ time ./verbal_arithmetic.rb 'send more' + money Found mapping: m: 1 y: 2 n: 6 o: 0 d: 7 e: 5 r: 8 s: 9 real 0m1.074s user 0m0.993s sys 0m0.019s $ ./verbal_arithmetic.rb 'forty ten ten' + sixty Found mapping: x: 4 y: 6 n: 0 o: 9 e: 5 f: 2 r: 7 s: 3 t: 8 i: 1 $ ./verbal_arithmetic.rb 'foo bar' - 'zag' Found mapping: a: 0 b: 3 z: 4 o: 1 f: 7 r: 9 g: 2 $ ./verbal_arithmetic.rb 'fo ba' \* 'wag' Found mapping: w: 4 a: 2 b: 1 o: 5 f: 3 g: 0

on 2007-06-17 23:32

This program solves addition problems with any number of terms. It finds and displays all solutions to the problem. The solving process is broken up into a sequence of simple steps all derived from class Step. A Step can be something such as 1) choosing an available digit for a given letter or 2) summing up a column and seeing if the result matches an already-assigned letter. As steps succeed the process continues with the following steps. But if a step fails (i.e., there's a contradiction) then the system backs up to a point where another choice can be made. This is handled by recursing through the sequence of steps. In fact, even when a solution is found, the program still backtracks to find other solutions. The expectation is that by testing for contradictions as early as possible in the process we'll tend to avoid dead ends and the result will be much better than an exhaustive search. For example, here are the steps for a sample equation: send +more ----- money 1. Choose a digit for "d". 2. Choose a digit for "e". 3. Sum the column using letters "d", "e" (and include carry). 4. Set the digit for "y" based on last column summed. 5. Choose a digit for "n". 6. Choose a digit for "r". 7. Sum the column using letters "n", "r" (and include carry). 8. Verify that last column summed matches current digit for "e". 9. Choose a digit for "o". 10. Sum the column using letters "e", "o" (and include carry). 11. Verify that last column summed matches current digit for "n". 12. Choose a digit for "s". 13. Verify that "s" has not been assigned to zero. 14. Choose a digit for "m". 15. Verify that "m" has not been assigned to zero. 16. Sum the column using letters "s", "m" (and include carry). 17. Verify that last column summed matches current digit for "o". 18. Sum the column using letters (and include carry). 19. Verify that last column summed matches current digit for "m". 20. Display a solution (provided carry is zero)! Eric ---- Are you interested in on-site Ruby training that's been highly reviewed by former students? http://LearnRuby.com ==== # This is a solution to Ruby Q. #128. As input it takes a "word # equation" such as "send+more=money" and determines all possible # mappings of letters to digits that yield a correct result. # # The constraints are: 1) a given digit can only be mapped to a single # letter, 2) the first digit in any term cannot be zero. # # The solving process is broken up into a sequence of simple steps all # derived from class Step. A Step can be something such as 1) # choosing an available digit for a given letter or 2) summing up a # column and seeing if the result matches an already-assigned letter. # As steps succeed the process continues with the following steps. # But if a step fails (i.e., there's a contradiction) then the system # backs up to a point where another choice can be made. This is # handled by recursing through the sequence of steps. In fact, even # when a solution is found, the program still backtracks to find other # solutions. require 'set' # State represents the stage of a partially solved word equation. It # keeps track of what digits letters map to, which digits have not yet # been assigned to letters, and the results of the last summed column, # including the resulting digit and any carry if there is one. class State attr_accessor :sum, :carry attr_reader :letters def initialize() @available_digits = Set.new(0..9) @letters = Hash.new @sum, @carry = 0, 0 end # Return digit for letter. def [](letter) @letters[letter] end # The the digit for a letter. def []=(letter, digit) # if the letter is currently assigned, return its digit to the # available set @available_digits.add @letters[letter] if @letters[letter] @letters[letter] = digit @available_digits.delete digit end # Clear the digit for a letter. def clear(letter) @available_digits.add @letters[letter] @letters[letter] = nil end # Return the available digits as an array copied from the set. def available_digits @available_digits.to_a end # Tests whether a given digit is still available. def available?(digit) @available_digits.member? digit end # Receives the total for a column and keeps track of it as the # summed-to digit and any carry. def column_total=(total) @sum = total % 10 @carry = total / 10 end end # Step is an "abstract" base level class from which all the "concrete" # steps can be deriveds. It simply handles the storage of the next # step in the sequence. Subclasses should provide 1) a to_s method to # describe the step being performed and 2) a perform method to # actually perform the step. class Step attr_writer :next_step end # This step tries assigning each available digit to a given letter and # continuing from there. class ChooseStep < Step def initialize(letter) @letter = letter end def to_s "Choose a digit for \"#{@letter}\"." end def perform(state) state.available_digits.each do |v| state[@letter] = v @next_step.perform(state) end state.clear(@letter) end end # This step sums up the given letters and changes to state to reflect # the sum. Because we may have to backtrack, it stores the previous # saved sum and carry for later restoration. class SumColumnStep < Step def initialize(letters) @letters = letters end def to_s list = @letters.map { |l| "\"#{l}\"" }.join(', ') "Sum the column using letters #{list} (and include carry)." end def perform(state) # save sum and carry saved_sum, saved_carry = state.sum, state.carry state.column_total = state.carry + @letters.inject(0) { |sum, letter| sum + state[letter] } @next_step.perform(state) # restore sum and carry state.sum, state.carry = saved_sum, saved_carry end end # This step determines the digit for a letter given the last column # summed. If the digit is not available, then we cannot continue. class AssignOnSumStep < Step def initialize(letter) @letter = letter end def to_s "Set the digit for \"#{@letter}\" based on last column summed." end def perform(state) if state.available? state.sum state[@letter] = state.sum @next_step.perform(state) state.clear(@letter) end end end # This step will occur after a column is summed, and the result must # match a letter that's already been assigned. class CheckOnSumStep < Step def initialize(letter) @letter = letter end def to_s "Verify that last column summed matches current " + "digit for \"#{@letter}\"." end def perform(state) @next_step.perform(state) if state[@letter] == state.sum end end # This step will occur after a letter is assigned to a digit if the # letter is not allowed to be a zero, because one or more terms begins # with that letter. class CheckNotZeroStep < Step def initialize(letter) @letter = letter end def to_s "Verify that \"#{@letter}\" has not been assigned to zero." end def perform(state) @next_step.perform(state) unless state[@letter] == 0 end end # This step represents finishing the equation. The carry must be zero # for the perform to have found an actual result, so check that and # display a digit -> letter conversion table and dispaly the equation # with the digits substituted in for the letters. class FinishStep < Step def initialize(equation) @equation = equation end def to_s "Display a solution (provided carry is zero)!" end def perform(state) # we're supposedly done, so there can't be anything left in carry return unless state.carry == 0 # display a letter to digit table on a single line table = state.letters.invert puts puts table.keys.sort.map { |k| "#{table[k]}=#{k}" }.join(' ') # display the equation with digits substituted for the letters equation = @equation.dup state.letters.each { |k, v| equation.gsub!(k, v.to_s) } puts puts equation end end # Do a basic test for the command-line arguments validity. unless ARGV[0] =~ Regexp.new('^[a-z]+(\+[a-z]+)*=[a-z]+$') STDERR.puts "invalid argument" exit 1 end # Split the command-line argument into terms and figure out how many # columns we're dealing with. terms = ARGV[0].split(/\+|=/) column_count = terms.map { |e| e.size }.max # Build the display of the equation a line at a time. The line # containing the final term of the sum has to have room for the plus # sign. display_columns = [column_count, terms[-2].size + 1].max display = [] terms[0..-3].each do |term| display << term.rjust(display_columns) end display << "+" + terms[-2].rjust(display_columns - 1) display << "-" * display_columns display << terms[-1].rjust(display_columns) display = display.join("\n") puts display # AssignOnSumStep which letters cannot be zero since they're the first # letter of a term. nonzero_letters = Set.new terms.each { |e| nonzero_letters.add(e[0, 1]) } # A place to keep track of which letters have so-far been assigned. chosen_letters = Set.new # Build up the steps needed to solve the equation. steps = [] column_count.times do |column| index = -column - 1 letters = [] # letters for this column to be added terms[0..-2].each do |term| # for each term that's being added... letter = term[index, 1] next if letter.nil? # skip term if no letter in column letters << letter # note that this letter is part of sum # if the letter does not have a digit, create a ChooseStep unless chosen_letters.member? letter steps << ChooseStep.new(letter) chosen_letters.add(letter) steps << CheckNotZeroStep.new(letter) if nonzero_letters.member? letter end end # create a SumColumnStep for the column steps << SumColumnStep.new(letters) summed_letter = terms[-1][index, 1] # the letter being summed to # check whether the summed to letter should already have a digit if chosen_letters.member? summed_letter # should already have a digit, check that summed digit matches it steps << CheckOnSumStep.new(summed_letter) else # doesn't already have digit, so create a AssignOnSumStep for # letter steps << AssignOnSumStep.new(summed_letter) chosen_letters.add(summed_letter) # check whether this letter cannot be zero and if so add a # CheckNotZeroStep steps << CheckNotZeroStep.new(summed_letter) if nonzero_letters.member? summed_letter end end # should be done, so add a FinishStep steps << FinishStep.new(display) # print out all the steps # steps.each_with_index { |step, i| puts "#{i + 1}. #{step}" } # let each step know about the one that follows it. steps.each_with_index { |step, i| step.next_step = steps[i + 1] } # start performing with the first step. steps.first.perform(State.new)

on 2007-06-18 07:29

Hello everyone, At the outset, I had no idea how difficult this problem would be. In fact, it turns out that this is actually an NP-complete problem if the number base is not fixed at 10. Anyway... Initially I experimented with a backtracking solution, but ran into trouble getting it to work properly. So for now I am submitting this solution. Although less than perfect, it does work for the test inputs, as well as additional solutions I have tried. The idea is to brute force a solution by trying all possible combinations of solutions until one works. For this solution, I used the Permutation Gem available here: http://permutation.rubyforge.org/doc/index.html require 'Permutation' To avoid having to (re)write the combination code myself, I also used the Combinations class from: http://butunclebob.com/ArticleS.UncleBob.RubyCombinations This is great code, someone really should create a gem for it! (see link for the code) I packaged all of my code inside the following class: class VerbalArithmetic # Parse given equation into lvalues (words on the left-hand side of the '=' that # are to be added together) and an rvalue (the single word on the right-hand side) def parse_equation (equation) lvalues = equation.split("+") rvalue = lvalues[-1].split("=") lvalues[-1] = rvalue[0] # Get last lvalue rvalue = rvalue[1] # Get rvalue return lvalues, rvalue end # Brute force a solution by trying all possible combinations def find_solution(lvalues, rvalue) # Form a list of all letters words = Marshal::load(Marshal::dump(lvalues)) words.push(rvalue) letters = {} words.each do |word| word.split("").each do |letter| letters[letter] = letter if letters[letter] == nil end end # Format l/r values to ease solution analysis below lvalues_formatted = [] lvalues.each {|lval| lvalues_formatted.push(lval.reverse.split(""))} rvalue_formatted = rvalue.reverse.split("") # For all unordered combinations of numbers... for i in Combinations.get(10, letters.values.size) # For all permutations of each combination... perm = Permutation.for(i) perm.each do |p| # Map each combination of numbers to the underlying letters map = {} parry = p.project for i in 0...letters.size map[letters.values[i]] = parry[i] end # Does this mapping yield a solution? if is_solution?(lvalues_formatted, rvalue_formatted, map) return map end end end nil end # Determines if the given equation may be solved by # substituting the given number for its letters def is_solution?(lvalues, rvalue, map) # Make sure there are no leading zero's for lval in lvalues return false if map[lval[-1]] == 0 end return false if map[rvalue[-1]] == 0 # Perform arithmetic using the mappings, and make sure they are valid remainder = 0 for i in 0...rvalue.size lvalues.each do |lval| remainder = remainder + map[lval[i]] if map[lval[i]] != nil # Sum values end return false if (remainder % 10) != map[rvalue[i]] # Validate digit remainder = remainder / 10 # Truncate value at this place end true end end Finally, this code puts everything together: va = VerbalArithmetic.new lvalues, rvalue = va.parse_equation("send+more=money") map = va.find_solution(lvalues, rvalue) puts "Solution: ", map if map != nil And here is the output: Solution: m1n6y2d7o0e5r8s9 Thanks, Justin

on 2007-06-18 15:33

#! /usr/bin/env ruby # # quiz-128 -- Ruby Q. #128 -- Verbal Arithmetic. # # Usage: quiz-128 [ -v ] '<equation string>' # or: quiz-128 [ -v ] <addend> <addend> [ <addend> ... ] <sum> # # See the Ruby Q. #128 documentation for more information # (http://www.rubyquiz.com/quiz128.html). # # Glen Pankow 06/17/07 Original version. # # Licensed under the Ruby License. # #----------------------------------------------------------------------------- # # I take a slightly different approach to this quiz: I build up code that # kinda-sorta models addition as taught in U.S. elementary schools (taking # into account the various constraints of the problem), then eval it. # # And instead of generating permutations of unassigned digits and using # recursion for backtracking, I have a single array of digits that is used # sort of as a mask (nil entries mark assigned digits) and use Ruby's # wonderful magic iterator facility to scan through it. # # $ uname -srmpio # Linux 2.6.9-55.ELsmp i686 i686 i386 GNU/Linux # # $ time ./quiz-128 'send + more = money' # d = 7, e = 5, y = 2, n = 6, r = 8, o = 0, s = 9, m = 1 # 1 0 1 1 # s:9 e:5 n:6 d:7 # + m:1 o:0 r:8 e:5 # --------------------- # m:1 o:0 n:6 e:5 y:2 # 0.065u 0.003s 0:00.07 85.7% 0+0k 0+0io 0pf+0w # # $ time ./quiz-128 forty ten ten sixty # y = 6, n = 0, t = 8, e = 5, r = 7, x = 4, o = 9, i = 1, f = 2, s = 3 # 1 2 1 0 # f:2 o:9 r:7 t:8 y:6 # t:8 e:5 n:0 # + t:8 e:5 n:0 # --------------------- # s:3 i:1 x:4 t:8 y:6 # 0.029u 0.002s 0:00.03 66.6% 0+0k 0+0io 0pf+0w # # $ time ./quiz-128 eat+that=apple # t = 9, e = 8, a = 1, l = 3, h = 2, p = 0 # 1 1 0 1 # e:8 a:1 t:9 # + t:9 h:2 a:1 t:9 # --------------------- # a:1 p:0 p:0 l:3 e:8 # 0.008u 0.002s 0:00.01 0.0% 0+0k 0+0io 0pf+0w # # $ time ./quiz-128 ruby rubber baby buggy bumper # y = 0, r = 7, b = 8, e = 1, g = 4, u = 2, a = 3, p = 9, m = 6 # ... # y = 0, r = 7, b = 8, e = 5, g = 4, u = 2, a = 3, p = 9, m = 6 # 1 2 1 2 0 # r:7 u:2 b:8 y:0 # r:7 u:2 b:8 b:8 e:5 r:7 # b:8 a:3 b:8 y:0 # + b:8 u:2 g:4 g:4 y:0 # ------------------------- # b:8 u:2 m:6 p:9 e:5 r:7 # 0.120u 0.002s 0:00.13 92.3% 0+0k 0+0io 0pf+0w # class Array # # For each non-nil element of the current array, (destructively) set it to # nil (i.e., 'marking' it), yield the original value (to the assumed block), # and restore it back to what it originally was (i.e., 'unmarking' it). # # For example, the code: # array = [ 0, nil, 2, 3 ] # p "before: array = #{array.inspect}" # array.unmarkeds { |elem| p "elem = #{elem}, array = #{array.inspect}" } # p "after: array = #{array.inspect}" # would print: # before: array = [0, nil, 2, 3] # elem = 0, array = [nil, nil, 2, 3] # elem = 2, array = [0, nil, nil, 3] # elem = 3, array = [0, nil, 2, nil] # after: array = [0, nil, 2, 3] # def unmarkeds (0...size).each do |i| next if (at(i).nil?) elem = at(i) ; self[i] = nil yield elem self[i] = elem end end # # If the <i>-th element of the current array is nil, do nothing. Otherwise # (destructively) set it to nil (i.e., 'marking' it), yield (to the assumed # block), and restore it back to what it originally was (i.e., 'unmarking' # it). This is basically the guts of unmarkeds(), and is provided for safe # manual element marking. # def if_unmarked(i) return if (at(i).nil?) elem = at(i) ; self[i] = nil yield self[i] = elem end # # Note: typically one might say 'yield elem, self' in these methods, but # I don't need them for this application due to Ruby's scoping mechanism. # end # # Process the command-line arguments of addend strings and the sum string. # verbose = false addend_strs = [ ] ARGV.each do |arg| if (arg == '-v') verbose = true else arg.split(/[\s+=]+/).each { |term| addend_strs << term } end end addend_strs = [ 'send', 'more', 'money' ] if (addend_strs.empty?) sum_str = addend_strs.pop # # Split the strings up into their component letters; create some (ugly) code to # eventually print out a nice table of the addition. # table_print_code = 'print " ' (sum_str.length - 1).downto(1) do |i| table_print_code << " \#{carry#{i}}" end table_print_code << "\\n\"\n" addends = [ ] first_letters = { } (0...addend_strs.size).each do |i| addend_str = addend_strs[i] addend_chars = addend_str.split(//).reverse addends << addend_chars first_letters[addend_chars[-1]] = 1 table_print_code \ << 'print "' << ((i < addend_strs.size - 1)? ' ' : '+') \ << ' ' * (sum_str.length - addend_str.length) \ << addend_str.gsub(/([a-z])/, ' \1:#{\1}') << "\\n\"\n" end table_print_code \ << 'print "-' << ('----' * sum_str.length) << "\\n\"\n" \ << 'print " ' << sum_str.gsub(/([a-z])/, ' \1:#{\1}') << "\\n\"\n" sum_chars = sum_str.split(//).reverse first_letters[sum_chars[-1]] = 1 # # Build the addition code. # # This, too, is quite ugly and I don't bother to document it, as printing out # the generated code will probably give one a better idea of how it works than # my usual verbose documentation (i.e., run this script with -v). # seen_chars = { } code_head = "rem_digs = (0..9).to_a\n" code_tail = '' answer_print_code = 'print "' indent = '' (0...sum_chars.size).each do |col| sum_char = sum_chars[col] col_sum_code = "#{indent}carry#{col+1}, #{sum_char} = (carry#{col}" addends.inject([ ]) { |dc, addend| dc << addend[col] }.each do |dig_char| next if (dig_char.nil?) if (seen_chars[dig_char].nil?) code_head << "#{indent}rem_digs.unmarkeds do |#{dig_char}|\n" code_tail[0,0] = "#{indent}end\n" indent << ' ' seen_chars[dig_char] = 1 code_head << "#{indent}next if (#{dig_char} == 0) # leading 0?\n" \ if (first_letters.has_key?(dig_char)) col_sum_code[0,0] = ' ' # fix indentation answer_print_code << ", #{dig_char} = \#{#{dig_char}}" end col_sum_code << " + #{dig_char}" end col_sum_code << ").divmod(10)\n" col_sum_code.sub!(/\(carry0 \+ /, '(') if (seen_chars[sum_char].nil?) code_head << col_sum_code code_head << "#{indent}next if (#{sum_char} == 0) # leading 0?\n" \ if (first_letters.has_key?(sum_char)) code_head << "#{indent}rem_digs.if_unmarked(#{sum_char}) do\n" code_tail[0,0] = "#{indent}end\n" indent << ' ' seen_chars[sum_char] = 1 answer_print_code << ", #{sum_char} = \#{#{sum_char}}" else col_sum_code.sub!(/ = /, '2 = ') code_head \ << col_sum_code \ << "#{indent}next unless (#{sum_char}2 == #{sum_char}) # inconsistent?\n" end end answer_print_code.sub!(/\", /, '"\n') answer_print_code << "\\n\"\n" # # And print out the code (if verbose) and run it! # code = code_head + answer_print_code + table_print_code + code_tail print code, "\n" if (verbose) eval(code)

on 2007-06-18 18:56

Here is my solution for Ruby Q. 128. It's a little rough, but I can't afford to spend any more time working on it. One thing I didn't have time to do was to provide a user interface. Another thing I didn't do was proper commenting. I apologize for the latter. I thought a Darwinian search (aka genetic algorithm) would be an interesting way to tackle this quiz. I have been looking for a excuse to write such a search in Ruby for quite awhile and this seemed to be it. Here are is the output from one run of my quiz solution: <result> Solution found after 15 steps SEND+MORE=MONEY 9567+1085=10652 Solution found after 27 steps FORTY+TEN+TEN=SIXTY 29786+850+850=31486 </result> And here is the code: <code> #! /usr/bin/env ruby -w # # solution.rb # Quiz 128 # # Created by Morton G. on 2007-06-18. # Assumption: equations take the form: term + term + ... term = sum ROOT_DIR = File.dirname(__FILE__) $LOAD_PATH << File.join(ROOT_DIR, "lib") require "cryptarithm" require "solver" EQUATION_1 = "SEND+MORE=MONEY" EQUATION_2 = "FORTY+TEN+TEN=SIXTY" POP_SIZE = 400 FECUNDITY = 2 STEPS = 50 Cryptarithm.equation(EQUATION_1) s = Solver.new(POP_SIZE, FECUNDITY, STEPS) s.run puts s.show Cryptarithm.equation(EQUATION_2) s = Solver.new(POP_SIZE, FECUNDITY, STEPS) s.run puts s.show </code> And here are the library classes: <code> # lib/cryptarithm.rb # Quiz 128 # # Created by Morton G. on 2007-06-18. DIGITS = (0..9).to_a class Cryptarithm @@equation = "" @@max_rank = -1 def self.equation(str=nil) if str @@equation = str.upcase lhs, rhs = @@equation.gsub(/[A-Z]/, "9").split("=") @@max_rank = [eval(lhs), eval(rhs)].max else @@equation end end attr_accessor :ranking, :solution def initialize @solution = @@equation.delete("+-=").split("").uniq @solution = @solution.zip((DIGITS.sort_by {rand})[0, @solution.size]) rank end def mutate(where=rand(@solution.size)) raise RangeError unless (removed_email_address@domain.invalid).include?(where) digits = @solution.collect { |pair| pair[1] } digits = DIGITS - digits return if digits.empty? @solution[where][1] = digits[rand(digits.size)] end def swap m = rand(@solution.size) n = m while n == m n = rand(@solution.size) end @solution[m][1], @solution[n][1] = @solution[n][1], @solution [m][1] end def rank sum = @@equation.dup solution.each { |chr, num| sum.gsub!(chr, num.to_s) } lhs, rhs = sum.split("=") terms = lhs.split("+") << rhs if terms.any? { |t| t[0] == ?0 } @ranking = @@max_rank else @ranking = eval("#{lhs} - #{rhs}").abs end end def initialize_copy(original) @solution = original.solution.collect { |pair| pair.dup } end def inspect [@ranking, @solution].inspect end def to_s sum = @@equation.dup solution.each { |chr, num| sum.gsub!(chr, num.to_s) } "#{@@equation}\n#{sum}" end end </code> <code> # lib/solver.rb # Quiz 128 # # Created by Morton G. on 2007-06-18. # # Attempts to a solve cryptarithm puzzle by applying a Darwinian search # (aka genetic algorithm). It can thought of as a stochastic breadth- first # search. Although this method doesn't guarantee a solution will be # found, it often finds one quite quickly. MUTATION = 0.5 SWAP = 1.0 class Solver attr_reader :best, :population, :step def initialize(pop_size, fecundity, steps) @pop_size = pop_size @fecundity = fecundity @steps = steps @mid_step = steps / 2 @step = 1 @population = [] @pop_size.times { @population << Cryptarithm.new } select end def run @steps.times do replicate select break if @best.rank.zero? @step += 1 end @best end def replicate @pop_size.times do |n| crypt = @population[n] # mate = crypt # while mate.equal?(crypt) # mate = @population[rand(@pop_size)] # end @fecundity.times do child = crypt.dup child.mutate if crypt.solution.size < 10 && rand <= MUTATION child.swap if rand <= SWAP @population << child end end end def select @population = @population.sort_by { |crypt| crypt.rank } @population = @population[0, @pop_size] @best = @population.first end def show if @step > @steps "No solution found after #{step} steps" else "Solution found after #{step} steps\n" + @best.to_s end end end </code> Regards, Morton

on 2007-06-18 20:46

If anyone would like more samples to try their code with, I found the following on the web: * http://www.gtoal.com/wordgames/alphametic/testscript * http://www.gtoal.com/wordgames/alphametic/testscript.out The first is a script which poses the problems to their own solving- program. The second shows the solutions that their program produces. Some of the sample equations have multiple solutions, such as 'eins +eins=zwei'. 'united+states=america' has no solutions. And of the few that I've tested with my own solution, 'saturn+pluto+uranus +neptune=planets' takes the longest to solve. Eric ---- Interested in hands-on, on-site Ruby training? See http://LearnRuby.com for information about a well-reviewed class.

on 2007-06-19 04:01

On Jun 17, 2007, at 9:03 AM, Raf C. wrote: > Here's my solution: http://pastie.caboo.se/71188 > > It's of the dumb-brute-force-slow-as-hell variety: Mine was too: #!/usr/bin/env ruby -wKU EQUATION = ARGV.shift.to_s.downcase.sub("=", "==") LETTERS = EQUATION.scan(/[a-z]/).uniq CHOICES = LETTERS.inject(Hash.new) do |all, letter| all.merge(letter => EQUATION =~ /\b#{letter}/ ? 1..9 : 0..9) end def search(choices, mapping = Hash.new) if choices.empty? letters, digits = mapping.to_a.flatten.partition { |e| e.is_a? String } return mapping if eval(EQUATION.tr(letters.join, digits.join)) else new_choices = choices.dup letter = new_choices.keys.first digits = new_choices.delete(letter).to_a - mapping.values digits.each do |choice| if result = search(new_choices, mapping.merge(letter => choice)) return result end end return nil end end if solution = search(CHOICES) LETTERS.each { |letter| puts "#{letter}: #{solution[letter]}" } else puts "No solution found." end __END__ James Edward G. II

on 2007-06-19 08:40

Sorry for the second post, but I just noticed that I pasted a wrong (earlier) version of solver.rb into my first post. This is what I intended to post. The difference between the two versions is minor, but this version eliminates some commented-out experimental code and corrects a minor bug. Sample output: <result> Solution found after 15 steps SEND+MORE=MONEY 9567+1085=10652 Solution found after 27 steps FORTY+TEN+TEN=SIXTY 29786+850+850=31486 </result> And here is the corrected code: <code> #! /usr/bin/env ruby -w # # solution.rb # Quiz 128 # # Created by Morton G. on 2007-06-18. # Assumption: equations take the form: term + term + ... term = sum ROOT_DIR = File.dirname(__FILE__) $LOAD_PATH << File.join(ROOT_DIR, "lib") require "cryptarithm" require "solver" EQUATION_1 = "SEND+MORE=MONEY" EQUATION_2 = "FORTY+TEN+TEN=SIXTY" POP_SIZE = 400 FECUNDITY = 2 STEPS = 50 Cryptarithm.equation(EQUATION_1) s = Solver.new(POP_SIZE, FECUNDITY, STEPS) s.run puts s.show Cryptarithm.equation(EQUATION_2) s = Solver.new(POP_SIZE, FECUNDITY, STEPS) s.run puts s.show </code> And here are the library classes: <code> # lib/cryptarithm.rb # Quiz 128 # # Created by Morton G. on 2007-06-18. DIGITS = (0..9).to_a class Cryptarithm @@equation = "" @@max_rank = -1 def self.equation(str=nil) if str @@equation = str.upcase lhs, rhs = @@equation.gsub(/[A-Z]/, "9").split("=") @@max_rank = [eval(lhs), eval(rhs)].max else @@equation end end attr_accessor :ranking, :solution def initialize @solution = @@equation.delete("+-=").split("").uniq @solution = @solution.zip((DIGITS.sort_by {rand})[0, @solution.size]) rank end def mutate(where=rand(@solution.size)) raise RangeError unless (removed_email_address@domain.invalid).include?(where) digits = @solution.collect { |pair| pair[1] } digits = DIGITS - digits return if digits.empty? @solution[where][1] = digits[rand(digits.size)] end def swap m = rand(@solution.size) n = m while n == m n = rand(@solution.size) end @solution[m][1], @solution[n][1] = @solution[n][1], @solution [m][1] end def rank sum = @@equation.dup solution.each { |chr, num| sum.gsub!(chr, num.to_s) } lhs, rhs = sum.split("=") terms = lhs.split("+") << rhs if terms.any? { |t| t[0] == ?0 } @ranking = @@max_rank else @ranking = eval("#{lhs} - #{rhs}").abs end end def initialize_copy(original) @solution = original.solution.collect { |pair| pair.dup } rank end def inspect [@ranking, @solution].inspect end def to_s sum = @@equation.dup solution.each { |chr, num| sum.gsub!(chr, num.to_s) } "#{@@equation}\n#{sum}" end end </code> <code> # lib/solver.rb # Quiz 128 # # Created by Morton G. on 2007-06-18. # # Attempts to a solve cryptarithm puzzle by applying a Darwinian search # (aka genetic algorithm). It can thought of as a stochastic breadth- first # search. Although this method doesn't guarantee a solution will be # found, it often finds one quite quickly. MUTATION = 0.5 SWAP = 1.0 class Solver attr_reader :best, :population, :step def initialize(pop_size, fecundity, steps) @pop_size = pop_size @fecundity = fecundity @steps = steps @mid_step = steps / 2 @step = 1 @population = [] @pop_size.times { @population << Cryptarithm.new } select end def run @steps.times do replicate select break if @best.ranking.zero? @step += 1 end @best end def replicate @pop_size.times do |n| crypt = @population[n] @fecundity.times do child = crypt.dup child.mutate if crypt.solution.size < 10 && rand <= MUTATION child.swap if rand <= SWAP @population << child end end end def select @population = @population.sort_by { |crypt| crypt.rank } @population = @population[0, @pop_size] @best = @population.first end def show if @step > @steps "No solution found after #{step} steps" else "Solution found after #{step} steps\n" + @best.to_s end end end </code> Regards, Morton

on 2007-06-19 18:30

On Jun 18, 4:59 pm, James Edward G. II <removed_email_address@domain.invalid> wrote: > def search(choices, mapping = Hash.new) > if result = search(new_choices, mapping.merge(letter => choice)) > else > puts "No solution found." > end > > __END__ > I really like it - your solution covers any type of expression :). Perhaps some preprocessing of CHOICES can save the performance.... The only thing, assuming that all examples are always integers, I'd add .sub(/([a-z])([^a-z])/,'\1.0\2') to EQUATION to cover examples with division (to drop the assumption, before adding .0 one can regexp it for '.'s) for 'boaogr/foo=bar' (arbitrary example) your code gives b: 1 o: 0 a: 2 g: 3 r: 7 f: 8 and modified one corrects it to b: 1 o: 0 a: 2 g: 3 r: 7 f: 8 I can expect that this modification may break on rounding errors, but in this task domain I do not think it will

on 2007-06-19 21:20

On Jun 19, 2007, at 9:30 AM, Eugene K. wrote: >> end >> >> > Perhaps some preprocessing of CHOICES can save the performance.... Thanks. There have been far more clever submissions though. > > o: 0 > a: 2 > g: 3 > r: 7 > f: 8 > > I can expect that this modification may break on rounding errors, > but in > this task domain I do not think it will Good points. James Edward G. II

on 2007-06-19 22:49

Ruby Q. wrote: > This week's quiz is to build a program that reads in equations and outputs > solutions. You can decide how complex of an equation you want to support, with > the examples above being the minimum implementation. > This solution requires Gecode/R[1] 0.2.0 (need to install Gecode[3,4] followed by Gecode/R[2]). Gecode does the actual search, so we just specify the constraints. Two important aspects that are used to make the solution quick to compute are: == Propagation rather than branching (deduction rather than exploration) Branching (i.e. testing a guess, i.e. exploring the search space) costs since we have to save the old state and start a new one. Rather Gecode tries to prune as much of the search space as it can before resorting to exploration. In the case of linear constraints (e.g. a + 10*b + c= 10*d + e, which corresponds to the problem a + bc ---- de ) it takes each variable and considers the smallest and largest values it can possibly take. E.g. if we consider a in the above example we can rewrite the linear equation to a = 10*(d - b) + e - c From that equation we can check how large and small the right hand side can be and then remove all other possibilities from the domain of a. We can end up pruning quite a lot if we do this for each variable until there are no more possibilities left to remove (coupled with the distinct constraint). In fact when we use this for send+more=money, without doing any sort of exploration we can directly reduce the domains of the variables down to s=9, e=4..7, n=5..8, d=2..8, m=1, o=0, r=2..8, y=2..8 So without any guessing at all we already know the value of three letters and have pruned the domains of the others (i.e. pruned the number of possibilities left, i.e. reduced the search space). Since we now have to start exploring the search space we make a guess that e=4. Propagating the constraints once again with e given the domain 4 will directly result in a failure, so we backtrack and now know that e!=4. With that information we redo the propagation and directly end up at the solution with no need to explore any further. So in total we only need to explore 4 out of 9^2*10^6 nodes in the search space. == Branching with a good heuristic We don't randomly select where to go next in the search space, rather we use a fail-first heuristic to try to cut down the search space faster. The heuristic is to simply try to fixate a value for the variable with the smallest domain. The reason that it works well is that we exhaust domains quicker, hence forcing failures quicker (hence fail-first) == The code require 'rubygems' require 'gecoder' # Solves verbal arithmetic problems # ( http://en.wikipedia.org/wiki/Verbal_arithmetic ). Only supports # addition. class VerbalArithmetic < Gecode::Model # Creates a model for the problem where the left and right hand sides # are given as an array with one element per term. The terms are given # as strings def initialize(lhs_terms, rhs_terms) super() # Set up the variables needed as a hash mapping the letter to its # variable. lhs_terms.map!{ |term| term.split(//) } rhs_terms.map!{ |term| term.split(//) } all_terms = (lhs_terms + rhs_terms) unique_letters = all_terms.flatten.uniq letter_vars = int_var_array(unique_letters.size, 0..9) @letters = Hash[*unique_letters.zip(letter_vars).flatten!] # Must satisfy the equation. sum_terms(lhs_terms).must == sum_terms(rhs_terms) # Must be distinct. letter_vars.must_be.distinct # Must not begin with a 0. all_terms.map{ |term| term.first }.uniq.each do |letter| @letters[letter].must_not == 0 end # Select a branching, we go for fail first. branch_on letter_vars, :variable => :smallest_size, :value => :min end def to_s @letters.map{ |letter, var| "#{letter}: #{var.val}" }.join("\n") end private # A helper to make the linear equation a bit tidier. Takes an array of # variables and computes the linear combination as if the variable # were digits in a base 10 number. E.g. x,y,z becomes # 100*x + 10*y + z . def equation_row(variables) variables.inject{ |result, variable| variable + result*10 } end # Computes the sum of the specified terms (given as an array of arrays # of characters). def sum_terms(terms) rows = terms.map{ |term| equation_row(@letters.values_at(*term)) } rows.inject{ |sum, term| sum + term } end end if ARGV.empty? abort "Usage: #{$0} '<word_1>+<word_2>+...+<word_n>=<word_res>'" end lhs, rhs = ARGV[0].split('=').map{ |eq_side| eq_side.split('+') } solution = VerbalArithmetic.new(lhs, rhs).solve! if solution.nil? puts 'Failed' else puts solution.to_s end [1] http://gecoder.rubyforge.org/ [2] http://gecoder.rubyforge.org/installation.html [3] http://www.gecode.org/download.html [4] http://www.gecode.org/gecode-doc-latest/PageComp.html

on 2007-06-20 02:34

Andreas L. wrote: > Since we now have to start exploring the search space we make a guess > that e=4. Propagating the constraints once again with e given the domain > 4 will directly result in a failure, so we backtrack and now know that > e!=4. With that information we redo the propagation and directly end up > at the solution with no need to explore any further. So in total we only > need to explore 4 out of 9^2*10^6 nodes in the search space. > The last part is probably a bit misleading/incorrect. We are not directly exploring the space of all possible assignments when we branch, so saying that we have explored 4 nodes in that space is incorrect (i.e. we have not just explored 4 possible assignments, but we have visited just 4 nodes in our search space).

on 2007-06-20 17:14

On 6/15/07, Ruby Q. <removed_email_address@domain.invalid> wrote: > > This week's quiz is to build a program that reads in equations and outputs > solutions. You can decide how complex of an equation you want to support, with > the examples above being the minimum implementation. > Here's my solution: http://pastie.caboo.se/72030 It's a brute-force search, but has reasonable speed. +, -, and * operators are supported.

on 2007-06-21 09:55

Andreas L. wrote: > Since we now have to start exploring the search space we make a guess > that e=4. Propagating the constraints once again with e given the domain > 4 will directly result in a failure, so we backtrack and now know that > e!=4. With that information we redo the propagation and directly end up > at the solution with no need to explore any further. I forgot a node in the middle there (since there are 4 nodes). The complete search is: 1) Root: s=9, e=4..7, n=5..8, d=2..8, m=1, o=0, r=2..8, y=2..8 2) Tries e = 4: fails. 3) Knows e != 4: s=9, e=5..7, n=6..8, d=2..8, m=1, o=0, r=2..8, y=2..8 4) Tries e = 5: s=9, e=5, n=6, d=7, m=1, o=0, r=8, y=2 (success)