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. -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= by Jay A. For this quiz the goal is to make a constraint processing library for ruby. A Constraint Satisfaction Problem consists of variables, domains for each variable, and constraints among the variables. Here's a sample (Solutions DO NOT need to follow this syntax, this is just an example): a = IntVar.new(:a, (0..4).to_a) #Set up the variables and their domains. b = IntVar.new(:b, (0..4).to_a) c = IntVar.new(:c, (0..4).to_a) con1 = a < b #Create constraints on the problem. con2 = a + b == c prob = Problem.new(con1, con2) #Create a problem with the constraints solution = prob.solve #Find a solution p solution There are many solutions. It could return any (or all) of the following: {:a => 0, :b => 1, :c => 1} {:a => 0, :b => 2, :c => 2} {:a => 0, :b => 3, :c => 3} {:a => 0, :b => 4, :c => 4} {:a => 1, :b => 2, :c => 3} {:a => 1, :b => 3, :c => 4} Another example would be to solve the magic square: SIDE = 3 MAX = SIDE**2 SUM = (MAX*(MAX+1))/(2*SIDE) square = Array.new(SIDE) do |x| Array.new(SIDE) {|y| IntVar.new("#{x},#{y}", (1..MAX).to_a ) } end cons = [] zero = IntVar.new(:zero, [0]) SIDE.times do |row| Ã?Ã?Ã? sum = zero Ã?Ã?Ã? SIDE.times {|col| sum += square[col][row] } Ã?Ã?Ã? cons << sum == SUM end SIDE.times do |col| Ã?Ã?Ã? sum = zero Ã?Ã?Ã? SIDE.times {|row| sum += square[col][row] } Ã?Ã?Ã? cons << sum == SUM end #A constraint to ensure no two variables have the same value in a solution. cons << AllDistinct.new(*square.flatten) prob = Problem.new(*cons) solution = prob.solve p solution There are many problems that can be solved through constraint programming (even some past quizzes): gift exchanges, sudoku, magic squares, N queens, cryptoarithmetics, scheduling problems, etc... So be creative here. Pick a simple problem to solve with your Constraint Programming Engine. Good luck! For more information see: http://ktiml.mff.cuni.cz/~bartak/constraints/ and: http://en.wikipedia.org/wiki/Constraint_programming

on 2006-03-10 15:51

on 2006-03-14 00:15

Here is my own solution to this week's Ruby Q.. I built a simple constraint library and used it to solve a sudoku puzzle. First the library (constraint.rb): #!/usr/local/bin/ruby -w class Problem def initialize @vars = Hash.new { |vars, name| vars[name] = Array.new } @rules = Hash.new { |rules, var| rules[var] = Array.new } yield self if block_given? end def var( name, *choices ) if choices.empty? values = @vars[name] values.size == 1 ? values.first : values else @vars[name].push(*choices) end end def rule( name, &test ) @rules[name] << test end def solve loop do changed = false @vars.each do |name, choices| next if choices.size < 2 failures = choices.select do |choice| @rules[name].any? { |rule| !rule[choice] } end unless failures.empty? @vars[name] -= failures changed = true end end break unless changed end self end end def problem( &init ) Problem.new(&init).solve end __END__ And here is my sudoku solver (sudoku.rb), using that library: #!/usr/local/bin/ruby -w require "constraint" # sudoku conveniences indices = (0..8).to_a boxes = Hash.new [(0..2), (3..5), (6..8)].each do |across| [(0..2), (3..5), (6..8)].each do |down| squares = across.map { |x| down.map { |y| "#{x}#{y}" } }.flatten squares.each { |square| boxes[square] = squares - [square] } end end solution = problem do |prob| # read puzzle, setting problem variables from data (ARGV.empty? ? DATA : ARGF).each_with_index do |line, y| line.split.each_with_index do |square, x| prob.var("#{x}#{y}", *(square =~ /\d/ ? [square.to_i] : (1..9))) end end # apply the rules of the game indices.each do |x| indices.each do |y| col = (indices - [y]).map { |c| "#{x}#{c}" } # other cells in column row = (indices - [x]).map { |r| "#{r}#{y}" } # other cells in row box = boxes["#{x}#{y}"] # other cells in box [col, row, box].each do |set| # set rules requiring a cell to be unique prob.rule("#{x}#{y}") { |n| !set.map { |s| prob.var (s) }.include?(n) } end end end end # pretty print the results puts "+#{'-------+' * 3}" indices.each do |y| print "| " indices.each do |x| print "#{solution.var("#{x}#{y}").inspect} " print "| " if [2, 5].include? x end puts "|" puts "+#{'-------+' * 3}" if [2, 5, 8].include? y end __END__ 7 _ 1 _ _ _ 3 _ 5 _ 8 _ 1 _ 5 _ 6 _ 2 _ _ _ _ _ _ _ 9 _ _ 6 5 _ 1 2 _ _ _ 3 _ _ _ _ _ 1 _ _ _ 8 3 _ 4 9 _ _ 9 _ _ _ _ _ _ _ 8 _ 2 _ 6 _ 9 _ 4 _ 6 _ 5 _ _ _ 7 _ 1 Running that on the included puzzle produces the following output: +-------+-------+-------+ | 7 6 1 | 9 4 2 | 3 8 5 | | 3 8 9 | 1 7 5 | 4 6 2 | | 2 5 4 | 8 6 3 | 1 7 9 | +-------+-------+-------+ | 4 9 6 | 5 8 1 | 2 3 7 | | 5 3 2 | 7 9 6 | 8 1 4 | | 1 7 8 | 3 2 4 | 9 5 6 | +-------+-------+-------+ | 9 1 3 | 4 5 7 | 6 2 8 | | 8 2 7 | 6 1 9 | 5 4 3 | | 6 4 5 | 2 3 8 | 7 9 1 | +-------+-------+-------+ Thanks for the quiz, Jav. I learned a lot! James Edward G. II

on 2006-03-15 04:10

Hi, Thanks for another quiz. I read it and thought "that looks difficult and ugly", but I had a little spare time, and, to my own surprise, was able to solve it rather quickly. My naive solution just makes big nested loops over the domains of all the variables, so it's not going to solve a sudoku this century, but you wouldn't have to change the constraint code to speed it up - the engine can be fixed keeping the same interface. The interface is DSP-ish and blocky as seems to be the fashion in Ruby these days. Here's N-Queens (n=4 so it works with my naive constraint processor): http://www.dave.burt.id.au/ruby/constraints/nqueens.rb And here's the thing itself: http://www.dave.burt.id.au/ruby/constraints/constr... Cheers, Dave

on 2006-03-16 12:04

Hi Here is my CSP language. I have actually been doing this for a class, so I got an extra week to work on it. As test cases, I have been modeling typical CSP problems so right now, I can do cryptarthemtic, sudoku, mastermind, map coloring, and the zebra problem. I use forward checking with the MRV heuristic and the variable with the most constraints heuristic for tie breaking. I have also been working on a domain specific language that uses my CSP library. Some of the test cases use the language and some of them use the library. The language uses generic variables and requires user defined domain constricting functions for non-trival constraints. I would love some feedback on what I have done so far, including the syntax of the domain language and methods for the library. I plan to put the library on RubyForge at the end of the class. Here is the link to the files needed to try it out: http://reducto.case.edu/projects/team2/attachment/... Chris Parker

on 2006-03-16 15:19

I've been having a lot of fun with this quiz. Pit C. sent me a improved version of Amb that is more "Rubyish" and much easier to read (my original was a direct translation of the scheme version). I've added comments and a bit of polish to his cleanup, so here's the new and improved version of Amb along with another puzzle (just for fun). -- BEGIN AMB --------------------------------------------------------------- #!/usr/bin/env ruby # Copyright 2006 by Jim W. (removed_email_address@domain.invalid). All rights reserved. # Permission is granted for use, modification and distribution as # long as the above copyright notice is included. # Amb is an ambiguous choice maker. You can ask an Amb object to # select from a discrete list of choices, and then specify a set of # constraints on those choices. After the constraints have been # specified, you are guaranteed that the choices made earlier by amb # will obey the constraints. # # For example, consider the following code: # # amb = Amb.new # x = amb.choose(1,2,3,4) # # At this point, amb may have chosen any of the four numbers (1 # through 4) to be assigned to x. But, now we can assert some # conditions: # # amb.assert (x % 2) == 0 # # This asserts that x must be even, so we know that the choice made by # amb will be either 2 or 4. Next we assert: # # amb.assert x >= 3 # # This further constrains our choice to 4. # # puts x # prints '4' # # Amb works by saving a contination at each choice point and # backtracking to previousl choices if the contraints are not # satisfied. In actual terms, the choice reconsidered and all the # code following the choice is re-run after failed assertion. # # You can print out all the solutions by printing the solution and # then explicitly failing to force another choice. For example: # # amb = Amb.new # x = Amb.choose(*(1..10)) # y = Amb.choose(*(1..10)) # amb.assert x + y == 15 # # puts "x = #{x}, y = #{y}" # # amb.failure # # The above code will print all the solutions to the equation x + y == # 15 where x and y are integers between 1 and 10. # # The Amb class has two convience functions, solve and solve_all for # encapsulating the use of Amb. # # This example finds the first solution to a set of constraints: # # Amb.solve do |amb| # x = amb.choose(1,2,3,4) # amb.assert (x % 2) == 0 # puts x # end # # This example finds all the solutions to a set of constraints: # # Amb.solve_all do |amb| # x = amb.choose(1,2,3,4) # amb.assert (x % 2) == 0 # puts x # end # class Amb class ExhaustedError < RuntimeError; end # Initialize the ambiguity chooser. def initialize @back = [ lambda { fail ExhaustedError, "amb tree exhausted" } ] end # Make a choice amoung a set of discrete values. def choose(*choices) choices.each { |choice| callcc { |fk| @back << fk return choice } } failure end # Unconditional failure of a constraint, causing the last choice to # be retried. This is equivalent to saying # <code>assert(false)</tt>. def failure @back.pop.call end # Assert the given condition is true. If the condition is false, # cause a failure and retry the last choice. def assert(cond) failure unless cond end # Report the given failure message. This is called by solve in the # event that no solutions are found, and by +solve_all+ when no more # solutions are to be found. Report will simply display the message # to standard output, but you may override this method in a derived # class if you need different behavior. def report(failure_message) puts failure_message end # Class convenience method to search for the first solution to the # constraints. def Amb.solve(failure_message="No Solution") amb = self.new yield(amb) rescue Amb::ExhaustedError => ex amb.report(failure_message) end # Class convenience method to search for all the solutions to the # constraints. def Amb.solve_all(failure_message="No More Solutions") amb = self.new yield(amb) amb.failure rescue Amb::ExhaustedError => ex amb.report(failure_message) end end -- END AMB --------------------------------------------------------------- And a new puzzle to go along with the new version of Amb: -- BEGIN PUZZLE ----------------------------------------------------------- #!/usr/bin/env ruby # Two thieves have being working together for years. Nobody knows # their identities, but one is known to be a Liar and the other a # Knave. The local sheriff gets a tip that the bandits are about to # commit another crime. When the sheriff arrives at the seen of the # crime, he finds three men, A, B, and C. C has been stabbed with a # dagger. He cries out, "A stabbed me" before anybody can say anything # else; then, he falls down dead from the stabbing. # # Not sure which of the three men are the crooks, the sheriff takes # the two suspects to the jail and interrogates them. He gets the # following information. # # A's statements: # 1. B is one of the crooks. # 2. B?s second statement is true. # 3. C was telling the truth. # # B's statements: # 1. A killed the other guy. # 2. C was killed by one of the thieves. # 3. C?s next statement would have been a lie. # # C's statement: # 1. A stabbed me. # # The sheriff knows that the murderer is among these three people. Who # should the sheriff arrest for killing C? # # NOTE: Liars always lie, knights always tell the truth and knaves # strictly alternate between truth and lies. require 'amb' # Some helper methods for logic class Object def implies(bool) self ? bool : true end def xor(bool) self ? !bool : bool end end # True if the given list of boolean values alternate between true and # false. def alternating(*bools) expected = bools.first bools.each do |item| if item != expected return false end expected = !expected end true end # A person class to keep track of the information about a single # person in our puzzle. class Person attr_reader :name, :type, :murderer, :thief attr_accessor :statements def initialize(amb, name) @name = name @type = amb.choose(:liar, :knave, :knight) @murderer = amb.choose(true, false) @thief = amb.choose(true, false) @statements = [] end def to_s "#{name} is a #{type} " + (thief ? "and a thief." : "but not a thief.") + (murderer ? " He is also the murderer." : "") end end # Some lists used to do collective assertions. people = Array.new(3) thieves = Array.new(2) # Find all the solutions. Amb.solve_all do |amb| count = 0 # Create the three people in our puzzle. a = Person.new(amb, "A") b = Person.new(amb, "B") c = Person.new(amb, "C") people = [a, b, c] # Basic assertions about the thieves. thieves = people.select { |p| p.thief } amb.assert thieves.size == 2 # Only two thieves amb.assert thieves.collect { |p| # One is a knave, the other a liar p.type.to_s }.sort == ["knave", "liar"] # Basic assertions about the murderer. amb.assert people.select { |p| # There is only one murderer p.murderer }.size == 1 murderer = people.find { |p| p.murderer } amb.assert ! c.murderer # No suicides # Create the logic statements of each of the people involved. Note # we are just creating them here. We won't assert the truth of them # until a bit later. c1 = a.murderer # A stabbed me c2 = case c.type # (hypothetical next statement) when :knight false when :liar true when :knave !c1 end c.statements = [c1, c2] b1 = a.murderer # A killed the other guy b2 = murderer.thief # C was killed by one of the thieves b3 = ! c2 # C's next statement would have been true b.statements = [b1, b2, b3] a1 = b.thief # B is one of the crooks a2 = b2 # B's second statement is true a3 = c1 # C was telling the truth. a.statements = [a1, a2, a3] # Now we make assertions on the truthfulness of each of persons # statements based on whether they are a Knight, a Knave or a Liar. people.each do |p| amb.assert( (p.type == :knight).implies( p.statements.all? { |s| s } )) amb.assert( (p.type == :liar).implies( p.statements.all? { |s| !s } )) amb.assert( (p.type == :knave).implies( alternating(*p.statements) )) end # Now we print out the solution. count += 1 puts "Solution #{count}:" people.each do |p| puts p end puts end -- END PUZZLE ------------------------------------------------------------- -- Jim W.

on 2006-03-16 16:25

On Mar 16, 2006, at 7:20 AM, Jim W. wrote: > I've been having a lot of fun with this quiz. Pit C. sent me a > improved version of Amb that is more "Rubyish" and much easier to read > (my original was a direct translation of the scheme version). Oh sure, you would do that after I summarized the trickier version. ;) > # Make a choice amoung a set of discrete values. > def choose(*choices) > choices.each { |choice| > callcc { |fk| > @back << fk > return choice > } > } > failure > end I tried to eliminate the outer continuation when I was summarizing, convinced it wasn't needed. My attempt wasn't successful though. : ( It's good to know I wasn't completely wrong. James Edward G. II

on 2006-03-16 16:31

On Mar 16, 2006, at 4:03 AM, Chris Parker wrote: > Here is my CSP language. I have actually been doing this for a class, > so I got an extra week to work on it. This is a great glimpse at a more robust solution. Thanks for sending it in! > As test cases, I have been > modeling typical CSP problems so right now, I can do cryptarthemtic, > sudoku, mastermind, map coloring, and the zebra problem. Ooo, I liked the Mastermind example. We need to do that as a Ruby Quiz at some point... James Edward G. II

on 2006-03-16 16:42

James G. wrote: > Oh sure, you would do that after I summarized the trickier version. ;) ;) > I tried to eliminate the outer continuation when I was summarizing, > convinced it wasn't needed. My attempt wasn't successful though. : > ( It's good to know I wasn't completely wrong. Yes, the new version is SO much easier on the eyes. Eliminating the extra continuation not only makes it easier to read, but a bit faster as well. The other simplification is that the original version supported delayed choices, i.e. passing in a lambda as a choice where that lambda wouldn't get evaluated until the choice was needed. Although a cool idea, I don't think I ever used it in any of the puzzles. So, it was extra baggage that wasn't really needed. -- -- Jim W.