I had to give this a couple tries before I just broke down and went for what amounts to brute force. This solution can get slow pretty quickly; I'm sure there are some easy speedups (i.e. narrow the search space) that can be done, but I figured to put this up for now. # Helpers class Integer def even? (self % 2).zero? end end class Symbol def <=> other self.to_s <=> other.to_s end end # Constraint Solver class class Problem def initialize(&block) @domain = {} @consts = Hash.new { [] } instance_eval(&block) end def variable(var, domain) raise ArgumentError, "Cannot specify variable #{var} more than once." if @domain.has_key?(var) @domain[var] = domain.to_a end def constrain(*vars, &foo) raise ArgumentError, 'Constraint requires at least one variable.' if vars.size.zero? vars.each do |var| raise ArgumentError, "Unknown variable: #{var}" unless @domain.has_key?(var) end @consts[vars] = @consts[vars] << foo end def solve # Separate constraint keys into unary and non-unary. unary, multi = @consts.keys.partition{ |vars| vars.size == 1 } # Process unary constraints first to narrow variable domains. unary.each do |vars| a = vars.first @consts[vars].each do |foo| @domain[a] = @domain[a].select { |d| foo.call(d) } end end # Build fully-expanded domain (i.e. across all variables). full = @domain.keys.map do |var| @domain[var].map do |val| { var => val } end end.inject do |m, n| m.map do |a| n.map do |b| a.merge(b) end end.flatten end # Process non-unary constraints on full domain. full.select do |d| multi.all? do |vars| @consts[vars].all? do |foo| foo.call( vars.map { |v| d[v] } ) end end end end end # A simple example problem = Problem.new do variable(:a, 0..10) variable(:b, 0..10) variable(:c, 0..10) constrain(:a) { |a| a.even? } constrain(:a, :b) { |a, b| b == 2 * a } constrain(:b, :c) { |b, c| c == b - 3 } end puts "Simple example solutions:" problem.solve.each { |sol| p sol } # Calculate some primes... The constraint problem actually finds # the non-primes, which we remove from our range afterward to get # the primes. problem = Problem.new do variable(:a, 2..25) variable(:b, 2..25) variable(:c, 2..50) constrain(:a, :b) { |a, b| a <= b } constrain(:a, :b, :c) { |a, b, c| a * b == c } end puts "The primes up to 50:" puts ((2..50).to_a - problem.solve.map { |s| s[:c] }).join(", ") puts

on 2006-03-14 05:29

on 2006-03-14 19:19

I didn't write this for the quiz, but here is a simple csp library written in ruby: http://sillito.ca/ruby-csp It has a few features to speed up solving including forward checking, dynamic variable ordering and support for specialized propagation, but it remains very basic. There is lots more that could be done with it, and I plan to release an update when I find the time. Feedback is definitely welcome. Here is how the famous N-Queens problem can be modeled and solved with the library: require 'ai/csp' include AI::CSP def problem(n) # variables are columns and values are rows, so assigning # the first variable the value 2 corresponds to placing a # queen on the board at col 0 and row 2. variables = (0...n).collect {|i| Variable.new(i, (0...n)) } problem = Problem.new(variables) # None of the queens can share a row. AllDifferent is a # built in constraint type. problem.add_constraint(AllDifferent.new(*variables)) # No pair of queens can be on the same diagonal. variables.each_with_index {|v1,i| variables[(i+1)..-1].each_with_index{ |v2,j| problem.add_constraint(v1, v2) { |row1,row2| (j+1) != (row1-row2).abs } } } problem end solver = Backtracking.new(true, FAIL_FIRST) solver.each_solution(problem(8)) { |solution| puts solution } puts solver # prints some statistics Cheers, Jonathan

on 2006-03-14 19:46

Here's some more puzzle solutions using Amb. This one is the one described in the "Learn Scheme in Fixnum days" online book (where I found the original Amb implementation). (see http://www.ccs.neu.edu/home/dorai/t-y-scheme/t-y-s...) However, this is not an exact transcription of the book's logic, I was able to simplify a number of the assertions. Enjoy! -- Jim Weirich # The Kalotans are a tribe with a peculiar quirk. Their males always # tell the truth. Their females never make two consecutive true # statements, or two consecutive untrue statements. # # An anthropologist (let's call him Worf) has begun to study # them. Worf does not yet know the Kalotan language. One day, he meets # a Kalotan (heterosexual) couple and their child Kibi. Worf asks # Kibi: ``Are you a boy?'' Kibi answers in Kalotan, which of course # Worf doesn't understand. # # Worf turns to the parents (who know English) for explanation. One of # them says: ``Kibi said: `I am a boy.' '' The other adds: ``Kibi is a # girl. Kibi lied.'' # # Solve for the sex of the parents and Kibi. require 'amb' # Some helper methods for logic class Object def implies(bool) self ? bool : true end def xor(bool) self ? !bool : bool end end count = 0 A = Amb.new # Begin the solution begin # Kibi's parents are either male or female, but must be distinct. parent1 = A.choose(:male, :female) parent2 = A.choose(:male, :female) A.assert parent1 != parent2 # Kibi sex, and Kibi's self description are separate facts kibi = A.choose(:male, :female) kibi_said = A.choose(:male, :female) # We will capture whether kibi lied in a local variable. This will # make some later logic conditions a bit easier. (Note: the Scheme # implementation sets the kibi_lied variable to a choice of true or # false and then uses assertions to make all three variables # consistent. This way however, is just so much easier.) kibi_lied = kibi != kibi_said # Now we look at what the parents said. If the first parent was # male, then kibi must have described itself as male. A.assert( (parent1==:male).implies( (kibi_said == :male ) ) ) # If the first parent is female, then there are no futher deductions # to make. Their statement could either be true or false. # If the second parent is male, then both its statements must be # true. A.assert( (parent2 == :male).implies( kibi==:female )) A.assert( (parent2 == :male).implies( kibi_lied )) # If the second parent is female, then the condition is more # complex. In this case, one or the other of the parent 2's # statements are false, but not both are false. Let's introduce # some variables for statements 1 and 2 just to make this a bit # clearer. s1 = kibi_lied s2 = (kibi == :female) A.assert( (parent2 == :female).implies( (s1 && !s2).xor(!s1 && s2) ) ) # Now just print out the solution. count += 1 puts "Solution #{count}" puts "The first parent is #{parent1}." puts "The second parent is #{parent2}." puts "Kibi is #{kibi}." puts "Kibi said #{kibi_said} and #{kibi_lied ? 'lied' : 'told the truth'}." puts A.failure # Force a search for another solution. rescue Amb::ExhaustedError puts "No More Solutions" end