Constraint Processing


#1

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


#2

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


#3

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-scheme-Z-H-16.html#node_chap_14)

However, this is not an exact transcription of the book’s logic, I was
able to simplify a number of the assertions.

Enjoy!

– Jim W.

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