# Forum: Ruby Constraint Processing (#70)

Announcement (2017-05-07): www.ruby-forum.com is now read-only since I unfortunately do not have the time to support and maintain the forum any more. Please see rubyonrails.org/community and ruby-lang.org/en/community for other Rails- und Ruby-related community platforms.
on 2006-03-10 14:51
```The three rules of Ruby Quiz:

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 Quiz 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 Talk follow the discussion.

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

by Jay Anderson

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!

http://ktiml.mff.cuni.cz/~bartak/constraints/

and:

http://en.wikipedia.org/wiki/Constraint_programming```
on 2006-03-13 23:15
```Here is my own solution to this week's Ruby Quiz.  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 Gray II```
on 2006-03-15 03:10
```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 11:04
```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 14:19
```I've been having a lot of fun with this quiz.  Pit Capitain 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 Weirich (jim@weirichhouse.org).  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_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 Weirich```
on 2006-03-16 15:25
```On Mar 16, 2006, at 7:20 AM, Jim Weirich wrote:

> I've been having a lot of fun with this quiz.  Pit Capitain 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 Gray II```
on 2006-03-16 15:31
```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 Gray II```
on 2006-03-16 15:42
```James Gray 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 Weirich```
This topic is locked and can not be replied to.