[Quiz 70 Solution] -- Using Amb

The Amb library is part of the continuations talk that Chad and I gave
at last years Ruby conference. Amb is a library that allows you to
choose from a number of possibilities, e.g.:

A = Amb.new
x = A.choose(1,2,3,4)
# at this point, x may be 1, 2, 3, or possibly 4

And then assert things that must be true about the choice:

A.assert (x % 2)
# at this point, x can only be 2 or 4.

If the assertion is not true, Amb will rewind the program back to the
point of choosing, and make a different choice.

Here is the solution to part one of the quiz using the Amb library:

–BEGIN-----------------------------------------------------------
require ‘amb’

A = Amb.new

begin
a = A.choose((0…4))
b = A.choose(
(0…4))
c = A.choose(*(0…4))

A.assert a < b
A.assert a + b == c

puts “a=#{a}, b=#{b}, c=#{c}”

A.failure

rescue Amb::ExhaustedError
puts “No More Solutions”
end
–END-------------------------------------------------------------

Here’s the output of the solution:

$ ruby quiz70.rb
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
No More Solutions

The magic square solution is a bit more involved for two reasons. (1) I
created a magic square object that allowed me to express the constraints
easily, and the class turned out to be a bit wordy. (2) In order to
reduced the number of states searched for a solution, it is necessary to
assert constraints about the magic square as early as possible. For
example, we assert that a new number is unique as soon as we add it to
the square, rather than building the entire square and asserted that all
members are unique. This greatly reduced the number of possibilities
that needed to be tested and allowed the program to finish quite
quickly.

Here is the magic square solution:

–BEGIN------------------------------------------------------
#!/usr/bin/env ruby

Ruby Q. # 70 – Constraint Solving

Copyright 2006 by Jim W.

Again, using Amb to solve the magic square constraint problem. The

key to getting this to run half-way efficiently is to do the

assertions as soon as possible, before additional choices have been

made. This minimizes the amount of backtracking needed.

See the following for details about Amb:

http://www.ccs.neu.edu/home/dorai/t-y-scheme/t-y-scheme-Z-H-16.html#node_chap_14

require ‘amb’

A Magic Square class, mostly to get convenient column/row/diagonal

summations and detecting previously used values.

class MagicSquare
def initialize(side)
@side = side
@values = Array.new(side**2) { 0 }
end

Access square value

def [](col, row)
@values[col + @side*row]
end

Set square value

def []=(col, row, value)
@values[index(col, row)] = value
end

Calculate the sum of value along row +row+.

def row_sum(row)
(0…SIDE).inject(0) { |sum, col| sum + self[col,row] }
end

Calculate the sum of values in column +col+.

def col_sum(col)
(0…SIDE).inject(0) { |sum, row| sum + self[col,row] }
end

Calculate the sum of values along the major diagonal (i.e. r ==

c).

def diagonal_sum
(0…SIDE).inject(0) { |sum, i| sum + self[i,i] }
end

Calculate the sum of values along the other diagnol (i.e. r ==

SIDE-c-1).

def other_diagonal_sum
(0…SIDE).inject(0) { |sum, i| sum + self[i,SIDE-i-1] }
end

Has this value been previously used in the square in any previous

row, or any previous column of the current row?

def previously_used?(col, row, value)
(0…index(col, row)).any? { |i| @values[i] == value }
end

Convert to string for display.

def to_s
(0…SIDE).collect { |r|
(0…SIDE).collect { |c| self[c,r] }.join(’ ')
}.join(“\n”)
end

private

def index(col, row)
col + @side*row
end
end

SIDE = 3
MAX = SIDE**2
SUM = (MAX*(MAX+1))/(2*SIDE)

A = Amb.new
square = MagicSquare.new(SIDE)

Build up the square position by position. To minimize backtracking,

assert everything you know about a position as early as possible.

count = 0

begin
(0…SIDE).each do |r|
(0…SIDE).each do |c|
value = A.choose(*(1…MAX))
A.assert ! square.previously_used?(c, r, value)
square[c,r] = value
A.assert( (r < SIDE-1) || square.col_sum(c) == SUM)
end
A.assert square.row_sum(r) == SUM
end
A.assert square.diagonal_sum == SUM
A.assert square.other_diagonal_sum == SUM

count += 1
puts “SOLUTION #{count}:”
puts square

Explicitly force a failure and make the program search for another

solution.

A.failure

rescue Amb::ExhaustedError
puts “No More Solutions”
end
–END--------------------------------------------------------

And the output of the magic squares program:

$ ruby magic_square.rb
SOLUTION 1:
2 7 6
9 5 1
4 3 8
SOLUTION 2:
2 9 4
7 5 3
6 1 8
SOLUTION 3:
4 3 8
9 5 1
2 7 6
SOLUTION 4:
4 9 2
3 5 7
8 1 6
SOLUTION 5:
6 1 8
7 5 3
2 9 4
SOLUTION 6:
6 7 2
1 5 9
8 3 4
SOLUTION 7:
8 1 6
3 5 7
4 9 2
SOLUTION 8:
8 3 4
1 5 9
6 7 2
No More Solutions

Finally, we get to the Amb library. It is a direct Ruby port of the
Scheme amb function described in Teach yourself Scheme in fixnum days.
In brief, Amb saves a continuation whenever a choice has to be made.
Whenever an Amb assertion fails, the last choice is redone with a
different result by invoking the latest continuation. If all choices
are used up, then we go to the previous choice and undo retry it.

If you want a more indepth description of Amb, see
http://www.ccs.neu.edu/home/dorai/t-y-scheme/t-y-scheme-Z-H-16.html#node_chap_14

Here is the amb.rb library:

–BEGIN (amb.rb) ------------------------------------------------
class Amb
class ExhaustedError < RuntimeError; end

def initialize
@fail = proc { fail ExhaustedError, “amb tree exhausted” }
end

def choose(*choices)
prev_fail = @fail
callcc { |sk|
choices.each { |choice|
callcc { |fk|
@fail = proc {
@fail = prev_fail
fk.call(:fail)
}
if choice.respond_to? :call
sk.call(choice.call)
else
sk.call(choice)
end
}
}
@fail.call
}
end

def failure
choose
end

def assert(cond)
failure unless cond
end
end
–END------------------------------------------------------------


– Jim W.

Oops, typo!

Jim W. wrote:
[…]

And then assert things that must be true about the choice:

A.assert (x % 2)
# at this point, x can only be 2 or 4.

The assertion should read:

A.assert (x % 2) == 0

Sorry.


– Jim W.