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The power of depth-first search coupled with backtracking becomes obvious when applied to solving logic puzzles. These problems are extraordinarily difficult to solve procedurally, but can be solved concisely and declaratively with amb, without taking anything away from the charm of solving the puzzle.

14.4.1  The Kalotan puzzle

The Kalotans are a tribe with a peculiar quirk.6 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.


The solution consists in introducing a bunch of variables, allowing them to take a choice of values, and enumerating the conditions on them as a sequence of assert expressions.

The variables: parent1, parent2, and kibi are the sexes of the parents (in order of appearance) and Kibi; kibi-self-desc is the sex Kibi claimed to be (in Kalotan); kibi-lied? is the boolean on whether Kibi's claim was a lie.

(define solve-kalotan-puzzle
  (lambda ()
    (let ((parent1 (amb 'm 'f))
          (parent2 (amb 'm 'f))
          (kibi (amb 'm 'f))
          (kibi-self-desc (amb 'm 'f))
          (kibi-lied? (amb #t #f)))
       (distinct? (list parent1 parent2)))
       (if (eqv? kibi 'm)
           (not kibi-lied?)))
       (if kibi-lied?
            (and (eqv? kibi-self-desc 'm)
                 (eqv? kibi 'f))
            (and (eqv? kibi-self-desc 'f)
                 (eqv? kibi 'm)))))
       (if (not kibi-lied?)
            (and (eqv? kibi-self-desc 'm)
                 (eqv? kibi 'm))
            (and (eqv? kibi-self-desc 'f)
                 (eqv? kibi 'f)))))
       (if (eqv? parent1 'm)
            (eqv? kibi-self-desc 'm)
             (and (eqv? kibi 'f)
                  (eqv? kibi-lied? #f))
             (and (eqv? kibi 'm)
                  (eqv? kibi-lied? #t))))))
       (if (eqv? parent1 'f)
            (eqv? kibi 'f)
            (eqv? kibi-lied? #t))))
      (list parent1 parent2 kibi))))

A note on the helper procedures: The procedure distinct? returns true if all the elements in its argument list are distinct, and false otherwise. The procedure xor returns true if only one of its two arguments is true, and false otherwise.

Typing (solve-kalotan-puzzle) will solve the puzzle.

14.4.2  Map coloring

It has been known for some time (but only lately proven) that four colors suffice to color a terrestrial map -- ie, to color the countries so that neighbors are distinguished. To actually assign the colors is still an undertaking, and the following program shows how nondeterministic programming can help.

The following program solves the problem of coloring a map of Western Europe. The problem and a Prolog solution are given in The Art of Prolog (It is instructive to compare our solution with the book's.)

The procedure choose-color nondeterministically returns one of four colors:

(define choose-color
  (lambda ()
    (amb 'red 'yellow 'blue 'white)))

In our solution, we create for each country a data structure. The data structure is a 3-element list: The first element of the list is the country's name; the second element is its assigned color; and the third element is the colors of its neighbors. Note we use the initial of the country for its color variable.7 Eg, the list for Belgium is (list 'belgium b (list f h l g)), because -- per the problem statement -- the neighbors of Belgium are France, Holland, Luxembourg, and Germany.

Once we create the lists for each country, we state the (single!) condition they should satisfy, viz, no country should have the color of its neighbors. In other words, for every country list, the second element should not be a member of the third element.

(define color-europe
  (lambda ()

    ;choose colors for each country
    (let ((p (choose-color)) ;Portugal
          (e (choose-color)) ;Spain
          (f (choose-color)) ;France
          (b (choose-color)) ;Belgium
          (h (choose-color)) ;Holland
          (g (choose-color)) ;Germany
          (l (choose-color)) ;Luxemb
          (i (choose-color)) ;Italy
          (s (choose-color)) ;Switz
          (a (choose-color)) ;Austria

      ;construct the adjacency list for
      ;each country: the 1st element is
      ;the name of the country; the 2nd
      ;element is its color; the 3rd
      ;element is the list of its
      ;neighbors' colors
      (let ((portugal
             (list 'portugal p
                   (list e)))
             (list 'spain e
                   (list f p)))
             (list 'france f
                   (list e i s b g l)))
             (list 'belgium b
                   (list f h l g)))
             (list 'holland h
                   (list b g)))
             (list 'germany g
                   (list f a s h b l)))
             (list 'luxembourg l
                   (list f b g)))
             (list 'italy i
                   (list f a s)))
             (list 'switzerland s
                   (list f i a g)))
             (list 'austria a
                   (list i s g))))
        (let ((countries
               (list portugal spain
                     france belgium
                     holland germany
                     italy switzerland

          ;the color of a country
          ;should not be the color of
          ;any of its neighbors
           (lambda (c)
              (not (memq (cadr c)
                         (caddr c)))))

          ;output the color
           (lambda (c)
             (display (car c))
             (display " ")
             (display (cadr c))

Type (color-europe) to get a color assignment.

6 This puzzle is due to Hunter.

7 Spain (Espana) has e so as not to clash with Switzerland.

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