Nondeterminism

McCarthy’s nondeterministic operator amb [25433] is as old as Lisp itself, although it is present in no Lisp. amb takes zero or more expressions, and makes a nondeterministic (or “ambiguous”) choice among them, preferring those choices that cause the program to converge meaningfully. Here we will explore an embedding of amb in Scheme that makes a depth-first selection of the ambiguous choices, and uses Scheme’s control operator call/cc to backtrack for alternate choices. The result is an elegant backtracking strategy that can be used for searching problem spaces directly in Scheme without recourse to an extended language. The embedding recalls the continuation strategies used to implement Prolog-style logic programming [167], but is sparer because the operator provided is much like a Scheme boolean operator, does not require special contexts for its use, and does not rely on linguistic infrastructure such as logic variables and unification.

14.1  Description of amb

An accessible description of amb and many example uses are found in the premier Scheme textbook SICP [1]. Informally, amb takes zero or more expressions and nondeterministically returns the value of one of them. Thus,

(amb 1 2)

may evaluate to 1 or 2.

amb called with no expressions has no value to return, and is considered to fail. Thus,

(amb)
-->ERROR!!! amb tree exhausted

(We will examine the wording of the error message later.)

In particular, amb is required to return a value if at least one its subexpressions converges, ie, doesn’t fail. Thus,

(amb 1 (amb))

and

(amb (amb) 1)

both return 1.

Clearly, amb cannot simply be equated to its first subexpression, since it has to return a non-failing value, if this is at all possible. However, this is not all: The bias for convergence is more stringent than a merely local choice of amb’s subexpressions. amb should furthermore return that convergent value that makes the entire program converge. In denotational parlance, amb is an angelic operator.

For example,

(amb #f #t)

may return either #f or #t, but in the program

(if (amb #f #t)
    1
    (amb))

the first amb-expression must return #t. If it returned #f, the if’s “else” branch would be chosen, which causes the entire program to fail.

14.2  Implementing amb in Scheme

In our implementation of amb, we will favor amb’s subexpressions from left to right. Ie, the first subexpression is chosen, and if it leads to overall failure, the second is picked, and so on. ambs occurring later in the control flow of the program are searched for alternates before backtracking to previous ambs. In other words, we perform a depth-first search of the amb choice tree, and whenever we brush against failure, we backtrack to the most recent node of the tree that offers a further choice. (This is called chronological backtracking.)

We first define a mechanism for setting the base failure continuation:

(define amb-fail '*)

(define initialize-amb-fail
  (lambda ()
    (set! amb-fail
      (lambda ()
        (error "amb tree exhausted")))))

(initialize-amb-fail)

When amb fails, it invokes the continuation bound at the time to amb‑fail. This is the continuation invoked when all the alternates in the amb choice tree have been tried and were found to fail.

We define amb as a macro that accepts an indefinite number of subexpressions.

(define-macro amb
  (lambda alts...
    `(let ((+prev-amb-fail amb-fail))
       (call/cc
        (lambda (+sk)

          ,@(map (lambda (alt)
                   `(call/cc
                     (lambda (+fk)
                       (set! amb-fail
                         (lambda ()
                           (set! amb-fail +prev-amb-fail)
                           (+fk 'fail)))
                       (+sk ,alt))))
                 alts...)

          (+prev-amb-fail))))))

A call to amb first stores away, in +prev‑amb‑fail, the amb‑fail value that was current at the time of entry. This is because the amb‑fail variable will be set to different failure continuations as the various alternates are tried.

We then capture the amb’s entry continuation +sk, so that when one of the alternates evaluates to a non-failing value, it can immediately exit the amb.

Each alternate alt is tried in sequence (the implicit-begin sequence of Scheme).

First, we capture the current continuation +fk, wrap it in a procedure and set amb‑fail to that procedure. The alternate is then evaluated as (+sk alt). If alt evaluates without failure, its return value is fed to the continuation +sk, which immediately exits the amb call. If alt fails, it calls amb‑fail. The first duty of amb‑fail is to reset amb‑fail to the value it had at the time of entry. It then invokes the failure continuation +fk, which causes the next alternate, if any, to be tried.

If all alternates fail, the amb‑fail at amb entry, which we had stored in +prev‑amb‑fail, is called.

14.3  Using amb in Scheme

To choose a number between 1 and 10, one could say

(amb 1 2 3 4 5 6 7 8 9 10)

To be sure, as a program, this will give 1, but depending on the context, it could return any of the mentioned numbers.

The procedure number‑between is a more abstract way to generate numbers from a given lo to a given hi (inclusive):

(define number-between
  (lambda (lo hi)
    (let loop ((i lo))
      (if (> i hi) (amb)
          (amb i (loop (+ i 1)))))))

Thus (number‑between 1 6) will first generate 1. Should that fail, the loop iterates, producing 2. Should that fail, we get 3, and so on, until 6. After 6, loop is called with the number 7, which being more than 6, invokes (amb), which causes final failure. (Recall that (amb) by itself guarantees failure.) At this point, the program containing the call to (number‑between 1 6) will backtrack to the chronologically previous amb-call, and try to satisfy that call in another fashion.

The guaranteed failure of (amb) can be used to program assertions.

(define assert
  (lambda (pred)
    (if (not pred) (amb))))

The call (assert pred) insists that pred be true. Otherwise it will cause the current amb choice point to fail.1

Here is a procedure using assert that generates a prime less than or equal to its argument hi:

(define gen-prime
  (lambda (hi)
    (let ((i (number-between 2 hi)))
      (assert (prime? i))
      i)))

This seems devilishly simple, except that when called as a program with any number (say 20), it will produce the uninteresting first solution, ie, 2.

We would certainly like to get all the solutions, not just the first. In this case, we may want all the primes below 20. One way is to explicitly call the failure continuation left after the program has produced its first solution. Thus,

(amb)
=> 3

This leaves yet another failure continuation, which can be called again for yet another solution:

(amb)
=> 5

The problem with this method is that the program is initially called at the Scheme prompt, and successive solutions are also obtained by calling amb at the Scheme prompt. In effect, we are using different programs (we cannot predict how many!), carrying over information from a previous program to the next. Instead, we would like to be able to get these solutions as the return value of a form that we can call in any context. To this end, we define the bag‑of macro, which returns all the successful instantiations of its argument. (If the argument never succeeds, bag‑of returns the empty list.) Thus, we could say,

(bag-of
  (gen-prime 20))

and it would return

(2 3 5 7 11 13 17 19)

The bag‑of macro is defined as follows:

(define-macro bag-of
  (lambda (e)
    `(let ((+prev-amb-fail amb-fail)
           (+results '()))
       (if (call/cc
            (lambda (+k)
              (set! amb-fail (lambda () (+k #f)))
              (let ((+v ,e))
                (set! +results (cons +v +results))
                (+k #t))))
           (amb-fail))
       (set! amb-fail +prev-amb-fail)
       (reverse! +results))))

bag‑of first saves away its entry amb‑fail. It redefines amb‑fail to a local continuation +k created within an if-test. Inside the test, the bag‑of argument e is evaluated. If e succeeds, its result is collected into a list called +results, and the local continuation is called with the value #t. This causes the if-test to succeed, causing e to be retried at its next backtrack point. More results for e are obtained this way, and they are all collected into +results.

Finally, when e fails, it will call the base amb‑fail, which is simply a call to the local continuation with the value #f. This pushes control past the if. We restore amb‑fail to its pre-entry value, and return the +results. (The reverse! is simply to produce the results in the order in which they were generated.)

14.4  Logic puzzles

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.2 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)))
      (assert
       (distinct? (list parent1 parent2)))
      (assert
       (if (eqv? kibi 'm)
           (not kibi-lied?)))
      (assert
       (if kibi-lied?
           (xor
            (and (eqv? kibi-self-desc 'm)
                 (eqv? kibi 'f))
            (and (eqv? kibi-self-desc 'f)
                 (eqv? kibi 'm)))))
      (assert
       (if (not kibi-lied?)
           (xor
            (and (eqv? kibi-self-desc 'm)
                 (eqv? kibi 'm))
            (and (eqv? kibi-self-desc 'f)
                 (eqv? kibi 'f)))))
      (assert
       (if (eqv? parent1 'm)
           (and
            (eqv? kibi-self-desc 'm)
            (xor
             (and (eqv? kibi 'f)
                  (eqv? kibi-lied? #f))
             (and (eqv? kibi 'm)
                  (eqv? kibi-lied? #t))))))
      (assert
       (if (eqv? parent1 'f)
           (and
            (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 not proven until 1976 [29]) 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 [31]. (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.3 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)))
            (spain
             (list 'spain e
                   (list f p)))
            (france
             (list 'france f
                   (list e i s b g l)))
            (belgium
             (list 'belgium b
                   (list f h l g)))
            (holland
             (list 'holland h
                   (list b g)))
            (germany
             (list 'germany g
                   (list f a s h b l)))
            (luxembourg
             (list 'luxembourg l
                   (list f b g)))
            (italy
             (list 'italy i
                   (list f a s)))
            (switzerland
             (list 'switzerland s
                   (list f i a g)))
            (austria
             (list 'austria a
                   (list i s g))))
        (let ((countries
               (list portugal spain
                     france belgium
                     holland germany
                     luxembourg
                     italy switzerland
                     austria)))

          ;the color of a country
          ;should not be the color of
          ;any of its neighbors
          (for-each
           (lambda (c)
             (assert
              (not (memq (cadr c)
                         (caddr c)))))
           countries)

          ;output the color
          ;assignment
          (for-each
           (lambda (c)
             (display (car c))
             (display " ")
             (display (cadr c))
             (newline))
           countries))))))

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


1 SICP names this procedure require. We use the identifier assert in order to avoid confusion with the popular if informal use of the identifier require for something else, viz, an operator that loads code modules on a per-need basis.

2 This puzzle is due to Hunter [19].

3 Spain (España) has e so as not to clash with Switzerland.