Question

I'm trying to implement back-tracking search in Scheme. So far, I have the following:

(define (backtrack n graph assignment)  
    (cond (assignment-complete n assignment) (assignment) )

    (define u (select-u graph assignment))

    (define c 1)
    (define result 0)

    (let forLoop ()
        (when (valid-choice graph assignment c)
             (hash-set! assignment u c)

             (set! result (backtrack n graph assignment))

             (cond ((not (eq? result #f)) result))

             (hash-remove! assignment u)            
        )

        (set! c (+ c 1))
        (when (>= n c) (forLoop))
    )

   #f ; I believe this is where I'm having problems
)

My functions assignment-complete and select-u pass unit tests. The argument assignment is a hash-table make with (make-hash), so it should be fine.

I believe the problem I have is related to returning false at the end of the loop, if no recursive returns a non-false value (which should be a valid assignment). Is there a Scheme equivalent of an explicit return statement?

Was it helpful?

Solution

the answer to your question is yes:

(define (foo ...)
  (call-with-current-continuation
    (lambda (return)
      ...... ; here anywhere inside any sub-expression 
      ...... ; you can call (return 42)
      ...... ; to return 42 from `foo` right away
    )))

This sets up an exit continuation so that you can return a result value from inside a function's body. The usual Scheme way is to put your return form as the last one, so its value is returned:

    (let forLoop ()
        (when (valid-choice graph assignment c)
             (hash-set! assignment u c)
             (set! result (backtrack n graph assignment))
             (cond
                 ((not (eq? result #f))
                   result))       ; the value of `cond` form is ignored
             (hash-remove! assignment u))
                                  ; the value of `when` form is ignored
        (set! c (+ c 1))
        (if (>= n c)     ; `if` must be the last form 
           (forLoop)     ; so that `forLoop` is tail-recursive
           ;; else:
           return-value) ; <<------ the last form's value 
    )                    ; is returned from `let` form

   ;; (let forLoop ...) must be the last form in your function
   ;;                   so its value is returned from the function
)

You also have a problem here:

(cond (assignment-complete n assignment) (assignment) )

this code does not make a call (assignment-complete n assignment). Rather, it checks whether a variable assignment-complete has a non-null value, and if not it checks assignment variable, but in any case its returned value is just ignored anyway. Perhaps some more parentheses are missing there, and/or an else clause.

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