SICP: Why does the forget-value-process call the new-value process?

This code is from SICP 3.3.5 Restriction Propagation . I can't figure out why process-forget-value

it needs to be called process-new-value

as the last step.

The text says, "The reason for this last step is that one or more connectors may still have a value (that is, a connector may have a value that was not originally set by the adder), and those values ​​may need to be multiplied back across adder".

What is the simplest web of constraints that can show why it is required (process-new-value)

? Thank!

(define (adder a1 a2 sum)
  (define (process-new-value)
    (cond ((and (has-value? a1) (has-value? a2))
           (set-value! sum
                       (+ (get-value a1) (get-value a2))
          ((and (has-value? a1) (has-value? sum))
           (set-value! a2
                       (- (get-value sum) (get-value a1))
          ((and (has-value? a2) (has-value? sum))
           (set-value! a1
                       (- (get-value sum) (get-value a2))

  (define (process-forget-value)
    (forget-value! sum me)
    (forget-value! a1 me)
    (forget-value! a2 me)
    (process-new-value))  ;;; * WHY * ???

  (define (me request)
    (cond ((eq? request 'I-have-a-value)  
          ((eq? request 'I-lost-my-value) 
           (error "Unknown request -- ADDER" request))))

  (connect a1 me)
  (connect a2 me)
  (connect sum me)



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1 answer

This is the test I removed process-new-value

from the adder. You will see that the behavior is different.

(define c (make-connector))
(define a (make-connector))
(define b (make-connector))
(define d (make-connector))

(constant 10 a)
(constant 10 c)
(constant 10 d)

(define adder1 (adder a b c))
(define adder2 (adder a b d))

> (has-value? b)

> (get-value b)

> (forget-value! b adder1)

> (has-value? b)


If you do it with the correct version.

> (has-value? b)


Second time. As they say, when adder1

informs b

forget its meaning. a

and c

will be permanent, and the last process-new-value

to adder2

re-install b

to 0. This will also work if you are using set-value!

for a

and c




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