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Published byMarian Cook Modified over 8 years ago
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PPL CPS
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Moed A 2007
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Solution (define scale-tree (λ (tree factor) (map (λ (sub-tree) (if (list? sub-tree) (scale-tree sub-tree factor) (* sub-tree factor))) tree))) (scale-tree '(((1 4) 2)) 5) (scale-tree '( ) 2)
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Continuation Passing Style Main idea: instead of returning a value, you pass it as a parameter to another function More specific: every user defined procedure f$ gets another parameter called continuation. When f$ ends we apply the continuation Distinction between creating the continuation and applying it All user defined function are in tail-position
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Simple Examples: Normal (define square (lambda (x) (* x x))) (define add1 (lambda (x) (+ x 1))) CPS (define square$ (lambda (x cont) (cont (* x x))) (define add1$ (lambda (x cont) (cont (+ x 1))))
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> (add1$ 5 (λ(x) x)) ( (λ(x) x) (+ 5 1) ) 6 > (square$ 5 (λ(x) x)) ( (λ(x) x) (* 5 5) ) 25 > (add1$ 5 (λ(x) (square$ x (λ(x) x)))) (define square$ (lambda (x cont) (cont (* x x))) (define add1$ (lambda (x cont) (cont (+ x 1)))) ( (λ(x) (square$ x (λ(x) x)))) (+ 5 1) ) ( (λ (x) x) 36 ) 36
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Evaluation Order Order unknown (define h2 (λ (x y) (mult (square x) (add1 y)))) (define mult (λ (x y) (* x y))) We set the order (define h2$ (λ (x y cont) (square$ x (λ (square-res) (add1$ y (λ (add1-res) (mult$ square-res add1-res cont))))))) (define square$ (lambda (x cont) (cont (* x x))) (define add1$ (lambda (x cont) (cont (+ x 1))))
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CPS is Good For: Order of computation (just seen) Turning recursion into iteration (seen in the past, see more now) Controlling multiple future computations (the true power of CPS)
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A Word about Typing… You can avoid complicated typing by using letrec (see notes)
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Recursion Into Iteration (define fact (λ (n) (if (= n 0) 1 (* n (fact (- n 1)))))) (fact 3)
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(define fact$ (λ (n cont) (if (= n 0) (cont 1) (fact$ (- n 1) (λ (res) (cont (* n res)))))))
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(fact$ 3 (λ (x) x)) ; ==> (fact$ 2 (λ (res 1 ) ((λ (x) x) (* 3 res 1 )))) ; ==> (fact$ 1 (λ (res 2 ) ((λ (res 1 ) ((λ (x) x) (* 3 res 1 ))) (* 2 res 2 )))) ; ==> (fact$ 0 (λ (res 3 ) ((λ (res 2 ) ((λ (res 1 ) ((λ (x) x) (* 3 res 1 ))) (* 2 res 2 ))) (* 1 res 3 )))) ( (λ (res 3 ) ( (λ (res 2 ) ( (λ (res 1 ) ( (λ (x) x) (* 3 res 1 ))) (* 2 res 2 ))) (* 1 res 3 ))) 1) ; ==> ( (λ (res 2 ) ( (λ (res 1 ) ( (λ (x) x) (* 3 res 1 ))) (* 2 res 2 ))) 1) ; ==> ( (λ (res 1 ) ( (λ (x) x) (* 3 res 1 ))) 2) ; ==> ( (λ (x) x) 6) 6
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CPS Map (define map (λ (f lst) (if (null? lst) lst (cons (f (car lst)) (map f (cdr lst)))))) (define map$ (λ (f$ list c) (if (null? list) (c list) (f$ (car list) (λ (f-res) (map$ f$ (cdr list) (λ (map-cdr) (c (cons f-res map-cdr)))))))))
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> (map$ (λ (x c) (c (* x x))) ‘(1 3) (λ (x) x)) ((λ (x c) (c (* x x))) 1 (λ (f-res) (map$ (λ (x c) (c (* x x)) ‘(3) (λ (map-res) ( (λ (x) x) (cons f-res map-res)) )))))) (define map$ (λ (f$ list c) (if (null? list) (c list) (f$ (car list) (λ (f-res) (map$ f$ (cdr list) (λ (map-res) (c (cons f-res map-res)))))))))
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Map$ Another Version (define map$ (λ (f$ list cont) (if (null? list) (cont list) (map$ f$ (cdr list) (λ (map-res) (f$ (car list) (λ (f-res) (cont (cons f-res map-res)))))))))
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Multiple Future Computation The true power of CPS Most useful example: errors – Errors are unplanned future – The primitive error breaks the calculation and returns void We want more control, and we can do it with CPS
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Sum List with Error (define sumlist (lambda (li) (cond ((null? li) 0) ((not (number? (car li))) (error "non numeric value!")) (else (+ (car li) (sumlist (cdr li))))))) (sumlist '(1 2 a))
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Try 1 (define sumlist (lambda (li) (cond ((null? li) 0) ((not (number? (car li))) (error "non numeric value!") 0) (else (+ (car li) (sumlist (cdr li)))))))
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Try 2 (define sumlist (lambda (li) (cond ((null? li) 0) ((not (number? (car li))) #f) (else (let ((sum-cdr (sumlist (cdr li)))) (if sum-cdr (+ (car li) sum-cdr) #f))))))
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Why is it so Complicated We are deep inside the recursion: the stack is full with frames and we need to “close” every one of them If only there was a way NOT to open frames on the stack…
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Sum List with CPS (define sumlist$ (λ (l succ fail) (cond ( (null? l) (succ l) ) ( (number? (car l)) (sumlist$ (cdr l) (λ (sum-cdr-l) (succ (+ (car l) sum-cdr-l))) fail)) (else (fail l)))))
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Run Example (sumlist$ '(1 2 3 a) (lambda (x) x) (lambda (x) (display x) (display " ") (display 'not-a-num)))
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Fail Continuation For Backtracking ;; Purpose: Find the left most even leaf of a binary ;; tree whose leaves are labeled by numbers. ;; Type: [LIST -> Number union Boolean] ;; Examples: (leftmost-even ’((1 2) (3 4))) ==> 2 ;; (leftmost-even ’((1 1) (3 3))) ==> #f (define leftmost-even (λ (tree) (letrec ((iter (λ (tree) (cond ((null? tree) #f) ((not (list? tree)) (if (even? tree) tree #f)) (else (let ((res-car (iter (car tree)))) (if res-car res-car (iter (cdr tree))))))))) (iter tree)))) No CPS
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Fail Continuation For Backtracking (define leftmost-even$ (λ (tree succ-cont fail-cont) (cond ((null? tree) (fail-cont)) ; Empty tree ((not (list? tree)) ; Leaf tree (if (even? tree) (succ-cont tree) (fail-cont))) (else ; Composite tree (leftmost-even$ (car tree) succ-cont (λ () (leftmost-even$ (cdr tree) succ-cont fail-cont)))))))
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(leftmost-even$ ((1 2) (3 4)) (λ (x) x) (λ () #f)) ==> (leftmost-even$ (1 2) (λ (x) x) (λ () (leftmost-even$ ((3 4)) (λ (x) x) (λ () #f)))) ;==> (leftmost-even$ 1 (λ (x) x) (λ () (leftmost-even$ (2) (λ (x) x) (λ () (leftmost-even$ ((3 4)) (λ (x) x) (λ () #f)))))) ;==>* (leftmost-even$ (2) (λ (x) x) (λ () (leftmost-even$ ((3 4)) (λ (x) x) (λ () #f)))) ;==>* ( (λ (x) x) 2) ;==> 2
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Construct Tree with CPS (define replace-leftmost (λ (tree old new succ-cont fail-cont) (cond ((null? tree) (fail-cont)) ; Empty tree ((not (list? tree)) ; Leaf tree (if (eq? tree old) (succ-cont new) (fail-cont))) (else ; Composite tree (replace-leftmost$ (car tree) (λ (car-res) (succ-cont (cons car-res (cdr tree)))) (λ () (replace-leftmost$ (cdr tree) (λ (cdr-res) (succ-cont (cons (car tree) cdr-res))) fail-cont)))))))
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Moed A 2007
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Solution (define scale-tree (λ (tree factor) (map (λ (sub-tree) (if (list? sub-tree) (scale-tree sub-tree factor) (* sub-tree factor))) tree))) (scale-tree '(((1 4) 2)) 5) (scale-tree '( ) 2)
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Solution (define scale-tree$ (λ (tree factor c) (map$ (λ (sub-tree c) (if (list? sub-tree) (scale-tree$ sub-tree factor (λ (scale-sub-tree) (c scale-sub-tree))) (c (* sub-tree factor)))) tree c)))
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