15 hours ago

my attempt to do the exercises in sicp.

## Thursday, July 17, 2008

### sicp exercise 2.36

; Exercise 2.36. The procedure accumulate-n is similar to accumulate except that it takes as its third argument a sequence of sequences, which are all assumed to have the same number of elements. It applies the designated accumulation procedure to combine all the first elements of the sequences, all the second elements of the sequences, and so on, and returns a sequence of the results. For instance, if s is a sequence containing four sequences, ((1 2 3) (4 5 6) (7 8 9) (10 11 12)), then the value of (accumulate-n + 0 s) should be the sequence (22 26 30). Fill in the missing expressions in the following definition of accumulate-n:

; (define (accumulate-n op init seqs)

; (if (null? (car seqs))

; nil

; (cons (accumulate op init <??>)

; (accumulate-n op init <??>))))

(define (accumulate op init seq)

(cond ((null? seq) init)

(else (op (car seq) (accumulate op init (cdr seq))))))

(define (accumulate-n op init seqs)

(if (null? (car seqs)) (list)

(cons (accumulate op init (map car seqs)) (accumulate-n op init (map cdr seqs)))))

(display (accumulate + 0 (list 3 4 5 6))) (newline)

(display (accumulate-n + 0 (list (list 1 2 3) (list 4 5 6) (list 7 8 9) (list 10 11 12)))) (newline)

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