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But (lambda (s) (car s)) does not return car (the procedure) but the car of a pair. The notation (lambda (s) (car s)) shows the used operation in the mapping process more explicit.

(define (accumulate-n op init seqs) (if (null? (car seqs)) nil (cons (accumulate op init (map (lambda (s) (car s)) seqs)) (accumulate-n op init (map (lambda (s) (cdr s)) seqs))))) ;Maybe a helpful stepping stone. ;Example in Mit-Scheme: ;(user) => (map (lambda (s) (cdr s)) '((1 2 3) (4 5 6))) ;Value: ((2 3) (5 6))

I would first like to set the record straight: (lambda (s) (car s)) is just a substitute for car (the procedure), it is not an expression in and of itself which is evaluated otherwise this would violate the requirement of the map function, the first argument of which has to be proc (a procedure).

Anyway, using a map was not intuitive to me in this case while writing the solution (though it is in retrospect), hence the following solution:

(define (accumulate-n op init seqs) (if (null? (car seqs)) '() (cons (accumulate op init (append-first seqs)) (accumulate-n op init (append-rest seqs))))) ;auxiliary functions ;append the first element of each sub-list in seqs (define (append-first seqs) (define (iter items res) (if (null? items) res (iter (cdr items) (cons (car (car items)) res)))) (reverse (iter seqs '()))) ;append the cdr of each sub-list in seqs (define (append-rest seqs) (define (iter items res) (if (null? items) res (iter (cdr items) (cons (cdr (car items)) res))) ) (reverse (iter seqs '()))) (define (reverse items) (define (iter z res) (if (null? z) res (iter (cdr z) (cons (car z) res)))) (iter items '()))

As `map`, here recursive implementation should be better.

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I don't know what you mean by "it is *not* an expression in and of itself which is *evaluated*" since all expressions except special forms will be evaluated by evaluator. See "A lambda expression evaluates to a procedure" in https://standards.scheme.org/corrected-r7rs/r7rs-Z-H-6.html#TAG:__tex2page_index_116.

Maybe jz made a small mistake for "but (lambda (s) (car s)) is just a function that returns car ...". In a nutshell, `(lambda (s) (car s))` just does `(car s)` for *each element* when map, so it is same as `car`.

Accumulate-n.

Given a sequence of sequences, applies accumulate to the first item from each, then the next item of each, etc.

Subproblem: define a proc that returns the first item from each nested sequence, and another that returns the remaining parts (accumulate-n will accumulate the former, and call itself with the latter):

Originally I had (map (lambda (s) (car s)) sequence), but (lambda (s) (car s)) is just a function that returns car ...