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copied!<p>I just today finished this question in SICP, so I'll try to help here:</p> <p>I tested your code with this my-remove:</p> <pre><code>(define (my-remove num num-list) (cond ((null? num-list) nil) ((= num (car num-list)) (cdr num-list)) (else (cons (car num-list) (my-remove num (cdr num-list)))))) </code></pre> <p>tested with:</p> <pre><code>(let ((result (queens 8))) (display (length result))(newline) (map (lambda (possible-solution) (display possible-solution)(newline)) result) ) </code></pre> <p>returned 92 . . (3 7 2 8 5 1 4 6) (3 7 2 8 6 4 1 5) . .</p> <p>92 results, one of which is the solution in the book</p> <p>Here's a part of your code (slightly) refactored, I got rid of my-remove, and used a let*. I'm not going to refactor the whole thing, but I atleast tried to add meaning in the variable names. with these changes it was somewhat faster: (also added some comments)</p> <pre><code>;no major refactoring, essentially removed my-remove and some refactors (define (queens board-size) (define (safe? k position) (define (iter last-element front-lst col-num k) ;what's ok? (define (ok? l-e car-lst) (and (not (= l-e car-lst)) (not (= (abs (- l-e car-lst)) (abs (- k col-num)))))) (if (null? front-lst) true (and (ok? last-element (car front-lst)) (iter last-element (cdr front-lst) (++ col-num) k)))) ;queens means a list of rows, each representing a queen position ;that is, when queens[col] = row that means there's a queen in row 'row and ;column 'col ;reimplementing removing the queen, and the let turned to let* (let* ((reverse-column-queens (my-reverse position)) (last-queen (car reverse-column-queens)) ;remove that queen from the list ;btw 'position is actually 'positions (all-other-queens (my-reverse (cdr reverse-column-queens)))) (iter last-queen all-other-queens 1 k))) . . . </code></pre> <p>so, I guess the my-remove you used might have been wrong.</p> <p>Now, the names of your variables are not very discriptive, so it would add to an already confusing code (no offence meant to the author of this exercise or you).</p>

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