问题
The following is code to a problem that I have been working on for a few days. The problem I encountered is that for some reason when I call:
(apa-multi '(7 3 1 2) '(6 1 4))
the return is:
'(4 8 9 5 6 8)
The answer that it should output is
'(4 4 8 9 5 6 8)
When I call:
(apa-multi '(3 1 2) '(6 1 4))
The output is:
'(1 9 1 5 6 8)
which is correct.
I have debugged my code multiple times, and I can't seem to find out what the problem is (by the way, I know that the "remove-empty" function that I wrote is most likely unnecessary). Can anyone tell me where I am going wrong here? (My goal for this problem is to keep the arbitrary precision numbers in list format, and I can not create a function that converts numbers from list->num or num->list.) I believe that I have provided all of the necessary code for someone to work out what I was going for, but if not please let me know. The hint that I have for this is that " Multiplication of d = dndn−1 ...d1 by e = emem−1 ...e1 can be carried out by the rule de=d∗e1 +10∗(d∗em em−1...e2).)"
(define (remove-empty L)
(define (remove-empty-h L accum)
(cond ((null? L) accum)
((null? (car L))
(remove-empty (cdr L)))
(else (cons (car L) (remove-empty-h (cdr L) accum)))))
(remove-empty-h L '()))
(define (apa-add lst1 lst2)
(define (apa-add-h lst1 lst2 carry)
(cond ((and (null? lst1) (null? lst2))
(if (not (= 0 carry))
(list carry)
'()))
((null? lst1) (append (apa-add-h lst1 '() carry)
(list (+ (car (reverse-l lst2)) carry))
(reverse-l(cdr (reverse-l lst2)))))
((null? lst2) (append (apa-add-h '() lst2 carry)
(list (+ (car (reverse-l lst1)) carry)))
(reverse-l(cdr (reverse-l lst1))))
(else
(append (apa-add-h (cdr lst1) (cdr lst2) (quotient (+ (car lst1) (car lst2) carry) 10))
(list (modulo (+ (car lst1) (car lst2) carry) 10))))))
(apa-add-h (reverse-l lst1) (reverse-l lst2) 0))
(define (d-multiply lst factor)
(define (d-multiply-h lst factor carry)
(cond ((null? lst) (if (= carry 0)
'()
(list carry)))
((>= (+ (* (car lst) factor) carry) 10)
(append ;(list (check-null-and-carry-mult lst carry))
(d-multiply-h (cdr lst) factor (quotient (+ (* (car lst) factor) carry) 10))
(list (modulo (+ (* (car lst) factor) carry) 10))))
(else (append ;(list (check-null-and-carry-mult lst carry))
(d-multiply-h (cdr lst) factor (quotient(+ (* (car lst) factor) carry) 10))
(list (+ (* (car lst) factor) carry))))))
(remove-empty (d-multiply-h (reverse-l lst) factor 0)))
(define (nlength l)
(if (null? l)
0
(+ 1 (nlength (cdr l)))))
(define (apa-multi d e)
(define temp '())
(cond ((= (max (nlength e) (nlength d)) (nlength e))
(set! temp e)
(set! e d)
(set! d temp))
(else
(set! temp d)
(set! d e)
(set! e temp)))
(define (apa-multi-h d e)
(cond ((null? e) (list 0))
(else (append (apa-add (d-multiply d (car e))
(append (apa-multi-h d (cdr e)) (list 0)))))))
(apa-multi-h d (reverse-l e)))
回答1:
Not sure why it doesn't work, all those appends and reverses are hard to follow, and not sure what's going on with all that set! stuff. Putting the state into a tail call is a lot easier to follow and usually more efficient to boot.
(define (apa-add . Lists)
(define (cdrs-no-null L)
(cond ((null? L) '())
((null? (cdar l)) (cdrs-no-null (cdr L)))
(else (cons (cdar l) (cdrs-no-null (cdr l))))))
(let loop ((carry 0) (Lists (map reverse Lists)) (sum '()))
(if (null? Lists)
(if (zero? carry) sum (cons carry sum))
(loop (quotient (fold + carry (map car Lists)) 10)
(cdrs-no-null Lists)
(cons (modulo (fold + carry (map car Lists)) 10) sum)))))
(define (apa-mult . Lists)
(define (mult-by-factor n order L)
(let loop ((order order) (L (reverse L)) (carry 0) (sum '()))
(cond ((> order 0) (loop (- order 1) L carry (cons 0 sum)))
((null? L) (if (zero? carry)
sum
(cons carry sum))) ;;bug here if carry > 9
(else (loop 0
(cdr L)
(quotient (+ carry (* n (car L))) 10)
(cons (modulo (+ carry (* n (car L))) 10) sum))))))
(define (apa-mult2 L1 L2)
(let ((rL1 (reverse L1))
(rL2 (reverse L2))
(zip-with-order
(lambda (L)
(let loop ((order 0) (L L) (accum '()))
(if (null? L)
accum
(loop (+ 1 order)
(cdr L)
(cons (cons (car L) order) accum)))))))
(fold apa-add '(0) (map (lambda (x)
(mult-by-factor (car x) (cdr x) L2))
(zip-with-order rl1)))))
(fold apa-mult2 '(1) Lists)))
(apa-mult '(3 1 2) '(6 1 4)))
;Value 7: (1 9 1 5 6 8)
(apa-mult '(2 0 0) '(3 1 2) '(6 1 4))
;Value 8: (3 8 3 1 3 6 0 0)
(apa-mult '(7 3 1 2) '(6 1 4))
;Value 9: (4 4 8 9 5 6 8)
回答2:
The reason your code does not work is because your apa-add is broken. For example:
> (apa-add '(7 3 1 2) '(6 1 4))
'(9 2 6)
> (+ 7312 614)
7926
The rest of your code seems to work, at least for your 2 examples, if you use a working apa-add.
I admit I did not try to understand your code fully; the poor formatting and the set! procedures at the end made me want to start from scratch. So even if you could simple correct your apa-add, maybe have a look at my version anyway, because it is way shorter and probably easier to understand.
Building on my previous answer for apa-add multiplication is a matter of apa-adding, multiplying one list by a digit at a time, and adding zeroes at the end of the intermediary multiplications just as you'd do it manually:
(define (car0 lst) (if (empty? lst) 0 (car lst)))
(define (cdr0 lst) (if (empty? lst) empty (cdr lst)))
(define (apa-add l1 l2) ; apa-add (see https://stackoverflow.com/a/19597007/1193075)
(let loop ((l1 (reverse l1)) (l2 (reverse l2)) (carry 0) (res '()))
(if (and (null? l1) (null? l2) (= 0 carry))
res
(let* ((d1 (car0 l1)) (d2 (car0 l2)) (ad (+ d1 d2 carry)) (dn (modulo ad 10)))
(loop (cdr0 l1) (cdr0 l2) (quotient (- ad dn) 10) (cons dn res))))))
(define (mult1 n lst) ; multiply a list by one digit
(let loop ((lst (reverse lst)) (carry 0) (res '()))
(if (and (null? lst) (= 0 carry))
res
(let* ((c (car0 lst)) (m (+ (* n c) carry)) (m0 (modulo m 10)))
(loop (cdr0 lst) (quotient (- m m0) 10) (cons m0 res))))))
(define (apa-multi l1 l2) ; full multiplication
(let loop ((l2 (reverse l2)) (app '()) (res '()))
(if (null? l2)
res
(let* ((d2 (car l2)) (m (mult1 d2 l1)) (r (append m app)))
(loop (cdr l2) (cons '0 app) (apa-add r res))))))
来源：https://stackoverflow.com/questions/19609519/multiplication-of-arbitrary-precision-numbers-in-scheme