Representing an amount of money with specific bill

Question

I want to write a function in Racket which takes an amount of money and a list of specific bill-values, and then returns a list with the amount of bills used of every type to make the given amount in total. For example (calc 415 (list 100 10 5 2 1)) should return '(4 1 1 0 0).

I tried it this way but this doesn't work :/ I think I haven't fully understood what you can / can't do with set! in Racket, to be honest.

(define (calc n xs)
  (cond ((null? xs) (list))
        ((not (pair? xs)) 
          (define y n) 
          (begin (set! n (- n (* xs (floor (/ n xs)))))
                 (list (floor (/ y xs))) ))
        (else (append (calc n (car xs))
                      (calc n (cdr xs))))))

Answer 1

Your procedure does too much and you use mutation which is uneccesary. If you split the problem up.

(define (calc-one-bill n bill)
  ...)

;; test 
(calc-one-bill 450 100) ; ==> 4
(calc-one-bill 450 50)  ; ==> 9

Then you can make:

(define (calc-new-n n bill amount)
  ...)

(calc-new-n 450 100 4) ; ==> 50
(calc-new-n 450 50 9)  ; ==> 0

Then you can reduce your original implememntation like this:

(define (calc n bills)
  (if (null? bills)
      (if (zero? n)
          '()
          (error "The unit needs to be the last element in the bills list"))
      (let* ((bill (car bills))
             (amount (calc-one-bill n bill)))
        (cons amount
              (calc (calc-new-n n bill amount)
                    (cdr bills))))))

This will always choose the solution with fewest bills, just as your version seems to do. Both versions requires that the last element in the bill passed is the unit 1. For a more complex method, that works with (calc 406 (list 100 10 5 2)) and that potentially can find all combinations of solutions, see Will's answer.

Answer 2

This problem calls for some straightforward recursive non-deterministic programming.

• We start with a given amount, and a given list of bill denominations, with unlimited amounts of each bill, apparently (otherwise, it'd be a different problem).
• At each point in time, we can either use the biggest bill, or not.
• If we use it, the total sum lessens by the bill's value.
• If the total is 0, we've got our solution!
• If the total is negative, it is invalid, so we should abandon this path.

The code here will follow another answer of mine, which finds out the total amount of solutions (which are more than one, for your example as well). We will just have to maintain the solutions themselves as well, whereas the code mentioned above only counted them.

We can code this one as a recursively backtracking procedure, with a callback to be called with each successfully found solution:

(define (change sum bills callback)
  (let loop ((sum sum) (sol '()) (bills bills))     ; "sol" for "solution"
    (cond
       ((zero? sum) (callback sol))
       ((< sum 0) #f)
       ((null? bills) #f)
       (else (apply
              (lambda (b . bs)
                (loop (- sum b) (cons b sol) bills) ; either use the first denomination,
                (loop sum sol bs))                  ;  or backtrack, and don't!
              bills)))))

It is to be called through one of

(define (first-solution proc . params)
  (call/cc (lambda (return)
             (apply proc (append params
                                 (list (lambda (sol) 
                                         (return sol))))))))

(define (n-solutions n proc . params)     ; n assumed an integer
  (let ([res '()])                        ; n <= 0 gets ALL solutions
    (call/cc (lambda (break)
               (apply proc (append params
                                   (list (lambda (sol)
                                           (set! res (cons sol res))
                                           (set! n (- n 1))
                                           (cond ((zero? n) (break)))))))))
    (reverse res)))

Testing,

> (first-solution change 406 (list 100 10 5 2))

'(2 2 2 100 100 100 100)

> (n-solutions 7 change 415 (list 100 10 5 2 1))

'((5 10 100 100 100 100)
  (1 2 2 10 100 100 100 100)
  (1 1 1 2 10 100 100 100 100)
  (1 1 1 1 1 10 100 100 100 100)
  (5 5 5 100 100 100 100)
  (1 2 2 5 5 100 100 100 100)
  (1 1 1 2 5 5 100 100 100 100))

---

Regarding how this code is structured, cf. How to generate all the permutations of elements in a list one at a time in Lisp? It creates nested loops with the solution being accessible in the innermost loop's body.

Regarding how to code up a non-deterministic algorithm (making all possible choices at once) in a proper functional way, see How to do a powerset in DrRacket? and How to find partitions of a list in Scheme.

Answer 3

I think this is the first program I wrote when learning FORTRAN! Here is a version which makes no bones about using everything Racket has to offer (or, at least, everything I know about). As such it's probably a terrible homework solution, and it's certainly prettier than the FORTRAN I wrote in 1984.

Note that this version doesn't search, so it will get remainders even when it does not need to. It never gets a remainder if the lowest denomination is 1, of course.

(define/contract (denominations-of amount denominations)
  ;; split amount into units of denominations, returning the split
  ;; in descending order of denomination, and any remainder (if there is
  ;; no 1 denomination there will generally be a remainder).
  (-> natural-number/c (listof (integer-in 1 #f))
      (values (listof natural-number/c) natural-number/c))
  (let handle-one-denomination ([current amount]
                                [remaining-denominations (sort denominations >)]
                                [so-far '()])
    ;; handle a single denomination: current is the balance,
    ;; remaining-denominations is the denominations left (descending order)
    ;; so-far is the list of amounts of each denomination we've accumulated
    ;; so far, which is in ascending order of denomination
    (if (null? remaining-denominations)
        ;; we are done: return the reversed accumulator and anything left over
        (values (reverse so-far) current)
        (match-let ([(cons first-denomination rest-of-the-denominations)
                     remaining-denominations])
          (if (> first-denomination current)
              ;; if the first denomination is more than the balance, just
              ;; accumulate a 0 for it and loop on the rest
              (handle-one-denomination current rest-of-the-denominations
                                       (cons 0 so-far))
              ;; otherwise work out how much of it we need and how much is left
              (let-values ([(q r)
                            (quotient/remainder current first-denomination)])
                ;; and loop on the remainder accumulating the number of bills
                ;; we needed
                (handle-one-denomination r rest-of-the-denominations
                                         (cons q so-far))))))))

Answer 4

I solved it this way now :)

(define (calc n xs)
  (define (calcAssist n xs usedBills)
    (cond ((null? xs) usedBills)
          ((pair? xs) 
            (calcAssist (- n (* (car xs) (floor (/ n (car xs))))) 
                        (cdr xs) 
                        (append usedBills 
                                (list (floor (/ n (car xs)))))))
          (else 
            (if ((= (- n (* xs (floor (/ n xs)))) 0)) 
               (append usedBills (list (floor (/ n xs))))
               (display "No solution")))))

  (calcAssist n xs (list)))

Testing:

> (calc 415 (list 100 10 5 2 1))
'(4 1 1 0 0)