How to find all combinations of coins when given some dollar value

I looked into this once a long time ago, and you can read my little write-up on it. Here’s the Mathematica source.

By using generating functions, you can get a closed-form constant-time solution to the problem. Graham, Knuth, and Patashnik’s Concrete Mathematics is the book for this, and contains a fairly extensive discussion of the problem. Essentially you define a polynomial where the nth coefficient is the number of ways of making change for n dollars.

Pages 4-5 of the writeup show how you can use Mathematica (or any other convenient computer algebra system) to compute the answer for 10^10^6 dollars in a couple seconds in three lines of code.

(And this was long enough ago that that’s a couple of seconds on a 75Mhz Pentium...)

I would favor a recursive solution. You have some list of denominations, if the smallest one can evenly divide any remaining currency amount, this should work fine.

Basically, you move from largest to smallest denominations.
Recursively,

1. You have a current total to fill, and a largest denomination (with more than 1 left). If there is only 1 denomination left, there is only one way to fill the total. You can use 0 to k copies of your current denomination such that k * cur denomination <= total.
2. For 0 to k, call the function with the modified total and new largest denomination.
3. Add up the results from 0 to k. That's how many ways you can fill your total from the current denomination on down. Return this number.

Here's my python version of your stated problem, for 200 cents. I get 1463 ways. This version prints all the combinations and the final count total.

#!/usr/bin/python

# find the number of ways to reach a total with the given number of combinations

cents = 200
denominations = [25, 10, 5, 1]
names = {25: "quarter(s)", 10: "dime(s)", 5 : "nickel(s)", 1 : "pennies"}

def count_combs(left, i, comb, add):
    if add: comb.append(add)
    if left == 0 or (i+1) == len(denominations):
        if (i+1) == len(denominations) and left > 0:
           if left % denominations[i]:
               return 0
           comb.append( (left/denominations[i], demoninations[i]) )
           i += 1
        while i < len(denominations):
            comb.append( (0, denominations[i]) )
            i += 1
        print(" ".join("%d %s" % (n,names[c]) for (n,c) in comb))
        return 1
    cur = denominations[i]
    return sum(count_combs(left-x*cur, i+1, comb[:], (x,cur)) for x in range(0, int(left/cur)+1))

count_combs(cents, 0, [], None)

Here's some absolutely straightforward C++ code to solve the problem which did ask for all the combinations to be shown.

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
    if (argc != 2)
    {
        printf("usage: change amount-in-cents\n");
        return 1;
    }

    int total = atoi(argv[1]);

    printf("quarter\tdime\tnickle\tpenny\tto make %d\n", total);

    int combos = 0;

    for (int q = 0; q <= total / 25; q++)
    {
        int total_less_q = total - q * 25;
        for (int d = 0; d <= total_less_q / 10; d++)
        {
            int total_less_q_d = total_less_q - d * 10;
            for (int n = 0; n <= total_less_q_d / 5; n++)
            {
                int p = total_less_q_d - n * 5;
                printf("%d\t%d\t%d\t%d\n", q, d, n, p);
                combos++;
            }
        }
    }

    printf("%d combinations\n", combos);

    return 0;
}

But I'm quite intrigued about the sub problem of just calculating the number of combinations. I suspect there's a closed-form equation for it.

Let C(i,J) the set of combinations of making i cents using the values in the set J.

You can define C as that:

(first(J) takes in a deterministic way an element of a set)

It turns out a pretty recursive function... and reasonably efficient if you use memoization ;)

semi-hack to get around the unique combination problem - force descending order:

$denoms = [1,5,10,25]
def all_combs(sum,last) 
  return 1 if sum == 0
  return $denoms.select{|d| d &le sum && d &le last}.inject(0) {|total,denom|
           total+all_combs(sum-denom,denom)}
end

This will run slow since it won't be memoized, but you get the idea.

Both: iterate through all denominations from high to low, take one of denomination, subtract from requried total, then recurse on remainder (constraining avilable denominations to be equal or lower to current iteration value.)

If the currency system allows it, a simple greedy algorithm that takes as many of each coin as possible, starting with the highest value currency.

Otherwise, dynamic programming is required to find an optimal solution quickly since this problem is essentially the knapsack problem.

For example, if a currency system has the coins: {13, 8, 1}, the greedy solution would make change for 24 as {13, 8, 1, 1, 1}, but the true optimal solution is {8, 8, 8}

Edit: I thought we were making change optimally, not listing all the ways to make change for a dollar. My recent interview asked how to make change so I jumped ahead before finishing to read the question.

Duh, I feel stupid right now. Below there is an overly complicated solution, which I'll preserve because it is a solution, after all. A simple solution would be this:

// Generate a pretty string
val coinNames = List(("quarter", "quarters"), 
                     ("dime", "dimes"), 
                     ("nickel", "nickels"), 
                     ("penny", "pennies"))
def coinsString = 
  Function.tupled((quarters: Int, dimes: Int, nickels:Int, pennies: Int) => (
    List(quarters, dimes, nickels, pennies) 
    zip coinNames // join with names
    map (t => (if (t._1 != 1) (t._1, t._2._2) else (t._1, t._2._1))) // correct for number
    map (t => t._1 + " " + t._2) // qty name
    mkString " "
  ))

def allCombinations(amount: Int) = 
 (for{quarters <- 0 to (amount / 25)
      dimes <- 0 to ((amount - 25*quarters) / 10)
      nickels <- 0 to ((amount - 25*quarters - 10*dimes) / 5)
  } yield (quarters, dimes, nickels, amount - 25*quarters - 10*dimes - 5*nickels)
 ) map coinsString mkString "\n"

Here is the other solution. This solution is based on the observation that each coin is a multiple of the others, so they can be represented in terms of them.

// Just to make things a bit more readable, as these routines will access
// arrays a lot
val coinValues = List(25, 10, 5, 1)
val coinNames = List(("quarter", "quarters"), 
                     ("dime", "dimes"), 
                     ("nickel", "nickels"), 
                     ("penny", "pennies"))
val List(quarter, dime, nickel, penny) = coinValues.indices.toList


// Find the combination that uses the least amount of coins
def leastCoins(amount: Int): Array[Int] =
  ((List(amount) /: coinValues) {(list, coinValue) =>
    val currentAmount = list.head
    val numberOfCoins = currentAmount / coinValue
    val remainingAmount = currentAmount % coinValue
    remainingAmount :: numberOfCoins :: list.tail
  }).tail.reverse.toArray

// Helper function. Adjust a certain amount of coins by
// adding or subtracting coins of each type; this could
// be made to receive a list of adjustments, but for so
// few types of coins, it's not worth it.
def adjust(base: Array[Int], 
           quarters: Int, 
           dimes: Int, 
           nickels: Int, 
           pennies: Int): Array[Int] =
  Array(base(quarter) + quarters, 
        base(dime) + dimes, 
        base(nickel) + nickels, 
        base(penny) + pennies)

// We decrease the amount of quarters by one this way
def decreaseQuarter(base: Array[Int]): Array[Int] =
  adjust(base, -1, +2, +1, 0)

// Dimes are decreased this way
def decreaseDime(base: Array[Int]): Array[Int] =
  adjust(base, 0, -1, +2, 0)

// And here is how we decrease Nickels
def decreaseNickel(base: Array[Int]): Array[Int] =
  adjust(base, 0, 0, -1, +5)

// This will help us find the proper decrease function
val decrease = Map(quarter -> decreaseQuarter _,
                   dime -> decreaseDime _,
                   nickel -> decreaseNickel _)

// Given a base amount of coins of each type, and the type of coin,
// we'll produce a list of coin amounts for each quantity of that particular
// coin type, up to the "base" amount
def coinSpan(base: Array[Int], whichCoin: Int) = 
  (List(base) /: (0 until base(whichCoin)).toList) { (list, _) =>
    decrease(whichCoin)(list.head) :: list
  }

// Generate a pretty string
def coinsString(base: Array[Int]) = (
  base 
  zip coinNames // join with names
  map (t => (if (t._1 != 1) (t._1, t._2._2) else (t._1, t._2._1))) // correct for number
  map (t => t._1 + " " + t._2)
  mkString " "
)

// So, get a base amount, compute a list for all quarters variations of that base,
// then, for each combination, compute all variations of dimes, and then repeat
// for all variations of nickels.
def allCombinations(amount: Int) = {
  val base = leastCoins(amount)
  val allQuarters = coinSpan(base, quarter)
  val allDimes = allQuarters flatMap (base => coinSpan(base, dime))
  val allNickels = allDimes flatMap (base => coinSpan(base, nickel))
  allNickels map coinsString mkString "\n"
}

So, for 37 coins, for example:

scala> println(allCombinations(37))
0 quarter 0 dimes 0 nickels 37 pennies
0 quarter 0 dimes 1 nickel 32 pennies
0 quarter 0 dimes 2 nickels 27 pennies
0 quarter 0 dimes 3 nickels 22 pennies
0 quarter 0 dimes 4 nickels 17 pennies
0 quarter 0 dimes 5 nickels 12 pennies
0 quarter 0 dimes 6 nickels 7 pennies
0 quarter 0 dimes 7 nickels 2 pennies
0 quarter 1 dime 0 nickels 27 pennies
0 quarter 1 dime 1 nickel 22 pennies
0 quarter 1 dime 2 nickels 17 pennies
0 quarter 1 dime 3 nickels 12 pennies
0 quarter 1 dime 4 nickels 7 pennies
0 quarter 1 dime 5 nickels 2 pennies
0 quarter 2 dimes 0 nickels 17 pennies
0 quarter 2 dimes 1 nickel 12 pennies
0 quarter 2 dimes 2 nickels 7 pennies
0 quarter 2 dimes 3 nickels 2 pennies
0 quarter 3 dimes 0 nickels 7 pennies
0 quarter 3 dimes 1 nickel 2 pennies
1 quarter 0 dimes 0 nickels 12 pennies
1 quarter 0 dimes 1 nickel 7 pennies
1 quarter 0 dimes 2 nickels 2 pennies
1 quarter 1 dime 0 nickels 2 pennies

This blog entry of mine solves this knapsack like problem for the figures from an XKCD comic. A simple change to the items dict and the exactcost value will yield all solutions for your problem too.

If the problem were to find the change that used the least cost, then a naive greedy algorithm that used as much of the highest value coin might well fail for some combinations of coins and target amount. For example if there are coins with values 1, 3, and 4; and the target amount is 6 then the greedy algorithm might suggest three coins of value 4, 1, and 1 when it is easy to see that you could use two coins each of value 3.

• Paddy.

/*
* make a list of all distinct sets of coins of from the set of coins to
* sum up to the given target amount.
* Here the input set of coins is assumed yo be {1, 2, 4}, this set MUST
* have the coins sorted in ascending order.
* Outline of the algorithm:
* 
* Keep track of what the current coin is, say ccn; current number of coins
* in the partial solution, say k; current sum, say sum, obtained by adding
* ccn; sum sofar, say accsum:
*  1) Use ccn as long as it can be added without exceeding the target
*     a) if current sum equals target, add cc to solution coin set, increase
*     coin coin in the solution by 1, and print it and return
*     b) if current sum exceeds target, ccn can't be in the solution, so
*        return
*     c) if neither of the above, add current coin to partial solution,
*        increase k by 1 (number of coins in partial solution), and recuse
*  2) When current denomination can no longer be used, start using the
*     next higher denomination coins, just like in (1)
*  3) When all denominations have been used, we are done
*/

#include <iostream>
#include <cstdlib>

using namespace std;

// int num_calls = 0;
// int num_ways = 0;

void print(const int coins[], int n);

void combine_coins(
                   const int denoms[], // coins sorted in ascending order
                   int n,              // number of denominations
                   int target,         // target sum
                   int accsum,         // accumulated sum
                   int coins[],        // solution set, MUST equal
                                       // target / lowest denom coin
                   int k               // number of coins in coins[]
                  )
{

    int  ccn;   // current coin
    int  sum;   // current sum

    // ++num_calls;

    for (int i = 0; i < n; ++i) {
        /*
         * skip coins of lesser denomination: This is to be efficient
         * and also avoid generating duplicate sequences. What we need
         * is combinations and without this check we will generate
         * permutations.
         */
        if (k > 0 && denoms[i] < coins[k - 1])
            continue;   // skip coins of lesser denomination

        ccn = denoms[i];

        if ((sum = accsum + ccn) > target)
            return;     // no point trying higher denominations now


        if (sum == target) {
            // found yet another solution
            coins[k] = ccn;
            print(coins, k + 1);
            // ++num_ways;
            return;
        }

        coins[k] = ccn;
        combine_coins(denoms, n, target, sum, coins, k + 1);
    }
}

void print(const int coins[], int n)
{
    int s = 0;
    for (int i = 0; i < n; ++i) {
        cout << coins[i] << " ";
        s += coins[i];
    }
    cout << "\t = \t" << s << "\n";

}

int main(int argc, const char *argv[])
{

    int denoms[] = {1, 2, 4};
    int dsize = sizeof(denoms) / sizeof(denoms[0]);
    int target;

    if (argv[1])
        target = atoi(argv[1]);
    else
        target = 8;

    int *coins = new int[target];


    combine_coins(denoms, dsize, target, 0, coins, 0);

    // cout << "num calls = " << num_calls << ", num ways = " << num_ways << "\n";

    return 0;
}

Here's a C# function:

    public static void change(int money, List<int> coins, List<int> combination)
    {
        if(money < 0 || coins.Count == 0) return;
        if (money == 0)
        {
            Console.WriteLine((String.Join("; ", combination)));
            return;
        }

        List<int> copy = new List<int>(coins);
        copy.RemoveAt(0);
        change(money, copy, combination);

        combination = new List<int>(combination) { coins[0] };
        change(money - coins[0], coins, new List<int>(combination));
    }

Use it like this:

change(100, new List<int>() {5, 10, 25}, new List<int>());

It prints:

25; 25; 25; 25
10; 10; 10; 10; 10; 25; 25
10; 10; 10; 10; 10; 10; 10; 10; 10; 10
5; 10; 10; 25; 25; 25
5; 10; 10; 10; 10; 10; 10; 10; 25
5; 5; 10; 10; 10; 10; 25; 25
5; 5; 10; 10; 10; 10; 10; 10; 10; 10; 10
5; 5; 5; 10; 25; 25; 25
5; 5; 5; 10; 10; 10; 10; 10; 10; 25
5; 5; 5; 5; 10; 10; 10; 25; 25
5; 5; 5; 5; 10; 10; 10; 10; 10; 10; 10; 10
5; 5; 5; 5; 5; 25; 25; 25
5; 5; 5; 5; 5; 10; 10; 10; 10; 10; 25
5; 5; 5; 5; 5; 5; 10; 10; 25; 25
5; 5; 5; 5; 5; 5; 10; 10; 10; 10; 10; 10; 10
5; 5; 5; 5; 5; 5; 5; 10; 10; 10; 10; 25
5; 5; 5; 5; 5; 5; 5; 5; 10; 25; 25
5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 10; 10; 10; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 10; 25
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 25; 25
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 10; 10; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 25
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 10; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 25
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 25
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5

Below is a python program to find all combinations of money. This is a dynamic programming solution with order(n) time. Money is 1,5,10,25

We traverse from row money 1 to row money 25 (4 rows). Row money 1 contains the count if we only consider money 1 in calculating the number of combinations. Row money 5 produces each column by taking the count in row money r for the same final money plus the previous 5 count in its own row (current position minus 5). Row money 10 uses row money 5, which contains counts for both 1,5 and adds in the previous 10 count (current position minus 10). Row money 25 uses row money 10, which contains counts for row money 1,5,10 plus the previous 25 count.

For example, numbers[1][12] = numbers[0][12] + numbers[1][7] (7 = 12-5) which results in 3 = 1 + 2; numbers[3][12] = numbers[2][12] + numbers[3][9] (-13 = 12-25) which results in 4 = 0 + 4, since -13 is less than 0.

def cntMoney(num):
    mSz = len(money)
    numbers = [[0]*(1+num) for _ in range(mSz)]
    for mI in range(mSz): numbers[mI][0] = 1
    for mI,m in enumerate(money):
        for i in range(1,num+1):
            numbers[mI][i] = numbers[mI][i-m] if i >= m else 0
            if mI != 0: numbers[mI][i] += numbers[mI-1][i]
        print('m,numbers',m,numbers[mI])
    return numbers[mSz-1][num]

money = [1,5,10,25]
    num = 12
    print('money,combinations',num,cntMoney(num))

output:    
('m,numbers', 1, [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1])
('m,numbers', 5, [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3])
('m,numbers', 10, [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 4, 4, 4])
('m,numbers', 25, [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 4, 4, 4])
('money,combinations', 12, 4)

The following one-liner in Mathematica produces the desired result

parts[n_]:=Flatten[Table[{(n-i)/25,(i-j)/10,(j-k)/5,k},{i,n,0,-25},{j,i,0,-10},{k,j,0,-5}],2]

The table trivially builds up all combinations by iterating through all ways of how many coins of a bigger kind fit into n, while repeating the same steps with coins of smaller kind for the remainder. The implementation should be similarly trivial in any other programming language.

The output is a list of elements with the notation:

{number of quarters, number of dimes, number of nickels, number of cents}

The intuition behind the construction can be read off of an example:

parts[51]

I implemented this puzzle in C#, I think it's kind of different from other answers. In addition I added comments to make my algorithm more understandable. It was a very nice puzzle.

Start from larger coin and look on all Q,D,N and fill up the remainder with pennies.

So I can start with having 0 quarter, 0 dime and 0 nickel and the rest with pennies

For each coin, I have ($Amount / $Change) + 1, which $Amount is the input parameter and $Change is [Q]uarter or [D]ime or [N]ickel. So, assuming the input parameter is $1,

• I have ($1 / 0.25) + 1 = 5 choices for Quarter (0, 1, 2, 3, 4) -- qLoop variable in my code
• For the same $1, I will have ($1 / 0.10) + 1 = 11 options for Dime (0, 1, 2 ,3, 4, 5, 6, 7, 8, 9, 10) -- dLoop variable in my code
• The same for Nickel, I will have ($1 / 0.05) + 1 = 21 option (0 through 20) -- nLoop variable in my code

Note that in each loop, I have to use the reminder from outer loop.

As an example, if I pick 1 quarter, (where q is 1), I have $1 - 0.25 left for the inner loop (Dime loop) and so on

static void CoinCombination(decimal A)
{
    decimal[] coins = new decimal[] { 0.25M, 0.10M, 0.05M, 0.01M };

    // Loop for Quarters
    int qLoop = (int)(A / coins[0]) + 1;
    for (int q = 0; q < qLoop; q++)
    {
        string qS = $"{q} quarter(s), ";
        decimal qAmount = A - (q * coins[0]);
        Console.Write(qS);

        // Loop for Dimes
        int dLoop = (int)(qAmount / coins[1]) + 1;
        for (int d = 0; d < dLoop; d++)
        {
            if (d > 0)
                Console.Write(qS);
            string dS = $"{d} dime(s), ";
            decimal dAmount = qAmount - (d * coins[1]);
            Console.Write(dS);

            // Loop for Nickels
            int nLoop = (int)(dAmount / coins[2]) + 1;
            for (int n = 0; n < nLoop; n++)
            {
                if (n > 0)
                    Console.Write($"{qS}{dS}");
                string nS = $"{n} nickel(s), ";
                decimal nAmount = dAmount - (n * coins[2]);
                Console.Write(nS);

                // Fill up with pennies the remainder
                int p = (int)(nAmount / coins[3]);
                string pS = $"{p} penny(s)";
                Console.Write(pS);

                Console.WriteLine();
            }
            Console.WriteLine();
        }
        Console.WriteLine();
    }
}

Output