# Solving the change-counting problem using generating functions

The change-counting problem is a standard example of a dynamic programming problem and frequently appears in undergraduate computer science courses and in job interviews. One setup is that there is some number of distinct denominations of coins, and, given a particular amount of change to dispense, we would like to count the number of ways to make that change. For instance, if there are 1- and 3-cent pieces, there are three ways to make 6 cents in change: all 1-cent pieces, one 3-cent piece and three 1-cent pieces, and two 3-cent pieces. A common variation is only allowing up to a certain number of a particular denomination, and another part which varies is whether only a particular amount is considered, or whether it is all amounts up to a given amount. And one other variation is whether we want to find the way to use the least number of coins, but we will not consider this.

This also happens to be a standard example for generating functions, and one thing I have not seen before is to make a program which implements the generating function. (Source code: coins.py)

For the simple case of a single coin with denomination `d`, the
series `f`_{d}(`x`) = 1 + `x`^{d} + `x`^{2d} + ⋯ = 1/(1 − `x`^{d}) counts the number of
ways to dispense change. The coefficient in front of `x`^{n} gives
the number of ways of dispensing `n` cents, and here there are zero
ways unless `n` is a multiple of `d`, in which case the only way to
make change is with `n`/`d` of the `d`-cent pieces.

A nice fact about generating functions is that to count the number
of ways to make a particular sum `a` + `b` = `n`, where `a` and `b` are
counted by respective generating functions `f`(`x`) and `g`(`x`), you
just multiply the generating functions. (This is because
`x`^{a}`x`^{b} = `x`^{a + b}.) So, the generating function for the
change-counting problem is

`f`(

`x`) =

1 |

1 − x^{d1} |

1 |

1 − x^{d2} |

1 |

1 − x^{dk} |

where `d`_{1}, ⋯, `d`_{k} are the denominations of the coins.

This can be calculated by repeated division of 1 by each of the denominators, and this can be effectively implemented using Python generators (or Haskell lists). We will represent an infinite series as an infinite generator consisting of the series coefficients. The first ingredient is a way to convert polynomials (finite lists) into series.

def series_from_poly(p): """Takes a polynomial (a list of coefficients, least to greatest power) and returns a series representation (an infinite generator of coefficients).""" yield from p while True: yield 0The next ingredient is a function which produces the series obtained by dividing a series by a polynomial, which is a straightforward implementation of long division. Our series involve only integer coefficients, so to get full precision we use integer division (and the assert can be uncommented to verify it).

def pdivide(f, g) : """f is a generator of series coefficients. g is a polynomial.""" g = list(g) r = [next(f) for i in range(len(g)-1)] r.append(0) # this is the location "bringing down" the next term of f goes for fj in f: r[-1] = fj q0 = r[0]//g[0] #assert r[0]-q0*g[0] == 0 # we know all terms _must_ be integers for this problem yield q0 # now do r[0:-2] = r[1:] - q0*g[1:]. # (note r[0]-q0*g[0] == 0, and gets dropped off front of remainder) for i in range(len(g)-1): r[i] = r[i+1]-q0*g[i+1]

To test this, here is a utility function to take a certain number of terms from a series.

def take(n, f): """Gives an nth order polynomial approximation of the series f.""" from itertools import islice return list(islice(f, 0, n+1))Some useful series to test are geometric series and the Fibonacci sequence, whose generating function is 1/(1 −

`x`−

`x`

^{2}).

def geom_series(a): """return 1/(1-ax)""" return pdivide(series_from_poly([1]), [1,-a]) def geom_series_pow(a,n): """return 1/(1-ax^n)""" g = [1]+[0]*(n-1)+[-a] return pdivide(series_from_poly([1]), g) def fib_series(): return pdivide(series_from_poly([1]), [1,-1,-1])For example,

>>> take(10, geom_series(2)) [1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024] >>> take(10, geom_series_pow(2,2)) [1, 0, 2, 0, 4, 0, 8, 0, 16, 0, 32] >>> take(10, fib_series()) [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]

The coin series is just a “right fold,” but instead of messing
around with `reduce`, we’ll just implement it directly.

def coin_series(coins): """Coins is a list of denominations. Returns the generating function for the number of ways to dispense a particular amount of change. 1/(1-x^d1) * 1/(1-x^d2) * ... * 1/(1-x^dn).""" if not coins: return series_from_poly([1]) else: d = coins[0] return pdivide(coin_series(coins[1:]), [1]+[0]*(d-1)+[-1])The

`[1]+[0]*(d-1)+[-1]`gives the coefficients to the polynomial 1 −

`x`

^{d}. Finally, the solution to the change-counting problem is to obtain the

`n`th coefficient of the series. The use of

`islice`is just a way to skip to the term we care about.

def dispense(coins, amount): """Returns the number of ways to dispense a certain amount given a list of allowed denominations.""" from itertools import islice return next(islice(coin_series(coins), amount, None))So, the number of ways to dispense 100 cents using U.S. currency (excluding the dollar coin) is

>>> dispense([1,5,10,25,50], 100) 292

Taking stock in the solution, we can see that each divider keeps track
of `d`_{i} numbers for the current remainder, and that for each term of
`coin_series` they are all iterated over once, so the running
time is `O`(`n``d`_{1}`d`_{2}⋯`d`_{k}), where `n` is desired amount of change.

## 1. Questions

- Can this approach be used to find the largest amount which cannot be dispensed with the given coins? For example, 5-cent and 7-cent coins, 23 cents cannot be dispensed, but you can always dispense 24 cents or more. (This is known as the Frobenius coin problem, or the McNuggest problem. It seems the problem is NP-hard.)
- Can a more traditional dynamic programming solution do better
than
`O`(`n``d`_{1}`d`_{2}⋯`d`_{k})? Is it easier to understand? - It is possible to calculate a particular Fibonacci number in
logarithmic time by doing fast exponentiation of the transition
matrix
`[[0,1],[1,1]]`. Does the coin dispensing problem admit a similar speedup? (My guess for the worst case is`O`(`μ`(`m`)log`n`), where`n`is the amount,`m`=`d`_{1}⋯`d`_{k}, and`μ`(`m`) is the running time for`m`×`m`matrix multiplication.) - Can generating functions be used to find the way to dispense
`n`coins with the least number of coins?

## 2. In Mathematica

Of course, Mathematica is particularly well-disposed for such problems, since we can just ask it for a particular coefficient in the series.

changegen[denoms_] := Times @@ ((1/(1 - x^#)) & /@ denoms); changeways[n_, denoms_] := SeriesCoefficient[changegen[denoms], {x, 0, n}]; changeways[100, {1, 5, 10, 25, 50}]And Mr. Wolfram’s Language agrees with the previous calculation: there are 292 ways to make change.