What This Country Needs is an 18 Piece∗
Jeffrey Shallit
Department of Computer Science
University of Waterloo
Waterloo, Ontario, Canada N2L 3G1
shallit@graceland.uwaterloo.ca
May 26, 2003

Abstract
We consider sets of coin denominations which permit change to be made using as
few coins as possible, on average, and explain why the United States should adopt an18 piece.
1Introduction

Most businesses in the United States currently make change using four different types ofcoins: 1 (cent),1 5 (nickel), 10 (dime), and 25 (quarter). For people who make change
on a daily basis, it is desirable to make change in as efficient a manner as possible. One
criterion for efficiency is to use the smallest number of coins. For example, to make change
for 30 , one could, at least in principle, give a customer 30 1-cent coins, but most would
probably prefer receiving a quarter and a nickel.
Formally, we can define the optimal representation problem as follows: given a set of D
integer denominations e1 < e2 < · · · < eD and an
P integer N ≥ 0, we wish to express N asa non-negative
integer linear combination N = 1≤i≤D ai ei such that the number of coinsPS = 1≤i≤D ai is minimized. In order that every number actually have a representation, wedemand that e1 = 1. If (a1 , a2 , . . . , aD ) is the D-tuple that minimizes S, then we say it is an
optimal representation, and we define opt(N ; e1 , e2 , . . . , eD ) := S.
The optimal denomination problem is to find denominations that minimize the average
cost of making change. We assume that every amount of change between 0 and 99 isequally likely.2 We then ask, what choice of D denominations minimizes the average number



∗



“What this country needs is a really good five-cent cigar.” T. R. Marshall (US Vice-President), NewYork Tribune, January 4, 1920.
1
Informally, a 1-cent coin is usually called a “penny”, but this usage is frowned upon by numismatists.
2This assumption is probably inaccurate for several reasons, not least being the fact that many items
have prices that end in the digit 9. Also, Benford’s law may play a role; see Raimi [7].

1of coins needed to make change? More formally, solving the optimal denomination problem
for D denominations up to the limit L means determining the denominations e1 , e2 , . . . , eD
which minimize
1 X
opt(i; e1 , e2 , . . . , eD ).
cost(L; e1 , e2 , . . . , eL ) :=L 0≤i<L
For the current system, where (e1 , e2 , e3 , e4 ) = (1, 5, 10, 25), a simple computation determines that cost(100; 1, 5, 10, 25) = 4.7. In other words, on average a change-maker must
return 4.7 coins with every transaction.
Can we do better? Indeed we can. There are exactly two sets of four denominations thatminimize cost(100; e1 , e2 , e3 , e4 ); namely, (1, 5, 18, 25) and (1, 5, 18, 29). Both have an average
cost of 3.89. We would therefore gain about 17% efficiency in change-making by switching
to either of these four-coin systems. The first system, (1, 5, 18, 25), possesses the notableadvantage that we only need make one small alteration in the current system: replace thecurrent 10 coin with a new 18 coin. This explains the title of this article.
Figure 1 gives the optimal denominations of size D for 1 ≤ D ≤ 7, and their associatedcosts.


D
1234

(e1 , . . . , eD )(1)
(1, 10)(1, 11)(1, 12, 19)(1, 5, 18, 25)(1, 5, 18, 29)(1, 5, 16, 23, 33)(1, 4, 6, 21, 30, 37)(1, 5, 8, 20, 31, 33)(1, 4, 9, 11, 26, 38, 44)567

cost(100; e1 , . . . , eD )
49.5
9
5.15
3.89
3.29
2.92
2.65Figure 1: Optimal denominations for change-making for 1 ≤ D ≤ 7 denominations
Although the system (1, 5, 18, 25) would be superior to the current (1, 5, 10, 25) for changemaking, it may be difficult to convince people to accept the removal of the popular dime.
Thus it may be worthwhile to consider a different question: what single denomination could
we add to (1, 5, 10, 25) to achieve the maximum improvement in cost? The unique answer
is 32 ; this improves cost(100; 1, 5, 10, 25) = 4.7 to cost(100; 1, 5, 10, 25, 32) = 3.46. If wealso allow the infrequently-used 50 piece as a legitimate denomination, then the maximum
improvement comes from adding an 18 piece; this improves cost(100; 1, 5, 10, 25, 50) = 4.2
to cost(100; 1, 5, 10, 18, 25, 50) = 3.18. Yet another reason to add an 18 piece to US coinage!
Other countries provide different problems. In Canada, the coin denominations currently
in wide circulation are 1 , 5 , 10 , 25 , 100 (called a “loonie” for the loon on the reverse),and 200 (called a “toonie”). The smallest denomination of paper money in wide circulation

2

is a 5 bill. Assuming each amount of change between 0 and 499 is equally likely, the average cost of making change in Canada is cost(500; 1, 5, 10, 25, 100, 200) = 5.9. This can bebest improved by adding an 83 coin; we have cost(500; 1, 5, 10, 25, 83, 100, 200) = 4.578. On
the other hand, the new system of Euros introduced in Europe provides coins of denomination.01, .02, .05, .1, .2, .5, 1, and 2 Euros. For this system we have cost(500; 1, 2, 5, 10, 20, 50, 100, 200) =4.6. This can be best improved (to average cost 3.92) by adding a coin of denomination 1.33
or 1.37 Euros.




2

Greedy methods for change-making

One nice feature of the current set of US denominations (1, 5, 10, 25) is that the greedy
algorithm determines the representation with the minimum number of coins. By the greedy
algorithm, I mean the following procedure: given a number N to be represented as a nonnegative integer linear combination of denominations e1 < e2 < · · · < eD , take as many copies
aD of the largest denomination eD as possible, so that aD eD ≤ N . Then set N := N − aD eD
and continue the procedure with the remaining smaller denominations. Use of the greedy
algorithm provides a simple, easily-remembered method for making change. Not all sets ofdenominations have the property that the greedy method always determines the optimalrepresentation. For example, with denominations (1, 7, 10) the greedy algorithm gives therepresentation 14 = 10 + 1 + 1 + 1 + 1, whereas 7 + 7 uses fewer coins.
Unfortunately, none of the optimal sets of denominations in Figure 1 for D ≥ 3 give
optimal representations when used greedily. For example, when we try to greedily makechange for 24 using the system (1, 12, 19), we get 19 + 1 + 1 + 1 + 1 + 1, a far cry from theoptimal representation 12 + 12.
This suggests considering a variation on the optimal denomination problem, where costis replaced by the analogous function gcost, and we count only the cost of greedy representations. For the current system we still have gcost(100; 1, 5, 10, 25) = 4.7. Figure 2 displays
the results for optimal sets of denominations. An asterisk denotes an optimal set for which
the greedy representation is always an optimal representation.
We also might consider what single denomination could be added to the current USsystem (1, 5, 10, 25) to best improve the greedy cost. It turns out that adding either a 2 piece or a 3 -piece improves gcost(100; −) from 4.7 to 3.9, and this is the best possible 1-coinimprovement. It is interesting to note that the US actually had a 2 -piece from 1864 to 1873,and two different 3 -pieces: one made in silver from 1851 to 1873, and one made in nickelfrom 1865 to 1889.








3

Computational questions

So far we have focused on systems particular to the US, Canada, and Europe, but a goodmathematician will want more general results. Let us examine the computational complexityof the problems we have studied, and some related ones.
1. Suppose we are given an amount of change to make, say N , and a system of denomi3

D1234567(e1 , . . . , eD )(1) ∗(1, 10) ∗(1, 11) ∗(1, 5, 22) ∗(1, 5, 23) ∗(1, 3, 11, 37) ∗(1, 3, 11, 38) ∗(1, 3, 7, 16, 40)(1, 3, 7, 16, 41)(1, 3, 7, 18, 44) ∗(1, 3, 7, 18, 45)(1, 3, 8, 20, 44) ∗(1, 3, 8, 20, 45)(1, 2, 5, 11, 25, 62)(1, 2, 5, 11, 25, 63)(1, 2, 5, 13, 29, 64)(1, 2, 5, 13, 29, 65)(1, 2, 5, 8, 17, 27, 63)..
.

gcost(100; e1 , . . . , eD )
49.5
9
5.26
4.1
3.46

3.13

2.86

(27 other sets omitted)..
.
(1, 2, 5, 8, 19, 30, 63) ∗(1, 2, 5, 8, 19, 30, 64)(1, 2, 5, 8, 19, 30, 66) ∗(1, 2, 5, 8, 19, 30, 67)
Figure 2: Optimal denominations for greedy change-making
nations, say 1 = e1 < e2 < · · · <PeD . How easy is it to compute opt(N ; eP
1 , e2 , . . . , eD ) or findan optimal representation N = 1≤i≤D ai ei , i.e., one which minimizes 1≤i≤D ai ?
The answer depends on how N and the ei are written down. If they are written in
ordinary decimal notation, or in binary, then there is no fast algorithm known to solve thisproblem. In fact, it follows easily from results of Lueker [5] that this problem is NP-hard;roughly speaking this means it is at least as hard as many famous combinatorial problems,such as the travelling salesman problem, for which no polynomial-time algorithm is currently
known.
If, on the other hand, N and the ei are represented in unary, then a simple dynamic
programming algorithm (e.g., [10]) solves the optimal representation problem in polynomialtime.
2. Suppose we are given N and a system of denominations. How easy is it to determine
if the greedy representation for N is actually optimal? Kozen and Zaks [4] have shown that4this problem is co-NP-complete if the data is provided in ordinary decimal, or binary. Thisstrongly suggests there is no efficient algorithm for this problem.
3. Suppose we are given a system of denominations. How easy is it to decide whether thegreedy algorithm always produces an optimal representation, for all values of N ? It turns
out that this problem can be solved efficiently; this surprising result is due to Pearson [6].
Since Pearson’s result appeared only in an obscure technical report, we give a few details.
Suppose the greedy algorithm for the system of denominations 1 = e1 < e2 < · · · < eD is
not always optimal. Pearson showed there exist integers i, j with 1 ≤ j ≤ i < D such thatthe minimal representation of the minimal counterexample is of the form
0 · e1 + 0 · e2 + · · · + 0 · ej−1 + (aj + 1)ej + aj+1 ej+1 + · · · + aD eD ,where the greedy representation of ei+1 − 1 isa1 e 1 + a 2 e 2 + · · · + a D e D .
This gives the following algorithm for finding the smallest number such that the greedy
algorithm fails to be optimal (or ∞ if no such number exists):
PearsonTest(e1 , e2 , . . . , eD )m := ∞for j := 1 to D − 1 dofor i := Pj to D − 1 do
Let 1≤i≤D ai ei be the greedy representation of ei+1 − 1aj := aj + 1for k := 1 to j − 1 do
akP:= 0r := 1≤i≤D ai ei
if r < mPthen
LetP 1≤i≤D bi eiPbe the greedy representation of r
if 1≤i≤D bi > 1≤i≤D ai thenm := rreturn(m)(Here the scope of the loops is denoted by indentation.) It is easy to see that this algorithm
performs O(n3 ) arithmetic operations on numbers of size O(eD ).
4. Suppose we are given N and a system of denominations. How easy is it to computecost(N ; e1 , e2 , . . . , eD )? Since
opt(N ; e1 , e2 , . . . , eD ) = (N + 1)cost(N + 1; e1 , e2 , . . . , eD ) − N cost(N ; e1 , e2 , . . . , eD ),any algorithm to compute cost would also provide an algorithm to compute opt. It follows
that computing cost is NP-hard under Turing reductions.
55. Suppose we are given L and D and want to find an optimal set of denominations thatminimizes the average cost of making change for all amounts from 0 to L − 1? I don’t know
the computational complexity of this problem, but it seems quite hard. The data presented
in Figure 1 were computed using a brute-force enumeration of possibilities, but with sometricks to speed up the computation.
6. A related problem is the Frobenius problem. Here we are given a set of D denominations e1 < e2 < · · · < eD with gcd(e1 , e2 , . . . , PeD ) = 1, and we want to find the largest integer
N which cannot be expressed in the form 1≤i≤D ai ei with the ai non-negative integers.
There is a huge literature on this problem (see, for example, Guy [2, pp. 113–114]), but only
recently have researchers considered its computational complexity. Kannan [3] gave an algorithm for the Frobenius problem that runs in polynomial time if the dimension D is fixed.
On the other hand, Ramı́rez-Alfonsı́n [8] has shown that the general Frobenius problem isNP-hard.
7. Another related problem is the postage stamp problem. There are two flavors. The“local” problem asks, given a set of D denominations 1 = e1 < e2 < · · · < eD andP a boundh, what is the smallest integer N which cannot
P be represented in the form N = 1≤i≤D ai ei
where the ai are non-negative integers and 1≤i≤D ai ≤ h? In the “global” version, we aregiven D and h and want to find the set of denominations that maximizes N . There is alarge literature on these two problems (see Guy [2, pp. 123–127]), with much effort devoted
to finding efficient algorithms for small D.
However, I recently showed [9] that the local postage stamp problem is NP-hard under
Turing reductions, and that there is a polynomial-time algorithm for every fixed D.

4

Asymptotic results

Now we turn to some asymptotic estimates.
Let optcost(L, D) denote the minimum value of cost(L; e1 , e2 , . . . , eD ) over all suitablevalues of e1 , . . . , eD . Can we find good upper and lower bounds on optcost(L, D)?
One way to find an upper bound is as follows: let k = dL1/D e, and define ei = k i−1 for
1 ≤ i ≤ D. In this case, the greedy algorithm always finds the optimal representation for
any N , and it turns out to be the base-k expansion of N . Letting sk (N ) denote the sum of
the digits in the base-k expansion of N , we find
1 Xsk (i).
cost(L; e1 , e2 , . . . , eD ) = gcost(L; e1 , e2 , . . . , eD ) =L 0≤i≤L−1Hence
optcost(L, D) ≤1S 1/D (L)L dL eP
where Sk (N ) := 0≤i<N sk (i).
Now the quantity Sk (N ) has a long history; it is known thatµ¶log Nk−1
N log N + N Fk,Sk (N ) =2 log klog k
6(1)

(2)

where Fk is a continuous, nonpositive, nowhere differentiable function of period 1; see, forexample, [1]. Combining (1) and (2), we obtain the upper boundoptcost(L, D) ≤D 1/DL .
2

Furthermore, using the identity
Sk (kN + a) = kSk (N ) +k(k − 1)Na(a − 1)
+ ask (N ) +2
2

one can compute Sk (N ) in time polynomial in the number of digits in k and N . This provides
a fast way to compute the upper bound (1).
For a lower bound, one may reason as follows: fix a set of D denominations
P e 1 , e2 , . . . , e D ,and consider the number of different D-tuples (a1 , a2 , . . . , aD ) such that 1≤i≤D ai ≤ k. A¡¢
¡D+k¢simple combinatorial argument shows that this number is D+k.
Nowif
≤ L/2, itD
Dfollows that for at least L/2 choices of N , 1 ≤ N ≤ L, any representation for N must use¡¢ (k+D)D≤ D! ; if D is fixedat least k + 1 coins, and hence optcost(L, D) ≥ 21 (k + 1). Now D+kD
1/Dand L → ∞, this gives the lower bound of optcost(L, D) = Ω(L ).

5Acknowledgments

I obtained the results in Tables 1 and 2 in October 1999. Erik Demaine kindly pointed outthat similar results were posted to the Usenet newsgroup sci.math by Jeffry Johnston andBill Kinnersley in April 2000.
I thank Troy Vasiga, Ming-wei Wang, and Erik Demaine for their helpful comments.

References[1] H. Delange. Sur la fonction sommatoire de la fonction “somme des chiffres”. Enseign.
Math. 21 (1975), 31–47.
[2] R. K. Guy. Unsolved Problems in Number Theory. Springer-Verlag, 2nd edition, 1994.
[3] R. Kannan. Lattice translates of a polytope and the Frobenius problem. Combinatorica
12 (1992), 161–177.
[4] D. Kozen and S. Zaks. Optimal bounds for the change-making problem. Theoret.
Comput. Sci. 123 (1994), 377–388.
[5] G. S. Lueker. Two NP-complete problems in nonnegative integer programming. Technical Report TR-178, Computer Science Laboratory, Department of Electrical Engineering, Princeton University, March 1975.
[6] D. Pearson. A polynomial-time algorithm for the change-making problem. TechnicalReport TR 94-1433, Department of Computer Science, Cornell University, June 1994.
Available from http://citeseer.nj.nec.com/pearson94polynomialtime.html.
7[7] R. A. Raimi. The first digit problem. Amer. Math. Monthly 83 (1976), 521–538.
[8] J. L. Ramı́rez-Alfonsı́n. Complexity of the Frobenius problem. Combinatorica 16 (1996),143–147.
[9] J. Shallit. The computational complexity of the local postage stamp problem. SIGACT
News 33(1) (2002), 90–94.
[10] J. W. Wright. The change-making problem. J. Assoc. Comput. Mach. 22 (1975), 125–128.

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