What This Country Needs is an 18¢ Piee
Jerey Shallit
Department of Computer Siene
University of Waterloo
Waterloo, Ontario, Canada N2L 3G1
shallitgraeland.uwaterloo.a
November 8, 2002
Abstrat
We onsider sets of oin denominations whih p ermithange to b emade using as
fewoins as p ossible, on average,and explain why theUnited States shouldadopt an
18¢piee.
1 Introdution
Most businesses in the United States urrently make hange using four dierent typ es of
oins: 1¢ (ent), 1
5¢ (nikel),10¢ (dime),and 25¢ (quarter). For p eople who make hange
on a daily basis, it is desirable to make hange in as eÆient a manner as p ossible. One
riterion for eÆienyis to use the smallestnumb erof oins. For example,to make hange
for 30¢, one ould, at least in priniple, give a ustomer 30 1-ent oins, but most would
probably preferreeiving aquarter and a nikel.
Formally,we an dene the optimalrepresentationproblem as follows: givena set of D
integer denominationse
1
<e
2
< < e
D
and an integer N 0, we wish to express N as
a non-negative integer linear ombination N = P
1iD a
i e
i
suh that the numb er of oins
S = P
1iD a
i
isminimized. Inorder that everynumb eratuallyhave arepresentation,we
demandthat e
1
=1. If(a
1
;a
2
;:::;a
D
) istheD -tuple thatminimizesS,then wesay itis an
optimalrepresentation,and we dene opt(N;e
1
;e
2
;:::;e
D
):=S.
The optimal denominationproblem is to nd denominationsthat minimizethe average
ost of making hange. We assume that every amount of hange b etween 0¢ and 99¢ is
equallylikely.
2
Wethenask, whathoieofD denominationsminimizestheaveragenumb er
\Whatthis ountry needs is areally go o d ve-ent igar." T. R. Marshall (US Vie-President), New
YorkTribune,January4,1920.
1
Informally,a1-entoinisusuallyalled a\p enny",but thisusageisfrownedup onbynumismatists.
2
This assumption is probablyinaurate for several reasons, notleast b eing the fat that manyitems
have priesthatend inthedigit9. Also,Benford'slawmayplayarole;seeRaimi[7℄.
for D denominationsup to the limit L means determiningthe denominations e
1
;e
2
;:::;e
D
whihminimize
ost(L;e
1
;e
2
;:::;e
L ):=
1
L X
0i<L
opt (i;e
1
;e
2
;:::;e
D ):
For the urrent system, where (e
1
;e
2
;e
3
;e
4
) = (1;5;10;25), a simple omputation de-
termines that ost(100;1;5;10;25) =4:7. In other words, on average a hange-makermust
return 4:7 oins with everytransation.
Can wedo b etter? Indeedwean. Thereare exatlytwosetsof fourdenominationsthat
minimizeost (100;e
1
;e
2
;e
3
;e
4
);namely,(1;5;18;25) and(1;5;18;29). Bothhaveanaverage
ost of 3:89. We would therefore gain ab out 17% eÆienyin hange-making by swithing
to either of these four-oin systems. The rst system, (1;5;18;25), p ossesses the notable
advantage that we only need makeone smallalteration in the urrent system: replae the
urrent10¢ oin with a new 18¢ oin. This explains the titleof this artile.
Figure 1 givesthe optimaldenominationsof sizeD for 1D 7, and their asso iated
osts.
D (e
1
;:::;e
D
) ost(100;e
1
;:::;e
D )
1 (1) 49.5
2 (1;10) 9
(1;11)
3 (1;12;19) 5.15
4 (1;5;18;25) 3.89
(1;5;18;29)
5 (1;5;16;23;33) 3.29
6 (1;4;6;21;30;37) 2.92
(1;5;8;20;31;33)
7 (1;4;9;11;26;38;44) 2.65
Figure1: Optimaldenominationsfor hange-makingfor 1D 7 denominations
Althoughthesystem(1;5;18;25)wouldb esup eriortotheurrent(1;5;10;25)forhange-
making, it may b e diÆult to onvine p eople to aept the removal of the p opular dime.
Thusitmayb eworthwhileto onsideradierentquestion: what singledenominationould
we add to (1;5;10;25) to ahieve the maximumimprovement in ost? The unique answer
is 32¢; this improves ost(100;1;5;10;25) = 4:7 to ost(100;1;5;10;25;32 ) = 3:46. If we
also allowthe infrequently-used 50¢ pieeas alegitimatedenomination,then the maximum
improvementomes from adding an 18¢ piee;this improvesost(100;1;5;10;25;50) = 4:2
to ost(100;1;5;10;18;25 ;50) =3:18. Yetanotherreasonto addan18¢pieeto USoinage!
Otherountriesprovidedierentproblems. In Canada,theoin denominationsurrently
in wideirulation are 1¢, 5¢, 10¢, 25¢, 100¢ (alled a\lo onie" forthe lo on on the reverse),
and200¢(alleda\to onie"). Thesmallestdenominationof pap ermoneyinwideirulation
erage ost of makinghange inCanada is ost(500;1;5;10;25;10 0;20 0) =5:9. This an b e
b estimprovedbyadding an 83¢oin;wehaveost(500;1;5;10;25;83 ;100 ;200 ) =4:578. On
theotherhand,thenewsystemofEurosintro duedinEurop eprovidesoinsofdenomination
:01;:02;:05;:1;:2;:5;1,and2Euros. Forthissystemwehaveost(500;1;2;5;10;20;5 0;10 0;20 0) =
4:6. This anb eb estimproved(to averageost 3:92)by addingaoin ofdenomination1:33
or 1:37 Euros.
2 Greedy methods for hange-making
One nie feature of the urrent set of US denominations (1;5;10;25) is that the greedy
algorithmdeterminesthe representationwith the minimumnumb erof oins. Bythe greedy
algorithm, I mean the following pro edure: given a numb erN to b e represented as a non-
negativeintegerlinearombinationofdenominationse
1
<e
2
<<e
D
,takeasmanyopies
a
D
of thelargestdenominatione
D
as p ossible, so that a
D e
D
N. ThensetN :=N a
D e
D
and ontinue the pro edure with the remaining smaller denominations. Use of the greedy
algorithmprovides a simple,easily-rememb eredmetho d for makinghange. Not all sets of
denominations have the prop erty that the greedy metho d always determines the optimal
representation. For example, with denominations (1;7;10) the greedy algorithm gives the
representation 14=10+1+1+1+1, whereas 7+7 uses feweroins.
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 make
hangefor 24¢ usingthe system (1;12;19), weget 19+1+1+1+1+1, afar ryfrom the
optimalrepresentation12+12.
This suggests onsidering a variation on the optimal denominationproblem, where ost
is replaedby the analogous funtion gost, and weount only the ost of greedy represen-
tations. For the urrent systemwestill havegost (100;1;5;10;25) =4:7. Figure2 displays
the results for optimal sets of denominations. An asterisk denotes an optimalset for whih
the greedy representationis always an optimal representation.
We also might onsider what single denomination ould b e added to the urrent US
system (1;5;10;25) to b est improve the greedy ost. It turns out that adding either a 2¢-
pieeora3¢-pieeimprovesgost (100; )from4.7 to3.9,and thisistheb estp ossible1-oin
improvement. Itisinterestingto notethattheUSatuallyhada2¢-pieefrom1864to1873,
and two dierent 3¢-piees: one made in silverfrom 1851 to 1873, and one made innikel
from1865 to 1889.
3 Computational questions
So far we have fo used on systems partiular to the US, Canada, and Europ e, but a go o d
mathematiianwillwantmoregeneralresults. Letusexaminetheomputationalomplexity
of the problemswehave studied,and somerelated ones.
1. Supp ose we are givenan amountof hangeto make,say N,and asystem ofdenomi-
1 D 1 D
1 (1) 49.5
2 (1;10) 9
(1;11)
3 (1;5;22) 5.26
(1;5;23)
4 (1;3;11;37) 4.1
(1;3;11;38)
5 (1;3;7;16;40) 3.46
(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)
6 (1;2;5;11;25;62) 3.13
(1;2;5;11;25;63)
(1;2;5;13;29;64)
(1;2;5;13;29;65)
7 (1;2;5;8;17;27;63) 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)
Figure2: Optimaldenominationsfor greedy hange-making
nations, say 1=e
1
<e
2
<<e
D
. Howeasyis ittoomputeopt(N;e
1
;e
2
;:::;e
D
)or nd
an optimal representationN = P
1iD a
i e
i
, i.e.,one whih minimizes P
1iD a
i
?
The answer dep ends on how N and the e
i
are written down. If they are written in
ordinary deimalnotation, or in binary, then there is no fast algorithmknown to solve this
problem. In fat, it follows easily from results of Lueker [5℄ that this problem is NP-hard;
roughly sp eaking this means it is at least as hard as manyfamous ombinatorial problems,
suhasthetravellingsalesmanproblem,forwhihno p olynomial-timealgorithmisurrently
known.
If, on the other hand, N and the e
i
are represented in unary, then a simple dynami
programmingalgorithm(e.g.,[10 ℄)solvesthe optimalrepresentationprobleminp olynomial
time.
2. Supp ose we are given N and asystem ofdenominations. How easyis itto determine
if the greedy representationfor N isatually optimal? Kozen and Zaks[4℄have shown that
strongly suggests there is no eÆientalgorithmfor this problem.
3. Supp oseweare givenasystemofdenominations. Howeasyisittodeidewhetherthe
greedy algorithm always pro dues an optimal representation, for all values of N? It turns
out that this problem an b e solved eÆiently; this surprising result is due to Pearson [6℄.
Sine Pearson's result app eared only inan obsuretehnialrep ort, we give afew details.
Supp ose the greedyalgorithmforthe systemof denominations1=e
1
<e
2
< <e
D is
not always optimal. Pearson showed there existintegers i;j with 1 j i <D suh that
the minimalrepresentationof the minimalounterexampleisof the form
0e
1
+0e
2
++0e
j 1 +(a
j +1)e
j +a
j+1 e
j+1
++a
D e
D
;
where the greedy representationof e
i+1 1 is
a
1 e
1 +a
2 e
2
++a
D e
D :
This gives the following algorithmfor nding the smallest numb ersuh that the greedy
algorithmfails to b e optimal(or 1 ifno suh numb er exists):
PearsonTest(e
1
;e
2
;:::;e
D )
m :=1
for j :=1to D 1do
for i:=j to D 1do
Let P
1iD a
i e
i
b e the greedy representationof e
i+1 1
a
j :=a
j +1
for k:=1to j 1 do
a
k :=0
r:=
P
1iD a
i e
i
if r<m then
Let P
1iD b
i e
i
b e the greedy representationof r
if P
1iD b
i
>
P
1iD a
i then
m :=r
return(m)
(Herethe sop e ofthe lo ops is denotedby indentation.) It iseasyto seethat this algorithm
p erformsO (n 3
)arithmetiop erations on numb ersof sizeO (e
D ).
4. Supp ose we are givenN and a systemof denominations. How easy is it to ompute
ost(N;e
1
;e
2
;:::;e
D
)? Sine
opt (N;e
1
;e
2
;:::;e
D
)=(N +1)ost (N+1;e
1
;e
2
;:::;e
D
) Nost (N;e
1
;e
2
;:::;e
D );
any algorithm to omputeost would also providean algorithm to ompute opt. It follows
that omputing ost is NP-hard underTuring redutions.
minimizesthe averageost of makinghangefor allamounts from0to L 1? I don't know
the omputationalomplexityof this problem,but itseems quitehard. The data presented
in Figure 1 were omputed using a brute-fore enumeration of p ossibilities, but with some
triks to sp eed up the omputation.
6. A relatedproblemis the Frob enius problem. Hereweare givena setof D denomina-
tionse
1
<e
2
<<e
D
with gd(e
1
;e
2
;:::;e
D
)=1,andwewantto ndthe largestinteger
N whih annot b e expressed in the form P
1iD a
i e
i
with the a
i
non-negative integers.
Thereisahugeliteratureonthis problem(see,for example,Guy[2,pp.113{114℄), butonly
reently have researhers onsidered its omputational omplexity. Kannan [3℄ gave an al-
gorithmforthe Frob eniusproblemthatruns inp olynomialtimeif the dimensionD is xed.
On the other hand, Ramrez-Alfonsn [8℄ has shown that the general Frob enius problem is
NP-hard.
7. Another related problem is the p ostage stamp problem. There are two avors. The
\lo al" problem asks, givena set of D denominations1= e
1
< e
2
< < e
D
and a b ound
h, whatisthe smallestintegerN whihannot b erepresentedinthe formN = P
1iD a
i e
i
where the a
i
are non-negative integers and P
1iD a
i
h? In the \global" version,we are
given D and h and want to nd the set of denominations that maximizes N. There is a
large literatureon these twoproblems(see Guy[2, pp. 123{127℄), with muheort devoted
to ndingeÆientalgorithms forsmall D .
However, I reently showed [9℄ that the lo al p ostage stamp problem is NP-hard under
Turing redutions, and that there is ap olynomial-timealgorithmfor everyxedD .
4 Asymptoti results
Nowwe turn to someasymptoti estimates.
Let optost(L;D ) denote the minimum value of ost(L;e
1
;e
2
;:::;e
D
) over all suitable
values of e
1
;:::;e
D
. Can we nd go o d upp er and lower b oundson optost (L;D )?
One way to nd an upp er b ound is as follows: let k =dL 1=D
e, and dene e
i
= k i 1
for
1 i D . In this ase, the greedy algorithm always nds the optimal representation for
any N, and it turns out to b e the base-k expansionof N. Letting s
k
(N) denotethe sumof
the digits in the base-k expansionof N, we nd
ost(L;e
1
;e
2
;:::;e
D
)=gost (L;e
1
;e
2
;:::;e
D )=
1
L X
0iL 1 s
k (i):
Hene
optost(L;D ) 1
L S
dL 1=D
e
(L) (1)
where S
k
(N):=
P
0i<N s
k (i).
Nowthe quantity S
k
(N) has a long history; itis known that
S
k (N)=
k 1
2logk
NlogN +NF
k
logN
logk
; (2)
k
example,[1℄. Combining(1) and (2), we obtainthe upp er b ound
optost (L;D ) D
2 L
1=D
:
Furthermore, using the identity
S
k
(kN +a)=kS
k (N)+
k(k 1)N
2
+as
k (N)+
a(a 1)
2
oneanomputeS
k
(N)intimep olynomialinthenumb erofdigitsinkandN. Thisprovides
a fastway to ompute the upp er b ound (1).
For alowerb ound,onemayreasonasfollows: xasetofDdenominationse
1
;e
2
;:::;e
D ,
and onsiderthe numb er of dierent D -tuples (a
1
;a
2
;:::;a
D
) suh that P
1iD a
i
k. A
simple ombinatorial argument shows that this numb er is D +k
D
. Now if D +k
D
L=2, it
follows that for at least L=2 hoies of N, 1 N L, any representation for N must use
at least k+1oins, and heneoptost (L;D ) 1
2
(k+1). Now D +k
D
(k +D )
D
D !
;if D isxed
and L!1, this gives the lowerb ound of optost (L;D )=(L 1=D
).
5 Aknowledgments
I obtained the results in Tables 1 and 2 inOtob er 1999. Erik Demainekindly p ointed out
that similarresults werep osted to the Usenetnewsgroup si.mathby Jery Johnston and
BillKinnersleyin April2000.
I thank Troy Vasiga, Ming-wei Wang, and Erik Demainefor their helpfulomments.
Referenes
[1℄ H. Delange. Sur la fontion sommatoirede la fontion \somme des hires". Enseign.
Math. 21 (1975), 31{47.
[2℄ R. K.Guy. Unsolved Problems in Number Theory. Springer-Verlag, 2nd edition, 1994.
[3℄ R.Kannan. Lattietranslatesofap olytop eandtheFrob eniusproblem. Combinatoria
12 (1992), 161{177.
[4℄ D. Kozen and S. Zaks. Optimal b ounds for the hange-making problem. Theoret.
Comput. Si. 123 (1994), 377{388.
[5℄ G.S.Lueker.TwoNP-ompleteproblemsinnonnegativeintegerprogramming.Tehni-
al Rep ort TR-178, Computer Siene Lab oratory, Departmentof Eletrial Engineer-
ing, PrinetonUniversity,Marh1975.
[6℄ D. Pearson. A p olynomial-time algorithm for the hange-making problem. Tehnial
Rep ort TR 94-1433, Department of Computer Siene, Cornell University, June 1994.
Availablefromhttp://iteseer.nj.ne.om/pearson94polynomialtime.html.
[8℄ J.L.Ramrez-Alfonsn.ComplexityoftheFrob eniusproblem.Combinatoria16(1996),
143{147.
[9℄ J.Shallit. The omputationalomplexityof thelo alp ostagestampproblem. SIGACT
News 33(1) (2002), 90{94.
[10℄ J. W. Wright. The hange-makingproblem. J. Asso. Comput.Mah. 22 (1975), 125{
128.
