Making Change for n Cents: Two Approaches

Transcription

1 Maing Change for n Cents: Two Approaches William Gasarch arxiv:406523v4 [mathho] 26 Jun 204 Univ of MD at College Par Abstract Given a dollar, how many ways are there to mae change using pennies, nicels, dimes, and quarters? What if you are given a different amount of money? What if you use different coin denominations? First we derive formulas for some problems of this type, including pennies, nicels, dimes, and quarters Second we derive, for every finite set of coins, a formula Introduction How many ways are there to mae change of a dollar using pennies, nicels, dimes, and quarters? This is a well nown question; however, the answers I found in the literature, and on the web 2 They were of two types: There are 242 ways to mae change The author then points to a program he wrote or to the actual list of ways to do it University of Maryland, College Par, MD 20742, gasarch@csumdedu Searching JSTOR for occurrences of the words change and coin in an article in a Mathematics Journal turned up only one that is relevant: Deborah Levine s article [4] has a formula for maing change with pennies, nicels, and quarters This article does not seem to be well nown 2 This entry on mathstacexchange maing-change-for-a-dollar-and-other-number-partitioning-problems offers programs and generating functions but no formula

2 2 The number of ways to mae change for n cents is the coefficient of z n in the power series for ( z( z 5 ( x 0 ( z 25 which can be wored out One possible exception: Graham, Knuth, Patashni [3] obtained a formula for pennies, nicels, dimes, quarters, and half-dollars that wored when n 0 (mod 50 A general formula (even for just pennies, nicels, dimes, and quarters would involve 4 cases and 3 constants (see [2] for an exposition The first answer yields an actual number but is not interesting mathematically The second answer is interesting mathematically but not does easily yield an actual number Def If S is a set of coin denominations then the change function for S is the function that, on input n, outputs the number of ways to mae change for n using the coins in S In the first part of this paper we derive simple formulas for the change function for two types of sets: ( S = {, s, s} where s, 2, and (2 S = {, s, s, rs} where x 2, and 2 < r As a corollary we obtain the case of pennies, nicels, dimes and quarters In passing we solve the change-for-a-dollar problem by hand In the second part of this paper obtain, for any finite set S, the change function The formulas obtained are rather complicated Hence we will also discuss what is meant by the informal term formula 2 General Definitions and Theorems Convention 2 Let S be a non empty set of coins The number of ways to mae 0 cents change is For all n the number of ways to mae n cents change is 0 Def 22 Let S = { < s < t < u} 2

3 a n is the number of ways to mae change of n cents using pennies Clearly ( n[a n = ] 2 b n is the number of ways to mae change of n cents using the first two coins (pennies and s-cent coins Clearly ( n[b n = a n + b n s ] We use that ( n [a n = 0] 3 c n is the number of ways to mae change of n cents using the first three coins (pennies, s-cent coins, and t-cent coins Clearly ( n[c n = b n + c n t ] 4 d n is the number of ways to mae change of n cents using all four coins (pennies, s-cent coins, t-cent coins, and u-cent coins Clearly ( n[d n = b n + c n u ] We do one example: Let S = {, 2, 4, 5} What is d 9? If one 5-cent coin is used then for the remaining four cents you must use either one 4-cent coin; two 2-cents coins; one 2-cent coin and two pennies; or four pennies 2 If no 5-cent coins and one 4-cent coin is used then for the remaining five cents you must use either two 2-cents coins and one penny; one 2-cent coin and three pennies; or five pennies 3 If no 5-cent coins and two 4-cent coins are used then for the remaining one cent you must use one penny 4 If no 5-cent coins and no 4-cent coins are used then you must use either zero, one, two, three, or four 2-cent coins and the appropriate number of pennies This results in five ways to mae change Hence d 9 = = 2 Using the recurrence for b n and ( n[a n = ], one can show the following Theorem 23 ( n[b n = n s + ] We can now solve the change-for-a-dollar problem by hand Let S = {, 5, 0, 25} We need to compute d 00 We use the exact formula for b n and the recurrences for c n and d n 3

4 d 00 = c 00 + c 75 + c 50 + c 25 + c 0 c 0 = c 25 = b 25 + b 5 + b 5 = = 2 c 50 = b 50 + b 40 + b 30 + b 20 + b 0 + b 0 = = 36 c 75 = b 75 + b 65 + b 55 + b 45 + b 35 + c 25 = = 72 c 00 = b 00 + b 90 + b 0 + b 70 + b 60 + c 50 = = 2 Hence d 00 = = The Coin Set {, s, s} Throughout this section we will be using the coin set S = {, s, s} where s, 2 are fixed natural numbers The quantities a n, b n, c n are as in Definition 22 with coin set S To determine the number of ways to mae change, you can always round down to the nearest multiple of s Formally c sl+l0 = c sl We use this without mention Let n = sl + L 0 where 0 L 0 s and L Using the recurrence for c n and the formula for b n (from Theorem 23 we have: c n = c sl = b sl + c s(l = b sl + b s(l + c s(l = b s(l 0 + b s(l + + b s(l i + c s(l i = (L + + (L (L i + + c s(l i Let L j (mod Let i = (L j Then the last term in the sum is c sj Since j, sj < s Hence c sj = b sj = j + The resulting sum is an arithmetic series with first term j +, last term j + + ( L j have the following, and number of terms L j + = L j+ Hence after easy algebra we Theorem 3 Let n = sl + L 0 where 0 L 0 s and L (So that L = n s 4

5 Let j be such that L j mod Then 2 c n = L2 + ( + 2L + n 2 + n c s 2 2s n n2 + (+2n + ( 22 + s 2 s + ( 2j j2 Proof: Part follows from our wor We prove Part 2 For the lower bound we use that L n s s, of For the upper bound we use that L n s 0 j, of ( 22 + and note that the last term has min value, as 0 j and note that the last term is max value, as Note 32 Theorem 3 for the special case of pennies, nicels, and dimes (s = 5, = 2 was proven by Deborah Levine s article [4] Note 33 One can derive c n = n2 + Θ( n from Schur s theorem [, 5, 6] s 2 4 The Coin Set {, s, s, rs} Throughout this section we will be using the coin set S = {, s, s, rs} where s,, r are fixed natural numbers with s,, r 2 and r > The quantities a n, b n, c n, d n are as in Definition 22 with coin set S To determine the number of ways to mae change, you can always round down to the nearest multiple of s Formally d sl+l0 = d sl We use this without mention Let n = s(rl + M + L 0 where 0 M r, 0 L 0 s Using the recurrence for d n we have: 5

6 d n = d s(rl+m = c s(rl+m + d s(rl+m r = c s(rl+m + c s(rl+m r + d s(rl+m r 2 = c s(rl+m + c s(rl+m r + c s(rl+m r c s(m+r + d sm = c s(rl+m + c s(rl+m r + c s(rl+m r c s(m+r + c sm = L c s(ri+m Using the formula for c n from Theorem 3 we obtain d n = L (M + ri 2 + ( + 2(M + ri + + L ( 2(ri + M mod (ri + M mod 2 We will evaluate the second sum later For now we name it: Notation 4 (L, M = L ( 2(ri+M mod (ri+m mod 2 Thus d n is L (M + ri 2 + ( + 2(M + ri + + (L, M = ( (L + (M 2 + M + 2M + + r(2m L i + r 2 =0 L i 2 + (L, M = ( (L + (2r 2 L 2 + (r 2 + 6Mr + 3r + 6rL + 6M 2 + (6 + 2M + + (L, M 6

7 Lemma 42 Let L, M and a 0 L (ri + M mod a = j=0 (rj + Ma L j+ 2 (L, M = ( j=0 ( L j + (( 2(rj + M mod (rj + M mod 2 3 If = 2 and r 0 (mod 2 then (L, M = (+( M+ (L+ 4 If = 2 and r (mod 2 then (L, M = 2L+(+( L (+( M+ +(+( L+ 6 Proof: We brea this sum into parts depending on what i is congruent to mod L (ri + M mod a = L j=0,i j mod = j=0 (ri + M mod a L,i j mod (rj + M mod a = j=0 (rj + M mod a L = j=0 (rj + M mod a L j+,i j mod 2 This follows from part using a = and a = 2 3 and 4 If = 2 then notice that the expression for (L, M is simplified considerably since 2 = 0 and (rj + M mod 2 2 = (rj + M mod 2 Also note that the summation only has two terms (j = 0 and j = Hence we obtain (L, M = 4 Case 0: r 0 (mod 2 ( L + 2 If M 0 (mod 2 then (L, M = 0 If M (mod 2 then 2 (M mod 2 + L + (L, M = ( L + 2 L (M mod 2 = L + 4

8 One can chec that (L, M = (+( M+ (L+ Case : r (mod 2 Then (L, M = 4 ( L (M mod 2 + L + 2 ( + M mod 2 The following table summarizes what (L, M is, given what L, M are mod 2 L mod 2 M mod 2 (L, M 0 0 L 0 L+2 0 L+ L+ One can chec that (L, M = 2L+(+( L (+( M+ +(+( L+ 6 Putting this all together we have the following Theorem 43 Let n = s(rl + M + L 0 where 0 M r, 0 L 0 s, and L (So L = n rs, M = n mod rs s, and n L0 (mod s Then the following are true d n is ( (L + (2r 2 L 2 + (r 2 + 6Mr + 3r + 6rL + 6M 2 + (6 + 2M + + ( L j ( j=0 2 d n = n3 6rs 3 + Θ(n 2 + (( 2(rj + M mod (rj + M mod 2

9 3 If = 2 and r 0 (mod 2 then d n is ( (L + (2r 2 L 2 + (r 2 + 6Mr + 2rL + 6M M + 24 ( + ( M+ (L If = 2 and r (mod 2 then d n is ( (L + (2r 2 L 2 + (r 2 + 6Mr + 2rL + 6M M L + ( + ( L ( + ( M+ + ( + ( L+ 6 Note 44 One can derive the asymptotic result (part 2 from Schur s theorem [, 5, 6] As a corollary of Theorem 43 we obtain a formula for maing change of n cents using pennies, nicels, dimes, and quarters Corollary 45 If s = 5, = 2, and r = 5 then d n is ( (L + (50L 2 + (530ML + 6M M Any Finite Coin Set + 2L + ( + ( L ( + ( M+ + ( + ( L+ 6 Throughout this section S = {t < t 2 < < t v } is our coin set We will derive the change function for S using generating functions While this is new many of the ideas are from Graham, Knuth, Patashni [3] 9

10 Lemma 5 For v ( x = v ( i + v v x i It is easy to see that the number of ways to mae change of n cents using the coins in S is the coefficient of z n in C(z = v i= ( z t i Let t be the least common multiple of {t,, t v } For i v let f i be the polynomial such that ( z t = ( z t i f i (z Then C(z = f (z f v (z ( z t v Let A(z = f (z f v (z Let M be the degree of A(z which is ((t t + (t t (t t v = tv Let the coefficient of z j in A(z be a j By Lemma 5 ( M ( ( i + v C(z = a j z z j ti = v j=0 M j=0 ( i + v a j z ti+j v The coefficient of z n is 0 j M:j n mod t ( n j + v a t j v Can this be considered a formula? We argue yes Given t,, t v one can find the a j s and the binomial coefficients needed for the formula Then, given n, one can find the answer in roughly M/t v times (One would need to code this up carefully and only visit those j n mod t Note that v does not depend on n so the number of steps is constant Hence the above is a formula There is one caveat Finding the a j s and the binomial coefficients can tae a lot of time (though still constant Nevertheless, we regard the above formula as a formula 0

11 6 Acnowledgment I would lie to than Tucer Bane, Adam Busis, and Clyde Krusal for proofreading and suggestions I would also lie to than Eric Weaver who wrote a program that helped give confidence in the formulas presented References [] W Gasarch Quantum techniques/gen functions-don t be afraid of new techniques, quantum-techniquesgen-functions-don t-behtml [2] W Gasarch An exposition of how GKP solved the change problem, umdedu/~gasarch/blogpapers/nuthchangepdf [3] R L Graham, D E Knuth, and O Patashni Concrete Mathematics: a foundation for computer science Addison-Wesley, 99 [4] D Levine Maing change: a problems solving activity The Mathematics Teacher, 75:4 7, 92 [5] Wiipedia Schur s theorem s_theorem [6] H Wilf Generatingfunctionology Academic Press, 994