Dynamic Programming Solution to the Coin Changing Problem

Topics: Metropolitana di Napoli, Madrid Metro, Beijing Subway Pages: 5 (1077 words) Published: January 12, 2013
Dynamic Programming Solution to the
Coin Changing Problem
(1) Characterize the Structure of an Optimal Solution. The Coin Changing problem exhibits optimal substructure in the following manner. Consider any optimal solution to making change for n cents using coins of denominations d1 , d2 , . . . , dk . Now consider breaking that solution into two diﬀerent pieces along any coin boundary. Suppose that the “left-half” of the solution amounts to b cents and the “right-half” of the solution amounts to n − b cents, as shown below.

n−b

b

d1

d3

d1

d2

d4

d5

d2

Claim 1 The left-half of the solution must be an optimal way to make change for b cents using coins of denominations d1 , d2 , . . . , dk , and the right-half of the solution must be an optimal way to make change for n − b cents using coins of denominations d1 , d2 , . . . , dk . Proof: By contradiction, suppose that there was a better solution to making change for b cents than the “left-half” of the optimal solution shown. Then the left-half of the optimal solution could be replaced with this better solution, yielding a valid solution to making change for n cents with fewer coins than the solution being considered. But this contradicts the supposed optimality of the given solution, →←. An identical argument applies to the “right-half” of the solution.

2
Thus, the optimal solution to the coin changing problem is composed of optimal solutions to smaller subproblems.
(2) Recursively Deﬁne the Value of the Optimal Solution. First, we deﬁne in English the quantity we shall later deﬁne recursively. Let C [p] be the minimum number of coins of denominations d1 , d2 , . . . , dk needed to make change for p cents. In the optimal solution to making change for p cents, there must exist some ﬁrst coin di , where di ≤ p. Furthermore, the remaining coins in the optimal solution must themselves be the optimal solution to making change for p − di cents, since coin changing exhibits optimal substructure as proven above. Thus, if di is the ﬁrst coin in the optimal solution to making change for p cents, then C [p] = 1 + C [p − di ]; i.e., one di coin plus C [p − di ] coins to optimally make change for p − di cents. We don’t know which coin di is the ﬁrst coin in the optimal solution to making change for p cents; however, we may check all k such possibilities (subject to the constraint that di ≤ p), and the value of the optimal solution must correspond to the minimum value of 1 + C [p − di ], by deﬁnition. Furthermore, when making change for 0 cents, the value of the optimal solution is clearly 0 coins. We thus have the following recurrence. Claim 2 C [p] =

0
mini:di ≤p {1 + C [p − di ]}

if p = 0
if p > 0

Proof: The correctness of this recursive deﬁnition is embodied in the paragraph which precedes it.

1

2

(3) Compute the Value of the Optimal Solution Bottom-up. Consider the following piece of pseudocode, where d is the array of denomination values, k is the number of denominations, and n is the amount for which change is to be made.

Change(d, k, n)
1 C [0] ← 0
2 for p ← 1 to n
3
min ← ∞
4
for i ← 1 to k
5
if d[i] ≤ p then
6
if 1 + C [p − d[i]] < min then
7
min ← 1 + C [p − d[i]]
8
coin ← i
9
C [p] ← min
10
S [p] ← coin
11 return C and S

Claim 3 When the above procedure terminates, for all 0 ≤ p ≤ n, C [p] will contain the correct minimum number of coins needed to make change for p cents, and S [p] will contain (the index of ) the ﬁrst coin in an optimal solution to making change for p cents.

Proof: The correctness of the above procedure is based on the fact that it correctly implements the recursive deﬁnition given above. The base case is properly handled in Line 1, and the recursive case is properly handled in Lines 2 to 10, as shown below. Note that since the loop deﬁned in Line 2 goes from 1 to n and since di ≥ 1 for all i, no array element C [·] is accessed in either Line 6 or 7 before it has been computed. Lines...