Greedy Algorithms (4)

COS 320 - Algorithm Design

Greedy Algorithms (4)

Coin Changing

coin changing

coin changing

Goal: Given U.S. currency denominations (\(\{1, 5, 10, 25, 100\}\)), devise a method to pay amount to customer using fewest coins.

Ex: 34¢

Cashier's Algorithm: At each iteration, add coin of the largest value that does not take us past the amount to be paid.

Ex: $2.89

cashier's algorithm

At each iteration, add coin of the largest value that does not take us past the amount to be paid.

Cashiers-Algorithm (x, c1, c2, ..., cn)
    Sort n coin denominations so that 0 < c1 < c2 < ... < cn.
    S <- { }  // multiset of coins selected
    While x > 0
        k <- largest coin denomination ck such that ck <= k
        If no such k, Return "no solution"
        Else
            x <- x - ck
            S <- S | { k };
    Return S

quiz 1: greedy algorithms

Is the cashier's algorithm optimal?

1. Yes, greedy algorithms are always optimal

2. Yes, for any set of coin denominations \(c_1 < c_2 < \ldots < c_n\) provided \(c_1 = 1\)

3. Yes, because of special properties of U.S. coin denominations

4. No.

Cashier's algorithm (for arbitrary coin denominations)

Q: Is Cashier's algorithm optimal for any set of denominations?

A: No. Consider U.S. postage: 1, 10, 21, 34, 70, 100, 350, 1225, 1500.

• Cashier's algorithm: 140¢ = 100 + 34 + 1 + 1 + 1 + 1 + 1 + 1
• Optimal: 140¢ = 70 + 70

A: No. It may not even lead to a feasible solution if \(c_1 > 1\): 7,8,9

• Cashier's algorithm: 15¢ = 9 + ?
• Optimal: 15¢ = 7 + 8

properties of any optimal solution (U.S. coin denominations)

Property: Number of pennies ≤ 4
Pf: Replace 5 pennies with 1 nickel

Property: Number of nickels ≤ 1

Property: Number of quarters ≤ 3

Property: Number of nickels + number of dimes ≤ 2
Pf:

• Recall: ≤ 1 nickel
• Replace 3 dimes and 0 nickels with 1 quarter and 1 nickel
• Replace 2 dimes and 1 nickel with 1 quarter

properties of any optimal solution (U.S. coin denominations)

Theorem: Cashier's Algorithm is optimal for U.S. coins {1,5,10,25,100}

Pf: (by induction on amount to be paid \(x\))

• Consider optimal way to change \(c_k \leq x \leq c_{k+1}\): greedy takes coin \(k\)
• We claim that any optimal solution must take coin \(k\)
  • If not, it needs enough coins of type \(c_1, \ldots, c_{k-1}\) to add up to \(x\)
  • Following table indicates no optimal solution can do this
• Problem reduces to coin-changing \(x-c_k\) cents, which, by induction, is optimally solved by cashier's algorithm. ∎

properties of any optimal solution (U.S. coin denominations)