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Q56E

Expert-verified
Found in: Page 204

### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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# Show that if there were a coin worth 12 cents, the greedy algorithm using quarters, 12 - cent coins, dimes, nickels, and pennies would not always produce change using the fewest coins possible

For 12 - cents, greedy algorithm will always not produce fewest number of coins.

See the step by step solution

## Step 1

As mentioned in the question we have to use the greedy algorithm using quarters for 12 cents.

As per given preferences rankings of the men for the three women are:

${m}_{1}:{w}_{3},{w}_{1},{w}_{2};{m}_{2}:{w}_{1},{w}_{2},{w}_{3};{m}_{3}:{w}_{2},{w}_{3},{w}_{1}$

## Step 2

Now, for the rankings of the women, we have:

${w}_{1}:{m}_{1},{m}_{2},{m}_{3};{w}_{2}:{m}_{2},{m}_{1},{m}_{3};{w}_{3}:{m}_{3},{m}_{2},{m}_{1}$.

This is because then no women will be matched with the men out of her preferences and also will not violate the condition of stability.

Therefore, we can conclude that for cents greedy algorithm will always not produce fewest number of coins.

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