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Q56E

Expert-verified

Discrete Mathematics and its Applications

Found in: Page 204

Book edition 7th

Author(s) Kenneth H. Rosen

Pages 808 pages

ISBN 9780073383095

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Short Answer

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 by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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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