hw1-sol

# hw1-sol - ECS 120 Theory of Computation Homework 1...

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ECS 120: Theory of Computation Homework 1 Solutions Problem 1. Show that at a party of 20 people, there are at least 2 people who have the same number of friends present at the party. Assume (however unrealistically) that friendship is symmetric and anti-reﬂexive. Hint: Carefully use the pigeonhole principle. Solution: There either is a person at the party that is friendless or there is no such person. Let us look at the Frst case Frst. Because friendship is assumed to be symmetric, then no one can be friends with everyone. Thus the number of friends a person may have is one of { 0 , 1 , 2 ,..., 18 } – 19 possibilities in all. But there are 20 people at this party who have their number of friends drawn from this set. Thus, some two people have the same number of friends. On the other hand, suppose there is no one at the party who is friendless. Then the number of possible friends one might have at the party is drawn from { 1 , 2 19 } , and the result follows as above.

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## This note was uploaded on 05/06/2008 for the course ECS 120 taught by Professor Filkov during the Fall '07 term at UC Davis.

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hw1-sol - ECS 120 Theory of Computation Homework 1...

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