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 antireﬂexive.
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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 Fall '07
 Filkov
 Proof by contradiction, 00 L

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