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Combinatorics Difficulty 9.0 Shortlist Prove it Estonia

There are 2020 inhabitants in a town. Before Christmas, they are all happy; but if an inhabitant does not receive any Christmas card from any other inhabitant, he or she will become sad. Unfortunately, there is only one post company which offers only one kind of service: before Christmas, each inhabitant may appoint two different other inhabitants, among which the company chooses one to whom to send a Christmas card on behalf of that inhabitant. It is known that the company makes the choices in such a way that as many inhabitants as possible will become sad. Find the least possible number of inhabitants who will become sad.

Solution

Partition 2019 inhabitants into 673 groups, each containing 3 inhabitants. Suppose that each inhabitant appoints two other members of the same group. As no group member is appointed thrice, the company cannot send three cards to one group member. Hence in every group, two different members get a card and at most one member will become sad. The inhabitant who does not belong to any group will become sad, too. Altogether, at most 674 inhabitants will become sad.

We show now that this is the least possible number. More precisely, we show that, in the case of nn inhabitants, the company can make ⌊n3⌋\lfloor \frac{n}{3} \rfloor inhabitants sad. Call inhabitants XX and YY competitors if somebody appoints them together to the company. By conditions of the problem, there are as many pairs of competitors as inhabitants or less (if several inhabitants appoint the same pair or some inhabitant appoints no pair). Thus it suffices to

* If every inhabitant has exactly 2 competitors then choose an arbitrary inhabitant XX and leave out XX along with both competitors. Among the remaining n−3n-3 inhabitants, there are at most n−3n-3 pairs of competitors (besides two pairs containing XX, removing either competitor canceled one more pair). By the induction hypothesis, one can find ⌈n−33⌉\lceil \frac{n-3}{3} \rceil remaining inhabitants with no pair of competitors. Adding XX to them results in ⌈n3⌉\lceil \frac{n}{3} \rceil inhabitants with no pair of competitors.

* If an inhabitant XX has at most 1 competitor then there must exist an inhabitant YY with at least 3 competitors. Let ZZ be the competitor of XX if XX has a competitor and an arbitrary inhabitant different from XX and YY otherwise. After leaving out XX, YY and ZZ, we have n−3n-3 inhabitants and at most n−3n-3 pairs of competitors (removing YY cancels at least 3 pairs of competitors). By the induction hypothesis, one can find ⌈n−33⌉\lceil \frac{n-3}{3} \rceil remaining inhabitants with no pair of competitors. Adding XX to this set results in a group of ⌈n3⌉\lceil \frac{n}{3} \rceil inhabitants with no pairs of competitors. This proves the desired claim.

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