朋友問題 (六人)

Proof by contradiction. Assume the 2 group types below do not exist.Group type I: a group of 3 and all of them are friendsGroup type II: a group of 3 and all of them are strangersStarting with A, A can have 0, 1, 2, 3, 4 or 5 friends in the group which corresponds to 5, 4, 3, 2, 1 or 0 strangers. So A has at least 3 friends or sees at least 3 strangersCase (a) A has at least 3 friends, say, B, C, D then no pair of them can be friends otherwise together with A, they form a group type I. But if they do not know each other, then they form a group type II.Case (b) A sees at least 3 strangers, say, B, C, D then they must know each other otherwise any pair of strangers together with A will form a group type II. But if they know each other, they form a group type I.Since there are contradictions in both cases (a) & (b), the assumption cannot be true.

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