[鴿籠] 6人中 必3人彼此相識或不相識
suppose that p1~p6 denote six people
where every two people are either familiar with or strange to each other
prove that at least three of p1~p6 can be found so that they are either
familiar with or strange to one another
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將p2~p6分兩堆，一堆為與p1相識，另一堆為與p1不相識
由鴿籠原理知，至少有一堆有至少3人
case 1: 與p1相識那堆至少3人
令p2~p4與p1相識
若p2~p4中有兩人相識，則此二人加上p1共三人彼此相識，得證
否則p2~p4三人彼此不認識，亦得證
case 2: 不與p1相識那堆至少3人
令p2~p4與p1不相識
若p2~p4有兩人不相識，則此二人加上p1共三人彼此不相識，得證
否則p2~p4三人彼此認識，亦得證 <=====我對這行有問題
請問為什麼會跑出p2~p4三人彼此認識@@?
想不通...
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