推 ilovecs34 :因為p2~p4中沒有任意兩人不相識 所以三個人互相認識 02/02 16:05
→ mqazz1 :thx!! 02/02 16:35
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
========================================================================
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...
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 61.228.25.176