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 ======================================================================== 将p2~p6分两堆，一堆为与p1相识，另一堆为与p1不相识 由鸽笼原理知，至少有一堆有至少3人 case 1: 与p1相识那堆至少3人 令p2~p4与p1相识 若p2~p4中有两人相识，则此二人加上p1共三人彼此相识，得证 否则p2~p4三人彼此不认识，亦得证 case 2: 不与p1相识那堆至少3人 令p2~p4与p1不相识 若p2~p4有两人不相识，则此二人加上p1共三人彼此不相识，得证 否则p2~p4三人彼此认识，亦得证 <=====我对这行有问题 请问为什麽会跑出p2~p4三人彼此认识@@? 想不通... --

※ 发信站: 批踢踢实业坊(ptt.cc) ◆ From: 61.228.25.176