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For an incidence matrix A with related matrix B defined by Bij = 1 if i is related to j and j is related to i, and Bij = 0 otherwise, prove that i belongs to a clique if and only if (B3)ii > 0. 解答: Let G = G(B) be the graph associated to the symmetric matric B. And (B3)ii is the number of walk of length 3 from i to i. If i is in some clique, then there must be a walk of length 3 from i back to i since a clique must have number of vertex greater than 3. 不是三個點就好了嗎? 為什麼是>3? Conversely, if (B3)ii is greater than zero, this means there is at least one walke of length 3 from i to i, say i → j → k → i. Note that i, j, and k should be different vertices since length is 3 and there is no loop. i → j → k → i → j → k → i ... 這樣不算loop? So i, j, and k must be a triangle, this means three vertices adjacent to each others. So i is contained in {i, j, k} and so contained in some clique. -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 220.136.42.73 (臺灣) ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1690807644.A.ABC.html
wrvuxci : clique只要三個點就好,我想是的,這樣邏輯前後是通 09/16 07:25
wrvuxci : 順的,只要你課堂上/書上是這樣定的就沒問題(我學的 09/16 07:26
wrvuxci : 定義沒有點數的限制,但不同書的定義略有不同不奇怪 09/16 07:27
wrvuxci : ),然後loop是只一個有向邊的兩端是同一點,也就是 09/16 07:28
wrvuxci : i→i 09/16 07:28