Re: [圖論] tree的pendant vertices

看板Math作者 (Mathkid)時間14年前 (2012/03/08 21:13), 編輯推噓1(100)
留言1則, 1人參與, 最新討論串1/1
※ 引述《wsx02 ()》之銘言: : Let T=(V,E) be a tree with V={v1, v2, ..., vn}, for n >= 2 : Prove that the number of pendant vertices in T is equal to : 2 + Σ (deg(vi)-2) : deg(vi)>=3 : 請問這題要怎麼證呢? : 謝謝 Let t be the number of leaves in T 2(n-1) = 2e(T) = \sum_{v_i\in V} deg(v_i) = \sum_{v_i:deg(v_i\ge 2)} deg(v_i) + t so t = 2t-t = \sum_{v_i:deg(v_i\ge 2)} deg(v_i) -2(n-t) + 2 = \sum_{v_i:deg(v_i\ge 2)} (deg(v_i)-2) = \sum_{v_i:deg(v_i\ge 3)} (deg(v_i)-2) -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 118.166.195.141

03/09 20:18, , 1F
謝謝
03/09 20:18, 1F
文章代碼(AID): #1FMB1Y5E (Math)