Contact us
Choose a category below so we can get back to you as quickly as possible.
反證法.假設G中不存在度數為1的結點,G是連通圖,所以G的結點的度數至少是2. G有3度節點,所以G的所有結點的度數之和大於等於2(n-1)+3=2n+1. 而G有n條邊,度數之和是2n. 衝突. 所以G中至少存在有一個度數為1的結點.
We and our partners use cookies and other technologies to analyze traffic and optimize your experience. View more info and control your cookies settings at any time in our Cookies Policy.