not sure about your solution, pls enlighten mecan you define so-called hamilton path?

ha? is the simplified version of Euler circle?if you want to have any EUla circle, then every points must have Even path linking out if you want just go through , no need to come back then can have one point with odd number of paths linking out...

dun have, becoz got points with odd linking

not sure about your solution, pls enlighten mecan you define so-called hamilton path?

ha? is the simplified version of Euler circle?if you want to have any EUla circle, then every points must have Even path linking out if you want just go through , no need to come back then can have one point with odd number of paths linking out...

some of your statement is wrongif you want just go through , no need to come back then can have one point with odd number of paths linking out... I think it's wrong. Make a simple example. 5 points in a straight line. got 2 points with odd number of path, right? From what I have known, to be a shape that can go through at once, it will have the property below. The point of odd number of paths linking out should be less than 2 or even number. However, not all the shapes fulfil this requirement can be a euler circle.

wrong...Hamilton Path和Euler circle完全是2个不同的概念 简单来说，Hamilton Path是包含所有点的一条路径，而且在这条路径上，每个点刚好出现一次

wrong...Also，you should express youself understood by people even at low level

wrong...Hamilton Path和Euler circle完全是2个不同的概念 简单来说，Hamilton Path是包含所有点的一条路径，而且在这条路径上，每个点刚好出现一次

ft看到这里才发现原来hamitton path是原来所学的哈密顿路径。恍然大悟ing

我想可否用博萨定律（sorry,不知道怎么用英语说这个定律），来判断是否是哈密顿图，如果是的话，就可以一笔画通了，可是，点数太多了，好麻烦啊。。。。可否略提点一二。。。。

还是哪位高人提出个答案统一一下吧hamilton path和euler circuit好像不是一回事吧，哪个高人统一一下答案。 有人说没有简单的证明途径，没有什么vertex之类的说法吧。

ft,我的第一万贴阿！！不小心用错id了斑竹大大可否给一个一分的小草草安慰安慰，这样我的经验值和发贴数就对得上了^_^

sorry, should be EXACTLY 2 odd vertex if it hasan open Euler path