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It does according to Dirac's Theorem. | It does according to Dirac's Theorem. | ||
a,b,c,f,d,e | a,b,c,f,d,e | ||
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| + | ---- | ||
| + | I don't necessarily agree with this. That is a Hamilton path described above. A circuit must get back to a causing it to touch c and f twice. It does not pass Dirac's theorem and it does not pass ore's theorem. | ||
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| + | Dirac | ||
| + | n=6 | ||
| + | degree of every vertex must be at least n/2, or 3, and a,d,b,e fail this. | ||
| + | --[[User:Podarcze|Podarcze]] 12:21, 19 November 2008 (UTC) | ||
Revision as of 07:21, 19 November 2008
It does according to Dirac's Theorem. a,b,c,f,d,e
I don't necessarily agree with this. That is a Hamilton path described above. A circuit must get back to a causing it to touch c and f twice. It does not pass Dirac's theorem and it does not pass ore's theorem.
Dirac n=6 degree of every vertex must be at least n/2, or 3, and a,d,b,e fail this. --Podarcze 12:21, 19 November 2008 (UTC)
