Overall comments - Your math looks fine. However, there are some things to keep in mind. This assignment is a mathematical paper rather than a homework assignment. In particular do no state a problem and jump right into a solution. Segue naturally into the problems.This is quite terse, so more words and explanations are needed and it should flow like a report or paper. Some background on theorems and properties used should be provided for the reader and flow into it the assigned problems one at a time with something like "Consider the Cayley graph of...Let's determine if this has a Hamilton circuit". When you provide background and theorems, use your own numbering. Do not say, for example "Theorem 14.4". Assume the reader does not own the textbook.
Last, but not least, your paper will need to be different than this, i.e. your own work. Points will be deducted if you simply use a slightly different version from this draft that does not indicate your own work.
Kumyiah McDonald
Scott Herwitt
Lyrica Collins
Melaney Ramirez
Show that the Cayley digraph given in Example 7 has a Hamiltonian path from e to
As you see in the digraph of example 7 above,there are a few paths that are available from the vertex e to vertex a. So we would consider e→a,it has a Hamiltonian path because the path there are no vertex that is repeated more than once.
Now you can see there is a Hamiltonian circuit as well,from vertex e to e,the path e→ab→a^2→a^3 b→a^3→a^2 b→a→b→e
Prove that the Cayley digraph given in Example 6 does not have a Hamiltonian circuit. Does it have a Hamiltonian path?
In example 6,there is no Hamiltonian circuit in any path,so maybe if we consider a path from
243→(143)→(124)→(14)(23)→(13)(24)→(142)
To prove the Hamiltonian paths,maybe we can use theorem: Let G be a finite Abelian group,and let S be any(nonempty)generating set for G. Then Cay(S:G) has a Hamiltonian path.
A necessary condiion for Hamiltonian circuits is:Cay({(1,0),(0,1) }:Zm ?Zn)does not have a Hamiltonian circuit when m and n are relatively prime and greater than 1.
A sufficient condition:Cay({(1,0),(0,1) }:Zm?Zn) has a Hamiltonian circuit when n divides m.