PS05-100406-solutions

# PS05-100406-solutions - MATH 682 Problem Set#5 Solutions...

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Unformatted text preview: MATH 682 Problem Set #5 Solutions 1. (10 points) Complete the proof that the Harary graphs are k-connected. You may use the case presented in class of even values of k , either by citation or imitation. (a) (5 points) Show that H n,k is k-connected for even n and odd k . This graph is symmetric, so you may specifically demonstrate connectedness between v 1 and an arbitrary v i in H n,k- S for v 1 ,v i 6∈ S and | S | < k . Let k = 2 r + 1, and consider | S | < k . The exact technique described in class can be used to “crawl over” any cluster of vertices in H n,k- S in which S does not have r consecutive elements. Thus, either { v 2 ,...,v i- 1 } is navigable this way, or { v i +1 ,...,v n } is, or S consists specifically of r consecutive elements of { v 2 ,...,v i- 1 } and r consecutive elements of { v i +1 ,...,v n } . Since only one configuration of S presents difficulties not addressed in a previous proof, let us consider specifically the case where S = { v 2 ,v 3 ,...,v r +1 ,v a ,v a +1 ,...,v a + r } , for r + 1 < b < b + r ≤ n . It is obvious by our limitations on the selection of S that this graph has at most two components: { v a + r +1 ,v b + r +2 ,...,v n ,v 1 } and { v r +2 ,v r +3 ,...,v a- 1 } ; any diameter between these two sets will imply that they are a single component and that H n,k- S is connected. Since r < k 2 ≤ n 2 , the diameter { v 1 ,v 1+ n 2 is such an edge unless either a + r < 1 + n 2 or a ≤ 1 + n 2 ≤ a + r . In the former case, the entirety of S lies in v 1 ,...,v n 2 , so every point in that range ont in S is still incident on a diameter, guaranteeing connectedness; in the latter case, we see that the diameter { v a- 1 ,v a- 1+ n 2 } must exist, since a + r < a- 1 + n 2 ≤ n . (b) (5 points) Show that H n,k is k-connected for odd n and odd k . This graph is not sym- metric, so you must distinguish between the cases where v 1 ∈ S and v 1 / ∈ S . As above, let k = 2 r + 1, and consider | S | < k , and since H n,k- 1 ⊂ H n,k , the same argument as seen in class and invoked above is true except when S consists of two strings of r consecutive vertices. Unfortunately, every vertex of H n,k is not identical in this case; while before we assumed one of these strings started at v 2 , here we must consider more possibilities. Let us con- sider S = { v a ,v a +1 ,...,v a + r ,v b ,v b +1 ,...,v b + r } , where we consider our addition modulo n to concisely state cyclicity results. It is obvious by our limitations on the selec- tion of S that this graph has at most two components: { v b + r +1 ,v b + r +2 ,...,v a- 1 } and { v a + r +1 ,v a + r +2 ,...,v b- 1 } ; any quasi-diameter between these two sets will imply that they are a single component and that H n,k- S is connected. Since r < k 2 ≤ n- 1 2 vertices are in S , we know that there is some pair of vertices adjacent via a quasidiameter neither of which lies in S (we can partition V ( H n,k ) into ( n- 1) adjacency classes and a leftover vertex; not every adjacency class can be represented in...
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PS05-100406-solutions - MATH 682 Problem Set#5 Solutions...

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