Download MC 302 – GRAPH THEORY – HW #5 – 11/17/11 Due Tuesday, 11

MC 302 – GRAPH THEORY – HW #5 – 11/17/11
Due Tuesday, 11/22/11, by 5 PM (paper) or midnight (electronic)
INSTRUCTIONS: This assignment is a team assignment for the members of each paper/talk
team to work on together. Submit one assignment with both team members’ names on it.
1. Fill in the left-hand column in the table below with the numbers from 1 to 20 to indicate
your rankings for choice of paper/talk topic. Number 1 indicates your top choice; number
20 indicates your bottom choice.
RANKING TOPIC
Arboricity [D]4.2
Block structure of a 1-connected graph: [BM] 3.2, [D] 3.1
Chromatic polynomial of a graph: [BM] 8.4-8.5
Counting trees and Cayley’s Formula [BM] 2.4, [CH] 2.3
Degree Sequences: Additional Results [BM] p. 11, [G] 1.1, [D] 10.2
Dual of a planar graph and its properties: [CH] 5.6, [D] 4.6, [BM]9.2
Edge coloring: [CH] 6.5, [BM] Chap. 6, [D] 5.3
Geometric drawings of graphs [HR]
Graphs on surfaces and genus [HR]
Infinite graphs [D]
Line graphs [D] 1.1
List-coloring: [D] 5.4
Menger’s Theorem (relating disjoint paths and cut sets) [CH] 8.3
Network flows and the Max-flow, Min-cut Theorem [CH] 8.1-8.2, [BM] 11.3
Perfect graphs [D] 5.5
Platonic Solids: [CH] 5.3
Ramsey numbers and Ramsey’s Theorem: [CH] Chap. 9, [BM] 7.2, [D] Chap. 9
Tournaments [CH] 7.3, [BM] 10.2
Reconstruction [CH] Chap. 10
Random graphs [D] Chap. 11
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2. Suppose G is a k-regular graph with n vertices, and suppose that both G and its
complement G are connected. Show that if G is not Eulerian, then G is Eulerian.
Answer: Since G is not Eulerian k must be odd. If k were even, G would be Eulerian since G is
Eulerian IFF every vertex of G has even degree. Thus every vertex of G has odd degree, k.
Assume n is odd. Then by the degree sum theorem, the number of edges in G = (n*k)/2.
Since n is odd and k is odd, n*k is odd and you end up with a half edge in your graph. Thus
you cannot have a k-regular graph with n vertices if n and k are both odd. Thus n has to be
even. So the number of edges in G = (n*k)/2 and the number of edges in the complete graph
of n vertices = (n*(n-1))/2. Thus G-complement has (n*(n-1))/2 – (n*k)/2 edges =
(n*(n – 1 – k))/2. Since n is even n – 1 is odd. Thus (n-1)-k is an odd number minus an odd
number, which is equal to an even number. By the degree-sum theorem, we see that in Gcomplement, n vertices have degree (n-1-k). Thus all vertices of G-complement have even
degree and thus G-complement is Eulerian.
3. [CH] 3.3.14 There are n guests at a dinner party, where n  4. Any two of these guests
know, between them, all the other n – 2 people. Prove that the guests can be seated
around a circular table so that each one is sitting between two people they know.
Hint: Show that the hypothesis of Dirac’s Theorem is satisfied. Be sure to say what
graph you are applying the theorem to, and how it is related to the conditions and the
claim of the problem.
Let the people of this graph be represented by vertices and an edge represent two people
knowing each other. Let v be a vertex, and therefore a person. Assume deg(v) <n/2. Clearly v
can be the only vertex with degree less than n/2. If there were two vertices with degree<n/2,
the sum of their degrees would be less than n and also, you would have to subtract two for
those vertices themselves. Thus the sum of their neighbors is less than n-2 and the given
condition is not satisfied. Let u be a vertex with degree greater than n/2.
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