TEN2 2015-11-02 — Solutions to problems

Lars Hellstr¨om MAA600 , networks and applications

Problem 1. Subproblem a. Shown right is the D¨urergraph (so named because the German renaissance artist Albrecht D¨urer in one of his engravings depicted a solid having this graph as its skele- ton). Find a Hamiltonian cycle in and a 3-edge-colouring of this graph.

Recall that a cycle is Hamiltonian if it visits all vertices. The theorems of Dirac (2.17) and Ore (2.16) would for this 12- graph require a vertex of degree at least 12/2 = 6 to apply, but the D¨urergraph is only 3-regular, so there’s no aid from that angle. We’ll simply have to find a cycle through some trial and error. For this, it is useful that the graph is 3-regular, since it means the Hamiltonian cycle will have to make use of two of the three edges at any vertex. One initial observation that can be made is that each of the two triangles will have to be covered in one pass; it is not possible to have the cycle pick up just two vertices on one pass and return for the final vertex in a second pass, since there would then be nowhere for the cycle to continue afterwards:

not !

This means the central triangles end up behaving like jumps of length 2 visavi

1 the outer cycle:

Exactly which two jumps they correspond to is arbitrary, but there is no escaping that the length of the jump (as measured along the outer cycle) is 2, so it will be a jump past one of the outer vertices. This means the only way for the cycle to pick up that vertex is that after one jump forward you go back to the intermediate vertex and use the other triangle for jumping out from there:

This results in the following Hamiltonian cycle in the full graph:

As for the 3-edge-colouring, one possibility is to observe (as in Exercise 5.31) that since the Hamiltonian cycle has even length, it can be edge-coloured with 2 colours. Since the graph is 3-regular, the edges outside the Hamiltonian cycle constitute a matching, and that can be coloured with the third colour. This

2 results in the 3-edge-colouring shown left below.

3 1 1 2 3 2 2 1 2 3 3 2 3 2 1 3 1 3 1 1 1 1 2 1 2 3 3 2 3 2 2 3 2 3 1 1

Alternatively, one can cyclically colour the spokes 1, 2, 3, 1, 2, 3—this more sym- metric colouring turns out to extend to all edges, as shown above to the right. It does however not correspond to a Hamiltonian cycle. Problem 1. Subproblem b. Give an example of a bridgeless graph with mini- mum degree at least 3 which is not Hamiltonian.

(The reason graphs with bridges are excluded is that these trivially prevent the graph from being Hamiltonian. A cycle cannot cross a bridge—a single-edge cut in the graph—since once it would have crossed the bridge there is no way for it to get back.) One approach for preventing Hamiltonian cycles in a bridgeless graph is to start with a theta-shaped graph, such as the following:

a c

A cycle can pick up two paths between a and c, but not the third, so as long as there is an intermideate vertex on every a–c path, the graph cannot be Hamiltonian. A problem is however that this graph does not have minimum degree 3, due to the intermediate vertices all having degree 2. This can be fixed by replacing each such vertex with the 3-regular gadget

3 which results in the non-Hamiltonian 3-regular graph

A different solution is to remember that the 5-sided counterpart of the D¨urer graph—the far more famous Petersen graph—is non-Hamiltonian:

A third solution is to observe that (really by mistake in the phrasing of the problem) it actually wasn’t required that the graph is connected. This yields the smallest solution: the graph with two K4 components. Problem 2. Subproblem a. Define network and flow on network. A network (G, c) is a directed graph G together with a function c: E(G) −→ R giving the capacity c(e) of each edge e. A general flow on the network is a function f : E(G) −→ R satisfying 0 6 f(e) 6 c(e) for all e ∈ E(G). For the more specific concept of an s–t-flow, one also requires that, for all vertices except s and t, the sum of the flow on all incoming edges equals the sum of the flow on all outgoing edges. The value of such a flow (which we’ll need below) is the sum of the flow on all edges from s minus the sum of the flow on all edges to s.

Problem 2. Subproblem b. Define the capacity of an s–t-cut in a network. An s–t-cut M in a network is a set of edges such that every path from vertex s to vertex t in that network contains some edge in M. The capacity of the cut P M is the sum e∈M c(e) of the capacities of all edges in it. Problem 2. Subproblem c. State the max-flow-min-cut theorem.

4 The max-flow-min-cut theorem states that the maximum value of an s–t-flow in a network is equal to the minimum capacity of an s–t-flow in that network. Problem 3. The concept of ‘bipartite’ has a synonym related to colouring. Which term is that? Bipartite is the same as 2-colourable; if a graph has been given a 2-colouring, then the vertices in one colour class constitutes one part in the partition, and the vertices in the other colour class is the other part.

Problem 4. The ladder graph Ln has 2n vertices x1, . . . , xn and y1, . . . , yn. Vertices xi and yi are adjacent for all 1 6 i 6 n. For 1 6 i < n, vertices xi and xi+1 are adjacent, as are vertices yi and yi+1. Finally vertex xn is adjacent to vertex x1 and vertex yn is adjacent to vertex y1. Subproblem a. Draw L5.

The following is L5:

x1 x2 x3 x4 x5

y1 y2 y3 y4 y5

Problem 4. Subproblem b. For which n > 3 are Ln bipartite?

There is always a cycle through all xk vertices, and when n is odd that cycle will be odd too, which means Ln is not bipartite for any odd n. For even n, the function f : V(Ln) −→ {1, 2} defined by  1 if u = xk for some odd k,  2 if u = x for some even k, f(u) = k (1) 2 if u = yk for some odd k,  1 if u = yk for some even k constitutes a 2-colouring of Ln, which means it is bipartite. Hence Ln is bipartite if and only if n > 3 is even.

Problem 5. The M¨obiusladder graph Mn is similar to the ladder graph Ln, except that xn is adjacent to y1 rather than x1 and yn is adjacent to x1 rather than y1. Subproblem a. Draw M5.

The following is M5:

x1 x2 x3 x4 x5

y1 y2 y3 y4 y5

5 Problem 5. Subproblem b. Prove that Mn is bipartite if n > 3 is odd. With the same function f as in (1) above, we find that:

• f(xi) 6= f(xi+1) and f(yi) 6= f(yi+1) for all 1 6 i < n.

• f(xi) 6= f(yi) for all 1 6 i 6 n.

• f(xn) 6= f(y1) and f(yn) 6= f(x1); here we make use of the fact that n is odd.

Hence f is a 2-colouring of Mn, and thus Mn is bipartite.

Problem 5. Subproblem c. M3 is isomorphic to a graph in one of the standard families. Which graph is that?

The graph in question is the complete K3,3, which becomes clear if we rearrange the vertices a bit:

x3 y3

y2 x2

x1 y1

Problem 5. Subproblem d. Prove that Mn is not planar for any n > 3.

From subproblem c, we know that M3 is K3,3. Subdividing the edge from x3 to y1 with vertices x4, x5, etc. and the edge from y3 to x1 with vertices y4, y5, etc. produces a subgraph of Mn for arbitrarily high n (namely that subgraph which is missing the xiyi edges for i > 4). This means every Mn for n > 3 has a subdivision of K3,3 as a subgraph, which by Kuratowski’s theorem implies that they are not planar.

Problem 6. Prove that a connected planar simple graph with n > 3 vertices can have at most 3n − 6 edges.

Proof. Let G be an arbitrary connected plane simple graph with n > 3 vertices. By Euler’s formula, n − m + f = 2, where m is the number of edges in G and f is the number of faces of G. [Remark: Here the argument is slightly simplified by the condition that G is connected, since otherwise the right hand side would be 1 plus the number of components of G.] Since G is simple, there are no 2-cycles; the walk around the boundary of a face has to have length at least 3. On the other hand, every edge is either incident with 2 different faces, or incident with one face on both sides. This means 2m/f is the average length of the walk around the boundary of a face, and since no such walk has length 2 shorter than 3, we know that 3 6 2m/f, or equivalently f 6 3 m. Inserting that into Euler’s formula, we get

2 1 2 = n − m + f 6 n − m + 3 m = n − 3 m.

Solving for m, it follows that m 6 3n − 6.

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