Graph Theory - Soltutions to Problems | Lecture notes Mathematics

Graph theory - solutions to problem set 4

1. In this exercise we show that the sufficient conditions for Hamiltonicity that we saw in the lecture are

“tight” in some sense.

(a) For every n≥2, find a non-Hamiltonian graph on nvertices that has (n−1

2)+1 edges.

Solution: Consider the complete graph on n−1 vertices Kn−1. Add a new vertex vand connect

it to a vertex V(Kn−1). This graph has (n−1

2)+1 edges and it is non-Hamiltonian: every cycle

uses 2 edges at each vertex, but vhas only one adjacent edge.

(b) For every n≥2, find a non-Hamiltonian graph on nvertices that has minimum degree ⌈n

2⌉−1.

Solution: Let G1be a complete graph on ⌈n

2⌉vertices and G2be a complete graph on ⌊n

2⌋vertices

which is disjoint from G1. Fix a vertex v∈V(G1)and connect it to all the vertices of G2. Let

Gbe the resulting graph: it has minimum degree ⌈n

2⌉−1 and it is non-Hamiltonian, since every

cycle passing through all the vertices of Ghas to pass through vat least twice.

cycle longer than k+1. Solution: Let a=⌈n/k⌉, and take adisjoint copies of Kk+1. Then this

graph has a(k+1)>nvertices, each of them has degree k, but there is no cycle longer than k+1.

We can actually find a connected such graph if join all of these cliques at one vertex, creating a

star of cliques.

2. Check that the proof of Dirac’s Theorem also proves the following statement (called Ore’s theorem):

If for all non-adjacent vertices u,v in an n-vertex graph Gwe have d(u)+d(v)≥n, then Ghas a

Hamilton cycle.

Solution: There were two places in the proof of Dirac’s Theorem where we used the condition that

δ(G)≥n

2: To show that Gis connected, and to show that there is an edge in Pthat is both type-1

and type-2. The proof of the first statement is very similar in this case: if uand vwere in different

components, then the component of uwould contain at least d(u)+1 vertices, and the component of

vwould have at least d(v)+1 vertices, which would give more than nvertices in total. The second

statement follows because there are d(v1)type-1 edges and d(vk)type-2 edges. But then if v1and vk

are not adjacent, then Phas n−1<d(v1)+d(vk)edges by assumption, so some edge is both type-1

and type-2 and we can continue the argument. Otherwise, we get a cycle v1...vkv1that we can use

as Cfor the rest of the proof.

3. The graph below is called the Petersen graph. Does it have a Hamilton path? And a Hamilton cycle?

Solution: Here is a Hamilton path:

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Graph theory - solutions to problem set 4

In this exercise we show that the sufficient conditions for Hamiltonicity that we saw in the lecture are “tight” in some sense.

(a) For every n ≥ 2, find a non-Hamiltonian graph on n vertices that has n 2 − 1 + 1 edges. Solution: Consider the complete graph on n − 1 vertices Kn− 1. Add a new vertex v and connect it to a vertex V (Kn− 1 ). This graph has n 2 − 1 ^ + 1 edges and it is non-Hamiltonian: every cycle uses 2 edges at each vertex, but v has only one adjacent edge. (b) For every n ≥ 2, find a non-Hamiltonian graph on n vertices that has minimum degree ⌈ n 2 ⌉ − 1. Solution: Let G 1 be a complete graph on ⌈ n 2 ⌉ vertices and G 2 be a complete graph on ⌊ n 2 ⌋ vertices which is disjoint from G 1. Fix a vertex v ∈ V (G 1 ) and connect it to all the vertices of G 2. Let G be the resulting graph: it has minimum degree ⌈ n 2 ⌉ − 1 and it is non-Hamiltonian, since every cycle passing through all the vertices of G has to pass through v at least twice. (c) For every k, n ≥ 2, find a graph G on at least n vertices such that δ(G) = k but G contains no cycle longer than k + 1. Solution: Let a = ⌈n~k⌉, and take a disjoint copies of Kk+ 1. Then this graph has a(k + 1 ) > n vertices, each of them has degree k, but there is no cycle longer than k + 1. We can actually find a connected such graph if join all of these cliques at one vertex, creating a star of cliques.

Check that the proof of Dirac’s Theorem also proves the following statement (called Ore’s theorem): If for all non-adjacent vertices u, v in an n-vertex graph G we have d(u) + d(v) ≥ n, then G has a Hamilton cycle. Solution: There were two places in the proof of Dirac’s Theorem where we used the condition that δ(G) ≥ n 2 : To show that G is connected, and to show that there is an edge in P that is both type- and type-2. The proof of the first statement is very similar in this case: if u and v were in different components, then the component of u would contain at least d(u) + 1 vertices, and the component of v would have at least d(v) + 1 vertices, which would give more than n vertices in total. The second statement follows because there are d(v 1 ) type-1 edges and d(vk) type-2 edges. But then if v 1 and vk are not adjacent, then P has n − 1 < d(v 1 ) + d(vk) edges by assumption, so some edge is both type- and type-2 and we can continue the argument. Otherwise, we get a cycle v 1... vkv 1 that we can use as C for the rest of the proof.
The graph below is called the Petersen graph. Does it have a Hamilton path? And a Hamilton cycle?

Solution: Here is a Hamilton path:

Let us show that there is no Hamilton cycle in the Petersen graph P. One can check that the girth of P is 5 (i.e. P has no 3-cycle or 4-cycle). Assume there is a Hamilton cycle C in P. Since C must go through each vertex, C is actually C 10 (i.e. the Petersen graph contains C 10 ). Then there are five more edges in P. If each of the latter edges connects two opposite vertices on C, then there is a 4-cycle. Therefore, some edge e joins vertices at distance 4 in C (why cannot it be 2 or 3?). Let e be incident to vertices A and B, and D be the opposite vertex to A in C. The vertex D must be connected to one of the neighbours of A in C (Why?), let us call it F. Then ABDF A is a 4-cycle. This is a contradiction.

Show the following two properties of Minimum Spanning Trees (MST), under the assumption that no two edge weights are equal.

(a) Cut Property: the smallest edge crossing any cut must be in all MSTs. Reminder: a cut in a graph G = (V, E) is a partition A ⊍ B = V. (b) Cycle Property: The largest edge on any cycle is never in any MST.

Solution:

(a) Let G = (V, E) be a graph, A ⊍ B = V a cut of this graph and let T be a spanning tree of G. Let e = (u, v) be the edge of minimum weight that crosses from A to B. Suppose now that T does not contain e. Since T is connected, it contains a path P between u and v, and P must contain an edge f that crosses from A to B. But T ′^ = T − f + e is a tree of smaller weight, which is a contradiction. (b) Let C be a cycle in G and let e = (u, v) be the edge of maximum weight on C. Suppose it is contained in an MST T. Deleting e from T we get two connected components T 1 , T 2. But the cycle C must contain an edge f ≠ e with one end in T 1 and one end in T 2 , and T − e + f is a tree of smaller weight. Contradiction.

Let G = (V, E) be a graph with weights w ∶ E → R. Consider the problem of identifying a forest of maximum weight in G. Show that this problem can be reduced to the problem of computing a minimum weight spanning tree in a suitable graph G′^ with weights w′. Is your reduction efficient in the sense that G′^ is of polynomial size in G? Solution: We construct a new graph G′^ = (V ′, E′) with edge weight w′^ ∶ E′^ → R such that V ⊆ V ′, E ⊆ E′^ and a minimum weight spanning tree of G′^ restricted to G is a maximum weight forest of G. Let V ′^ = V ∪ {u} where u is a new vertex and let E′^ = E ∪ {(u, v) ∶ v ∈ V } be the new edge set. Define w′(e) = −w(e) for all edges e ∈ E and w′(e) = 0 otherwise. Let T be a spanning tree of G′^ and F its restriction to G. Then if F is not of maximal weight, there exists an other forest F ′^ with higher weight. Adding an edge from u to each connected component of F ′^ gives a tree in G′^ whose weight is lower than the weight of T. This implies that T was not a minimal weight spanning tree and thus each minimal weight spanning tree of G′^ gives a maximal weight forest in G.
(a) Show that a k-regular graph with girth 5 must have at least k^2 + 1 vertices.

(b) Find a k-regular graph with girth 5 and k^2 + 1 vertices for k = 2 , 3. (c) Show that a k-regular graph with girth 4 must have at least 2k vertices. Solution:

Construct a directed graph that has a directed Hamilton cycle if and only if the following formula has a satisfying assignment:

{x 1 , x¯ 3 , x 4 } ∧ {x 2 , x 3 , ¯x 4 } ∧ {¯x 2 , x¯ 3 }.

Solution: To keep the drawing of the construction simple, we solve the exercise for C 1 ∧ C 2 ∧ C 3 where C 1 ∶= {x 1 , x¯ 3 }, C 2 ∶= {x 2 , x 3 }, C 3 ∶= {¯x 2 , x¯ 3 }. The following figure shows the construction. Add a fourth module to extend the construction to the variable x 4.

Prove that if a tournament contains a directed cycle (i.e., it is not the transitive tournament) then it contains a directed triangle (3-cycle), as well. Solution: Take a shortest directed cycle in the tournament C = v 1... vk. If k > 3 then C has a “diagonal”: v 1 and v 3 are connected by an edge in some direction. If v 1 → v 3 then v 1 v 3 v 4... vk is a directed cycle of length k − 1. If v 3 → v 1 then v 1 v 2 v 3 is a directed cycle of length 3. Either way, there is a shorter cycle, contradicting our assumption.
We say that a vertex u in a tournament is almost central if for every other vertex v, there is a directed u-v path of length at most 2. Prove that every tournament has an almost central vertex. Solution: We proceed by induction on the number of vertices. For n = 2, it is clear. Let n > 2 and let G = (V, A) be a tournament on n vertices. Pick a vertex w and consider the tournament G′^ = G − w on the set V ′^ of n − 1 vertices. By induction, there exists an almost central vertex v in G′. The situation now breaks down in three cases. If there is an arc a ∈ A directed from v to w, then v is almost central in G. Otherwise, consider the sets V (^) i′ ∶= {u ∈ V ′^ S d(v, u) = i} , i = 1 , 2 , of vertices at distance i of v. Note that because v is almost central in G′, we have that {v} ∪ V 1 ′ ∪ V 2 ′ = V ′. If there is an arc directed from V 1 ′ to w, the v is almost central in G. Otherwise, all arcs between w and V 1 ′ are directed from w to V 1 ′ and there is a directed path of length at most 2 from w to all vertices in V 1 ′ ∪ V 2 ′. Furthermore, w and v are connected by an arc from w to v. Thus w is almost central in G.

Graph Theory - Soltutions to Problems, Lecture notes of Mathematics

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Graph theory - solutions to problem set 4