Selected Solutions to Graph Theory, 3rd Edition · many vertices are adjacent to second vertex except rst vertex, obviously answer is n 2. Similarly compute how many vertices are

Selected Solutions to GraphTheory, 3rd Edition

Reinhard Diestel::

Rakesh

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Research Scholar

Mathematics

IITGuwahati

Rakesh Jana

Department of Mathematics

IIT Guwahati

March 1, 2016

Acknowledgement

These solutions are the result of taking CS-520(Advanced Graph Theory) coursein the Jan-July semester of 2016 at Indian Institute of Technology Guwahati. Thisis not a complete set of solutions in that book. It may happen that solution ofsome problem may be wrong. I have not verified these problem from some expart.It is my kind request you that do not belive the answer blindly. If you found anymistake please inform me. I know these article must contain some typographicalerrors, in that case please inform me. If you have any better solution in any of theseproblem please let me know. I will upload that solution in this content with yourname. If you want to discuss any of these solution with me please ping me in mygiven email address or meet me in research scholar office (RS-E1-010) Departmentof Mathematics, IIT Guwahati.

You can find List of Solved Exercises at the end. Please e-mail [email protected] or [email protected] for any corrections and suggestions.

Copyright c©2015–2016, Rakesh Jana

mailto:[email protected]

mailto:[email protected]

[email protected]

Contents

1 The Basics 3

2 Matching, Covering and Packing 10

3 Connectivity 14

4 Planar Graphs 18

5 Colouring 19

6 Some Arbitary Problem 211 The Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 Matching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 Connectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234 Planar graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 Colouring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

7 Solved Exercise Reference 26

Rakesh Jana IIT Guwahati

1 The Basics

See some extra problem on basic in the end( problem-6.1.1 - problem-6.1.2).

Exercise 1.1. What is the number of edges in K n?

Proof. Notice that first vertex adjacent to other n − 1 vertices. Now compute howmany vertices are adjacent to second vertex except first vertex, obviously answer isn− 2. Similarly compute how many vertices are adjacent to third vertex except firstand second vertices, answer is n− 3, and so on. Thus total number of edge is K n is

(n− 1) + (n− 2) + · · ·+ 1 + 0 =n(n− 1)

2.

Exercise 1.2. Determine the average degree, number of edges, diameter, girth, andcircumference of the hypercube graph Qd.

Proof. Since V is the set of all 0 − 1 sequences of length d. Thus total number ofvertices is 2d, since in each place we can assign two number 0, 1. Since two suchsequence form an edge if and only if they differ in exactly one position. Thus eachvertices has degree d. Now we know that

2|E| =∑v∈V

d(v)

|E| = d× 2d

2= d× 2d−1.

• Thus average degree of Qd = 2|E||V | = d.

• Notice that the distance between any two vertices depends on the number ofdifferent bits, so diameter is d, i.e. diam G = d.

• Girth(Q1) =∞, because there are no cycles on hypercube graph Q1.Girth(Qd) = 4, where d ≥ 2, this is because Qd

∼= K2 ×Qd−1.

• Circumference of Qd is 2d.

Exercise 1.3. Let G be a graph containing a cycle C, and assume that G containsa path of length at least k between two vertices of C. Show that G contains a cycleof length at least

√k. Is this best possible?

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Proof. Let the path P start at x and end at y where x, y are lie on C. Suppose thatP leaves C for the ith time at vertex xi , and arrives at C for the ith time at vertexyi (it is possible for xi+1 = yi). Then the xi, yi -portion of P together with the xi, yi-portion of C forms a cycle of length at least 1 + li , where li = distP (xi, yi).

Let among xi’s and yi’s there are t many distict vertices. Without loss of anygenerality Now if t ≥

√k then we can take a path along C connecting these distinct

vertices and then traverse C then we will get a cycle of length atleast√k + 1.

Let t <√k. Let us consider subpath of path P , Pi = xiPyi and Pi = yiPxi+1.

Then there are atmost t many internally disjoint subpath of P . Let Pk be the subpathof P with maximum length, say l(pk) = m. Then

k ≤ l(P )

≤ tm

≤√km

Thus m >√k. Hence we have a subpath of the path P of lenght atleast

√k whose

end points are in C and also distinct. Hence We get a cycle of length atleast√k.

Notice in the solution that we can improve the size of cycle from√k to

√k + 1.

Exercise 1.4. We know that from proposition 1.3.2 that every graph containing acycle satisfying g(G) ≤ 2 diamG+ 1. Is the bound is best possible?

Proof. Yes. It is the best possible bound because equality occur when G = K3.

Exercise 1.5. Show that radG ≤ diamG ≤ 2 radG.

Proof. We know that diamG = maxx,y∈v(G) dG(x, y).

radG = minx∈V (G)

maxy∈V (G)

dG(x, y)

≤ minx∈V (G)

maxy∈V (G)

diamG

= diamG.

To show diamG ≤ 2 radG.Let a, b, v ∈ V (G) such that dG(a, b) = diamG and radG = maxy∈V (G) dG(v, y).

diamG = dG(a, b) ≤ dG(a, v) + dG(v, b)

≤ radG+ radG = 2 radG.

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Exercise 1.6. Prove the weakening of Theorem 1.3.4 obtained by replacing averagewith minimum degree. Deduce that |G| ≥ n0(d/2, g) for every graph G as given inthe theorem.

Proof. Case 1: Consider g = 2r + 1, r ∈ N. This proof is similar to proof of propo-sition 1.3.3. Let v ∈ V (G) be any vertex in G. Let us consider Di = {u ∈ V (G) :dG(u, v) = i}, for i ∈ N∪{0}. It is clear that Di∩Dj = ∅,∀i 6= j and V (G) = ∪i≥0Di.Since v can not contained in any cycle of length lesser then 2r + 1. Thus for anyu,w ∈ Di, 0 ≤ i ≤ r− 1, N(u)∩Di+1 is disjoint from N(w)∩Di+1, otherwise we canconstruct a cycle of length atmost r−1+1+1+r−1 = 2r < 2r+1, a contradiction.Hence each vertex in Di, 0 ≤ i ≤ r − 1 is connected to exactly one vertex in Di−1and atleast δ − 1 vertices in Di+1. Hence |D0| = 1, |D1| ≥ δ and |Di| ≥ δ(δ − 1)i−1,for 2 ≤ i ≤ r. Thus

|V (G)| ≥r−1⋃i=0

|Di| = 1 + δr−1∑i=0

(δ − 1)i−1.

Case-2: Consider g = 2r, r ∈ N. In this case proof is same as previous one insteadof a vertex we have to start with two adjacent vertices. Let uv ∈ E(G). Similar wayconsider for i ∈ N ∪ {0},

Dui = {y : d(u, y) = i}

Dvi = {y : d(v, y) = i}

where d(x, y) := dG−uv(x, y). Similar way as case-1, for any x ∈ {u, v}, Dxi ∩Dx

j = ∅for i 6= j and for any a, b ∈ Dx

i , N(a) ∩ Dxi+1 is disjoint from N(b) ∩ Dx

i+1 for1 ≤ i < r − 1. Also each vertex in Dx

i (0 ≤ i < r − 1) is connected to exactly onevertex in Dx

i−1 and atleast δ − 1 vertices in Dxi+1. Now |Dx

0 | = 1, |Dx1 | ≥ δ − 1 and

|Dxi | ≥ (δ − 1)i, for 1 ≤ i ≤ r − 1. Let us define for x ∈ {u, v},

Tx =⋃

0≤i≤r−1

Dxi .

Notice that |Tx| =∑r−1

i=0 (δ − 1)i for any x ∈ {u, v}.Claim. Tu ∩ Tv = ∅. To prove this claim first notice that Du

i ∩ Dvj = ∅, for all

0 ≤ i, j ≤ r− 1. If not let x ∈ Dui ∩Dv

j for some 0 ≤ i, j ≤ r− 1. Then there exist acycle in G of length atmost r − 1 + r − 1 + 1 = 2r − 1 < 2r, a contradiction. HenceTu ∩ Tv = ∅.

Hence |V (G)| ≥ |Tu|+ |Tv| = 2∑r−1

i=0 (δ − 1)i.

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Lemma 1.1. Let P be a path in a connected graph G. If there is u ∈ V (G) \ V (P ),then there exist v ∈ V (G) \ V (P ) adjacent to P .

Proof. Let u ∈ V (G)\V (P ) and w ∈ V (P ). Since G is connected there exist a u−wpath in G, say Q. Now consider last common vertex p ∈ V (P ) ∩ V (Q) (traversingfrom w to u) then there exist a vertex in v ∈ V (Q) \V (P ) such that vp ∈ E(G).

Exercise 1.7. Show that every connected graph G contains a path of length at leastmin{2δG, |G| − 1}.

Proof. Let us consider P = x0x1 · · ·xk be a longest path in G. We have to showk ≥ min{2δG, |G| − 1}. If possible let k < min{2δG, |G| − 1}. Now G is connected|E(G)| ≥ |G| − 1.Since k < |G| − 1 there exist u ∈ V (G) \ V (P ) and by lemma 1.1 there existy ∈ V (G) \ V (P ) such that yxi ∈ E(G), for some 0 ≤ i ≤ k. Now if x0xk ∈ E(G)we can get a path P ′ = yxiPxkx0Pxi which is longer then P , a contradiction. Thusx0xk /∈ E(G).Now N(x0) ⊆ V (P ) and N(xk) ⊆ V (P ), since P is the longest path. Let us considerS = {xj|xj+1 ∈ N(x0), 0 ≤ j ≤ k − 2}. Clearly |S| ≥ δ(G). Since k < 2δ(G) gives|V (P ) \ (S ∪ {xk})| ≤ δ(G). By pigeonhole principle there exist xj ∈ S such thatxjxk ∈ E(G). Hence we get a cycle C = x0PxjxkPxj+1x0. Now consider the pathP ∗ = yxiCxi which is longer then P , a contradiction. Hence k ≥ min{2δG, |G| −1}.

Exercise 1.8. Find a good lower bound for the order of a connected graph in termsof its diameter and minimum degree.

Proof. The following claim gives the lower bound for the order of connected graph.

Claim. Let G be any connected graph with diamG = k and δ(G) = d then |G| ≥kd/3.

Let dG(x, y) = k, for some x, y ∈ V (G) and distance achieve by the path P =x0x1 · · ·xk, where x0 = x, xk = y. Let u be a vertex not on P that is adjacent to somevertex on P . Let i be the smallest integer such that xi is adjacent to v. Notice thatif v is adjacent to xj for j > i+ 2 then we can get a path x0PxivxjPxk and which isshorter then P , a contradiction to dG(x, y) = k. Hence each v ∈ V (G) \ V (P ) canadjacent to at most 3 vertices on P .

Exercise 1.10. Show that every 2-connected graph conatains a cycle.

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Proof. Let G be a 2-connected graph. Then δ(G) ≥ 2, since if d(v) = 1 or 0, forsome v ∈ V (G) then that vertex will be either a cut vertex or isolated, in both caseit contradict that G is 2-connected. Hence by proposition-1.3.1, it has a cycle oflength atleast δ(G) + 1.

Exercise 1.11. Determine κ(G) and λ(G) for G = Pm, Cn, Kn, Km,n, and Qd;d,m, n ≥ 3.

Proof. Recall. κ(G) denote for connectivity(vertex) of a graph G and λ(G) denoteedge-connectivity of graph G. Also by proposition-1.4.2, κ(G) ≤ λ(G) ≤ δ(G).

Given graphs are all connected.κ(Pm) = 1, for m ≤ 2 it is clear, for m > 2 if you remove an interior vertex of Pm ,it becomes a disconnect graph.λ(Pm) = 1, as if you delete any edge, the graph becomes disconnected.κ(Cn) = 2, because if you remove any vertex you will get P n−1, hence if you deleteany two vertex from Cn then it becomes a disconnect graph.λ(Cn) = 2, as if you delete any edge, the graph becomes P n and if you delete onemore edge it become disconnected.Similarly, κ(Kn) = n− 1λ(Kn) = n− 1κ(Km,n) = min{m,n}λ(Km,n) = min{m,n}κ(Qd) = dλ(Qd) = d.

Exercise 1.12. Is there any function f : N → N such that, for all k ∈ N, everygraph of minimum degree atleast f(k) is k connected?

Proof. No. Suppose f(1) = t ∈ N. Then there exist graphs G and H with both haveminimum degree t but one of them is connected and other is not. For instance takeG = Kt+1 and take H be two disjoint component of Kt+1.

Exercise 1.16. Show that every tree T has atleast ∆(T ) leaves.

Proof.

Exercise 1.17. Show that a tree without a vertex of degree 2 has more leaves thanother vertices. Can you find a very short proof that does not use induction?

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Proof. Let T be a tree with no vertex of degree 2. Let Vi = {v ∈ V (G) : dG(v) = i}.Notice that V2 = ∅. Now average degree of a tree is 2|E|

|V | = 2(|V |−1)|V | < 2. Now

2 >

∑v∈V dG(v)

|V |=

∑|V |−1i=1 i|Vi||V |

=|V1|+

∑|V |−1i=3 i|Vi||V |

≥ |V1|+ 3∑|V |−1

i=3 |Vi||V |

=|V |+ 2

∑|V |−1i=3 |Vi|

|V |= 1 +

2∑|V |−1

i=3 |Vi||V |

Hence we get,

|V1|+|V |−1∑i=3

|Vi| = |V | > 2

|V |−1∑i=3

|Vi| =⇒ |V1| >|V |−1∑i=3

|Vi|.

This complete the answer.

Exercise 1.19. Let G be a connected graph, and let r ∈ G be a vertex. Starting fromr, move along the edges of G, going whenever possible to a vertex not visited so far.If there is no such vertex, go back along the edge by which the current vertex was firstreached (unless the current vertex is r; then stop).(This procedure has earned thosetrees the name of depth-first search trees.)

Show that the edges traversed in depth-first search form a normal spanning treein G with root r.

Proof. First notice that in dfs we always get a tree, since we always add a vertex tothe current subgraph if it is not end point of any edge of current subgraph, so cannotcreate a cycle. To show it is spanning. Suppose it not spanning tree of G then thereis a vertex v which is not in the tree but adjacent to a vertex u in the tree. But thenwhen we left u for the last time we would have visited v instead of returning to r. Sowe get a contradiction that depth-first search completed. Hence we get a spanningtree, say T . To show T is normal. Last part remaining

Exercise 1.23. Show that a graph is bipartite if and only if every induced cycle haseven length.

Proof. Recall. An induced cycle in G is a cycle in G forming an induced subgraphwithout any chords.

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If a graph is bipartite then it does not have any odd cycle by proposition-1.6.1,hence does not have any induced cycle of odd length. To prove reverse part. Let usassume G is not bipartite. Since G is not bipartite so it has an odd cycle. Let C bea smallest odd cycle in G. Then C can not be induced cycle, since all induced cycleare even lengths. Then there exist x, y ∈ V (C) but xy /∈ E(C). Thus we get twocycle C1 = xCyx,C2 = yCxy (traverse clockwise direction), among them one is oddand other is even. Hence we get a shorter odd cycle, a contradiction. This provesthe result.

Exercise 1.24. Find a function f : N→ N such that, for all k ∈ N , every graph ofaverage degree at least f(k) has a bipartite subgraph of minimum degree at least k.

Proof. Define a map f : N→ N by f(k) = 4k,∀k ∈ N. The idea behind to considerthis function is following: Every graph with an average degree of 4k have a subgraphH with minimum degree 2k, and we will lose another factor of 2 in moving H toits bipartite subgraph. Let H∗ be the bipartite subgraph of H with the maximalnumber of edges. My claim is that H∗ have minimum degree atleast k. If not, letv ∈ H∗ such that dH∗(v) < k. This means v lost more then half of its neighbours inthe process to form H to H∗. This means v is on the same partition with its loosesneighbours. But in that case if we consider v in the other partition we can able toconnect those previously looses vertices to v and form a new bipartite subgraph ofH with more edges then H∗ have, a contradiction. Hence it proves of my claim.

Exercise 1.26. Prove or Disprove that every connected graph contains a walk thattraverses each of its edges exactly once in each direction.

Proof. Let W = v0v1 · · · vk be a longest walk in G that traverses every edge exactlyonce in each direction. If possible let there exist a vertex v not visited by W . Withoutloss of any generality let us assume v ∈ N(vi) for some 0 ≤ i ≤ k. Now considera walk v0WvivviWvk which is longer then W and also traverses each of its edgesexactly once in each direction, a contradiction. So each vertex in G visited by W .

Again suppose that W doesn’t contain all the edges, since W visits every vertexin G so G has an edge e = vivj(i < j) not traversed by W . Consider a new walkv0WvivjviWvk which is longer then W and also traverses each of its edges exactlyonce in each direction, a contradiction. So each edge in G visited by W .

Hence every connected graph contains a walk that traverses each of its edgesexactly once in each direction.

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2 Matching, Covering and Packing

See some extra problem on Matching, Covering and Packing in the end( problem-6.2.3 - problem-6.2.7).

Exercise 2.2. Describe an algorithm that finds, as efficiently as possible, a matchingof maximum cardinality in any bipartite graph.

Proof. We already know that A matching M is maximum in a graph G if and onlyif there are no augmenting paths with respect to M . Let A and B be the bipartitionof G, and let M be the matching in G, initially M = ∅ . Following algorithm knownas Hungarian Method.

Algorithm 1 Maximum-Matching (G,A,B,M)

1: if M saturates every vertex in A then2: stop;

3: Let u be an an M -unsaturated vertex in A.4: Set S = {u}, T = ∅.5: if N(S) = T then |N(S)| < |S|, since |T | = |S| − 1 then by Hall’s theorem there

is no matching that saturates every vertex in A then6: stop;

7: Let v ∈ N(S) \ T .8: if v is M -saturated, let vw ∈M then9: Set S = S ∪ {w}, T = T ∪ {v}; (Observe that |T | = |S| − 1 is maintained).10: goto step-5.

11: Otherwise we get an M -augmenting u− v path.12: M = M4P = (M \P )∪ (P \M); (symmetric difference of the two sets of edges)13: Maximum-Matching (G,A,B,M)

Exercise 2.3. Show that if there exist injective function A→ B and B → A betweentwo infinite set A and B then there exist a bijection between A→ B.

Proof. The above problem is known as Cantor-Schrouder-Bernstein theorem. Al-though the statement seemingly obvious statement is surprisingly difficult to prove.The the strategy of the proof is following:

Let f : A→ B and g : B → A be two injective map. First, we apply f(A) = B1 ⊆B. Next, g(B1) = A2 ⊆ A. Iterating this, we keep bouncing back and forth between

10


smaller and smaller subsets of A and B until the process stabilizes and we end upwith some sets A ⊆ A and B ⊆ B for which f(A) = B and g(B) = A. This impliesthat A ∼ B. The next task is to show that A \ A ∼ B \ B. Finally, we concludethat A ∼ B. You can get complete proof of this result in the book Introductory RealAnalysis by A. N. Kolmogorov and S. V. Fomin, 1st edition, Dover Publications.

Exercise 2.4. Find an infinite counterexample to the statement of the marriagetheorem.

Proof. Let A = Z+ ∪ {a} and B = Z+, here a is an alphabet. Let us considera graph G with vertex set V (G) = A ∪ B, (consider A and B are different set).Let xy ∈ E(G), x ∈ A and y ∈ B if and only if x = y or x = a. Then for anyS ⊆ A, |N(S)| ≥ |S|. But a matching saturating A must saturate Z+ ⊆ A, andsince these vertices have degree 1 and they already matched with every vertex in B,it cannot saturate a. Hence there is no matching saturating A.

Exercise 2.5. Let k be an integer. Show that any two partitions of a finite set intok-sets admit a common choice of representatives.

Proof. Let k ∈ N . Let X be a set of n elements with k|n and m = nk, and let

A1, · · · , Am and B1, · · · , Bm are partitions of V into k-sets. Let us define a bipartitegraph G with the vertex set V (G) = A ∪ B where A = {A1, · · · , Am} and B ={B1, · · · , Bm}, viewing each set Ai as a vertex. Let AiBj ∈ E(G) if and only ifAi ∩ Bj 6= ∅. We want to apply Hall theorem. Let S ⊆ A and we have to estimate|N(S)|. Let S contain t many element of A. Then number of element of X contain inS is tk. Now number of element of B covering these tk element is atleast t, since eachelement Bj contains k elements. It follows that |N(S)| ≥ |S|. Hence Hall conditionholds. Now by Halls theorem, we have a matching saturaing A, that is, we have aperfect matching. Therefore, any two partitions of a finite set into k-sets admit acommon choice of representatives.

Exercise 2.6. Let A be a finite set with subsets A1, · · · , An, and let d1, · · · , dn ∈ N.Show that there are disjoint subsets Dk ⊆ Ak , with |Dk| = dk for all k ≤ n, if andonly if ∣∣∣∣∣⋃

i∈I

Ai

∣∣∣∣∣ ≥∑i∈I

di

for all I ⊆ {1, · · · , n}.

Proof. Suppose∣∣⋃

i∈I Ai∣∣ ≥∑i∈I di holds for all I ⊆ {1, · · · , n}. Let us denote the

elements of Ai as follows, Ai = {ai1, ai2, · · · , aiti}, for 1 ≤ i ≤ n. Clearly for each

11


i, di ≤ ti. Let us construct a bipartite graph G with bipartition {X,A}, whereX =

⋃ni=1{aij : 1 ≤ j ≤ di} and join aij ∈ X to a ∈ A if and only if a ∈ Ai.

Notice that in G for any S ⊆ X, |N(S)| ≥ |S|. Thus by Hall’s theorem G containsa matching saturaing X. That matching gives us disjoint subsets Dk ⊆ Ak , with|Dk| = dk for all k ≤ n.

If there exist disjoint subsets Dk ⊆ Ak , with |Dk| = dk for all k ≤ n thenthis is equivalent to that there exist a matching in the bipartite graph G (which isconstructed earlier). saturating X. Hence, by the Hall’s theorem, for any S ⊆ X,|N(S)| ≥ |S|. Now for any I ⊆ {1, · · · , n} consider S =

⋃i∈I Di gives,

∣∣⋃i∈I Ai

∣∣ ≥∑i∈I di.

Exercise 2.8. Find a bipartite graph and a set of preferences such that no matchingof maximum size is stable and no stable matching has maximal size.

Proof. Try C6.

Exercise 2.9. Find a non-bipartite graph and a set of preferences that has no stablematching.

Proof. Try K3.

Exercise 2.10. Show that all stable matchings of a given bipartite graph cover thesame vertices. (In particular, they have same size.)

Proof. Suppose M1 and M2 are two stable matchings of G that don’t cover the samevertices. Then there exist some vertex x which is matched under M1 but unmatchedunder M2. Let xy ∈M1. Notice that y must be matched in M2, otherwise there exista edge xy whose both end points are unmatched under M2, a contradiction that M2

is a stable matching. Since y is matched in M2 thus there exist z ∈ V (G) such thatyz ∈ M2. Thus we have xyz path with edges alternately in M1 and M2. Continuethis way we get a path P = v0v1 · · · vn for some n ≥ 2 where v0 = x, v1 = y, v2 = z.Notice that vn−2 <vn−1 vn in M2, but vn <vn−1 vn−2 in M1, a contradiction. Thus M1

and M2 cover same vertices.

Exercise 2.13. Show that a graph G contains k independent edges if and only ifq(G− S) ≤ |S|+ |G| − 2k for all sets S ⊆ V (G).

Exercise 2.14. Find a cubic graph without a 1-factor.

Proof. From Theorem 2.2.1 we know that A graph G has a 1-factor if and only ifg(G− S) ≤ |S| for all S ⊆ V (G). Consider the following graph G,

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Figure 1: Cubic graph G Figure 2: Graph G− SLet us consider S ∈ V (G) to be red colored vertices. Thus q(G− S) = 6 > |S| = 4.Hence G can not have 1−factor.

Exercise 2.15. Derive the merriage theorem from Tutte’s theorem.

Proof.

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3 Connectivity

See some extra problem on connectivity in the end( problem-6.3.8 - problem-6.3.10).

Exercise 3.4. Let X and X ′ be minimal separating vertex sets in G such that Xmeets at least two components of G−X ′. Show that X ′ meets all the components ofG−X, and that X meets all the components of G−X ′.

Proof. Suppose that X ′ meets G − X in only one component, say C. Then X ′ ⊆X ∪C. So the components of G−X ′ are components which come from C−X ′ and acomponent which contains the rest of G. So, X meets only one component of G−X ′. This is a contradiction. Hence, X ′ meets at least two components of G−X. Then,it follows from symmetry that X meets every component of G−X ′ .

Exercise 3.7. Show, without using Menger’s theorem, that any two vertices of a2-connected graph lie on a common cycle (that is there exist two internally vertexdisjoint path between those two vertices).

Proof. This problem is known as Whitney’s theorem which is solved in 1932. You canfind this theorem in many books such as Graph theory with application by Bondy &Murty. But third solution in this note which given by Kewen Zhao is simplier thenany other solutions.

First Solution1: I have to show between any two vertices in a biconnectedgraph there exist two internally vertex disjoint path. I will show it by inductivelyby using induction on their distance. Let us assume x, y be any two vertex set inG and d = distG(x, y). If d = 1, that is xy ∈ E(G). since G is biconnected thusthere exist another x− y path in G, otherwise xy is a bridge in G, a contradiction.Let us assume the statement is hold for any vertex of degree less than d ≥ 2. LetS be the shortest path between x and y. Let v ∈ V (S) with lesser distance to yin S.Then distG(x, v) = d − 1 < d. Thus by induction hypothesis there exist twointernally vertex disjoint x − v path in G, say P1, P2. Now if y is a vertex of anyone of these path, say P1 then xP1y and P2y gives us two different paths, as inxP1y path v will not come as v is the end vertex and y comes before v. Let usassume y /∈ V (P1) ∪ V (P2). Since G − {v} is connected there exist a x − y path inG − {v}, say Q. Now consider the last vertex w ∈ V (Q)(according to the distancefrom x in Q) such that w ∈ V (P1) ∪ V (P2). Since w 6= v thus either w ∈ V (P1) orw ∈ V (P2). Without loss of any generality assume w ∈ V (P1). Now consider a newpath P3 := xP1wQy. Notice that V (P3)∩ V (P2) = {x}. Thus consider another pathP4 := xP2vy. Hence P3, P4 are two internal vertex disjoint x− y path.

1Based on Theorem-3.2(pp. 44–45), Bondy & Murty Book.

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Second solution:2 Suppose that there are two vertices u, v such that they arenot on a cycle. Then we want to show that there is a cut vertex separating u andv. To argue about the paths from u to v, we want to first order the vertices. Oneway to do this is to do a depth first search of G from u, label all the vertices ina pre-order traversal, and for each vertex w let a(w) be the smallest ancestor thatcan be reached from w through one of its descendants; note that there can only beback edges in this traversal, since G is undirected. We want to characterize whenare two vertices u and v on a cycle in such a DFS. It is clear that if a(v) = u thenthey are on a cycle, however, this is not necessary. The only other possibility is thefollowing: consider the path between v and a(v); if there is a vertex w on this pathsuch that a(w) = u then u and v are also on a cycle. Thus two vertices are on acycle iff in the path P from v to a(v) there is a vertex w such that a(w) = u. In oursetting, however, u and v are not on a cycle. The cut vertex separating u and v isthe ancestor a(w), where w ∈ P , closest to u.

Third solution:3 Let G be 2-connected graph and assume there exist two ver-tices u and v without two internally-disjoint u− v paths. Let P and Q be two u− vpaths with the common vertex set S as small as possible. Let w ∈ S \ {u, v} andP1 := uPw, P2 = wPv and Q1 := uQw,Q2 := wQv. Since G is 2-connected, let Rdenote a shortest path from some vertex x ∈ (V (P1) ∪ V (Q1)) \ {w} to some vertexy ∈ (V (P2) ∪ V (Q2)) \ {w} without passing through w. We may assume, withoutloss of generality, that x is in P1 and y in Q2. Let T denote the u− v path composedof uP1xRyQ2v. Clearly the common vertices of T and the u − v path composed ofQ1P2 are all in S \ {w}. This contradicts the choice of both P and Q as having thesmallest number of vertices common.

Exercise 3.9. Let G be a 2-connected graph but not a triangle, and let e be an edgeof G. Show that either G − e or G/e is again 2-connected. Deduce a constructivecharacterization of a 2-connected graphs analogous to Theorem 3.2.2.

Proof.

Exercise 3.10. Let G be a 3-connected graph, and let xy be an edge of G. Show thatG/xy is 3-connected if and only if G− {x, y} is 2-connected.

Proof. Given thatG is 3-connected with an edge xy ∈ E(G). LetG/xy is 3−connected.To show G−{x, y} is 2−connected. Suppose if possible G−{x, y} is not 2−connected.Then there exist a vertex z in G − {xy} which separate G − {x, y}. Then {z, vxy}becomes a separating set of G/xy , a contradiction. Hence G−{x, y} is 2-connected.

2http://www.imsc.res.in/~vikram/DiscreteMaths/2011/connectivity.pdf3A simple proof of Whitney’s theorem on connectivity in a graph by Kewen Zhao.

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http://www.imsc.res.in/~vikram/DiscreteMaths/2011/connectivity.pdf


Conversely, suppose G − {x, y} is 2−connected. To show G/xy is 3−connected.If possible let G/xy is not 3-connected. Then there exist a separating set {u, v} inG/xy which separate G/xy. Now if u, v 6= vxy then {u, v} becomes a separating setof G, a contradiction. Suppose u = vxy. Then v separates G−{x, y}, a contradiction.Hence G/xy is 3−connected.

Exercise 3.11. Show that every cubic 3-edge-connected graph is 3-connected.

Proof.

Lemma 3.1. Let G be a k ≥ 2 connected graph. Let S be the set of k vertices andu ∈ V \S. Then there exist k many u−S path contianing u as only common vertex.

Proof of lemma: Let us consider a graph G′ such that V (G′) = V (G) ∪ {x} andE(G′) = E(G) ∪ {xy|y ∈ S}. Then G′ is k-connected graph. By Menger’s theoremthere exist k (internal) vertex disjoint u − x path in G′. These path must passesthrough vertices of S. Since |S| = k, every path contain exactly one element fromS. Consider corresponding path in G gives required k paths.

Exercise 3.16. Let k ≥ 2. Show that every k-connected graph of order at least 2kcontains a cycle of length at least 2k.

Proof. Let C be the longest cycle in G. If v ∈ V (C), for all v ∈ V (G) then we aredone. Let there exist v ∈ V (G) \ V (C). Then by lemma-3.1 there exist k vertexdisjoint (only common vertex is v) v−C paths. Now if |C| < 2k then by pigeonholeprinciple there exist two paths whose end points are adjacent on C. We can constructa longer cycle by detouring along these paths, a contradiction.

Exercise 3.17. Let k ≥ 2. Show that in a k-connected graph any k vertices lie on acommon cycle.

Proof. I will prove this statement using induction on k. The base case of inductionis k = 2. It is directly follows from Menger’s theorem although I have given aalternating proof of this result in problem-3.7. For induction hypothesis let us assumefor any k−1 connected graph the statment holds. Now let G be a k-connected graphand S = {v1, · · · , vk} are k many vertices in G. Now since G is k-connected, thenG− v1 is k− 1 connected. By induction hypothesis we get vertices v2, · · · , vk are lieon a common cycle in G− v1, say C. Now conside two vertex set {v1} and V (C).

Case-1: Let l(C) = k−1. Consider v ∈ V \V (C) and C∪{v}. Then by lemma-3.1there k − 1 many internal vertex disjoint v1 − C path in G haveing v1 is their onlycommon vertex. Hence we can construct a new path containing all those k vertices.

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Case-2: Let l(C) ≥ k. Then there k many internal vertex disjoint v1−C path inG, say Pi, 1 ≤ i ≤ k such that if xi be the end points of Pi then V (Pi) ∩ C = {xi}and xi 6= xj, for each i 6= j. Such type of path exist by lemma-3.1. Without loss ofgenerality assume that xi are present in C in anticlockwise order. Then these manypaths devide C into k segment. Let Ci = xiCxi+1 for 1 ≤ i ≤ k (consider k+ 1 = 1)are the segment of C. Since |S − {v1}| = k − 1 then by pigeonhole principle atleastone segment does not contain any vertices of S−{v1}, say Ci. Then we can get a newcycle C = v1Pi+1xi+1CxiPiv1. Here xi+1Cxi taken anticlockwise direction. Hence Ccontain all k vertices of S.

Exercise 3.18.

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4 Planar Graphs

See some extra problem apart from Diestel’ books on planar graphs in the end(problem-6.4.11 - problem-6.4.15).

Exercise 4.5. Show that every planar graph is unioin of three forests.

Exercise 4.23. For every k ∈ N , construct a triangle free k-chromatic graph.

Proof. The following construction is known as Tutte’s construction of triangle-freek-chromatic graph.

G ⊕H to be the graph obtained by joining every vertex in G with every vertexin H. For k ≤ 5 it is easy to construct.

Now let k > 5. Clearly χ(G⊕H) = χ(G) +χ(H). (Exercise-6.5.18). Notice thatχ(C5⊕C5) = 6. Consider G6 = C5⊕C5 and for k > 6, consider Gk = Gk−1⊕Gk−1.

Claim. Gk is triangle free.

Proof. We will prove it by induction on k. Notice that G2 = K2, triangle free. LetGk−1 triangle free.

Claim. χ(Gk) = k.

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5 Colouring

See some extra problem on colouring in the end( problem-6.5.16 - problem-6.5.22).

Exercise 4.5. Show that every graph G has a vertex ordering for which the greedyalgorithm uses only χ(G) colours.

Proof. Since G is χ(G) colourable there exist a colouring of G which takes χ(G)colours. According to this colouring partition the vertex set V (G) into χ(G) parts.Let these parts are V1, · · · , Vχ(G). Notice that all vertices belongs to same partitionare independent. Now apply Greddy algorithm as follows: first colour all verticesof V1 by colour 1, after colour all vertices of V2 by colour 1 or 2. This is becauseeach such vertex may not have any neighbours among vertices from V1. In general,color all vertices of Vi with the colour from {1, 2, · · · , i} because some vertices of Vimay not adjacent to any already-colored vertices. Hence Greddy algorithm will useatmost χ(G) colour. Since G can not be colour by lower then χ(G) colour. HenceGreddy algorithm will use exactly χ(G) colour.

Exercise 4.6. For evry n > 1, find a bipartite graph on 2n vertices, ordered in sucha way that the greddy algorithm uses n rather than 2 colours.

Proof. Consider complete Kn,n graph with bipartition A,B. Let A = {a1, · · · , an}and B = {b1, · · · , bn}. Suppose M = {aibi : 1 ≤ i ≤ n} be the perfect matching ofKn,n. Let us consider a new bipartite graph G = Kn,n \M (removing only edges).Now apply Greddy algorithm in G with the vertex order {a1, b1, · · · , an, bn}. Thena1 and b1 will be assigned color 1, since they are not adjacent to any already-coloredvertices. Since a2 and b2 are each adjacent to a vertex already assigned color 1 theywill be assigned color 2. Agian a3 and b3, each adjacent to vertices of colors 1 and2, must be assigned color 3 and so on. Since an and bn each adjacent to vertices ofevery color from 1 to n− 1 it must be assigned color n.

Exercise 4.9. Find a lower bound for the colouring number in terms of averagedegree.

Proof. Recall proposition-1.2.2: Every graph G with atleast one degree has a sub-graph H∗ with δ(H∗) > ε(H∗) ≥ ε(G) = d(G)

2.

It is clear that,

χ(G) = maxH⊆G

δ(H) + 1

≥ δ(H∗) + 1 =d(G)

2+ 1.

Hence d(G)2

+ 1 ≤ χ(G).

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Exercise 4.10. A k-chromatic graph is called critically k-chromatic, or just critical,if χ(G − v) < k for every v ∈ V (G). Show that every k-chromatic graph has acritical k-chromatic induced subgraph, and that any such subgraph has minimumdegree atleast k − 1

Proof. For first part use induction. If the graph G itself is critically k-chromatic,then we are done, otherwise there exists a vertex v in G such that χ(G − v) = k.Then apply induction hypothesis to the graph G− v to find a critically k-chromaticsubgraph of G.

Let G be itself a critically k-chromatic graph. It is enough to show that, if thereexist a vertex v ∈ G such that deg(v) < χ(G)− 1 then χ(G) = χ(G− v).

Let there exist a vertex v ∈ G such that deg(v) = χ(G)− 2. Since χ(G− v) < kthus we can color G − v using k − 1 color. Since v is adjacent to k − 2 edges so itsneighbours will get atmost k − 2 colors so we can color v with remaining one color.Thus G is k − 1 colourable, a contradiction.

Exercise 4.11. Determine the critical 3-chromatic graphs.

Proof. We will prove that a graph is critical 3-chromatic graph if and only if it is anodd cycle.

Since we know that every odd cycle is 3-chromatic and if we delete any one vertexof this cycle, we will get a path which is 2-colourable. Hence every odd cycle is critical3-chromatic.

Let G be a critical 3-chromatic graph. Then G can not be a bipartite graph.Thus G had an odd cycle,say C. Claim is G = C. If not then there exist a vertexv ∈ G− C. Now notice that G− v contain an odd cycle C so its chromatic numberis 3, but since G is crticial 3-chromatic graph, χ(G− v) < 3, a contradiction. HenceG = C.

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6 Some Arbitary Problem

1 The Basics

Exercise 6.1.1. Prove that the number of simple even graphs (degree of all vertices

is even) with n vertices is 2(n−12 )

Proof. There is a bijection between simple graphs with n − 1 vertices and evensimple graphs on n vertices. Given a simple graph G with V (G) = {v1, · · · , vn}we can construct a even simple graph of n vertices. We know that no of verticesof odd degree is even. Construct a new graph G′ with V (G′) = V (G) ∪ {vn} andE(G′) = E(G)∪{vivn : vi ∈ V (G), degG(vi) is odd}. Then G′ is a even simple graph.Conversely given a even simple graph G′ we will get back G by G′ − vn. Since in asimple graph of n − 1 vertices can have atmost

(n−12

)edges thus no of even simple

graph of n vertices is 2(n−12 ).

Exercise 6.1.2.

2 Matching

Exercise 6.2.3. Prove that a nonempty bipartite graph has a matching such that allvertices of maximal degree are saturated.

Exercise 6.2.4. Show that the following ’obvious’ algorithm need not produce astable matching in a bipartite graph. Starting with any matching. If the currentmatching is not maximal, add an edge. If it is maximal but not stable, insert an edgethat creates instability, deleting any current matching edges at its ends.

Proof. Consider a bipartite graph G with bipartition {A,B} where A = {1, 2, 3, 4}and B = {a, b}.

4

3

2

1 a

b

Consider preferences:

• a : 3 > 2 > 1 > 4

• b : 2 > 3 > 4 > 1

• 1 : b > a

• 2 : a > b

• 3 : b > a

• 4 : a > bLet us consider the following matching:

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4

3

2

1 a

bNotice that a prefers 2 over 1, and 2 prefersa, so this is not a stable matching. Deletethe two edges consider 2a as new matchingedge with another independent edge.

4

3

2

1 a

b Notice that a prefers 3 over 2, and 3 is un-matched, so this is not a stable matching.Again doing same operation.

4

3

2

1 a

b Notice that b prefers 3 over 4, and 3 alsoprefer b, so this is not a stable matching.Again doing same operation.

4

3

2

1 a

b Notice that b prefers 2 over 3, and 2 is un-matched, so this is not a stable matching.Again doing same operation.

Hence we get looped back to where we started.

4

3

2

1 a

b

Thus given algorithm need not produce a stable matching.

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Exercise 6.2.5. A matching M in a graph is of maximal cardinality if and only ifthe graph has no augmenting path with respect to M .

Exercise 6.2.6. Prove that every tree has atmost one perfect martching.

Exercise 6.2.7.

3 Connectivity

Exercise 6.3.8.

Exercise 6.3.9. Let G be a biconnected graph with δ(G) ≥ 3. Prove that there exista vertex v ∈ V (G) such that G− v is also biconnected.

Exercise 6.3.10.

4 Planar graph

Exercise 6.4.11. Prove that every planar graph has a vertex of degree atmost 5.

Exercise 6.4.12. Prove that there does not exist any 6-connected planar graph.

Exercise 6.4.13. Prove that every planar 5-connected graph has atleast 12 vertices.

Exercise 6.4.14. For which r there exist a planar r-regular graph.

Exercise 6.4.15. Show that the petersen graph is not planar.

5 Colouring

Exercise 6.5.16. Find chromatic number of graphs a) Kn; b) Kn,m; c) Pn; d) Cn;e) Petersen graph.

Exercise 6.5.17. Prove that if every block of a graph is k-colorable then the graphis k-colorable.

Exercise 6.5.18. Let G,H be any two graph. Show that χG⊕H) = χ(G) + χ(H).

Exercise 6.5.19. Prove that difference ∆(G)− χ(G) maybe arbitarily large.

Exercise 6.5.20. Prove that the number of edges of a graph G is atleast χ(G)(χ(G)−1)/2.

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Proof. Let 0 < i, j < χ(G) and i 6= j. Then there exist an edge whose end verticescolor with i and j, otherwise we can color G with lesser number of color then χ(G).

Hence total no of edge is atleast(χ(G)2

)= χ(G)(χ(G)−1)

2.

Exercise 6.5.21. Find all counterexample to this statement: every connected graphG contains a vertex such that deg v ≥ χ(G)

Exercise 6.5.22.

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T1

T3 T5

T2

T4

a

ac d

d

b

b−1

c

Figure 3: Labeling of edges and identification of vertices in P .

a c

b

dx0 x1

Figure 4: Sketch of A = π(BdP ).

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7 Solved Exercise Reference

List of Solved Exercises

1 The Basics 3Exercise 1.1 . . . . . . . . . . . . 3Exercise 1.2 . . . . . . . . . . . . 3Exercise 1.3 . . . . . . . . . . . . 3Exercise 1.4 . . . . . . . . . . . . 4Exercise 1.5 . . . . . . . . . . . . 4Exercise 1.6 . . . . . . . . . . . . 5Exercise 1.7 . . . . . . . . . . . . 6Exercise 1.8 . . . . . . . . . . . . 6Exercise 1.10 . . . . . . . . . . . 6Exercise 1.11 . . . . . . . . . . . 7Exercise 1.12 . . . . . . . . . . . 7Exercise 1.16 . . . . . . . . . . . 7Exercise 1.17 . . . . . . . . . . . 7Exercise 1.19 . . . . . . . . . . . 8Exercise 1.23 . . . . . . . . . . . 8Exercise 1.24 . . . . . . . . . . . 9Exercise 1.26 . . . . . . . . . . . 9

2 Matching, Covering and Packing 10Exercise 2.2 . . . . . . . . . . . . 10Exercise 2.3 . . . . . . . . . . . . 10Exercise 2.4 . . . . . . . . . . . . 11Exercise 2.5 . . . . . . . . . . . . 11Exercise 2.6 . . . . . . . . . . . . 11Exercise 2.8 . . . . . . . . . . . . 12Exercise 2.9 . . . . . . . . . . . . 12Exercise 2.10 . . . . . . . . . . . 12Exercise 2.13 . . . . . . . . . . . 12Exercise 2.14 . . . . . . . . . . . 12

Exercise 2.15 . . . . . . . . . . . 13

3 Connectivity 14Exercise 3.4 . . . . . . . . . . . . 14Exercise 3.7 . . . . . . . . . . . . 14Exercise 3.9 . . . . . . . . . . . . 15Exercise 3.10 . . . . . . . . . . . 15Exercise 3.11 . . . . . . . . . . . 16Exercise 3.16 . . . . . . . . . . . 16Exercise 3.17 . . . . . . . . . . . 16Exercise 3.18 . . . . . . . . . . . 17

4 Planar Graphs 18Exercise 4.5 . . . . . . . . . . . . 18Exercise 4.23 . . . . . . . . . . . 18

5 Colouring 19Exercise 4.5 . . . . . . . . . . . . 19Exercise 4.6 . . . . . . . . . . . . 19Exercise 4.9 . . . . . . . . . . . . 19Exercise 4.10 . . . . . . . . . . . 20Exercise 4.11 . . . . . . . . . . . 20

6 Some Arbitary Problem 211 The Basics . . . . . . . . . . . . . 212 Matching . . . . . . . . . . . . . 213 Connectivity . . . . . . . . . . . . 234 Planar graph . . . . . . . . . . . 235 Colouring . . . . . . . . . . . . . 23

7 Solved Exercise Reference 26

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Documents

Selected Solutions to Graph Theory, 3rd Edition · many vertices are adjacent to second vertex except rst vertex, obviously answer is n 2. Similarly compute how many vertices are