Selected Solutions to Graph Theory, 3rd Edition log hno y Gu ec h n ati :: Rake ah s r ola Sch h s rc tic sea ma ati Re the h a M wa Gu T II w titute of Ins T Reinhard Diestel :: Jana India Rakesh Jana Department of Mathematics IIT Guwahati May 1, 2016 Acknowledgement These solutions are the result of taking CS-520(Advanced Graph Theory) course in the Jan-July semester of 2016 at Indian Institute of Technology Guwahati. This is not a complete set of solutions in that book. It may happen that solution of some 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 any mistake please inform me. I know these article must contain some typographical errors, in that case please inform me. If you have any better solution in any of these problem please let me know. I will upload that solution in this content with your name. If you want to discuss any of these solution with me please ping me in my given email address or meet me in research scholar office (RS-E1-010) Department of Mathematics, IIT Guwahati. You can find List of Solved Exercises at the end. Please e-mail jana.rakesh. [email protected] or [email protected] for any corrections and suggestions. c Copyright 2015–2016, Rakesh Jana Contents 1 The Basics 3 2 Matching, Covering and Packing 10 3 Connectivity 14 4 Planar Graphs 18 5 Colouring 19 9 Ramsey Theory for Graphs 21 10 Some Arbitary Problem 1 The Basics . . . . . . 2 Matching . . . . . . . 3 Connectivity . . . . . 4 Planar graph . . . . . 5 Colouring . . . . . . . 22 22 22 24 24 24 . . . . . 11 Solved Exercise Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 Rakesh Jana 1 IIT Guwahati The Basics See some extra problem on basic in the end( problem-10.1.1 - problem-10.1.2). 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 how many vertices are adjacent to second vertex except first vertex, obviously answer is n − 2. Similarly compute how many vertices are adjacent to third vertex except first and 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, and circumference of the hypercube graph Qd . Proof. Since V is the set of all 0 − 1 sequences of length d. Thus total number of vertices is 2d , since in each place we can assign two number 0, 1. Since two such sequence form an edge if and only if they differ in exactly one position. Thus each vertices has degree d. Now we know that X 2|E| = d(v) v∈V |E| = • Thus average degree of Qd = d × 2d = d × 2d−1 . 2 2|E| |V | = d. • Notice that the distance between any two vertices depends on the number of different 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 ∼ = K 2 × Qd−1 . • Circumference of Qd is 2d . Exercise 1.3. Let G be a graph containing a cycle C, and assume that G contains a path of length at√ least k between two vertices of C. Show that G contains a cycle of length at least k. Is this best possible? 3 Rakesh Jana IIT Guwahati Proof. Let the path P start at x and end at y where x, y are lie on C. Suppose that P leaves C for the ith time at vertex xi , and arrives at C for the ith time at vertex yi (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 any these distinct generality Now if t ≥ k then we can take a path along C connecting √ 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 = xi P yi and P̄i = yi P xi+1 . Then there are atmost t many internally disjoint subpath of P . Let Pk be the subpath of 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 √ k. end points are in C and also distinct. Hence We get a cycle of length atleast √ √ 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 a cycle satisfying g(G) ≤ 2 diam G + 1. Is the bound is best possible? Proof. Yes. It is the best possible bound because equality occur when G = K 3 . Exercise 1.5. Show that rad G ≤ diam G ≤ 2 rad G. Proof. We know that diam G = maxx,y∈v(G) dG (x, y). rad G = min max dG (x, y) x∈V (G) y∈V (G) ≤ min max diam G x∈V (G) y∈V (G) = diam G. To show diam G ≤ 2 rad G. Let a, b, v ∈ V (G) such that dG (a, b) = diam G and rad G = maxy∈V (G) dG (v, y). diam G = dG (a, b) ≤ dG (a, v) + dG (v, b) ≤ rad G + rad G = 2 rad G. 4 Rakesh Jana IIT Guwahati Exercise 1.6. Prove the weakening of Theorem 1.3.4 obtained by replacing average with minimum degree. Deduce that |G| ≥ n0 (d/2, g) for every graph G as given in the theorem. Proof. Case 1: Consider g = 2r + 1, r ∈ N. This proof is similar to proof of proposition 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≥0 Di . Since v can not contained in any cycle of length lesser then 2r + 1. Thus for any u, w ∈ Di , 0 ≤ i ≤ r − 1, N (u) ∩ Di+1 is disjoint from N (w) ∩ Di+1 , otherwise we can construct 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−1 and 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 [ |Di | = 1 + δ i=0 r−1 X (δ − 1)i−1 . i=0 Case-2: Consider g = 2r, r ∈ N. In this case proof is same as previous one instead of a vertex we have to start with two adjacent vertices. Let uv ∈ E(G). Similar way consider for i ∈ N ∪ {0}, ˆ y) = i} Diu = {y : d(u, ˆ y) = i} Dv = {y : d(v, i ˆ y) := dG−uv (x, y). Similar way as case-1, for any x ∈ {u, v}, Dx ∩ Dx = ∅ where d(x, i j x x for is disjoint from N (b) ∩ Di+1 for i 6= j and for any a, b ∈ Dix , N (a) ∩ Di+1 x 1 ≤ i < r − 1. Also each vertex in Di (0 ≤ i < r − 1) is connected to exactly one x x . Now |D0x | = 1, |D1x | ≥ δ − 1 and and atleast δ − 1 vertices in Di+1 vertex in Di−1 |Dix | ≥ (δ − 1)i , for 1 ≤ i ≤ r − 1. Let us define for x ∈ {u, v}, [ Dix . Tx = 0≤i≤r−1 P i Notice that |Tx | = r−1 i=0 (δ − 1) for any x ∈ {u, v}. Claim. Tu ∩ Tv = ∅. To prove this claim first notice that Diu ∩ Djv = ∅, for all 0 ≤ i, j ≤ r − 1. If not let x ∈ Diu ∩ Djv for some 0 ≤ i, j ≤ r − 1. Then there exist a cycle in G of length atmost r − 1 + r − 1 + 1 = 2r − 1 < 2r, a contradiction. Hence Tu ∩ Tv = ∅. P i Hence |V (G)| ≥ |Tu | + |Tv | = 2 r−1 i=0 (δ − 1) . 5 Rakesh Jana IIT Guwahati 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 − w path in G, say Q. Now consider last common vertex p ∈ V (P ) ∩ V (Q) (traversing from 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 least min{2δG, |G| − 1}. Proof. Let us consider P = x0 x1 · · · xk be a longest path in G. We have to show k ≥ 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 exist y ∈ V (G) \ V (P ) such that yxi ∈ E(G), for some 0 ≤ i ≤ k. Now if x0 xk ∈ E(G) we can get a path P 0 = yxi P xk x0 P x̊i which is longer then P , a contradiction. Thus x0 xk ∈ / E(G). Now N (x0 ) ⊆ V (P ) and N (xk ) ⊆ V (P ), since P is the longest path. Let us consider S = {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 that xj xk ∈ E(G). Hence we get a cycle C = x0 P xj xk P xj+1 x0 . Now consider the path P ∗ = yxi C x̊i 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 terms of 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 diam G = 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 = x0 x1 · · · xk , where x0 = x, xk = y. Let u be a vertex not on P that is adjacent to some vertex on P . Let i be the smallest integer such that xi is adjacent to v. Notice that if v is adjacent to xj for j > i + 2 then we can get a path x0 P xi vxj P xk and which is shorter then P , a contradiction to dG (x, y) = k. Hence each v ∈ V (G) \ V (P ) can adjacent to at most 3 vertices on P . Exercise 1.10. Show that every 2-connected graph conatains a cycle. 6 Rakesh Jana IIT Guwahati Proof. Let G be a 2-connected graph. Then δ(G) ≥ 2, since if d(v) = 1 or 0, for some v ∈ V (G) then that vertex will be either a cut vertex or isolated, in both case it contradict that G is 2-connected. Hence by proposition-1.3.1, it has a cycle of length atleast δ(G) + 1. Exercise 1.11. Determine κ(G) and λ(G) for G = P m , C n , K n , Km,n , and Qd ; d, m, n ≥ 3. Proof. Recall. κ(G) denote for connectivity(vertex) of a graph G and λ(G) denote edge-connectivity of graph G. Also by proposition-1.4.2, κ(G) ≤ λ(G) ≤ δ(G). Given graphs are all connected. κ(P m ) = 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(C n ) = 2, because if you remove any vertex you will get P n−1 , hence if you delete any two vertex from Cn then it becomes a disconnect graph. λ(C n ) = 2, as if you delete any edge, the graph becomes P n and if you delete one more edge it become disconnected. Similarly, κ(K n ) = n − 1 λ(K n ) = n − 1 κ(Km,n ) = min{m, n} λ(Km,n ) = min{m, n} κ(Qd ) = d λ(Qd ) = d. Exercise 1.12. Is there any function f : N → N such that, for all k ∈ N, every graph 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 have minimum degree t but one of them is connected and other is not. For instance take G = K t+1 and take H be two disjoint component of K t+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 than other vertices. Can you find a very short proof that does not use induction? 7 Rakesh Jana IIT Guwahati 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| = 2(|V|V|−1) < 2. Now |V | | P|V |−1 P |−1 dG (v) |V1 | + |V i=1 i|Vi | i=3 i|Vi | 2> = = |V | |V | |V | P|V |−1 |V1 | + 3 i=3 |Vi | ≥ |V | P|V |−1 P |−1 |V | + 2 i=3 |Vi | 2 |V i=3 |Vi | = =1+ |V | |V | P v∈V Hence we get, |V1 | + |V |−1 |V |−1 X X |Vi | = |V | > 2 i=3 |V |−1 |Vi | =⇒ |V1 | > i=3 X |Vi |. i=3 This complete the answer. Exercise 1.19. Let G be a connected graph, and let r ∈ G be a vertex. Starting from r, 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 first reached (unless the current vertex is r; then stop).(This procedure has earned those trees the name of depth-first search trees.) Show that the edges traversed in depth-first search form a normal spanning tree in G with root r. Proof. First notice that in dfs we always get a tree, since we always add a vertex to the current subgraph if it is not end point of any edge of current subgraph, so cannot create a cycle. To show it is spanning. Suppose it not spanning tree of G then there is a vertex v which is not in the tree but adjacent to a vertex u in the tree. But then when we left u for the last time we would have visited v instead of returning to r. So we get a contradiction that depth-first search completed. Hence we get a spanning tree, 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 has even length. Proof. Recall. An induced cycle in G is a cycle in G forming an induced subgraph without any chords. 8 Rakesh Jana IIT Guwahati 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 us assume G is not bipartite. Since G is not bipartite so it has an odd cycle. Let C be a smallest odd cycle in G. Then C can not be induced cycle, since all induced cycle are even lengths. Then there exist x, y ∈ V (C) but xy ∈ / E(C). Thus we get two cycle C1 = xCyx, C2 = yCxy (traverse clockwise direction), among them one is odd and other is even. Hence we get a shorter odd cycle, a contradiction. This proves the result. Exercise 1.24. Find a function f : N → N such that, for all k ∈ N , every graph of average 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 consider this function is following: Every graph with an average degree of 4k have a subgraph H with minimum degree 2k, and we will lose another factor of 2 in moving H to its bipartite subgraph. Let H ∗ be the bipartite subgraph of H with the maximal number of edges. My claim is that H ∗ have minimum degree atleast k. If not, let v ∈ H ∗ such that dH ∗ (v) < k. This means v lost more then half of its neighbours in the process to form H to H ∗ . This means v is on the same partition with its looses neighbours. But in that case if we consider v in the other partition we can able to connect those previously looses vertices to v and form a new bipartite subgraph of H 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 that traverses each of its edges exactly once in each direction. Proof. Let W = v0 v1 · · · vk be a longest walk in G that traverses every edge exactly once in each direction. If possible let there exist a vertex v not visited by W . Without loss of any generality let us assume v ∈ N (vi ) for some 0 ≤ i ≤ k. Now consider a walk v0 W vi vvi W vk which is longer then W and also traverses each of its edges exactly 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 vertex in G so G has an edge e = vi vj (i < j) not traversed by W . Consider a new walk v0 W vi vj vi W vk which is longer then W and also traverses each of its edges exactly once 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 edges exactly once in each direction. 9 Rakesh Jana 2 IIT Guwahati Matching, Covering and Packing See some extra problem on Matching, Covering and Packing in the end( problem10.2.3 - problem-10.2.7). Exercise 2.2. Describe an algorithm that finds, as efficiently as possible, a matching of maximum cardinality in any bipartite graph. Proof. We already know that A matching M is maximum in a graph G if and only if there are no augmenting paths with respect to M . Let A and B be the bipartition of G, and let M be the matching in G, initially M = ∅ . Following algorithm known as Hungarian Method. Algorithm 1 Maximum-Matching (G, A, B, M ) 1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: if M saturates every vertex in A then stop; Let u be an an M -unsaturated vertex in A. Set S = {u}, T = ∅. 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 then stop; Let v ∈ N (S) \ T . if v is M -saturated, let vw ∈ M then Set S = S ∪ {w}, T = T ∪ {v}; (Observe that |T | = |S| − 1 is maintained). goto step-5. Otherwise we get an M -augmenting u − v path. M = M 4P = (M \ P ) ∪ (P \ M ); (symmetric difference of the two sets of edges) Maximum-Matching (G, A, B, M ) Exercise 2.3. Show that if there exist injective function A → B and B → A between two infinite set A and B then there exist a bijection between A → B. Proof. The above problem is known as Cantor-Schrouder-Bernstein theorem. Although 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 Rakesh Jana IIT Guwahati smaller and smaller subsets of A and B until the process stabilizes and we end up with some sets Ā ⊆ A and B̄ ⊆ B for which f (Ā) = B̄ and g(B̄) = Ā. This implies that Ā ∼ B̄. The next task is to show that A \ Ā ∼ B \ B̄. Finally, we conclude that A ∼ B. You can get complete proof of this result in the book Introductory Real Analysis by A. N. Kolmogorov and S. V. Fomin, 1st edition, Dover Publications. Exercise 2.4. Find an infinite counterexample to the statement of the marriage theorem. Proof. Let A = Z + ∪ {a} and B = Z + , here a is an alphabet. Let us consider a 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 any S ⊆ A, |N (S)| ≥ |S|. But a matching saturating A must saturate Z + ⊆ A, and since 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 into k-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 bipartite graph 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 Ai Bj ∈ E(G) if and only if Ai ∩ 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 in S is tk. Now number of element of B covering these tk element is atleast t, since each element Bj contains k elements. It follows that |N (S)| ≥ |S|. Hence Hall condition holds. Now by Halls theorem, we have a matching saturaing A, that is, we have a perfect matching. Therefore, any two partitions of a finite set into k-sets admit a common 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 and only if [ X di Ai ≥ i∈I i∈I for all I ⊆ {1, · · · , n}. S P 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 Rakesh Jana IIT Guwahati i, di ≤ Snti . Let us construct a bipartite graph G with bipartition {X, A}, where X = i=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 contains a 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 then this is equivalent to that there exist a matching in the bipartite graph G (which is constructed earlier). saturating X. Hence, by the Hall’s S theorem, for any S S ⊆ X, |N (S)| ≥ |S|. Now for any I ⊆ {1, · · · , n} consider S = D gives, i i∈I i∈I Ai ≥ P i∈I di . Exercise 2.8. Find a bipartite graph and a set of preferences such that no matching of maximum size is stable and no stable matching has maximal size. Proof. Try C 6 . Exercise 2.9. Find a non-bipartite graph and a set of preferences that has no stable matching. Proof. Try K 3 . Exercise 2.10. Show that all stable matchings of a given bipartite graph cover the same vertices. (In particular, they have same size.) Proof. Suppose M1 and M2 are two stable matchings of G that don’t cover the same vertices. Then there exist some vertex x which is matched under M1 but unmatched under M2 . Let xy ∈ M1 . Notice that y must be matched in M2 , otherwise there exist a 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 that yz ∈ M2 . Thus we have xyz path with edges alternately in M1 and M2 . Continue this way we get a path P = v0 v1 · · · 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 if q(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 if g(G − S) ≤ |S| for all S ⊆ V (G). Consider the following graph G, 12 Rakesh Jana IIT Guwahati Figure 2: Graph G − S Figure 1: Cubic graph G Let 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. 13 Rakesh Jana 3 IIT Guwahati Connectivity See some extra problem on connectivity in the end( problem-10.3.8 - problem-10.3.10). Exercise 3.4. Let X and X 0 be minimal separating vertex sets in G such that X meets at least two components of G − X 0 . Show that X 0 meets all the components of G − X, and that X meets all the components of G − X 0 . Proof. Suppose that X 0 meets G − X in only one component, say C. Then X 0 ⊆ X ∪ C. So the components of G − X 0 are components which come from C − X 0 and a component which contains the rest of G. So, X meets only one component of G − X 0 . This is a contradiction. Hence, X 0 meets at least two components of G − X. Then, it follows from symmetry that X meets every component of G − X 0 . Exercise 3.7. Show, without using Menger’s theorem, that any two vertices of a 2-connected graph lie on a common cycle (that is there exist two internally vertex disjoint path between those two vertices). Proof. This problem is known as Whitney’s theorem which is solved in 1932. You can find 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 then any other solutions. First Solution1 : I have to show between any two vertices in a biconnected graph there exist two internally vertex disjoint path. I will show it by inductively by using induction on their distance. Let us assume x, y be any two vertex set in G and d = distG (x, y). If d = 1, that is xy ∈ E(G). since G is biconnected thus there 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. Let S be the shortest path between x and y. Let v ∈ V (S) with lesser distance to y in S.Then distG (x, v) = d − 1 < d. Thus by induction hypothesis there exist two internally vertex disjoint x − v path in G, say P1 , P2 . Now if y is a vertex of any one of these path, say P1 then xP1 y and P2 y gives us two different paths, as in xP1 y path v will not come as v is the end vertex and y comes before v. Let us assume y ∈ / V (P1 ) ∪ V (P2 ). Since G − {v} is connected there exist a x − y path in G − {v}, say Q. Now consider the last vertex w ∈ V (Q)(according to the distance from x in Q) such that w ∈ V (P1 ) ∪ V (P2 ). Since w 6= v thus either w ∈ V (P1 ) or w ∈ V (P2 ). Without loss of any generality assume w ∈ V (P1 ). Now consider a new path P3 := xP1 wQy. Notice that V (P3 ) ∩ V (P2 ) = {x}. Thus consider another path P4 := xP2 vy. Hence P3 , P4 are two internal vertex disjoint x − y path. 1 Based on Theorem-3.2(pp. 44–45), Bondy & Murty Book. 14 Rakesh Jana IIT Guwahati Second solution:2 Suppose that there are two vertices u, v such that they are not on a cycle. Then we want to show that there is a cut vertex separating u and v. To argue about the paths from u to v, we want to first order the vertices. One way to do this is to do a depth first search of G from u, label all the vertices in a pre-order traversal, and for each vertex w let a(w) be the smallest ancestor that can be reached from w through one of its descendants; note that there can only be back edges in this traversal, since G is undirected. We want to characterize when are two vertices u and v on a cycle in such a DFS. It is clear that if a(v) = u then they are on a cycle, however, this is not necessary. The only other possibility is the following: consider the path between v and a(v); if there is a vertex w on this path such that a(w) = u then u and v are also on a cycle. Thus two vertices are on a cycle iff in the path P from v to a(v) there is a vertex w such that a(w) = u. In our setting, however, u and v are not on a cycle. The cut vertex separating u and v is the ancestor a(w), where w ∈ P , closest to u. Third solution:3 Let G be 2-connected graph and assume there exist two vertices u and v without two internally-disjoint u − v paths. Let P and Q be two u − v paths with the common vertex set S as small as possible. Let w ∈ S \ {u, v} and P1 := uP w, P2 = wP v and Q1 := uQw, Q2 := wQv. Since G is 2-connected, let R denote a shortest path from some vertex x ∈ (V (P1 ) ∪ V (Q1 )) \ {w} to some vertex y ∈ (V (P2 ) ∪ V (Q2 )) \ {w} without passing through w. We may assume, without loss of generality, that x is in P1 and y in Q2 . Let T denote the u − v path composed of uP1 xRyQ2 v. Clearly the common vertices of T and the u − v path composed of Q1 P2 are all in S \ {w}. This contradicts the choice of both P and Q as having the smallest number of vertices common. Exercise 3.9. Let G be a 2-connected graph but not a triangle, and let e be an edge of G. Show that either G − e or G/e is again 2-connected. Deduce a constructive characterization 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 that G/xy is 3-connected if and only if G − {x, y} is 2-connected. Proof. Given that G is 3-connected with an edge xy ∈ E(G). Let G/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. 2 3 http://www.imsc.res.in/~vikram/DiscreteMaths/2011/connectivity.pdf A simple proof of Whitney’s theorem on connectivity in a graph by Kewen Zhao. 15 Rakesh Jana IIT Guwahati 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} in G/xy which separate G/xy. Now if u, v 6= vxy then {u, v} becomes a separating set of 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 and u ∈ V \ S. Then there exist k many u − S path contianing u as only common vertex. Proof of lemma: Let us consider a graph G0 such that V (G0 ) = V (G) ∪ {x} and E(G0 ) = E(G) ∪ {xy|y ∈ S}. Then G0 is k-connected graph. By Menger’s theorem there exist k (internal) vertex disjoint u − x path in G0 . These path must passes through vertices of S. Since |S| = k, every path contain exactly one element from S. 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 2k contains 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 are done. Let there exist v ∈ V (G) \ V (C). Then by lemma-3.1 there exist k vertex disjoint (only common vertex is v) v − C paths. Now if |C| < 2k then by pigeonhole principle there exist two paths whose end points are adjacent on C. We can construct a 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 a common cycle. Proof. I will prove this statement using induction on k. The base case of induction is k = 2. It is directly follows from Menger’s theorem although I have given a alternating proof of this result in problem-3.7. For induction hypothesis let us assume for any k − 1 connected graph the statment holds. Now let G be a k-connected graph and S = {v1 , · · · , vk } are k many vertices in G. Now since G is k-connected, then G − v1 is k − 1 connected. By induction hypothesis we get vertices v2 , · · · , vk are lie on 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-3.1 there k − 1 many internal vertex disjoint v1 − C path in G haveing v1 is their only common vertex. Hence we can construct a new path containing all those k vertices. 16 Rakesh Jana IIT Guwahati Case-2: Let l(C) ≥ k. Then there k many internal vertex disjoint v1 − C path in G, 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 of generality assume that xi are present in C in anticlockwise order. Then these many paths devide C into k segment. Let Ci = xi Cxi+1 for 1 ≤ i ≤ k (consider k + 1 = 1) are the segment of C. Since |S − {v1 }| = k − 1 then by pigeonhole principle atleast one segment does not contain any vertices of S −{v1 }, say Ci . Then we can get a new cycle C̄ = v1 Pi+1 xi+1 Cxi Pi v1 . Here xi+1 Cxi taken anticlockwise direction. Hence C̄ contain all k vertices of S. Exercise 3.18. 17 Rakesh Jana 4 IIT Guwahati Planar Graphs See some extra problem apart from Diestel’ books on planar graphs in the end( problem-10.4.11 - problem-10.4.15). 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-free k-chromatic graph. G ⊕ H to be the graph obtained by joining every vertex in G with every vertex in H. For k ≤ 5 it is easy to construct. Now let k > 5. Clearly χ(G⊕H) = χ(G)+χ(H). (Exercise-10.5.18). Notice that χ(C 5 ⊕ C 5 ) = 6. Consider G6 = C 5 ⊕ C 5 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 = K 2 , triangle free. Let Gk−1 triangle free. Claim. χ(Gk ) = k. 18 Rakesh Jana 5 IIT Guwahati Colouring See some extra problem on colouring in the end( problem-10.5.16 - problem-10.5.22). Exercise 4.5. Show that every graph G has a vertex ordering for which the greedy algorithm 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 partition are independent. Now apply Greddy algorithm as follows: first colour all vertices of V1 by colour 1, after colour all vertices of V2 by colour 1 or 2. This is because each 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 Vi may not adjacent to any already-colored vertices. Hence Greddy algorithm will use atmost χ(G) colour. Since G can not be colour by lower then χ(G) colour. Hence Greddy algorithm will use exactly χ(G) colour. Exercise 4.6. For evry n > 1, find a bipartite graph on 2n vertices, ordered in such a 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 = {ai bi : 1 ≤ i ≤ n} be the perfect matching of Kn,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 }. Then a1 and b1 will be assigned color 1, since they are not adjacent to any already-colored vertices. Since a2 and b2 are each adjacent to a vertex already assigned color 1 they will be assigned color 2. Agian a3 and b3 , each adjacent to vertices of colors 1 and 2, must be assigned color 3 and so on. Since an and bn each adjacent to vertices of every 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 average degree. Proof. Recall proposition-1.2.2: Every graph G with atleast one degree has a subgraph H ∗ with δ(H ∗ ) > (H ∗ ) ≥ (G) = d(G) . 2 It is clear that, χ(G) = max δ(H) + 1 H⊆G ≥ δ(H ∗ ) + 1 = Hence d(G) 2 + 1 ≤ χ(G). 19 d(G) + 1. 2 Rakesh Jana IIT Guwahati 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 a critical k-chromatic induced subgraph, and that any such subgraph has minimum degree 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-chromatic subgraph of G. Let G be itself a critically k-chromatic graph. It is enough to show that, if there exist 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) < k thus we can color G − v using k − 1 color. Since v is adjacent to k − 2 edges so its neighbours 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 an odd cycle. Since we know that every odd cycle is 3-chromatic and if we delete any one vertex of this cycle, we will get a path which is 2-colourable. Hence every odd cycle is critical 3-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 vertex v ∈ G − C. Now notice that G − v contain an odd cycle C so its chromatic number is 3, but since G is crticial 3-chromatic graph, χ(G − v) < 3, a contradiction. Hence G = C. 20 Rakesh Jana 9 IIT Guwahati Ramsey Theory for Graphs Exercise 9.6. Use Ramsey’s theorem to show that for any k, l ∈ N there is an n ∈ N such that every sequence of n distinct integers contains an increasing subsequence of length k + 1 or a decreasing subsequence of length l + 1. Find an example showing that n > kl. Then prove the theorem of Erdos and Szekeres that n = kl + 1 will do. Proof. Exercise 9.7. Sketch a proof of the following theorem of Erdos and Szekeres: for every k ∈ N there is an n ∈ N such that among any n points in the plane, no three of them collinear, there are k points spanning a convex k-gon, i.e. such that none of them lies in the convex hull of the others. Proof. Exercise 9.8. Prove the following result of Schur: for every k ∈ N there is an n ∈ N such that, for every partition of {1, · · · , n} into k sets, at least one of the subsets contains numbers x, y, z such that x + y = z. Proof. 21 Rakesh Jana 10 1 IIT Guwahati Some Arbitary Problem The Basics Exercise 10.1.1. Prove that the number of simple even graphs (degree of all vertices n−1 is even) with n vertices is 2( 2 ) Proof. There is a bijection between simple graphs with n − 1 vertices and even simple 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 vertices of odd degree is even. Construct a new graph G0 with V (G0 ) = V (G) ∪ {vn } and E(G0 ) = E(G) ∪ {vi vn : vi ∈ V (G), degG (vi ) is odd}. Then G0 is a even simple graph. Conversely given a even simple graph G0 we will get back G by G0 − vn . Since in a simple graph of n − 1 vertices can have atmost n−1 edges thus no of even simple 2 n−1 graph of n vertices is 2( 2 ) . Exercise 10.1.2. 2 Matching Exercise 10.2.3. Prove that a nonempty bipartite graph has a matching such that all vertices of maximal degree are saturated. Exercise 10.2.4. Show that the following ’obvious’ algorithm need not produce a stable matching in a bipartite graph. Starting with any matching. If the current matching is not maximal, add an edge. If it is maximal but not stable, insert an edge that 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}. Consider preferences: 1 a • a:3>2>1>4 2 b • b:2>3>4>1 • 1:b>a 3 • 2:a>b 4 • 3:b>a • 4:a>b Let us consider the following matching: 22 Rakesh Jana IIT Guwahati 1 a 2 b Notice that a prefers 2 over 1, and 2 prefers a, so this is not a stable matching. Delete the two edges consider 2a as new matching edge with another independent edge. 3 4 1 a 2 b Notice that a prefers 3 over 2, and 3 is unmatched, so this is not a stable matching. Again doing same operation. 3 4 1 a 2 b Notice that b prefers 3 over 4, and 3 also prefer b, so this is not a stable matching. Again doing same operation. 3 4 1 a 2 b Notice that b prefers 2 over 3, and 2 is unmatched, so this is not a stable matching. Again doing same operation. 3 4 Hence we get looped back to where we started. 1 a 2 b 3 4 Thus given algorithm need not produce a stable matching. 23 Rakesh Jana IIT Guwahati Exercise 10.2.5. A matching M in a graph is of maximal cardinality if and only if the graph has no augmenting path with respect to M . Exercise 10.2.6. Prove that every tree has atmost one perfect martching. Exercise 10.2.7. 3 Connectivity Exercise 10.3.8. Exercise 10.3.9. Let G be a biconnected graph with δ(G) ≥ 3. Prove that there exist a vertex v ∈ V (G) such that G − v is also biconnected. Exercise 10.3.10. 4 Planar graph Exercise 10.4.11. Prove that every planar graph has a vertex of degree atmost 5. Exercise 10.4.12. Prove that there does not exist any 6-connected planar graph. Exercise 10.4.13. Prove that every planar 5-connected graph has atleast 12 vertices. Exercise 10.4.14. For which r there exist a planar r-regular graph. Exercise 10.4.15. Show that the petersen graph is not planar. 5 Colouring Exercise 10.5.16. Find chromatic number of graphs a) K n ; b) Kn,m ; c) Pn ; d) Cn ; e) Petersen graph. Exercise 10.5.17. Prove that if every block of a graph is k-colorable then the graph is k-colorable. Exercise 10.5.18. Let G, H be any two graph. Show that χG ⊕ H) = χ(G) + χ(H). Exercise 10.5.19. Prove that difference ∆(G) − χ(G) maybe arbitarily large. Exercise 10.5.20. Prove that the number of edges of a graph G is atleast χ(G)(χ(G)− 1)/2. 24 Rakesh Jana IIT Guwahati Proof. Let 0 < i, j < χ(G) and i 6= j. Then there exist an edge whose end vertices color with i and j, otherwise we can color G with lesser number of color then χ(G). χ(G)(χ(G)−1) χ(G) . Hence total no of edge is atleast 2 = 2 Exercise 10.5.21. Find all counterexample to this statement: every connected graph G contains a vertex such that deg v ≥ χ(G) Exercise 10.5.22. 25 Rakesh Jana IIT Guwahati T4 T2 T1 T3 T5 a c d b a c d b−1 Figure 3: Labeling of edges and identification of vertices in P . b a x0 x1 c d Figure 4: Sketch of A = π(Bd P ). 26 11 Solved Exercise Reference List of Solved Exercises 1 2 3 The Basics Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise 1.1 . 1.2 . 1.3 . 1.4 . 1.5 . 1.6 . 1.7 . 1.8 . 1.10 1.11 1.12 1.16 1.17 1.19 1.23 1.24 1.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Matching, Covering and Exercise 2.2 . . . . . Exercise 2.3 . . . . . Exercise 2.4 . . . . . Exercise 2.5 . . . . . Exercise 2.6 . . . . . Exercise 2.8 . . . . . Exercise 2.9 . . . . . Exercise 2.10 . . . . Exercise 2.13 . . . . Exercise 2.14 . . . . Exercise 2.15 . . . . Connectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Packing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 3 3 3 4 4 5 6 6 6 7 7 7 7 8 8 9 9 Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise 3.4 . 3.7 . 3.9 . 3.10 3.11 3.16 3.17 3.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 14 15 15 16 16 16 17 4 Planar Graphs 18 Exercise 4.5 . . . . . . . . . . . . 18 Exercise 4.23 . . . . . . . . . . . 18 5 Colouring Exercise Exercise Exercise Exercise Exercise 10 10 9 10 11 11 11 12 10 12 12 12 12 13 4.5 . 4.6 . 4.9 . 4.10 4.11 Ramsey Theory Exercise 9.6 Exercise 9.7 Exercise 9.8 . . . . . for . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 19 19 19 20 20 Graphs 21 . . . . . . . . . . 21 . . . . . . . . . . 21 . . . . . . . . . . 21 Some Arbitary Problem 1 The Basics . . . . . . 2 Matching . . . . . . 3 Connectivity . . . . . 4 Planar graph . . . . 5 Colouring . . . . . . . . . . . 14 11 Solved Exercise Reference 27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 22 22 24 24 24 27