Selected Solutions to Graph Theory, 3rd Edition
Reinhard Reinhard Diestel Diestel
Rakesh Rakesh Jana Depar Department tment of Mathema Mathematics tics
IIT Guwaha Guwahati ti
May 1, 2016
Acknowledgement These solutions are the result of taking CS-520(Advanced Graph Theory) course in the Jan-July Jan-July semester semester of 2016 at Indian Institute Institute of Technology echnology Guwahati. Guwahati. This is not a complete complete set of solu soluti tion onss in that book. It may happen happen that that soluti solution on of some problem problem may be wrong. wrong. I have not verified verified these problem problem from some expart. It is my kind request request you that do not belive belive the answer answer blindly blindly.. If you found any mistak mistakee please please inform inform me. I know know these these articl articlee must must contai contain n some some typogra typographi phical cal errors, in that case please inform me. If you have any better solution in any of these proble problem m please please let me know. know. I will will upload upload that soluti solution on in this this conten contentt with with your your name. name. If you want want to discuss discuss any of these these solution solution with me please 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. end. Please Please e-mail e-mail jana.rakesh.
[email protected] [email protected] or o r
[email protected] for any corrections and suggestions.
Copyright c 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
22
1 2 3 4 5
The Basics . . Matching . . . Connectivity . Planar graph . Colouring . . .
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11 Solved Exercise Reference
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22 22 24 24 24 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(n − 1) (n − 1) + (n − 2) + ··· + 1 + 0 = . 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
| |
2 E =
d(v)
v ∈V
|E | =
d
d
×2 2
= d
d−1
×2
.
2|E | |V |
• Thus average degree of Q = = d. • Notice that the distance between any two vertices depends on the number of d
different bits, so diameter is d, i.e. diam G = d.
• Girth(Q ) = ∞, because there are no cycles on hypercube graph Q . Girth(Q ) = 4, where d ≥ 2, this is because Q ∼ = K × Q . • Circumference of Q is 2 . 1
1
d
d
d
2
d−1
d
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
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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 i th time at vertex x i , and arrives at C for the i th time at vertex yi (it is possible for x i+1 = y i ). Then the x i , yi -portion of P together with the x i , 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 generality 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. ¯i = yi P xi+1 . Let t < k. Let us consider subpath of path P , P i = xi P yi and P Then there are atmost t many internally disjoint subpath of P . Let P k be the subpath of P with maximum length, say l( pk ) = m. Then
≥ √
√
√
≤ l(P ) ≤ tm √ ≤ km
k
√
√ √ √ √
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 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) d G (x, y). rad G = min max d G (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 d G (a, b) = diam G and rad G = maxy∈V (G) d G (v, y).
∈
diam G = d G (a, b)
≤ d (a, v) + d (v, b) ≤ rad G + rad G = 2 rad G. G
4
G
Rakesh Jana
IIT Guwahati
Exercise 1.6. Prove the weakening of Theorem 1.3.4 obtained by replacing average
| | ≥ n (d/2, g) for every graph G as given in
with minimum degree. Deduce that G the theorem.
0
∈
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 D j = , i = j and V (G) = i≥0 Di . Since v can not contained in any cycle of length lesser then 2 r + 1. Thus for any u, w D i , 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
∈
≤ ≤
∈ { ∈ } ∈ ∪{ } ∩ ∅∀ ∪ ≤ ≤ − ∩ ∩ − − ≤ ≤ − − | | | | ≥ | | ≥ − r −1
r −1
|V (G)| ≥ |D | = 1 + δ i
i=0
(δ
i=0
− 1)
i−1
.
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) := d G−uv (x, y). Similar way as case-1, for any x where d(x, u, v , D ix D jx = for i = j and for any a, b Dix , N (a) D ix+1 is disjoint from N (b) D ix+1 for 1 i < r 1. Also each vertex in Dix (0 i < r 1) is connected to exactly one vertex in Dix−1 and atleast δ 1 vertices in Dix+1 . Now D0x = 1, D1x δ 1 and x i (δ 1) , for 1 i r 1. Let us define for x Di u, v ,
≤ − | |≥ −
∈ − ≤ ≤ −
∈ { } ∩ ∩ − | | | | ≥ − ∈{ }
∩ ≤
T x =
∅
Dix .
0≤i≤r−1 −1 Notice that T x = ri=0 (δ 1)i for any x u, v . Claim. T u T v = . To prove this claim first notice that Diu D jv = , for all 0 i, j r 1. If not let x D iu D jv 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 T u T v = . −1 Hence V (G) (δ 1)i . T u + T v = 2 ri=0
| | − ∈{ } ∩ ∅ ∩ ≤ ≤ − ∈ ∩ ≤ ≤ − − − − ∩ ∅ | | ≥ | | | | − 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 ),
∈ 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). then there exist v
Exercise 1.7. Show that every connected graph G contains a path of length at least
| | − 1}.
min 2δG, G
{
···
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 = yxi P xk x0 P ˚ xi 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 = x j x j +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 x j S such that x j xk E (G). Hence we get a cycle C = x0 P x j xk P x j +1 x0 . Now consider the path min 2δG, G P ∗ = yx i C ˚ xi which is longer then P , a contradiction. Hence k 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 d G (x, y) = k, for some x, y V (G) and distance achieve by the path P = x0 x1 xk , where x 0 = 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 x j for j > i + 2 then we can get a path x 0 P xi vx j P xk and which is shorter then P , a contradiction to d G (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 , K m,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 P m , it becomes a disconnect graph. λ(P m ) = 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 C n 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(K m,n ) = min m, n λ(K m,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?
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Rakesh Jana
IIT Guwahati
{ ∈
}
Proof. Let T be a tree with no vertex of degree 2. Let V i = v V (G) : d G (v) = i . |−1) Notice that V 2 = . Now average degree of a tree is 2||V E || = 2(|V < 2. Now |V |
∅
2 >
|V |−1
v∈V d G (v) = V
|V |−1 i=3
| | = |V | + i|V | |V | |V |
i=1
i V i
1
i
| | ≥ |V | + 3 |V | |V | | | |V | 2 V | + 2 V | = = 1+ |V | |V | 1
|V |−1 i=3
|V |−1 i=3
i
|V |−1 i=3
i
i
Hence we get,
|V | + 1
|V |−1
|V |−1
|V |−1
i=3
|V | = |V | > 2 i
i=3
|V | =⇒ |V | > i
1
i=3
|V |. i
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
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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 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 C 1 = xCyx, C 2 = 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 4 k 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 = v 0 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 v j (i < j) not traversed by W . Consider a new walk v0 W vi v j 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.
∈
≤ ≤
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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: if M saturates 2: stop;
every vertex in A then
3: Let u be an an M -unsaturated vertex 4: Set S = u , T = . 5: if N (S ) = T then N (S ) < S , since
{}
∅ |
| ||
in A.
|T | = |S | − 1 then by Hall’s theorem there
is no matching that saturates every vertex in A then 6: stop; 7: Let v N (S ) T . 8: if v is M -saturated, let vw M then Set S = S w , T = T 9: v ; (Observe that T = S 1 is maintained). goto step-5. 10: 11: Otherwise we get an M -augmenting u v path. 12: M = M P = (M P ) (P M ); (symmetric difference of the two sets of edges) 13: Maximum-Matching (G,A,B,M )
∈
\ ∪{ }
∈ ∪ { }
\ ∪ \
| | | | −
−
→ B and B → A between two infinite set A and B then there exist a bijection between A → B. Exercise 2.3. Show that if there exist injective function A
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 ) = A 2 A. Iterating this, we keep bouncing back and forth between
→
⊆
→
⊆
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smaller and smaller subsets of A and B until the process stabilizes and we end up ¯ B for which f (A) ¯ = B ¯ and g(B) ¯ = A. ¯ This implies with some sets A¯ A and B ¯ The next task is to show that A A¯ ¯ Finally, we conclude that A¯ B. B B. 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 B 1 , , 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 B j E (G) if and only if Ai B j = . 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 B j 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.
··· { ··· } ∩ ∅ | |
∈
···
∪
⊆
|
| { ··· ∈
}
| ≥| |
··· , A , and let d , ··· , d ∈ N. Show that there are disjoint subsets D ⊆ A , with |D | = d for all k ≤ n, if and Exercise 2.6. Let A be a finite set with subsets A 1 , k
only if
k
1
n
k
n
k
Ai ≥ di i∈I
⊆ {1, ··· , n}.
for all I
i∈I
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 ,
··· , a }, for 1 ≤ i ≤ n.
11
iti
Clearly for each
Rakesh Jana
≤
IIT Guwahati
{
}
i, di ti . Let us construct a bipartite graph G with bipartition X, A , where X = 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 contains a matching saturaing X . That matching gives us disjoint subsets Dk Ak , with Dk = d k 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 theorem, for any S X , 1, N (S ) S . Now for any I , n consider S = i∈I Di gives, i∈I A i i∈I d i .
{
| | |
≤ ≤ } ⊆ | ≤
|≥| |
∈ | ≥| |
∈
∈
⊆
⊆
| |
≤
⊆
⊆ { ··· }
≥
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 M 1 and M 2 are two stable matchings of G that don’t cover the same vertices. Then there exist some vertex x which is matched under M 1 but unmatched under M 2 . Let xy M 1 . Notice that y must be matched in M 2 , otherwise there exist a edge xy whose both end points are unmatched under M 2 , a contradiction that M 2 is a stable matching. Since y is matched in M 2 thus there exist z V (G) such that yz M 2 . Thus we have xyz path with edges alternately in M 1 and M 2 . Continue this way we get a path P = v 0 v1 vn for some n 2 where v 0 = x, v1 = y, v2 = z . Notice that v n−2 < v vn in M 2 , but v n < v vn−2 in M 1 , a contradiction. Thus M 1 and M 2 cover same vertices.
∈
∈
∈
···
n−1
≥
n−1
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,
− ≤| |
⊆
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− S 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. Figure 1: Cubic graph G
Figure 2: Graph G
Exercise 2.15. Derive the merriage theorem from Tutte’s theorem.
Proof.
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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 be minimal separating vertex sets in G such that X
meets at least two components of G X . Show that X meets all the components of G 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 a component 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 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 P 1 , P 2 . Now if y is a vertex of any one of these path, say P 1 then xP 1 y and P 2 y gives us two different paths, as in xP 1 y path v will not come as v is the end vertex and y comes before v. Let us assume y / V (P 1 ) V (P 2 ). 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 (P 1 ) V (P 2 ). Since w = v thus either w V (P 1 ) or w V (P 2 ). Without loss of any generality assume w V (P 1 ). Now consider a new path P 3 := xP 1 wQy. Notice that V (P 3 ) V (P 2 ) = x . Thus consider another path P 4 := xP 2 vy. Hence P 3 , P 4 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 P 1 := uPw, P 2 = wP v and Q1 := uQw,Q2 := wQv. Since G is 2-connected, let R denote a shortest path from some vertex x (V (P 1 ) V (Q1 )) w to some vertex y (V (P 2 ) V (Q2 )) w without passing through w. We may assume, without loss of generality, that x is in P 1 and y in Q 2 . Let T denote the u v path composed of uP 1 xRyQ2 v. Clearly the common vertices of T and the u v path composed of Q1 P 2 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, v } becomes a separating set of G/xy , a contradiction. Hence G −{x, y} is 2-connected. xy
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 = v xy then u, v becomes a separating set of G, a contradiction. Suppose u = v xy . 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.
≥ 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 G such that V (G ) = V (G) ∪ {x} and E (G ) = E (G) ∪ {xy|y ∈ S }. Then G is k-connected graph. By Menger’s theorem there exist k (internal) vertex disjoint u − x path in G . These path must passes through vertices of S . Since |S | = k, every path contain exactly one element from Lemma 3.1. Let G be a k
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.
∈
\
−
| |
≥ 2. Show that in a k-connected graph any k vertices lie on a
Exercise 3.17. Let k
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 v 2 , , 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 v 1 C path in G, say P i , 1 i k such that if xi be the end points of P i then V (P i ) C = xi and xi = x j , for each i = j . Such type of path exist by lemma-3.1. Without loss of generality assume that x i are present in C in anticlockwise order. Then these many paths devide C into k segment. Let C i = x i 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 C i . Then we can get a new ¯ = v 1 P i+1 xi+1 Cxi P i v1 . Here x i+1 Cxi taken anticlockwise direction. Hence C ¯ cycle C 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.
∈ N , construct a triangle free k-chromatic graph.
Exercise 4.23. For every k
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 G 6 = C 5 C 5 and for k > 6, consider G k = G k−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 V 1 , , V χ(G) . Notice that all vertices belongs to same partition are independent. Now apply Greddy algorithm as follows: first colour all vertices of V 1 by colour 1, after colour all vertices of V 2 by colour 1 or 2. This is because each such vertex may not have any neighbours among vertices from V 1 . In general, color all vertices of V i with the colour from 1, 2, , i because some vertices of V i 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 K n,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 K n,n . Let us consider a new bipartite graph G = K n,n M (removing only edges). Now apply Greddy algorithm in G with the vertex order a1 , b1 , , an , bn . Then a1 and b 1 will be assigned color 1, since they are not adjacent to any already-colored vertices. Since a 2 and b 2 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(2G) . It is clear that,
≥
χ(G) = max δ (H ) + 1 H ⊆G
≥ δ (H ) + 1 = d(G) + 1. 2 ∗
Hence
d(G)
2
+1
≤ χ(G). 19
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
∈ N there is an n ∈ N
Exercise 9.6. Use Ramsey’s theorem to show that for any k, l
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
is even) with n vertices is 2(
)
n−1 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 G with V (G ) = V (G) vn and E (G ) = E (G) vi vn : v i V (G),degG (vi ) is odd . Then G is a even simple graph. Conversely given a even simple graph G we willget back G by G vn . Since in a simple graph of n 1 vertices can have atmost n−1 edges thus no of even simple 2 graph of n vertices is 2 ( ) .
∪{
∈
−
}
{ ··· } ∪{ }
−
n−1 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 pro- duce
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
2
b
• a : 3 > 2 > 1 > 4 • b : 2 > 3 > 4 > 1 • 1 : b > a • 2 : a > b • 3 : b > a • 4 : a > b
3 4 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)
∈ V (G) such that G − v is also biconnected.
exist a vertex v
≥ 3.
Prove that there
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) K n,m ; c) P n ; d) C n ;
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 Exercise 10.5.19. Prove that difference ∆(G)
⊕ H ) = χ(G) + χ(H ).
− 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 = j . Then there exist an edge whose end vertices color with i and j , otherwise we can with lesser number of color then χ(G). χ(Gcolor G χ(G)(χ(G)−1) ) Hence total no of edge is atleast 2 = . 2 Exercise 10.5.21. Find all counterexample to this statement: every connected graph
≥ χ(G)
G contains a vertex such that deg v Exercise 10.5.22.
25
Rakesh Jana
IIT Guwahati
T T T
4
2
1
T
T
3
5
a c
d b
a d
c b−1
Figure 3: Labeling of edges and identification of vertices in P .
b a
x0
d
x1
c
Figure 4: Sketch of A = π(Bd P ).
26
11
Solved Exercise Reference
List of Solved Exercises 1
The Basics
Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise 2
3
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
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Matching, Covering and Packing
Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Connectivity
Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise
3
2.2 . 2.3 . 2.4 . 2.5 . 2.6 . 2.8 . 2.9 . 2.10 2.13 2.14 2.15
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Planar Graphs
14 14 15 15 16 16 16 17 18
Exercise 4.5 . . . . . . . . . . . . 18 Exercise 4.23 . . . . . . . . . . . 18 Colouring
Exercise Exercise Exercise Exercise Exercise
19
4.5 . 4.6 . 4.9 . 4.10 4.11
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19 19 19 20 20
10
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10 9 10 11 11 11 12 10 12 12 12 12 13 14
Ramsey Theory for Graphs
Exercise 9.6 . . . . . . . . . . . . 21 Exercise 9.7 . . . . . . . . . . . . 21 Exercise 9.8 . . . . . . . . . . . . 21 Some Arbitary Problem
1 2 3 4 5
The Basics . . Matching . . Connectivity . Planar graph Colouring . .
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22
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11 Solved Exercise Reference
27
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22 22 24 24 24 27
