Solutions to Homework of Graph Theory

Li Yi∗

Last updated on 9th January

Exercise 2 A graph is even if and only if it has no edge cuts of odd size; a graph is bipartite if andonly if it has no odd cycle.

Proof. (i) ‘Only If’: Suppose [S, S] is a cut of G. Then we claim that∑

v∈S deg(v) = |[S : S]| +2E(G[S]), because the edges in G[S] contributes to the summation twice and the ones in [S : S]only once. Thus it follows easily that |[S : S]| is even.

‘If’: If G has an vertex, say v, of odd degree, then the cut [{v} : V (G) \ {v}] is odd.

(ii) ‘Only If’: Any cycle alternates between the two vertex classes, so it has even length.

‘If’: We may assume that the graph G is connected, since a graph is bipartite if its components arebipartite. Let the distance d(u, v) be the length of shortest path between u and v. Fix u. DefineUi = {v : d(u, v) = i}. Note that an edge of G can join vertices in Ui and Uj only if j = i orj = i + 1 or j = i− 1. We claim that there is no edge between vertices in Ui.

If xy ∈ Ui then the path x-u and y-u have the same length of i. Let z be the last common vertex.Then z-x, z-y and xy form a cycle of length 2(i− d(u, v))+1, which contradicts the absence of oddcycles.

Thus V (G) can be divided into two parts, U0 ∪ U2 ∪ U4 ∪ · · · and U1 ∪ U3 ∪ U5 ∪ · · · . And it isa bipartition of G.

Exercise 4 Prove Turan’s Theorem: Let n ≥ p ≥ 3. The unique simple graph G on n vertices withoutp-cliques and the maximum number of edges is the complete multipartite graph Kn1 , · · · ,Knp−1 , where∑

ni = n and |ni − nj | ≤ 1.

Proof. The proof here is from Exercise 5.2.23 of [1].Firstly, we claim that a maximal simple graph having no r +1-clique has an r-clique. Since adding an

edge to the graph may form an r +1-clique, hence the graph contains Kr+1 \ {e}, thus it has an r-clique.Now we establish an indentity about number of edges of Tn,r. Let a = [n/r], then Tn,r has n − ar

partite sets of size a + 1 and (a + 1)r − n partite sets of size a. Then Tn−r,r has n − ar partite sets ofsize a and (a + 1)r − n partite sets of a− 1. Hence

e(Tn,r)− e(Tn−r,r) =(

n− ar

2

)(2a + 1) +

((a + 1)r − n

2

)(2a− 1) + (n− ar)((a + 1)r − n)2a

=12(2n− r)(r − 1) =

(r

2

)+ (n− r)(r − 1).

∗Student ID 5040309741, Class F0403026

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Now we prove Turan’s Theorem by induction on n. It is trivial (including the uniqueness) when n < ras Tn,r = Kn. Inductive step. Suppose Turan’s Theorem is true for n, now we consider a simple graphwith n + r vertices. Since G has no r + 1-clique, it has an r-clique. Removing an r-clique from G yieldsa graph G′ with n vertices. Adding the r-clique back to G′ brings

(r2

)edges. Now we consider the edges

between G′ and the Kr. Since G has no r + 1-clique, it is clear that there is no vertex in G′ is connectedto all vertices in Kr. Hence there are at most n(r − 1) edges between G′ and Kr. We have that G hasat most e(G′) +

(r2

)+ n(r− 1) = e(Tn+r,r) edges. It is clear that Tn+r,r satisfies the condition of Turan’s

Theorem. The uniqueness is not hard to see from the construction of G.

Exercise 6 If mS(G) is odd for each two element set S ⊂ E(G), can we conclude that G is even?

Proof. No. An obvious counterexample is a graph with two vertices and three edges connecting them.But the statement is true for other cases, see Theorem 28.

Exercise 8 Prove that the number of paths between any two distinct vertices in an Eulerian graph iseven.

Proof. See Exercise 34, because that Theorem 7 implies the ‘only if’ part of Collorary 32.

Exercise 9 Generalize Theorem 7 to be a theorem on linear space.

Proof. We generalize it as follows: W is a linear space of Fn2 and W contains (1 1 · · · 1)T . Let S ⊆

{1, . . . , n}. We define WS = {x ∈ W : suppx ⊇ S} and m(WS) be the set of minimal vectors in WS .Then |m(WS)| is odd.

Exercise 10 Prove that for any graph G and S ⊂ E(G), |MS(G)| − |mS(G)| is an even number.

Exercise 13 Prove that each simple graph G contains at leastE(4E − V 2)

3Vtriangles.

Proof. Let (u v) be an edge of G. There are at least deg(u) + deg(v) − V vertices which are adjacentto both u and v. Thus the number of triangles t satisfies t ≥ 1

3

∑(deg(u) + deg(v) − V ), where the

summation is taken over the edges of G. For each vertex u, the number deg(u) appears exactly deg(u)times in this sum, so T ≥ 1

3 (∑

deg(u)2 − V E). Now we apply Cauchy-Schwarz inequality,

T ≥ 13

((∑

u deg(u))2

V− V E

)=

4E2 − EV 2

3V.

Exercise 14 Suppose G is a simple graph with bm2

4 c − n edges where n, m are positive integers. Provethat if G contains a triangle then it contains at least bm

2 c − n− 1 triangles.Exercise 17 Let e be an edge of G with u, v as its endpoints. If every vertex of G other than u, v hasodd degree, then there is an even number of Hamiltonian circuits passing through e.

Proof. We construct an auxiliary graph G′. Its vertex set is all Hamiltonian paths starting with the edgeuv. Two vertices in G′ are joined if one of them can be obtained from the other by adding an edge to theend and removing a disjoint edge, namely, H1 = uv . . . wx . . . y and H2 are joined if H2 = uv . . . wy . . . x.Let P = uv . . . w be a Hamiltonian path. Then

deg(P ) =

{deg(w)− 2, if uw ∈ E(G),deg(w)− 1, otherwise.

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So P has odd degree if and only if wu ∈ E(G). The set of Hamiltonian cycles containing uv is exactlythe set of Hamiltonian paths which have odd degree in G′. The conclusion follows from Theorem 14.

Exercise 18 Denote by tn the maximum number k such that each bi-coloring of E(Kn) contains kmonochromatic triangles. Prove the following.

t2n = 2(

n

3

),

t2n+1 =

{(n−2)n(2n+1)

6 , if n ≡ 0 (mod 2),(n−2)n(2n+1)

6 + 12 , if n ≡ 1 (mod 2).

Proof. Let us label the vertices from 1 and n and colour the graph in red and blue. Suppose i, j, k aredistinct vertices. Define f(i, j, k) = 2 if (i j) and (i k) have the same colour and f(i, j, k) = −1 otherwise.Now we see that it holds that

f(i, j, k) + f(j, k, i) + f(k, i, j) = 6

for a monochromatic triangle ijk and

f(i, j, k) + f(j, k, i) + f(k, i, j) = 0

otherwise.Let si =

∑j<k f(i, j, k) and s =

∑si. Suppose there are r red triangles and b blue ones. Then we

haves = 6(r + b).

On the other hand, let ri denote the number of red edges with endpoint i. Thus there are n− 1− ri blueedges with endpoint i. Hence

si = 2(

ri

2

)+ 2(

n− 1− ri

2

)− ri(n− 1− ri) = 3r2

i − 3(n− 1)ri + n2 − 3n + 2.

If n = 2m, then si ≥ m2 − 3m + 2 = (m − 1)(m − 2). Hence s ≥ 2m(m − 1)(m − 2), so b + r ≥2m(m−1)(m−2)

6 = 2(m3

)and t2m ≥ 2

(m3

). Conversely, let us consider the complete graph whose edges (i j)

are coloured red if i and j have the same parity, and blue otherwise. It is not difficult to verify that thereare exactly 2

(m3

)monochromatic triangles, so t2m ≤ 2

(m3

). Thus we conclude that t2m = 2

(m3

)as desired.

If n = 2m + 1, then si ≥ m(m − 2) and s ≥ m(m − 2)(2m + 1). So b + r ≥ m(m−2)(2m+1)6 , and

t2m+1 ≥⌈

m(m−2)(2m+1)6

⌉. This is equivalent to

t2m+1 ≥

{(m−2)m(2m+1)

6 , if m ≡ 0 (mod 2),(m−2)m(2m+1)

6 + 12 , if m ≡ 1 (mod 2).

Case n = 2m+1 = 4k +1. We are going to find balanced-coloured graph. A balanced-coloured graphhere means that from each vertex there are the same number of red edges and blue ones. We constructthem by induction. The graph K5 can be drawn as all the line segments joining two of the vertices ofa regular convex pentagon. We colour the edges of the pentagon in read and others in blue. We get abalanced-coloured graph for k = 1. And in this case, there is no monochromatic triangles. Suppose we

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have had a balanced-coloured graph for k. Now we are going to colour K4(k+1)+1 in a balanced way. Wecolour the complete subgraph with vertices 1 to 5 and the subgraph with vertices 5 to 4(k + 1) + 1 inbalanced ways. Then we colour 1i and 2i in the same colour of 5i, and colour 3i and 4i in the othercolour for all i ≥ 6. Now we get a balanced graph for k + 1. For such a complete graph of size 4k + 1,ri = 2k for each i. Thus si = 4k(k − 1), s = 4(4k + 1)k(k − 1), b + r = s

6 = (m−2)m(2m+1)6 . This implies

that t4k+1 ≤ (m−2)m(2m+1)6 and therefore t4k+1 = (m−2)m(2m+1)

6 .Case n = 2m + 1 = 2(2k + 1) + 1 = 4k + 3. First we colour the complete subgraph with vertices from

3 to 4k + 3 in a balanced way as described above. Then we colour 1i in the same colour of 3i and 2i isthe other colour for all i ≥ 3. Colour 12 in blue, 13 and 23 in red. Now ri = 2k + 1 for all i 6= 3 andr3 = 2k. Some similar computations give the result.

Exercise 20 Any bipartite graph G with n(G) + e(G) even has an even number of spanning trees.

Proof. We define the same auxiliary graph G′ to Theorem 18. Let T be a spanning tree in G. Any edgenot in T induces a circuit with the edges of T . Since it is a bipartite graph, the circuit is even (Exercise 2),so any edge not in T induces an odd number of swaps of edge, namely, an odd number of other spanningtrees. Because n(G) + e(G) is even, there is an odd number of edges not in T . So each vertex in G′ hasodd degree. The conclusion follows immediately.

Exercise 21 Any bipartite Eulerian graph has an even number of spanning trees.

Proof. We use Matrix-Tree Theorem here. Let L be the Laplacian matrix of G = (X, Y )(|X| ≤ |Y |).We arrange V (G) such that V (X) comes before V (Y ). Deleting the first row and the first column, and

making all the entries modulo 2, we get a matrix of the form of Then L has the form of(

0 AAT 0

).

The size of the upper-left zero minor is smaller than the bottom-right one. So in the expansion of thedeterminant of the matrix, every term contains a term from one of the zero minors. Therefore, thedeterminant is zero. The conclusion follows from Matrix-Tree Theorem.

Exercise 23 Illustrate that Theorem 15 follows from Exercise 16 while Exercise 16 follows from Theorem22.

Proof. It is obvious that Theorem 15 follows from Exercise 16.In Exercise 16, let h(u) = h(v) = 1 and h(w) = 2 for all w 6= u, v. Since deg(w) is odd, k(w) is odd

for w 6= u, v. Theorem 22 tells us that the number of good spanning trees is even, and the good spanningtrees here are Hamiltonian paths from u to v.

Exercise 24 Can you find a generalization of Theorem 22 for linear space?Exercise 26 Deduce Theorem 7 from Theorem 25 and vice versa.

Proof. Deduce Theorem 7 from Theorem 25: Let S be a subset of E(G). Let C be a cycle in the cyclespace C, and B = E(G) − C. Then B is orthogonal to C, and B is maximal in B ∩ SC , where B is thebond space. The conclusion follows easily from Theorem 25.

Exercise 29 Work out a generalization of Theorem 28 in the context of linear space (binary matroid).Exercise 34 Use Corollary 32 to give a proof of Exercise 8.

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Proof. Let u and v be two distinct vertices in an even graph. If they are adjacent, then there is an oddnumber of circuits passing through the edge. Hence the number of paths connecting them is even. If uand v are not connected by an edge, we add a couple of edges, say e1 and e2, connecting u and v to keepthe graph Eulerian. There is an odd number of circuits passing through e1, and there is an odd numberof paths connecting u and v except e1. But e2 is included there, so there is an even number of paths ofu-v in the original graph.

Exercise 35 Use Theorem 30 to give a proof of Exercise 10.

Proof. Let the linear space V = FE(G)2 and W be the cycle space. Define VS = {x ∈ V : x(i) =

x(j), ∀i, j ∈ S} and WS = VS ∩ W . Define an inner product 〈·, ·〉 on VS properly such that S andei(i /∈ S) is an orthogonal basis of VS . Then the parity of MS(G) is the same to

∑x∈M(WS)〈x, S〉 =

〈∑

x∈M(WS) x, S〉. Similarly, the parity of mS(G) is the same to 〈∑

x∈m(WS) x, S〉. The conclusion followsfrom Theorem 30 immediately.

Exercise 36 Compare the proofs of Theorems 7, 25, 28 and 30.Exercise 37 For any graph G, E(G) is a disjoint union of circuits and cocircuits.

Proof. Let N be the cycle space, which is the set of all boolean sums of circuits. Then N⊥ is the cutspace, which is the set of all boolean sums of cocircuits. If x ∈ N ∩N⊥, then |E(G) ∩ x| is even, hencex ⊥ E(G). Thus E(G) ⊂ (N ∩N⊥)⊥, and in fact, E(G) = (N ∩N⊥)⊥ = N⊥ + N .

Exercise 39 Let C be a k-dimensional subspace of Fn2 . Let H be an (n − k) × n matrix whose rows

constitute a basis of C⊥. Prove that the covering radius of C is the smallest number s such that everyvector in Fn−k

2 is a linear combination of s columns of H. In particular, ρ(C) ≤ n− k.

Proof. If s is an integer such that every vector in Fn−k2 is a linear combination of s columns of H, then

for every v ∈ Fn−k2 , there exists z ∈ Fn

2 such that | supp z| ≤ s and Hz = v. Let x ∈ Fn2 , and it has

decomposition of x = y + y′, where y ∈ L and y′ ∈ L⊥. Then Hx = Hy′ = Hz for some z with| supp z| ≤ s. Then x + z ∈ L and d(x, x + z) = d(z, 0) = | supp z| ≤ s.

Consider the quotient space Fn2/L, it has 2n−k elements. H is well-defined on the quotient space, and

it gives different values on each element in the quotient space. Hence the image of H is exactly Fn−k2 .

Conversely, let r be an integer such that O(x, r) ∩ L 6= ∅ for all x ∈ Fn2 . Let v ∈ Fn−k

2 . There existsx ∈ Fn−k

2 such that Hx = v as H is surjective. There exists y ∈ L such that d(x, y) = | supp(x + y)| ≤ r.Suppose x + y =

∑i χki

and v = Hx = H(x + y) = H∑

χki=∑

i βki, where β1, . . . , βn are columns of

H and n ≤ r.

Exercise 40 Prove that the covering radius of a path of odd length is 0 and the covering radius of apath of even length is 1. Please estimate the covering radius of CT for a general tree T . Is it true thatthe covering radius of T is always less than or equal to b `

2c?Exercise 41 An acyclic graph G is a tree if and only if e(G) = n(G)− 1.

Proof. ‘Only if’ part: Trivial by induction.‘If’ part: If G1, G2 be the components of G, n(G1) + n(G2) = n. Both G1 and G2 have no cycles and

they are connected, hence e(G1) = n(G1)− 1 and e(G2) = n(G2)− 1. Sum they sum and we arrive at acontradiction. Hence G is connected, it is also acyclic and hence a tree.

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Exercise 42 A binary phylogenetic n-tree is an ordered pair (T, φ), where T is a tree with only degreeone and degree three nodes and φ is a bijection from {1, . . . , n} to the leaves of T . Prove that the totalnumber of binary phylogenetic n-trees is (2n−4)!

(n−2)!2n−2 .

Proof. Let Tn be the number of binary phylogenetic trees with n leaves. Given a tree with n leaves, anextra leaf can be added on any branch to make a tree with (n+1) leaves. Note that a tree with n leaves has(2n− 3) branches, yielding Tn+1 = (2n− 3)Tn. Finally we obtain that Tn = (2n− 5)!! = (2n−4)!

(n−2)!2n−2 .

Exercise 43 Let T be the set of spanning forest of a graph G. Let H be the graph with vertex set Tin which T1, T2 ∈ T are joined by an edge if and only if |E(T1)4E(T2)| = 2. (i) H has diameter at mostn(G) − c, where c is the number of components of G; (ii) Every edge of H is contained in a Hamiltoncircuit of H.

Proof. (i) It suffices to prove it in the case of c = 1. Let T1 and T2 be two spanning trees of aconnected graph G. We can pick an edge in E(T1) \E(T2), delete it, and replace it with some edgein E(T2) \ E(T1), and the conclusion will follow.

Now we justify our operation. First we remove an edge in E(T1) \ E(T2). This disconnects thetree, and the new graph is a forest with two connected components. There must be an edge inE(T2) which connects them, and it must be different from the one we removed, and thus differentfrom any edge in E(T1), meaning that it belongs to E(T2) \ E(T1). Add this edge to the forest tocomplete a spanning tree, and we are done.

Exercise 44

(i) Suppose D is a digraph with chromatic number χ. Then its line digraph L(D) has chromaticnumber at least log2(χ).

(ii) Let Gn be the graph with V (Gn) = {(i, j) : 0 ≤ i < j ≤ n} and E(Gn) = {((i, j), (j, k)) : 0 ≤ i <j < k ≤ n}. Prove that χ(Gn) = dlog2(n)e.

Exercise 45 There are n girls g1, . . . , gn and m boys b1, . . . , bm such that each gi knows hi < m boysand each bj knows aj < n girls. We adopt the convention that b knows g if and only if g knows b. Supposethat whenever bj does not know gi, we have aj ≥ hi. Show that m ≤ n.

Proof. Let G = (S, T ) be a bipartite graph. For u ∈ S, v ∈ T not adjacent, it holds degG(u) ≥ degG(v).We shall prove that |S| ≤ |T |.

Suppose that |S| > |T |. Let G′ = (S, T ) with (u, v) ∈ E(G′) iff (u, v) /∈ E(G). Then degG′(u) =|T | − degG(u) and degG′(v) = |S| − degG(v). Then for an edge (u, v) ∈ E(G′) we have

degG′(u) ≤ |T | − |S|+ degG′(v) < degG′(v).

from the assumptions. We shall prove that G′ satisfies Hall’s Condition.If not, then we pick a minimal set U ⊆ T which does not satisfy Hall’s Condition. Let u ∈ U and

U ′ = U \ {u}. Then from the minimality, we have |U ′ ≤ |N(U ′)| ≤ |N(U)| < |U | = |U ′| + 1, whence|U ′| = |N(U ′)| and N(U) = N(U ′).

Let G′′ = (N(U ′), U ′) be the subgraph in G′. It is easy to see that for (u, v) ∈ E(G′′), degG′′(v) =degG′(v) > degG′(u) ≥ degG′′(u). Hall’s condition is satisfied in G′′ and by Hall’s theorem there exists amatching saturating U ′. Since |N(U ′) = |U ′|, the matching also saturates N(U ′). Suppose the matching

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is {(ui, vi)}. It follows from degG′′(vi) > degG′′(ui), we get |E(G′′)| =∑

degG′′(vi) >∑

degG′′(ui) =|E(G′′)|, contradiction.

Therefore, G′ satisfies Hall’s Condition and it has a matching saturating T , say {(ui, vi)}. Thusthere exists x ∈ S which is unmatched. Each pair of (ui, vi) is not adjacent in G, hence

∑i degG(ui) ≥∑

i degG(vi). Note that deg x > 0, we have∑u∈S

degG(u) ≥ deg x +∑

i

degG(ui) >∑

i

degG(vi) =∑v∈S

degG(v),

which is a contradiction.

An alternative proof. This proof based on double-counting is much more elegant. We go on using thenotations above.

Suppose |S| > |T |, we have

1 =∑u∈S

∑v∈T

(u,v)/∈E

1|S|(|T | − deg u)

>∑v∈T

∑u∈S

(u,v)/∈E

1|T |(|S| − deg v)

= 1,

which is a contradiction.

Exercise 47 Construct natural isomorphisms from ker∇T to RE/ im∇ and from ker∇ to RV / im∇T .

Proof. By Theorem 46, it suffices to show that a subspace W of V is isomorphic to V/W⊥.Define σ : W → V/W⊥ by σ(x) = W⊥ + x. It is clear that σ is injective. We shall show that it is

surjective. Let x ∈ V , according to Projection Theorem, x has the unique decomposition x = v1 + v2,where v1 ∈ W and w2 ∈ W⊥. Thus σ(v1) = W⊥ + v1 = W⊥ + x, which implies that σ is surjective. It iseasy to verify that σ is a homomorphism, hence it is an isomorphism.

Exercise 49 Prove the following for a graph with c components: dim B(G) = rank(∇) = rank(L) =n(G)− c; dim C(G) = e(G)− n(G) + c; dim ker∇ = c; dim im∇⊥ = n(G)− c.

Proof. From our knowledge of linear algebra, it is clear that dim im∇ = rank(∇) = rank(∇T∇) =rank(L). We need only to show rank(∇) = n(G) − c. We can rearrange the columns of ∇ properlysuch that ∇ has the form of blocks on the diagonal (in fact, the blocks are ∇Ki

, where Ki are thecomponents of G). So it suffices to show that rank(∇K) = n(K) − 1 for a connected graph K. Let Tbe a spanning tree of K (ignoring the directions of edges), then rank∇T = n(T )− 1 = n(K)− 1 by thepart of (iii)⇒(i) in Exercise 50. Consider an edge e /∈ T , then it forms a cycle(ignoring the directions)if we add it to T . Thus e is a linear combination of edges in T . Hence rank∇K = rank∇T = n(K)− 1and dim ker∇ = dim V − dim im∇ = c. From Theorem 46, we also have dim C(G) = dim ker∇T =e(G)− dim im∇ = e(G)− n(G) + c and dim im∇T = n(G)− dim ker∇ = n(G)− c.

Exercise 50 Let G be a graph with e(G) = n(G) − 1. Prove the equivalence of the following: (i) ∇G

has full row rank; (ii) Any (n(G)− 1)-minor of ∇G is invertible; (iii) G is a tree.

Proof. (iii)⇒(i): Let us use Theorem 2.1.3 in [1]. Let the rows of ∇G be r1, . . . , re(G). By Theorem 2.1.3,there exist two leaves, say v1 and v2. The column of vi(i = 1, 2) has exactly one non-zero element, andsuppose they are in ri and rj , respectively. Compare the v1-component and v2-component of both sides ink1r1 +k2r2 + · · ·+re(G)ve(G) = 0, we obtain that ki = kj = 0. Thus remove them and we get a new graph

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G′ which satisfies that n(G′) = n(G)− 2 and e(G′) = e(G)− 2, thus it remains that e(G′) = n(G′)− 1.The proof can be done by induction.

(i)⇒(iii): It follows immediately from Exercise 49(note that we use (iii)⇒(i) only in the proof ofExercise 49)

(ii)⇒(i): It is obvious.(i),(iii)⇒(ii): The proof is similar to (iii)⇒(i), which is based on the same idea of deleting rows.

Exercise 52 There are n parking spots 1, 2, . . . , n on a one-way street. Cars 1, 2, . . . , n arrive in thisorder. Each car i has a favorite parking spot f(i). When a car arrives, it first goes to its favorite spot.If the spot is free, the car will take it, if not, it goes to the next spot. Again, if that spot is free, the carwill take it and will move further otherwise. If a car had to leave even the last spot and did not find thespace, then its parking attempt is unsuccessful. If, at the end of this procedure, all cars have a parkingspot, we say that f is a parking function on [n]. Prove that the number of parking functions on [n] is(n + 1)n−1.

Proof. Let us add an additional space n + 1, and arrange the spaces in a circle. Now all cars can parkand there is only one empty space. f is a parking function if and only if the empty space is n + 1. Andit is not difficult to see that exactly one of (a1 + i, a2 + i, . . . , an + i) (mod n + 1) is a parking function,so the number of parking functions is (n+1)n

n+1 = (n + 1)n−1.

Exercise 60 Let T be a tree with V (T ) = {v1, . . . , vn} and X be the (n− 1)× (n− 1) matrix obtainedfrom ∇T by deleting the column indexed by vn. Then X is nonsingular and the (vi, ej) entry of X−1 isT

ejvnvi . Especially, deduce that detX ∈ {1,−1}.

Proof. Let A be a matrix whose entries is given by Aij = Tejvivn . Let us compute XA. (XA)i,j =∑

k Xi,kAk,j = −Ae−i ,j + Ae+i ,j = −T

ej

vn,e−i+ T

ej

vn,e+i

. Note that the path from e−i to vn and from e+i to vn

differ in ei only, then it is not difficult to see that

Aij ={

1, i = j0, i 6= j

Thus XA = E, and X−1 = A. Since the entries of X and A are all integers, both detX and detA areintegers. Thus |det X| = 1.

Exercise 61 Let M ∈ Rm×p, N ∈ Rm×(m−p) be two matrices of full column rank. Suppose thatMT N = 0, namely the column space of M is the orthogonal complement of that of N in Rm. Prove thatI − N(NT N)−1NT = M(MT M)−1MT is the orthogonal projection to the column space of M . Try tofind a basis for the cocycle space and a basis for the cycle space of a graph G, respectively, and deducematrix solutions for the projection operator P . Compare with the tree solution for P as displayed in Eq.(9) and see what can be asserted.

Proof. Let us first prove that I −N(NT N)−1NT = M(MT M)−1MT , or,

M(MT M)−1MT + N(NT N)−1NT = I.

Let S = M(MT M)−1MT + N(NT N)−1NT . We see that SM = M and SN = N , hence Sβ = β, whereβ is a column of M or N . Because all the columns of M and N form a basis of the whole space, we seethat Sx = x for all x. Hence S = E.

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Now a decomposition of x is induced: x = s+t, where s = M(MT M)−1MT x and t = N(NT N)−1NT x.It is obvious that 〈s, t〉 = 0, hence it is an orthogonal decomposition.

The rest of problem refers to page 107 to 117 of the slide.

Exercise 62 Let G be a connected graph on n vertices and L its Laplacian. Prove that there isU ∈ Rn×(n−1) such that L = UUT . Prove that the Steiner circumscribed ellipsoid of the simplex spannedby the n columns of UT in Rn−1 is defined by the equation xT (UT U)−1x = n−1

n for an unknown x ∈ Rn−1.Discuss the relationship between the eigenvectors of L and this ellipsoid. What can be said when G isnot connected?

Proof. Since G is connected, rankL = n−1. And L is semi-positive definite, there exists orthogonal matrix

K such that L = K diag{λ1, · · · , λn, 0}K−1 (λi > 0). Let R =(

D0

), where D = diag{

√λ1, · · · ,

√λn}.

Then diag{λ1, · · · , λn, 0} = DDT , L = (KD)(KD)T . Let U = KD, we have L = UUT and rank U =n− 1.

Suppose the rows of U are xi. Since rank U = n − 1, we have∑

xi = 0 by observing that L is theGram matrix of {ui}. We denote the column vector with all 1 entries by ι, then we can rewrite it asUT ι = 0.

Consider S = U(UT U)−1UT . We observe that S2 = S, hence S is a projector, so is I − S. Fromrank S = n− 1, we know that rank(I − S) = 1. I − S sends a vector to an 1-dimensional space. Noticethat (I − S)ι = ι − Sι = ι, whence the 1-dimensional space is spanned by {ι}. Thus (I − S)x = 〈x,ι〉

‖ι‖ ι.If follows easily that I − S = 1

nE and S = E − 1nJ , where E is the matrix with all entries of 1. Thus

xi(UT U)−1xTi = n−1

n . Hence the quadratic contains xi.The tangent hyperplane at the vertex xk has the equation

xk(UT U)−1xT =n− 1

n.

The hyperplane containing all xi 6= xk has the equation

xk(UT U)−1xT +1n

= 0,

becausexk(UT U)−1xT

i +1n

= (S)ik +1n

= 0.

The two hyperplanes are parallel. So the quadric is precisely the Steiner circumscribed ellipsoid.Suppose that α is an eigenvector of L, i.e., Lα = UUT α = λα, whence we see that UT α is an

eigenvector of UT U as (UT U)UT α = λ(UT α). If α is not perpendicular to the space , or, α and ι arelinearly independent, then λ 6= 0, thus UT α is an eigenvector with eigenvalue 1

λ of (UT U)−1, whichimplies that it represents the direction of one axis of the quadratic.

Exercise 63 For any two spanning forests T and T ′ of a graph G, prove that fT ′fT = fT ′ and fT P = fT .

Proof. In order to prove that fFT ′fT = fT ′ , it is sufficient to show that fT ′fT (e) = fT ′(e) for everye ∈ E(G). Suppose T (e− → e+) = {e1, e2, . . . , en} sequentially.

fT ′fT (e) = fT ′(T (e− → e+)) = fT ′

(∑i

ei

)=∑

i

fT ′(ei) =∑

i

T ′(e−i → e+i )

= T ′(e− → e+) = fT ′(e),

9

Hence

fT P = fT

∑T ′ fT ′

k(G)=∑

T ′ fT fT ′

k(G)=∑

T ′ fT

k(G)=

k(G)fT

k(G)= fT .

Exercise 64 Let G be a graph and T, T ′ two of its spanning forests. For a matrix M whose columnsare indexed by E(G), let MT be the matrix consisting of the columns of M indexed by E(T ) and MN bethe matrix obtained from M be deleting columns indexed by E(T ). Prove that (fT ′)N = (fT ′)T (fT )N

and PN = PT (fT )N .

Proof. This problem is quite similar to the previous one. The only difference in the proof is that we usematrix here. Let us prove the first identity for example.

First, we clarify that (fT ′)N has size E(T ′)× (E(G)−E(T )), (fT ′)T has size E(T ′)×E(T ) and (fT )N

has size E(T ) × (E(G) − E(T )). Let f be an edge outside T . Let the column in (fT )N of index f isf = (b1 b2 · · · bE(T ))T =

∑χ(ei), which represents the path from f− to f+ in T . Hence the column of

index f in (fT ′)N , by Exercise 63, is (fT ′)T f =∑

(fT ′)T

∑χ(ei). Each (fT ′)T

∑χ(ei) represents a path

from e−i to e+i (ei ∈ T ) in T ′, hence the sum of them represents the path from f− to f+ in T ′. That is,

(fT ′)N = (fT ′)T (fT )N , as required.

Exercise 70 There is a nonzero R-bicycle on G if and only if there is β ∈ C0(G, R) such that Lβ = 0,but ∇β 6= 0.

Proof. If part: Let γ = ∇β, then γ is a cut. But Lβ = ∇T γ = 0, whence γ belongs to the cycle space.Hence γ is a bicycle.

Only if part: Let γ be a bicycle. γ is a cut, so γ = ∇β for some β, and Lβ = ∇T γ = 0 since γ is acycle.

Exercise 71 Prove that det C =√

det CCT = det BfBTf = det CfCT

f = k(G), where

C =(

Bf

Cf

)=(

In−c Y−Y T Im−n+c

).

Proof. First we have

detBfBTf = det

(I Y

)( IY T

)= det(I + Y Y T ),

and

detCfCTf = det

(−Y T I

)(−YI

)= det(I + Y T Y ),

it follows that detC and they are equal from our knowledge of linear algebra.(proof is uncompleted)

Exercise 72 Let G be a graph and F a field of characteristic p. Prove that there is a nonzero F -bicycleon G if and only if p|k(G).Exercise 73 There exists a nonzero Zd-bicycle f on a graph G if and only if gcd(d, k(G)) 6= 1.Exercise 74 Show that Z +

√2Z is a free abelian group of rank two but is not a lattice in the real line.

10

Proof. First it is clear that Z +√

2Z is a free abelian group. It is clear that {1,√

2} is a basis, so therank is two.

If it is a lattice on the real line, the generator matrix must be (1√

2), however, it does not havefull column rank. So Z +

√2Z is not a lattice in the real line.

Exercise 75 Prove that detL = vol(FB).

Proof. We choose an orthonormal basis from M⊥ and extend M to be M = (M M ′) by adding the basis

as column vectors. Thus detMTM =(

MT M 00 E

). Thus detMTM = det MT M and det M = detM.

detM is exactly the volume of the generated fundamental domain, and it is numerically equal to thevolume of the original fundamental domain.

Exercise 76 Let L1 be a sublattice of L0. Show that for any generator matrix Mi of Li(i = 0, 1),there is a nonsingular integer matrix N such that M1 = M0N . Prove that [L0 : L1] = det L0

det L1= det N .

Also illustrate that each fundamental domain of L1 is the disjoint union of [L0 : L1] translations of somefundamental domain of L0.

Proof. Since L1 is a sublattice, hence each column of M1 is a combination of columns of M0 with integercoefficients. We arranging these integer coefficients vertically and let this column vector be a column ofN . We obtain a matrix N of integer entries in this way, which satisfies M1 = M0N . Since M0 and M1

are both of full column rank, N must be non-singular.Now consider the Smith canonical form of N : N ′ = PNQ = diag{h1, h2, . . . , hn}, where h1|h2| · · · |hn

and detP = detQ = 1. Then M1Q = M0P−1N ′. Observe that M0P

−1 and M1Q are new gen-erator matrices of L0 and L1, respectively; we can say that w1, . . . , wn is a new basis of L0 andh1w1, h2w2, . . . , hnwn is a new basis of L1. It is quite clear now that [L0 : L1] = h1h2 · · ·hn = det N ′ =

detN =det L1

det L0.

Exercise 79 The dual basis of the dual basis of a basis is the basis itself.

Proof. Suppose the generator matrix of L is M , then by Theorem 77, the generator matrix of L# isM(MT M)−1. The generator matrix of (L#)# is M(MT M)−1 · ((M(MT M)−1)T ·M(MT M)−1)−1 = M ,thus (L#)# = L.

Exercise 80 The dual lattice of a lattice is still a lattice.

Proof. It follows immediately from Theorem 77.

Exercise 81 Prove that rankL = rank L# and detLdetL# = 1. If L is integral, then we have L is asublattice of L# with index det(MT M).

Proof. rank L# = rankM(MT M)−1 = rankM = rankL.

detLdetL# =√

det MT M√

det(M(MT M)−1)T ·M(MT M)−1

=√

det MT M√

det((MT M)−1(MT M)(MT M)−1)

=√

det(MT M)(MT M)−1(MT M)(MT M)−1)

=√

det I = 1

11

Let us denote by M ′ the generator matrix of L#, then M ′ = M(MT M)−1 and hence M = M ′(MT M).Hence the columns of M are linear combinations of columns of M ′, which implies that L ≤ L#. Combiningit with rankL = rankL#, we see that L is a sublattice of L#. The rest follows from Exercise 75.

Exercise 87 Let T and T ′ be two spanning forests of G. Let B = Bf,T ′ be the fundamental cut matrix ofG with respect to T ′. Then there is an integer matrix C with det C = ±1 such that PT = BT (BBT )−1C.Exercise 88 Finish the proof of Eq. (19).Exercise 89 Prove that the integral cohomology group H1(G, Z) = CI(G)

BI(G) is isomorphic with ZI(G)#.Exercise 90 Determine the Jacobian of the following graphs: Complete graphs, the line graphs ofcomplete graphs, the Petersen graph, the hypercubes.Exercise 91 Let G be a graph with c components and let v1, . . . , vc be c vertices in pairwise differentcomponents. Show that {∇v : v ∈ V (G) \ {v1, . . . , vc}} is a basis for BI(G).Exercise 92 Write a survey on various proofs of Matrix-Tree theorems and their generalizations andvariations.Exercise 95 A rooted tree is a tree with one of its vertices specified as the root. A rooted forest isthe disjoint union of a set of rooted trees. Show that the number of spanning rooted forests of Kn is(n + 1)n−1. Recall Exercise 52. Can you find any natural bijection between the set of spanning rootedforests and the set of parking functions?

Proof. For a rooted forest in Kn, we put it into Kn+1 and join the roots of the forest to the n + 1-thvertex, yielding a rooted spanning tree in Kn+1. This process is reversible, but only the tree rooted atn + 1 induce a rooted forest of Kn. And we see that there are n + 1 rooted spanning trees in Kn+1 sharea common underlying tree, which is a spanning tree. So the number of rooted forests of Kn equals thenumber of rooted spanning tree of Kn+1 divided by n + 1, and this is exactly the number of spanningtrees of Kn+1, which is (n + 1)n−1, given by Cayley’s Theorem.

Exercise 97 For each arborescence T ∈ Arbo1(Kn), put xdeg−(T ) = xdeg−T (1)1 · · ·xdeg−T (n)

n . Show that∑T∈Arob1(Kn)

xdeg−(T ) = (x1 + · · ·+ xn)n−2x1. (1)

Proof. Denote by f the summation∑

xi f . Let X =(x2 x3 · · · xn

). Take aij = xj in Eq. (22),

then it is verified that xdeg−(T ) =∏

ij∈E(T ) aij . By Theorem 96(i), the left hand side of (1) is D =det(fE − 1X) = fn−2 det(fE −X1) = fn−2x1. The proof is complete.

Exercise 98 Let Cn be the n-cube, namely the graph with vertex set Zn2 and two vertices being adjacent

if and only if the Hamming distance between them is one. Show that k(Cn) = 22n−n−1∏n

i=1 i(ni).

Proof. Firstly it is observed that L(G1⊗G2) = L(G1)⊗E+E⊗L(G2), where E is the identity matrix andL(G) is the Laplacian matrix of G. Hence the eigenvalues of L(G1⊗G2) are λ+µ, where λ is an eigenvalueof L(G1) and µ of L(G2). We iterate this, and conclude that the eigenvalue of L(Kn1 × · · · ×Knk

) hasthe eigenvalue

∑i∈S ni with multiplicity

∏i∈S(ni − 1) for each subset S ⊆ [k] since the eigenvalues of

L(Kn) is n with multiplicity of n − 1 and 0. Apply this result to Cn = K2 ×K2 × · · · ×K2︸ ︷︷ ︸n times

, we know

12

the eigenvalues of L(Cn) are 2k with multiplicity(nk

). By Matrix-Tree theorem, the number of spanning

trees is given by12n

n∏k=1

(2k)(nk) =

2Pn

k=1 (nk)

2n

n∏k=1

k(nk) =

22n−1

2n

n∏k=1

k(nk),

as desired.

Exercise 99 Let G be a connected plane graph and let G∗ be its dual. Prove that k(G) = k(G∗).

Proof. If T is a spanning tree of G and T ∗ is the subgraph of G∗ containing the edges corresponding toedges of G \ T . The conclusion follows immediately if we prove that T ∗ is a spanning tree of G∗.

First we claim that T ∗ does not contain a circuit. If it does so, then some vertex of G lies in thecircuit. But there is another vertex lying outside the circuit. Hence the path connecting the two verticesin T must across the circuit, which contradicts to the construction of T ∗.

Now let u, v ∈ V (G∗). There is no circuit in T ∗, so u and v remain connected on the plane after wedelete the edges of T . Thus we can find a polygonal curve joining them without touching the edges of T .The polygonal curve corresponds to a path in T ∗ which connects u and v. Hence T ∗ is connected andcontains all the vertices of G∗.

Remark. It also follows immediately from Exercise 134(i).

Exercise 101 For any f ∈ [m][m]k , define a sequence as follows: X1 = X2 = · · · = Xk = 0; Xn+k+1 =f(Xn+k, Xn+k−1, · · · , Xn+1) when n ≥ 0. For how many of the mmk

functions f is this sequence periodicwith a period of the maximum length mk?Exercise 102 Show that when specifying the matroid structure on an infinite set by requiring thehereditary system has the augmentation property, it may happen that there is no basis, and thus suchan object may lose the aspect of bases.

Proof. The hereditary system consists of the sets which has finitely many elements. It is easy to see thatthe hereditary system fulfils the augmentation property. However, the maximal set under inclusion doesnot exist.

Exercise 103 Let E be a finite set and I a nonempty hereditary collection of subsets of E. Prove thatI is the set of independent sets of a matroid on E if and only if for every A ⊆ E, any two maximalmembers of I contained in A have equal size.

Proof. ‘If’ part: We need only to show the augmentation property. Let I1 and I2 be two elements in thehereditary system with I1 ⊆ A, I2 ⊆ A. Suppose |I1| < |I2|. If I1 ∪ {e} /∈ I for all e ∈ I2 \ I1, then I1

is a maximal element. Combining with |I1| < |I2|, we arrive at a contradiction, as all maximal memberscontained in A have the equal size.

‘Only if’ part: Suppose B1, B2 are two maximal elements in I and B1 ⊆ A, B2 ⊆ A, |B1| 6= |B2|.Without loss of generality, suppose |B1| < |B2|. By augmentation property, there exists e ∈ B1 \ B2

such that B1 ∪ {e} ∈ I. Notice that it is also contained in A, which leads to a contradiction with themaximality of B1.

Exercise 107 Every graphic matroid is a vectorial matroid.

13

Proof. We prove that every graphic matroid is binary, thus vectorial over GF (2). Let A be the incidentmatrix of graph G.

If C ⊆ E(G) contains a cycle c1, . . . , ck, then the sum of the corresponding columns is zero, as ineach row there are two columns with entries of 1. Thus the set of columns corresponding to C is linearlydependent.

Conversely, if some columns are linearly dependent, there is a subset of the columns C = {c1, . . . , ck}such that

∑ci = 0. Thus the degree of corresponding vertices C is even, and the corresponding subgraph

of the columns contains a cycle.

Exercise 109 Let R be an integral domain and let N ⊆ RS be a module over R. Prove that thesupports of elements of N which are minimal under inclusion satisfy the circuit weak elimination axiomand hence they are the circuits of a matroid on S as long as S is finite. This matroid is called the matroidof the chain group N and denoted M(N).

Proof. Let x and y be two elements in N with suppx 6= supp y. Suppose k ∈ supp x∩ supp y. Then let usconsider z = xky− ykx ∈ N . Let j ∈ suppx\ supp y then zj = xkyj − ykxj = 0− ykxj 6= 0 because R hasno zero divisors. Hence z 6= 0 and it is clear that supp z ⊆ suppx∪ supp y. The circuit weak eliminationis satisfied.

Exercise 110 Assume that R is a field and N is a vector subspace of RS . Show that M(N) is a vectorialmatroid with the representation f : S → RS/N given by f(s) = χ(s) + N for any s ∈ S. Prove thatU ⊆ S contains a basis of M(N) if and only if S \ U is an independent set in M(N⊥).

Proof. (i) We show that M(N) is a vectorial matroid. Suppose A ⊆ S and A /∈ IM , then A contains acircuit C. Hence there is a minimal vector in N , say n, with its support contained in A. Supposen =

∑i∈supp n tiχ(i). Notice that

∑tif(i) = 0 in RS/N , that means n is dependent under the

vectorial representation, so is A.

Conversely, if A is dependent under the vectorial representation, then it holds∑i∈supp A

tif(i) = 0. (2)

Let us denote by I the set of indices i such that ti 6= 0. Then∑

i∈I tiχ(i) ∈ N . Thus amongall such I which is induced from the tuples of ti satisfying (2), we can find a minimal set underinclusion, and it is correspondent to a minimal element in N . So A contains a circuit in M(N) andit is dependent.

(ii) ‘Only if’ part: As U contains a basis of M(N), spanU = M(N). Hence span(U + N) = RS ,or RU + N = RS . Then {0} = (RU + N)⊥ = (RU )⊥ ∩ N⊥ = RS\U ∩ N⊥ = {0}, thus S \ Uis independent under the vectorial representation in RS/U , equivalently, M(N⊥). Since the stepsabove are invertible, so ‘if’ part holds, too.

Exercise 111 Assume that R is a field and N is a d-dimensional subspace of RS . Let Q ∈ R(n−d)×n

be a generator matrix of the code N⊥, where n = |S|. Show that M(N) is a vectorial matroid withthe representation f : S → Rn−d where f(s) is the s-th column of Q for any s ∈ S. Deduce from thisrepresentation and the last claim of Exercise 110 that for any generator matrix P ∈ Rd×n and any A ⊆ S,the submatrix of Q formed by taking columns labelled by A is a nonsingular matrix if and only if thesubmatrix of P formed by deleting those columns labelled by A is a nonsingular matrix.

14

Exercise 112 Suppose that a matroid M on a finite set S has a vectorial representation f over a fieldF. Consider the matrix A whose columns are labelled by S and whose s-th column is f(s). Assume thatthe row space of this matrix is N ′ ≤ FS . Let N = N ′⊥. Prove that M = M(N).

Proof. Let D be a circuit in M(N) and suppose suppx = D. Then it is clear that∑s∈D

x(s)f(s) = 0, (3)

which implies that {f(s)} (x ∈ D) are linearly dependent, hence D is a dependent set in M . We claimthat D is minimal, or it contains a circuit C in M . The columns f(s) (s ∈ C) are linearly dependent,hence there exists x(s) such that it holds (3), where x(s) are not all zero. Hence it induces a vector v,supp v 6= ∅, supp v ⊆ C ⊂ D and v is orthogonal to N ′. Thus supp v is a dependent set in M(N), and itis smaller than D under inclusion, which contracts with the assumption of maximality of D.

A similar argument shows that a circuit in M is also a circuit in M(N).

Exercise 113 Let E be a finite set and f : 2E → {0, 1, . . . } be a submodular function satisfyingf(∅) = 0. Then the function ρ defined by ρ(A) = minB⊆A(f(B) + |A \ B|) is the rank function of somematroid on E.

Proof. We need to verify

(i) ρ(A) ≤ |A|: Trivial.

(ii) ρ is order-preserving because that f is.

(iii) ρ is submodular: Given X, Y ⊆ E, the formula for ρ yields U ⊆ X and V ⊆ Y such that

ρ(X) = f(U) + |X \ U |, ρ(Y ) = f(V ) + |Y \ V |,

and

ρ(X ∩ Y ) ≤ f(U ∩ V ) + |(X ∩ Y ) \ (U ∩ V )|, ρ(X ∪ Y ) ≤ f(U ∪ V ) + |(X ∪ Y ) \ (U ∪ V )|.

Applying the submodularity of f and

|(X ∩ Y ) \ (U ∩ V )|+ |(X ∪ Y ) \ (U ∪ V )| = |X \ U |+ |Y \ V |,

we reach ρ(X ∩ Y ) + ρ(X ∪ Y ) ≤ ρ(X) + ρ(Y ).

Exercise 114 Let G be a finite graph with vertex set V and edge set E. Define a function f(g) from2E(2V ) to Z+ by putting f(A)(g(B)), A ∈ 2E(B ∈ 2V ), to be the number of vertices (edges) which areincident to at least one edge (vertex) in A (B). Prove that f is a submodular function on E and g is asubmodular function on V . Try to give some descriptions of the matroid whose rank function is generatedby f or g as described in Exercise 106.

Proof. Since the ground set is finite, the submodularity of function r

∀A,B ⊆ E, r(A ∩B) + r(A ∪B) ≤ r(A) + r(B)

is equivalent to∀A ⊆ B ⊆ E,∀x ∈ E, r(A + x)− r(A) ≥ r(B + x)− r(B). (4)

The submodularity of f and g can be easily verified using (4).

15

Exercise 118 All heaviest bases of a matroid are possible to be obtained from the greedy algorithm.

Proof. It is sufficient to prove that all heaviest bases have the same set consisting of weight of eachelement in a base. Suppose A = {a1, . . . , am}, let W (A) = {w(a1), . . . , w(am)}.

Assume it is not true. We have two bases, B1 and B2. W (B1) \ W (B2) = {w(e1), . . . , w(en)} andW (B2)\W (B1) = {w(f1), . . . , w(fn)} (both are sorted by weight in a descending roder). If w(fn) < w(ek)for all k, then by base exchange property, B2 − fn + ek ∈ B is a base but it has a larger weight than B1

and B2. Contradiction. Thus w(fn) > w(ej) for some j, and hence w(en) < w(fk) for all k and it alsoyields a contradiction in a similar way.

Exercise 120 X is an independent set (subbase) of M∗ if and only if E\X is a spanning set (superbase)of M .

Proof. ‘If’ part: Suppose E \X contains a base, say B. Then X ⊆ E \B, thus X is contained in a baseof M∗ and X is an independent set in M∗.

‘Only if’ part: Suppose X is an independent in M∗ then we can extend it to B′, which is a base inM∗. Thus E \X ⊇ E \B′ and E \B′ is a base in M . Thus E \X is a superbase in M .

Exercise 121 For any hereditary system M we have (M∗)∗ = M .

Proof. It follows immediately from the definition of B∗ of M∗.

Exercise 122 If M is a matroid, then M∗ is a matroid.

Proof. We have known that M∗ is a hereditary system and now we prove the base exchange propertyfor M∗. If B1, B2 ∈ B∗ and e ∈ B1 \ B2, then B1, B2 ∈ B and e ∈ B2 \ B1. It is clear that there existsf ∈ B1 \B2 such that B1 ∪ {e} \ {f} ∈ B and thus B1 \ {e} ∪ {f} ∈ B∗.

Exercise 124 Let M be a hereditary system on E. Prove that X ∈ CM\F if and only if X ∈ CM andX ⊆ E \ F while X is independent in M \ F if and only if X ⊆ E \ F and X is independent in M .

Proof. From the definition of M \ F , it is clear that there is a one-to-one correspondence between thedependent sets of M \ F and the dependents sets of M which are disjoint with F . The first assertionfollows immediately. The second one is exactly the definition.

Exercise 125 (i) Let M be a matroid on E. Show that (i) X ⊆ E \F is a circuit of M/F if and only ifX = Y ∩ (E \F ) for a circuit Y of M and is minimal with respect to these two properties; (ii) Y ⊆ E \Fis independent in M/F if and only if there is a maximum independent subset X ⊆ F such that X ∪ Y isindependent in M .

Proof. (i) Firstly it is checked that the hypobases of M \ F are the maximal proper subsets of F ofthe form H ∩ F , where H is a hypobase of M . Then,

X ∈ CM/F ⇐⇒ X ∩ F ∈ H(M/F )∗ ⇐⇒ X ∩ F ∈ HM∗\F

⇐⇒ X ∩ F is maximal proper subset of F of the form H ∩ F , H ∈ HM∗

⇐⇒ X ∩ F is maximal proper subset of F of the form C ∩ F , C ∈ CM

⇐⇒ X is minimal non-empty of the form C ∩ F , C ∈ CM .

16

(ii) ‘If’: Suppose Y ⊆ E \ F and X is a maximum independent subset properly contained in F suchthat X ∪ Y is independent. We extend X ∪ Y to a base B of M , thus B \ F ∈ SM/F . It is clearfrom the maximality of X that B∩F = X and Y ⊆ B \F . Suppose Y ′ is a minimal subset of B \Fsuch that Y ′ ⊇ Y and Y ′ ∈ SM/F . Then the base contained in Y ′ ∪ F contains Y ′, suppose thebase is X ′ ∪ Y ′ where X ′ ⊆ F . Since bases have the same size, we have |X ′|+ |Y ′| = |X|+ |B \F |.Notice that |X ′| ≤ |X|, it must hold that |Y ′| ≥ |B \ F | thus B \ F = Y ′ is a base in B/F . ThenY is independent since it is a subset of a base.

‘Only if’: Suppose Y is independent in M/F . It can be extended to a base B′ in M/F , and B′ isthe minimal element in SM/F , which implies that B′ ∪ F contains a base in M , say B. We musthave B′ ⊆ B. Let us prove that B ∩ F is a maximal independent set in F . If not, there existsA ) B∩F which is independent, then A∪B contains a base. From the uniformity of M , it containsa base B ⊇ A and B ∩B′ ( B′, which contradicts with the minimality of B′.

Exercise 126 If e is either a loop or a coloop, then M/e = M \ e.

Proof. First suppose that e is a loop. The underlying set of M/e and M \ e are the same. From Exercise117, we see that the circuits in M \ e are the circuits in M not containing e. Since {e} is a circuit, theyare exactly all the other circuits of M than {e}. From Exercise 118, the circuits in M/e are minimalnon-empty sets of the form C \ e, where C is a circuit of M . Since e is a circuit, they are exactly{C : C ∈ CM and C 6= {e}}. So the circuits of M/e and M \ e are the same, and M/e = M \ e.

If e is a coloop, then M∗/e = M∗ \ e. Apply Theorem 117 to it, we obtain that (M \ e)ast = (M/e)∗,or, M \ e = M/e.

Exercise 127 (i) For L ≤ Rn, define L \ n ⊆ Rn−1 to be {(x1, . . . , xn−1) : (x1, . . . , xn−1, 0) ∈ L},and define L/n ⊆ Rn−1 to be {(x1, . . . , xn−1) : ∃r ∈ R such that (x1, . . . , xn−1, r) ∈ L. Prove thatL⊥/n = (L \ n)⊥ and L⊥ \ n = (L/n)⊥. (ii) Prove Farkas Lemma: For any subspace L of Rn, eitherthere is x ∈ L satisfying x ≥ 0 and x1 > 0, or there is y ∈ L⊥ satisfying y ≥ 0 and y1 > 0, but not both.

Proof. (i) Let y ∈ L⊥/n and x ∈ L \ n. It is easy to verify that 〈x, y〉 = 0. Let x run over the wholeL \ n and we see that y ⊥ (L \ n) and L⊥/n ⊆ (L \ n)⊥.

Now we prove that (L \ n)⊥ ⊆ L⊥/n. Let x = (x1, . . . , xn−1) ∈ (L \ n)⊥ and β1, . . . , βs be abasis of L. We can choose the basis such that the last entry of βi(1 ≤ i ≤ s − 1) are all zeroes (ifthere are v1 and v2 such that (v1)n 6= 0 and (v2)n 6= 0 then we can replace v1 by (v2)nv1− (v1)nv2).Now consider the system of equations

βT

1...

βTs−1

βTs

x1

x2

...xn−1

r

=

00...00

. (5)

The first n− 1 equations are identities because x is orthogonal to L⊥/n, and the last equation is

(βs)1x1 + · · ·+ (βs)n−1xn−1 + (βs)nr = 0.

If (βs)n 6= 0, then the equation above has a solution r, thus (5) implies that x ∈ L⊥/n. If (βs)n = 0,then (5) is hold for any r, and thus x ∈ L⊥/n, too.

17

Combining the two results above, we conclude that L⊥/n = (L \ n)⊥ and the other identityfollows immediately, replacing L by L⊥ in the first identity.

(ii) Let us prove it by induction on n.

It is easy to see when n = 1. The subspace of R is {0} or R itself. So the statements holds.

Inductive Step. Suppose the statement holds for n, we are going to show that it holds for n + 1,too. If the statement is false, then there exists neither (1 x1 · · ·xn−1) ∈ L \ n (xi ≥ 0) nor(1 y1 · · · yn−1) ∈ L⊥ \ n (yi ≥ 0). So by the induction hypothesis, there exist (1 a1 · · · an−1) ∈(L\n)⊥ = L⊥/n (ai ≥ 0) and (1 b1 · · · bn−1) ∈ (L⊥ \n)⊥ = L/n (bi ≥ 0). Thus (1 b1 · · · bn r1) ∈ Land (1 a1 · · · an r2) ∈ L⊥ for some r1, r2. But they are orthogonal, or 1 +

∑aibi + r1r2 = 0,

thus one of r1, r2 must be positive, which implies a required vector in L or L⊥ and leads us to acontradiction. So the statement holds for n + 1, too.

Finally, it can’t hold both clearly.

Exercise 131 Let G = (V,E) be a graph with V = V1 ∪ V2 such that G〈V1 ∩ V2〉 is a complete graph

without loops and there is no edge connecting V1 \V2 to V2 \V1. Prove that pG(x) =pG〈V1〉(x)pG〈V2〉(x)

pG〈V1∩V2〉(x).

Exercise 133 Let F be a ring. For any matroid M there is a unique function f(M ;x, y, σ, τ) belongingto the polynomial ring F [x, y, σ, τ ] and having the following properties: (i) f(∅;x, y, σ, τ) = 1; (ii) Ife is a loop, then f(M ;x, y, σ, τ) = yf(M \ e;x, y, σ, τ); (iii) If e is a coloop, then f(M ;x, y, σ, τ) =xf(M/e;x, y, σ, τ); (iv) If e is neither a loop nor a coloop, then f(M ;x, y, σ, τ) = σf(M \ e;x, y, σ, τ) +τf(M/e;x, y, σ, τ). Furthermore, if E is the underlying set of M , the function f is given by

T ′(M ;x, y, σ, τ) = σ|E|−ρ(E)|τρ(E)T (M ;x

τ,y

σ) = σn(E)τρ(E)T (M ;

x

τ,y

σ).

Exercise 134 Let M be a matroid. Show that: (i) T (M ; 1, 1) is the number of bases of M ; (ii) T (M ; 2, 1)is the number of independent sets of M ; (iii) T (M ; 1, 2) is the number of spanning subsets of M .

Proof. (i) It is not difficult to verify the recurrent relations in (25) for the number of bases. We also havefrom Theorem 126 that T ({e}, 1, 1) = T (∅, 1, 1) = 1 if e is a loop, and T ({e}, 1, 1) = T (∅, 1, 1) = 1if e is a coloop. Hence the initial values are the same with the number of the bases. Therefore,T (M, 1, 1) is the number of bases of M .

(ii) It is verified that the recurrent relations in (25) for the number of independent sets. From Theorem126, it holds that T ({e}, 2, 1) = T (∅, 2, 1) = 1 if e is a loop, and T ({e}, 2, 1) = 2T (∅, 2, 1) = 2 if e isa coloop. The initial values are the same with the number of the independent sets. So T (M, 2, 1)is the number of independent sets of M .

(iii) It is not difficult to verify the recurrent relations in (25) for the number of spanning subsets. We havefrom Theorem 126 that T ({e}, 1, 2) = 2T (∅, 1, 1) = 2 if e is a loop, and T ({e}, 1, 2) = T (∅, 1, 2) = 1if e is a coloop. We know that the initial values are the same with the number of the spanningsubsets. Therefore, T (M, 1, 2) is the number of spanning subsets of M .

Remark. It is also possible to prove this problem by calculating out the expression of T (M ;x, y) with theassumption of 00 = 1. For (i), all contributions are from such subsets : ρ(E) = ρ(A) = |A|, thus A is abase. Hence the Tutte polynomial T (M, 1, 1) gives the number of bases of M . For (ii), the contributions

18

are from such subsets: ρ(A) = |A|, thus A is an independent set. For (iii), the contributions are fromsuch subsets: ρ(E) = ρ(A) and A is a spanning set.

Exercise 135 (i) For a binary matroid M , |T (M,−1,−1)| = |T (M, 1, 1)| equals the number of bicyclesin M . (ii) A binary matroid has an odd number of bases if and only if it has no nontrivial bicycle.Exercise 137 Write down explicitly the recurrence relation for the q-state Potts model partition func-tion and then use it to prove that PG(q, β) = e−β|E(G)|ZG(q,v), where v = eβ − 1 and

ZG(q, v) = qk(G)vρ(G)T (G; 1 +q

v, 1 + v).

Exercise 138 (i) ZG(q, v) is the unique polynomial such that ZEn(q, v) = qn for every integer n ≥ 1

and ZG = ZG−e + vZG/e for every e ∈ E(G). (ii) ZG(q, v) is the polynomial U(G;x, y, α, σ, τ) evaluatedat α = q, σ = 1, τ = v, x = v

q + 1 and y = v + 1.Exercise 139 Make use of Eqs. (23) and (35) to show that

pG(x) = ZG(x,−1) = xk(G)(−1)|V (G)|−k(G)T (G; 1− x, 0).

Proof. In fact (23) says that pG(x) = pG−e(x)− pG/e(x). Comparing it with (35), it yields v = −1. Theconclusion follows immediately.

Exercise 140 Deduce Theorem 123 from the symmetry of the Tutte polynomial as indicated in Eqs.(26) and (27).Exercise 143 Let B and B′ be two bases of a matroid M . Prove that there is a bijection f from B toB′ such that (B \ {e}) ∪ {f(e)} is a basis for each e ∈ B. (This is Exercise 8.2.23 in [1])

Proof. Let G be a B,B′-bigraph with e ∈ B adjacent to f ∈ B′ when (B \ {e}) ∪ {f} is basis. It followseasily from base exchange property (by considering e ∈ B ∩B′ and e ∈ B \B′) that |N(S)| ≥ |S| for allS ⊆ B1. Then the condition of Hall’s Theorem is satisfied, and from the theorem we conclude that G hasa perfect matching, that is, if we view the matching as a map from B to B′, then the map is injective.Since |B| = |B′|, the matching is bijective.

Exercise 144 Let B be a basis of a matroid M and suppose y ∈ M \B, x ∈ B. Prove that

(i) B ∪ {y} contains a unique circuit Cy, which is known as a fundamental circuit of B;

(ii) y ∈ Cy;

(iii) (B ∪ {y}) \ {x′} is a basis of M if and only if x′ ∈ Cy;

(iv) (B \ {x})∪ {y′} is a basis of M if and only if there is a circuit C containing {x, y′} such that thereis no circuit C ′ satisfying x ∈ C ′ and C ′ \B ( C \B.

Proof. (i) B ∪ {y} must be dependent from the maximality of B in independent sets, and thus itcontains a circuit Cy. It is also trivial that Cy must contain y. If there are two circuits contained inB ∪ {y}, then both of them contain Cy, the weak elimination property leads us to a contradictionimmediately. Hence the circuit is unique.

(ii) is proved in (i).

19

(iii) ‘If’ part: Let x′ ∈ Cy. If (B ∪ {y}) \ {x′} is dependent, then we may obtain a circuit contained init. It is clear that the circuit contains y. Thus we get two distinct circuits contained in B ∩ {y}.Contradiction. Hence (B ∪{y}) \ {x′} must be independent, and it is maximal base, because it hasthe same number of elements of B. So it is a base.

‘Only if’ part: If x′ /∈ Cy then C ⊆ (B \ {x}) ∪ {y′} and thus (B \ {x}) ∪ {y′} is not a base.Contradiction.

(It is Exercise 8.2.20 (a) in [1])

(iv) ‘If’ part: Suppose there is a circuit C containing x, y′ and there is no circuit C ′ satisfying x ∈ C ′

and C ′ \ B ( C \ B. If B − x + y′ is dependent, then y′ /∈ x and it contains a circuit C∗ whichcontains y′. Then by weak elimination property, there exists a circuit C ′ ⊆ (C ∪ C∗) \ {y′}. It isclear that C ′ \B ⊆ C \B, and the equality can not hold because y ∈ C \B and y /∈ C ′ \B. We meeta contradiction. Hence B − x + y′ must be independent, and it is a base from |B − x + y′| = |B|.

‘Only if’ part: Since B − x + y′ is a basis, B + y′(adding x back to it) contains a circuit, say C.It is clear that x ∈ C. It is also clear that y′ ∈ C, too. If there exists a circuit C ′ satisfying x′ ∈ Cand C \B ( C \B = {y′}. Thus C \B = ∅ and C = B is independent. Contradiction.

Exercise 145 Let B and B′ be two bases of a matroid M . For any x ∈ B, let Sym(x, B,B′) be theset of elements y ∈ B′ such that both (B ∪ {y}) \ {x} and (B′ ∪ {x}) \ {y} are bases of M . Prove thatSym(x,B,B′) 6= ∅. (This is Exercise 8.2.24 in [1])

Proof. Suppose x ∈ B and y /∈ B. It is clear that the following are equivalent:

(i) B + y − x is a base;

(ii) x is in the fundamental circuit of B and y;

(iii) y is in the fundamental cocircuit of B and x. (the concept of fundamental cocircuit is in Exercise8.2.38(a) in [1])

The equivalence of (i) and (ii) is proved in the previous exercise. Applying it in M∗, the dual matroid ofM , we get the equivalence of (i) and (iii), as (i) is equivalent to B + x− y is a base in M∗.

On one hand, Sym(x,B,B′) contains elements y ∈ B′ lying in the fundamental circuit C of B′ andx. On the other hand, Sym(x, B,B′) contains the elements y ∈ B′ lying in the fundamental cocircuit ofB and x. Thus Sym(x,B,B′) = C ∩ C∗ − x. Since |C ∩ C∗| 6= 1 by Theorem 8.2.39 in [1], it holds thatSym(x, B,B′) 6= ∅.

Exercise 146 For a matroid M and an integer k ≥ 2, there are bases B, B′ and x ∈ B with|Sym(x,B,B′)| = k if and only if there is a circuit C and a cocircuit C∗ of M such that |C ∩C∗| = k +1.

Proof. ‘If ’ part: Let x ∈ C ∩ C∗, then C − x is independent. Extend it to a basis B in M . Since C∗

is a hypobases in M(Lemma 8.2.31 in [1]), we can find a base B′ ⊆ C∗ ∪ {x} containing x. Now let usexamine Sym(x,B′, B).

Consider y ∈ B ∩ C∗. There exists a circuit Cy ⊆ B′ ∪ {y} containing y. Notice that y ∈ C∗ ∩ Cy.Since |C∗ ∩ Cy| 6= 1 (Theorem 8.2.29 in [1]), so C∗ ∩ Cy = {x, y}. Hence (B′ \ {x}) ∪ {y} is a base.Similarly we can prove that (B′ \ {y}) ∪ {x} is a base, too.

If y ∈ B \ C∗, then y ∈ B′ \ {x}, so (B′ \ {x}) ∪ {y} is not a base because its cardinality is smallerthan |B′|.

20

Hence Sym(x,B′, B) = (C ∩ C∗) \ {x} and |Sym(x, B′, B)| = k.‘Only If ’ part: Notice that |Sym(x, B,B′)| > 1, we know that x /∈ B′. Hence Sym(x,B,B′) =

(C ∩ C∗) \ {x}(from Exercise 116) and |C ∩ C∗| = k + 1.

Exercise 147 Let B and B′ be two bases of a matroid M and suppose X ⊂ B. Prove that there existsX ′ ⊂ B′ such that both (B ∪X ′) \X and (B′ ∪X) \X ′ are bases of M . (This is Exercise 8.2.22 in [1])Exercise 149 Prove the following form of the Schroder-Bernstein theorem. Let G be a bipartite graphwith vertex classes X and Y having arbitrary cardinalities. Let A ⊆ X and B ⊆ Y . Suppose there arecomplete matchings from A into Y and from B into X. Prove that G contains a set of independent edgescovering all the vertices of A ∪B.Exercise 155 Let H be minimally imperfect. Then, we have the followings: (i) n(H) = 1 + α(H)ω(H)(ii) H has exactly n(H) α-independent sets Qi, i ∈ [n], and exactly n(H) ω-cliques Si, i ∈ [n], such that|Si ∩Qj | = 1− δ(i, j).

Proof. (i) By Lemma 144, the condition for β-perfection fails only for the full vertex set V (H). Hence,for each x in V (H) it holds that

n(H)− 1 ≤ α(H − v)ω(H − v) = α(H)ω(H) ≤ n(H)− 1.

Therefore, n(H) = α(H)ω(H) + 1.

(ii) Notice that χ(H − x) = ω(H − x) = ω(H) and θ(H − x) = α(H − x) = α(H) for each x ∈ V (H).Choose a clique S of size ω(H). Since n(H−x) = α(H)ω(H), for each x ∈ S, H−x has a partitioninto α(H) cliques of size ω(H). Then S and those ω(H) partitions form a list of n = α(H)ω(H)+1maximum cliques S1, . . . , Sn. Each vertex in S appears in S and once in ω(H)− 1 partitions. Eachvertex outside S appears in ω(H) partitions. Hence every vertex appears in exactly ω(H) cliquesin the list.

Each Si induces an α-independent set Qi disjoint from Si as follows. Let x ∈ Si. The χ(H−x) =ω(H) maximum independent sets that partition V (H−x) can meet Si only at the ω(H)−1 verticesother than x and it is clear that each independent sets intersect Si − x at exactly one vertex. Oneof the independent sets misses Si and call it Qi.

Let A be the incidence matrix with ai,j = 1 if xj ∈ Si and ai,j = 0 otherwise. Let B be thematrix with bi,j = 1 if xj ∈ Qi and bi,j = 0 otherwise. We shall prove that ABT = J − I, whereJ is the matrix of all 1s. We need only to show that each column of ABT sums to n − 1 since(ABT )i,i = |Si ∩Qi| = 0 by construction. Since every column of A sums to ω(H) and every row ofB sums to α(H), we have

1T (ABT ) = (1T A)BT = ω1T BT = ωα1T = (n− 1)1,

whence we conclude that ABT = J − I and it is invertible, so are A and B. Their rows are distinctand thus S1, . . . , Sn and Q1, . . . , Qn are distinct. It also follows |Si∩Qi| = 1−δi,j from ABT = J−I.

To complete the proof, we need to show that the two lists contain all the cliques and independentsets. First we find A−1 and B−1. Clearly,

A(ω−1J −BT ) = ω−1(AJ)−ABT = ω−1ωJ − (J − I) = I

and henceA−1 = ω−1J −BT .

21

Similarly we have B−1 = α−1J −AT .

Let s be the incidence vector of a maximum clique S and we shall show that s is a row of A.Recall that rankA = n and its rows span Rn, so s is a linear combination of rows of A and we canwrite s = tA. Thus t = sA−1 = ω−1sJ − sBT = 1T − sBT . The i-th component of sBT is |S ∩Qi|,which is 1 or 0. Hence t is a 0,1-vector and s is a sum of rows of A. Since the sum of entries of s isexactly ω(H), it must be a row of A. Therefore, S1, . . . , Sn are the only maximum cliques.

A similar argument shows that Q1, . . . , Qn are the only maximum independent sets.

Exercise 156 An irreflexive and transitive relation on a set is called a partial ordering relation. Aset together with a partial ordering relation on it is called a poset. The width of a poset is the largestsize of any antichain of the poset and the dimension of a poset is the minimum linear orderings whoseintersection is the poset. (i) Show that the comparability graph of a poset is perfect; (ii) Show that theincomparability graph of a poset is perfect; (iii) Prove that the dimension of any finite poset is less thanor equal to its width.Exercise 157 Let S be the set of all interval digraphs and T the set of all Weiner digraphs. Prove thatT ( S.Exercise 162 Construct an infinite families of score sequences which do not uniquely determine atournament.Exercise 163 Construct a way of arranging a Round Robin Tournament with n players with as fewrounds as possible.

Proof. Notice that if n is even, then n − 1 rounds are required; if n is odd, then n rounds are requiredand each team is idle in exactly one round.

Firstly let us assume that n is odd. Make an n × n array called A, where ar,i = j iff Team i playsTeam j in Round r. And ar,i = i means that Team i is idle in the round. We define A as

ar,i = (r − i) mod n,

where m mod n gives the remainder of m/n if n - m and n if n | m. It is easy to verify the followingthree conditions:

(i) It is a well-defined schedule, that is, if ar,i = j then ar,j = i.

(ii) Every team must play every other team. It follows from that all of the numbers from 1 to n appearin each column.

(iii) Each team is in at most one game per round. It follows from that all of the numbers from 1 to nappear in each row exactly once.

Secondly, let n be even. Let A′(n−1)×(n−1) be the schedule of n− 1 teams. We define A by joining a

column full of n to A′. Now, for i from 1 to n− 1, let r = 2i mod (n− 1), and swap the numbers at Ar,i

and Ar,n−1. We shall prove that the three conditions are satisfied.

Exercise 165 Theorem 157 still holds when Eq. (37) is replaced by∑

x∈S sx ≥(|S|

2

).

Proof. We verify that it holds Eq. (37), too. For the S in Eq. (37), we have∑

x∈V−S sx ≥(|V−S|

2

)=(

n−|S|2

). Since

∑x∈S +

∑x∈V−S sx =

∑x∈V sx = |E| =

(n2

), we have

∑x∈S sx ≤

(n2

)−(n−|S|

2

).

22

Exercise 166 Show that we can delete a set of 2n − n inequalities among all those 2n inequalities inEq. (37) and still guarantee the truth of Theorem 157.

Proof. Let us arrange the sequence in increasing order: s1 ≤ s2 ≤ · · · sn. Then we need only the followingn equations

n∑i=n−k+1

si ≤(

n

2

)−(

n− k

2

), k = 1, 2, . . . , n.

Exercise 168 A digraph is a semiorder digraph if and only if it has no induced subdigraph isomorphicto either 2+2 or 1+3.

Proof. ‘Only if’: Firstly, a semiorder digraph is a Weiner digraph, thus it does not contain 2+2. Considerthe interval representation of 1 + 3. The interval of the three vertices are distinct and from left to right.And the interval of the first vertex – since the vertex is disconnected from the three ones – must intersectwith each of the three intervals. However, it is impossible when the presentation is a unit intervalrepresentation. Hence a semiorder digraph does not contain 1 + 3, either.

‘If’: Since G has no 2 + 2, it is a Weiner digraph and has an interval representation. First, we modifyit to be a proper representation. Suppose Iu ⊂ Iv. Since G has no 1 + 3, Iv does not intersect with theone of the intervals to the left and right of Iu that do not intersect Iu. Suppose Ix = [a, b] and Iv = [c, d]with c < a < b < d. Then [c, a] or [b, d] contains at least one endpoint of interval that does not intersectIu, take [c, a] for example. So we can shorten Iv by putting c > a a little without changing the graphobtained from the representation. Repeating this until no more pairs of intervals are related by inclusionand we arrive at a proper interval representation.

Now we shall modify the proper representation into a unit representation. An interval is said to beleft to another if its left endpoint is left to the other one’s endpoint. In fact, the right endpoint has thesame property because no interval properly contains another. Processing from left to right, we adjust allintervals to unit length. Let Iu = [a, b] be the current leftmost unadjusted interval. If Ix contains theright endpoint of some other interval, then let x be the rightmost one, or let x = a otherwise. In thefirst case, the interval that x corresponds to has been adjusted to unit length. Hence x < min{a + 1, b}.Now, adjust the portion of the representation in [a,∞) by shrinking or expanding [x, b] to [x, a + 1]and translating [b,∞) to [a + 1,∞). The order of endpoints does not change, intervals earlier than Ix

still have length 1, and Ix also now has length 1. Iterating this operation produces the unit intervalrepresentation.

Exercise 170 Let Xn be the set of all(2nn

)lattice paths from (0, 0) to (n, n) with steps (0, 1) and (1, 0).

Define the excedance of a path P ∈ Xn to be the number of i such that at least one point (i, i′) of P liesabove the diagonal x1 = x2. Show that the number of paths in Xn with excedance j is independent of j.

Proof. An arbitrary path can be uniquely identified by a sequence (a0, a1, . . . , an), where ai describesthe number of horizontal steps on x = i. We claim that all the cyclic permutations of the sequence aredistinct, if not, the length of repetend, say k < n + 1, divides n + 1. But n =

∑ai = n+1

k m, where mis the sum of elements of a repetend. We see that n+1

k divides n, but it also divides n + 1. Notice thatn and n + 1 are relatively prime, thus k = n + 1, a contradiction. Hence all the cyclic permutations aredistinct, and the excedances of them are exactly 0, . . . , n. The conclusion follows.

23

Exercise 171 Let Sn denote the set of n(n− 1) hyperplanes xi − xj = 0, 1, 1 ≤ i < j ≤ n in Rn. Showthat the number of connected regions of Rn \

⋃H∈S H is the same as the number of spanning trees of

Kn+1.Exercise 173 Every interval graph is chordal.

Proof. It is contained in the solution to Exercise 5.3.28 of [1].

Exercise 174 Each simple graph has a finite boxity.Exercise 176 A simple graph has treewidth at most one if and only if it is a forest.

Proof. ‘If’: Suppose T is a forest and assume that |T | ≥ 2(It is trivial when T contains a single vertex).Let the bags be {(u, v) : (u, v) ∈ E(T )} and two bags are join if and only if the edges they correspondto have a common vertex. It is easy to verify that this gives a tree decomposition of T , and clearly, thetreewidth is 1.

‘Only if’: Let T be the tree decomposition of graph G. We may assume that no two vertices i, j ∈ Tsuch that Xi ⊆ Xj . (Otherwise we can merge i and j if they are adjacent, or i and its nearest neighbouron the path to j if they are not.)

Consider a leaf l in T . By the assumption, Xl must contain a vertex v which is not contained in Xw,where w is the neighbour of l in T , and hence not by any other vertex in the tree either. As |Xl| ≤ 2,it follows that degG v ≤ 1. Removing this vertex from G, the treewidth cannot increase. Hence we canrepeatedly remove vertices of degree at most 1, until an empty graph is left. This implies that G is aforest.

Exercise 177 For any graph G it holds tw(G) ≤ |V (G)| − 1.

Proof. It G is not a complete graph, then some edge (u, v) is missing. Taking X1 = V (G) \ {u} andX2 = V (G) \ {v}, we set up a tree decomposition (which has two nodes only) with treewidth |V (G)| − 2.Hence tw(G) ≤ |V (G)| − 2.

If G is a complete graph, we claim that tw(G) = |V (G)| − 1. If tw(G) < |V (G)| − 1, then there existone vertex whose degree is at most |V (G)| − 1 and G is not complete, contradiction.

Exercise 180 A chordal graph with at least one isolated vertex is a competition graph.Exercise 185 If Σ is pairwise compatible, then Σ = Σ(T ), where T = (G(Σ), φΣ). Compare this resultwith Theorem 175 (iv).Exercise 186 Let Σ be a nonempty set of X-splits. Denote by I(Σ) the collection of subsets of Σthat are either pairwise incomparable or have cardinality at most one. Prove that |V (Σ)| = |I(Σ)| and|E(Σ)| =

∑I∈I(Σ) |I|.

Exercise 187 Let T and T ′ be two X-trees. Then d(T , T ′) is equal to the minimum number k forwhich there is a sequence T0, . . . , Tk of X-trees such that T0 = T, Tk = T ′, and Ti is obtained from Ti−1

be either a contraction or an expansion of an edge.Exercise 188 Use probabilistic method to prove Theorem 27: Every set B = {b1, . . . , bn} of nonzerointegers contains a sum-free subset of size > n/3.

Proof. Pick up a prime p = 3k + 2, where k is large enough such that p > max |bi|. Let A = {k + 1, k +2, . . . , 2k + 1}, then A is sum-free and |A| > p−1

3 . Choose a number from {1, 2, . . . , p − 1} randomly.

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Define X = (xB mod p) ∩ A. Then the set of those integers b ∈ B such that xb mod p ∈ X is alsosum-free. It is sufficient to show that E(|X|) > n

3 . From

|X| =n∑

i=1

IA(xbi mod p)

(where IA in the indicator function) and the linearity of expectation, we have

E(|X|) =n∑

i=1

E(IA(xbi mod p)).

Since |A| > p−13 and x is chosen uniformly, E(IA(xbi mod p)) = |A|

p−1 > 13 . Hence E(|X|) > n

3 , concludingthe proof.

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Exercises from Introduction to Graph Theory

Exercise 1.2.42 Let G be a connected simple graph that does not have P4 or C4 as an induced subgraph.Prove that G has a vertex adjacent to all other vertices.

Proof. Let u be a vertex of maximum degree. Assume that deg u > 1. (The case deg u = 1 is trivial)We are going to prove that u is adjacent to all other vertices. Let v1, . . . , vk be the adjacent verticesof u. If k < V (G) − 1 then there exists w not connected with u such that w is connected with vi forsome i because the graph is connected. Thus a path of length 4 is obtained: wviuvj (j 6= i), this is acontradiction with the assumption.

Exercise 1.4.40 A directed graph is unipathic if for every pair of vertices x, y there is at most one(directed) x, y-path. Let Tn be the tournament on n vertices with the edge between vi and vj directedtoward the vertex with larger index. What is the maximum number of edges in a unipathic subgraph ofTn? How many unipathic subgraphs are there with the maximum number of edges?

Proof. We claim that the maximum number of edges in a unipathic subgraph of Tn is n−1. If a subgraph ofTn contains n edges, then it contains a cycle C (ignoring the directions). Let S = {i : vi ∈ S}, m = minSand M = max S. Then there are two paths from vm to vM , so it is not unipathic. Note that the path12 . . . n has n− 1 edges, we have proved that the maximum number of edges in a unipathic subgraph ofTn is n− 1.

In fact, any spanning tree (ignoring directions) of Tn is a unipathic subgraph. It is known that Kn

has nn−2 spanning trees(Cayley’s Formula, Theorem 2.2.3), so there are nn−2 unipathic subgraphs withn− 1 edges.

Exercise 2.1.29 Every tree is bipartite. Prove that every tree has a leaf in its larger partite set (inboth if they have equal size).

Proof. For the first part, we prove it by induction on n.Base : n = 1.Inductive step. Suppose that we know every tree on n nodes is bipartite for some n ≥ 1. Let T be

a tree on n + 1 nodes. Since n + 1 ≥ 2, we know that it has a leaf v from Lemma 2.1.3. Let u be theneighbour of v and T ′ = T − {v}. T ′ is a tree and hence it is bipartite, suppose T ′ = (X ′, Y ). Supposeu ∈ Y , let X = X ′ ∪ {v}, and (X, Y ) gives a bipartition of T .

Now we prove the second part of the problem. Suppose T = (X, Y ). If |X| ≥ |Y | and there are noleaves in X, then each node in X is adjacent to at least nodes in Y , thus there are more than 2|X| edges.However, it is a tree and it has exactly |X| + |Y | − 1 < 2|X| edges. We arrive at a contradiction. Asimilar argument works for the other case.

Remark. If T = (X, Y ) is a bipartition and |X| = |Y |+ k, then there are at least k + 1 leaves in X.

Exercise 2.1.54 Centers of trees. Let T be a tree.

(i) Give a noninductive proof that the center of T is a vertex or an edge.

(ii) Prove that the center of T is one vertex if and only if diam T = 2 radT .

(iii) Use part 1 to prove that if n(T ) is odd, then every automorphism of T maps some vertex to itself.

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Proof. (i) Suppose the center contains more than one vertex. Let u, v be such vertices. Supposed(x, u) = ε(x). It is not difficult to see that v must be on the path of u-x, just prove it bycontradiction and note that ε(u) = ε(v). Let y be the vertex satisfying d(v, y) = ε(y). Then ushould be on the path of v-y. If there is another vertex z in the center of the tree, then z is on bothpath u-x and v-y. Thus it is on the path u-v. We claim that ε(z) < ε(x). If d(z, s) ≥ d(u, x) forsome s, then both u, v should be on the path z-s, which is impossible. Now we see that the centercontains at most two vertices, furthermore, uv must be an edge. (Otherwise, a same contradictioncan be easily derived)

(ii) ‘Only If’ part: Suppose the center is v, then radT = ε(v). Suppose d(u, v) = ε(v) and d(u, w) =ε(u). It suffices to show that d(v, w) = ε(v) and d(u, w) = diam T .

Without loss of generality, suppose ε(v) > 1. First, d(v, w) ≤ ε(v). Let x be the vertex adjacentto v and x is on the path v-u. If d(v, w) < ε(v), then d(x,w) ≤ ε(v). Suppose d(x, s) = ε(x) then vmust be on the path x-s (otherwise d(v, s) > d(x, s) = ε(x) thus ε(x) < ε(v), which contradicts withthe assumption that v is the center). Then d(u, s) = d(u, x) + d(x, s) ≥ d(u, x) + d(x,w) = d(u, w),so d(u, s) = d(u, w) and d(v, s) = d(v, w) < ε(v), we obtain that ε(s) ≤ ε(v), which contradicts withthe choice of v again. Now we have proved that d(v, w) = ε(v).

Suppose d(x, y) = diam T and we know that v must lie on the path x-y (otherwise v-x-y will bea longer path). Now we have d(v, x) ≤ ε(v) and d(v, y) ≤ ε(v), and diam T = d(x, y) ≤ 2ε(v) =d(u, w). Therefore, diam T = d(u, w) = 2ε(v) = 2 rad T .

‘If ’ part: Suppose d(x, y) = diam T . Let v be a vertex on the path such that d(x, v) = radT . Weare going to show that v is the center.

It is easy to see that ε(v) = d(x, v) = radT . Suppose there is another vertex, say u, satisfyingε(u) ≤ ε(v). Since d(x, y) = diam T ≥ d(u, y), we know that u must lie on the path x-y, otherwiseu-x-y would be a longer path. If u is on the path x-v, then d(u, y) > d(x, y) = radT = ε(v) whichcontradicts to the hypothesis of ε(u) ≤ ε(v). Similarly, u cannot be on the path v-y. We arrive ata contradiction, which implies that v is the center.

(iii) Let K be the center of T and f be an automorphism. Because an automorphism preserves distance,it is clear that f(K) is the center of f(T ). So f(K) = K, or f preserves the center. If the center ofthe tree is a single vertex, we are done. Assume the center is an edge, say uv, then the automorphismpreserves it. Suppose its endpoints are swapped under the automorphism. Let the original tree beT and denote the automorphism by f . If we remove uv from T , then it breaks into two components,say T1 and T2, where T1 is rooted by u and T2 is rooted by v. Then f(T1) and f(T2) are two disjointcomponents. f(T ) is decomposed into two components by removing uv, too. Hence f(T1) = T2

and f(T2) = T1, whence we find out that n(T ) is even, which is a contradiction to the assumption.Thus f fixes u and v.

Exercise 2.1.58 Let S and T be trees with leaves {x1, . . . , xk} and {y1, . . . , yk}, respectively. Supposethat dS(xi, xj) = dT (yi, yj) for each pair i, j. Prove that S and T are isomorphic.

Proof. We claim that the isomorphism maps xi to yi. We prove it by induction k, the number of leaves.Base : The case k = 1 is trivial. For k = 2, a tree with two leaves is a path. Since two paths have the

same length, they are isomorphic.

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Inductive Step. Suppose the statement is true for all tree with at most k(≥ 2) leaves. Now considertwo trees S and T with leaves {x1, . . . , xk+1} and {y1, . . . , yk+1} satisfying dS(xi, xj) = dT (xi, xj) foreach pair 1 ≤ i, j ≤ k + 1.

Let P be a maximal path in S from xk to v (v is determined according to the maximality of this path,and v is not a leaf, and the removal of P−{v} remains a tree) such that every intermediate node has degreeof 2. We obtain a new tree S′ by deleting P −{v} from S. Similarly we define Q be a maximal in T andobtain T ′ from T . Because S and T have at least three leaves, x′ and y′ are not leaves, then S′ and T ′ aretrees with k−1 leaves. And we have that dS′(xi, xj) = dS(xi, xj) = dT (xi, xj) = dT ′(xi, xj)(i, j ≤ k−1).By the induction hypothesis, S′ and T ′ are isomorphic.

We now claim that x′ is mapped to y′. If f(x′) 6= y′ then we pick a path yi . . . f(x′) . . . y′ . . . yj in T .In tree S, we have

dS(xi, xk) = dS(xi, x′) + dS(x′, xk)

anddS(xj , xk) = dS(xj , x

′) + dS(x′, xk).

Subtracting the second one from the first one, we get

dS(xi, xk)− dS(xj , xk) = dS(xi, x′)− dS(xj , x

′). (6)

In tree T , we havedT (yi, yk) = dT (yi, f(x′)) + dT (f(x′), y′) + dT (y′, yk)

anddT (yj , yk) = dT (yj , f(x′))− dT (f(x′), y′) + dT (y′, yk).

Subtracting the second one from the first one, we get

dT (yi, yk)− dT (yj , yk) = dT (yi, f(x′))− dT (yj , f(x′)) + 2dT (f(x′), y′).

Comparing it with (6), we conclude that 2dT (f(x′), y′) = 0, which contradicts with f(x′) 6= y′. Therefore,f(x′) = y′.

Now we come to the last part of our proof. Since x1 is mapped to y1 by the induction hypothesis, andx′ is mapped to y′, we have dS′(x1, x

′) = dT ′(y1, y′). We also have dS(x1, xk) = dS(x1, x

′) + dS(x′, xk)and dT (y1, yk) = dS(y1, y

′) + dS(y′, yk). Then it yields that dS(x′, xk) = dS(y′, yk). Hence P and Q havethe same length, S and T are isomorphic by adding P to S′ and Q to T ′.

Exercise 2.1.72 Prove that if G1, . . . , Gk are pairwise-intersecting subtrees of a tree G, then G has avertex that belongs to all G1, . . . , Gk. Comment: This result is the Helly property for trees.

Proof. We prove it by induction on k. It is obvious for k = 1, 2.Inductive Step. Suppose the statement is true for k. Suppose we have a family of subtrees G1, . . . , Gk+1

such that Gi∩Gj 6= 0 all for i 6= j. Let u ∈ G1∩Gk+1 and by induction hypothesis, there exists v ∈k⋂

i=1

Gi

and w ∈k+1⋂i=2

Gi. If two of u, v, w coincides, we are done; otherwise, we have |Gi∩{u, v, w}| ≥ 2 for all i. In

this case, each Ti contains at least one of the paths Puv, Puw, Pvw. Thus⋂

Ti ⊃ Puv ∩Puw ∩Pvw 6= ∅.

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Exercise 2.3.30 Consider n messages occurring with probabilities p1, . . . , pn such that each pi is apower of 1/2 (each pi ≥ 0 and

∑pi = 1).

(i) Prove that the two least likely messages have equal probability.

(ii) Prove that expected message length of the Huffman code for this distribution is −∑

pi log pi.

Proof. (i) Suppose pi = 2−mi , and without loss of generality, suppose m1 ≤ m2 ≤ · · · ≤ mn. We shall

prove mn−1 = mn. Sincen∑

i=1

12mi

= 1, we get1

2mn= 1−

n−1∑i=1

12mi

. It is clear that the right side has

the form ofN

2mn−1(1 ≤ N < 2mn−1). So

12mn

=N

2mn−1. Note that mn ≥ mn−1, it follows easily

that mn = mn−1.

(ii) It suffices to show that the message of probability pi = 2−mi is on the mi-th level of the Huffmantree. It is not difficult to see that for such distribution p1, . . . , pn given in the problem, there are2q numbers equal to min pi for some q.

At first, suppose there are 2q messages of probability min pi. After q steps, we get 2q−1 messagesof probability 2pi and there remain n − q probabilities p1, . . . , pn−q. So we get a new distributionp1, . . . , pn−q, 2pi which also satisfies all the assumptions. Thus we can make an inductive proof.

Exercise 2.3.31 Suppose that n messages occur with probabilities p1, . . . , pn and that the words areassigned distinct binary code words. Prove that for every code, the expected length of a code wordrespect to this distribution is at least −

∑pi log pi.

Exercise 5.3.28 Let G be an interval graph. Prove that G is a chordal graph and that G is a compar-ability graph.

Proof. Firstly we show that G is chordal. It suffices to show that Cn is not an interval graph for n ≥ 4.Suppose Cn = (v1, . . . , vn) is chordal and [ai, bi] corresponds to vi.

Since v2 is adjacent to v1 and v3 only and v1, v3 are not adjacent, we have a2 ≤ b1 < a3 ≤ b2 ora2 ≤ b3 < a1 ≤ b2. Without loss of generality, we suppose a2 ≤ b1 < a3 ≤ b2. Since v4 is adjacent tov3 and not adjacent to v2, we must have b2 < a4 ≤ b3. In this way, we reach that bi−1 < ai ≤ bi−1 for4 ≤ i ≤ n. And finally we see that [a1, b1] and [an, bn] are disjoint, which contradicts to v1 and vn areadjacent. Therefore, Cn(n ≥ 4) is not chordal.

Secondly let us show that G is a comparability graph. We orientate the edges of G by letting vi → vj

if bi < aj . It is easy to verify the transitivity.

Exercise 5.3.29 Determine the smallest imperfect graph G such that χ(G) = ω(G).

Proof. Since G is imperfect, it must contain an induced proper subgraph H such that χ(H) > ω(H).The smallest imperfect graph is C5(χ(C5) = 3 > ω(C5) = 2), so n(G) > 5(whether it contains C5 or not).We construct a imperfect graph G with six vertices such that χ(G) = ω(G) and it must be the smallestone. Let G contains C5 and call the sixth vertex v. Join v to the two adjacent vertices in C5 and thereforms a triangle. It is clear that ω(G) = 3 and χ(G) = 3.

Exercise 5.3.30 An edge in an acyclic orientation of G is dependent if reversing it yields a cycle.

(i) Prove that every acyclic orientation of a connected n-vertex graph has at least n − 1 independentedges.

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(ii) Prove that if χ(G) is less than the girth of G, then G has an orientation with no dependent edges.

Proof. (i) Prove it by induction. It is true when n ≤ 3.

Inductive Step. Assume that the statement is true for n(≥ 4). Consider a graph G with n + 1vertices. Since the orientation is acyclic, G must have at least a sink, say w. (consider the longestdirected path). G− w has at least n− 2 independent edges, and clearly they are also independentin G. At least one of the edges incident with w is independent by noticing that there is at mostone edge in a cycle. Thus G has at least n independent edges. The statements is true for n + 1.

(ii) Colour the graph in χ(G) colours and orient each edge from the lower colored node to the higher.Clearly this orientation is acyclic. Suppose it has a dependent edge. Let C be the cycle formed byreversing the edge. Then the colours lie on C in increasing order. Thus C has at most χ(G) edges,which contradicts to the hypothesis that C has more vertices than χ(G).

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References

[1] West, D.B., Introduction to Graph Theory, 2nd edition, Prentice Hall, 2001.

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