Incidence matrix games - Graphs and Matrices

12.4 Incidence matrix games

Let G be a directed graph with V (G) = {1, . . . , n} and E(G) = {e₁, . . . , e_m}. Con-sider the following two-person zero-sum game. The pure strategy sets of Players I and II are V (G) and E(G), respectively. If Player I selects i and Player II selects e_j, then the payoff to Player I from Player II is defined as follows. If i and ej are not incident then the payoff is zero. If e_j originates at i then the payoff is 1, while if ej terminates at i then the payoff is −1. Clearly the payoff matrix of this game is the (vertex-edge) incidence matrix Q of G. We refer to this game as the incidence matrix game corresponding to G.

Lemma 12.16. Let G be a directed graph with V (G) = {1, . . . , n} and E(G) = {e₁, . . . , e_m}. Let Q be the n × m incidence matrix of G. Then 0 ≤ v(Q) ≤ 1. Fur-thermore, v(Q) = 0 if G has a directed cycle, and v(Q) = 1 if G is the star K1,n−1, with the central vertex being a source.

Proof. The strategy z =¹_n1 for Player I satisfies z⁰Q= 0. Let y be optimal for Player II so that Qy ≤ v(Q)1. Then v(Q) ≥ z⁰Qy= 0. Since q_{i j}≤ 1 for all i, j, it follows that v(Q) ≤ 1.

Suppose G has a directed cycle with k vertices. Consider the strategy z for Player II, who chooses each edge of the cycle with probability ¹_k. Then Qz = 0. Let x be optimal for Player I, so that x⁰Q≥ v(Q)1⁰. Hence, v(Q) ≤ x⁰Qz= 0. Since we have shown earlier that v(Q) ≥ 0, it follows that v(Q) = 0.

Now suppose G = K_1,n−1, and let 1 be the central vertex, which is assumed to be a source. It can be verified that the pure strategy 1 for Player I and any pure strategy

for Player II are optimal and v(Q) = 1. ut

It is evident from Lemma 12.16 that if G has a directed cycle, then the inci-dence matrix corresponding to G has value 0, and the optimal strategies are easily determined. We now assume that G is acyclic. As usual, let V (G) = {1, . . . , n} and E(G) = {e₁, . . . , e_m}. For each i ∈ V (G) let P(i) denote a path originating at i and having maximum length. Let ρ(i) denote the length (the number of edges) in P(i).

If i is a sink then we set ρ(i) = 0. For each edge ej∈ E(G), let η(ej) denote the number of vertices i such that e_jis on the path P(i). With this notation we have the following.

Lemma 12.17.

n i=1

∑

ρ (i) =

∑

j=1

η (ej).

Proof. Let B be the n × m matrix defined as follows. The rows of B are indexed by V(G), and the columns of B are indexed by E(G). If i ∈ V (G) and e_j∈ E(G) then the (i, j)-entry of B is 1 if e_j∈ P(i) and 0, otherwise. Observe that the row sums of Bare ρ(1), . . . , ρ(n) and the column sums of B are η(e1), . . . , η(e_m). Since the sum of the row sums must equal that of the column sums, the result is proved. ut

Theorem 12.18. Let G be a directed graph with V (G) = {1, . . . , n} and E(G) =

Then v(Q) = θ . Furthermore, θ ρ and θ η are optimal strategies for Players I and II, respectively, where ρ is the n × 1 vector with components ρ(1), . . . , ρ(n) and η is the m× 1 vector with components η(e1), . . . , η(em).

Proof. First note that by Lemma 12.17,

Note that ρ(`) ≥ ρ(k) + 1 and therefore it follows from (12.7) that

θ non-positive. Suppose that U 6= φ . Observe that for any vertex s 6= i, the path P(s) either contains exactly one edge from U and one edge from W or has no intersection with either U or W. Thus, for any s 6= i, the path P(s) either makes a contribution of 1 to both

∑

j∈U

η (ej) and

∑

j∈W

η (ej), or does not contribute to either of these terms. Also, the path P(i) makes a contribution of 1 to

∑

j∈U

12.4 Incidence matrix games 153 In view of these observations, we conclude from (12.9) that for i ∈ {1, . . . , n},

∑

j=1

q_{i j}η (e_j) ≤ θ . (12.10)

The result is proved combining (12.8) and (12.10). ut Corollary 12.19. Let G be a directed graph with V (G) = {1, . . . , n} and E(G) = {e₁, . . . , e_m}, and let Q be the n × m incidence matrix of G. Then v(Q) = 0 if and only if G has a directed cycle, and v(Q) = 1 if and only if G is a star with the central vertex being a source.

Proof. The “if” parts were proved in Lemma 12.16, while the “only if” parts follows

from Theorem 12.18. ut

Theorem 12.20. Let G be a directed graph with V (G) = {1, . . . , n} and E(G) = {e₁, . . . , e_m}. Let Q be the n × m incidence matrix of G. Consider the incidence matrix game corresponding to G. Then Player I has a unique optimal strategy.

Proof. Suppose {φ (i), i ∈ V (G)} is optimal for Player I. Let k ∈ V (G) be a sink. Let y∈Pmbe optimal for Player II. If φ (k) > 0 then by Corollary 12.10, we must have

∑

j=1

q_{k j}y_j= v(Q). (12.11)

Since k is a sink, q_{k j}≤ 0, j = 1, . . . , m, whereas by Theorem 12.18 v(Q) > 0. This contradicts (12.11) and hence φ (k) = 0.

Let u ∈ V (G) be a vertex that is not a sink, and let u = u₀, u₁, . . . , u_k= w be a directed path of maximum length, originating at u. Since φ is optimal,

φ (u_i) − φ (ui+1) ≥ v(Q), i= 0, 1, . . . , k − 1.

Thus,

k−1 i=0

∑

(φ (ui) − φ (ui+1)) ≥ kv(Q), and hence

φ (u) − φ (w) ≥ ρ (u)v(Q).

Since w must necessarily be a sink, φ (w) = 0 by our earlier observation, and hence

φ (u) ≥ ρ (u)v(Q). (12.12)

Thus,

1 =

∑

u∈V (G)

φ (u) ≥ v(Q)

∑

u∈V (G)

ρ (u) = 1,

where the last equality follows from Theorem 12.18. Thus, equality must occur in (12.12), and

φ (u) = ρ (u)v(Q), u∈ V (G).

Therefore, the strategy of Player I is unique. ut

Example 12.21. Consider the directed, acyclic graph G:

•1 ^e¹ //

Longest paths emanating from each vertex are given below:

v P(v) multi-plied by 1/10, are the optimal strategies for Players I and II, respectively, in the incidence matrix game corresponding to G, and the value of the game is ₁₀¹.

We turn to the optimal strategy space for Player II. Let G be a directed graph with V(G) = {1, . . . , n} and E(G) = {e1, . . . , e_m}. Let Q be the n × m incidence matrix of G. Consider the incidence matrix game corresponding to G. By Theorem 12.20 the optimal strategy for Player I is unique. By Theorem 12.18 any vertex that not a sink is essential for Player I. Let s be the number of sinks in G and let t be the number of inessential strategies (that is, edges) for Player II. Using the notation of Theorem 12.5, we have f₁= n − s, f₂= m −t. Since dim(OptI(A)) = 0, we conclude by Theorem 12.5 that

dim(Opt_II(A)) = m − n − t + s.

We now consider the 0 − 1 incidence matrix of an undirected graph and discuss some results for the value of the corresponding matrix game. Let G be a graph with V(G) = {1, . . . , n} and E(G) = {e1, . . . , e_m}. We recall some terminology. A set of edges constitute a matching if no two edges in the set are incident with a common vertex. The maximum cardinality of a matching is called the matching number of G, denoted by ν(G). A set of vertices of G form a vertex cover if they are collectively incident with all the edges in G. The minimum cardinality of a vertex cover is the vertex covering number of G, denoted by τ(G).

12.4 Incidence matrix games 155 Lemma 12.22. Let G be a graph with n vertices and m edges. Let M be the n × m, 0 − 1 incidence matrix of G. Then

τ (G) ≤ v(M) ≤ 1 ν (G).

Proof. Let τ(G) = k, ν(G) = `, and suppose, without loss of generality, that the vertices 1, . . . , k form a vertex cover and that the edges e₁, . . . , e_`form a matching. If Player I chooses the vertices 1, . . . , k uniformly with probability ¹_k, then against any pure strategy of Player II, Player I is guaranteed a payoff of at least ¹_k. Similarly, if Player II chooses the edges e₁, . . . , e_`uniformly with probability¹_`, then against any pure strategy of Player I, Player II loses at most¹_`. These two observations together

give the result. ut

A graph is said to have a perfect matching if it has a matching in which the edges are collectively incident with all the vertices. A graph is Hamiltonian if it has a cycle, called a Hamiltonian cycle, containing every vertex exactly once.

In the next result we identify some classes of graphs for which the value of the corresponding game is easily determined.

Theorem 12.23. Let G be a graph with V (G) = {1, . . . , n} and E(G) = {e1, . . . , e_m}.

Let M be the n× m (0 − 1)-incidence matrix of G. Then the following assertions hold.

(i) If G is bipartite then v(M) = ¹

ν (G).

(ii) If G is the path then v(M) =²_nif n is even, and_n−1² if n is odd.

(iii) If G has a perfect matching then v(M) =²_n. (iv) If G is Hamiltonian then v(M) =²_n.

(v) If G= K_nthen v(M) =²_n.

Proof. If G is bipartite, then by the K¨onig–Egervary theorem, ν(G) = τ(G), and (i) follows by Lemma 12.22. Since a path is bipartite, (ii) follows from (i) and the fact that the matching number of a path on n vertices is ⁿ₂if n is even and ⁿ⁻¹₂ if n is odd.

If G has a perfect matching then ν(G) = τ(G) =ⁿ₂, and (iii) follows from (i).

To prove (iv), first suppose that G is the cycle on n vertices. Then n = m and the strategies for Players I and II, which choose all pure strategies uniformly with probability¹_n, are easily seen to be optimal. Thus, v(M) =²_n.

Suppose G is Hamiltonian. The value of M is at least equal to the value of the game corresponding to a Hamiltonian cycle in G and thus v(M) ≥²_n, in view of the preceding observation. If Player II chooses only the edges in the Hamiltonian cycle with equal probability, then against any pure strategy of Player I, Player II loses at most ²_n. Therefore, (iv) is proved.

Finally, (v) follows since a complete graph is clearly Hamiltonian. ut

Exercises

Hence, determine the value of a square diagonal matrix.

2. Let G be a directed graph and let A be the skew matrix of G. Consider the matrix game A. Show that the dimension of the optimal strategy set and the number of essential strategies of a player are of the same parity.

3. Let G be a directed graph and let A be the skew matrix of G. Consider the matrix game A. Suppose every pure strategy is essential. Show that the dimension of the optimal strategy set equals n − 1 − rank A.

4. Let G be an acyclic directed graph with n vertices, m edges, m ≥ 2, and let Q be the incidence matrix of G. Show that v(Q) ≥ _m(m−1)² .

5. Consider the graph G :

•1 ^e¹ // •2

Show that there are more than one optimal strategies for Player I in the corre-sponding incidence matrix game.

For an introduction to game theory, including matrix games, see [6,7]. Proofs of Theorems 12.4 and 12.5 can be found in [2,4]. Relevant references for various sec-tions are as follows: Section 12.2: [5], Section 12.3: [3], Section 12.4: [1].

References and Further Reading

1. R.B. Bapat and S. Tijs, Incidence matrix games. In Game Theoretical Applica-tions to Economics and OperaApplica-tions Research(Bangalore, 1996), Theory Decis.

Lib. Ser. C, Game Theory Math. Program. Oper. Res., 18, Kluwer Acad. Publ., Boston, 1997, 9–16.

2. H.F. Bohnenblust, S. Karlin and L.S. Shapley, Solutions of discrete two-person games. In Contrbutions to The Theory of Games, Volume 1, H.W. Kuhn and A.W.

Tucker, Eds., 1950, Princeton University Press, Princeton, 51–72.

12.4 Incidence matrix games 157 3. D.C. Fisher and J. Ryan, Optimal strategies for a generalized “scissors, paper,

and stone” game, American Mathematical Monthly, 99:935–942 (1992).

4. D. Gale and S. Sherman, Solutions of finite two-person games. In Contributions to the Theory of Games, Volume 1, H.W. Kuhn and A.W. Tucker Eds., 1950, Princeton University Press, Princeton, 37–49.

5. T.S. Michael and T. Quint, Optimal strategies for node selection games on oriented graphs: Skew matrices and symmetric games, Linear Algebra Appl., 412:77–92 (2006).

6. G. Owen, Game Theory, Second ed., Academic Press, New York, 1982.

7. S. Tijs, Introduction to Game Theory, Texts and Readings in Mathematics, 23.

Hindustan Book Agency, New Delhi, 2003.

Hints and Solutions to Selected Exercises

Chapter 1

1. Ax = 0 clearly implies A⁰Ax= 0. Conversely, if A⁰Ax= 0 then x⁰A⁰Ax= 0, which implies (Ax)⁰Ax= 0, and hence Ax = 0.

4. If G = A⁺then the two equations are easily verified. Conversely, suppose A⁰AG= A⁰and G⁰GA= G⁰. Since rank A⁰A= rank A, we may write A = X A⁰Afor some X . Then A = X A⁰A= X A⁰AGA= AGA. Also, A⁰AG= A⁰implies G⁰A⁰= G⁰A⁰AG= (AG)⁰AG, which is symmetric. Similarly, using G⁰GA= G⁰, we may conclude that GAG = G and that GA is symmetric.

5. A = xy⁰ for some column vectors x and y. First determine x⁺ and y⁺. α = (trace A⁰A)⁻¹.

Chapter 2

2. Suppose y_i= 1, yj= −1 and yk= 0, k 6= i, k 6= j. Consider an (i j)-pathP. Let xbe a vector with its coordinates indexed by E(G). Set x_k= 0 if ekis not inP.

Otherwise, set e_k= 1 or ek= −1 according as ekis directed in the same way as, or in the opposite way to,P, respectively. Verify that Qx = y.

3. Q⁺= Q⁰(QQ⁰)⁻¹. Note that QQ⁰has a simple structure.

4. If G is not bipartite, then it has an odd cycle. Consider the submatrix of M cor-responding to the cycle.

5. This is the well-known Frobenius–K¨onig theorem. Let G be the bipartite graph with bipartition (X ,Y ), where X = Y = {1, . . . , n}, and i and j are adjacent if and only if a_{i j}= 1. Condition (i) is equivalent to ν(G) < n. Use Theorem 2.22.

Chapter 3

1. The characteristic polynomial of either graph is λ⁶− 7λ⁴− 4λ³+ 7λ²+ 4λ − 1.

5. ε(Kn) = 2(n − 1), ε(K_mn) = 2√ mn.

6. Use Lemma 3.25.

7. Use the previous exercise to find the eigenvalues of the two graphs.

159

8. Note that A(G₁) = 0 1 1 0

⊗ A and A(G₂) = 1 1 1 1

⊗ A, respectively. Use Lemma 3.25

9. Let n be pendant and suppose it is adjacent to n − 1. Assume the result for T \ {n}

and proceed by induction on n.

Chapter 4

1. A repeated application of Laplace expansion shows that det(L + J) is equal to the sum of det L and the sum of all cofactors of L. (Also see Lemma 8.3.) Use Theorem 4.8.

2. Let |V (G)| = n and |V (H)| = m. Then L(G × H) = L(G) ⊗ I_m+ I_n⊗ L(H). If λ1, . . . , λn and µ1, . . . , µm are the eigenvalues of L(G) and L(H), respectively, then the eigenvalues of L(G × H) are λi+ µj; i = 1, . . . , n; j = 1, . . . , m.

3. Use Theorem 4.11 and the arithmetic mean-geometric mean inequality.

4. Use Theorem 4.13.

5. Let (X ,Y ) be a bipartition of T. Make all edges oriented from X to Y. The result holds for any bipartite graph.

6. For the first part, verify that (A⁺)⁰satisfies the definition of the Moore–Penrose inverse of A⁰. Then, for the second part note, using the first part, that

AA⁰(A⁰)⁺A⁺AA⁰= AA⁰(A⁺)⁰A⁺AA⁰.

Since the column space of A⁰is the same as that of A⁺, it follows that A⁺AA⁰= A⁰. Substituting in the previous equation and using the first part shows that (A⁰)⁺A⁺is a g-inverse of AA⁰. The other Moore–Penrose conditions are proved similarly.

Chapter 5 1. Note that B

I B_f

−B⁰_f I

. By the Schur complement formula the determi-nant of B

is seen to be nonzero.

2. Let B be the fundamental cut matrix. There exists an (n−1)×(n−1) nonsingular matrix Z such that X⁰= ZB. Use the fact that B is totally unimodular.

3. The proof is similar to that of Theorem 5.13.

4. First show that det BB⁰[E(T₁)|E(T₁)] is the number of spanning trees of G con-taining T₁as a subtree. Use this observation and Theorem 4.7.

Chapter 6

1. Let A be the adjacency matrix of G and suppose u ≥ 0 satisfies Au = µu. There exists x > 0 such that Ax = ρ(G)x. Consider u⁰Ax.

Hints and Solutions to Selected Exercises 161 2. The Perron eigenvalue of a cycle and of K_1,4is 2.

3. Use Corollary 6.14.

4. If G is strongly regular with parameters (n, k, a, c) then G^c is strongly regular with parameters (n, k₁, a₁, c₁), where k1= n − k − 1, a1= n − 2k − 2 + c and c₁= n − 2k + a.

5. For the first part use Theorem 6.25.

6. Let λ1, . . . , λnbe the eigenvalues of A. Since A is nonsingular, the eigenvalues are nonzero. By the arithmetic mean-geometric mean inequality,

1. The Laplacian L of K_1,n−1has I_n−1as a principal submatrix. Therefore, the rank of L − I_n−1is 2 and hence its nullity is n − 2. Thus, 1 is an eigenvalue of L with multiplicity n − 2. Clearly, 0 is an eigenvalue. The remaining eigenvalue, easily found using the trace, is n.

4. Use a symmetry argument.

5. The first part is an easy consequence of Theorem 7.16. For the second part, using the fact that Q_n is the n-fold Cartesian product of Q₂, show that the algebraic connectivity of Q_nis 2. Also see the remark following Corollary 7.18.

6. Let f : V (G) → {0, 1, −1} be defined by setting it equal to 0 on V₁, 1 on V2

and −1 on V₃. Use the inequality f⁰L f ≥ µ f⁰f where L is the Laplacian. For a generalization and an application to “competitive learning process”, see [3].

7. This is an easy consequence of Theorem 7.20.

8. Use Exercise 7, above .

9. Let fi= (n + 1) − 2i, i = 1, . . . , n. Note that f⁰1 = 0. Use (7.17).

Chapter 8

1 and 2. Follow an argument similar to that in the proof of Theorem 8.2.

3. For α 6= 0 evaluate

5. Suppose (D⁻¹− S)x = 0 for some vector x. Premultiply this equation by 1⁰ and use the formula for D⁻¹ given in Theorem 8.9 to conclude τ⁰x= 0 and hence that (−¹₂L− S)x = 0, where L is the Laplacian of T. Then x⁰(−¹₂L− S)x = 0, and since¹₂L+ S is positive semidefinite, conclude that x= 0.

7. For any i, j, k ∈ V (T ), d_{i j}= d_ik+ d_{k j}mod 2.

8. Use (8.26) and that L⁺, being positive semidefinite, is a Gram matrix, that is, there exist points x¹, . . . , xⁿin IRⁿsuch that `⁺_{i j}= (xⁱ)⁰x^j, i, j = 1, . . . , n.

10. Observe that I_kis a principal submatrix of the Laplacian matrix of T. Use interlacing and then apply Theorem 8.16.

Chapter 9

2. The resistance distance between any two vertices of the cycle is easily found by series-parallel reduction. Lemma 9.9 and a symmetry argument may also be used.

3. First prove the result when there is a cycle containing i and j. Then use the fact that if there are two (i j)-paths then there is an (i j)-path that meets a cycle.

4. By Theorem 9.12, if x is an n × 1 vector orthogonal to τ, then x⁰Rx≤ 0.

6. Use (9.3) and Theorem 4.7.

7. There is a one-to-one correspondence between the spanning trees of G not con-taining the edge e_kand the spanning trees of G^∗containing e⁰_k. Use the equation

χ⁰(G)

χ (G) +χ (G) − χ⁰(G)

χ (G) = 1

and the previous exercise.

9. Use Theorem 9.12, the multilinearity of the determinant and the fact that each cofactor of L equals χ(G).

10. Assume that n is a pendant vertex and that the formula holds for T \ {n}. Use induction on the number of vertices.

Chapter 10

1. Use the recursive definition of a threshold graph and induction.

2. We may encode a threshold graph by a binary sequence b₁, . . . , b_n, with b1= 1.

In the recursive procedure to obtain the graph we add an isolated vertex if b_i= 0, and a dominating vertex if b_i= 1.

3. Use the recursive definition of a threshold graph and induction.

4. Use the recursive definition of a cograph and the fact that the union of two Lapla-cian integral graphs is LaplaLapla-cian integral and the complement of a LaplaLapla-cian integral graph is Laplacian integral.

5. Whether a graph G is a cograph or not can be checked recursively. Take the complement of G. Then it should split into connected components, each of which must be a cograph. Thus, if we take components of G^cand repeat the procedure of taking complements, we must end up with isolated vertices if the graph is a cograph. The presence of P₄ will not lead to this situation since P₄ is self-complementary. Incidentally, it is known that the property of not containing a P₄ as an induced subgraph characterizes cographs.

6. The eigenvalues of L(G) are: n with multiplicity |V₁|, |V₁| with multiplicity |V₂|−

1; and 0. The number of spanning trees in K_m\ G is m^m−n−1(m − |V₁|)^|V²^|−1(m − n)^|V²^|.

7. The eigenvalues of L(G) are given by: 2r + 2, 2r + 1, r + 2 with multiplicity r, r+ 1 with multiplicity 2r − 2, r with multiplicity r, 1 and 0.

Hints and Solutions to Selected Exercises 163 8. The eigenvalues of L(K_n× K₂) are: n + 2 with multiplicity n − 1; n with

multi-plicity n − 1; 2; and 0. (see [2].)

Chapter 11

1. A graph is rank k completable if and only if its bipartite complement has a match-ing of size k.

2. Use the definition of chordal graph. If G is a split graph then so is G^c.

5. Let G be the graph with V (G) = {1, . . . , 7} and with i ∼ j if and only if a_{i j}6= 0. Then G is chordal and a perfect elimination ordering for G is given by 1, 2, 4, 5, 6, 7, 3. Perform Gaussian elimination using pivots according to this ordering. So, first subtract a suitable multiple of a first row from the other rows to reduce all entries in the first column to zero except the (1, 1)-entry. Then subtract a suitable multiple of the first column from the remaining columns to reduce all entries in the first row to zeros, except the (1, 1)-entry. Repeat the process with the second row and column, then with the fourth row and column, and so on. In the process, no zero entry will be changed to a nonzero entry.

6. Use the Jacobi identity.

Chapter 12

2. Use Theorem 12.4 and the fact that the rank of a skew-symmetric matrix is even.

3. The optimal strategy set comprises the vectors inPnthat are in the null space of A.

4. It is sufficient to show that ∑vρ (v) ≤^m(m−1)₂ . Let u be a source. Assume the result for G \ {u} and proceed by induction on the number of vertices.

5. [¹₃,¹₃,¹₃, 0]⁰and ¹₄1⁰are both optimal for Player I.

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在文檔中 Graphs and Matrices (頁 162-182)