Spanning Trees - BUKU GRAPHS & DIGRAPH PDF

12. Let T be a tree order 3 or more, and letT^′ be the tree obtained fromT by deleting its end-vertices.

(a) Show that diam(T) = diam(T^′) + 2, rad(T) = rad(T^′) + 1 and Cen(T) = Cen(T^′).

(b) Show that a treeT is central or bicentral (see Exercise 11) according to whether diam(T) = 2 rad(T) or diam(T) = 2 rad(T)−1.

13. Let Gbe a connected graph of order nand sizemsuch that V(G)={v₁, v₂, . . .,vn}, wherevi belongs tob(vi) blocks (1≤i≤n).

(a) Show thatPⁿ

i=1b(vi)≤2m. (b) Show thatPⁿ

i=1b(vi) = 2mif and only ifGis a tree.

14. Determine all treesT such thatT is also a tree.

15. Prove that ifT is a tree of order at least 3, thenT contains a cut-vertex v such that every vertex adjacent to v, with at most one exception, is an end-vertex.

16. Let T be a tree of order nwith degree sequenced₁, d₂, . . . , dn such that d₁ ≥ d₂ ≥ · · · ≥ dn. Prove that di ≤ n−1

for each integer i with 1≤i≤n.

17. Let G be a connected graph that is not a tree containing two distinct verticesu and v such thatG−uand G−v are both trees. Show that degu= degv.

18. Show that there exists no tree T containing two distinct edges e₁ and e₂ such that the two components of T−e₁ are isomorphic andthe two components ofT−e₂ are isomorphic.

19. Let T be a tree of ordern. Prove thatT is isomorphic to a subgraph of Cn+2.

20. Prove that if T is a tree of order n ≥ 2 that is not a star, then T is isomorphic to a subgraph of T.

21. Find all those graphsGof order n≥4 such that the subgraph induced by every three vertices ofGis a tree, or show that no such graph exists.

70 CHAPTER 2. TREES AND CONNECTIVITY

Cayley’s Tree Formula

Two labelings of the same graph from the same set of labels are considered distinct labelingsif they produce different edge sets. Figure 2.10 shows three labelings of a graph of order 9 from the set {1,2, . . . ,9}. Since the first two labelings produce the same edge set, these two labelings are considered the same. The third labeling is different from the first two, however, since 26 is an edge there while 26 is not an edge in either of the first two labelings.

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Figure 2.10: Labelings of a graph

There are three labeled trees of order 3 (whose vertices are labeled from the same set of three labels) and there are 16 labeled trees of order 4 (whose vertices are labeled from the set of four labels). These 19 trees are shown in Figure 2.11, where the vertex sets are{1,2,3}and{1,2,3,4}.

As we will see, the number of distinct labeled trees of ordernwhose vertices are labeled from the same set ofnlabels is nⁿ⁻². Thus the number of labeled trees of order 3 is 3¹= 3 and the number of labeled trees of order 4 is 4²= 16, as mentioned above. This formula for the number of labeled trees of order n is due to the famous British mathematician Arthur Cayley, whom we have encountered earlier and will encounter again. While Cayley discovered this result in 1889, the proof we present was given in 1918 by Heinz Pr¨ufer [175], a German mathematician known for his work on abelian groups.

Pr¨ufer’s proof of Cayley’s result (often calledCayley’s Tree Formula) con- sists of establishing a one-to-one correspondence between the trees of ordern with the same vertex set, say{1,2, . . . , n}, and the sequences (calledPr¨ufer codes) of lengthn−2 whose entries come from the set{1,2, . . . , n}. Since the number of such sequences is nⁿ⁻², the proof is complete once the one-to-one correspondence has been verified.

Before giving a proof of Cayley’s Tree Formula, we present an example to illustrate the technique employed. Consider the tree T of order n = 8 in Figure 2.12 whose vertices are labeled with elements of the set {1,2, . . . ,8}. The first term of the Pr¨ufer code of T₁=T is the neighbor of the end-vertex of T₁ having the smallest label. This end-vertex is then deleted, producing a new tree T₂. The second term of the Pr¨ufer code forT is the neighbor of the end-vertex of T₂ having the smallest label inT₂. We continue in this manner

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2 2

1 2 1 1 1

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Figure 2.11: Labeled trees of orders 3 and 4

until we arrive at Tn−2 =K₂. The resulting sequence of length n−2 is the Pr¨ufer code forT, which in this case is (1,8,1,5,2,5).

In this example, every vertex v of T appears in its Prüfer code degv−1 times. This is true in general. Therefore, no end-vertex of T appears in the Prüfer code forT. So ifT is a tree of ordernand sizem, then the number of terms in its Prüfer code is

v∈V(T)

(degv−1) = 2m−n= 2(n−1)−n=n−2.

We now consider the converse question. Suppose thats= (a₁, a₂, . . . , an−2) is a sequence of lengthn−2 whereai∈ {1,2, . . . , n}for eachi(1≤i≤n−2).

We construct a labeled tree T of order n with vertex set {1,2, . . . , n} such that the given sequences is the Pr¨ufer code ofT. For example, suppose that the given sequence is s= (1,8,1,5,2,5). This sequence has length n−2 = 6 and so n = 8. The smallest element of {1,2, . . . ,8} not appearing in this sequence is 3. We join vertex 3 to the first term (vertex) of the sequence, that is, vertex 3 is joined to vertex 1. The first term of the sequence is then deleted, producing the reduced sequence (8,1,5,2,5). Also, the element 3 is removed from {1,2, . . . ,8}. The smallest element of{1,2,4,5,6,7,8}not appearing in (8,1,5,2,5) is 4, which is joined to vertex 8, the first term of the sequence

72 CHAPTER 2. TREES AND CONNECTIVITY

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1 6

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Pr¨ufer code forT : (1,8,1,5,2,5)

T₁=T : T₂: T₃:

T₄: T₅: T₆: T₇:

Figure 2.12: Determining the Pr¨ufer codes for a tree

(8,1,5,2,5). This procedure is continued until only two elements of the set remain. These two vertices (5 and 8 in this case) are joined and a treeT has been constructed whose Pr¨ufer code iss. This is illustrated in Figure 2.13.

In general then, to construct the Pr¨ufer code (a₁, a₂, . . . , an−2) of a treeT = T₁ of ordernwhose vertices are labeled with elements of the set{1,2, . . . , n}, we apply the following algorithm:

Algorithm 2.21 Construct the Pr¨ufer code of a labeled tree with vertex set {1,2, . . . , n}.

Input: An integern≥3 and a treeT =T₁ with vertex set{1,2, . . . , n}. Output: The Pr¨ufer code (a₁, a₂, . . . , an−2) ofT.

1. Fori= 1 ton−2

1.1. Locate the smallest leaf vi ofTi and let ai be the neighbor of vi in Ti.

1.2. LetTi+1=Ti−vi. 2. Output (a₁, a₂, . . . , an−2).

For a fixed setS =S₁ of n positive integers and for a sequences =s₁ of lengthn−2 whose terms belong to S₁, a graphGof order nwithV(G) =S₁ is constructed by following the algorithm below. We may assume that S₁ = {1,2, . . . , n} ands₁= (a₁, a₂, . . . , an−2), whereai∈S₁ for 1≤i≤n−2.

Algorithm 2.22 Given a sequence s₁ of length n−2 whose terms belong to the setS₁={1,2, . . . , n}, construct a tree with vertex setS₁ whose Pr¨ufer code iss₁.

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1 3 6 (1, 8, 1, 5, 2, 5) 1, 2, 3, 4, 5, 6, 7, 8

6 3 6

8 4

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1 1

8 4 5 2 7 (8, 1, 5, 2, 5)

1, 2, 5, 6, 7, 8

(1, 5, 2, 5) (5, 2, 5)

1, 2, 5, 7, 8 1, 2, 4, 5, 6, 7, 8

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1 (2, 5)

2, 5, 7, 8

(5)

6 3

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T :

5, 8

2, 5, 8

Figure 2.13: Constructing a tree with a given Pr¨ufer code

Input: An integern≥3,S₁={1,2, . . . , n} and a sequence s₁= (a₁,a₂,. . ., an−2), where ai∈ {1,2, . . . , n}for 1≤i≤n−2.

Output: A graphGwith vertex set{1,2, . . . , n}. 1. Fori= 1 ton−2,

1.1. Let ki be the smallest element of Si not appearing in si and let ei=kiai.

1.2. LetSi+1 =Si− {ki}.

1.3. Fori≤n−3, let si+1= (aⁱ₊₁, ai+2, . . . , an−2).

2. For the two elementsx, y ∈Sn−1, leten−1=xy.

3. Output E(G) ={e₁, e₂, . . . , en−2}.

Algorithm 2.21 shows that the Pr¨ufer code of a tree with vertex set{1, 2, . . .,n}is a sequence of lengthn−2 whose terms belong to{1,2, . . . , n}, while Algorithm 2.22 constructs a graph of order n (which will be shown to be a

74 CHAPTER 2. TREES AND CONNECTIVITY tree) whose Pr¨ufer code is a given sequence. We now use Pr¨ufer codes to verify Cayley’s Tree Formula [35].

Theorem 2.23 (Cayley’s Tree Formula) For each positive integern, there arenⁿ⁻² distinct labeled trees of order nhaving the same vertex set.

Proof. The result is obvious forn = 1 and n = 2. For n≥ 3, we show by induction that there is a one-to-one correspondence between the set of distinct labeled trees of ordernhaving a fixed vertex setSofnpositive integers and the set of sequences of lengthn−2 whose terms belong toS. By Algorithm 2.21, the Pr¨ufer code of every tree of order n with vertex set S is a sequence of lengthn−2 whose terms belong toS. The proof will be complete once we show that for each such sequence only one tree of ordernwith vertex setS has this sequence as its Pr¨ufer code.

Forn= 3, letSbe a set of three positive integers, sayS ={1,2,3}. There are three trees with vertex setS, which are shown in Figure 2.11. These trees have Pr¨ufer codes (1), (2) and (3), respectively. Thus the result is true for n= 3.

Assume forn ≥4, that for each sequence s₀ of length n−3 whose terms belong to a fixed (n−1)-element setS₀of positive integers that Algorithm 2.22 constructs a unique tree of ordern−1 with vertex setS₀ whose Pr¨ufer code is s₀.

Now let S = {1,2, . . . , n} and let s be a sequence of length n−2 whose terms belong to S. LetG be the graph constructed by Algorithm 2.22. The goal then is to show thatGis the unique tree with Pr¨ufer codes.

Let k be the smallest element of S that does not belong to s. By Al- gorithm 2.22, ka₁ is an edge of the graph G constructed. By the induction hypothesis, there is a unique treeT^′with vertex setS^′ =S− {k}having Pr¨ufer codes^′= (a₂, a₃, . . . , an−2). SinceGis obtained fromT^′by adding a new vertex kto T^′ and joiningkto a₁, the graphGis a tree that is uniquely determined.

Furthermore, the vertex set ofGisS and its Pr¨ufer code iss.

Aspanning treeof a graphGis a spanning subgraph ofGthat is a tree.

A graphGcan only contain a spanning tree ifGis connected. Since the size of every tree of ordernisn−1, it follows that ifGis a connected graph of order nand size m, thenm≥n−1.

Spanning trees of a connected graphGcan be obtained in a variety of ways.

One possibility is to begin with V(G). To construct the edge set E(T) of a spanning tree T of G, we begin by selecting an edge e₁ of G. Next select an edge e₂ ofGthat is distinct from e₁. This is followed by selecting an edgee₃ distinct from those previously selected and that does not produce a cycle with those previously selected. We continue in this manner until an edge en−1 is selected. ThenE(T) ={e₁, e₂, . . . , en−1}.

Another way to produce a spanning treeT of a connected graph G is to begin withG. IfGis a tree, then Gitself is a spanning tree. Otherwise, letf₁ be an edge on a cycle ofG and remove it. Let G₁ = G−f₁. If G₁ is a tree,

thenG₁is a spanning tree ofG. Otherwise, letf₂ be an edge on a cycle ofG₁ and remove it. LetG₂ =G₁−f₂ and continue in this manner until no cycle remains. The spanning subgraphT resulting in this manner is a spanning tree ofG. Hencem−(n−1) =m−n+ 1 edges must be deleted from Gto obtain T. The numberm−n+ 1 is referred to as thecycle rank ofG. Thus a tree has cycle rank 0.

IfGis a connected graph of order n and sizem having cycle rank 1, then m−n+ 1 = 1 and so n =m. Such a graph is therefore a connected graph with exactly one cycle. These graphs are calledunicyclic graphs. All of the graphs in Figure 2.14 are unicyclic.

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Figure 2.14: Unicyclic graphs

Still another way to construct a spanning tree ofGis to begin with a vertex v ofG. Suppose that the eccentricitye(v) ofv iskand let

Ai={u∈V(G) :d(v, u) =i} for 0≤i≤k.

Let T be a spanning subgraph of G such that V(T) = V(G) and for each w∈Ai (1≤i≤k), letwx∈E(T) for exactly one vertexx∈Ai−1. ThenT is a spanning tree with the property thatdT(v, u) =dG(v, u) for eachu∈V(G).

Such a spanning treeT is calleddistance-preservingfromv.

For the graphGof Figure 2.15, the trees T₁ and T₂ are spanning trees of G. The treeT₂ is distance-preserving fromu. While T₁ has three leaves and T₂has five leaves, no spanning tree ofGhas fewer than three or more than five leaves. According to the next result by Seymour Schuster [201] there must be a spanning tree ofGwith four leaves.

Ifuand vare two nonadjacent vertices of a treeT₀, thenT₀+uvcontains exactly one cycle C and the cycle C contains the edge uv. For any edge xy belonging toC, the graphT₁=T₀+uv−xyis again a tree. The treeT₀is said to be transformed intoT₁by anedge exchange. This transformation fromT₀ toT₁ is illustrated in Figure 2.16.

Theorem 2.24 Let G be a connected graph. If G contains a spanning tree with exactlyrleaves and a spanning tree with exactly tleaves wherer < t, then for every integer swith r < s < t, there is a spanning tree ofG with exactlys leaves.

Proof. LetT₀ be a spanning tree ofGwithrleaves and let T be a spanning tree of G with t leaves. Clearly, E(T₀) 6= E(T). Let e be an edge of T not

76 CHAPTER 2. TREES AND CONNECTIVITY

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v w t u

z x

z x u

v w u

x v

y z

y y

T₁

G T₂

Figure 2.15: Two spanning trees T₁ andT₂ in a graph

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.... ............................. ............................ ............................ ..

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........ ............................. ........................... ............................. ..

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.............................

...................... .

x y v

y v x

T₀: T₁:

Figure 2.16: Transforming a treeT₀ into a treeT₁ by an edge exchange

belonging to T₀. Then T₀+e is a unicyclic graph whose cycle C contains an edge f not belonging to T. Thus T₀ is transformed into the spanning tree T₁=T₀+e−f by an edge exchange whereT₁ has one more edge in common withT thanT₀does. IfT₁=T, then the transformation is complete; otherwise, T₁can be transformed into a spanning treeT₂having one more edge in common withT thanT₁does. We continue this until we arrive at T, that is, we have a sequence

T₀, T₁, T₂, . . . , Tk =T

of spanning trees ofGwhere for eachi(1≤i≤k),Ti−1 is transformed intoTi

by an edge exchange and Ti has one more edge in common with T thanTi−1

does. Suppose that the tree Ti (0 ≤ i ≤ k) has ai leaves. Thusa₀ = r and ak =t.

When the treeTj−1is transformed intoTj(1≤j≤k) by an edge exchange, suppose that the edge uv is added to Tj−1 and the edge xy is removed from Tj−1+uv, that is, Tj =Tj−1+uv−xy. (It is possible that one of uandv is the same as one ofxandy.) Depending on the degrees ofuandv inTj−1 and the degrees of xand y in Tj−1+uv, the number aj−aj−1 can be any of the integers−2,−1,0,1 or 2.

If the number s does not appear in the sequence r = a₀, a₁, . . . , ak = t, then it must occur thataj−1 =s−1 and aj =s+ 1 for some integerj with 1≤j ≤k. Suppose thatTj =Tj−1+uv−xy. Sinceaj =aj−1+ 2, it follows that the verticesu, v, xand yare distinct and that

degTj−1u≥2, degTj−1v≥2, degTj−1x= 2 and degTj−1y= 2.

Observe that the unicyclic graphTj−1+uvhasaj−1leaves. Now bothuandx lie on the cycleCinH=Tj−1+uvand degHu≥3 and degHx= 2. Hence on a u−xpath onConH, there are adjacent verticeswandzwith degHw≥3 and degHz= 2. DeletingwzfromH produces a spanning treeT^′ withaj−1+ 1 =s leaves.

The Matrix-Tree Theorem

Cayley’s Tree Formula, which gives the number of labeled trees of a given order, has another interpretation. As a consequence of this formula, there are nⁿ⁻² distinct spanning trees of the labeled graph Kn. This brings up the question of determining the number of distinct spanning trees of labeled graphs in general. An answer to this question has been given as a determinant of a matrix. This result, implicit in the work [133] of Gustav Kirchhoff, is known as theMatrix-Tree Theorem.

Kirchhoff is well known for his research on electrical currents, which he announced in 1845. This led to Kirchhoff’s laws, the first of which states that the sum of the currents into a vertex equals the sum of the currents out of the vertex. Two years later, in 1847, he graduated from the University of K¨onigsberg. It was during that year that he published the paper that led to his theorem on counting spanning trees. Kirchhoff spent much of his life working on experimental physics.

The proof we give of the Matrix-Tree Theorem will employ several results from matrix theory. LetM be anr×s matrix and M^′ an s×r matrix with r ≤ s. The product M ·M^′ is therefore an r×r matrix. Since M ·M^′ is a square matrix, its determinant det(M ·M^′) exists. Anr×rsubmatrix M₀ of M is said to correspond to ther×rsubmatrixM₀^′ ofM^′ if the column numbers ofM determiningM₀are the same as the row numbers ofM^′ determiningM₀^′. For example, suppose thatr= 2 ands= 3 and

M =

1 −2 3

2 0 4

and M^′ =



 2 −1

3 1

0 2



. (2.2)

Then the following 2×2 matrices M₀ and M₀^′ of M and M^′, respectively, correspond to each other:

M₀=

1 −2

2 0

and M₀^′ =

2 −1

3 1

. A result from matrix theory states that

det(M ·M^′) =X

(detM₀)(detM₀^′), (2.3)

78 CHAPTER 2. TREES AND CONNECTIVITY where the sum is taken over all r×r submatricesM₀ of M and whereM₀^′ is ther×rsubmatrix corresponding to M₀. The numbers det(M₀) and det(M₀^′) are referred to asmajor determinantsofM andM^′, respectively.

For the matricesM andM^′ in (2.2), M·M^′=

1 −2 3

2 0 4



 2 −1

3 1

0 2



=

−4 3 4 6

Thus det(M ·M^′) = (−4)·6−4·3 = −36.From the result in matrix theory mentioned above, we also have

1 −2

2 0

2 −1

3 1

1 3 2 4

2 −1

0 2

−2 3 0 4

3 1 0 2

=−36.

Suppose that A is an n×n matrix for some n ≥ 3. Let A^′ be the (n− 1)×(n−1) submatrix of Aobtained by deleting row iand columnj from A, where 1≤ i, j ≤ n. Then (−1)ⁱ⁺^jdet(A^′) is a cofactor of A. For the 3×3 matrixAin Figure 2.17,A^′ is the 2×2 submatrix obtained by deleting row 3 and column 3 ofA, whileA^′′is the 2×2 submatrix obtained by deleting row 1 and column 2 ofA. Then the cofactors ofA^′ andA^′′ are (−1)³⁺³det(A^′) and (−1)¹⁺²det(A^′′).





3 −1 −2

1 2 −3

−4 −1 5





A^′=

3 −1

1 2

(−1)³⁺³det(A^′) = 3(2)−(−1)1 = 7

A^′′=

1 −3

−4 5

(−1)¹⁺²det(A^′′) =−[1(5)−(−3)(−4)] = 7 Figure 2.17: Computing cofactors

Observe that both cofactors of A in Figure 2.17 have the value 7. Also observe that every row sum and column sum of A is 0. It is a theorem of matrix theory that whenever each row sum and column sum of a square matrix M is 0, then all cofactors ofM have the same value. IfM is a square matrix whose rows (or columns) are linearly dependent, then det(M) = 0.

Let G be a graph with V(G) = {v₁, v₂, . . . , vn}. The degree matrix D(G) = [d^ij] is then×nmatrix with

dij =

degvi ifi=j 0 ifi6=j.

Theorem 2.25 (The Matrix-Tree Theorem) IfGis a nontrivial labeled graph with adjacency matrixAand degree matrixD, then the number of distinct spanning trees ofGis the value of any cofactor of the matrixD−A.

Proof. LetV(G) ={v₁, v₂, . . . , vn}. First, observe that the row sum of rowi (or column sum of columni) in A is degvi, so every row sum or column sum ofD−Ais 0. Consequently, the cofactors ofD−Ahave the same value.

Assume first thatGis a disconnected graph. Of course in this case,Ghas no spanning trees. Let G₁ be a component of Gand suppose that V(G₁) = {v₁, v₂, . . . , vr}, where 1≤r < n. Let M be the (n−1)×(n−1) submatrix of D−A obtained by deleting rown and columnnfrom D−A. Since the sum of the first rrows of M is the zero vector of length n−1, the rows ofM are linearly dependent and so det(M) = 0, as desired.

Henceforth, we assume that G is a connected graph of order n and size m where E(G) = {e₁, e₂, . . . , em}. Thus m ≥ n−1. Let C = [c^ij] be an n×m matrix wherecij = 1 orcij =−1 ifvi is incident with ej and such that each column has one entry that is 1 and one entry that is−1, while all other entries in the column are 0. We show that for the transposeC^t ofC, we have C·C^t=D−A. The (i, j)-entry ofC·C^tis

k=1

cikcjk,

which has the value degvi ifi=j, the value−1 ifi6=j and vivj ∈E(G) and the value 0 ifi6=j andvivj ∈/ E(G). Hence, as claimed,C·C^t=D−A.

Consider a spanning subgraphH ofGcontainingn−1 edges ofG. LetC^′ be the (n−1)×(n−1) submatrix ofC determined by the columns associated with the edges ofH and by all rows of Cwith one exception, say rowk.

We now determine the absolute value|det(C^′)|of the determinant ofC^′. If H is disconnected, then H has a component H₁ not containing vk. The sum of the row vectors ofC^′ corresponding to the vertices ofH₁ is the zero vector of length n−1. Hence the row vectors in C^′ are linearly dependent and so

|det(C^′)|= 0.

Next assume thatH is connected. ThusH is a spanning tree ofG. Letu₁ be an end-vertex ofH that is distinct fromvk and letf₁ be the edge ofH that is incident with u₁. In the treeH −u₁, let u₂ be an end-vertex distinct from vk and letf₂ be the edge ofH−u₁that is incident withu₂. This procedure is continued until only the vertexvk remains.

A matrixC^′′= c^′′ij

can now be obtained by a permutation of the rows and columns of C^′ such that|c^′′ij| = 1 if and only if ui and fj are incident. From the manner in which C^′′ is defined, any vertex ui is incident only with edges fj with j ≤ i. This, however, implies that C^′′ is a lower triangular matrix and since |c^′′ii| = 1 for all i, we conclude that |det(C^′′)| = 1. Consequently,

|det(C^′)|=|det(C^′′)|= 1.

Since every cofactor of D−A has the same value, it suffices to evaluate the determinant of the matrix obtained by deleting both row i and column i from D−A for some i(1 ≤i≤n). Let Ci denote the matrix obtained from C by removing rowi. Then the cofactor mentioned above equals det(Ci·Ci^t), which implies by (2.3) that this number is the sum of the products of the

Dalam dokumen BUKU GRAPHS & DIGRAPH PDF (Halaman 82-102)