CAPITULO II DESARROLLO EXPERIMENTAL
3.2 Caracterización estructural de polímeros heteroarómaticos sintetizados con
3.2.2 Espectroscopía FT-IR de PTh-Cl
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 [142] of Gustav Kirchhoff, is known as the Matrix-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. Let M be an r × s matrix and M0 an s × r matrix with r ≤ s. The product M · M0 is therefore an r × r matrix. Since M · M0 is a square matrix, its determinant det(M · M0) exists. An r × r submatrix M0 of M is said to correspond to the r × r submatrix M00 of M0if the column numbers of M determining M0are the same as the row numbers of M0determining M00.
For example, suppose that r = 2 and s = 3 and
M =
1 −2 3
2 0 4
and M0 =
2 −1
3 1
0 2
. (3.2)
Then the following 2 × 2 matrices M0 and M00 of M and M0, respectively, correspond to each other:
M0=
1 −2
2 0
and M00 =
2 −1
3 1
. A result from matrix theory states that
det(M · M0) =X
(det M0)(det M00), (3.3) where the sum is taken over all r × r submatrices M0 of M and where M00 is the r × r submatrix corresponding to M0. The numbers det(M0) and det(M00) are referred to as major determinants of M and M0, respectively.
For the matrices M and M0 in (3.2),
M · M0=
1 −2 3
2 0 4
2 −1
3 1
0 2
=
−4 3 4 6
.
Thus det(M · M0) = (−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 A0 be the (n − 1) × (n − 1) submatrix of A obtained by deleting row i and column j from A, where 1 ≤ i, j ≤ n. Then (−1)i+jdet(A0) is a cofactor of A. For the 3×3 matrix A in Figure 3.17, A0 is the 2 × 2 submatrix obtained by deleting row 3 and column 3 of A, while A00is the 2 × 2 submatrix obtained by deleting row 1 and column 2 of A. Then these two cofactors of A are (−1)3+3det(A0) and (−1)1+2det(A00).
Observe that both cofactors of A in Figure 3.17 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 of M have the same value. If M 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) = [dij] is the n × n matrix with
dij =
deg vi if i = j 0 if i 6= j.
3.3. SPANNING TREES 79
A =
3 −1 −2
1 2 −3
−4 −1 5
A0=
3 −1
1 2
(−1)3+3det(A0) = 3(2) − (−1)1 = 7
A00=
1 −3
−4 5
(−1)1+2det(A00) = −[1(5) − (−3)(−4)] = 7 Figure 3.17: Computing cofactors
Theorem 3.26 (The Matrix-Tree Theorem) If G is a nontrivial labeled graph with adjacency matrix A and degree matrix D, then the number of distinct spanning trees of G is the value of any cofactor of the matrix D − A.
Proof. Let V (G) = {v1, v2, . . . , vn}. First, observe that the row sum of row i (or column sum of column i) in A is deg vi, so every row sum or column sum of D − A is 0. Consequently, the cofactors of D − A have the same value.
Assume first that G is a disconnected graph. Of course in this case, G has no spanning trees. Let G1 be a component of G and 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 row n and column n from D − A. Since the sum of the first r rows of M is the zero vector of length n − 1, the rows of M 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 = [cij] be an n × m matrix where cij = 1 or cij= −1 if vi 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 transpose Ctof C, we have C · Ct= D − A. The (i, j)-entry of C · Ct is
m
X
k=1
cikcjk,
which has the value deg vi if i = j, the value −1 if i 6= j and vivj∈ E(G) and the value 0 if i 6= j and vivj ∈ E(G). Hence, as claimed, C · C/ t= D − A.
Consider a spanning subgraph H of G containing n − 1 edges of G. Let C0 be the (n − 1) × (n − 1) submatrix of C determined by the columns associated with the edges of H and by all rows of C with one exception, say row k.
We now determine the absolute value | det(C0)| of the determinant of C0. If H is disconnected, then H has a component H1 not containing vk. The sum of the row vectors of C0 corresponding to the vertices of H1 is the zero vector of length n − 1. Hence the row vectors in C0 are linearly dependent and so
| det(C0)| = 0.
Next, assume that H is connected. Thus, H is a spanning tree of G. Let u1
be an end-vertex of H that is distinct from vk and let f1 be the edge of H that is incident with u1. In the tree H − u1, let u2 be an end-vertex distinct from vk and let f2 be the edge of H − u1that is incident with u2. This procedure is continued until only the vertex vk remains.
A matrix C00=c00ij can now be obtained by a permutation of the rows and columns of C0 such that |c00ij| = 1 if and only if ui and fj are incident. From the manner in which C00 is defined, any vertex ui is incident only with edges fj with j ≤ i. This, however, implies that C00 is a lower triangular matrix and since |c00ii| = 1 for all i, we conclude that | det(C00)| = 1. Consequently,
| det(C0)| = | det(C00)| = 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 row i. Then the cofactor mentioned above equals det(Ci· Cit), which implies by (3.3) that this number is the sum of the products of the corresponding major determinants of Ciand Cit. However, these corresponding major determinants have the same value and so their product is 1 if the defining columns correspond to a spanning tree and 0 otherwise.
Cayley’s Tree Formula (Theorem 3.23) is then a corollary of the Matrix-Tree Theorem (see Exercise 58). We illustrate the Matrix-Matrix-Tree Theorem for the graph G of Figure 3.18 where the matrices D and D − A are also shown.
....
. .. . . . . . . .. ....
....
. .. . . . . . . . .. ...
.... . . .. . . . . . .. ....
....
.. .. . . . . .. . ....
...................................................
G :
v1 v2
v4
v3
D =
2 0 0 0
0 3 0 0
0 0 3 0
0 0 0 2
D − A =
2 −1 −1 0
−1 3 −1 −1
−1 −1 3 −1
0 −1 −1 2
Figure 3.18: Illustrating the Matrix-Tree Theorem
To calculate a cofactor of D − A, we delete the entries in row i and column j for some i and j with 1 ≤ i, j ≤ 4 and compute the product of (−1)i+j and the determinant of the resulting submatrix. For example, the cofactor of the (2, 3)-entry in the matrix D − A in Figure 3.18 is
(−1)2+3
2 −1 0
−1 −1 −1
0 −1 2
.
3.4. THE MINIMUM SPANNING TREE PROBLEM 81 Expanding this determinant along the first row, we obtain
−
Consequently, there are eight distinct spanning trees of the labeled graph G of Figure 3.18, all of which are shown in Figure 3.19.
....
Figure 3.19: The distinct spanning trees of a graph
3.4 The Minimum Spanning Tree Problem
Let G be a connected graph each of whose edges is assigned a real number (called the cost or weight of the edge). We denote the weight of an edge e of G by w(e). Such a graph is called a weighted graph. For each subgraph H of G, the weight w(H) of H is defined as the sum of the weights of its edges, that is,
w(H) = X
e∈E(H)
w(e).
We seek a spanning tree of G whose weight is minimum among all spanning trees of G. Such a spanning tree is called a minimum spanning tree. The problem of finding a minimum spanning tree in a connected weighted graph is called the Minimum Spanning Tree Problem.
The importance of the Minimum Spanning Tree Problem is due to its ap-plications in the design of computer, communications and transportation net-works. The history of this problem was researched by Ronald L. Graham and Pavol Hell [107] in 1985. They concluded that the Minimum Spanning Tree Problem was initially formulated by Otakar Bor ˙uvka [35] in 1926 because of his interest in the most economical layout of a power-line network. He also gave the first solution of the problem. Prior to Bor ˙uvka, however, the anthropolo-gist Jan Czekanowski’s work on classification schemes led him to consider ideas closely related to the Minimum Spanning Tree Problem.