The Number of Spanning Trees in a Graph
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1 Course Trees The Ubiquitous Structure in Computer Science and Mathematics, JASS 08 The Number of Spanning Trees in a Graph Konstantin Pieper Fakultät für Mathematik TU München April 28, 2008 Konstantin Pieper: Counting Spanning Trees 1/ 29
2 Preliminaries Definition 1 Let G = (V, E) with V = {1,..., n} and E = {e 1,..., e m } be a directed graph. Then the incidence matrix S G M(n, m) of G is defined as: 1 if e j ends in i (S G ) i,j := 1 if e j starts in i 0 else Konstantin Pieper: Counting Spanning Trees 2/ 29
3 Preliminaries Definition 1 Let G = (V, E) with V = {1,..., n} and E = {e 1,..., e m } be a directed graph. Then the incidence matrix S G M(n, m) of G is defined as: 1 if e j ends in i (S G ) i,j := 1 if e j starts in i 0 else Remark 1 For an undirected graph G every SḠ of some arbitrarily oriented directed variant Ḡ of G can be taken as the incidence matrix. Konstantin Pieper: Counting Spanning Trees 2/ 29
4 Preliminaries Example 2 S G = Konstantin Pieper: Counting Spanning Trees 3/ 29
5 Preliminaries Theorem 3 The rank of the incidence matrix of a graph on n vertices is: Konstantin Pieper: Counting Spanning Trees 4/ 29
6 Preliminaries Theorem 3 The rank of the incidence matrix of a graph on n vertices is: rank(s G ) = n weakly connected components of G Konstantin Pieper: Counting Spanning Trees 4/ 29
7 Preliminaries Theorem 3 The rank of the incidence matrix of a graph on n vertices is: rank(s G ) = n weakly connected components of G Proof. Reorder the edges and vertices so that: S G S G = S G S Gr Konstantin Pieper: Counting Spanning Trees 4/ 29
8 Preliminaries Remark 2 Since (1,..., 1) S G = 0, we can remove an arbitrary row from S G without losing information. Konstantin Pieper: Counting Spanning Trees 5/ 29
9 Preliminaries Remark 2 Since (1,..., 1) S G = 0, we can remove an arbitrary row from S G without losing information. Definition 4 For every A M(n, m) define à M(n 1, m) as A without the n-th row. Konstantin Pieper: Counting Spanning Trees 5/ 29
10 Example 5 S G S T G = Konstantin Pieper: Counting Spanning Trees 6/ 29
11 Example 5 S G S T G = ( ) = ( ) Konstantin Pieper: Counting Spanning Trees 6/ 29
12 Kirchhoff s theorem Theorem 6 (Kirchhoff) The number of spanning trees of a graph G can be calculated as: det(d G ) where D G = S G S T G Konstantin Pieper: Counting Spanning Trees 7/ 29
13 Kirchhoff s theorem Theorem 6 (Kirchhoff) The number of spanning trees of a graph G can be calculated as: det(d G ) where D G = S G S T G Remark 3 deg(i) if i = j (D G ) i,j = 1 if {i, j} E 0 else Konstantin Pieper: Counting Spanning Trees 7/ 29
14 Kirchhoff s theorem Lemma 7 Let T = (V, E) be a directed tree that is rooted at n. We can order E so that e i ends in i. Konstantin Pieper: Counting Spanning Trees 8/ 29
15 Kirchhoff s theorem Lemma 7 Let T = (V, E) be a directed tree that is rooted at n. We can order E so that e i ends in i. Proof. Take e i := (p(i), i), where p(i) is the parent of i. Konstantin Pieper: Counting Spanning Trees 8/ 29
16 Kirchhoff s theorem Lemma 7 Let T = (V, E) be a directed tree that is rooted at n. We can order E so that e i ends in i. Proof. Take e i := (p(i), i), where p(i) is the parent of i. Remark 4 Every undirected tree on V has exactly one undirected variant that is rooted at n. So for constructing/counting spanning trees we only have to consider graphs with i e i. Konstantin Pieper: Counting Spanning Trees 8/ 29
17 Kirchhoff s theorem Lemma 8 Let G = (V, E) with E = n 1 be a directed graph which is not a tree. Then det( S G ) = 0. Konstantin Pieper: Counting Spanning Trees 9/ 29
18 Kirchhoff s theorem Lemma 8 Let G = (V, E) with E = n 1 be a directed graph which is not a tree. Then det( S G ) = 0. Proof. Since E = n 1, G is not weakly connected. So rank( S G ) = rank(s G ) n 2. Konstantin Pieper: Counting Spanning Trees 9/ 29
19 Kirchhoff s theorem Lemma 9 Let T = (V, E) be a tree with e i E ending in i V. Konstantin Pieper: Counting Spanning Trees 10/ 29
20 Kirchhoff s theorem Lemma 9 Let T = (V, E) be a tree with e i E ending in i V. Then det( S T ) = 1. Konstantin Pieper: Counting Spanning Trees 10/ 29
21 Kirchhoff s theorem Lemma 9 Let T = (V, E) be a tree with e i E ending in i V. Then det( S T ) = 1. Proof. Order the vertices (and edges simultaneously e i has to end in i) so that p(i) > i. Then, S T = Konstantin Pieper: Counting Spanning Trees 10/ 29
22 Kirchhoff s theorem Proof of Kirchhoff s theorem. We observe that the i-th column of D G is the sum of incidence vectors that correspond to edges in G that have endpoints in the i-th vertex. Figure: First column of D G D.,1 = = Konstantin Pieper: Counting Spanning Trees 11/ 29
23 Kirchhoff s theorem Proof of Kirchhoff s theorem. We observe that the i-th column of D G is the sum of incidence vectors that correspond to edges in G that have endpoints in the i-th vertex. If we now use the linearity of the determinant in every column we obtain: Figure: Expansion of the determinant = det(d) = =... Konstantin Pieper: Counting Spanning Trees 11/ 29
24 Kirchhoff s theorem Proof of Kirchhoff s theorem. We observe that the i-th column of D G is the sum of incidence vectors that correspond to edges in G that have endpoints in the i-th vertex. If we now use the linearity of the determinant in every column we obtain: det(d G ) = H H det( S H ) where H is the set of all subgraphs of G which correspond to a selection of n 1 edges, with e i ending in i. Konstantin Pieper: Counting Spanning Trees 11/ 29
25 Kirchhoff s theorem Proof of Kirchhoff s theorem. We observe that the i-th column of D G is the sum of incidence vectors that correspond to edges in G that have endpoints in the i-th vertex. If we now use the linearity of the determinant in every column we obtain: det(d G ) = H H det( S H ) where H is the set of all subgraphs of G which correspond to a selection of n 1 edges, with e i ending in i. To prove the theorem it is now sufficient to use the preceding lemmata. Konstantin Pieper: Counting Spanning Trees 11/ 29
26 Extension towards MST Definition 10 For a weighted graph G with edge weights w i,k let S G (x) be the incidence matrix S G with every column corresponding to (i, k) E rescaled by x w i,k. Konstantin Pieper: Counting Spanning Trees 12/ 29
27 Extension towards MST Definition 10 For a weighted graph G with edge weights w i,k let S G (x) be the incidence matrix S G with every column corresponding to (i, k) E rescaled by x w i,k. x w k,i if e j = (k, i) (S G (x)) i,j = x w i,k if e j = (i, k) 0 else Konstantin Pieper: Counting Spanning Trees 12/ 29
28 Extension towards MST Example 11 (Toy graph H) S G (x) = x 2 x 2 x x x x x 0 x 0 0 x x x x x x 3 Konstantin Pieper: Counting Spanning Trees 13/ 29
29 Extension towards MST Theorem 12 (Matrix-Tree Theorem for weighted graphs) The generating function of the number of spanning trees by weight w is the determinant of D G (x) = S G (x) S T G where S G (x) and S G are defined as above. Konstantin Pieper: Counting Spanning Trees 14/ 29
30 Extension towards MST Theorem 12 (Matrix-Tree Theorem for weighted graphs) The generating function of the number of spanning trees by weight w is the determinant of D G (x) = S G (x) S T G where S G (x) and S G are defined as above. In other words: det (D G (x)) = ST by weight w x w w=0 Konstantin Pieper: Counting Spanning Trees 14/ 29
31 Extension towards MST Example 13 D G (x) = x 4 + 2x 2 x 2 0 x 2 x 2 x 2 + 2x x x 0 x 2x x x 2 x x x 3 + x 2 + 2x det (D G (x)) = 2x x 9 + 8x 8 + 6x 7 Konstantin Pieper: Counting Spanning Trees 15/ 29
32 Extension towards MST Remark 5 It is easy to check that we can also write down D G (x) for any given graph G directly: (D G (x)) i,j = {i,k} E x w i,k x w i,j if i = j if {i, j} E 0 else Konstantin Pieper: Counting Spanning Trees 16/ 29
33 Extension towards MST Proof. As above. Here each spanning tree by weight w contributes x w to the determinant of D G (x). Konstantin Pieper: Counting Spanning Trees 17/ 29
34 Computing the number of MSTs Should we try to calculate det(d G (x))? Konstantin Pieper: Counting Spanning Trees 18/ 29
35 Computing the number of MSTs Should we try to calculate det(d G (x))? Consider the following graph on 2n vertices: Konstantin Pieper: Counting Spanning Trees 18/ 29
36 Computing the number of MSTs Should we try to calculate det(d G (x))? Consider the following graph on 2n vertices: det(d G (x)) can have Ω(2 n ) coefficients Konstantin Pieper: Counting Spanning Trees 18/ 29
37 Computing the number of MSTs Compute only the number of minimal spanning trees (the coefficient of the minimum degree monomial). Konstantin Pieper: Counting Spanning Trees 19/ 29
38 Computing the number of MSTs Compute only the number of minimal spanning trees (the coefficient of the minimum degree monomial). If w min is the minimal weight for a spanning tree, clearly x w min must divide det(d G (x)) Konstantin Pieper: Counting Spanning Trees 19/ 29
39 Computing the number of MSTs Compute only the number of minimal spanning trees (the coefficient of the minimum degree monomial). If w min is the minimal weight for a spanning tree, clearly x w min must divide det(d G (x)) Try to factor out the minimum degree monomial of each column and use the linearity of the determinant. det(d G (x)) = x 4 + 2x 2 x 2 0 x 2 x 2 x 2 + 2x x x 0 x 2x x x 2 x x x 3 + x 2 + 2x = x 5 x x 0 x 1 x x 2 + x + 2 Konstantin Pieper: Counting Spanning Trees 19/ 29
40 Computing the number of MSTs Example 14 (Factor det (D G (x))) det(d G (x)) = x 4 + 2x 2 x 2 0 x 2 x 2 x 2 + 2x x x 0 x 2x x x 2 x x x 3 + x 2 + 2x = x 5 x x 0 x 1 x x 2 + x + 2 Konstantin Pieper: Counting Spanning Trees 20/ 29
41 Computing the number of MSTs The entries of the i th column of D G (x) correspond to edges in the cut (i, N (i)): Konstantin Pieper: Counting Spanning Trees 21/ 29
42 Computing the number of MSTs The entries of the i th column of D G (x) correspond to edges in the cut (i, N (i)): Change the cuts? Konstantin Pieper: Counting Spanning Trees 21/ 29
43 Computing the number of MSTs Theorem 15 If we have a minimal spanning tree T of G, we can modify D G (x) (without changing the determinant) so that the product of the minimum degree monomials of each column is w min. Konstantin Pieper: Counting Spanning Trees 22/ 29
44 Computing the number of MSTs Theorem 15 If we have a minimal spanning tree T of G, we can modify D G (x) (without changing the determinant) so that the product of the minimum degree monomials of each column is w min. Algorithm A: Modify D G (x) T := mst(g) D (x) := D G (x) while T {} do i := arbitrary leaf of T with i n p := parent of i add the i-th column in D to the p-th column T := T \ {i} od Konstantin Pieper: Counting Spanning Trees 22/ 29
45 Computing the number of MSTs Example 16 (MST of H and corresponding σ(i)) det(d (x)) = x 4 + 2x 2 x 4 + x 2 0 x 4 x 2 x x 0 0 x 2x 0 x 2 x 2 2x x x 3 = x 7 x x 3 + x 0 x x Konstantin Pieper: Counting Spanning Trees 23/ 29
46 Computing the number of MSTs Lemma 17 The i-th column of D (x) contains the sum of the columns in D G (x) corresponding to σ(i). Konstantin Pieper: Counting Spanning Trees 24/ 29
47 Computing the number of MSTs Lemma 17 The i-th column of D (x) contains the sum of the columns in D G (x) corresponding to σ(i). Writing out the entries of D (x) we get: For i / σ(j) D i,j(x) = x w k,i {i,k} E:k σ(j) Konstantin Pieper: Counting Spanning Trees 24/ 29
48 Computing the number of MSTs Lemma 17 The i-th column of D (x) contains the sum of the columns in D G (x) corresponding to σ(i). Writing out the entries of D (x) we get: For i / σ(j) D i,j(x) = x w k,i {i,k} E:k σ(j) For i σ(j) the above cancels with D i,i D i,j(x) = {i,k} E:k / σ(j) x w k,i Konstantin Pieper: Counting Spanning Trees 24/ 29
49 Computing the number of MSTs Proof. So the j th column of D contains only terms corresponding to edges in the cut (σ(j), V \ σ(j)). Konstantin Pieper: Counting Spanning Trees 25/ 29
50 Computing the number of MSTs Proof. So the j th column of D contains only terms corresponding to edges in the cut (σ(j), V \ σ(j)). The only edge of T in the j th cut is (p(j), j). Konstantin Pieper: Counting Spanning Trees 25/ 29
51 Computing the number of MSTs Proof. So the j th column of D contains only terms corresponding to edges in the cut (σ(j), V \ σ(j)). The only edge of T in the j th cut is (p(j), j). All the other edges from the cut must have higher weight because T is minimal. Konstantin Pieper: Counting Spanning Trees 25/ 29
52 Computing the number of MSTs Proof. So the j th column of D contains only terms corresponding to edges in the cut (σ(j), V \ σ(j)). The only edge of T in the j th cut is (p(j), j). All the other edges from the cut must have higher weight because T is minimal. n 1 i=1 w p(i),i = w min Konstantin Pieper: Counting Spanning Trees 25/ 29
53 Optimizations The runtime of our implementation so far is O (mn + M(n)) where M(n) is the time required to multiply two n nmatrices. O (M(n)) can be thought of as O(n 2+ɛ ). Konstantin Pieper: Counting Spanning Trees 26/ 29
54 Optimizations The runtime of our implementation so far is O (mn + M(n)) where M(n) is the time required to multiply two n nmatrices. O (M(n)) can be thought of as O(n 2+ɛ ). We can further optimize the O(mn) part by only calculating the minimum degree monomials for each entry. Konstantin Pieper: Counting Spanning Trees 26/ 29
55 Optimizations The runtime of our implementation so far is O (mn + M(n)) where M(n) is the time required to multiply two n nmatrices. O (M(n)) can be thought of as O(n 2+ɛ ). We can further optimize the O(mn) part by only calculating the minimum degree monomials for each entry. calculating the negative and positive entries separately. Konstantin Pieper: Counting Spanning Trees 26/ 29
56 Optimizations The runtime of our implementation so far is O (mn + M(n)) where M(n) is the time required to multiply two n nmatrices. O (M(n)) can be thought of as O(n 2+ɛ ). We can further optimize the O(mn) part by only calculating the minimum degree monomials for each entry. calculating the negative and positive entries separately. In the following E is sorted so that p(j) > j. Konstantin Pieper: Counting Spanning Trees 26/ 29
57 Optimizations For i / σ(j) no entries cancel the naive approach works. Konstantin Pieper: Counting Spanning Trees 27/ 29
58 Optimizations For i / σ(j) no entries cancel the naive approach works. Algorithm B1: Negative Entries of D for j = 1 to n 1 do for i / σ(j) do D i,j := min_d(d G (x) i,j + k child of j in T D i,k ) /* min_d computes the minimum degree monomial */ od od Konstantin Pieper: Counting Spanning Trees 27/ 29
59 Optimizations For i / σ(j) no entries cancel the naive approach works. Algorithm B1: Negative Entries of D for j = 1 to n 1 do for i / σ(j) do D i,j := min_d(d G (x) i,j + k child of j in T D i,k ) /* min_d computes the minimum degree monomial */ od od This is O(n 2 ). Konstantin Pieper: Counting Spanning Trees 27/ 29
60 Optimizations For i σ(j) we use the explicit formula D i,j = and run over the rows first. {i,k} E:k / σ(j) x w k,i Konstantin Pieper: Counting Spanning Trees 28/ 29
61 Optimizations For i σ(j) we use the explicit formula D i,j = and run over the rows first. {i,k} E:k / σ(j) x w k,i Algorithm B2: Positive Entries of D for i = 1 to n 1 do L := sort(n (i)) for j = 1 to n 1, i σ(j) do L := L \ σ(j) if L {} k := first_element(l) D i,j := x w i,k s L : w i,k = w i,s fi od od Konstantin Pieper: Counting Spanning Trees 28/ 29
62 Optimizations For i σ(j) we use the explicit formula D i,j = and run over the rows first. {i,k} E:k / σ(j) x w k,i Algorithm B2: Positive Entries of D for i = 1 to n 1 do L := sort(n (i)) for j = 1 to n 1, i σ(j) do L := L \ σ(j) if L {} k := first_element(l) D i,j := x w i,k s L : w i,k = w i,s fi od od This is O(n 2 log n). Konstantin Pieper: Counting Spanning Trees 28/ 29
63 Optimizations Conclusion: We could calculate the number of spanning trees by arbitrary weight. Konstantin Pieper: Counting Spanning Trees 29/ 29
64 Optimizations Conclusion: We could calculate the number of spanning trees by arbitrary weight. We can find the number of minimal spanning trees in O(n 2 + m log n + M(n)) = O(M(n)) Konstantin Pieper: Counting Spanning Trees 29/ 29
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