Spanning Trees Exercises 2. Solutions

Transcription

1 Spanning Trees Exercises 2. Solutions Instructor: Mednykh I. A. Sobolev Institute of Mathematics Novosibirsk State University Winter School in Harmonic Functions on Graphs and Combinatorial Designs January, 2014 Mednykh I. A. (Sobolev Institute of Math) Spanning Trees January / 14

2 Exercises Exercise 2.1. Prove that the number of spanning trees for the path graph P n is 1. Exercise 2.2. Prove that the number of spanning trees for the cyclic graph C n is n. Mednykh I. A. (Sobolev Institute of Math) Spanning Trees January / 14

3 Exercise 2.3. Prove the Cayley formula for the number of spanning trees for the complete graph K n : t(k n ) = n n 2. Exercise 2.4. Prove that the number of spanning trees for the complete bipartite graph K n,m is given by the formula t(k n,m ) = m n m 1. Mednykh I. A. (Sobolev Institute of Math) Spanning Trees January / 14

4 Exercise 2.5. Denote by G e the graph obtained by removing edge e from the graph G. Let G\e be the graph obtained from graph G by contracting edge e. In other words, G\e is obtained by deleting edge e and identifying its ends. Prove the following formula t(g) = t(g e) + t(g\e). Exercise 2.6. Denote by G s,e the graph resulting from subdivision of an edge e of a graph G. Then t(g s,e ) = t(g\e) + 2t(G e) = t(g) + t(g e). Mednykh I. A. (Sobolev Institute of Math) Spanning Trees January / 14

5 Exercise 2.7. Find the number of spanning trees for the wheel graph W n = K 1 C n. Answer: If n is odd then t(w n ) = l 2 k, if n is even then t(w n) = 5fk 2, where l j is j-th Lukas number and f k is k-th Fibonacci number. Note: l 1 = 1, l 2 = 3, l k+2 = l k+1 + l k, k 1. f 1 = 1, f 2 = 1, f k+2 = f k+1 + f k, k 1. f 2n = l n f n and l n = f + f n+1. Mednykh I. A. (Sobolev Institute of Math) Spanning Trees January / 14

6 Exercise 2.8. Find the number of spanning trees for the fan graph F n = K 1 P n. Solution: By Exercise 1.7 and Kel mans-chelnokov theorem we obtain t(f n ) = (3 2 cos πk n ) = 2 ( 3 2 k=1 = U ( 3 2 ) = 1 5 ( = 1 5 ( ( ( k=1 cos πk n ). ) n ( 3 5 ) n) 2 ) 2n ( 1 5 ) 2n) = f 2n. 2 Mednykh I. A. (Sobolev Institute of Math) Spanning Trees January / 14

7 Exercise 2.9. Find the number of spanning trees for the lattice graph L m,n = K m K n. Solution: The Laplace spectrums of the graphs K m and K n are µ 0 = 0, µ i = m, i = 1,..., m 1 and λ 0 = 0, λ j = n, j = 1,..., n 1. Then t(k m K n ) = 1 m n 1 m n m 1 λ j i=1 m 1 µ i i=1 m 1 i=0 j=0 i+j>0 (µ i + λ j ) = (λ j + µ i ) = m m 2 n n 2 (m + n) (m 1)(). Mednykh I. A. (Sobolev Institute of Math) Spanning Trees January / 14

8 Exercise Prove that the following result by Boesch and Prodinger t(k m C n ) = n m 2m 1 (T n (1 + m 2 ) 1), where T n (x) = cos(n arccos x) is the Chebyshev polynomial of the first kind. Solution: t(k m C n ) = 1 mn = t(c n )t(k m ) m 1 i=1 m 1 i=0 j=0 i+j>0 (µ i + λ j ) = 1 mn (m cos(2πj/n)) m 1 λ j i=1 m 1 µ i i=1 (λ j + µ i ) = n m m 2 ( [ ] (m cos 2 (πj/n))) m 1 = n m m 2 U 2 ( m+4 m 1. 4 ) Mednykh I. A. (Sobolev Institute of Math) Spanning Trees January / 14

9 From elementary identities sin 2 (u) = 1 cos(2u) 2, cos(2u) = 2 cos 2 (u) 1 and basic definitions of the Chebyshev polynomials one can derive the following relations U 2 (x) = 1 2(1 x 2 ) (1 T 2n(x)) 1 = 2(1 x 2 ) (1 T 1 n(t 2 (x))) = 2(1 x 2 ) (1 T n(2x 2 1)). m+4 Putting x = 4 we get the formula by Boesch and Prodinger t(k m C n ) = n m 2m 1 (T n (1 + m 2 ) 1)m 1. Mednykh I. A. (Sobolev Institute of Math) Spanning Trees January / 14

10 Exercise Prove that the number of spanning trees for the prism P 2 C n is given by the formula t(p 2 C n ) = n 2 ((2 + 3) n + (2 3) n 2). Solution: Since P 2 = K 2, we put m = 2 in the solution of Exercise 2.10 to obtain t(p 2 C n ) = n(t n (2) 1) = n 2 ((2 + 3) n + (2 3) n 2). Mednykh I. A. (Sobolev Institute of Math) Spanning Trees January / 14

11 Exercise Prove that the number of spanning trees for the Moebius ladder graph M n is given by the formula t(m n ) = n 2 ((2 + 3) n + (2 3) n + 2). Solution: Let us note that the Laplacian matrix for M n is circulant circ{v 0..., v 2 }, where v 0 = 3, v 1 = 1, v 2 =... = v = 0, v n = 1, v n+1 =... = v 2n 2 = 0, v 2 = 1. Let ε = e 2πi 2n primitive root of unity. Then L(M n ) has the following spectrum λ k = 2 j=0 be the 2n-th ε k j v j = 3 + ( 1) k+1 2 cos πk, k = 0,..., 2n 1. n Mednykh I. A. (Sobolev Institute of Math) Spanning Trees January / 14

12 We have We note that t(m n ) = 1 2 (3 + ( 1) k+1 2 cos πk 2n n ) k=1 = 1 (4 2 cos 2n (2j 1)π ) n (2 2 cos 2jπ n ). (2 2 cos 2jπ n ) = (4 4 cos 2 jπ n ) = U2 (1) = n 2. Remark: U (1) = sin(n arccos 1) sin(arccos 1) = lim sin(n u) u 0 sin(u) = n. Mednykh I. A. (Sobolev Institute of Math) Spanning Trees January / 14

13 Now we simplify the first product (4 2 cos We get 2 (2j 1)π ) = n (4 2 cos 2jπ / 2n ) (4 2 cos 2jπ n ). (4 2 cos 2jπ n ) = (6 4 cos 2 jπ 3 n ) = U2 ( 2 ). From the properties of Chebyshev polynomias we have U 2 (x) = 1 2(1 x 2 ) (1 T 2n(x)) = 1 2(1 x 2 ) (1 T n(2x 2 1)). In particular, for x = 3/2 we obtain U 2 ( 3 2 ) = T n(2) 1. Simillary, 3 U2( 2 2 ) = T 2n(2) 1. Mednykh I. A. (Sobolev Institute of Math) Spanning Trees January / 14

14 By making use of the identity Equivalently, sin 2n u sin n u = 2 cos n u, we have t(m n ) = 1 2n T2n(2) 1 T n (2) 1 n2 = n(t n (2) + 1). t(m n ) = n 2 ((2 + 3) n + (2 3) n + 2). Mednykh I. A. (Sobolev Institute of Math) Spanning Trees January / 14