Counting Spanning Trees by Martin Rubey

By Martin Rubey

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Then t(G) = t(G0 ): 1. REDUCTION PROCEDURES 37 i;j ai aj Unfortunately, we do not have a combinatorial proof for this, so we have to use the Matrix-Tree-Theorem proved in Chapter 5, Section 1 on page 53. It expresses the number of spanning trees of a graph as the determinant of any principal minor of its Laplacian matrix. The Laplacian matrix of G can be written as 0 0 : : : 01 a1 a2 : : : an BB a1 a1 + d1 0 : : : 0 C C .. BB a2 0 a2 + d2 C A . C BB .. C .. C . . 0 . B C B C CG = B an 0 ::: 0 an + d n C; BB BB B@ 0 ..

Codes In 1918, Prufer 34] constructed a correspondence between the trees of the complete graph Kp for p > 1, and words of p 2 letters from a p-element set, showing t(Kp ) = p p 2 . 3. 1 (Prufer 1918). The following two maps de ne a correspondence between labelled forests on p vertices with roots in R (p > jRj) and words of p jRj letters from a p-element set, the last letter being an element of R. ) ) Let FR be a forest on p vertices labelled with numbers from 1 to p with roots in R. Produce the corresponding word as follows: WHILE there is at least one edge in the forest Write down the label of the vertex adjacent to the leaf with the smallest label.

15. The number of spanning trees of the fan Fn can be calculated as follows: t(K2 K1 ; Pn ]) = = Replacing i with n i we get t(Fn ) = = n X i=1 n X i=1 nX1 i=0 nX1 fi (Pn ) n+i 1 : 2i 1 2n 1 i 2n 1 2i 2n 1 i i=0 = f2n 1 ; i where fn is the nth Fibonacci number. 16. The wheel Wn has t(K2 K1 ; Cn ]) = = n X i=1 n X fi(Cn ) n n + i 1 spanning trees. i=1 i 2i 1 CHAPTER 5 Algebraic Proofs This chapter covers the most powerful methods for determining the number of spanning trees. The famous Matrix-Tree-Theorem and the theory of graph spectra enables us to obtain very general theorems.

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