By Martin Rubey

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**Sample text**

C .. C . . 0 . B C B C CG = B an 0 ::: 0 an + d n C; BB BB B@ 0 .. 0 At B C C C C C A where di = dG (i) ai . By the Matrix-Tree-Theorem, t(G) equals the determinant of any principal minor of CG : t(G) = det(CG )rc : For our purposes we demand r > n. We can then transform CG into the Laplacian matrix of G0 by adding ai times the rst row to each row i, i 2 f1; 2; : : : ; ng. The result of these operations is the matrix 0 a1 a2 : : : an BB0 a1 (1 a ) + d1 1;2 : : : 1;n .. BB0 . 1 ; 2 BB .. . .

If, for 0 < i < n, the arc ei is in E , then all arcs incident to vi must be in E , because ei 2 E(T ) is the last arc incident to vi added to E . Therefore, all arcs incident from vi are also in E . This applies in particular to ei+1 . Hence, as v was arbitrary, all arcs of G are in E , that is, E is Eulerian. Q Finally, let T be an arborescence of G and let E be one of the v2V(G) (diG (v) 1)! Eulerian tours produced by the second construction given T . It remains to show that the arborescence produced by the rst construction given E is equal to T .

Hence, before the last letter of the word wr can be removed, all words except wr and all forests except Fr must be empty. The word wr cannot contain letters of Hr , hence it must be empty, too. ) Now we can distinguish between two possible cases: Suppose wr = ry, which is not the root. Then ry belongs { as we required { to the component of Fr that contains the root. But edges from this component can be removed only after the removal of ry. Now suppose that wr contains only the root and there is an edge in a component T0 of Fr which does not contain the root.

### Counting Spanning Trees by Martin Rubey

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