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By Korzyuk V. I.

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140. [2+] A spanning tree of a graph G is a subgraph of G which is a tree and which uses every vertex of G. The number of spanning trees of G is denoted c(G) and is called the complexity of G. Thus Problem 126 is equivalent to the statement that c(Kn ) = nn−2 , where Kn is the complete graph on n vertices (one edge between every two distinct vertices). The complete bipartite graph Kmn has vertex set A ∪ B, where #A = m and #B = n, with an edge between every vertex of A and every vertex of B (so mn edges in all).

An of integers such that a1 = 0 and 0 ≤ ai+1 ≤ ai + 1 000 001 010 011 012 165. [1+] sequences a1 , a2 , . . , an−1 of integers such that ai ≤ 1 and all partial sums are nonnegative 0, 0 0, 1 1, −1 1, 0 1, 1 166. [2–] sequences a1 , a2 , . . , an of integers such that ai ≥ −1, all partial sums are nonnegative, and a1 + a2 + · · · + an = 0 0, 0, 0 0, 1, −1 1, 0, −1 39 1, −1, 0 2, −1, −1 167. [2–] Sequences of n − 1 1’s and any number of −1’s such that every partial sum is nonnegative 1, 1 1, 1, −1 1, −1, 1 1, 1, −1, −1 1, −1, 1, −1 168.

For such a simply stated problem, this seems remarkably difficult to prove. 212. [2] Let f (n) be the number of ways to write the permutation n, n − 1, n−2, . . , 1 ∈ Sn as a product of n2 (the minimum possible) adjacent transpositions si = (i, i + 1), 1 ≤ i ≤ n − 1. For instance, f (3) = 2, corresponding to s1 s2 s1 and s2 s1 s2 . Then f (n) is equal to the number of balanced tableaux (as defined in the previous problem) of shape (n − 1, n − 2, . . , 1). 51 Note. ,1) . Any bijective proof of this difficult result would be an impressive achievement.

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A Boundary Value Problem for a Hyperbolic Equation with a Third-Order Wave Operator by Korzyuk V. I.


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