By Guo L.-T.

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**Sample text**

1(iii) and by induction, we may assume that D has no cut vertices and that n ~ 3. H every directed cycle of D has length two, then since D is minimally directed and by n ~ 3, at least one of the two vertices in a directed cycle of length two is a cut vertex of D. Therefore, D must have a directed cycle C of length m ~ 3. We may assume that D =F C, and son- m ~ 1. Thus D/C has n- m + 1 ~ 2 vertices, and so by induction, D I a has at least two cyclic vertices, 11t and 112 (say). H both 11t, V2 E v (D) - V( C)' then they are both cyclic vertices of D.

Then f(n,e) = k -1. 12) Proof Let A= (ai;) E ~·(n,e). 2, it suffices to show that p(A) ~ k-1, and that p(A) = k -1 if and only if A is similar to the matrix in the theorem. Since A E ~*(n,e), we may assume that A= ( z ~t), where r < k -1 and where all the entries in the first column of A 1 are 1's. Since ~ is symmetric, there is an orthonormal matrix U such that UT ~U is a diagonalized matrix. Let V be the direct sum of U and In-ro and let R be the r by r matrix whose (1, 1) entry is an rand whose other entries are zero.

Hence each A; has equal row sums, so p(A) is the maximum row sum of A. j1 + 8e is an integer, and e =(k 2 ). Then p(A) = k -1, and it follows that there is one nonzero block At=~· This completes the proof. O = Let M (m;J) EM,. and let r; and 8; be the ith row sum and the ith column sum of M, respectively, for each i 1,2, · · · , n. n. 5 (Gregory, Shen and Liu [100]) Let M (m;J) E M,. 14) My=pxo Let x = (x 1 . )T and y = (y1 ° • oy; o. 21, we = E 1s;s,. 14) and Cauchy-Schwarz inequality, Matrices and Graphs for each i 31 = 1,2, · · · ,n, p2x~ (E m;m)2 = ( E = i E ~ m;m)2 J:mwFO E yJ m~; j:m;s¢0 j:m;s¢0 = r1(1- :E yJ).

### 3-restricted connectivity of graphs with given girth by Guo L.-T.

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