By Zhang Q.

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**Example text**

Ulv of B such that every row corresponding to an edge of H’ has a 1 in at least one of them. , j ” , Y t = [ 0 otherwise, yo = 0%.. , ys). Then yoB z= Wo, yo 3 0, y o . 1 = v. On the other hand, there are u rows b,,, . . ,b,” of B such that they correspond to edges of H’ and every column has at most one 1 in them. , i,, 0 otherwise, xo = (XI,.. ,X k ) we have Bxo 1, xo 3 0, woxo = V , showing that xo, yo are solution vectors of (12) and (13). ( 5 ) Finally, let w n = (w,, . . , w k )be an arbitrary non-negative integer vector.

Then Au, S 1, u, 2 0 , and wou, = 8(H’). (14) On the other hand, let the edges of H‘ be colored by p = p ( H ’ ) colors. This means that there are p (0,1)-vectors u l , . . ,up such that a l + . . + up = w0 and Ax s 1, x 3 0 implies arx S 1 for any 1 S 1 S p. Hence there is a (0, 1)-vector yl by part (2) of the proof such that yrA 3 a r , y l 3 0 , yl . 1 = 1. +y,, this vector satisfies yA 2 w,,, y 2 0, y . , by (14) the theorem is proved. 0 Appendix. A characterization of perfect graphs Let be the complement of G.

By the minimality of T,-, there is Ep-lE such that E,-, fl T,-, = {x,}. Since is red in c,-~ by ( 2 . p) we have n E, Tp-I,hence E,-, n E, C EP-,n Tn-,= { x p } , and thus E,-, n E, = { x , } . Repeating this procedure with instead of E,, and inductively, we get a sequence E P , ~ P , E p -,,. ~. ,xI P = z, Eo with E, E g,, x, E 7'-, and E, n = {x,-J of pairwise distinct edges and pairwise distinct vertices, hence an odd cycle (xu, Eo,x I , E l , .. ,x,, E,, xo). By the hypothesis of the lemma this cycle is not of maximal degree two there are three edges E,, El, E,, 0 s i < j < k G p, such that E, f l El n E, # @.