D. Marx, Similarly, we can show that there is a value x + y ? t 2 < x + 2y such that ?(S 2 (t 2 )) ? 2c(c + 1) + 1. Denote by T (t 1 , t 2 ) the vertices of M (t 1 , t 2 ) on level less than d 0 We claim that ?(T (t 1 , t 2 )) ? 3. First, T (t 1 , t 2 ) can contain at most 3 vertices on each level 3·2 d?d 0 ? 3y ? t 2 ?t 1 , contradicting the assumption on the ?-values. Every v i,j ? T (t 1 , t 2 ) has a descendant v i ? ,j ? ? T (t 1 , t 2 ) for every i < i ? < d 0 , namely v i ? ,j ? with j ? = j2 i ? ?i . Thus by covering the at most 3 vertices of T (t 1 , t 2 ) on level d 0 ? 1 by at most 3 large edges, we can cover T (t 1 , t 2 ), and ?(T (t 1 , t 2 )) ? 3 follows. Define S := S(t 1 )?S(t 2 )?T (t 1 , t 2 ) Clearly, ?(S) ? 2(2c(c+1)+1)+3 = 4c(c+1)+5. We show that S separates M (t 1 , t 2 ) from the rest of the vertices. Suppose that v i,j , v i ? ,j ? ? S are adjacent vertices such that v i,j ? M (t 1 , t 2 ) and v i ? ,j ? ? M (t 1 , t 2 ) We have i ? d 0 (otherwise v i,j ? T (t 1 , t 2 )), hence v i,j ? A(t 1 ) If ?(v i ? ,j ? ) < t 1 , then v i ? ,j ? ? S(t 1 ) ? S, a contradiction. Moreover, if ?(v i ? ,j ? ) > t 2 , then i ? ? d 0 as v i,j and v i ? ,j ? are not neighbors if ?(v i,j ) < ?(v i ? ,j ? ) and i > i ? . Thus v i ? ,j ? ? A(t 2 ) and v i,j ? S(t 2 ) ? S, a contradiction, By the definition of x and y, we have ?(M (t 1 , t 2 )?W ) ? ?(M (x, x+y)?W ) ? ?(W )/4. To complete the proof that ?(W ? C) ? 3?(W )/4 for every component C of H(d, c) \ S, we show that ?(M (t 1 , t 2 ) ? W ) ? 3?(W )/4: as we have seen that every such component C is fully contained in either M (t 1 , t 2 ) or V \ M (t 1 this means that no component C can have ?(W ? C) > 3?(W )/4. Since x? t 1 < y, the minimality of y implies ?(M (t 1 , x)? W ) ? ?(W )/4. Similarly, it follows from t 2 ?(x+y) < y that ?(M (x+y, t 2 )?W ) ? ?(W )/4. Now ?(M (t 1 , t 2 )? W ) = ?(M (t 1 , x)? W )+ ?(M (x, x+ y)? W )+ ?(M (x+ y, t 2 )? W ) ? 3 4 ?(W )

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DOI : 10.1109/FOCS.2007.27