By Papula

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We can answer this in the negative; all others are obtained from nonnegativity constraints by scaling. 11. If a x s a is a facet inducing inequality for P ( G ) having aj < 0 for some j E J, then this inequality must be ajxj s 0. Proof. Since P ( G ) is of full dimension, if we let M be the set of {0,2}matchings x of G satisfying ax = a and let T be the set of tours t of G satisfying at = a then the affine rank of X = M U T must be IEI. Therefore ax = a is the unique hyperplane containing all elements of X.

First we mention two preliminary results. 21). The dimension of IE,,I - I V,,lfor n 2 3. 2. The minimal afine space containing QF is equal to { x E R E n : x ( 6 ( i ) )= 2 for all i E V,,}. The importance of this corollary is that it completely characterizes which inequalities induce the same facet as some prescribed facet inducing inequality for QF. We summarize this as follows. 3. Let ax S a be a valid inequality for QF. (Ai E W : i E V,,)and any p > 0, the inequality ( p a ) x+ Then for any A = Aix(6(i))S pa + 2 A ( V,,) i E V.

Unfortunately, this transformation is not polynomial, and consequently, we shall study the more general optimization problems. Let K be the matrix of all maximal cliques of G = (V,E), where I = n, that is, each line of K is the characteristic vector of a maximal clique of G in (0, lp. Similarly, S is the matrix of all maximal stable sets of G. The four problems paired by duality are as follows: max z = 2 cixi , i subject to Kx < 1 , xiso, max z = 2 cixi , i subject to Sx c 1 , I min w = 2 yi , i subject to yk 3 c, yi”07 yi , min w = i subject to y S 3 c , We solve each of the following two pairs simultaneously: (i) Maximum weighted stable set and minimum covering by cliques, and (ii) Maximum weighted clique and minimum covering by stable sets.

### Mathematik fuer Ingenieure und Naturwissenschaftler by Papula

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