 Centre National de la Recherche Scientifique  Institut National Polytechnique de Grenoble  Université Joseph Fourier Laboratoire G-SCOP 46, av Félix.

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 Centre National de la Recherche Scientifique  Institut National Polytechnique de Grenoble  Université Joseph Fourier Laboratoire G-SCOP 46, av Félix Viallet Grenoble Cedex Lovász and Lehman Theorems on Clutters a common generalization Grigor GASPARYAN, Myriam PREISSMANN, András SEBŐ

Some definitions V={1,2,…, n} A set A of subsets of V is a clutter if ∄ A i, A j in A such that A i ⊂ A j We associate to A a matrix A of size m x n : the rows are the characteristic vectors of the elements of A the antiblocking polyhedron P ≤ ( A ) = {x ∈ R n ; A x≤1and x≥0} the antiblocker b ≤ ( A ) = {B; B ⊆ V maximal such that  B  A  ≤1  A  A } : a clutter on V

A particular case G =(V,E) Clutter A (G) ={maximal cliques of G} b ≤ ( A (G) ) = {B; B ⊆ V maximal such that  B  K  ≤1  K  A } = {maximal stable set of G} Theorem (Lovász 1972) : P ≤ ( A (G)) = {x ∈ R n ; Ax≤1and x≥0}= b ≤ ( A ) iff G is perfect. So if G is minimal imperfect then P ≤ ( A (G)) has some non integer vertex, but P ≤ ( A (G’)) any proper induced subgraph G’ of G.

Lovász Theorem and Padberg Corollaries Theorem : Let G=(V,E) be a minimal imperfect graph,  =maximum size of clique  =maximum size of a stable set then G -has n=  +1 vertices, -contains exactly n  -cliques K 1, …, K n and n  -stable sets S 1, …, S n and K i  S j =1 if i≠j and K i  S i =0, -every vertex v of G belongs to exactly   -cliques K i1, K i2, …, K i ,   -stable sets S j1, S j2, …, S j  and S i1, S i2, …, S i  is a partition of V\v K j1, K j2, …, K j  is a partition of V\v

Lovász Theorem and Padberg Corollaries another formulation Theorem : Let A G ={maximal cliques of a minimal imperfect graph G} then - P ≤ ( A G ) is non integer, (1/ , …, 1/  ) is its unique fractional vertex, and P ≤ ( A G’ ) is integer for every proper induced subgraph G’ of G, and there exists - X nxn matrix, rows =char. vectors of elements of A G - Y nxn matrix columns=char. vectors of elements of b ≤ ( A G ) such that X and Y are uniform and XY=YX=J-I J = nxn all one matrix, I= nxn identity matrix, uniform = same number of 1 in each row and column,  =max size of a clique,  = max size of a stable set.

Some other definitions Given a clutter A on V the matrix A of size m x n : rows = the characteristic vectors of the elements of A The blocking polyhedron P ≥ ( A ) = {x ∈ R n ; Ax≥1and x≥0} the blocker b ≥ ( A ) = {B; B ⊆ V maximal such that  B  A  ≥1  A  A } : a clutter Theorem (Edmonds-Fulkerson1970) : b ≥ (b ≥ ( A )). A is said to be ideal if P ≥ ( A ) = b ≥ ( A )

More definitions Let x in R n, i in V, P a polyhedron in R n The projection of x parallel to the ith coordinate is x i = (x 1,…,, x i-1,x i+1,…,x n ) Deletion : P\i={x i ;x ∈ P} Contraction : P/i= {x i ;x ∈ P and x i =0} A is minimally non ideal if P ≥ ( A ) = {x ∈ R n ; Ax≥1} has at least one non-integer vertex but  i  V all vertices of P\i and P/i are integer

Some minimally non-ideal clutters The degenerative projective plane clutter F n (n≥3) : F n = {1, 2, …,n-1}, {1,n}, {2,n}, …{n-1,n}} P ≥ ( F n ) has the fractional vertex (1/n-1, 1/n-1, …, n-2/n-1)

Lehman Theorem Theorem (Lehman 1990) Let A be a minimally non ideal clutter, either A = F n or there exists - nxn matrix X, rows =char. vectors of elements of A - nxn matrix Y columns=char. vectors of elements of B ≥ ( A ) such that X and Y are uniform and XY=YX=J +(  -1)I for some  ≥2

Our theorem Let A ≤ and A ≥ two clutters and P:= P ≤ ( A ≤ )  P ≥ ( A ≥ ) be minimally non integer, then either A ≤ = , A = F n and w=(1/n-1, 1/n-1, …, n-2/n-1) isa unique fractionnal vertex of P Or one or both of the following hold : A ≤ is as in the case of Lovasz theorem and (1/r ≤, 1/r ≤, …, 1/r ≤ ) is a vertex of P A ≥ is as in the case of Lehman theorem and (1/r ≥, 1/r ≥, …, 1/r ≥ )

One key element of the proofs The commutativity Lemma : If X and Y are two nxn (0,1) matrices and XY are such that all non diagonal elements are equal to 1 and the diagonal elements are either all equal to 0 or all >1 then X uniform  Y is uniform too, all diagonal elements are equal and XY=YX

Happy birthday Jack