Linear Complementarity and Oriented Matroids Linear Complementarity and Oriented Matroids Robin Leroy Laurin Stenz December 29, 2015 Contents 1 LCP Duality Theorem 2 2 Oriented Matroids 4 2.1 Definitions . .4 2.2 Dual . .6 2.3 OMCP Duality Theorem . .7 2.4 Basis . .9 2.5 Tableau . 10 3 Criss-Cross Method 11 3.1 Basis-Form Duality Theorem . 12 3.2 Algorithm . 16 This report relies heavily on the paper Linear Complementarity and Oriented Matroids by Komei Fukuda and Tam´asTerlaky. If not mentioned otherwise, one can assume that the ideas originated in this paper. Robin Leroy, Laurin Stenz 1 Linear Complementarity and Oriented Matroids 1 LCP Duality Theorem 2n+1 2n n In the following, n is fixed, vectors in R are indexed from 0, vectors in R or R are indexed from 1. Recall the linear complementarity problem, namely, given an n by n matrix A and n n b 2 R , to find w and z in R with nonnegative entries satisfying w = Az + b, zT w = 0. This problem can be reformulated as follows. Define the vector space 2n+1 V (A; b) = x 2 R (−b|−Aj1)x = 0 = ker(−b|−Aj1). The above linear complementarity problem is then equivalent to finding x 2 V (A; b) with nonnegative entries satisfying x0 = 1 and xixn+i = 0 for i = 1; : : : ; n. A vector satisfying xixn+i = 0 for i = 1; : : : ; n is called complementary. The equivalence is given by 0 1 1 B C x = @z A . w Indeed, substituting yields 0 1 1 1 B C (−b|−Aj ) @z A = −b − Az + w = 0, w ziwi = 0 for i = 1; : : : ; n, T where second condition is equivalent to z w = 0 since the ziwi are nonnegative. We can generalize the problem further to finding a complementary vector in any given 2n+1 subspace V of R with nonnegative entries and its first entry equal to one. 2n+1 2n We call a vector x 2 R or x 2 R strictly sign preserving (s.s.p.) if xixn+i ≥ 0 for i = 1; : : : ; n xjxn+j > 0 for some j > 0. Robin Leroy, Laurin Stenz 2 Linear Complementarity and Oriented Matroids We call it strictly sign reversing (s.s.r.) if xixn+i ≤ 0 for i = 1; : : : ; n xjxn+j < 0 for some j > 0. We will then prove the following result about this problem. Theorem 1.1 2n+1 Let V be a subspace of R satisfying the following conditions: either V contains no s.s.r. vector x with x0 = 0 (1) or V contains no s.s.p. vector x with x0 6= 0 and ? either V contains no s.s.r. vector y with y0 = 0 (2) ? or V contains no s.s.p. vector y with y0 6= 0: Then exactly one of the following statements holds: (a) There exists a nonnegative complementary vector x 2 V with x0 = 1. ? (b) There exists a nonnegative complementary vector y 2 V with y0 = 1. The above theorem suggests a dual problem to that of finding a nonnegative comple- mentary vector in V with its first entry equal to one, namely finding a nonnegative complementary vector in V ? with its first entry equal to one. Going back to the original LCP where V = V (A; b), the dual problem becomes to find a nonnegative complementary y 2 V (A; b)? with its first entry equal to 1. Since 0 1 −bT ? 1 ? 1 1 T B T C V (A; b) = (ker(−b|−Aj )) = coim(−b|−Aj ) = im(−b|−Aj ) = im @−A A , 1 n this is equivalent to finding an u 2 R such that 0 1 0 1 −bT −bT u B T C B T C @−A A u = @−A uA 1 u has nonnegative entries, first entry 1, and is complementary. This is, in turn, equivalent to u having nonnegative entries, AT u having nonpositive entries, bT u = 1, and uT AT u = 0. Robin Leroy, Laurin Stenz 3 Linear Complementarity and Oriented Matroids We note that when written directly in terms of the initial A and b, the dual of the LCP looks very different from the primal LCP; the more general linear algebraic approach yields a much more symmetrical statement. The actual proof of this result will use an even more general formulation, using combi- natorial objects instead of vector spaces. Before we get to the proof, note that the matrices such that for all b, V (A; b) contains no s.s.r vector whose first entry vanishes and V (A; b)? contains no s.s.p vector whose first entry does not vanish are exactly the sufficient matrices (as shown in the talk from November 17th). 2 Oriented Matroids Remark that the theorem as formulated in the previous section made assumptions and statements only about the signs of entries of vectors. It is possible to come up with objets more general than vectors that express only those signs. 2.1 Definitions Definition Notations Let E be a finite set. Let X; Y 2 f+; −; 0gE be so called sign vectors or just vectors. Let S ⊂ E. • E ground set E • 0 := (0 0 ··· 0)> 2 f0g zero sign vector •− X is defined componentwise by negative 8 >+ if Xe = − <> (−X)e := −Xe := − if Xe = + > :0 if Xe = 0: • X ◦ Y is defined componentwise by composition ( Xe if Xe 6= 0 (X ◦ Y )e = Ye otherwise. • D(X; Y ) := fe 2 E j Xe = −Ye 6= 0g set of separating elements Robin Leroy, Laurin Stenz 4 Linear Complementarity and Oriented Matroids • e 2 D(X; Y ) e separates X and Y • The support of a signed vector X is called support X := fj 2 E j Xj 6= 0g : • Restriction of a signed vector is a subvector, namely restriction omitting S EnS X nS 2 f+; −; 0g such that 8 e 62 S (X nS)e = Xe: Definition Oriented Matroid, Signed Vector An oriented matroid M is a pair (E; F), where E is a finite set and F ⊂ f+; −; 0gE is a set of sign vectors (or just vectors) for which the following axioms are valid. (OM1) 0 2 F (OM2) X 2 F then −X 2 F (symmetry) (OM3) X; Y 2 F then X ◦ Y 2 F (composition) (OM4) X; Y 2 F and f 2 D(X; Y ) (covector elimination) then 9 Z 2 F such that Zf = 0 8 j 2 E n D(X; Y ) holds Zj = (X ◦ Y )j: Intuitively, sign vectors X will correspond to vectors x in a subspace of R whose entries have the corresponding signs. Under that interpretation, −X corresponds to −x, and the composition X ◦ Y corresponds to \x + "y for some " > 0". E Formally, define δ : R −! f+; −; 0g by δ(x)j = sign(xj). δ(x) is called the incidence vector of x. Then, for any vector space V , the axioms (OM1{4) hold for δ(V ). The first two axioms are immediate: 0 = δ(0), and −δ(x) = δ(−x). For (OM3), given x; x0 2 V , let " > 0 be 0 0 0 such that jxij > "jx ij for all i such that xi 6= 0, then δ(x) ◦ δ(x ) = δ(x + "x ). Finally, Robin Leroy, Laurin Stenz 5 Linear Complementarity and Oriented Matroids 0 0 to prove (OM4), note that f 2 D(δ(x); δ(x )) implies xf = −λxf for some λ > 0. Let Z = δ(x + λx0), Z then has the required properties. 2.2 Dual We can now transfer the notion of orthogonal complement to the world of oriented matroids. P Two vectors v and w are orthogonal if i viwi = 0. A necessary (but not sufficient) condition for that is that the nonzero terms of this sum, if there are any, should not all be of the same sign. P Moreover, if i viwi has nonzero terms of both signs, then the sum can be made 0 by rescaling some entries of v by a positive factor, so that there is a w0 orthogonal to v with the same incidence vector as w. This motivates the following definition: two sign vectors X; Y 2 f+; −; 0gE are orthog- onal, denoted X ∗ Y , if either 8g 2 E; XgYg = 0 or 9f; g 2 E; Xf Yf = −XgYg 6= 0. The dual of a matroid (E; F) is (E; F ∗) defined by n o ∗ E F = Y 2 f+; −; 0g 8X 2 F;X ∗ Y . It can seen that this is a matroid, and that F ∗∗ = F. The considerations above show that δ(V )∗ = δ(V ?), so that this does indeed generalize orthogonal complements to oriented matroids. Let E2n = f1;:::; 2ng and E^2n = f0; 1;:::; 2ng. Moreover, let 8 <i + n 1 ≤ i ≤ n ¯i = :i − n n + 1 ≤ i ≤ 2n: More generally, we will use E2n for a set of 2n elements partitioned into n pairs, and E^2n for E2n with an additional element g. For a signed vector X 2 F of an oriented matroid (E^2n; F), X is said to be complemen- tary under the same condition as before, namely if for all i 2 E2n, XiX¯i = 0. Note that 2n+1 x 2 R is complementary if and only if δ(x) is. Similarly, define simply sign preserving and simply sign reversing sign vectors as for 2n+1 vectors in R , and note that these properties are preserved by δ. We now have the formalism needed to state the complementarity problem and the duality 2n+1 theorem using oriented matroids instead of subspaces of R .