Clique Facets of the Orthogonal Latin Squares Polytope

Transcription

1 Clique Facets of the Orthogoal Lati Squares Polytope By G Appa 1, D Magos 2, Mourtos 1, JCM Jasse 3 1 Operatioal Research Departmet, Lodo School of Ecoomics ad Political Sciece, Houghto Street, Lodo WC2A 2AE, UK (gappa, jmourtos)@lseacuk 2 Departmet of formatics, Techological Educatioal stitute (TE) of Athes, Ag Spyridoos Str, Egaleo, Athes, Greece dmagos@teiathgr 3 Departmet of Mathematics ad Statistics, Dalhousie Uiversity, Halifax NS, B3H 3J5, Caada jasse@mathstatdalca CDAM Research Report LSE-CDAM November 14, 2001 Abstract Sice 1782, whe Euler addressed the questio of existece of a pair of Orthogoal Lati Squares (OLS) by statig his famous cojecture ([8, 9, 13]), these structures have remaied a active area of research due to their theoretical properties as well as their applicatios i a variety of fields the curret work we cosider the polyhedral aspects of OLS particular we establish the dimesio of the OLS polytope, describe all cliques of the uderlyig itersectio graph ad categorize them ito three classes For two of these classes we show that the related iequalities have Chvátal rak two ad both are facet defiig For each such class, we give a separatio algorithm of the lowest possible complexity, ie liear i the umber of variables Keywords: Orthogoal Lati Squares, Polyhedral Combiatorics, Clique Facets Supported by the TMR project o Discrete Optimizatio (DONET) fuded by the EU uder grat o FMRX-CT

2 1 troductio A Lati square L of order is a square matrix havig 2 etries of differet elemets each occurrig exactly oce i every row ad colum Wlog we ca assume that the differet elemets are the itegers 0, 1,, 1 Two lati squares L 1 = a ij, L 2 = b ij o symbols are called orthogoal if every ordered pair of symbols occurs exactly oce amog 2 pairs (a ij, b ij )i, j = 0, 1,, 1 A example of a pair of orthogoal lati squares (OLS) of order 4 is illustrated i Table Table 1: A OLS cofiguratio of order A alterative defiitio of OLS ca be give with respect to disjoit sets ad trasversals A trasversal of a lati square of order is a set of cells, o two of which are i the same row or the same colum or cotai the same symbol Cosider three disjoit sets, J, K, = J = K = Let be the set of rows, J the set of colums ad K the set of symbols i the cells of a lati square The the lati square has a orthogoal mate if ad oly if it has disjoit trasversals ([8, Theorem 511]) Each trasversal has triplets, each represetig a differet row, colum, ad symbol Therefore the trasversal partitios the uio of the -sets of rows, colums, ad symbols For example, the four trasversals, umbered from 0 to 3, of the first lati square illustrated i Table 1 are: ((0, 0, 0), (1, 2, 3), (2, 3, 1), (3, 1, 2)) 0 ((0, 1, 1), (1, 3, 2), (2, 2, 0), (3, 0, 3)) 1 ((0, 2, 2), (1, 0, 1), (2, 1, 3), (3, 3, 0)) 2 ((0, 3, 3), (1, 1, 0), (2, 0, 2), (3, 2, 1)) 3 Let L, ( L = ) deote the idex-set of the trasversals The orthogoal mate ca be costructed if we set at each cell, defied by a row idex ad a colum idex, the value of the l idex which idicates the trasversal that this cell belogs to This defiitio reveals aother property of OLS: Lati squares L 1, L 2 of order are orthogoal if ad oly if for each symbol the cells that cotai it i L 1 correspod to cells of a trasversal i L 2 ad vice-versa This implies that each pair of symbols (a ij, b ij ) i, j = 0,, 1 appears exactly oce (the orthogoal property) Now cosider all possible triplets with oe elemet from each of the three disjoit sets, J, K ad a weight associated with each triplet The plaar four-idex assigmet problem (4P AP ) refers to idetifyig a miimum weight collectio of 2 triplets, which form disjoit subsets (trasversals) of disjoit triplets Each such subset forms a partitio of the uio of the three -sets This is the weighted OLS problem t is aalogous to the plaar three-idex assigmet problem 2

3 (3P AP ) which asks for a miimum weight collectio of 2 pairs which form disjoit subsets of disjoit pairs The 3P AP is equivalet to the problem of fidig a miimum weight lati square (see [14]), while 4P AP is equivalet to idetifyig a miimum weight pair of OLS OLS have attracted substatial attetio from very early o As oted i [8, p 156], Euler i 1779 had proposed the 36-officers problem which asked for a pair of OLS of order 6 (see also [13]) Havig failed to fid such a cofiguratio ad probably misguided by the oexistece of a pair of OLS for = 2, he cojectured that there exists o pair of OLS of order = 2(mod 4) ([8, 13]), ie odd multiple of two At the begiig of the 20th cetury his cojecture was prove for = 6 ([18]) However, it took sixty more years to prove that his cojecture for > 6 was wrog ([5]) Eve today OLS remai a very active area of research due to their theoretical properties ad their applicatios i diverse fields terms of theoretical iterest, these structures are strogly related to the theory of affie ad projective plaes ad to orthogoal hypercubes ad (t, m, s)-ets Their practical applicatios iclude the problem of multi-variate experimetal desig, the problem of desigig optimal error correctig codes ad that of ecryptio A extesive study of the theory of OLS ad related structures as well as a variety of applicatios ca be foud i [8, 9, 13] spite of the early availability of a 0-1 iteger programmig formulatio (due to D Gale as oted i [7]), OLS have ot bee aalyzed through mathematical programmig The curret work is a step i that directio this paper we focus o the OLS polytope, study its itersectio graph ad obtai facets iduced by clique costraits Sectio 2 we preset the mathematical formulatio of the problem ad discuss related problems The associated itersectio graph ad its cliques are described i Sectio 3 Sectio 4 we prove the dimesio of the uderlyig polytope ad show that two of the three classes of clique iequalities iduce facets of this polytope We also provide proofs that these iequalities are of Chvátal rak 2 A separatio algorithm for each of these two facet-defiig clique classes is give i Sectio 5 Throughout the rest of the paper we will assume > 1 sice for = 1 OLS reduces to a trivial sigle-variable problem 3

4 2 The OLS polytope ad related structures Appa ([1]) gives several differet mathematical programmig formulatios for the OLS problem ad idicates that the oe attributed to D Gale (i [7]) ad reproduced below is the most suitable for theoretical ad computatioal work: {xijkl : i, j J} = 1, k K, l L (21) {xijkl : j J, k K} = 1, i, l L (22) {xijkl : i, k K} = 1, j J, l L (23) {xijkl : k K, l L} = 1, i, j J (24) {xijkl : i, l L} = 1, j J, k K (25) {xijkl : j J, l L} = 1, i, k K (26) x ijkl {0, 1} i, j J, k K, l L (27) where, J, K, L are disjoit sets with = J = K = L = Give real weights c ijkl for every (i, j, k, l, ) J K L the problem of miimizig (maximizig) the fuctio {cijkl x ijkl : i, j J, k K, l L} over the polytope described by costraits (21),, (27) is the 4P AP This formulatio requires 4 biary variables ad 6 2 equality costraits Let A deote the coefficiet matrix of costraits (21),, (26) The we defie the polytope P L as P L = {x R 4 Ax = e, x 0} where e = {1, 1,, 1} T The covex hull of iteger poits of P L is defied as P = cov{x {0, 1} 4 Ax = e} This is the OLS polytope sice every iteger poit x P is a OLS P L is also called the liear relaxatio of P Clearly P P L We will sometimes refer to P as P so as to iclude the cocept of order i the otatio Thus P 6 = is aother way of statig Euler s cojecture for = 6 Substitutig (=) by ( ) i costraits (21),, (26) yields the polytope P = cov{x {0, 1} 4 : Ax e} The polytopes P, P are related sice P P Let D deote a matrix of zeros ad oes The P is a special case of the set partitioig polytope P SP P = {x {0, 1} q : Dx = e} whereas P is a special case of the set packig polytope P SP = {x {0, 1} q : Dx e} (see [2, 15] for details) There are two problems, each ivolvig three disjoit sets, that are highly related to the OLS problem: (3P AP ) ad the axial three-idex assigmet problem (3AAP ) We have referred to the former i the previous sectio The latter is defied with respect to three disjoit sets, amely, J, K, ad a weight coefficiet c ijk for each triplet (i, j, k) J K 3AAP is the problem of fidig disjoit triplets of miimum weight, ie fidig a sigle trasversal of 4

5 miimum weight The costraits of 3AAP are {xijk : i, j J} = 1, k K {xijk : j J, k K} = 1, i {xijk : i, k K} = 1, j J x ijk {0, 1}, i, j J, k K Remark 21 Costraits (21), (22), (23), for a give value of the l idex, are equivalet to the costraits of a 3AAP The above remark will be very useful for establishig the dimesio of P L ad P Research work o 3AAP ad 3P AP polytopes ca be foud i [3, 4, 11, 17] ad [10] 3 The itersectio graph ad its cliques Let R ad C deote the idex sets of rows ad colums respectively of the 0-1 A matrix We refer to a colum of the A matrix as a c for c C The itersectio graph G A (V, E) has a ode c for every colum a c of A ad a edge (c s, c t ) if ad oly if a cs a c t 1, ie, both colums c s ad c t of A have a +1 etry i at least oe commo row Let G A (C, E C ) deote the itersectio graph of OLS, where C = J K L t is coveiet to label the 4 colums of the OLS A matrix, ot from 1 to 4, but with four idices i, j, k ad l ragig from 1 to This leads to a equivalet defiitio of G A (C, E C ) where c s represets the idex set of colum s Defiitio 31 The itersectio graph of OLS G A (C, E C ) has a ode c, for every c C, ad a edge (c s, c t ) for every pair of odes c s, c t C such that c s c t = 2 or 3 Note that a edge (c s, c t ) E C correspods to colums a c s, a ct with a cs a ct = 1 or 3 The row set of the OLS A matrix is defied as R = (K L) ( L) (J L) ( J) (J K) ( K) Sice = J = K = L =, C = 4 ad R = 6 2 Propositio 32 The graph G A (C, E C ) is regular of degree 2(3 1)( 1) Proof Cosider ay c C There are ( 1) 4 elemets of C, which have o idex i commo with c For each of the four idices of c there are ( 1) 3 elemets of C, which share the same value for this idex but have differet values for the other three Therefore, there are 4( 1) 3 elemets of C which have exactly oe idex i commo with c By defiitio 31, c is coected oly to odes that have two or three idices i commo with it, so it is coected to all but ( 1) 4 + 4( 1) 3 odes Therefore the degree of each c C is 4 1 (( 1) 4 + 4( 1) 3 ) = = 2(3 1)( 1) 5

6 Corollary 33 E C = 4 (3 1)( 1) Proof Sice the umber of edges of a graph equals the sum of the degrees of its odes divided by 2, we have E C = (3 1)( 1) = 4 (3 1)( 1) A maximal complete subgraph of a graph G(V, E) is called a clique ([3, 10, 15]) Let Q V deote the ode set of a clique The cardiality of a clique is the cardiality of its ode set Q, deoted Q Cliques of the itersectio graph G A (V, E) defie iequalities of the form {x q : q Q} 1, which are highly relevat to the descriptio of the set packig polytope Next we will examie the cliques of G A (C, E C ) Let a c r deote the etry of the A matrix at row r ad colum c The we defie the set R(r) = {c C : ar c = 1} So R(r) deotes the set of colums with a o-zero etry i row r Propositio 34 For each r R, the ode set R(r) iduces a clique i G A (C, E C ) of cardiality 2 There are 6 2 cliques of this type Proof The subgraph iduced by the ode set R(r) is complete sice all its elemets have two idices i commo To prove that it is also maximal wlog assume that r = (i 1, j 1 ) J ad cosider c 0 = (i 0, j 0, k 0, l 0 ) C \ R(r) where i 0 = i 1 ad j 0 = j 1 Sice R(r) cotais all 2 elemets of C whose first two idices are i 1 ad j 1, it cotais a elemet c 1 = (i 1, j 1, k 1, l 1 ) with c 0 c 1 = 0 Next cosider c 0 = (i 1, j 0, k 0, l 0 ) C \ R(r) But the there exists c 1 C (for example c 1 = (i 1, j 1, k 1, l 1 )) so that c 0 c 1 = 1 The same happes if c 0 = (i 0, j 1, k 0, l 0 ) Therefore there is o c 0 such that the subgraph iduced by R(r) {c 0 } is complete Cosequetly, the subgraph havig R(r) as its ode set is maximal There are as may cliques of this type as the umber of rows of the A matrix, ie 6 2 Propositio 35 For each c C the set Q(c) = {c} {s C : c s = 3} iduces a clique of cardiality 4 3 i G A (C, E C ) There are 4 cliques of this type Proof Wlog cosider c = c 0 = (i 0, j 0, k 0, l 0 ) C ad c 1, c 2 Q(c 0 ) with c 1 c 2 c 0 c 1 Sice c 1, c 2 have three idices i commo with c 0, at least two of their idices coicide Therefore (c 1, c 2 ) E C for ay c 1, c 2 Q(c 0 ), thus Q(c 0 ) is complete To show that the subgraph is also maximal, cosider c 3 = (i 3, j 3, k 3, l 3 ) C \Q(c 0 ) ad (c 0, c 3 ) E C The c 3 has exactly two idices i commo with c 0, by defiitio f c 0 c 3 = 2, wlog cosider i 0 = i 3, j 0 = j 3 ad k 0 k 3, l 0 = l 3 By defiitio, Q(c 0 ) cotais two elemets, amely c s = (i s, j 0, k 0, l 0 ) ad c t = (i 0, j t, k 0, l 0 ) such that i 0 = i s ad j 0 j t But the c 3 c s = c 3 c t = 1, thus the graph with ode set Q(c 0 ) {c 3 } is ot complete So Q(c 0 ) is maximal The set Q(c 0 ) icludes ode c 0 = (i 0, j 0, k 0, l 0 ) ad all odes with exactly oe idex differet from c 0 So Q(c 0 ) = 4( 1) + 1 = 4 3 6

7 There are 4 elemets belogig to the set C, each of which ca play the role of c 0 Therefore, there are 4 cliques of this type Propositio 36 Let c, s C such that c s = 1 The the set Q(c, s) = {c} {t C : c t = 2, s t = 3} iduces a 4-clique i G A (C, E C ) Proof Wlog let c = c 0 = (i 0, j 0, k 0, l 0 ) ad s = (i 0, j 1, k 1, l 1 ) We ca uiquely defie three elemets t 1 = (i 0, j 0, k 1, l 1 ), t 2 = (i 0, j 1, k 0, l 1 ), t 3 = (i 0, j 1, k 1, l 0 ), satisfyig c t i = 2 ad s t i = 3 for i = 1, 2, 3 t is obvious that the ode set {c, t 1, t 2, t 3 } iduces a complete subgraph of G A (C, E C ) To show that it is also maximal, cosider c 2 = {i 2, j 2, k 2, l 2 } C \ Q(c, s) f i 2 i 0 the for a edge (c, c 2 ) to exist i G A (C, E C ) we must have c c 2 2, which implies that c 2 t i 1 for i = 1, 2, 3 Therefore Q(c, s) caot be exteded to iclude c 2, sice the resultig graph is ot complete f i 2 = i 0 either c 2 has aother elemet commo with c ad the remaiig two with s, i which case it coicides with oe of the t i s, or it has three elemets i commo with c ad oe with s the latter case, wlog let j 2 = j 0 ad k 2 = k 1 The we have c 2 t 2 = 1 Hece, i this case as well Q(c, s) caot be exteded Therefore the subgraph iduced by Q(c, s) is complete ad maximal Cocerig the cardiality of the set of cliques of this type, every ordered pair (c, s) such that c s = 1 ca be used to create a clique of this type Cosiderig that C = 4 ad that for each c C there are 4( 1) 3 possible s such that c s = 1, the umber of such ordered pairs is 4 4 ( 1) 3 Note, however, that the 4-clique Q(c, s) = (c, t 1, t 2, t 3 ) is also geerated as Q(c i, s i ) for i = 1, 2, 3 where c 1 = t 1 = (i 0, j 0, k 1, l 1 ) ad s 1 = (i 0, j 1, k 0, l 0 ), c 2 = t 2 = (i 0, j 1, k 0, l 1 ) ad s 2 = (i 0, j 0, k 1, l 0 ), c 3 = t 3 = (i 0, j 1, k 1, l 0 ) ad s 3 = (i 0, j 0, k 0, l 1 ) Propositio 37 Q(c, s) = Q(c i, s i ), i = 1, 2, 3 t is also obvious that the 4-clique Q(c, s) = (c, t 1, t 2, t 3 ) caot arise from ay other choice of c ad s Corollary 38 The umber of distict 4-cliques is 4 ( 1) 3 Proof Each 4-clique arises from four differet ordered pairs of C ad there exist 4 4 ( 1) 3 such pairs Cliques described i Propositios 34, 35 ad 36 will be called cliques of type, ad respectively Next we will show that these are the oly types of cliques i G A (C, E C ) 7

8 Theorem 39 The cliques of type, ad are the oly cliques i G A (C, E C ) Proof Let Q be the ode set of a clique i G A (C, E C ) Let c = (i 0, j 0, k 0, l 0 ) Q Every other q Q must have at least two idices i commo with c So there has to be a q s Q such that c q s = 2 Wlog let q s = (i 0, j 0, k 1, l 1 ) f every other elemet of Q has the same values i 0, j 0 for the first two idices the Q is a ode set of a clique of type f ot, the there exists a q t = (i t, j t, k t, l t ) Q which must satisfy the followig relatioships: (i) Either i t = i 0 or j t = j 0 f both i t = i 0 ad j t j 0 the k t = k 0 ad l t = l 0 i order for q t to be coected to c But the q s q t = 0, which meas that Q does ot iduce a clique (ii) Either k t = k 0 ad l t = l 1 or k t = k 1 ad l t = l 0 the, together with (i), c q t = 1 while if k t = k 0 ad l t = l 0 the q s q t = 1 both cases Q does ot iduce a clique Wlog assume that q t = (i 0, j 1, k 0, l 1 ) f (i 0, j 1, k 1, l 0 ) Q the Q Q(c, (i 0, j 1, k 1, l 1 )), i which case Q is a ode set of a clique of type f (i 0, j 1, k 1, l 0 ) / Q the there is a q r Q such that q r (i 0, j 1, k 1, l 0 ) 1, q r c 2, q r q s 2 ad q r q t 2 The oe ca check by eumeratio of case that every such q r must have at least three idices i commo with (i 0, j 0, k 0, l 1 ), i which case Q Q(i 0, j 0, k 0, l 1 ) ie Q iduces a clique of type Corollary 310 The total umber of cliques i G A (C, E C ) is 4 (( 1) 3 + 1) Proof As show above, there are 6 2, 4, ad 4 ( 1) 3 cliques of type, ad respectively 4 Facets iduced by clique iequalities We briefly summarize some basic cocepts ad defiitios of polyhedral theory (for a short but succict presetatio of this theory see [12, 16]) A polyhedro is the itersectio of a fiite set of half spaces A polytope is a bouded polyhedro A polytope P is of dimesio, deoted as dim(p ) =, if it cotais + 1 affiely idepedet poits By covetio, if P = the dim(p ) = 1 f P = {x R : B = x = b =, B x b } the dim(p ) = rak(b = ) Here B = ad B deote the matrix of co-efficiets of equality ad less tha or equal to type iequality costraits respectively, while b = ad b deote the correspodig right-had side vectors for the liear system defiig P A iequality ax a 0 is called valid for P if it is satisfied by all x P t is called supportig if it is valid ad there exist some ˆx P satisfyig aˆx = a 0 The set of poits which satisfy ax a 0 as equality (F = {x P : ax = a 0 }) is called a face of P A face F of a polytope P is said to be improper if ax = a 0 for all x P A proper, o-empty face F of P is called a facet if dim(f ) = dim(p ) 1 Facets are importat sice they provide a miimal iequality represetatio of a polyhedro Our 8

9 mai iterest here is i the facets of the covex hull P of iteger poits i P L defied i Sectio 2 Coditios (c) ad (d) of the followig theorem usually provide the two basic tools for provig that a give iequality ax a 0 iduces a facet Theorem 41 (see [16, Theorem 316], [12, Theorem1]) Let P R be a polyhedro ad assume that B is a real valued m matrix ad b R m such that P = {x R : B = x = b =, B x b } where B = (B =, B ) T ad b = (b =, b ) T Let F be a o-empty face of P, the the followig statemets are equivalet: (a) F is a facet of P (b) F is a maximal proper face of P (c) dim(f ) = dim(p ) 1 (d) There exists a iequality ax a 0 valid with respect to P with the followig three properties: (i) F = {x P : ax = a 0 } (ii) There exists x P with a x < a 0, ie the iequality is proper (iii) f ay other iequality dx d 0, valid with respect to P satisfies F = {x P : dx = d 0 }, the (d, d 0 ) ca be expressed as a liear (affie) combiatio of (B =, b = ) ad (a, a 0 ) this paper, we will use (d) to prove that cliques of type ad iduce facets of P The dimesio of P will also be established through the same approach The same techique has bee used for provig facet-defiig iequalities ad the dimesio of the 3AAP, ad 3P AP polytopes i [3] ad [10] respectively First, we discuss some properties of P [15] it is show that for the geeral packig polytope (P SP ) the cliques of the uderlyig itersectio graph iduce facet-defiig iequalities However, o similar result has bee prove for P SP P Sice the OLS packig polytope P is a special case of P SP, the iequalities {x q : q Q} 1, where Q is the ode set of a clique of G A (C, E C ), defie facets of the polytope P Other properties of P arisig from its relatio to P SP are: (i) P is full-dimesioal, ie dim( P ) = 4 The idepedet poits of P are the zero vector ad all the 4 uit vectors (ii) P is dow mootoe, ie x P y P for all y such that 0 y x (iii) The o-egativity costraits x ijkl 0 defie facets of P Although we kow quite a few thigs about the facial structure of P, the same caot be said with respect to P Sice P is a face of P we kow that dim(p ) dim( P ) However, the structure of P presets irregularities that do ot appear i P For example, we kow that P = for = 2 ad = 6 P 2 = ca be easily verified sice there are oly two lati 9

10 squares for = 2 As stated previously, P 6 = was prove i [18] Fortuately, P also [13, Theorem 29]) Before establishig the dimesio P we prove the dimesio of P L for 2, 6 as show i [6] (see Theorem 42 The rak of the system Ax = e is Proof Order the 4 colums of the A matrix, deoted by x ijkl, so that idices k, j, i ad l vary i that order For = 2, the order of the colum idices is: (1, 1, 1, 1), (1, 1, 2, 1), (1, 2, 1, 1), (1, 2, 2, 1), (2, 1, 1, 1), (2, 1, 2, 1), (2, 2, 1, 1), (2, 2, 2, 1), (1, 1, 1, 2), (1, 1, 2, 2), (1, 2, 1, 2), (1, 2, 2, 2), (2, 1, 1, 2), (2, 1, 2, 2), (2, 2, 1, 2), (2, 2, 2, 2) As to the 6 2 rows, we divide them ito six sets of 2 rows each, as defied by equalities (21)(26) Figure 1 illustrates the matrix for = 2 (each costrait set is separated from the ext by a horizotal lie) Figure 1: OLS A matrix for = To fid the rak of the A matrix, we follow five steps, the last four of which idetify a 3AAP substructure, exactly as at Remark 21 Step : t is obvious that the sum of all the rows of each set is the same, ie {x ijkl : i, j J, k K, l L} = 2 Therefore, ay oe costrait ca be removed from ay of the six sets as beig liearly depedet We choose to keep the row set (21) itact ad remove the first row of all the remaiig sets Table 2 shows the outcome 10

11 Table 2: Liearly depedet rows removed at Step Row set Rows removed Rows removed for =2 (21) (22) (23) (24) (25) (26) Step : Cosider row sets (21), (22) ad (23) Observe that, as oted at Remark 21, they form idepedet 3AAP problems, oe for each value of the idex l For l = l 0, the correspodig 3AAP ivolves the 3 variables x ijkl0, for i, j, k = 1,,, ad the 3 rows (l 0 1) + t, 2 + (l 0 1) + t, (l 0 1) + t, for t = 1,, Balas ad Saltzma show i [3] that the rak of a 3 3 3AAP matrix is 3 2 So, we ca remove up to 2 rows from each of the 3AAP problems Note that, havig removed rows 2 + 1, at Step, for l 0 = 1, the correspodig 3AAP icludes o liearly depedet rows For the remaiig 1 idepedet 3AAP problems, we ca remove the two liearly depedet rows We choose to remove rows umbered 2 +(t 1) +1, 2 2 +(t 1) +1, for t = 2,,, a total of 2( 1) rows Table 3 gives a complete list of rows removed so far Table 3: Liearly depedet rows removed at Steps & Row set Rows removed Rows removed for =2 (21) (22) { 2 + (t 1) + 1, t = 1} 5, 7 (23) {2 2 + (t 1) + 1, t = 1} 9, 11 (24) (25) (26) Step : Cosider row sets (21), (25) ad (26) Observe that they form idepedet 3AAP problems, oe for each value of the idex k For k = k 0, the correspodig 3AAP ivolves variables x ijk0 l, for i, j, l = 1, ad rows k 0 + (t 1), k 0 + (t 1), k 0 + (t 1), for t = 1,, Agai, for k 0 = 1, the correspodig 3AAP icludes o liearly depedet rows, sice rows , have already bee removed All the rows of the remaiig 1 idepedet 3AAP problems are preset We choose to remove two liearly depedet rows from each problem, amely rows t, t, for t = 2,,,ie 2( 1) rows i total Table 4 gives a complete list of rows removed so far Step V: Cosider row sets (23), (24) ad (25) Observe that they form idepedet 3AAP problems, oe for each value of the idex j For j = j 0, the correspodig 3AAP ivolves variables x ij0kl, for i, k, l = 1,, ad rows j 0 + (t 1), j 0 + (t 1), (j 0 1) + t, for t = 1,, For j 0 = 1, the correspodig 3AAP icludes o liearly depedet rows, sice rows , t, t, for t = 1,,, have already bee removed The 11

12 Table 4: Liearly depedet rows removed at Steps - Row set Rows removed Rows removed for =2 (21) (22) { 2 + (t 1) + 1, t = 1} 5, 7 (23) {2 2 + (t 1) + 1, t = 1} 9, 11 (24) (25) , {4 2 + t, t = 1} 17, 18 (26) , {5 2 + t, t = 1} 21, 22 remaiig 1 idepedet 3AAP problems have bee left itact We choose to remove two liearly depedet rows from each problem, amely rows t, (t 1), for t = 2, ie 2( 1) rows Table 5 gives a complete list of rows removed so far Table 5: Liearly depedet rows removed at Steps - V Row set Rows removed Rows removed for =2 (21) (22) { 2 + (t 1) + 1, t = 1} 5, 7 (23) {2 2 + (t 1) + 1, t = 1} 9, 11 (24) , {3 2 + t, t = 2} 13, 14 (25) {4 2 + (t 1) + 1, t = 1}, {4 2 + t, t = 2} 17, 18, 19 (26) , {5 2 + t, t = 2} 21, 22 Step V: Cosider row sets (22), (24) ad (26) Observe that they form idepedet 3AAP problems oe for each value of the idex i For i = i 0, the correspodig 3AAP ivolves variables x i0jkl, for j, k, l = 1,,, ad rows 2 +i 0 +(t 1), 3 2 +(i 0 1) +t, 5 2 +(i 0 1) +t, for t = 1,, Agai for i 0 = 1, the correspodig 3AAP icludes o liearly depedet rows, sice rows 2 + (t 1) + 1, t, for t = 1,, ad have already bee removed All the rows of the remaiig 1 idepedet 3AAP problems are preset We choose to remove two liearly depedet rows from each problem, amely rows (t 1), (t 1), for t = 2,, ie 2( 1) rows Table 6 gives a complete list of rows removed so far Table 6: Liearly depedet rows removed at Steps - V Row set Rows removed Rows removed for =2 (21) (22) { 2 + (t 1) + 1, t = 1} 5, 7 (23) {2 2 + (t 1) + 1, t = 1} 9, 11 (24) {3 2 + (t 1) + 1, t = 1}, {3 2 + t, t = 2} 13, 14, 15 (25) {4 2 + (t 1) + 1, t = 1}, {4 2 + t, t = 2} 17, 18, 19 (26) {5 2 + (t 1) + 1, t = 1}, {5 2 + t, t = 2} 21, 22, 23 total, 4 2 ( 1) + 5 = 8 3 rows have bee removed Therefore, is a upper boud o the rak of A We will complete the proof by exhibitig affiely idepedet colums Cosider the colums: 12

13 (1, 1, 1, 1),, (1, 1,, 1),, (1, 1, 1, ),, (1, 1,, ) ( 2 colums) (2, 1, 1, 1),, (, 1, 1, 1),, (2, 1, 1, ),, (, 1, 1, ) (( 1) colums) (1, 2, 1, 1),, (1,, 1, 1),, (1, 2, 1, ),, (1,, 1, ) (( 1) colums) (2, 2, 1, 1),, (,, 1, 1),, (2, 2, 1, 1),, (,, 1, 1) (( 1) 2 colums) (1, 2, 2, 1),, (1,,, 1),, (1, 2, 2, 1),, (1,,, 1) (( 1) 2 colums) (2, 1, 2, 1),, (, 1,, 1),, (2, 1, 2, 1),, (, 1,, 1) (( 1) 2 colums) The matrix formed by these colums ad the remaiig rows of A is upper triagular, with each diagoal elemet equal to oe Corollary 43 dim(p L ) = We ow describe the tools eeded to obtai the dimesio ad the clique facets of P Uless otherwise stated, we will illustrate a pair of OLS as poits of P expressed i terms of four sets of idices, viz for the row set, J for the colum set, ad K ad L for the set of elemets of the first ad the secod lati square respectively The elemets of all four sets are the itegers from 1 to We will further use k(i, j) (respectively l(i, j)) to deote the value of the cell i row i, colum j of the first (secod) lati square Thus k(i, j) K ad l(i, j) L The followig remark reveals a very useful property of the poits of P correspodig to a pair of OLS Remark 44 Give a OLS structure ad m 1, m 2 M, where M ca be ay oe of the disjoit sets, J, K, L the (iter)chagig all m 1 values to m 2 ad all m 2 values to m 1 yields aother OLS structure These two structures are called equivalet ([8, p 168]) f the iterchage is carried out for elemets of set (ie M = ) we will call it a first idex iterchage, for elemets of set J a secod idex iterchage, etc To facilitate a study of iterchages, we defie the iterchage operator ( ) Thus, by settig x = x(i 1 i 2 ) 1 we imply that at poit x P we apply a first idex iterchage betwee rows i 1, i 2 derivig poit x P Note that brackets must also have a idex for deotig the set of the idices that are iterchaged Notatio without this subscript i cases like (1 ) becomes ambiguous t is easy to see that x(m 1 m 2 ) t = x(m 2 m 1 ) t ad x = x(m 1 m 2 ) t (m 1 m 2 ) t A series of iterchages at a poit x P is expressed by usig the operator ( ) as may times as the umber of iterchages with priority from left to right For example, x = x(i 1 i 2 ) 1 (1 ) 3 is take to mea that at poit x we apply a first idex iterchage betwee i 1, i 2 ad the at the derived poit we apply a third idex iterchage betwee 1, K, thus yieldig poit x We additioally defie a coditioal iterchage as the iterchage to be performed oly whe a certai coditio is met The coditio refers to a logical expressio cosistig of values of a idex set, at a give poit x f this expressio evaluates true the the iterchage will be applied to poit x Sice we are goig to use oly coditioal iterchages for which both the logical expressio ad the iterchage refer to elemets of the same set we will use a commo subscript 13

14 for both Thus to deote this type of iterchage at poit x we will use the expressio x(coditio?iterchage) t, where the subscript refers to the idex set For example, x 2 = x 1 (i 1 =?1 i 1 ) 1 implies that if i 1 = at poit x 1 the we derive poit x 2 by applyig at poit x 1 the first idex iterchage betwee 1, i 1 f i 1 = the x 2 = x 1 As i the previous case we ca have more tha oe coditioal iterchage i the same expressio Agai priority is cosidered from left to right As we will see shortly, the iterchages will be used extesively for provig the dimesio of P ad facet defiig iequalities A additioal complicatio for provig the dimesio of P comes from the fact that it is ot easy to exhibit a pair of OLS for every value of, ie for every it is difficult to demostrate a 0-1 vector feasible wrt costraits (21),,(26) that will have specific variables set to oe cotrast, both for 3AAP ad 3P AP such trivial poits exist For the 3AAP the trivial poit has x ijk = 1 for i = j = k ad x ijk = 0 for i j k i For the 3P AP the trivial solutio is defied wrt k = i + j 1 mod, ie the trivial poit has x ij k = 1 for k 0 > 0, x ij = 1 for k 0 = 0 ad all other variables set to zero To overcome this difficulty, i the followig lemma we establish, for 4 ad = 6, the existece of a OLS structure with four specific variables set to oe Lemma 45 For 4 ad 6 let i 0 \ {1}, j 0 J \ {1}, k 0, k 1, k 2 K \ {1} with k 2 = k 0, k 1, l 0, l 1 L \ {1} The there exists a poit x 0 P with four particular variables takig value oe as illustrated i Table 7 1 j k 1 i 0 k 0 k 2 Table 7: Poit x 0 1 j l 1 i 0 l 0 1 Proof Cosider a arbitrary poit x P as illustrated i Table 8 Table 8: A arbitrary poit x P (Lemma 45) 1 j 0 1 j 0 1 k b k c i 0 k d k e 1 l b l c i 0 l d l e where k b, k c, k d, k e K, l b, l c, l d, l e L For x to be a valid OLS structure we must have k b = k c, k d ; k e k c, k d ; l b l c, l d ; l e l c, l d We cosider two cases: case 1: k b k e 14

15 Let x = x(k b 1?1 k b ) 3 (l b 1?1 l b ) 4 f l e = 1 the we are doe, ie x is x 0 if we deote k d as k 0, k c as k 1, l c as l 1 ad l d as l 0 f l e 1 the let i 1 (i 1 1) be the row for which l(i 1, j 0 ) = 1 at poit x By labelig k(i 1, 1) as k 0, k(i 1, j 0 ) as k 2, k c as k 1, l(i 1, 1) as l 0 ad l c as l 1 the x 0 = x (i 0 i 1 ) 1 case 2: k b = k e this case l b l e (orthogoal property) Agai we set x = x(k b 1?1 k b ) 3 (l b 1?1 l b ) 4 At poit x if we deote k d as k 0, k c as k 1, l d as l 0, l c as l 1 ad l e as l 2 ad exchage the roles of sets K ad L we have poit x 0 There might be more tha oe poits of P with these four variables set to oe However, oe poit suffices to carry out the proofs that follow Sice P P L, dim(p ) dim(p L ) Moreover, dim(p ) < dim(p L ) if ad oly if there exists a equatio ax = a 0 satisfied by all x P such that it is ot implied by (ie caot be expressed as a liear combiatio of) the equatios Ax = e We ow show that o such equatio exists Theorem 46 Let 4 ad 6, ad suppose every x P satisfies ax = a 0 for some a R 4, a 0 R The there exist scalars λkl 1, λ2 il, λ3 jl, λ4 ij, λ5 jk, λ6 ik, i, j J, k K, l L, satisfyig a ijkl = λ 1 kl + λ 2 il + λ 3 jl + λ 4 ij + λ 5 jk + λ 6 ik (41) a 0 = {λ 1 kl : k K, l L} + {λ 2 il : i, l L} + {λ 3 jl : j J, l L} + {λ 4 ij : i, j J} (42) + {λ 5 jk : j J, k K} + {λ 6 ik : i, k K} Proof Defie 15

16 λ 1 kl = a 11kl λ 2 il = a i11l a 111l λ 3 jl = a 1j1l a 111l λ 4 ij = a ij11 a i111 a 1j11 + a 1111 λ 5 jk = a 1jk1 a 1j11 a 11k1 + a 1111 λ 6 ik = a i1k1 a i111 a 11k1 + a 1111 Note that these 6 2 scalars are defied i such a way so that exactly 8 3 of them, correspodig to the depedet rows of the A matrix, equal zero By substitutig the λs i equatio (41) we get: a ijkl = a 11kl + a i11l + a 1j1l + a ij11 + a 1jk1 + a i1k1 2a i111 2a 1j11 2a 11k1 2a 111l + 3a 1111 (43) Substitutio aloe is eough to show that (43) is true for a 1111 ad for all cases where at least two of the idices are equal to oe For all cases where oly oe of the idices equals oe, equatio (43) becomes a ijk1 = a ij11 + a i1k1 + a 1jk1 a i111 a 1j11 a 11k1 + a 1111 (44) a ij1l = a ij11 + a i11l + a 1j1l a i111 a 1j11 a 111l + a 1111 (45) a i1kl = a i1k1 + a i11l + a 11kl a i111 a 11k1 a 111l + a 1111 (46) a 1jkl = a 1jk1 + a 1j1l + a 11kl a 1j11 a 11k1 a 111l + a 1111 (47) Before provig that (43) to (47) hold, we prove the followig propositio 16

17 Propositio 47 For 3 ad 6, it ca be show that a i1 j 1k(i 1,j 1)l(i 1,j 1) + a i1j 2k(i 1,j 2 )l(i 1,j 2 ) + a i2 j 1 k(i 2,j 1 )l(i 2,j 1) + a i2 j 2k(i 2,j 2 )l(i 2,j 2 ) +a i1 j 1 k(i 2,j 2)l(i 2,j 2 ) + a i1 j 2k(i 2,j 1)l(i 2,j 1 ) + a i2j 1 k(i 1,j 2)l(i 1,j 2 ) + a i2j 2k(i 1,j 1 )l(i 1,j 1) = a i1j 1k(i 2,j 1 )l(i 2,j 1) + a i1 j 2 k(i 2,j 2 )l(i 2,j 2 ) + a i2j 1k(i 1,j 1)l(i 1,j 1 ) + a i2j 2k(i 1,j 2 )l(i 1,j 2 ) +a i1 j 1k(i 1,j 2)l(i 1,j 2 ) + a i1j 2 k(i 1,j 1)l(i 1,j 1) + a i2 j 1 k(i 2,j 2 )l(i 2,j 2) + a i2 j 2 k(i 2,j 1)l(i 2,j 1) (48) for i 1, i 2, i 1 i 2 ad j 1, j 2 J, j 1 = j 2 Proof Let b, c, d, e K ad p, q, r, s L ad a arbitrary poit x P as illustrated i Table 9 Table 9: Poit x (Propositio 47) j 1 j 2 i 1 b c i 2 d e j 1 j 2 i 1 p q i 2 r s Let x = x(i 1 i 2 ) 1 (Table 10) Table 10: Poit x (Propositio 47) j 1 j 2 i 1 d e i 2 b c j 1 j 2 i 1 r s i 2 p q Let x = x(j 1 j 2 ) 2 (Table 11) Let x = x(i 1 i 2 ) 1 (Table 12) Let k x (i s, j t ) (l x (i s, j t )) deote the value of the k(l) idex for i = i s, j = j t at poit x k x (i s, j t ), k x (i s, j t ), k x (i s, j t ) ad l x (i s, j t ), l x (i s, j t ), l x (i s, j t ) are defied accordigly for poits x, x ad x Sice x, x P we have ax = ax By observig that all a ijkl terms for every i \ {i 1, i 2 } are caceled out, 17

18 Table 11: Poit x(propositio 47) j 1 j 2 i 1 c b i 2 e d j 1 j 2 i 1 q p i 2 s r Table 12: Poit x (Propositio 47) j 1 j 2 i 1 e d i 2 c b j 1 j 2 i 1 s r i 2 q p ax = ax becomes a i1j 1bp + a i1j 2cq + {a i1jk x (i 1,j)l x (i 1,j) : j J \ {j 1, j 2 }} + a i2 j 1 dr + a i2 j 2es + {a i2jk x (i 2,j)l x (i 2,j) : j J \ {j 1, j 2 }} = a i1 j 1 dr + a i1 j 2es + {a i1 jk x (i 1,j)l x (i : j J \ {j 1,j) 1, j 2 }} + a i2j 1bp + a i2 j 2cq + {a i2jk x (i 2,j)l x (i : j J \ {j 2,j) 1, j 2 }} (49) We observe that k x (i 2, j) = k x (i 1, j), k x (i 1, j) = k x (i 2, j) ad l x (i 2, j) = l x (i 1, j), l x (i 1, j) = l x (i 2, j) for j J \ {j 1, j 2 } Writig (49) i terms of poit x, we derive a i1j 1bp + a i1 j 2 cq + {a i1jk x (i 1,j)l x (i 1,j) : j J \ {j 1, j 2 }} + a i2 j 1 dr + a i2 j 2 es + {a i2 jk x (i 2,j)l x (i 2,j) : j J \ {j 1, j 2 }} = a i1j 1 dr + a i1 j 2 es + {a i1jk x (i 2,j)l x (i 2,j) : j J \ {j 1, j 2 }} + a i2j 1 bp + a i2j 2cq + {a i2 jk x (i 1,j)l x (i 1,j) : j J \ {j 1, j 2 }} (410) Similarly, sice x, x P we have a x = a x Terms a ijkl, for every i \ {i 1, i 2 }, are caceled out, so a x = a x becomes 18

19 a i1j 1 cq + a i1 j 2bp + {a i1 jk x (i 1,j)l x (i 1,j) : j J \ {j 1, j 2 }} + a i2 j 1es + a i2 j 2dr + {a i2jk x (i,j)l x 2 (i 2,j) : j J \ {j 1, j 2 }} = a i1 j 1es + a i1 j 2 dr + {a i1 jk x (i,j)l x 1 (i 1,j) : j J \ {j 1, j 2 }} + a i2j 1cq + a i2 j 2 bp + {a i2jk x (i 2,j)l x (i 2,j) : j J \ {j 1, j 2 }} (411) We observe that k x (i 2, j) = k x (i 1, j), k x (i 1, j) = k x (i 2, j) ad l x (i 2, j) = l x (i 1, j), l x (i 1, j) = l x (i 2, j) for j J \ {j 1, j 2 } Writig (411) i terms of poit x, we derive a i1 j 1 cq + a i1 j 2bp + {a i1jk x (i,j)l x 1 (i 1,j) : j J \ {j 1, j 2 }} + a i2 j 1es + a i2j 2 dr + {a i2jk x (i,j)l x 2 (i 2,j) : j J \ {j 1, j 2 }} = a i1j 1 es + a i1j 2dr + {a i1jk x (i,j)l x 2 (i 2,j) : j J \ {j 1, j 2 }} + a i2 j 1 cq + a i2j 2 bp + {a i2 jk x (i,j)l x 1 (i 1,j) : j J \ {j 1, j 2 }} (412) Subtractig (410) from (412) ad observig that for i {i 1, i 2 } ad j J \ {j 1, j 2 } we have k x (i, j) = k x (i, j) ad l x (i, j) = l x (i, j), we obtai a i1j 1bp + a i1 j 2 cq + a i2 j 1 dr + a i2 j 2 es (a i1j 1 cq + a i1j 2bp + a i2j 1es + a i2 j 2 dr) = a i1j 1 dr + a i1j 2 es + a i2 j 1bp + a i2 j 2cq (a i1 j 1es + a i1 j 2dr + a i2j 1 cq + a i2j 2 bp) f we elimiate the egative sig by movig terms i brackets to the other side of the equatio ad write the elemets of sets K ad L usig the otatio k(i, j) ad l(i, j) respectively, we obtai equatio (48) Propositio 47, the role of the sets, J for the row ad colum set, respectively, is purely covetioal Ay pair of sets from, J, K, L ca be used for the role of row/colum set Hece, for the rest of the paper, the otatio x((m 1, m 2 ) t1 ; ( 1, 2 ) t2 ) implies equatio (48), derived by applyig Propositio 47 at poit x, for rows m 1, m 2 ad colums 1, 2 this expressio, the first pair deotes the rows whereas the secod deotes the colums The subscripts t 1, t 2 declare the sets that idex the rows ad the colums, respectively Followig the same covetio as for the iterchages, 1 is used to deote set, 2 is used to deote set J, etc For example, x((1, i 1 ) 1 ; (1, ) 2 ) deotes equatio (48) 19

20 writte for rows 1, i 1 ad colums 1, at poit x, where elemets of the first pair belog to set ad of the secod pair to set J (Back to the proof of Theorem 46) We will show (43),,(47) for i = i 0, j = j 0, k = k 0, l = l 0 At poit x 0 of Lemma 45 we distiguish two cases, viz k 0 = k 1, k 0 k 1 case 1: k 0 = k 1 Let x 1 = x 0 (see Table 13) this case l 0 l 1 (orthogoal property) For 4 there exists (aother) k 1 Table 13: Poit x 1(Theorem 46, case 1) 1 j 0 1 j k 0 i 0 k 0 k l 1 i 0 l 0 1 K \ {1, k 0, k 2 } Let x 1 = x 1(k 0 k 1 ) 3 (see Table 14) Table 14: Poit x 1 (Theorem 46, case 1) 1 j 0 1 j k 1 i 0 k 1 k l 1 i 0 l 0 1 x 1 ((1, i 0 ) 1 ; (1, j 0 ) 2 ) a a 1j0k 1l 1 + a i0 1k 1 l 0 + a i0j 0 k a 11k21 + a 1j0 k 1l 0 + a i0 1k 1l 1 + a i0 j 0 11 = a 11k1l 0 + a 1j0k a i a i0 j 0k 1l 1 + a 11k1l 1 + a 1j a i0 1k 21 + a i0j 0k 1 l 0 Let x 2 = x 1 (1 k 0 ) 3 (see Table 15) Table 15: Poit x 2 (Theorem 46, case 1) 1 j 0 1 j 0 1 k 0 k 1 i 0 k 1 k l 1 i 0 l

21 x 2 ((1, i 0 ) 1 ; (1, j 0 ) 2 ) a 11k0 1 + a 1j0k 1 l 1 + a i01k 1l 0 + a i0 j 0 k 21 + a 11k2 1 + a 1j0 k 1 l 0 + a i0 1k 1 l 1 + a i0j 0 k 01 = a 11k1 l 0 + a 1j0 k a i01k 01 + a i0 j 0 k 1 l 1 + a 11k1l 1 + a 1j0 k 01 + a i01k 21 + a i0 j 0 k 1 l 0 x 2 ((1, i 0 ) 1 ; (1, j 0 ) 2 )-x 1 ((1, i 0 ) 1 ; (1, j 0 ) 2 ) yields (44) This completes the proof of case 1 for (44) Before proceedig to case 2, we derive a further relatioship for case 1 which will be used later o, for provig (43) Let x 3 = x 1 (1 l 0 ) 4 ad x 4 = x 3 (1 k 0 ) 3 Poits x 3 ad x 4 are illustrated i Tables 16, 17 respectively Table 16: Poit x 3 (Theorem 46, case 1) 1 j 0 1 j 0 1 k 0 k 1 i 0 k 1 k 2 1 l 0 l 1 i 0 1 l 0 Table 17: Poit x 4 (Theorem 46, case 1) 1 j 0 1 j k 1 i 0 k 1 k 2 1 l 0 l 1 i 0 1 l 0 x 3 ((1, i 0 ) 1 ; (1, j 0 ) 2 ) a 11k0 l 0 + a 1j0 k 1l 1 + a i01k 11 + a i0 j 0k 2l 0 + a 11k2 l 0 + a 1j0k 11 + a i0 1k 1l 1 + a i0j 0 k 0 l 0 = a 11k1 1 + a 1j0k 2l 0 + a i0 1k 0l 0 + a i0 j 0 k 1l 1 + a 11k1 l 1 + a 1j0 k 0l 0 + a i01k 2 l 0 + a i0 j 0 k 11 x 4 ((1, i 0 ) 1 ; (1, j 0 ) 2 ) a 111l0 + a 1j0 k 1 l 1 + a i01k a i0 j 0 k 2 l 0 + a 11k2 l 0 + a 1j0 k a i0 1k 1 l 1 + a i0 j 0 1l 0 = a 11k1 1 + a 1j0 k 2l 0 + a i011l 0 + a i0 j 0 k 1l 1 + a 11k1 l 1 + a 1j01l 0 + a i0 1k 2l 0 + a i0j 0k

22 x 3 ((1, i 0 ) 1 ; (1, j 0 ) 2 )-x 4 ((1, i 0 ) 1 ; (1, j 0 ) 2 ) a i0 j 0 k 0 l 0 = (a i0j 0 1l 0 + a i0 1k 0l 0 + a 1j0k 0l 0 ) a i011l 0 a 1j01l 0 a 11k0 l 0 + a 111l0 (413) (413) will be used later for provig (43) case 2: k 0 k 1 Let x 1 = x 0 Thus poit x 1 is poit x 0 of Lemma 45, exactly as illustrated i Table 7 x 1 ((1, i 0 ) 1 ; (1, j 0 ) 2 ) a a 1j0 k 1 l 1 + a i0 1k 0 l 0 + a i0j 0 k 21 + a 11k21 + a 1j0 k 0l 0 + a i0 1k 1l 1 + a i0j 0 11 = a 11k0l 0 + a 1j0 k 21 + a i a i0 j 0k 1 l 1 + a 11k1l 1 + a 1j a i01k 21 + a i0 j 0 k 0 l 0 Let x 2 = x 1 (1 k 0 ) 3 (see Table 18) Table 18: Poit x 2 (Theorem 46, case 2) 1 j 0 1 j 0 1 k 0 k 1 i 0 1 k l 1 i 0 l 0 1 x 2 ((1, i 0 ) 1 ; (1, j 0 ) 2 ) a 11k0 1 + a 1j0k 1 l 1 + a i0 1k 0l 0 + a i0 j 0 k a 11k2 1 + a 1j0 k 0l 0 + a i0 1k 1l 1 + a i0j 0k 01 = a 111l0 + a 1j0k 21 + a i0 1k 01 + a i0j 0k 1 l 1 + a 11k1 l 1 + a 1j0k 01 + a i01k 21 + a i0j 01l 0 x 1 ((1, i 0 ) 1 ; (1, j 0 ) 2 )-x 2 ((1, i 0 ) 1 ; (1, j 0 ) 2 ) a a 1j0k 01 + a i0 1k a i0 j 0 11 (a i0 j 0 k a i a 1j a 11k01) = a i0j 0k 0l 0 + a i0 11l 0 + a 1j0 1l 0 + a 11k0l 0 (a i0j 0 1l 0 + a i01k 0l 0 + a 1j0 k 0l 0 + a 111l0 ) (414) We refer to equatio (414) as (414) l0 to distiguish it from (414) with l 1 i the place of l 0 deoted as (414) l1 We observe that we ca derive (414) l1 by applyig Propositio 47 at poits ẋ 1 = x 1 (l 1 l 0 ) 4 ad ẋ 2 = x 2 (l 1 l 0 ) 4 22

23 This is also true for the case where l(1, j 0 ) = l 0 (ie l 1 = l 0 ) at poits x 0, x 1 ad x 2 sice for 3 there exists (aother) l 1 L \ {1, l 0 } such that we ca derive ẋ 1, ẋ 2 both cases, (414) l0 +(414) l1 2(a a 1j0k 01 + a i0 1k 01 + a i0j 0 11 (a i0 j 0 k 01 + a i a 1j a 11k0 1)) = {a i0j 0k 0l + a i0 11l + a 1j01l + a 11k0 l (a i0j 0 1l + a i0 1k 0l + a 1j0 k 0 l + a 111l ) : l {l 0, l 1 }} (415) At poit x 2 we distiguish two cases viz l(1, j 0 ) = l 0, l(1, j 0 ) = l 1 l 0 case 21: l(1, j 0 ) = l 0 For 3 we have already establish the existece of l 1 L \ {1, l 0 } Let x 3 = x 2 (1 l 0 ) 4 (1 l 1 ) 4 (Table 19) ad x 4 = x 3 (1 k 0 ) 3 (Table 20) Table 19: Poit x 3 (Theorem 46, case 21) 1 j 0 1 j 0 1 k 0 k 1 i 0 1 k 2 1 l 0 l 1 i 0 l 1 l 0 Table 20: Poit x 4 (Theorem 46, case 21) 1 j 0 1 j k 1 i 0 k 0 k 2 1 l 0 l 1 i 0 l 1 l 0 x 3 ((1, i 0 ) 1 ; (1, j 0 ) 2 ) a 11k0l 0 + a 1j0 k 1l 1 + a i0 11l 1 + a i0 j 0 k 2 l 0 + a 11k2l 0 + a 1j01l 1 + a i0 1k 1 l 1 + a i0j 0k 0l 0 = a 111l1 + a 1j0k 2 l 0 + a i01k 0 l 0 + a i0j 0k 1l 1 + a 11k1l 1 + a 1j0 k 0l 0 + a i0 1k 2l 0 + a i0 j 01l 1 x 4 ((1, i 0 ) 1 ; (1, j 0 ) 2 ) a 111l0 + a 1j0 k 1 l 1 + a i0 1k 0l 1 + a i0 j 0 k 2l 0 + a 11k2l 0 + a 1j0 k 0l 1 + a i0 1k 1l 1 + a i0 j 01l 0 = a 11k0 l 1 + a 1j0k 2 l 0 + a i011l 0 + a i0j 0k 1 l 1 + a 11k1l 1 + a 1j0 1l 0 + a i01k 2 l 0 + a i0j 0 k 0 l 1 23

24 x 4 ((1, i 0 ) 1 ; (1, j 0 ) 2 )-x 3 ((1, i 0 ) 1 ; (1, j 0 ) 2 ) {ai0 j 0k 0l + a i011l + a 1j0 1l + a 11k0 l (a i0 j 0 1l + a i0 1k 0l + a 1j0k 0l + a 111l ) : l {l 0, l 1 }} = 0 (416) case 22: l(1, j 0 ) = l 1 l 0 Let x 3 = x 2 (1 l 0 ) 4 (1 l 1 ) 4 (Table 21) ad x 4 = x 3 (1 k 0 ) 3 (Table 22) Table 21: Poit x 3 (Theorem 46, case 22) 1 j 0 1 k 0 k 1 i 0 1 k 2 1 j 0 1 l 0 1 i 0 l 1 l 0 Table 22: Poit x 4 (Theorem 46, case 22) 1 j k 1 i 0 k 0 k 2 1 j 0 1 l 0 1 i 0 l 1 l 0 x 3 ((1, i 0 ) 1 ; (1, j 0 ) 2 ) a 11k0l 0 + a 1j0 k 11 + a i0 11l 1 + a i0 j 0k 2 l 0 + a 11k2 l 0 + a 1j01l 1 + a i01k a i0 j 0k 0 l 0 = a 111l1 + a 1j0k 2 l 0 + a i0 1k 0l 0 + a i0 j 0k 11 + a 11k1 1 + a 1j0 k 0 l 0 + a i0 1k 2 l 0 + a i0 j 0 1l 1 x 4 ((1, i 0 ) 1 ; (1, j 0 ) 2 ) a 111l0 + a 1j0 k 11 + a i01k 0 l 1 + a i0j 0k 2 l 0 + a 11k2l 0 + a 1j0k 0l 1 + a i01k 11 + a i0 j 01l 0 = a 11k0l 1 + a 1j0k 2 l 0 + a i0 11l 0 + a i0j 0 k 11 + a 11k1 1 + a 1j0 1l 0 + a i01k 2 l 0 + a i0j 0 k 0 l 1 x 4 ((1, i 0 ) 1 ; (1, j 0 ) 2 )-x 3 ((1, i 0 ) 1 ; (1, j 0 ) 2 ) yields (416) Hece (416) is valid for case 2 (both for case 21 ad 22) Substitutig the right-had side of equatio (415) from (416) yields (44) 24

25 Our proof with respect to (44) is ow complete Equatios (45),(46) ad (47) follow by symmetry To show (43) we cosider (413) This equatio is valid for case 2 as well; the right-had side of equatio (414) is equal to 0 due to equatio (44) thus yieldig (413) Therefore, if we substitute i equatio (413) terms i brackets from (45), (46) ad (47) we get (43) Fially, (42) is true sice for 3 ad differet tha 6, P (see [13, Theorem 29]) This implies that there exists at least oe 0-1 vector x for which (21),,(26) are satisfied, thus by multiplyig these equatios with the correspodig scalars ad summig over all rows we get: ax = {λ 1 kl : k K, l L} + {λ 2 il : i, l L} + {λ 3 jl : j J, l L} + {λ 4 ij : i, j J} + {λ 5 jk : j J, k K} + {λ 6 ik : i, k K} This completes the proof for geeral The theorem holds for 4 ad 6 sice the proof requires four distict values for oe of the idices (idex k) particular, we have used values (1, i 0 ) for idex i, (1, j 0 ) for idex j, (1, k 0, k 1, k 2 ) for idex k, (1, l 0, l 1 ) for idex l Hece, the result holds for 4 ad = 6 Corollary 48 For 4 ad 6, dim(p ) = Next we will examie which of the costraits defiig P L are facet defiig for P Propositio 49 For 4 ad = 6 every iequality of the type x c 0 for c C defies a facet of P Proof For ay c C cosider the polytope P c = {x P : x c = 0} We eed to show that dim(p c) = dim(p ) 1 Evidetly, dim(p c) 4 1 rak(a c ) where A c is the matrix obtaied from A if we remove colum a c t is ot hard to see that the rak of A c is equal to the rak of A This is immediate, if the colum a c is ot amog the colums of the upper triagular matrix described i Theorem 42, otherwise it follows by symmetry Therefore, dim(p c) To prove that this boud is attaied, we use the same approach as i the proof of Theorem 46, ie show that ay equatio ax = a 0 (differet tha x c = 0) satisfied for every x P c is a liear combiatio of Ac x = e The proof goes through essetially uchaged Propositio 410 For 3 ad 6 every iequality x c 1 for c C does ot defie a facet of P Proof For ay c C cosider the polytope P c = {x P : x c = 1} We will show that dim(p c ) < dim(p ) 1 We kow that dim(p c ) dim(p c L ) where P c L is the liear relaxatio of P c Settig x c to oe is equivalet to settig 25

26 the variables belogig to the commo costraits with x c to zero The umber of these variables is 2(3 1)( 1) (Propositio 32) Thus, dim(pl c) = 4 2(3 1)( 1) rak(a c ) where A c is the matrix obtaied from A by removig the colums correspodig to the variables set to zero Obviously rak(a 1 ) rak(a c ) where A 1 is the costrait matrix of the OLS of order 1 By Theorem 42 rak(a 1 ) = 6( 1) 2 8( 1) + 3 so dim(p c ) dim(p c L) 4 2(3 1)( 1) (6( 1) 2 8( 1) + 3) = which is less tha dim(p ) 1 = for 3 t is easy to see that cliques of type do ot iduce facets of P For each of these cliques the coefficiet vector of the correspodig iequality is idetical to a row of the A matrix Thus, each of this iequalities is satisfied as equality by all x P ad therefore defies a improper face of P Next we cosider the iequalities iduced by cliques of type Theorem 411 Let Q(c) deote the ode set of a clique of type The for 5 ad = 6 the iequality {xq : q Q(c)} 1 (417) defies a facet of P for every c C Proof As usual, we will assume that 2, 6 Let P Q(c) is a facet of P = {x P : {x q : q Q(c)} = 1} We will show that P Q(c) First we ote that (417) is a valid iequality for all x P because Q(c) is the ode set of a clique i the itersectio graph G A (C, E C ) Wlog let c = (,,, ) The Q(,,, ) = {(,,, ), (1,,, ),, ( 1,,, ), (, 1,, ),, (, 1,, ), (,, 1, ),, (,, 1, ), (,,, 1),, (,,, 1)} t is easy to show that the face P Q(,,,) is ot empty Clearly for ay poit x P there exist i 0 ad j 0 J such that x i0j 0 = 1 Let x = x(i 0?i 0 ) 1 (j 0 =?j 0 ) 2 Poit x belogs to P Q(,,,) x = 1 sice it has Now cosider a arbitrary poit x P Q(,,,) We will show that there exists at least oe poit i P \ P Q(,,,) 26

27 Wlog suppose that at poit x we have x j = 1, j = The we apply a first idex iterchage betwee ad i for ay i \ {} The resultig poit belogs to P but ot to P Q(,,,) Similarly at poit x x i = 1 ad i = or x k = 1 ad k or x l = 1 ad l = the a correspodig oe idex iterchage will give a poit i P \P Q(,,,) Fially, if at poit x we have x = 1 the we apply a first idex iterchage betwee ad i \{} ad the a secod idex iterchage betwee ad j J \ {} The resultig poit belogs to P but ot to P Q(,,,) Thus P Q(,,,) is a proper face of P To show that P Q(,,,) is a facet of P we will use the same approach as i Theorem 46 Thus we will exhibit 6 2 scalars λ 1 kl, λ2 il, λ3 jl, λ4 ij, λ5 jk, λ6 ik for i, j J, k K, l L ad a additioal scalar π for the clique iequality, such that if ax = a 0 for all x P Q(,,,), the λkl 1 a ijkl = + λ2 il + λ3 jl + λ4 ij + λ5 jk + λ6 ik, (i, j, k, l) C \ Q(,,, ) (418) λ 1 kl + λ2 il + λ3 jl + λ4 ij + λ5 jk + λ6 ik + π, (i, j, k, l) Q(,,, ) ad a 0 = {λ 1 kl : k K, l L} + {λ 2 il : i, l L} + {λ 3 jl : j J, l L} + {λ 4 ij : i, j J} (419) + {λ 5 jk : j J, k K} + {λ 6 ik : i, k K} + π Agai we defie λ 1 kl = a 11kl λ 2 il = a i11l a 111l λ 3 jl = a 1j1l a 111l λ 4 ij = a ij11 a i111 a 1j11 + a 1111 λ 5 jk = a 1jk1 a 1j11 a 11k1 + a 1111 λ 6 ik = a i1k1 a i111 a 11k1 + a 1111 Thus for (i, j, k, l) C \ Q(,,, ) we have to show a ijkl = a 11kl + a i11l + a 1j1l + a ij11 + a 1jk1 + a i1k1 2a i111 2a 1j11 2a 11k1 2a 111l + 3a 1111 (420) 27

28 This is clearly true for a 1111 ad for all cases where at least two of the idices equal oe Whe oe of the idices equals oe, we deote equatio (420) as (420) m=1 where m is ay of i, j, k, l For example equatio (420) l=1 is: a ijk1 = a ij11 + a i1k1 + a 1jk1 a i111 a 1j11 a 11k1 + a 1111 For each of the cases to be examied ext, whe at most oe of the idices is equal to oe, the proof of (420) will be carried out i a maer aalogous to that of Theorem 46 except that this time we will exclusively use poits belogig to P Q(,,,) For ay poit x P we deote as X the collectio of poits (x, x, x, x ) (otatio ad derivatio of poits from x as itroduced i Propositio 47) X P Q(,,,) implies that x, x, x, x P Q(,,,) Thus for all of the followig cases, equatio (420) l=1 is derived explicitly by the applicatio of Propositio 47 to poits x t X t where X t P Q(,,,) for t = 1,, 4 Equatios (420) i=1, (420) j=1, (420) k=1 follow by symmetry O the other had, whe oe of the idices is equal to oe, (420) is show by substitutig these equatios to a equatio, correspodig to (413) of Theorem 46, which is derived i the course of provig (420) l=1 The idices i, j, k, l of a ijkl give rise to the cases where oe, oe or two of them are equal to each of these cases we make use of the poit x q which is established i Lemma 412 Lemma 412 For 5 ad = 6 let i q \ {1, }, j q J \ {1, }, k q, k 1, k 2 K \ {1, } with k 2 = k 1, k q, l q, l 1 L \ {1, } with l q = l 1 The there exists the poit x q P as illustrated i Table 23 1 j q 1 1 k 1 i q k q k 2 Table 23: Poit x q (Lemma 412) 1 j q 1 1 l 1 i q l q Proof Cosider the OLS that has i atural order the elemets of the a) first row of the two lati squares ad b) the elemets of the first colum of oe of the two squares A pair of OLS with this property is said to be a stadardized set f P the there always exists a OLS structure of this type ([8, p 159]) Wlog assume the elemets of the first colum of the secod lati square (lati square cosistig of the values of set L) to be i atural order At this poit we have k(, 1) = k q Let j 3 J \ {1} be such that k(, j 3 ) = 1 For 5 there exist l q L \ {1, } ad l 1 L \ {1, l q, } such that l(1, j 1 ) = l q, l(1, j 2 ) = l 1 ad l(, j 2 ) = l q where j 1, j 2 L \ {1, } with j 1 = j 2, j 2 = j 3 We deote k(1, j 2 ) as k 1 ad k(, j 2 ) as k 2 Note that sice k 1, k 2 lie i the same colum we have k 1 k 2 Combiig this with k(1, 1) = 1, k(1, ) =, we obtai that k 1 K \ {1, k 2, } a similar maer, because k(, 1) = k q, k(, j 3 ) = 1 ad j 3 = j 2, k 1 = k 2 we derive that k 2 K \ {1, k 1, k q } We apply a fourth idex iterchage betwee l q ad The resultig poit amely x q is illustrated i 28