ON THE LINEAR PROGRAMMING AND GROUP RELAXATIONS OF THE UNCAPACITATED FACILITY LOCATION PROBLEM

Transcription

1 ON THE LINEAR PROGRAMMING AND GROUP RELAXATIONS O THE UNAPAITATED AILITY LOATION PROLEM MOHAMMAD KHALIL A THESIS PRESENTED TO THE GRADUATE SHOOL O THE UNIVERSITY O LORIDA IN PARTIAL ULILLMENT O THE REQUIREMENTS OR THE DEGREE O MASTER O SIENE UNIVERSITY O LORIDA 200

2 200 Mohammad Khall 2

3 Ths work s dedcaed o m parens m wfe Rana m daughers Habba and Jana and m sblngs Shamaa Sara and Ahmed for her connuous love and suppor 3

4 AKNOWLEDGMENTS rs of all I would lke o epress m sncere apprecaon o m advsor Dr Jean- Phlppe Rchard for hs nvaluable gudance In addon o provdng vson and encouragemen Dr Rchard gave me unlmed freedom o eplore new avenues for research He was alwas avalable o dscuss m deas and gve me ecellen feedback Hs enhusasm encouraged me o produce beer work He has augh me no onl Operaons Research bu also how o communcae effecvel and how o wre echncal work Hs devoon o each me o pa close aenon o m houghs (even when he are wrong) and o make hs hess as bes as can has been and wll connue o be an nspraon o me Whou hs help hs work would no have been possble Second I would lke o hank Dr Youngpe Guan he member of m eamnng commee for he me he spen revewng hs hess I am graeful o Dr Aa Gomaa he former char of Indusral Engneerng Deparmen a aoum Unvers who was he frs person o nroduce Operaons Research o me and encouraged me o pursue an academc career Hs suppor s endurng I am also ndebed o Mr Mahmoud Talaa who bul m basc mahemacal sklls and nspred me o sud engneerng nall I would lke o hank m wfe and daughers for her paence kndness and connuous suppor durng m sudes a he Unvers of lorda 4

5 TALE O ONTENTS page AKNOWLEDGMENTS 4 LIST O TALES 7 LIST O IGURES 8 LIST O ALGORITHMS 9 LIST O AREVIATIONS 0 LIST O NOTATIONS ASTRAT 2 HAPTER INTRODUTION 4 The acl Locaon Problem 5 2 Leraure Revew 6 3 The Uncapacaed acl Locaon Problem 7 4 Group Relaaons of Med Ineger Programmng 20 2 ON THE POLYHEDRAL STRUTURE O THE LINEAR PROGRAMMING RELAXATION O UNAPAITATED AILITY LOATION PROLEM 28 2 Inroducon ase : m + m < n 3 23 ase 2: m + m = n ase 3: m + m > n onsrucng ULP Insances of Desred Deermnan 62 3 MAXIMUM POSSILE DETERMINANT O ASES O THE LINEAR PROGRAMMING RELAXATION O UNAPAITATED AILITY LOATION PROLEM 79 3 ompung he MPD of ± Marces ompung he MPD of (0) Marces ompung he MPD of ases of he LPR of ULP On The easbl of The LP Soluon o ULP ha has The MPD Solvng Group Relaaons of ULP 95 5

6 4 SPEIAL ASES 02 4 ase : Two usomers and/or Two acles ase 2: Three usomers and Three acles 02 5 EXPERIMENTAL RESULTS 6 ONLUSION AND UTURE RESEARH 6 LIST O REERENES 7 IOGRAPHIAL SKETH 20 6

7 LIST O TALES Table page 3- Mamum possble deermnan of a ± square mar of sze ( hh ) MPD of a (0) square mar of sze ( hh ) and number of square (0) marces ha have he MPD Mamum possble deermnan of and T k for gven number of columns correspondng o and varables Probabl ha he MPD of pseudo bases of he LPR of he ULP for gven n and n s less han or equal U ( h ) Selecon of he parameers n and n n he consrucon of ULP epermens 5-2 Epermenal Resuls 4 7

8 LIST O IGURES gure page - onsran mar A of he LPR of he ULP formulaon shown n (3) 9-2 The group nework assocaed wh he MIP problem n Eample Mar A n Eample ass ha was obaned from mar A n Eample Illusraon of he dfferen cases encounered n he proof of Lemma nal arrangemen of columns ncluded n f m + m > n Illusraon of how o appl ERO (25) usng (27) n Eample n Eample 22 afer column permuaons n accordance wh gure Submar G reflecs he seleced columns n Eample Illusraon of Eample 25 a) Mar and b) Mar The s bases of he LPR of ULP (wh n = n = 3 ) ha have deermnan absolue values equal o A complee bpare graph beween he se of facles and he se of cusomers for ULP wh n = n = A complee bpare graph beween he se of facles and he se of cusomers ha corresponds o he machng assocaed wh each of he nequales n (40) 09 8

9 LIST O ALGORITHMS Algorhm page ULP-UNI- 44 ULP-UNI-2 44 ULP-UNI-3 44 ULP-DET 63 ULP-ASIS 64 ULP-INSTANE 78 HADAMARD 80 INARY 83 9

10 LIST O AREVIATIONS LP IP MIP MILP LPR LP ULP MPD ERO EO Lnear Programmng Ineger Programmng Med Ineger Programmng Med Ineger Lnear Programmng Lnear Programmng Relaaon acl Locaon Problem Uncapacaed acl Locaon Problem Mamum Possble Deermnan Elemenar Row Operaon Elemenar olumn Operaon 0

11 LIST O NOTATIONS Se of poenal locaons where facles can be opened Se of cusomers O Se of opened facles where O n = n = { } n = mn n n f os of openng facl c os of assgnng cusomer j o facl j c A b I The cos vecor consrans mar and he rgh-hand-sde vecor n he sandard LP form { mn c subjec o : A = b 0} Iden mar 0 Mar of zeros E Mar of ones n n -combnaon of r r A hk Mar A wh h rows and k columns A( h k ) omponen n row h and column k of mar A Ah ( ) A( k ) h h column of mar A h k row of mar A

12 Absrac of Thess Presened o he Graduae School of he Unvers of lorda n Paral ulfllmen of he Requremens for he Degree of Maser of Scence ON THE LINEAR PROGRAMMING AND GROUP RELAXATIONS O THE UNAPAITATED AILITY LOATION PROLEM har: Jean-Phlppe Rchard Major: Indusral and Ssems Engneerng Mohammad Khall Augus 200 The uncapacaed facl locaon problem (ULP) s a classcal problem n he Operaons Research commun ha has been suded eensvel Varous approaches have been proposed for s soluon In hs hess we seek o deermne f he group approach o med neger lnear programmng (MILP) could lead o new advances n he soluon of ULP To deermne wheher group relaaons of ULP can be solved effcenl we frs deermne he mamum possble deermnan (MPD) of a bass of s lnear programmng relaaon (LPR) Ths movaes he nvesgaon of he bases of he LPR of ULP Alhough we show ha he MPD s eponenal (n erms of he number of cusomers and he number of facles) we also show ha mos bases have small deermnans We gve several algorhms o consruc bases of he LPR of ULP In parcular we presen hree algorhms o consruc unmodular bases one algorhm o consruc bases wh desred deermnan and one algorhm o consruc bases wh he MPD We show ha he soluons correspondng o he bases of he LPR of ULP wh MPD 2

13 we descrbe are feasble We also show ha he correspondng lnear programmng (LP) soluon s no ver fraconal nall we use he above resuls o sud wo small nsances of ULP In he frs we assume ha we have wo cusomers and/or wo facles and we show ha he LPR of ULP alwas descrbes he conve hull of s neger soluons In he second we assume ha we have hree cusomers and hree facles and we show ha he conve hull of neger soluons can be obaned b addng s nequales o s LPR Ths hess s organzed as follows In haper we gve a bref nroducon o ULP and o group relaaons n MILP In haper 2 we oban some resuls abou he polhedral srucure of he LPR of ULP In haper 3 we deermne he MPD of a bass of he LPR of ULP we dscuss he feasbl of he soluons correspondng o he bases of he LPR of ULP wh MPD we consruc and conclude wh commens abou he effcenc of usng group relaaons o solve ULP In haper 4 we sud wo nsances of ULP wh small number of cusomers and/or facles In he frs we assume ha we have wo cusomers and/or wo facles whle n he second we assume ha we have hree cusomers and hree facles We oban conve hull descrpons for he se of neger soluons o hese problems In haper 5 we descrbe epermenal resuls on solvng he group relaaons of ULP nall we gve a concluson and dscuss fuure research drecons n haper 6 3

14 HAPTER INTRODUTION Man commercal companes servce organzaons and publc secor agences face he problem of decdng where o locae her facles The facles can be facores warehouses realers hospals schools ec or even rouers and caches for frms workng n web servces Ths problem s known n he Operaons Research commun as he facl locaon problem (LP) I s consdered o be a long erm decson problem ha can affec he success of an organzaon LP has vas applcaons and s ofen consdered o be a pllar of suppl chan managemen As a resul has been suded eensvel n he leraure Numerous varans of he problem have been consdered and a vas arra of soluon mehodologes has been proposed Varans of he problem nclude suaons where cusomers demands are deermnsc or sochasc he capaces of facles are fne or nfne he poenal locaons are connuous or dscree ec In hs hess we sud a varan of he problem ha s known as uncapacaed facl locaon problem (ULP) n whch demand s deermnsc he locaon of facles mus be chosen from a gven se and he capac of hese facles s suffcenl large o handle all cusomers demands Soluon mehodologes for ULP are dverse and nclude enumeraon echnques cung plane approaches appromaon algorhms and heurscs In hs hess our goal s o deermne wheher he group approach o med neger lnear programmng (MILP) could be useful n sudng ULP The remander of hs chaper s organzed as follows In Secon we descrbe dfferen varans of LP In Secon 2 we gve a bref leraure revew of sudes of 4

15 LP In Secon 3 we presen he classcal med neger programmng formulaon of ULP and s lnear programmng relaaon (LPR) nall n Secon 4 we gve an nroducon o group relaaons n med neger programmng The acl Locaon Problem Gven a se of poenal locaons and a se of cusomers he facl locaon problem (LP) seeks o deermne: how man facles should be opened where he open facles should be allocaed wha cusomers should be assgned o wha open facl so ha he oal cos assocaed wh openng facles and assgnng cusomers o open facles s mnmzed The sud of LP daes back o he 960s [23456] Dfferen varans have been consdered over he ears We ne revew some of he common varans of he problem ha have been suded We refer o [789] for dealed dscussons The uncapacaed facl locaon problem (ULP) assumes ha he capaces of he facles are nfne whle he capacaed facl locaon problem (LP) assgns a mamum capac lm for each facl The p-cener problem s smlar ULP ecep ha n he p-cener problem he number of facles ha should be opened s fed and he problem mnmzes he mamum dsance beween a cusomer and s assgned facl We sa ha a cusomer s covered f he dsance beween hs cusomer and s assgned facl s less han or equal a gven dsance The p-medan problem s a varan ha mnmzes he number of uncovered cusomers under he resrcon ha he number of opened facles s a mos equal o p If each cusomer s demand mus be enrel sasfed from a sngle facl he problem s called a sngle-sourcng facl locaon problem Oherwse cusomer s 5

16 demands can be sasfed from mulple facles eldng mul-sourcng facl locaon problems The dscree facl locaon problem s a varan of he problem where facl locaons mus be seleced from a dscree se However f he locaons are gven as coordnaes ha are connuous he problem s called a planar facl locaon problem The deermnsc facl locaon problem assumes ha he cusomers demands are deermnsc and known beforehand whle he sochasc facl locaon problem assumes ha he cusomers demand s known onl hrough a probabl dsrbuon 2 Leraure Revew The use of eac soluon algorhms (branch-and-bound and cung plane algorhms) o solve LP s dscussed n [0] The polhedral srucure of he LP s suded n [ ] ho Padberg and Rao [5] showed ha for wo cusomers and/or wo facles he LPR of ULP alwas has an neger opmal soluon In haper 3 we prove he same resul b dfferen means As far as appromaon algorhms are concerned dfferen greed algorhms have been developed n [79] lnear programmng (LP) roundng algorhms were provded n [20222] and prmal dual algorhms were gven n [2324] Guha and Khuller [25] proved ha here s no appromaon algorhm for he ULP ha can gve beer appromaon facor han 463 (unless P=NP) The bes appromaon facor known s 52 and was obaned b Mahdan and Zhang [26] Oher heurscs have also been proposed for hs problem ncludng local search smulaed annealng and varable neghborhood search; see [2728] 6

17 3 The Uncapacaed acl Locaon Problem Gven a se of n poenal locaons and a se of n cusomers ULP s concerned wh fndng a subse O of facles o open and an assgnmen of cusomers o open facles ha mnmzes oal cos Each facl s assumed o have an nfne capac; hence an open facl can sasf he demand of all cusomers assgned o urher we assume ha each cusomer s demand s enrel sasfed from a sngle facl We ne presen he classcal med neger programmng (MIP) formulaon of he ULP and s LPR [7] Inpus: f : cos of openng facl c : cos of assgnng cusomer j o facl j Decson varables: f facl s opened = 0 oherwse f cusomer j s assgned o facl j = 0 oherwse ULP can be formulaed usng MIP as follows mn j subjec o : ( a) = j ( b) () j ( c) { 0 } ( d) { 0 } j ( e) c + f 7

18 The objecve funcon (a) mnmzes he oal cos assocaed wh openng facles (such as consrucon coss) and assgnng cusomers o facles (such as ransporaon or roung coss) onsran (b) requres ha each cusomer j s assgned o eacl one facl j The second consran (c) ensures ha cusomer j s assgned o facl onl f facl s open (n oher words consran (c) mpose ha = 0 whenever = 0 ) onsrans (d) and (e) enforce he bnar naure of varables mposes snglesourcng of cusomers demand In parcular he fac ha { 0} The LPR of he prevous formulaon () can be obaned b relang he decson varables from bnar o connuous as shown n (2) mn j subjec o : ( 2a) = j ( 2b) (2) j ( 2c) ( 2d) 0 j ( 2e) c + f In (2) he relaed consran was removed snce s mpled b and LPR (2) can be ransformed o a sandard LP mn { c subjec o : A b 0} = (3) b nroducng slack varables s n (2c) and n (2d) eldng 8

19 mn j subjec o : ( 4a) = j ( 4b) (4) + s = 0 j ( 4c) + = ( 4d) 0 j ( 4e) c + f The LPR n a sandardzed form (4) can be used o descrbe mar A and vecors b and c n (3) Le nn I denoe he den mar of sze ( ) denoe he mar of all ones of sze ( nn ) The srucure of he mar nn and E nn A s shown n gure - where all emp spaces are zeros n nn n 2nn 2n n 2 2n n nn s s n s 2 s2n s n snn n n I n n n n I I n n s I n n I -E n n n I n n n n I I n n I -E n n n I n n I n n gure - onsran mar A of he LPR of he ULP formulaon shown n (3) The objecve and he rgh-hand-sde are gven as follows [ ] c = c f 2n n + 2n 2n 2n n (5) and 9

20 b + + = E 0 E (6) n nn n n nn n T 4 Group Relaaons of Med Ineger Programmng Man praccal applcaons of Operaons Research can be modeled usng MIP Two man famles of mehods are used o solve MIP problems n pracce; branch-andbound and cung planes algorhms The wo algorhms rel on solvng he LPR of a MIP problem The group relaaon approach (also known as corner relaaon) was nroduced b Gomor [29] Group relaaons can be used o replace he LPR n he branch-andbound algorhm because s smple o opmze lnear funcons over s feasble regon [30] We ne descrbe how o consruc he group relaaon of a MIP problem onsder he followng MIP n mn { c subjec o Amn n b n + } : = (7) To oban s corner relaaon we frs remove he negral consran and hence oban s lnear relaaon n mn { c subjec o Amn n b n + } : = (8) We ne solve he LPR n (8) usng he smple algorhm Le and N denoe he basc varables and he non-basc varables of he opmal LP soluon obaned b smple Also le varables and A denoe he submar of A correspondng o he basc A N he submar of A correspondng o he non-basc varables Smlarl denoe b c and c N he subvecors of c correspondng o he cos elemens of and We rewre (8) as N 20

21 N mn { c + cn N subjec o A + AN N = b + N + } The opmal soluon : (9) * of he LPR and he opmal value * z are gven b = A b = 0 (0) * * N and z * = c () To oban he corner relaaon of (7) we creae he problem mn { : N c + cn N subjec o A + AN N = b N + } where he nonnegav of he basc varables (2) has been relaed I s possble o reformulae he problem (2) usng he Smh Normal orm of A Lemma [3]: Gven a nonsngular neger mar A of sze ( nn ) here es ( nn ) unmodular marces U and 2 U such ha SN ( A) U AU 2 posve elemens d d n such ha d dvdes called he Smh Normal orm of A A = s a dagonal mar wh d + for { n } SN ( ) A s To reformulae (2) we compue he Smh Normal orm of he opmal LP bass ( ) 2 S = SN A = U AU (3) Then we mulpl he consrans n (2) b U on he lef and wre * mn { } N z + cn N subjec o : UAU 2U 2 + UAN N = Ub N + where c N s he reduced cos of varables N n (9) e (4) c c c A A N = N N 2

22 Afer subsung UAU 2 b S n (3) we oban : (5) * N mn { z + cn N subjec o SU 2 + UAN N = Ub N + } Snce Lemma saes ha U 2 s a unmodular mar U 2 s also a unmodular m m mar I follows ha U2 for ever Therefore (5) reduces o N { z cn N subjec o UAN N Ub S N + } * mn ( ) + : = mod (6) Le d be he dagonal elemens of Then he h row of (6) S for { n} s smpl consdered modulo d or ever dagonal elemen n he h row of S ha s equal o one s clear ha he correspondng h row of (6) can be removed Afer removng hese rows (6) represens he corner relaaon problem or group mnmzaon problem assocaed wh he MIP n (7) We ne presen an eample o llusrae he aforemenoned seps Eample : onsder he followng MIP mn subjec o : = = = (7) The consrans mar A he objecve c and he rgh-hand-sde b are A = c = ( ) and b = (8) Relang he negral consran we oban he followng LP 22

23 mn subjec o : = = = (9) or hs lnear program he basc and non-basc varables n he opmal soluon are gven b Therefore { } { } = and = (20) 2 3 N A = 3 AN = 4 0 c = ( ) and cn = ( ) (2) The opmal soluon and he opmal value of he LP (9) are 55 * * * = A b = 55 N = 0 and z = c = (22) The smple ableau correspondng o hs soluon s mn subjec o : = = = (23) The Smh Normal orm for he opmal bass A can be verfed o be U= S = and U2 = (24) 23

24 We hen compue UA Ub ( S) N N = mod as n (6) o oban mn subjec o : 4 5 ( ) ( ) ( ) mod 8 5 ( 4 5) = b = 3 mod mod4 (25) The frs row of (25) s removed snce S ( ) = and (25) reduces o mn subjec o : 4 5 ( mod 2) ( mod 4) 0 ( 4 5) = b = (26) We refer o (26) as he corner relaaon problem or group mnmzaon problem assocaed wh (7) To solve he corner relaaon problem assocaed wh a MIP problem a dreced graph GVE ( ) s creaed ha s called he group nework Ths nework represenaon was frs nroduced b Shapro [32] Each vere of he nework corresponds o a vecor { } δ δ δ m where 2 for all { m} δ + and δ d The number of verces n he group nework s herefore equal o he absolue value of he deermnan of he opmal bass ( A ) m de = d or each column j correspondng o a non-basc = varable and for ever vere g V k where ( ) { } k 0 de A we creae a dreced arc from he vere g o he vere ( g U A )( S) k k + j mod The cos assocaed wh hs arc s equal o c j Problem (6) hen reduces o fndng he shores pah 24

25 from he vere (0 00) o he vere correspondng o ( Ub )( mod S ) An approprae shores pah algorhm can be used o solve hs problem The soluon of he shores pah algorhm elds opmal values for he non-basc varables n an opmal soluon of (6) We hen subsue hese values no (5) o oban an neger soluon of (7) Alhough hs soluon s neger s no necessar nonnegave However f he obaned neger soluon s ndeed feasble (e nonnegave) hen s an opmal soluon for (7) We refer o [30] for more nformaon and show ne an eample ha llusraes he use of a shores pah algorhm o solve he group mnmzaon problem Eample -connued: Snce ( ) de A = 8 he group nework has 8 verces ha are arranged along 2 dmensons (because we have onl wo elemens dfferen from one on he dagonal of mar S ) or each of he egh verces we draw wo arcs correspondng o non-basc varables 4 and We represen he arcs ha are correspondng o 5 4 b connuous lnes and he arcs correspondng o 5 b doed lnes The cos of all he arcs drawn wh sold lnes s c 4 = 4 whle he cos of he arcs drawn wh dashed lnes s c 5 = 50 In hs nework we are lookng for he shores pah from node (00) o node (2) gure -2 llusraes he group nework where he source and desnaon nodes are hghlghed n black Solvng he shores pah problem we oban a shores pah from (00) o (2) ha vss vere (0) Ths soluon s represened b heav lnes n gure -2 The frs arc corresponds o non-basc varable 4 whle he second arc corresponds o nonbasc varable 5 As a resul an opmal soluon o he group mnmzaon problem s 25

26 gure -2 The group nework assocaed wh he MIP problem n Eample 4 = and 5 = Subsung no (23) we oban = 2 = 9 = = and = (27) We have an neger soluon o he MIP n (7) whose objecve value s equal o 829 Alhough hs soluon s neger s no opmal for (7) snce s nfeasble as 3 < 0 urher snce he group mnmzaon problem s a relaaon of (7) we have ha he opmal value of (7) s a leas 829 Gven a MIP problem he number of arcs ha orgnaes from each vere n he group nework of s corner relaaon s equal o he number of non-basc varables of s LPR urher he number of verces n he group nework s equal o he absolue value of he deermnan of he opmal bass of s LPR Therefore he sze of he group nework s a drec funcon of he deermnan of he opmal bass of he LPR of hs MIP problem Snce he runnng me of shores pah algorhms s a funcon of he number of verces and arcs of he nework hen follows ha he dffcul of solvng 26

27 he group relaaons of MIP problem s nmael relaed o he mamum possble deermnan (MPD) of he bases of s LPR Therefore o deermne he dffcul of solvng group relaaons of ULP s mporan o deermne he MPD of he bases of s LPR 27

28 HAPTER 2 ON THE POLYHEDRAL STRUTURE O THE LINEAR PROGRAMMING RELAXATION O UNAPAITATED AILITY LOATION PROLEM 2 Inroducon As menoned n Secon 4 he dffcul of solvng he group relaaon of a MIP problem s drecl relaed o he deermnan of he opmal bass of s LPR Therefore s mporan o deermne he bases wh MPD I s also mporan o oban nformaon abou s unmodular bases snce he eld neger soluons Ths movaes he nvesgaon of he polhedral srucure of he LPR of ULP Assumpons: The formulaon shown n (4) s he one for whch we sud bases Unless oherwse menoned he varables n he dfferen bases we dscuss are arranged from lef o rgh n he followng order: s hen j More precsel comes before * * f j * < or f * * = and j < j The same orderng s * appled o s Smlarl comes before * and comes before * f < rs we nroduce wo lemmas ha wll be used n he remander of hs chaper o compue deermnans and o oban he nverse of block marces q q p q q p p p Lemma 2 [33]: Le P R Q and V If V s an nverble mar hen P Q de = de( V ) de ( P QV R) R V (2) q q p q q p p p Lemma 22 [34]: Le P R Q and V If P and V are nverble marces hen 28

29 ( ) ( ) ( V RP Q) RP ( V RP Q) P Q P QV R P QV R QV = R V (22) A bass of A n + n n + n 2n n + 2n s a square submar of A n + n n + n 2n n + 2n ha s nverble and ha has ( n n n n ) + + rows and columns The oal number T of bases ha ma be obaned from mar A s bounded above b T ( n n + n ) ( n + n n + n ) ( n n + n n ) 2n 2 2 2! n + n n + nn + n =!! (23) Eample 2: or ULP wh n = 4 and n = 4 he oal number of dfferen bases of LPR s less han or equal o = Le denoe an arbrar square submar of A of sze ( n + n n + n ) or all we defne M and M s o be he ses of ndces j of varables assocaed columns are ncluded n Smlarl we denoe b ndces of varables M and and s whose M he ses of and whose assocaed columns are ncluded n nall le m m m and m denoe he cardnal of ses M M M and M and le s m s be he sums of m and s s m respecvel for e m = m = M m = m = M m = M and m = M s s s m and (24) Eample 22: or ULP wh n = 4 and n = 4 he consran mar A s gven n gure 2- The bass presened n gure 2-2 s obaned b selecng he columns marked wh ( ) n gure 2- Usng our noaon we wre 29

30 = = 2 = 3 = 4 = = 2 = 3 = s gure 2- Mar A n Eample 22 = s gure 2-2 ass ha was obaned from mar A n Eample 22 30

31 2 3 4 M = { } M = { } M = { } M = { } M { } M { } M { } M { } = 4 = 3 = 2 = and s s s s M = M = { } 234 urher m = m = { } 3 4 m = 2 and m = m = m = 4 s s To smplf he sud of he dfferen bases of he LPR of ULP we dvde he dscusson no hree man secons; m + m < n m + m = n and m + m > n 22 ase : m + m < n Lemma 23: Le be an arbrar square submar of A of sze ( n n n n ) s such ha m m n + < Then ( ) de = ha Proof: Le be an submar of A of sze ( n + n n + n ) We consder now he las n rows of Denoe b r he number of elemens ha are equal o one n he h of hese rows learl r { 02 } { } n see gure - urher n r = m + m < n Therefore here ess wh r = 0 e he h of he las n rows = of s dencall zero I follows ha de( ) = 0 snce conans a zero row Eample 23: When n = 4 and n = 4 an submar of A of sze (2424) ha has M = { 2} and { 4} M = s sngular because he rd 23 row s dencall zero e 3

32 ( + + ) = ( ) = 24 n n n Smlarl we oban he followng resul Lemma 24: Le be an arbrar square submar of A of sze ( n n n n ) ha ( M M ) Then ( ) de = 0 Proof: We consder he las n rows of { } hen If ( M M ) ( + + ) s dencall zero showng ha ( ) n n n de = such If he condon of Lemma 23 m + m < n holds hen he condon of Lemma 24 ( M M ) also holds Therefore Lemma 24 s a src generalzaon of Lemma 23 because can also be appled when m + m n Eample 23-conned: An submar of A of sze (2424) ha has { 2} M = { 4} s sngular because ( n + nn + ) = ( ) = ase 2: m + m = n M = and efore proceedng wh he dscusson of he case where m + m = n we nroduce a new noaon In parcular we nroduce an elemenar row operaon (ERO) ha wll modf he srucure of b elmnang some of s elemens (converng nonzero elemens o zero) 32

33 In he mar A ever column correspondng o varable has eacl wo componens equal o one; one n he h j row and one n he ( n ) h j + row; see gure - olumns correspondng o varables s however have eacl one componen equal o one whch s locaed n he same row as he second componen equal o one n he column; e s locaed n he ( n + j ) h row In summar; f ( ) A and A( s ) denoe he columns n mar A correspondng o varables and s hen { } ( ) ( ) f h j n + j 25a j A( h ) = 0 oherwse 25b (25) and ( ) ( ) f h = n + j 26a j A( h s ) = 0 oherwse 26b (26) Eample 22-connued: olumn 24 has eacl wo componens equal o one; n he 4 h and he 2 h row of A Also column s 24 has a sngle one n he 2 h row or ever column ha has been seleced n j we ma subrac from he row ha has he frs one componen he row ha has he second one componen The correspondng (ERO) s hen descrbed as ( ) ( ) ( ) h M : h h n + h (27) ERO (27) can also be obaned b mulplng mar on he lef wh a smple mar More precsel we mulpl wh mar ER o perform elmnaon n he frs upper n componens of column where 33

34 ( ) f ( 28a) ( ) ( ) ( ) ER h h = h n + n n + n ER = ER h h + n = - f h M 28b 0 oherwse 28c (28) de ER = urher applng hs ransformaon for all we Noe ha ( ) oban = ER (29) learl ( ) ( ) de = de (20) ecause ERO (27) s made onl o rows ( ) h where h M onl affecs he frs upper n rows whle he remanng nn + n rows reman unchanged As a resul ever one componen n he frs n rows and n he columns assocaed wh he wll be elmnaed and a block of zeros wll be obaned 0 or he columns n m assocaed wh he varables s f has a s column ha has he same ndces and j as a column ha also belongs o hen (27) wll cause ( ) A j s o ake on a coeffcen - nsead of zero Oherwse A( j s ) remans unchanged and so equal o zero In summar and ( : : ) n m = 0 (2) n m ( ) ( ) j M 22a j Ms ( j s ) = 0 j M 22b (22) 34

35 Lemma 25: Le be an arbrar square submar of A of sze ( n n n n ) + = and ( M M ) = ha m m n hen ( ) { } de 0 Proof: If we consder he las n rows of snce m m n hen we know ha ever row ( n n n h ) + + such = + = and ( M M ) + + has eacl one componen ha s equal o one h Wh column permuaons we can oban an den mar of sze ( ) n n n he lower rgh corner of (noe ha he column permuaons onl affec he sgn of deermnan whle our dscusson s concerned wh he absolue value of he deermnan) e ( ) n + n n + : n + n n + n n + n n + : n + n n + n = I (23) n n Moreover he frs n + nn columns of are a combnaon of and s columns (snce he and columns form he las n columns of as m + m = n ) I follows ha he las n rows of he frs n + nn columns are all zeros n oher words ( ) n + nn + : n + nn + n: n + nn = 0 n n + n n (24) We now use hese observaons o decompose no blocks of marces o ease he calculaon of s deermnan 2 n nn n nn + + n + nn n de ( ) = de 3 4 n n + nn n n (25) 35

36 Vercall corresponds o and s columns whle 3 T 2 4 T s assocaed wh and columns The rows are dvded no wo blocks one ha has he upper n + nn rows and he oher conans he lower n rows Usng (23) and (24) we can subsue 3 and 4 as follows 2 n nn n nn + + n + nn n de ( ) = de 0n n + nn I n n (26) Usng Lemma 2 we oban ha 2 ( ) ( In n ) ( n + n n n + n n n + n n n In n 0n n + n n ) de = de de (27) Ths epresson reduces o ( ) ( n + n n n + n n ) de = de (28) In (28) s composed of he frs n nn + rows of a se of and s columns onl j ecause he oal number of columns n A s nn and he oal number of s columns n A s also equal o nn (whch s less han n + n n ) hen he columns of canno be composed solel of or of s columns Ne we sud dfferen cases based on he choce of ase : All columns are ncluded n We decompose deermnan as follows: and s columns n = = e m nn ms n no blocks of marces o ease he calculaon of s : s n nn n n de ( ) = de s n n nn nn n (29) 36

37 In (29) he marces T conan all columns whle marces s s T conan all s columns The upper block s conans he upper n rows of Snce all columns are ncluded n hen appl ERO (27) on (29) As a resul we oban de ( ) de( ) = see (25) We n n nn I nn nn nsead of wh = Snce ERO (27) ransform n n n no 0 n n n we oban s s n nn n n 0 n nn n n de ( ) = de s = de s n n nn nn n I nn nn nn n (220) We hen permue he posons of he blocks of block I n n n n s moved o he lower rgh corner of so ha he nverble square e s n n 0 n nn de ( ) = de s n n n I nn nn (22) Usng Lemma 2 we oban ha s s s ( ) ( In n n n ) ( n n 0n n n In n n nn n n ) ( n n ) de = de de = de (222) Snce all he same ndces as some of he columns are ncluded n hen all he s columns have he columns I follows ha from (22a) s a mar where all elemens are zeros ecep for eacl n componens ha are equal o - one n each column I s easl verfed ha dependng on he arrangemens of he - s n n componens n s s de ( n n ) wll be eher 0 or In parcular de ( n n ) s n n wll be equal o onl f here s a - componen n ever row and n ever column of s n n 37

38 Ths wll happen when he ndces j of he Usng (20) (28) and (222) we herefore conclude ha s columns covers he range { n } ( ) f - Ms = 223a s de( ) = de( ) = de( ) n n = 0 oherwse 223b ( ) (223) ase 2: All s columns are ncluded n e ms nn = m = n Decomposng as descrbed n ase we oban s n n n nn de ( ) = de s n n n nn nn (224) where he blocks have szes dfferen from hose n (29) Snce ever s column has a sngle componen equal o one and all s columns are ncluded n hen from (26) we know ha = 0 and s n nn n nn s n n nn I nn nn = Usng hese observaons n (224) we oban s n n n nn n n 0 n nn de ( ) = de s = de n n n nn nn nn n I nn nn (225) We hen appl Lemma 2 o oban ( ) ( In n n n ) ( n n 0n n n In n n nn n n ) ( n n ) de = de de = de (226) Now observe ha s a square mar ha has eacl n componens ha n n are equal o one one n each column Smlar o (223) n ase we wre ha ( ) f - M = 227a de( ) = de( ) = de( ) n n = 0 oherwse 227b ( ) (227) 38

39 ase 3: No all he nor all he s columns are ncluded n e ms + mx = n + nn n ms < nn n m < nn and here s n he s columns ha can be combned wh he of sze ( ) n n n n n he lower lef square corner of mar n ( n : n n n n : n n n ) columns a subse of columns o oban an den mar e we can oban an den ha s solel composed of and s As menoned before ever column has wo componens ha are equal o one Denoe b V he se of ndces of rows ha have he second one componen among he columns n elemen among he s columns n ndces j of Also le V s denoes he se of ndces of rows ha have he one and s ha belong o Snce for an gven M and M s have all he hen usng (25) and (26) we oban j M { } V = n + j (228) and V = { n + j} s j Ms (229) An den mar of sze ( ) n n n n n he lower lef corner of obaned onl f we have eacl one componen n each row k for { } k n n n n + + (regardless of wheher s assocaed wh words we wll see such an den mar f { n n n n } ( V V ) can be or s ) In oher + + = (230) s 39

40 presen n When creang he above-menoned den mar we frs selec all columns and hen add s columns when needed Defne V s o be he se of ndces of he rows ha have no one componen assocaed wh an of he We need a s column for each of hese rows where { } columns V = n + n + n n V (23) s or all we defne M s M o be he se of ndces j of he s columns ha s wll be combned wh he denoe b columns o oban he aforemenoned den mar and M s he se of ndces j of he remanng s columns Noe ha Ms = Ms Ms urher le m s denoe he sum of he cardnal of M s e m s = M We have ha m + m = n n s s Le { m } v v Vs for { } s k m s v k denoe an nde of a row ha has no one componen assocaed wh an of he columns or ever v we can use (26) k o deermne he ndces and j of a s column ha gves one componen n he row wh nde v k Those s columns wll be seleced o complee he desred den mar and hence he are assocaed wh M Le s v j M s where = j = v n v Vs n (232) We perform column permuaons and decompose blocks conan he s columns of composed of he remanng s columns n such a wa ha he lef M s and all he columns whle he rgh block s 40

41 de s s n nn n n ( ) = de s s nn nn nn n (233) We hen appl ERO (27) on (233) so as o oban nsead of Gven he condon ha was se n he defnon of ase 3 and our above selecon of he columns we have ha = I and s nn nn nn nn = 0 and so s n nn n nn s 0 n nn n n de ( ) = de s In n nn nn n (234) We hen follow he seps ha we used n ase The resul s smlar ecep ha because no all of he columns are ncluded n we canno clam ha (22a) hold I ma be ha some of he s columns share ndces wh he columns (22a) or he reverse case (22b) In he number of componens ha are equal o - s n n can onl be shown o be less han or equal o n and no more han one nonzero s elemen s presen n a sngle column I follows ha he value de ( n n ) greaer han one If all he s columns ncluded n M s ) share he ndces j and wh columns n s n n canno be (he columns assocaed wh hen we have eacl n s n nn s componens ha are equal o - Then wheher ( n n n ) arrangemens of hese componens e de ( ) de = depends on he f & & 235a = 0 oherwse 235b M = ( j Ms j M ) Ms = ( ) (235) ( ) 4

42 ase 4: No all he nor all he s columns are ncluded n e ms + mx = n + nn n ms < nn n m < nn and here s no wa o fnd n a subse of he s columns ha can be combned wh he den mar of sze ( ) nn nn n he lower lef square corner of oban an den mar n ( n : n n n n : n n n ) composed of and s columns: columns o oban an e we canno ha s solel ompared o ase 3 we have a leas one row ha has no one componens correspondng o an of he nor he s columns { n n n n } ( V V ) + + (236) s In hs case because here s no wa o make = I here s a leas s n n nn nn nn one row n ha has all of s elemens equal o zero and hence s nn nn s sngular e de( ) = 0 Eample 24: or n = 4 and n = 4 we llusrae n gure 2-3 he dfferen cases consdered n he proof of Lemma 25 or each of hese cases we gve an eample of a submar of A sasfng he correspondng condons Submar s obaned b selecng he columns marked wh ( ) Proposon 2: In he frs hree cases of he proof of Lemma 25 we obaned condons ha make de ( ) = ; (223a) (227a) and (235a) We can use hese condons o consruc unmodular bases of he LPR of ULP Algorhms ULP-UNI- 42

43 (a) = = 2 = 3 = 4 = = 2 = 3 = 4 s (b) = = 2 = 3 = 4 = = 2 = 3 = 4 s (c) = = 2 = 3 = 4 = = 2 = 3 = 4 s (d) = = 2 = 3 = 4 = = 2 = 3 = 4 s (e) = = 2 = 3 = 4 = = 2 = 3 = 4 s (f) = = 2 = 3 = 4 = = 2 = 3 = 4 s (g) = = 2 = 3 = 4 = = 2 = 3 = 4 s gure 2-3 Illusraon of he dfferen cases encounered n he proof of Lemma 25 a) de de 0 de = d) ase ase ( ) = b) ase ( ) = c) ase 2 ( ) 2 de ( ) = 0 e) ase 3 de ( ) = f) ase 3 de ( ) 0 de ( ) = 0 = and g) ase 4 43

44 ULP-UNI-2 and ULP-UNI-3 (correspondng o ases o 3) descrbe how o selec varables s and wh ndces and j ha correspond o M Ms M and M o oban unmodular bases Algorhm ULP-UNI- Inpu: n and n Oupu: Ses of ndces of varables s and ( M Ms M and M ) ha eld a bass wh de ( ) = : M = 2: le 3: M 4: M = M Ms Algorhm ULP-UNI-2 Inpu: n and n such ha Ms = n and Oupu: Ses of ndces of varables s and bass wh de ( ) = : le M 2: M = 3: M 4: M = M Algorhm ULP-UNI-3 Inpu: n and n s such ha M = n and Oupu: Ses of ndces of varables s and bass wh de ( ) = : le M such ha 2: Ms = M 3: le Ms M such ha 4: M = M M 5: M 6: M = M s s s M = Ms = and M = s ( M M M and M ) ha eld a s M = ( M M M and M ) ha eld a M s s = n 44

45 24 ase 3: m + m > n Lemmas and 25 dscuss suaons where m + m n We now nvesgae submarces for whch m + m > n When performng hs sud suffces o consder suaons where ( ) M M = oherwse s sngular; see Lemma 24 In he followng dscusson we wll have o consder era columns correspondng o he and varables as compared o he cases we consdered above We frs consder he effec of ERO (27) on such columns and When m + m > n we wll frs permue he columns o oban an den mar of sze ( ) n n n he lower rgh square corner of We chose hs den mar o be composed of all he columns supplemened b columns when necessar Le M denoe he se of ndces of he columns ha wll supplemen he columns and denoe b M he se of ndces of he remanng m denoe he cardnal of and Ne we selec all he M and M e an den mar of sze ( ) columns nall le m and M = M (237) M = M M (238) m = M m = M (239) columns and add f needed some s columns o oban n n n n as we dd n ase 3 of he proof of Lemma 25 (we use he same noaon here) Then we permue he columns agan so ha he fnal 45

46 arrangemen of he columns from lef o rgh s as follows; he columns ha are assocaed wh M he s columns ha are assocaed wh M s he and he s columns correspondng o M s hen he columns and he columns correspondng o M gure 2-4 descrbes he fnal arrangemen of columns The ses of columns and he number of columns n each se are gven n he frs and second rows respecvel assocaed wh M hen s correspondng o ( ) m m m n M s s correspondng o and all + s s = s M s assocaed wh and all m + m = n n m + m = n gure 2-4 nal arrangemen of columns ncluded n f m + m > n The frs n componens n ever column are zeros There are eacl n componens equal o - and eacl one componen equal o one n he res of he column; e M { } ( ) f h n + k k 240a ( h) = f h= n + nn + 240b 0 oherwse 240c ( ) ( ) (240) or ever column n ERO (27) subracs a row ha has - componen (240a) from a row ha has a zero componen (240c) ERO (27) s appled onl o he rows wh ndces j ha corresponds o leads o a specfc srucure of he frs n rows of he he submar G n m represen he upper n rows of he columns n e j M see (28) Ths columns afer ERO (27) Le columns n or all M he srucure of he submar G n m wll reflec he columns ha have been 46

47 ncluded n as follows; for ever for ever ha s no n e j M column ha s n e j M ( ) ( ) G j = and G j = 0 and n hs case s s n e ( ) M G h ( ) ( ) f h M 24a = 0 oherwse 24b (24) varables I should be noed ha he column arrangemen gven n gure 2-4 locaes he n wo dfferen pars of he mar Therefore a vercal decomposon s 2 also appled o G We denoe b G and G he submarces made of he frs n rows of he n m n m n m columns correspondng o M and o M respecvel urher he frs n columns of see gure 2-4 are composed of and s columns ha are assocaed wh M We defne submar Ln m m o denoe he frs n rows of he s columns correspondng o s s s M s (he s columns par of he frs n columns of ) learl he upper lef square corner of of sze ( ) as ( ) n m n ms ms n n wll be wren : n : n = G L (242) rom (22) we know ha afer applng ERO (27) he s columns ma now have a - or zero componen n he frs n rows I follows ha he number of - componens n L sa r wll be less han or equal o ms ms (he number of s n ms ms columns) snce no more han one - componen can be presen per column If r < m s ms hen a leas one column n Ln m m s dencall zero and herefore ( : : ) s s n n s sngular Smlarl f r = ms ms bu we fnd more han one - 47

48 componen n he same row hen ( : n: n ) s sngular Therefore ( : n: n ) can onl be nonsngular f r = ms ms and ever - componen has a unque row nde Observe ha M s gves he ndces j ha correspond o he s columns n L and hence he ndces of he rows of Ln m m ha have - componens n ms ms s s (dependng on he case ha holds 22a or 22b) Therefore = H M s represens he ndces of he rows of Ln m m ha do no have - componens Snce s s we are onl neresed n he case where ( : : ) n n s nonsngular we know ha r = m m and follows ha s s Ms = ms ms and ( ) H = M = n m m = m s s s Ne we decompose he submar ( : : ) n n n (242) horzonall such ha s upper rows are assocaed wh M s and he lower rows are composed of he rows correspondng o H e ( : : ) n n G L Ms m Ms ms ms Gms ms m L ms ms ms ms = = Gmm Lmm s m G L s H m H ms m s (243) where we denoe b D mm he submar composed of he rows of ndces n H We ne llusrae our noaon on an eample G ha have Eample 22-connued: Applng ERO (27) on he bass gven n gure 2-2 produces he followng sequence of mar operaons n m 48

49 = ( ) ( ) ( ) 5 = ( ) ( ) ( ) = ( ) ( ) ( ) = ( ) ( ) ( ) 2 9 = ( ) ( ) ( ) = ( ) ( ) ( ) = ( ) ( ) ( ) 3 3 = ( ) ( ) ( ) = ( ) ( ) ( ) = ( ) ( ) ( ) = ( ) ( ) ( ) = ( ) ( ) ( ) The resul can also be obaned b applng (29) Marces ER o 4 ER are shown n gure 2-5 (a) o (d) Afer column permuaon s done accordng o he order of gure 2-4 s gven n gure 2-6 where he shaded area represens submar G gure 2-7 hen llusraes how he srucure of he submar G n m reflecs he n m selecon of he columns n Usng he smbols nroduced above we wre ha M = { } M = M = M M = { } s 234 s 234 and M = s M = { } M = { } H = M = { } s 234 ecause M = hen G = G urher M = s ( ) : n : n = G = G and D mm 0 0 = G =

50 (a) ER = (b) 2 ER =

51 (c) 3 ER = (d) 4 ER = gure 2-5 Illusraon of how o appl ERO (25) usng (27) n Eample 22 a) ER c) ER 3 and d) ER 4 2 ER b) 5

52 = s gure 2-6 n Eample 22 afer column permuaons n accordance wh gure G 0 0 = 0 0 = G = = 2 = 3 = gure 2-7 Submar G reflecs he seleced columns n Eample 22 We llusrae he same resul on anoher eample where he number of facles and he number of cusomers are dfferen Eample 25: or n = 5 and n = 3 we show n gure 2-8 (b) he mar correspondng o he submar obaned b selecng he columns marked wh ( ) n gure 2-8 (a) Usng he smbols nroduced above we have 52

53 M = { } M = { } M = M M = { } M s = { 345 } and M = { } s H = M = { } s The upper n rows of he frs n columns of can be compued o be ( : : ) ( : n: n ) = n n decomposes no he submarces G and L correspondng o he and he s columns G = and L = In gure 2-8 (b) he submar G s shaded nsde of We know ha n m G G G = 2 where G conans he columns assocaed wh he varables wh ndces for M and varables e 2 G conans he columns assocaed wh he remanng G = 0 0 and G 2 = 0 53

54 nall he submar D mm s composed of rows of G ha have ndces for H G G ms ms m = = 0 Gmm 0 and D = G = 0 mm mm We now use he above dervaon o compue he deermnan of bases of he LPR of ULP Lemma 26: Le be an arbrar square subma of A of sze ( n n n n ) ha m m n + > and ( M M ) ( n + n n + n n + n n + n ) = ( Dmm) de de Proof: The condon ( M M ) mar of sze ( ) = If s nonsngular hen + + such = ensures ha we can oban an den n n n he lower rgh corner of possbl afer permung columns Ne we permue he columns n accordance o gure 2-4 and appl ERO (27) There are wo cases ha are smlar o ases 3 and 4 n he proof of Lemma 25: ase : There s n a subse of he s columns ha can be combned wh he columns o oban an den mar of sze ( ) ( ) n n n n n n + : n + n n n + : n + n n e we can oban an den mar n ( : : ) n + n + nn n + n + nn ha s solel composed of and s columns 54

55 (a) = = 2 = 3 = = 2 = 3 s (b) = s s gure 2-8 Illusraon of Eample 25 a) Mar and b) Mar We decompose he columns of no he hree blocks descrbed n gure 2-4 We also decompose he rows of no hree blocks; he frs n rows he mddle nn rows hen he las n rows We oban ha s s n n n nn n n s s de ( ) = de n n n nn nn nn n s s n n n nn n n (244) 55

56 sze ( ) ecause our frs sep was o permue he columns o oban an den mar of n n n he lower rgh square corner of we know ha = I Also n n n n s s snce he block n n n s composed of and s columns onl hen n n n = 0n n n ; see (25) and (26) urher gven he condon ha was se b he defnon of ase (refer also o s (230) o verf when hs condon holds) hen = I The s columns n nn nn nn nn he mddle vercal block of ha are correspondng o M have no ndces shared wh an of he columns (oherwse we would no be able o oban he den s s mar = I ) and herefore we know from (22b) ha n n n = 0n n n nn nn nn nn The epresson (244) smplfes o s s n n 0 n nn n n s de ( ) = de n n n I nn nn nn n s n n 0 n nn I n n (245) Now we group some of he blocks and nroduce he followng noaon o ease he compuaon of he deermnan of We wre s n n 0 n nn n n 2 n n s n nn + n de ( ) = den de 3 4 n n I nn nn nn n = n n + n n nn + n nn + n s n n 0 n nn I n n (246) where = (247) s n n n n 56

57 ( ) + = 0 (248) 2 n nn n n nn n n 3 nn + n n s nn n = s n n (249) and 4 nn + n nn + n In n nn nn n = 0n nn I n n (250) Applng Lemma 2 on (250) we oban 4 ( n ) ( ) ( ) n n nn n I n n I nn nn nn ni n n0 n nn de + + = de de = (25) Ne we appl Lemma 22 on (250) o oban 4 ( n n + n n n + n ) In n nn nn n I nn nn I nn nn nn ni n n = = 0 I 0 I n nn n n n nn n n (252) We ne appl Lemma 2 on (246) o wre ( ) ( ) ( ) ( ) de = de de (253) nn + n nn + n n n n nn + n nn + n nn + n nn + n n 4 rom (25) we know ha ( n n n n n n ) de + + = Hence (253) reduces o ( ) ( ) de = de n n n nn n nn n nn n nn + n n (254) We now use (248) and (252) o calculae he par of he epresson enclosed n square brackes n (254) Specfcall I 2 4 nn nn I nn nn nn ni n n n ( ) ( ) nn n + nn + n nn + n = 0 n nn n n 0n nn I n n 2 ( 0n n n n n ) n n n + n = = (255) 57

58 Subsung (255) no (254) we oban 2 3 ( ) ( n n n n n + n n n + n n ) de = de (256) rom (248) and (249) we can rewre (256) as s ( ) ( ) ( ) n de de n n s = n de n 0 n nn n n = n n n n n n s n n (257) To compue de ( ) s we ne nvesgae he srucure of n n and n n Submar n n represens he upper n rows of he rgh vercal block n gure 2-4 whch s a combnaon of he columns and he columns assocaed wh M We know ha he frs n componens of he columns are all zeros The are unaffeced b ERO (27) see gure - Alhough he frs n componens of columns are all zeros some of hese componens ma be changed o ones afer applng ERO (27) see (24) In summar n n s composed of m columns ha are dencall zero and have ndces M and of m columns ha ma have nonzero componens and h have ndces M Le u be a (0) column of h n hen can be presened as n n n n n n ( ) h ( ) ( ) 0n h M 258a = un h M 258b (258) s Submar n n s formed from he lower n rows of he lef vercal block n gure 2-4 I s composed of he columns assocaed wh M and he s columns correspondng o M s orm (26) we know ha he las n componens of he s 58

59 columns are zero and wll reman unchanged afer applng ERO (27) Also referrng o (240b) we observe ha he las n componens of he columns onl have one nonzero componen ha s equal o one and wll no be affeced b ERO (27) In s concluson n has ( m m ) n s columns ha are dencall zero and have ndces j s j M and also has m columns ha have eacl one componen equal o one s and have ndces M If h s e denoe a column of n n h n whose componens are all zero ecep for one componen ha s equal o one hen we wre s n n ( ) h ( ) ( ) en h M 259a = 0 n h M 259b (259) Usng he nformaon of (258) and (259) we now compue he produc rom he srucure shown n (259) f { } s n n n n h h2 hm M we conclude ha ( ) ( ) ( ) s n 2 n n n = n n h n n h n n h m 0n 0 n (260) e he produc mar s obaned b selecng he approprae columns of Snce { } n n h h2 hm M and we know from (237) ha M M = s clear ha { h h h } M urher because { } 2 m h h h M hen usng (258) s 2 smple o verf ha ( k ) 0 k { h h h } = We conclude ha n n n 2 m m s n n n n = 0n 0 n 0 n 0 n 0 n 0 = n n (26) Subsung (26) no (257) we oban ( ) ( n n 0n n ) ( n n ) de = de = de (262) 59

60 Mar s formed from he frs n rows of he lef vercal block of gure 2- n n 4 The columns of are a combnaon of he columns assocaed wh M and n n he s columns correspondng o M We dscussed he srucure of s n he n n secon precedng hs proof We appl he decomposon n (24) usng he same noaon here As menoned prevousl afer applng ERO (27) ever s column ma have a - or zero componen n he frs n rows; see (22) We defne r o be he number of - componens n Ln m m If r < ms ms hen a leas one column n Ln m m ha s s s dencall zero and herefore ( : : ) s s n n s sngular Smlarl f r = ms ms bu we fnd more han one - componen n he same row ( : : ) ( : : ) n n s sngular Therefore n n can onl be nonsngular f r = ms ms and ever - componen has a unque row nde Ths case holds onl f all he s columns ha are correspondng o ndces wh he ( m m m m ) M s share columns The s columns wll have an den mar of sze wh - coeffcens n he rows ha have ndces j j M s s s s herefore L = I The remanng rows (ha have ndces j ms ms ms ms ms ms ms ms s j H M = ) are dencall zero L 0 s = Now (243) can be rewren as mms ms mms ms G I = ms ms m ms ms ms ms n n Gmm 0mm s ms (263) 60

61 We permue he blocks of n (263) so ha he nverble square mar n n s now locaed n he lower rgh corner Ims ms ms ms n n Gmm 0 mm s ms = Gms ms m Ims ms ms m s (264) wre We ne appl Lemma 2 on (264) o oban ( n n ) ( Im m m m ) ( Gm m 0m m m Im m m m Gm m m) de = de de + (265) s s s s s s s s s s s s Noe ha we prevousl denoed G mm b D mm ( ) ( n n ) ( Gm m) ( Dm m) Then usng (262) and (265) we de = de = de = de (366) ase 2: There s no wa o fnd n a subse of he s columns ha can be combned wh he columns o oban an den mar of sze ( ) n n n n n ( : : ) n + n + n n n + n + n n e we canno oban an den mar n ( : : ) n + n + nn n + n + nn ha s solel composed of and s columns In hs case we appl he same re-orderng of he columns as n ase bu we do no perform ERO (27) onsderng nsead of we oban s s n n n nn n n s s de ( ) = de n n n nn nn nn n s s n n n nn n n (267) Usng he same noaon as n (246) we oban 6

62 s s n n n nn n n 2 n n s s n nn + n de ( ) = den de 3 4 n n nn nn nn n = n n + n n nn + n nn + n s s n n n nn n n (268) s Snce we are no able o creae n n n n = In n n n (refer o (236) n ase 4 of 4 he proof of Lemma 25) hen clearl ( n n n n n n ) de + + = 0 urher because we dd s no proceed wh ERO (27) we know ha whch s composed of he frs n componens of he s or n n s columns wll be a mar of zeros e = 0 see n n n n (26b) and (240c) I follows ha n (268) s composed of blocks of marces where he blocks on he dagonal and n n + + have deermnan zero Therefore 4 n n n nn n s sngular Ths concludes he proof 25 onsrucng ULP Insances of Desred Deermnan Theorem 2: Le be an arbrar square subma of A of sze ( n n n n ) ha: m + m > n ( M M ) = + + such e s possble o oban an den mar of sze ( ) he lower rgh corner of and { n n n n } ( V V ) n n n + + = e we can oban an den mar of sze s ( n n n n ) n ( n + n + n n n + n + n n ) Then ( ) ( D) : : de = de urher Algorhm ULP-DET descrbes how o compue he mar D from he submar 62