Basic Polyhedral theory

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Bertha Hopkins
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1 Basic Polyhdral thory Th st P = { A b} is calld a polyhdron. Lmma 1. Eithr th systm A = b, b 0, 0 has a solution or thr is a vctorπ such that π A 0, πb < 0 Thr cass, if solution in top row dos not ist thn thr ists a π such that th bottom is tru. (in ach column, continuous). A = b A b A b π A 0, πb < 0 π A 0, πb < 0 π A 0, πb > 0 If { A b, intgr } = { C d, intgr} and { A b } { C d }, w say that th first formulation is Tightr (& bttr). Howvr w r intrstd in intgr solutions. Intgral polyhdra Givn a polyhdron P, its intgr hull P I is th conv hull of all intgral vctors in P. For any rational polyhdron P, th P I is also a polyhdron. Th following mthod gnrats valid inqualitis for P I (Chvatal s thorm) Lt P= { A b, 0}, whr A and b intgral, thn µ A µb is valid for P, for any µ 0, and µ A µ b is valid for P, bcaus 0, thn µ A µ b (1) is valid for P I bcaus is intgral. Lt us call S 1 th st of inqualitis using (1), and st S t+1 th st obtaind by applying this opration to S t 0 1. Ths sts gnrat a squnc P = P P... PI Eampl: X(1)+(4) +(6) =1 X(1)+(5)+(7) =1 X(3)+(4)+(8) =1 X(3)+(5)+(9) =1 X(2)+(7)+(9) =1 X(2)+(6)+(8) =1 Considr taking ½*(first four inqualitis) to gt (1/2+1/2)X(1)+ (1/2+1/2)(3)+ (1/2+1/2)(4)+ (1/2+1/2)(5) + 1/2 ((6)+(7)+(8)+(9))=2 1 1 X(1)+ 1 (3)+ 1 (4)+ 1 (5) + 2 ((6)+(7)+(8)+(9))= 2, 1 1 Whr 2 =0 so 2 ((6)+(7)+(8)+(9))=0 X(1)+(3)+(4)+(5)=2 Considr taking last thr inqualitis and first on with multiplir of ½

2 X(2)+(6)+(9)=2 Total unimodularity: If A is totally unimodular (TU) thn n P(b)= { R + m A b} is intgral b Z for which it is not mpty Eampl: A matri with (0,-1,1) lmnts, whr no mor than 2 non-zro lmnts in ach column and a 0 if column contains two non-zro lmnts, thn A is TU. i i, = Eampl: A is TU if dtrminant of ach submatri of A is (0, -1,1), and A is (0,1,-1) matri. Duality Primal problm: n+ z LP = ma{ c A b, R } Dual problm: m+ w LP = min{ ub ua c, u R } Wak duality * is primal fasibl, u * is dual fasibl, thn c * * zlp wlp u b Strong Duality If z LP or w LP is finit thn both P and D hav finit optimal valu z LP = w LP Wak dual of IP is any minimization problm zd = min{ zd( u) : u SD} that satisfis n zd( u) c, S, u SD, whrs = { Z, A b} and S D is dual intgr problm. If Dual problm is fasibl thn zip zd. Strong dual of IP is a wak dual that also satisfis if S 0, boundd from abov, u SD, S, zd( u ) = c Duality gap: for wak duals of IP, th sparation btwn primal and dual obctiv valus. zip Primal solution Wak duality Dual solution Primal solution Strong Duality

3 Totally Dual Intgrality Givn A and b with intgral ntris, w want to study cass whn th LP Ma c A <= b Has intgr optimum primal and intgr optimum dual solutions. Eampls: a) ntwork flow ma c A =b >=0 whr ach column of A has two non-zro ntris. On with valu 1 and othr with valu -1. ntwork flow matri b) ntwork flows with uppr bounds 0<=<=u c) flow problms in cross-fr familis [Nmhausr & Wolsly ttbook] d) submodular flows i. Dirctd cut covrs, matroids ( u) ( S ) = 1, u 1, S odd Prfct Matching problm: Th 2 nd inquality is ncssary in ordr to obtain intgral solutions. Without this inquality, fractional solutions ar possibl. Considr th graph : With only 1 st constraint you ll gt a possibl fractional solution which looks lik: Edgs shown hav 0.5 valus in solution. Dashd arrows ar th dgs which must b usd in inquality (2) to cut off this fractional solution. 0.5 on ach dg

4 But how do w automatically dtrmin th form of inquality (2) to cut off th currnt fractional solution? This is calld th Sparation Problm. For som problms lik matching, w know how to solv this problm using Branch and cut. For othr problms, w still do not know how to solv it, although som huristics ist but thy ar not guarantd to obtain a cut. For matching problm, w r givn a fractional vctor, w want on st S such that < 1, for S odd. ( S ) So for a st S, w can us th min cut problm: Givn a graph G=(V,E) and dg wights, w 0, w know how to solv th minimum cut problm: Min s w (S ) This can b solvd by ntwork flow tchniqus. If S is not odd, what do w do? Supos w solv a minimum cut problm with wights w =. Lt S b a solution. If S is odd, w ar don. If S is vn w hav to prov that thr is an odd st in S, includd in S or in V\S that solvs our problm. Using this proprty, w can dcompos th problm. Proof: W know that th following is tru. S T w ( S T ) T) o T long dg counts only in th right hand sid of th inquality. Suppos that S ( S is vn) givs a minimum cut and T givs a minimum cut with T odd. Now w ( S T ) T) is o+<=+o If w hav that w( ), w( T ) S T ), bcaus T and S T ar odd. This st of inqualitis imply w ( S T ) = w( T ) and w( = w( ), w( T ) = w( S T )

5 Proof. Thrfor S T is also a solution for th minimum odd cut problm. So tak union S T or tak complmnt of union V\ S T (that is includd in V\. Othr possibility: w ( S T ) T ) is + o + o. o o T Thn w ( S T ) = w( T ),bcaus w hav w( S T ), w( T) ) so thrfor w( T ) = w( ) Proof. In this cas is also a solution to th minimum odd cut problm. = if S dfins a minimum cut & S vn, thn solution ists in S or in V\S. W dcompos problm into two problms in th graphs S V\S W gnrat a nstd st of problms, that has at most 2n-1 lmnts. So in worst cas w hav to solv 2n-1 min cut problms. Finding on minimum cut taks O(n^3), so this is an O(n^4) algorithm. ( u) 1, u Non-Prfct Matching problm: S 1, S odd E ( S ) 2 whr E( ar th dgs with both nds in S. This can also b solvd similar to th prfct matching problm, using mincut algorithm. Othr problms, w know what som of th (facts) inqualitis look lik, but w can t solv th sparation problm: Nod or Vrt Packing Problms: Givn a graph G=(V,E) considr u + v 1 ( u, v) E 0 u V u

6 for dfinition of th problm. Howvr for obtaining intgral solutions w nd to add mor inqualitis (spcifically facts, which intrsct intgr vrtics of sarch spac). 1 K (if subgraph is complt) u V u C 1 u C (whr C is an odd cycl) u C 2 Howvr thr ar mor facts which ar difficult to obtain and for which w don t know all of thm. Thr ar lifting lmmas on can us to obtain mor facts of this problm. P = { a N b,0 Knapsack problm: 1},0 a b, N w nd facts of P I 0, 1 N dfin facts Lt b an intgr vctor in P, w say that st S = { = 1} is indpndnt. Lt C b a minimal dpndnt st C\{i} is indpndnt i C Th inquality C 1 (1) C is valid for P I Assum that a1 a2... an ar givn. A minimal dpndnt st C, dfin E(C) = C { k : ak a, C} Th inquality C 1 (2) E( C) is also valid for P I and strongr than (1). Th inquality (2) dfins a fact of P I undr any of th following conditions: a) C=N b) E(C)=N & (i) C\{ 1, 2 } {1 } is indpndnt c) C=E(C) and (ii) C\{ 1 } {p} is indpndnt, p = min{ N \ E( C)} d) C E( C) N & (i) & (ii) Whr C={ 1, 2,, k } EXAMPLE: P ,0 1} = { C 1 ={1,2,3} C 2 ={1,2,4,5} C 3 ={1,3,4,5} C 4 ={2,3,4,5} ar minimal dpndnt sts Now, E( C 1 )= C 1 E( C 2 )= C 2 E( C 3 )= C 3 E( C 4 )= C 4 {1 }

7 Lts s if any of th nw inqualitis of th form (2) ar facts: E( C 1 ): (1) is a fact if (ii) C\{ 1 } {p} = {2,3,4} is indpndnt, p = min{ N \ E( C)} Thrfor it s a fact *. E(C 2 ): (2) is a fact if (ii) C\{ 1 } {p} = {2,4,5,3} is indpndnt. It is not indpndnt, thrfor w do not know whthr it is a fact or not. E(C 3 ): (3) is a fact if (ii) C\{ 1 } {p} = {3,4,5,2} is indpndnt. It is not indpndnt, thrfor w do not know whthr it is a fact or not. E(C 4 ): (4) is a fact if (cas b)) (i) C\{ 1, 2 } {1 } = C\{2,3} {1 } = {4,5,1} is indpndnt. Is is indpndnt, thrfor it is a fact *. NOTE: any inquality which is a linar combination of facts, cannot b a fact. (3) = (4) + ( 2 0 ), thrfor (3) is not a fact. (bcaus it s a +v combination of 2 facts). (2) = (4) + ( 3 0 ), thrfor (2) dos not dfin a fact.

cycle that does not cross any edges (including its own), then it has at least

cycle that does not cross any edges (including its own), then it has at least W prov th following thorm: Thorm If a K n is drawn in th plan in such a way that it has a hamiltonian cycl that dos not cross any dgs (including its own, thn it has at last n ( 4 48 π + O(n crossings Th