CS 49 G Cobinatorial Optiization Lctur Nots Junhui Jia. Maiu Flow Probls Now lt us iscuss or tails on aiu low probls. Thor. A asibl low is aiu i an only i thr is no -augnting path. Proo: Lt P = A asibl low is aiu an Q = thr is no -augnting path. It is obvious that P Q. For th cas o Q P using th construction o proo o Ma- Flow an Min-Cut Thor w gt a cut δ(r) with () s = u( δ ( R)) thn cobin th Corollary o () ( )) Thor.2 s u( δ R w know is aiu. So Thor is prov. I u is intgral an thr ists a aiu low thn thr ists a aiu low that is intgral. Proo: St a asibl low which is an intgral low o aiu valu. Fro thor. i thr is no -augnting path is a aiu low. Suppos thr is an -augnting path th contraiction is that is not an intgral low o aiu valu bcaus th nw low coul b intgral whn an u ar intgral. Hnc thr is no -augnting path an ust b a aiu low. So Thor.2 is prov. Corollary.3 I is a asibl (r s)-low an δ(r) is an (r s)-cut thn is aiu an δ(r) is iniu i an only i = or all δ(r) an = 0 or all δ( R). u
2. Th Augnting Path Algorith 2. Th For-Fulkrson procur. Fin an -augnting path P. 2. Incrnt low along th path P. Th For-Fulkrson Algorith provis a tool to in a aiu low an a iniu cut. Whn w cut th stp 2 o th For-Fulkrson procur thr is a aiu valu pritt. It is ε = in(ε ε 2 ) w call ε th -with o P. That is: Th aiu valu pritt = in(ε ε 2 ) whr ε = in( u : orwar in P) ε 2 = in( : rvrs in P) 2.2 Th Auiliary or Rsiual Graph G( ) Dinition: 2. Givn a graph G = V E u an currnt low w in G( ) to b th auiliary graph whn put V ( G() ) = V an vw E( G() ) i an only i vw E an vw < u vw or wv E an wv > 0. I or vw E both vw < uvw an wv > 0 ar satisi thn put two paralll arcs into E( G() ). It is notic that an auiliary graph is un-wight. An apl is givn in Figur.. Obsrvation A (r s) ipath in G( ) is an augnting path. Dining th auiliary graph supplis us a tho or sarching augnting paths. It is obvious as wll that: E 2 ( G() ) E Bcaus thr ar only possibility o orwar an rvrs gs whn you obtain an auiliary graph G( ) ro a graph G th nubr o gs in an auiliary graph G( ) ust b at ost twic o that o original graph G. 2
In orr to iplnt th augnting path crat G( ). This can b on in ( ) having a path ro r to s with Brath-irst sarch tho. O ti p a p a 3 3 r s r s 42 42 q b q b 0 Figur. A asibl low o valu 3 an its auiliary igraph G( ) Notation: In igur. th lt han nubr stans or th capacity o that g an th right han nubr stans or th low through that g. Howvr th augnting path algorith cannot b consir accptabl. Figur.2 shows a ba apl or this algorith. I vry augnting path uss arc ab thn ach augntation will b o valu th algorith will nvr trinat an thr is no aiu low. r a s b Figur.2. A ba Eapl or th augnting path algorith Th running ti For gs an n vrtics Unior cost ol: 3
Input siz = + n + 3 It is unboun bcaus th ratio = a a 3 is inpnnt o. Th total ti takn = Log cost ol: Lt vry g has capacity o a thn w hav Input siz = + n + log a a 3 log 3 log a a whr a is th running ti choos a = 2 thn w hav It is an ponntial unction. 3. Eons-Karp Ia 3 2 2 O 2 3. Introuction to th Shortst -augnting path Dinition: 3. A shortst augnting path is an -augnting path which has th iniu possibl nubr o arcs. Dinition: 3.2 ( u v) is th lngth o th shortst path ro u to v in G( ). Consir a typical augntation ro low ' to trin by augnting path P r v b th last lngth path ro r to which hav no squnc as v 0 v... v k. Lt ( ) v in G( ) o a (r v) ipath thn w hav ( r v ) i an ( v s ) k i i o ( ' ) v w or so i. i = = v i G is not an arc o G( ) thn v = i Qustion: How to iplnt in O() ti? =. I an arc vw Brath-irst sarch will accoplish it in O() ti. So that it is asir to in a shortst augnting path than to in an augnting path. This givs us a polynoial-ti algorith or th aiu low probl. La 3.2 For ach v V w hav that 4
( r v) ( r v). ' Proo: Lt a no v b a vrt or which th la is als. That is ( r v) ( r v) ' < Choos th no v such that an v is th vrt or which ' ( r v) violating vrtics. Lt P ' b a (r v) ipath in G ( ' ) o lngth ' ( r v) no bor v. Thn ( r v) > ( r v) = ( r w) + ( r w) ' ' + is th sallst all such an lt w b th Now W can conclu that wv is not an arc o G( ) sinc i wv is an arc o G( ) thn Suppos thr is an arc in ( ' ) G( ). W hav G ( r v) ( r w) + but not in G( ) thn th arc ust hav bn rvrs in w = an v v v i = i A contraiction cos up iiatly sinc i i + So th La 3.2 is prov. Likwis w can prov that La 3.3 I ( r s) ' ( r s) ( v s) ( v s) ' = Lt E ~ () ={ E : is an arc o a shortst -augnting path} ~ ~ E ' E. Thn ( ) ( ) ~ ' an Proo: Pick so gs that satisy E ( ) ~ E Lt ( r s) = k inucs an arc vw in G ( ' ) an ' ( r v) = i ' ( w s) = k i () so i whr v = v i an w = vi as shown in Figur 3.. or 5
r v w P s G ( ' ) Figur 3. Illustration or th proo o E( ' ) E( ) ~ ~ Notation: In igur 3. th only g that can hav capacitis chang arc ust along path P. Th othr paths ar actly th sa as thos o G( ). Thror ( ) ( ) r v + w s k = ( r v ) ' + ' ( w s) = ( i ) + ( k i) ~ ' ( E() ) is not tru thn but P ~ Thror vry g in E ( ' ) ust b in E ~ () that is ~ ~ E ( ' ) E( ) Suppos this in G( ). This is a contraiction. ~ Now lt s prov E ( ' ) is th propr subst o E ~ (). Consir th g : vw that is saturat. I it ust b us on anothr shortst path th irction ust b rvrs at what it is on G( ). It is illustrat in Figur 3.2. what w n to in out is th critical g o th path P th lngth o any path using it is at last k+2 an it os not blong to ~ E '. Thn th La 3.3 woul b prov. ( ) It is obvious that or th original graph w hav ( r v) = i ( w s) = k i an r w ( ) = i ( r w ) i ' ( v s) = k i + ' ( v s) k i + So th lngth o P will b at last 6
whr ( r s) k =. = i = i i + k i + + = k + 2 0 ( r w) = i + ( v s) = k i P = V = r v " v v w v + " s = v So La 3.4 is prov. k r v w P s G ( ' ) ~ Figur 3.2 Illustration or th proo o E ( ' ) is th propr subst o E ~ (). Rrncs. Willia Cook Willia H. Cunningha Willia R. Pullyblank an Alanr Schrijvr. Cobinatorial Optiization. John Wily & Sons INC. 998. 7