Graph Searching & Perfect Graphs

Transcription

1 Grph Serhing & Perfet Grphs Lll Moutdid University of Toronto Astrt Perfet grphs, y definition, hve nie struture, tht grph serhing seems to extrt in, often non-inexpensive, mnner. We srth the surfe of this elegnt reserh re y giving two exmples: Lexiogrphi Bredth Serh on Chordl Grphs, nd Lexiogrphi Depth First Serh on Coomprility grphs. 1 Introdution Here s one prtiulr wy to ope with NP-hrdness: Suppose we know some struture out the ojet we re deling with, n we exploit sid struture to ome up with simple (hmm...?), effiient lgorithms to NP-hrd prolems. One exmple of this struture is the intervl representtion most people re fmilir with, where in suh setting we were le to solve the independent set prolem in liner time, wheres the independent set prolem on ritrry grphs nnot even e pproximted in polynomil time to onstnt ftor (unless P = NP). Before we delve into the topi, let s rell some definitions. Let G(V, E) e finite nd simple (no loops nd no multiple edges) grph: The neighourhood of vertex v is the set of verties djent to v. We write N(v) = {u uv E}. The losed neighourhood of v is N[v] = N(v) {v}. The omplement of G, denoted G(V, E) is the grph on the sme set of verties s G, where for ll u, v V, uv E uv / E. A indued sugrph of G is grph G (V, E ) where V V nd u, v V, uv E uv E. A yle C k = v 1, v 2,..., v k in G is n indued sugrph where for i [k], v i v i+1 edges present in C k. mod k re the only A lique is set S V where u, v S, uv E. The lique numer of G, denoted ω(g), is the size of the lrgest lique in G. An independent set is set S V where u, v S, uv / E. The independent set numer of G, denoted α(g), is the size of the lrgest independent set in G. These notes re of guest leture I gve for the Advned Algorithm Design ourse. 1

2 A proper olouring is funtion ol : V {1, 2,..., k} tht ssigns vlues i [k] suh tht uv E, ol(u) ol(v). The hromti numer of G, denoted χ(g), is the minimum numer of olours needed to properly olour G. A lique over of G, is prtioning P 1, P 2,..., P k of the verties of V where P i is lique. k i=1 P i = V nd i [k], The minimum lique numer of G, denoted κ(g), is the minimum numer of liques needed to over G. A hole is n odd yle on 5 or more verties. An ntihole is the omplement of hole. A grph G is omplete if G is lique. A vertex v V is simpliil if N(v) indues lique. A seprtor of grph G is suset of verties S V whose removl from G disonnets the grph into two or more onneted omponents. An -seprtor is seprtor of G tht disonnets from. A -seprtor S is miniml if no proper suset of S lso seprtes from. Some (ovious) remrks: The omplement of lique is n independent set nd vie vers. The lique numer of grph is lwys lower ound to its hromti numer; nd the lique over numer is lwys n upper ound to the indpendent set numer. To illustrte the definitions ove, onsider the grph elow: d χ(g) = ω(g) = 3 α(g) = κ(g) = 2 The sugrph H(V, E ) where V = {,, } nd E = {, } is not n indued sugrph of G, sine the edge / E. The set S = {,, } is lique in G, S = {,, d} is not. The set {, d} is n independent set, {, d, } is not. A proper olouring of G would e ol() = ol(d) = 1, ol() = 2, ol() = 3, therefore χ(g) 3. Sine ω(g) χ(g) nd ω(g) = 3, it follows χ(g) = 3 is optiml. Perfet Grphs: A grph fmily tht reeived signifint ttention euse of its nie struture is the lss of perfet grphs. Definition 1. A grph G(V, E) is perfet if for every indued sugrph H of G: χ(h) = ω(h) Perfet grphs were introdued y Clude Berge in the sixties, nd hve sine een well studied. In ft, mny NP-hrd prolems n e solved effiiently on perfet grphs using the ellipsoids method, however reserh is still eing developed to ome up with truly omintoril lgorithms. 2

3 Mny grph fmilies elong to the lss of perfet grphs; intervl grphs for instne, permuttion grphs - whih you my hve seen in the longest susequene prolem. We will fous on different grph lss, known s hordl grphs. First, we list two min theorems regrding perfet grphs. Theorem 1 (The Wek Perfet Grph Theorem). A grph is perfet if nd only if its omplement is perfet. Theorem 2 (The Strong Perfet Grph Theorem). A grph is perfet if nd only if it does not ontin n indued hole or n indued ntihole. Grph Serhing: Definition 2. Grph serhing is mehnism to trverse the grph one vertex t time in speifi mnner. Clssil grph serhes you ve seen efore re BFS nd DFS. In the reminder of this leture, we will look t two exmples of two grph serhes pplied on different grph lsses. Setion 2 fouses on Lexiogrphi Bredth First Serh nd Chordl grphs. Setion 3 fouses on Lexiogrphi Depth First Serh nd Coomprility grphs. 2 Lexiogrphi Bredth First Serh & Chordl Grphs 2.1 Chordl Grphs Definition 3. A grph G(V, E) is hordl if the lrgest indued yle in G is tringle. Chordl grphs re sometimes referred to s tringulted grphs s well. Convine yourself tht hordlity is hereditry property. This mens if G is hordl, so is every indued sugrph of G. Theorem 3. Chordl grphs re perfet. Proof. By the Strong Perfet Grph Theorem, it suffies to show tht if G hs no hole or ntihole. It is esy to see tht G hs no holes sine the lrgest yle is tringle. Suppose G ontins n ntihole C. Let C e the omplement of C in G. C is n odd yle with 5 or more verties. Notie first tht the omplement of C 5 is C 5 nd thus C C 5 sine G is hordl. Any other odd yle C k 7 in G must hve two edges, d where nd re oth not djent to nd d. In C, d forms C 4, therey ontrditing G eing hordl. Theorem 4. Every miniml seprtor of hordl grph is lique. Proof. Let G(V, E) e hordl grph. If G is omplete, then lim lerly holds. Suppose G is not omplete. Let S e miniml seprtor of G. Suppose S is not lique. This mens there exists two verties u, v S suh tht uv / E. Sine S is seprtor, G\S hs two or more onneted omponents. Let C 1, C 2 e two of these onneted omponents. Sine u S, there must exist two verties C 1, C 2 suh tht u elongs to n, pth, for otherwise S\{u} is smller seprtor thn S ( ontrdition to the minimlity of S). Similrly, there must exist two verties C 1, d C 2 suh tht v elongs to, d pth. Sine, C 1, let P 1 e n indued, pth in C 1 (onvine yourself suh pth must exist), nd let P 2 e n indued d, pth in C 2. Consider the sugrph indued y P 1, P 2, u nd v. This sugrph is yle of length t lest 4 (when =, = d, we hve C 4 =, u,, v). A ontrdition gin to G eing hordl. Therefore uv E nd S is lique. 3

4 Theorem 5. If G is hordl, then either G is omplete or G hs t lest two non-djent simpliil verties. Proof. The proof is y indution on the size of the grph, i.e. the numer of verties. If V = 1, then G is lerly omplete. Suppose V > 1 nd G is not omplete, then G hs miniml seprtor S. By Thereom 4, S is lique. Let C 1, C 2 e two onneted omponents of G\S. Consider the sugrph G 1 = C 1 S of G, sine hordlity is hereditry property, G 1 is hordl nd y indution hypothesis, G 1 is either omplete or hs t lest two non-djent simpliil verties. Either wy G 1 hs simpliil vertex tht remins simpliil in G (why?). Similrly G 2 = C 2 S is hordl nd hs t lest one simpliil vertex tht remins simpliil in G. Therefore G hs two non-djent simpliil verties. Corollry 1. G is hordl iff every indued sugrph of G hs simpliil vertex. Proof. left s n exerise. 2.2 Perfet Elimintion Orders Given grph G(V, E), nd totl ordering σ = v 1,..., v n of V. Let N (v i ) denote the neighours of v i tht pper to the left of v i in σ. Define N + (v i ) nlogously. N (v i ) = {v j : v j v i E nd j < i} N + (v i ) = {v j : v j v i E nd i < j} For two verties v i, v j suh tht i < j, we write v i σ v j to denote tht v i is the left of v j in σ. We drop the susript if σ is ler in the ontext, nd write v i v j. Definition 4. A vertex ordering σ = v 1, v 2,..., v n is perfet elimintion order - or PEO - if for ll i [n], v i is simpliil to the left in σ, i.e. N (v i ) is lique. σ =, e,, d,, f d e f Figure 1: Exmple of PEO ordering on grph G Surprisingly, one n use PEO to hrterize hordl grphs. Theorem 6. G is hordl if nd only if G hs perfet elimintion order. Proof. Suppose G hs PEO σ ut is not hordl. Let C e n indued yle in G on 4 or more verties. Let w e the vertex of C tht ppers lst in σ. Then w must hve t lest two predeessors in σ (nmely its neighours in C), ll them u, v. Then u, v σ w nd uv / E. A ontrdition to w eing simpliil to the left in σ. 4

5 Conversely, let G e hordl grph. By Theorem 5, G hs simpliil vertex. Cll it v n. Sine hordlity is hereditry property, G v n is lso hordl. An indution on the size of G now onludes the proof. In prtiulr, let σ = v 1, v 2,..., v n 1 e PEO of G v n, then σ = σ v n 1 is PEO of G sine v n is simpliil in G. Applitions: One of the min prolems in struturl grph theory is grph reognition: Given lss of grphs G tht stisfies some sort of struture, nd grph G, is it esy/hrd to hek if G G? Sine PEOs fully hrterize hordl grphs (Theorem 6), hordl grph reognition redues to omputing PEO (or t lest one wy to reognize this grph fmily would e to ompute PEO). Before looking t the omplexity of omputing suh orderings, let s see wht else we n use them for. Consider the following lgorithm: Algorithm 1 GreedyCOL Input: An ritrry grph G(V, E) nd n ordering σ of V. Output: A proper olouring of the verties of G. 1: for i = 1... n do 2: ssign v i the smllest olour not used mong N (v i ). 3: end for Running GreedyCOL on the ordering elow produes 3-olouring where 2-olouring exists. It s in ft esy to onstrut grphs where this lgorithm ehves ritrrily d. Surprisingly, if GreedyCOL is given good ordering, it produes n optiml olouring for ertin grph fmilies, hordl eing one of them, s shown in Theorem 7 elow. σ d ol( ) Theorem 7. If σ is PEO, lgorithm GreedyCOL gives n optiml olouring. Proof. Consider vertex v i in σ. Sine v i hs N (v i ) left neighours, t lest one of the olours 1,..., N (v i ) + 1 is not used mong N (v i ). Therefore the mximum numer of olours used y GreedyCOL is mx N (v i ) + 1. Let v i e the vertex tht hieves mx i N (v i ) + 1 = N (vi ) + 1. We therefore hve χ(g) N (v i ) + 1 (1) i Sine σ is PEO, v i is simpliil to its left in σ nd so N (vi ) forms lique, nd thus ω(g) N (v i ) + 1 (2) We know tht Comining (1), (2), nd (3) we get ω(g) χ(g) (3) χ(g) N (v i ) + 1 ω(g) χ(g) χ(g) = ω(g) = N (v i ) The ontention of σ nd v n 5

6 Therefore lgorithm GreedyCOL produes n optiml olouring on G. Corollry 2. Algorithm GreedyCOL n e tweked to ompute mximum lique on G if σ is PEO. Proof. The proof follows from perfetion. Exerise: Prove tht the following lgorithm omputes mximum independent set if G is hordl nd σ PEO. Algorithm 2 GreedyIS Input: An ritrry grph G(V, E) nd n ordering σ of V. Output: An independent set S 1: S = 2: for i = n... 1 do Notie we re snning σ in reverse order. 3: Add v i to S if none of its neighours pper in S lredy. 4: end for The question remins then: How do we ome up with PEO? Is it NP-hrd to ompute suh n ordering? No! In ft there is simple elegnt lgorithm to ompute suh n ordering in *drum roll* liner time *drum roll*! 2.3 Lexiogrphi Bredth First Serh Lexiogrphi redth first serh (LexBFS) is vrint of BFS, tht ssigns lexiogrphi lels to verties, nd uses sid lels to rek ny ties tht might our. Consider the grph elow for instne. e d σ =, d,,, e π =, d,,, e σ is BFS, verties,, d tied. π is LexBFS, verties nd d re not tied ny more. Figure 2: BFS vs. LexBFS LexBFS would modify BFS to visit verties with stronger previous pull first. In the exmple ove for instne, one, d were visited, we hve N () N (), nd thus we re fored to visit efore euse vertex is pulled y d. This ondition lone - on the size of left neighourhoods - does not tully hrterize LexBFS. In ddition to left neighourhoods, LexBFS tkes into ount the ordering of the verties in σ. In prtiulr, every LexBFS is first BFS, nd thus must respet the redth first ondition efore heking the size of left neighourhoods. Algorithm 3 elow formlly desries this grph serh. 6

7 Algorithm 3 LexBFS Input: A grph G(V, E) nd strt vertex s Output: An ordering σ of V 1: ssign the lel ɛ to ll verties, nd lel(s) {n + 1} 2: for i 1 to n do 3: pik n unnumered vertex v with lexiogrphilly lrgest lel 4: σ(i) v v is ssigned the numer i 5: foreh unnumered vertex w djent to v do 6: ppend(n i) to lel(w) 7: end for 8: end for e d f σ(i) Affeted Verties σ σ(1) = d lel() = lel() = lel(f) = d 5 σ(2) = lel() = 54 nd lel() = 4 d, σ(3) = lel() = 43 nd lel(e) = 3 d,, σ(4) = f lel(e) = 32 d,,, f σ(5) = d,,, f, σ(6) = e d,,, f,, e Figure 3: A step y step omputtion of LexBFS ordering on G strting t vertex d. LexBFS ws introdued y Rose, Trjn nd Lueker [6] to reognize hordl grphs. proved the following theorem: Theorem 8. Let G(V, E) e hordl grph, then LexFS(G) is PEO. In prtiulr they In order to prove Theorem 8, we use the following hrteriztion of LexBFS orderings given y Drgn, Flk nd Brndstädt [3]: Theorem 9. [The LexBFS 4 Point Condition] Let G(V, E) e n ritrry grph, nd σ n ordering of V. σ is LexBFS ordering if nd only if for every triple, if E, / E, then there exists vertex d suh tht d nd d E, d / E. The triple s defined ove is lled d (needy? :) LexBFS triple, nd vertex d privte neighour of with respet to. It is esy to see tht suh vertex d must exist nd e neighour of sine σ is BFS. The ft tht d / E, ensures tht must indeed e pulled first efore, despite the pull of on. Intuitively, this just ptures the lexiogrphi vrint. d Figure 4: The LexBFS 4 Point Condition Using the LexBFS 4 Point Condition, it is now esy to prove Theorem 8. 7

8 Proof of Theorem 8. Let G e hordl grph, nd σ = LexBFS(G). Suppose σ is not PEO. Choose vertex x to e the left most (the first) vertex in σ where N (x) is not simpliil. In prtiulr, let y, z x e two neighours of x suh tht yz / E. Without loss of generlity, suppose z y. By Theorem 9, there must exist vertex w suh tht w z nd wy E, wx / E. If wz E, the qudruple wzyx forms C 4, ontrdition to G eing hordl. Therefore wz / E, in whih se w z y forms needy/d LexBFS triple. Agin y Theorem 9, there must exist privte neighour q of z with respet to y, nd q w. Using the hordlity of G, it is esy to see tht qw / E, thus reting yet nother d triple. We repet the sme rgument over nd over gin, therey ontrditing the finiteness of G. Therefore x is simpliil, nd σ PEO. In the reminder of this setion, we riefly disuss one wy to implement LexBFS in liner time. 2.4 Prtition Refinement Definition 5. Let S = {u 1, u 2,..., u k } e set of elements. A olletion P = {P 1, P 2,..., P k } is prtition of S if for ll i j [k], P i P j = nd k i=1 P i = S. The P i s re lled prtition lsses. Definition 6. Given set S, prtition P = {P 1,..., P k } of S, nd suset T S, we sy tht T refines P if i [k], P i is repled with su-prtitions A i, B i in this order, where A i = P i T, B i = P i \T. This is known s prtition refinement, where T is used to refine the prtitions of S. One elegnt nd effiient wy to implement LexBFS in liner time for ritrry grphs (liner in the size of the grph, so O(m + n) time where m = E, n = V ) is y mens of prtition refinement [4]. The set S is V, nd T is the neighourhood of vertex p, lled pivot. The lgorithm goes s follows: Initilly σ is empty. We strt y letting P = {P 1 = V }. We hoose n ritrry strt vertex s V s pivot, nd use N(s) to refine P. We ppend s to σ. Initilly, we reple V with A 1 = V N(s), B 1 = V \N(s). The new prtition now looks like P = {A 1, B 1 }. Every time pivot p is seleted to refine, it is ppended to the ordering. In generl, given prtition P = {P 1, P 2,..., P k }, nd pivot p, we use N(p) to refine P y reting the following new prtition {P 1 N(p), P 1 \N(p), P 2 N(p), P 2 \N(p),..., P i N(p), P i \N(p),..., P k N(p), P k \N(p)}, while mintining the order of the prtition lsses, nd ppend p to σ. A vertex is eligile to e pivot if nd only if it elongs to the first prtition lss. The lgorithm stops when ll the prtition lsses re either empty or ontin single vertex. Exmple: Consider the grph in Figure 3, we will use prtition refinement to produe the sme LexBFS ordering in the tle of the sme figure. Initilly we strt with one prtition lss, V = {,,, d, e, f}. We hoose n ritrry strt vertex, d nd refine s follows: P 1 = V N(d) = {,, f} P 2 = V \N(d) = {, e} σ = d so fr. Verties in P 1 re ll eligile to e pivots, we hoose vertex s the next pivot nd refine 8

9 P = {P 1, P 2 } s follows: P 1 N() = {} P 1 \N() = {f} P 2 N() = {} P 2 \N() = {e} σ = d,. The new prtition now is P = {{}, {f}, {}, {e}} in this order. Sine ll the prtition lsses re singletons, the refinement is done, nd σ is d,,, f,, e, the sme LexBFS ordering produed in Figure 3. Try to onvine yourself (prove) tht every ordering produed y this refinement is indeed LexBFS ordering of n (ritrry) grph, in prtiulr notie the prllelism of mintining the ordering of the prtition lsses nd ppending lexiogrphi lels. 3 Lexiogrphi Depth First Serh & Coomprility Grphs Next, we ll look t nother exmple of different grph serh used on different grph lss. 3.1 Coomprility Grphs Coomprility grphs re lrge, well studied, grph fmily. It stritly ontins numer of grph lsses inluding intervl grphs nd permuttion grphs. In ft, lssil hrteriztion of intervl grph is the following: Theorem 10. A grph G(V, E) is n intervl grph if nd only if G is hordl nd oomprility. Coomprility grphs re the omplement of omprility grphs. A grph G(V, E) is omprility grph if its edge set dmits trnsitive orienttion. Tht is, there is wy to orient (single diretion) the edges in E, suh tht for every triple,, oriented,, there must exist n edge oriented. Coomprility grphs re perfet, nd thus y the Wek Perfet Grph Theorem, so re omprility grphs. Figure 5 elow gives n exmple of omprility nd non-omprility grph. Convine yourself tht the grph on the left is indeed omprility grph, y oming up with trnsitive orienttion of the its edges; wheres the grph on the right is not. d d e e Figure 5: A omprility grph (to the left) nd non-omprility grph (to the right). Coomprility grphs n e hrterized y oomprility orderings, lso known s umrell-free orderings. In prtiulr, grph G(V, E) is oomprility grph if nd only if there exists n ordering σ of V suh tht for every triple σ σ, if E then either E or E or oth. It is esy to see tht umrell free orderings re preisely trnsitive orienttions of the omprility grph. 9

10 Side note: There is lose reltionship etween oomprility/omprility grphs nd prtilly ordered sets. A prtilly ordered set, or poset, P (V, ) is n irreflexive, ntisymmetri nd trnsitive reltion on the set V. In prtiulr, two elements, V re omprle if or, otherwise they re inomprle, we write. euse of trnsitivity, if three elements re omprle, then. This is preisely wht trnsitive orienttion is on omprility grph. In ft, if G(V, E) is omprility grph, then G together with trnsitive orienttion of E n equivlently e represented y poset P (V, ) where E if nd only if nd re omprle in P. And thus umrell-free orderings n too e represented y posets. This equivlene gives nother wy to solve prolems for these grph fmilies. For more on posets nd order theory in generl, hek Tom Trotter s ourse [leture 13 nd eyond]. The Wiki pge is good strt too. 3.2 Lexiogrphi Depth First Serh One n extend DFS to lexiogrphi version s well, known s lexiogrphi depth first serh. LexDFS (for short) ws introdued in [2], nd hs sine led to numer of effiient lgorithms, espeilly on oomprility grphs. We egin y looking t properties of this grph serh, then give n exmple of its use; nmely ertifying lgorithm to ompute mximum independent set (MIS) on oomprility grphs. Formlly, LexDFS ssigns lels to verties s they re eing proessed, ties re roken using the lels, where verties with the highest lel re hosen first. A similr ide to wht LexBFS does, ut s we will see, the leling tkes into ount the depth spet of the serh s well. Algorithm 4 elow is forml desription of this proess. Algorithm 4 LexDFS Input: A grph G(V, E) nd strt vertex s Output: An ordering σ of V 1: ssign the lel ɛ to ll verties, nd lel(s) {0} 2: for i 1 to n do 3: pik n unnumered vertex v with lexiogrphilly lrgest lel 4: σ(i) v v is ssigned the numer i 5: foreh unnumered vertex w djent to v do 6: prepend i to lel(w) 7: end for 8: end for Running the lgorithm on the grph in Figure 2, gives the following ordering LexDFS(G) = σ =,,, e, d. (4) Notie in prtiulr how we re fored to visit vertex e efore vertex d. LexDFS n too e hrterized y 4 Point Condition, given y [2], whih sys: Theorem 11. [The LexDFS 4 Point Condition] Let G(V, E) e n ritrry grph, nd σ n ordering of V. σ is LexDFS ordering if nd only if for every triple, if E, / E, then there exists vertex d suh tht d nd d E, d / E. Vertex d is privte neighour of with respet to. Intuitively, the theorem shows tht despite vertex hving pull from vertex tht does not hve, sine σ is LexDFS, there must exist vertex lter (deeper?) in the ordering tht pulled first. This vertex is d. The forml proof of the sttement of the theorem is left s n exerise. 10

11 d Figure 6: The LexDFS 4 Point Condition Unfortuntely LexDFS nnot e implemented in liner time - yet - for ritrry grphs; prtition refinement is indeed one wy to do so, ut it requires the reshuffling/sorting of the prtitions, nd this sorting is the ottlenek to liner time lgorithm. However, linerity is hieved for speifi grph fmilies, oomprility eing one of them [5]. Multi-Sweep Algorithms: For the purpose of this (MIS) prolem, we need to introdue wht we ll multi-sweep lgorithms. These re lgorithms tht ompute sequene of orderings 2 σ 1, σ 2,... where σ i is used to ompute σ i+1. This mens, tht if there re dditionl ties etween verties when omputing σ i+1, the lgorithm uses σ i to rek them using some speifi rule. We will fous on the so lled + rule; where ties re roken y hoosing the right most vertex in σ i. Formlly, we write τ =LexDFS + (G, σ), where σ is some ordering of G, nd τ is LexDFS ordering of G tht uses σ to rek ties y lwys hoosing the right most vertex in σ. Consider for instne, the grph in Figure 2. Let σ =,,, e, d e the ordering omputed erlier in (4). A LexDFS + (G, σ) of G is the unique ordering τ = d,,,, e: vertex d ws hosen first euse it ws the right most vertex in σ, ws seond euse it ws the right most eligile vertex in σ, et. One nie property of + sweeps is the preservtion of umrell-free orderings; this mens: Theorem 12. Let G(V, E) e oomprility grph, nd σ n ritrry umrell-free ordering. The ordering τ =LexDFS + (G, σ) is n umrell-free ordering. We ll τ LexDFS umrell-free ordering. For proof of this result, see [1]. umrell-free orderings. In ft, the uthors hrterize ll the grph serhes tht preserve Comining Theorems 11 nd 12, one n dedue the LexDFS C 4 property of oomprility grphs (Figure 7). Let τ e n umrell-free ordering. Let e d triple in τ. This mens τ τ, where E, / E. Sine τ is n umrell-free ordering, it follows tht E. Sine is d LexDFS triple, there must exist privte neighour d of with respet to, suh tht τ d τ nd d E, d / E. The edge d E is now fored, for otherwise, we would hve τ d τ nd E nd oth d, d / E. A ontrdition to τ eing umrell-free. τ: d Figure 7: The LexDFS C 4 Property of oomprility grphs. 3.3 Mximum Independent Set on Coomprility Grphs We now present ertifying lgorithm to ompute mximum independent set (MIS) on oomprility grphs. A ertifying lgorithm is n lgorithm tht, long with solution, produes ertifite to hek if sid solution is indeed optiml. First, let s ome up with nturl ertifite for this prolem. 2 Either LexBFS, LexDFS or ny other type of orderings 11

12 We note tht omputing mximum independent set on oomprility grphs is equivlent to omputing mximum lique in the orresponding omprility grph. Sine oth grph fmilies re perfet, the lique numer is the hromti numer in the omprility grph. So one wy to ertify mximlity of the lique is to give proper olouring tht mthes the size of the lique. Let s look t the following exmple. Let G(V, E) e omprility grph, nd H(V, E ) = G the omplement of G. d d e f e f Figure 8: G omprility grph on the left, nd H the omplement of G on the right The grph in Figure 8 is omprility grph, s witnessed y the following trnsitive order τ =,, f,, e, d. Sine the lrgest lique in G is of size 3, nmely {,, d}, it follows tht the hromti numer of G, χ(g) = 3, nd proper olouring is given y the oloured verties ove. Every olour lss (the set of verties tht reeived the sme olour) forms n independent set in G. Therefore the olour lsses of G form liques in H. In ft, they form lique over of H. Sine χ(g) is minimum, the numer of olour lsses in minimum nd thus the lique over in H is minimum. We thus hve: χ(g) = ω(g) α(h) = κ(h) The lique over numer for ny grph is lwys n upper ound on the independent set - sine one n ollet t most one vertex from eh lique to ple in the independent set, nd these olleted verties re not neessrily pirwise independent. Therefore one wy to ertify the optimlity of the MIS lgorithm is to produe n independent set nd lique over of equl size. We n do this euse oomprility grphs re perfet. This is preisely wht we will do. Algorithm 5 Mximum Independent Set in Coomprility Grphs Input: A oomprility grph G(V, E) nd n umrell-free ordering σ. Output: An independent set S of G. 1: τ = LexDFS + (G, σ) 2: Sn τ right to left nd greedily dd verties to S tht do not lredy hve neighours in S. Rell tht greedy olletion of n independent set is the proess of snning the ordering -in this se right to left- nd pling vertex in S s long s none of its neighours re lredy in S. We illustrte Algorithm 5 on the grph H in Figure 8 (redrwn differently) elow (Figure 9). Notie (or hek for yourself) tht σ is indeed n umrell-free ordering, ut not LexDFS ordering (doesn t need to e) - in prtiulr σ σ f is d LexDFS triple. The lgorithm egins y pling (right most in τ) in S, this removes f nd, the right most ville vertex in τ is. Thus S, this removes e, then the right most eligile vertex is d. Thus S = {,, d}. 12

13 f e d σ =,, f,, e, d τ = LexDFS + (H, σ) = d, e,,, f, S = {,, d} Figure 9: An illustrtion of the lgorithm. Theorem 13. Algorithm 5 produes mximum independent set of G. Proof. Let τ = u 1, u 2,..., u n, nd let S = v 1, v 2,..., v k e the verties in the independent set where v k τ v k 1 τ... τ v 2 τ v 1 = u n. For every i [k], let T i e the set of verties in τ from v i (inluded) up to ut not inluding v i+1 ; T i = {u j v i+1 τ u j τ v i } {v i }. Clerly S is n independent set. Insted of proving the optimlity of the S, i.e. tht it is of mximum size, we ll produe lique over of equl rdinlity. In prtiulr, we lim tht for ny v i S, T i forms lique. If this is true, then eh v i elongs to lique in G, the T i s form lique over, nd thus S = κ(g). Therey ompleting the proof. Suppose there exists vertex v i suh T i is not lique. Let, T i e two verties suh tht τ τ v i nd / E. Notie tht v i is different thn oth nd, otherwise we ontrdit the hoie of v i+1. This triple v i forms d LexDFS triple in τ where / E, v i, v i E. By the LexDFS 4 Point Condition, there must exist vertex d suh tht τ d τ nd d E, dv i / E. However τ d nd dv i / E ontrdits the onstrution of T i nd thus the hoie of v i+1. Therefore E nd T i is lique. An lgorithm from the ook. A proof from the ook. :) To go k to the exmple in Figure 9, the lique over we get is {, f}, {,, e}, {d}. For the urious mind: Multi-sweep lgorithms hve led to numer of elegnt results in lgorithmi nd struturl grph theory. Below is list of some other prolems solved using grph serhing on vrious grph lsses - emil me for referenes. Grph reognition for numer of grph fmilies. Colouring, independent set, lique, lique over. Longest pth, Hmilton pth, nd minimum pth over (this ltter is n generliztion of the Hmilton pth prolem: Wht is the minimum numer (k) of pths neessry to over ll verties in G? For k = 1, we hve Hmilton pth.) Weighted Hmilton pth, i.e. TSP pth version, on proper intervl grphs. Mximum mthing in liner time. Domintion - vrious type of domintion prolems. One exmple is the dominting pir prolem: Given grph G(V, E), does there exist pir u, v V, suh tht every uv pth domintes G? A pth P domintes grph, if every vertex in G is either on P or hs neighour on P. Miniml/minimum tringultions: Given n ritrry grph G(V, E), wht is the minimum numer of edges one n dd to turn G into hordl grph? 13

14 Referenes [1] Derek G Corneil, Jérémie Dusrt, Mihel Hi, nd Ekkehrd Kohler. On the power of grph serhing for oomprility grphs. SIAM Journl on Disrete Mthemtis, 30(1): , [2] Derek G Corneil nd Rihrd M Krueger. A unified view of grph serhing. SIAM Journl on Disrete Mthemtis, 22(4): , [3] Feodor F Drgn, Flk Nioli, nd Andres Brndstädt. Lexfs-orderings nd powers of grphs. In Interntionl Workshop on Grph-Theoreti Conepts in Computer Siene, pges Springer, [4] Mihel Hi, Ross MConnell, Christophe Pul, nd Lurent Viennot. Lex-fs nd prtition refinement, with pplitions to trnsitive orienttion, intervl grph reognition nd onseutive ones testing. Theoretil Computer Siene, 234(1):59 84, [5] Ekkehrd Köhler nd Lll Moutdid. Liner time lexdfs on oomprility grphs. In Sndinvin Workshop on Algorithm Theory, pges Springer, [6] Donld J Rose, R Endre Trjn, nd George S Lueker. Algorithmi spets of vertex elimintion on grphs. SIAM Journl on omputing, 5(2): ,