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1 Lecture 5 Wlks, Trils, Pths nd Connectedness Reding: Some of the mteril in this lecture comes from Section 1.2 of Dieter Jungnickel (2008), Grphs, Networks nd Algorithms, 3rd edition, which is ville online vi SpringerLink. If you re t the university, either physiclly or vi the VPN, you cn downlod the chpters of this ook s PDFs. Severl of the exmples in the previous lectures for exmple two of the sugrphs in Figure 2.7 nd the grph in Figure 1.12 consist of two or more pieces. If one thinks out the definition of grph s pir of sets, these multiple pieces don t present ny mthemticl prolem, ut it proves useful to hve precise voculry to discuss them. 5.1 Wlks, trils nd pths The first definition we need involves sequence of edges Note tht some edges my pper more thn once. (e 1,e 2,...,e L ) (5.1) Definition 5.1. A sequence of edges such s the one in Eqn (5.1) is wlk if there exists corresponding sequence of vertices (v 0,v 1,...,v L ). (5.2) such tht e j =(v j 1,v j ): note tht the vertices don t hve to e distinct. A wlk for which v 0 = v L is closed wlk. This definition mkes sense in oth directed nd undirected grphs nd in the ltter cse corresponds to pth tht goes long the edges in the sense of the rrows tht represent them. 5.1

2 The wlk specified y the edge sequence (e 1,e 2,e 3,e 4 )=((1, 2), (2, 3), (3, 1), (1, 5)) hs corresponding vertex sequence (v 0,v 1,v 2,v 3,v 4 )=(1, 2, 3, 1, 5), while the vertex sequence (v 0,v 1,v 2,v 3 )=(1, 2, 3, 1) corresponds to closed wlk. Figure 5.1: Two exmples of wlks. Definition 5.2. The length of wlk is the numer of edges in the sequence. For the wlk in Eqn. 5.1 the length is thus L. It ll prove useful to define two more constrined sorts of wlk: Definition 5.3. A tril is wlk in which ll the edges e j re distinct nd closed tril is closed wlk tht is lso tril. Definition 5.4. A pth is tril in which ll the vertices in the sequence in Eqn (5.2) re distinct. Definition 5.5. A cycle is closed tril in which ll the vertices re distinct, except for the first nd lst, which re identicl. Remrk 5.6. In n undirected grph cycle is sugrph isomorphic to one of the cycle grphs C n nd must include t lest three edges, ut in directed grphs nd multigrphs it is possile to hve cycle with just two edges. Remrk 5.7. As the three terms wlk, tril nd pth men very similr things in ordinry speech, it cn e hrd to keep their grph-theoretic definitions stright, even though they mke useful distinctions. The following oservtions my help: All trils re wlks nd ll pths re trils. In set-theoretic nottion: Wlks Trils Pths There re trils tht ren t pths: see Figure Connectedness We wnt to e le to sy tht two vertices re connected if we cn get from one to the other y moving long the edges of the grph. Here s definition tht uilds on the terms defined in the previous section: 5.2

3 Definition 5.8. In grph G(V,E), two vertices nd re sid to e connected if there is wlk given y vertex sequence (v 0,...,v L ) where v 0 = nd v L =. Additionlly, we will sy tht vertex is connected to itself. Definition 5.9. A grph in which ech pir of vertices is connected is connected grph. See Figure 5.3 for n exmple of connected grph nd nother tht is not connected. Once we hve the definitions ove, it s possile to mke precise definition of the pieces of grph. It depends on the notion of n equivlence reltion, which you should hve met erlier your studies. Definition A reltion on set S is n equivlence reltion if it is: reflexive: for ll 2 S; symmetric: ) for ll, 2 S; trnsitive: nd c ) c for ll,, c 2 S. The min use of n equivlence reltion on S is tht it decomposes S into collection of disjoint equivlence clsses. Thtis,wecnwrite S = [ j S j where S j \ S k = ; if j 6= k nd if nd only if, 2 S j for some j Connectedness in undirected grphs The key ide is tht is-connected-to is n equivlence reltion on the vertex set of grph. To see this, we need only check the three properties: reflexive: This is true y definition, nd is the min reson why we sy tht vertex is lwys connected to itself. symmetric: If there is wlk from to then we cn simply reverse the corresponding sequence of edges to get wlk from to. c d f e Figure 5.2: The wlk specified y the vertex sequence (,, c, d, e,, f) is tril s ll the edges re distinct, ut it s not pth s the vertex is visited twice. 5.3

4 trnsitive: Suppose is connected to, so tht the grph contins wlk corresponding to some vertex sequence ( = u 0,u 1,...,u L1 1,u L1 = ) tht connects to. Ifthereislsowlkfrom to c given y some vertex sequence ( = v 0,v 1,...,v L2 1,v L2 = c) then we cn get wlk from to c y trcing over the two wlks listed ove, one fter the other. Tht is, there is wlk from to c given y the vertex sequence (, u 1,...,u L1 1,,v 1,...,v L2 1,c). We hve shown tht if is connected to nd is connected to c, then is connected to c nd this is precisely wht it mens for is-connected-to to e trnsitivereltion. The process of trversing one wlk fter nother, s we did in the proof of the trnsitive property, is sometimes clled conctention of wlks. Definition In n undirected grph G(V,E) connected component is n equivlence clss under the reltion is-connected-to on V. The disjointness of equivlence clsses mens tht ech vertex elongs to exctly one connected component nd so we will sometimes tlk out the connected component of vertex Connectedness in directed grphs In directed grphs is-connected-to isn t n equivlence reltion ecuse it s not symmetric. Tht is, even if we know tht there s wlk from some vertex to nother vertex, wehvenogurnteethtthere swlkfrom to : Figure 5.4 provides n exmple. None the less, there is n nlogue of connected component in directed grph tht s cptured y the following definitions: Definition In directed grph G(V,E) vertex is sid to e ccessile or rechle from nother vertex if G contins wlk from to. Additionlly, we ll sy tht ll vertices re ccessile (or rechle) from themselves. Figure 5.3: The grph t left is connected, ut the one t right is not, ecuse there is no wlk connecting the shded vertices lelled nd. 5.4

5 u v Figure 5.4: In directed grph it s possile to hve wlk from vertex to vertex without hving wlk from to, s in the digrph t left. In the digrph t right there re wlks from u to v nd from v to u so this pir is strongly connected. Definition Two vertices nd in directed grph re strongly connected if is ccessile from nd is ccessile from. Additionlly, we regrd vertex s strongly connected to itself. With these definitions it s esy to show (see the Prolem Sets) tht is-stronglyconnected-to is n equivlence reltion on the vertex set of directed grph nd so the vertex set decomposes into disjoint union of strongly connected components. This prompts the following definition: Definition A directed grph G(V,E) is strongly connected if every pir of its vertices is strongly connected. Equivlently, digrph is strongly connected if it contins exctly one strongly connected component. Finlly, there s one other notion of connectedness pplicle to directed grphs, wek connectedness: Definition A directed grph G(V,E) is wekly connected if, when one converts ll its edges to undirected ones, it ecomes connected, undirected grph. Figure 5.5 illustrtes the di erence etween strongly nd wekly connected grphs. Finlly, I d like to introduce piece of nottion for the grph tht one gets y ignoring the directedness of the edges in digrph: Definition If G(V,E) is directed multigrph then G is the undirected multigrph produced y ignoring the directedness of the edges. Note tht if oth the directed edges (, ) nd (, ) re present in digrph G(V,E), then two prllel copies of the undirected edge (, ) pper in G. 5.3 Afterword: useful proposition As with the Hndshking Lemm in Lecture 1, I d like to finish o long run of definitions y using them to formulte nd prove smll, useful result. Proposition 5.17 (Connected vertices re joined y pth). If two vertices nd re connected, so tht there is wlk from to, then there is lso pth from to. 5.5

6 convert directed edges to undirected ones Figure 5.5: The grph t the top is wekly connected, ut not strongly connected, while the one t the ottom is oth wekly nd strongly connected. Proof of Proposition 5.17 To sy tht two vertices nd in grph G(V,E) reconnectedmensthtthere is wlk given y vertex sequence where v 0 = nd v L =. Thereretwopossiilities: (i) ll the vertices in the sequence re distinct; (ii) some vertex or vertices pper more thn once. (v 0,...,v L ) (5.3) In the first cse the wlk is lso pth nd we re finished. In the second cse it is lwys possile to find pth from to y removing some edges from the wlk in Eqn. (5.3). This sort of pth surgery is outlined elow nd illustrted in Exmple We re free to ssume tht the set of repeted vertices doesn t include or s we cn esily mke this true y trimming some vertices o the two ends of the sequence. To e concrete, we cn define new wlk given y sequence ( = v 0 0,...,v 0 L 0 = ) =(v j,...,v k ) (5.4) where v j is the lst ppernce of in the originl sequence nd v k is the first ppernce of. To finish the proof we then need to del with the cse where the wlk in (5.4), which doesn t contin ny repets of or, still contins repets of one or more other vertices. Suppose tht c 2 V with c 6=, is such repeted vertex: we cn eliminte repeted visits to c y defining new wlk specified y the vertex sequence ( = u 0,...,u L 00 = ) =(v0,...,v 0 j,v 0 k+1, 0...,vL 0 0) (5.5) 5.6

7 q t w r s u v Figure 5.6: In the grph ove the shded vertices nd re connected y the pth (, r, s, u, v, ). where v 0 j is the first ppernce of c in the sequence t left in Eqn. (5.4) nd v 0 k is the lst. There cn only e finitely mny repeted vertices in the originl wlk (5.3) nd so, y using the pproch sketched ove repetedly, we cn eliminte them ll, leving pth from to. Exmple 5.18 (Connected vertices re connected y pth). Consider the grph is Figure 5.6. The vertices nd re connected y the wlk (v 0,...,v 15 )=(, q, r,, r, s, t, u, s, t, u, v,, w,v,) which contins mny repeted vertices. To trim it down to pth we strt y eliminting repets of nd using the pproch from Eqn. (5.4), which mounts to trimming o those vertices tht re underlined in the vertex sequence ove. To see how this works, notice tht the vertex ppers s v 0 nd v 3 in the originl wlk nd we wnt v0 0 = v j in (5.4) to e its lst ppernce, so we set v j = v 3. Similrly, ppers s v 12 nd v 15 in the originl wlk nd we wnt vl 0 = v 0 k to e s first ppernce, so we set v k = v 12. This leves us with (v 0 0,...,v 0 9)=(v 3,...,v 12 ) =(, r, s, t, u, s, t, u, v, ) (5.6) Finlly, we eliminte the remining repeted vertices y pplying the pproch from Eqn. (5.5) to the sequence (v0,...,v 0 9). 0 This mounts to chopping out the sequence of vertices underlined in Eqn. (5.6). To follow the detils, note tht ech of the vertices s, t nd u ppers twice in (5.6). To eliminte, sy, the repeted ppernces of s we should use (5.4) with u j = v2 0 s the first ppernce of s in Eqn. (5.6) nd u k = v5 0 s the lst. This leves us with the new wlk (u 0,...,u 6 ) = (v 0 0,v 0 1,v 0 2,v 0 6,v 0 7,v 0 8,v 0 9) = (, r, s, t, u, v, ) which is pth connecting to. 5.7