nenfains no odd epcles Proof. We have already seen that if a graph costaine as o

Question

nenfains no odd epcles Proof. We have already seen that if a graph costaine as odd qycle, then it's not bipartite, Tnprove the convene, let Gbe a-triral p'gh having that a graph is bipartite if and only if each of its componnts is bipactite, we vertices of G whose distance from u is even and et Wcoint of al vertices wb itfollows that u E U.We claim that every elp of Gyi·avert x of U sad Assume, to the contrary, that there exist two adjacent vertices in U or t that there are vertices e and u in W such that wEG Snce d,)snd distance from w is odd. Thos(0,W) is a partition of V G) Sine ds) vrtex of adjacent vertins ia W. Since these two situati as ue-,e will and P have their initial vertex s is common but they ay have other verti in common as well. Among the vertices P and P have n cmmon, let z be the last vertex. Perhaps z-uIn any case, z -integiti 0. Thus d(u,%) mi. Since z is oa P" and an, is y vertex of P" who" distance from u is i, it follows that So ) is a cycle of kent Howeve s adjaceat to every vertex of W. If this does happem bowe, then we eall a complete blpartite graph. A comglete bipanie gh n Kia. Observe drawn the same way that wedrew G=reer whe two pphs G and H are the same except oilbly for the wwy that t are labeled, then we write G H. (The techicall oe th a is the same graph as Co abogh ii certainly sot

Explanation / Answer

Let U is consist of all vertices of G whose distance from u is even

To prove by contradiction

If there exist two adjaccent vertices in U

Let v & w are two adjaccent vertices in U

therefore d(u,v) & d(u,w) are even

d(u,v)=2s aand d(u,ww)=2t

where s,t are non negative integer

Let p'=(u=v0,v1-----v2s=v) be u-v geodesic

P"=(u=w0,w1--------w2t) be u-w geodesic

Let x is least vertex

in any case x=vi   for some integer i>=0

d(u,vi)=i

since x in on P" & wi is the only vertex of P" whose distance from u is i

therfore x=vi=wi

than

C=(vi,vi+2,----v2s,w2t,w2t+2,-----wi=vi)

is cycle of length

[(2s-i)+(2t-i)]+1=2s+2t-2i+1

=2(s+t-i)+1

so C is odd cycle

which ih contadiction to G not cantain odd cycle

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