# Bipartite Graph

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NOTES FOR MATH 481 LECTURE 26 VIVEK DHAND 1. Bipartite Graphs A graph G = (V, E) is bipartite if we can split up the vertices into two sets X and Y so that every edge has one endpoint in X and the other endpoint in Y . In other words, no two vertices in X are adjacent, and no two vertices in Y are adjacent. We will refer to the sets X and Y as the partite sets of the bipartite graph. Theorem. Let G be a graph. G is bipartite if and only if it contains no odd cycles. Proof. Let G be a bipartite
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NOTES FOR MATH 481LECTURE 26 VIVEK DHAND 1. Bipartite Graphs A graph G = ( V,E  ) is bipartite if we can split up the vertices into two sets X  and Y  sothat every edge has one endpoint in X  and the other endpoint in Y  . In other words,no two vertices in X  are adjacent, and no two vertices in Y  are adjacent. We will referto the sets X  and Y  as the partite sets of the bipartite graph. Theorem. Let G be a graph. G is bipartite if and only if it contains no odd cy-cles. Proof. Let G be a bipartite graph. Suppose ( v 0 ,...,v k ) is a cycle in G . Since v i and v i +1 are adjacent, they must lie in diﬀerent partite sets. So v 1 ,v 3 ,v 5 ,... must lie inone partite set and v 0 ,v 2 ,v 4 ,... must lie in the other. Since v k − 1 and v 0 are adjacent,we see that k − 1 must be odd, so k must be even. This proves that bipartite graphscannot contain odd cycles.Now suppose G is a graph which does not contain any odd cycles. Note that G = K  1 is clearly bipartite and does not contain any odd cycles (a cycle must have at least3 vertices). Therefore, we may assume that G has at least two vertices. Also, G isbipartite if and only if each connected component of  G is bipartite, so we may assumethat G is connected. Pick a vertex v and deﬁne X  = { x ∈ V  | the shortest path from x to v has even length } Y  = { y ∈ V  | the shortest path from y to v has odd length } We claim that these sets X  and Y  make G into a bipartite graph. Let x 1 and x 2 be vertices of  X  and suppose they are adjacent. Note that v is not adjacent to x 1 ,(otherwise the shortest path between them would be length one), so v  = x 2 . Similarly, v is not adjacent to x 2 , so v  = x 1 . Let P  1 = ( v,v 1 ,...,v 2 k ) be a shortest path from v to v 2 k = x 1 and let P  2 = ( v,w 1 ,...,w 2 l ) be a shortest path from v to w 2 l = x 2 . Notethat P  1 and P  2 have even length.If  P  1 and P  2 have no vertices in common except v , then we have found an odd cycle:( v,v 1 ,...,x 1 ,x 2 ,...w 1 ,v ).If  P  1 and P  2 do have common vertices in addition to v , then let v  be the last suchvertex. Note that taking P  1 from x 1 to v  is the shortest path from x 1 to v  (if therewas a shorter path, we could just take that path and then follow the rest of  P  1 for a  2 VIVEK DHAND shorter path to v ). The same goes for P  2 : it is the shortest path from x 2 to v  . Now,the length of the routes that P  1 and P  2 take from v to v  must be the same. If one wasshorter than the other, we would be able to shorten either P  1 or P  2 . Therefore, we mayassume that v  = v i = w i for some i . This implies that ( v  ,v i +1 ,...,x 1 ,x 2 ,...,w i +1 ,v  )is an odd cycle.The same proof works for adjacent vertices y 1 ,y 2 ∈ Y  , except that P  1 and P  2 will haveodd length.We have shown that each connected component of  G is bipartite, so G is bipartite.  Example. K  n,m is the complete bipartite graph  . It is the bipartite graph of maximalsize with | X  | = n and | Y  | = m .

Dec 13, 2017

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Dec 13, 2017
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