Unformatted text preview: k+1 graph then assumed property (that G k+1 is an acyclic digraph with at least one node with indegree 0 does not hold). In order for G k+1 to exist with this property, one of the above edges must not exist. Again, we have to cases: a. If at least two nodes from the set {v i1 ,v i2 …,v ij } are equal and all the edges exist, then a cycle is created, which contradicts the hypotheses. b. If all the nodes {v i1 ,v i2 …,v ij } are different and all the edges exist then one of the nodes must be v k+1 which leads, again, to a cycle that contradicts the hypotheses. So, at least one edge must not exist, therefore at least one node has indegree 0 in the G k+1 graph. 2. G is not connected. Therefore, it has connected parts to which we can apply the same algorithm. All the connected parts are subgraphs of G k+1 and we prove for everyone that it has at least one node that has indegree 0, otherwise the parts would have cycles....
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 Spring '10
 M.K
 Graph Theory, Empty set, Topological space, Daniel Mihai

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