# p1.2 - k 1 graph then assumed property(that G k 1 is an...

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Daniel Mihai Problem 1.2 P(n):”Any acyclic digraph with non-empty set of nodes has at least one node with in-degree 0”. Base case: P(1): The graph contains just one node and no edges (due to the fact that is acyclic), therefore the in- degree is 0. Step case: We have to consider two cases: G is connected or G is not connected. 1. G is connected. Take an acyclic digraph G k+1 with k+1 nodes, {v 1 ,v 2 ,…,v k+1 }. Then take a subgraph of G k+1 with k nodes, {v 1 ,v 2 ,…,v k }, which is acyclic( due to the fact that we assumed G k+1 to be acyclic) and has at least one node with in-degree 0, say v i1 . We apply the same steps for another subgraph of G k+1 with n nodes, which does not contain v i1 . From this v i2 will result, which is a node with in-degree 0. We continue to apply this algorithm until we have all k+1 nodes or one node that is one we have already got. So, we have got the set {v i1 ,v i2 …,v ij } of nodes which have in-degree 0. If all the following edges {<v k+1 ,v i1 >,<v i1 ,v i2 >,..,<v ij-1 ,v ij >} exist in the G
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Unformatted text preview: k+1 graph then assumed property (that G k+1 is an acyclic digraph with at least one node with in-degree 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 in-degree 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 in-degree 0, otherwise the parts would have cycles....
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