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Prove that if

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Answer #1

Assuming G is a directed acyclic graph with n vertices and there are two vertices u and v. There is a directed path of length n-1 from u to v.  

Directed acyclic graph means there must not be cycle which means maximum length of any path from one vertex to another can be n-1 only. In our case length of directed path between u to v is n-1. Considering u has in-neighbor means incoming edge to it then there must be one vertex at-least from which edge is entering into u. If that vertex is v which means u-v-u will become a cycle and graph can't be DAG. If that vertex is other than v then also u-x-u will become cycle where x is the vertex which is part of u-v path.

in all cases graph will be having cycle which can't be true as Graph is DAG. Hence Proved.

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Answer 1

Assuming G is a directed acyclic graph with n vertices and there are two vertices u and v. There is a directed path of length n-1 from u to v.  

Directed acyclic graph means there must not be cycle which means maximum length of any path from one vertex to another can be n-1 only. In our case length of directed path between u to v is n-1. Considering u has in-neighbor means incoming edge to it then there must be one vertex at-least from which edge is entering into u. If that vertex is v which means u-v-u will become a cycle and graph can't be DAG. If that vertex is other than v then also u-x-u will become cycle where x is the vertex which is part of u-v path.

in all cases graph will be having cycle which can't be true as Graph is DAG. Hence Proved.