The claim is that there exists digraph with order two or bigger such that for every two distinct vertices they have unequal outdegree and indegree
This is obviously true for small orders but how to show that generally?
There exists digraph of order 2 or bigger having every two distinct vertices of different indegree and outdegree
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graph-theory
2 Answers
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Build them recursively:
the order 2 graph is just one directed edge, and the set of out/indegree are $(0,1), (1,0)$.
Once you have an order $n$ graph that satisfies the condition, add one node and connect it with an edge to every other node, directed such that the new node is the starting point. It's easy to see that the out/indegree are $$(0,n),(1,n-1),(2,n-2),\dots,(n-1,1),(n,0)$$
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0Why do we have one directed edge for order two ? how about another directed edge coming back? – 2017-02-24
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0we don't want to show that every graph has the property: we want to find a graph for each order that has the property. I'm just considering the graph of order two with only one edge. – 2017-02-24
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0Ok, I got the sequence but n decreased by 1 , so for n=4 .(0,3),(1,2),(2,1),(3,0) – 2017-02-24
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See the full graph withoot cycles.
indegee and outdegree of ith vertice is $i-1$ and $n-i-1$
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0ok for n=4 do we get (0,3),(1,2),(2,1),(3,0)? – 2017-02-24