first of all with a proper certificate its very easy to verify…

reduction from Independent Set to our problem -

given a graph G and a natural number k, we will return a graph G* - which is same as G only with one added vertex v*, connected to all other vertices in the graph. and k*=k+1 the reduction function is polynomial.

if we had an IS of size k, our new added vertex in graph G* will form a k-star induced subgraph from G*

if we didnt have an IS of size k, it means that we cant find k vertices that are independent from each other, so obviously there is no k+1 star.

obviously this graph has 50 2-star induced subgraphs. and the graph itself isnt a k-star graph.

another example:

let G be a graph, G* be a k-star subgraph of G, G' be a another subgraph of G, where v exists in G* iff v doesnt exists in G',

and G* and G' together, have all the vertices in the graph and both of them have at least one vertex.

G* and G' are connected by only one edge <Vi,Vj> where Vi exists in G* and Vj exists in G'.

for our induced subgraph we can choose G* with all its inner edges (WITHOUT the edge <Vi,Vj>, we are also not supposed to take it because we didnt take the vertex Vj).

so G has a k-star graph, and G itself isnt a k-star graph.

any other thoughts??

]]>Checking if a graph is of the k-star format is obviosly polinomial. ]]>

Determine if the problem is P or NP-Hard:

A) given graph G and a natural number k. does G has a k-star subgraph?

speculated answer - problem is in P,

for Graph G will create a table of size |V| (column i for vertex i), we will run through all of the edges of the graph, for an edge <Vi,Vj> and +1 to columns i and j.

such subgraph exists if one of the cells of the table is >=k. we run in time |E| (+|V| for counting the number if vertices) which is clearly in P.

B) given graph G and a natural number k, does G has an induced subgraph that has a k-star?

I have a (weird) feeling, this is also in P,

can someone please explain?