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Questions about the branch of combinatorics called graph theory (not to be used for questions concerning the graph of a function). This tag can be further specialized via using it in combination with more specialized tags such as extremal-graph-theory, spectral-graph-theory, algebraic-graph-theory, topological-graph-theory, random-graphs, graph-colorings and several others.

1 vote

Relationship between cycle length, number of chords, and number of induced $P_{4}$ subgraphs...

I agree with Cycle of length 5 with 0 chords: Number of P4 induced subgraphs: 5 Cycle of length 5 with 1 chord: Number of P4 induced subgraphs: 2 But I'm not sure how to interpret your statement: C …
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2 votes

Relaxing Meyniel graphs: condition for strongly perfect instead of very strongly perfect

You ask: Does this mean every induced subgraph of G which are cycles of odd length at least 5 has at least 2 chords? No, that doesn't make sense. If an induced subgraph is an induced cycle, then it …
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1 vote

Chordless cycles and planarity in graphs

I believe there is a characterization like the one you mention. The de Fraysseix–Rosenstiehl planarity criterion traverses a given graph with a depth-first search and characterizes the way edges can e …
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0 votes

In what types of graphs can the maximum independent set be found in polynomial time?

Considering your graphs seem to be defined sequentially (for numbers $N$ in $[2,15]$), if your graphs are constructed in a way that the graph at $N$ can be constructed from operations that duplicate ( …
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0 votes

maximal sets of vertices that avoids a clique

Given the fixed $k$, you could look at all $\binom{n}{k}$ subsets of vertices to see if they form a clique of size $k$ (or do whatever you like to enumerate all cliques of size $k$ -- there might be a …
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2 votes

Graph classes where Hamiltonian Cycle and Hamiltonian Path problems have different complexity

To answer your question: Is there a graph class, listed in the above database of graph classes, where Hamiltonian Cycle and Hamiltonian Path problems have different complexity? I think the answer is …
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