I have a graph and want to prove the sufficiency of this claim.
Is it sufficient that if the graph is bipartite and has some vertex degrees that correspond to any grid graph is a grid?
2026-03-26 09:27:02.1774517222
Is it sufficient that if graph is bipartite and have some vertex degrees that corresponds to any grid graph is a grid?
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The claim is false.
To give a specific example, the $4 \times 4$ grid graph is not isomorphic to the $4 \times 4$ knight graph:
However, they are both bipartite and both have the degree sequence $2,2,2,2,3,3,3,3,3,3,3,3,4,4,4,4$.
One way to distinguish the knight graph from any grid graph ($4\times 4$ or $2\times 8$ or $1\times 16$) is that the knight graph has degree-$2$ vertices adjacent to degree-$4$ vertices, which cannot happen in a grid graph.
But in general, we don't expect such a claim to be true, because even with a limitation like "bipartite" we expect lots and lots of realizations of most degree sequences. There's just lots and lots of graphs!