So I'm trying to find a biconnected bipartite non-Hamiltonian graph and here's what I found: https://i.stack.imgur.com/H3kHT.png There's no Hamiltonian cycle and we can split the vertices into two even parts. So does this graph fit all the properties and if it does then how can I prove it because I'm having trouble proving that it's biconnected.
2026-03-27 06:09:51.1774591791
A biconnected bipartite non-Hamiltonian graph?
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Why not $(\{1,2,3,4,5\}, \{\{1,2\}, \{1,3\}, \{1,4\}, \{2,5\}, \{3,5\}, \{4,5\}\})$? (This is $K_{2,3}$.) Biconnected is easy enough to see. The bipartition is $\{\{1,5\},\{2,3,4\}\}$. For Hamiltonian cycles, starting at $1$, you complete a cycle and then must leave $1$ to reach the remaining vertex and can never return to $1$. The analysis of $5$ is the same. For $2$, you must first go to $1$ (or $5$ which is the same under the relabelling "$1$" $\leftrightarrow$ "$5$"), then the analysis from $1$ applies.