Euler circuits use every edge once, and a quick degree count decides whether one exists. Ask instead for a route that visits every vertex once, and everything changes: no quick test is known, only good sufficient conditions and good reasons for failure.
Key idea
A Hamilton path visits every vertex exactly once; a Hamilton cycle also returns to its start. Dirac's theorem: if a simple graph has $n \ge 3$ vertices and every vertex has degree at least $n/2$, it has a Hamilton cycle. The condition is sufficient, not necessary.
Why it works
Why can't we just count degrees as for Euler? A Hamilton cycle is a global object: whether one exists depends on how the whole graph fits together, not on each vertex separately. Some local facts do block it. A vertex of degree $1$ cannot be inside a cycle. If removing one vertex splits the graph, a cycle through everything would need to pass through that vertex twice. In a bipartite graph a cycle alternates sides, so the two sides must have equal size. Dirac's theorem works in the other direction: with that many edges at every vertex, there is always room to reroute and close a cycle.
Worked examples
Worked example: The cube
Does the cube graph (vertices of a cube, edges along the cube's edges) have a Hamilton cycle?
- Label the top face $A, B, C, D$ in order and the bottom face $E, F, G, H$ below them.
- Try $A \to B \to C \to D \to H \to G \to F \to E \to A$.
- Every consecutive pair is an edge, each vertex appears once, and $E$ is joined to $A$. Yes.
Answer: Yes: $A, B, C, D, H, G, F, E$, back to $A$
Worked example: A graph that cannot
Does $K_{2,3}$ have a Hamilton cycle?
- A cycle in a bipartite graph alternates sides, so it uses equally many vertices from each side.
- The sides have $2$ and $3$ vertices, so no cycle can use all $5$.
Answer: No
Worked example: Using Dirac's theorem
A graph has $6$ vertices, and every vertex has degree at least $3$. Must it have a Hamilton cycle? What if the minimum degree is only $2$?
- $n/2 = 3$, and every degree is at least $3$: Dirac guarantees a Hamilton cycle.
- With minimum degree $2$, the theorem says nothing. The cycle $C_{6}$ has every degree $2$ and is itself a Hamilton cycle; two separate triangles also have every degree $2$ and have none.
Answer: Yes; with degree $2$ it may or may not
Common mistake
Treating Dirac's condition as necessary. "Not every degree is at least $n/2$, so there is no Hamilton cycle" is wrong: $C_{6}$ has all degrees $2 < 3$ and is a Hamilton cycle. Check: when Dirac's condition fails, try to build a cycle or look for a real obstacle (a degree-$1$ vertex, a cut vertex, unequal bipartite sides).