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A cut vertex, also called an articulation point, is a vertex $v$ of an undirected graph whose deletion increases the number of connected components of a graph. Write $k(G)$ for that count, including isolated vertices, and $G-v$ for the graph obtained by removing $v$ and every incident edge. Then the test is $k(G-v)>k(G)$, even when the original graph is disconnected. By contrast, deleting a bridge in a graph removes just one edge and keeps both endpoint vertices. These are different failure models. ^[ref]

First take two triangles in a graph, $ABC$ and $ADE$, sharing only the vertex $A$. This simple graph hasHere $XY$ abbreviates the unordered pair $\{X,Y\}$. Removing $A$ leaves the two connected components of a graph $\{B,C\}$ and $\{D,E\}$, so $k(G-A)=2>1=k(G)$. Removing any other vertex leaves a connected graph. Thus $A$ is the only cut vertex. Every edge lies on a triangle in a graph, so deleting any single edge still leaves a graph path between its endpoints along the other two edges. There are no bridges in a graph.

The generator computes the adjacency lists and checks every vertex deletion, giving the following complete table:

Now take two disjoint triangles in a graph, $ABC$ and $DEF$, joined by the edge $CD$:Deleting only $CD$ preserves all six vertices and leaves connected components of a graph $\{A,B,C\}$ and $\{D,E,F\}$: $CD$ is the only bridge in a graph. Deleting $C$ also removes $AC$ and $BC$, leaving connected components of a graph $\{A,B\}$ and $\{D,E,F\}$. Deleting $D$ instead leaves $\{A,B,C\}$ and $\{E,F\}$. Thus $C,D$ are exactly the cut vertices.

Again the adjacency lists and all vertex deletion counts are computed directly:

For a disconnected example, adjoin an isolated vertex $Z$ to the first graph. Now $k(G)=2$, deleting $A$ gives $k(G-A)=3$, and deleting $Z$ gives $k(G-Z)=1$. The original graph was already disconnected, but $A$ is still a cut vertex; $Z$ is not. The empty graph has count zero, so deleting the only vertex of a one-vertex graph does not create a cut vertex. Nor is either endpoint of a single isolated edge a cut vertex, although that edge is a bridge in a graph.

In a communications network, a cut vertex models a router whose failure separates surviving routers. The shared-$A$ example shows why having an alternative graph path around each individual edge does not protect against failure of the common router. An additional edge from $B$ to $D$ removes this particular vulnerability: after deleting $A$, $B,D$ still join the two sides. A bridge in a graph instead models a vulnerable link. This distinction tells us whether to add another cable between existing devices or a route avoiding an indispensable device; it describes connectivity, rather than traffic capacity or failure probabilities.

The beautiful structural consequence is that a graph splits into biconnected components joined at cut vertices. Within each nontrivial robust piece, a single vertex failure cannot separate the other vertices; between pieces, the shared cut vertices reveal the bottlenecks. Also, a Hamiltonian graph has no cut vertex: deleting any vertex from its Hamilton cycle leaves a Hamiltonian path through all remaining vertices.

Historically, Tarjan's June 1972 paper developed depth-first search as a tool for finding biconnected components in time linear in the numbers of vertices and edges. ^[ref] In his 1983 recollection, Tarjan credited Hopcroft with the original linear-time biconnected component algorithm during their collaboration beginning in summer 1970, and described the published version as a refinement. The historical lesson is how a carefully organized traversal turns many separate deletion experiments into one computation. ^[ref]

A Hamilton cycle, also called a Hamiltonian cycle, is a cycle in a graph containing every vertex of the graph. Here the convention is a finite simple graph whose edges are undirected, with $n=|V|\ge 3$. A traversal is a sequencewhere $v_0,\dots ,v_{n-1}$ are all the distinct vertices and $\{v_i,v_{i+1}\}\in E$ for $0\le i<n$. Thus "once" means once before the closing return: the start appears again at the end, and there are exactly $n$ traversed edges. The graph may contain additional unused edges.^[ref]

A Hamiltonian path visits all vertices once without requiring a return. Removing an edge from a Hamilton cycle gives a Hamiltonian path; a particular Hamiltonian path closes only if its endpoints are adjacent. An Euler circuit instead traverses every edge exactly once and can revisit vertices midway. The two requirements are independent, as these original labelled examples show.

For the complete graph $K_4$, use $V=\{A,B,C,D\}$ and $E=\{AB,AC,AD,BC,BD,CD\}$, where $AB$ abbreviates the unordered pair $\{A,B\}$. The route $A,B,C,D,A$ is a Hamilton cycle: it visits every vertex and uses four of the six edges, omitting $AC$ and $BD$. It is not an Euler circuit, and the graph admits no Euler circuit at all: every vertex degree is three, contradicting the Euler circuit criterion. The crossing of the drawn diagonals is not another vertex.

Conversely, take two triangles in a graph sharing only $X$, with $V=\{X,A,B,C,D\}$ and $E=\{XA,AB,BX,XC,CD,DX\}$. The route $X,A,B,X,C,D,X$ is an Euler circuit, using all six edges once; the vertex degrees are $4,2,2,2,2$. It cannot be a Hamilton cycle, since $X$ occurs midway as well as at the start and end. More strongly, no Hamilton cycle exists: $X$ is a cut vertex, and deleting it leaves the two separate edges $AB$ and $CD$. Nevertheless, $A,B,X,C,D$ is a Hamiltonian path, whose endpoints are not adjacent.

A Hamiltonian graph is a graph containing a Hamilton cycle. This describes existence of a spanning graph cycle, rather than requiring every edge of the graph to belong to that graph cycle. In the finite simple graph convention, $K_n$ is a Hamiltonian graph for $n\ge 3$, because any ordering of its vertices closes using available edges.

A Hamiltonian path is a path in a graph that visits every graph vertex exactly once. Removing any edge from a Hamilton cycle produces a Hamiltonian path. Conversely, a Hamiltonian path closes to a Hamilton cycle exactly when its endpoints are joined by an edge.

For a decreasing probability sequence $p(n)$, if $G(n,p(n))$ contains a Hamilton cycle with high probability, then $G(n,Cp(n))$ is pancyclic with high probability for every fixed sufficiently large $C$; three independent rounds suffice. One first exposes a Hamiltonian round and uses the independent sprinkled edges to create cycles of every shorter length.

The Icosian game made the vertex-visit puzzle tangible. The report of its 1857 presentation describes a plane projection of a regular dodecahedron, with holes at the vertices for ivory pins. The Trinity College Dublin archive connects this game with the search for closed routes visiting each vertex exactly once. This puzzle supplies a historical illustration of the Hamilton cycle requirement.^[ref]

A Hamiltonian graph must be a connected graph and have minimum degree $\delta (G)\ge 2$: each vertex has two distinct neighbours on its Hamilton cycle. It also has no cut vertex. Indeed, deleting any vertex from the Hamilton cycle leaves a graph path through every remaining vertex, so the remaining graph is connected. These are quick ways to rule out a Hamilton cycle.^[ref]

Even these conditions together are not sufficient: the complete bipartite graph $K_{2,3}$ is connected, has minimum degree two and no cut vertex, but has no Hamilton cycle. Every graph cycle in a bipartite graph alternates between its two parts, using equally many vertices from each; it cannot use all two and all three.

In the other direction, Dirac's theorem gives a sufficient condition: a finite simple graph with $n\ge 3$ and $\delta (G)\ge n/2$ has a Hamilton cycle. Failure of this inequality proves nothing by itself: the cycle graph $C_5$ has a Hamilton cycle despite $\delta (C_5)=2<5/2$.^[ref]

A Hamilton cycle is a cycle graph which visits every vertex of a finite graph exactly once, apart from returning to its starting vertex. A graph is Hamiltonian when it contains such a cycle.

Let $v_1\cdots v_r$ be a longest path in the finite simple graph. Every neighbour of either endpoint lies on it. Among indices $1\le i<r$, consider those with edges $v_1{v}_{i+1}$ and those with edges $v_i{v}_r$. Their cardinalities are the endpoint degrees, whose sum is at least $n\ge r$, exceeding the $r-1$ available indices. They intersect, producing the cycleEvery connected component has at least $\delta (G)+1>n/2$ vertices, so the graph is connected. If $r<n$, an edge joins a vertex outside this cycle to it, and breaking the cycle there gives a longer path, a contradiction. Hence $r=n$ and this is a Hamilton cycle, proving Dirac theorem.

The complete graph $K_4$ is planar, has chromatic number four and a Hamilton cycle. Adding one leaf attached to one of its vertices preserves planarity and chromatic number four but prevents a Hamilton cycle, since a leaf has degree one.

For the last claim, the given plane embedding has all vertices on the outer cycle. Add noncrossing diagonals until its polygonal interior is triangulated. A triangulated polygon has a vertex incident with only its two boundary neighbours: for example the dual graph of its triangles is a tree and a leaf triangle supplies an ear. Remove this vertex. Inductively three-colour the smaller triangulated polygon, starting with a triangle. The removed vertex's neighbours are adjacent and have different colours, so it can receive the third. Restriction to the original graph givesThis uses the specified embedding, not the false assertion that every planar Hamiltonian graph is three-colourable; $K_4$ is already a counterexample to that broader assertion.