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Two controlled walkers on the same connected undirected graph can each traverse one edge or wait in each unit-cost step. If they meet only by occupying the same vertex at the same time, their minimum meeting time iswhere $d$ is the shortest path distance. A common meeting point after $t$ steps implies $d(p,q)\le 2t$ by the triangle inequality. Conversely, choose a middle vertex of a shortest connecting graph path; both walkers reach it within $\lceil d/2\rceil$ steps, and the earlier arrival waits. The wait is essential when the path length is an odd integer. This distance on the joint Cartesian product state space is a consistent heuristic: each joint move changes the walkers’ separation by at most two. Replacing $d$ by an obstacle-free Manhattan distance preserves admissible heuristic and consistent heuristic guarantees.

The skeleton of a directed graph is the undirected graph obtained by replacing every directed edge with an undirected edge and forgetting its orientation. In causal structure learning, the PC algorithm first recovers this adjacency structure before learning orientations.

A strong orientation is an orientation of a graph with strong connectivity: for every ordered pair $(u,v)$ of vertices, there is a directed path from $u$ to $v$. A length-zero directed path permits $u=v$. Ordinary connectivity of the underlying undirected graph is necessary, but does not ensure this two-way reachability after directions are chosen. The acyclic $D_1$ is not a strong orientation; the cyclic $D_2$ is, since a journey can continue around its directed cycle.

Attach a new vertex $D$ to that triangle in a graph by the single edge $CD$. This edge is a bridge in a graph. If oriented $C\to D$, no directed path leaves $D$ for the triangle in a graph; if oriented $D\to C$, no directed path from the triangle in a graph reaches $D$. Neither choice gives a strong orientation, even when the triangle in a graph itself is cyclically oriented. By contrast, the four-vertex cycle graph has the strong orientationFollowing this directed cycle reaches any requested vertex in at most three steps. Removing any one of its undirected edges leaves a graph path through all four vertices, so it has no bridge in a graph.

For more than one vertex, a strong orientation cannot also be an acyclic orientation: directed paths from $u$ to $v$ and back give a nonempty closed directed walk, from which removing repeated portions leaves a directed cycle. A directed graph containing one directed cycle need not have strong connectivity, as the attached $D$ shows.

This decision problem asks whether a finite undirected graph has a graph colouring with three colours. It is NP-complete: a colouring certifies membership in NP, while 3-SAT reduces using Boolean-pair colouring gadgets and three-colour clause gadgets.

A trail is a walk whose edge sequence $e_1,\dots ,e_m$ has no repetitions. Its vertices may repeat, including its first and last vertices. The zero-length walk is a trail because it repeats no edge. Every graph path is a trail, but the converse fails: the trail in the shared example isIts six edges $BC,CA,AD,DE,EA,AF$ are pairwise distinct, although $A$ occurs twice. Visiting a junction again and traversing a road segment again are different events.^[ref]

In an undirected graph, traversing $AB$ and then $BA$ repeats the same edge; direction of travel does not give it a new identity. In a multigraph with distinct parallel edges $p,q$ joining $X$ and $Y$, the closed walk $(X,p,Y,q,X)$ is a trail, whereas $(X,p,Y,p,X)$ is not. Both have the same vertex sequence $X,Y,X$. In a directed graph, separate directed edges $X\to Y$ and $Y\to X$ may instead be used once each; record their identities and obey their directions. Thus testing a trail means checking the actual edge records, not merely checking consecutive endpoint labels.

For a finite graph, a trail has length at most $|E|$; unrestricted walks can have arbitrarily large length whenever a positive-length closed walk exists, since it can be repeated. An Eulerian trail is the special trail that uses every edge, so its length is exactly $|E|$. Our length-six trail omits $AB$ and is therefore not an Eulerian trail. In a road-inspection model, a trail describes servicing its selected segments once while allowing returns to junctions; requiring service of every segment leads to the Eulerian trail question. If extra travel between serviced segments is necessary, the complete journey can be a walk even when each segment is serviced only once. This distinction keeps the service requirement separate from the route actually travelled.

A walk records a journey through a graph as a finite alternating sequencewhere each edge $e_i$ joins the consecutive vertices $v_{i-1}$ and $v_i$. Both vertices and edges may repeat. Its length is $m$, counting edge traversals with repetitions, rather than counting different edges or measuring the drawing. Here a graph path means a walk with no repeated vertices; a trail instead forbids repeated edges. These conventions distinguish a journey's record from the subgraph consisting of the connections it used.^[ref]

Unless stated otherwise, the examples below use a finite simple graph that is an undirected graph, so an edge may be traversed in either direction. In a multigraph, retain each edge's individual identity as well as its endpoint vertices: two parallel edges are different choices for the same step. If self-loops are allowed, traversing one leaves the current vertex unchanged but adds one to the length. In a directed graph, each step must follow its directed edge from its initial to its terminal vertex, giving a directed walk; reversing a journey need not be possible.^[ref]

We allow the zero-length walk $(v)$ at every vertex $v$. It uses no edges, so it is also a trail and a graph path, and its coinciding endpoints make it a closed walk. This means staying put is a legitimate journey; a graph with no vertices has no such walk. Positive-length restrictions will be stated explicitly.^[ref]

Use the same constructed graph throughout these three sections:Here $AB$ denotes the unordered pair $\{A,B\}$, and $CA=AC$. Two triangles in a graph, $ABC$ and $ADE$, share $A$, with one extra edge $AF$. These are schematic connections, not measured roads; the highlighted connections in the diagram show which edges are used, while the numbered sequence records their order.

The length-six walk $B,C,A,B,C,A,F$ uses $BC$ and $CA$ twice, so it is not a trail. The length-six trail $B,C,A,D,E,A,F$ returns to $A$ but uses six different edges, so it is not a graph path. The length-three graph path $B,C,A,F$ repeats neither vertices nor edges. Thusfor this particular graph. The restrictions answer different routing questions: revisiting a junction is harmless for many journeys, whereas reusing a serviced road segment can matter for an inspection route. Travel costs require a weighted graph; a walk's cost adds the weight of each traversal, even when an edge is used again.

Every finite walk from $s$ to $t$ contains a graph path from $s$ to $t$ after deleting detours. For a proof outline, if $v_i=v_j$ with $i<j$, delete the intervening closed walk. The retained steps still meet at the same vertex, and the length strictly decreases. Repeating terminates with no repeated vertices, or with the zero-length graph path when $s=t$. This explains why reachability can be formulated using either walks or graph paths. With nonnegative edge weights, deletion also cannot increase cost, connecting the argument to the shortest path problem; with negative weights that cost conclusion fails.^[ref]

For a finite undirected graph without self-loops, give each edge a symmetric nonnegative weight $w_{ij}=w_{ji}$. In the weighted graph, set $(A_w{)}_{ij}=w_{ij}$ for an edge and zero otherwise, and $(D_w{)}_{ii}={\sum }_j{w}_{ij}$ with all other entries zero. The Weighted graph Laplacian isEach edge appears once in the quadratic form. For the path graph $A-B-C$ with weights $w_{AB}=2$ and $w_{BC}=3$,The Weighted graph Laplacian is still a positive semidefinite matrix. Its kernel consists of vectors constant on each connected component of a graph after zero-weight edges are removed: an edge of weight zero imposes no constraint. In an electrical network, taking the weights to be electrical conductances makes $x^T{L}_{w}x$ the total power dissipated by Joule heating when $x$ gives the voltages at the vertices. This physical interpretation explains why stronger connections penalize disagreement more heavily.^[ref]