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If a finite simple graph is a connected graph, a bipartite graph, and a planar graph, with $n\ge 3$ vertices, then $m\le 2n-4$. Every facial boundary walk has even length and length at least three, hence at least four. Thus $4f\le 2m$; substitute the Euler formula for a connected planar graph.^[ref]

The square cycle graph attains equality: $4=2\cdot 4-4$. The three-vertex graph path also attains equality: $2=2\cdot 3-4$. Its repeated bridges in a graph illustrate why counting facial boundary walk length correctly matters. Neither bound applies unchanged to a multigraph: two vertices may support arbitrarily many parallel edges in a plane graph.

For a nonempty finite simple graph $G$, the chromatic number $\chi (G)$ is the least positive integer $k$ for which $G$ has a proper vertex graph coloring with at most $k$ colours. To establish an exact chromatic number, give a colouring using that many colours and explain why fewer cannot suffice. A colouring alone gives only an upper bound.^[ref]

These small graphs show why the chromatic number measures more than the presence of triangles in a graph:

For a cycle graph of even length, alternating two colours closes consistently; for an odd cycle, it does not, and a third colour on the final vertex suffices. More generally, the two-colourability criterion for bipartite graphs says that $\chi (G)\le 2$ exactly when $G$ is a bipartite graph, and the odd-cycle characterization of bipartite graphs identifies the obstruction as an odd cycle. A nonempty graph with no edges has chromatic number one; a bipartite graph with at least one edge has chromatic number exactly two.

In the pairwise-conflict scheduling model of graph coloring, $\chi (G)$ is the least number of interchangeable slots needed. This turns a drawing into an optimization question: how many groups of compatible activities are unavoidable? Greedy coloring constructs a valid grouping quickly, but its number of colours can exceed the chromatic number.

A closed walk is a walk with equal endpoint vertices, $v_m=v_0$. Under our convention this includes the zero-length walk $(v)$; returning after traversing at least one edge requires $m>0$. Closure imposes no other restriction on repeated vertices or edges. In a directed graph, a closed directed walk must also follow every directed edge in its permitted direction.^[ref]

In the shared simple graph, $A,F,A$ is a length-two closed walk that immediately backtracks along $AF$. It is not a trail, since $AF$ is used twice, and it is not a graph cycle. The length-six closed walk $A,B,C,A,D,E,A$ is a trail, since its six edges are distinct, but is not a graph cycle: it returns to $A$ midway through the journey. The length-three closed walk $A,B,C,A$ is a graph cycle, with all vertices distinct apart from the required endpoint repetition. In a simple graph that is an undirected graph, a graph cycle has length at least three; the two-step backtrack is excluded.^[ref]

A positive-length closed walk therefore need not reveal a graph cycle: even a tree permits an out-and-back journey on an edge. A positive-length closed trail in a simple graph does contain a graph cycle. For a proof outline, choose a shortest positive-length consecutive portion of the trail whose endpoint vertices coincide. It cannot repeat an internal vertex, since that would give a shorter such portion; it cannot have length one or two because self-loops and repeated edges are excluded. It is consequently a graph cycle. Cutting it out leaves another closed trail, possibly of length zero, so repetition partitions the original edges into graph cycles. In our six-step example these are the two triangles in a graph. A closed trail using every edge would be an Euler circuit; here $AF$ is omitted.

Counting return journeys makes closed walks useful even when they repeat edges. For a finite simple graph with adjacency matrix $M$, the walk count from powers of an adjacency matrix gives $(M^k{)}_{uv}$ length-$k$ walks from $u$ to $v$, for every nonnegative integer $k$. A proof outline is that matrix multiplication appends one allowed step, summing over the previous endpoint; $M^0$ is the identity matrix, recording the unique zero-length walk at each vertex. ConsequentlyThe trace counts closed walks with a specified starting vertex and traversal order, not distinct graph cycles. These formulas use adjacency entries, not arbitrary travel costs in a weighted graph.^[ref]

In our example, $(M^2{)}_{AA}=5$: the five length-two closed walks are $A,B,A$, $A,C,A$, $A,D,A$, $A,E,A$, and $A,F,A$. Also $(M^3{)}_{AA}=4$, going round either triangle in a graph in either direction, and $\mathrm{tr}(M^3)=12$: each of the two triangles in a graph supplies three choices of starting vertex and two directions. These exact counts for the constructed graph illustrate why return counts and graph cycle counts differ.

A closed walk naturally models a delivery or inspection journey that returns to its depot. Repeated vertices let it combine several excursions; repeated edges let it serve a dead end and return. For an Euclidean travelling salesman tour, one can first construct a closed walk visiting all locations and then use the triangle inequality to shortcut repeated visits without increasing total length. That argument assumes direct travel between the retained locations is permitted, as in a complete graph weighted by Euclidean distance; an arbitrary road network need not contain those shortcut edges.

For a finite connected graph embedded as a plane graph, follow the edges around a face of a plane graph to obtain a closed walk. Count traversals, not distinct edges: a bridge in a graph occurs twice along one facial boundary walk. Each edge contributes two sides in total.^[ref]

For the three-vertex graph path $A-B-C$, the only facial boundary walk is $A,B,C,B,A$, of length four. Both edges are bridges in a graph; neither is a graph cycle. Thus a facial boundary walk need not be a graph cycle.

A finite simple graph that is a connected graph and a planar graph, with $n\ge 3$ vertices and $m$ edges, satisfies $m\le 3n-6$. Choose a plane graph: each facial boundary walk has length at least three, so $3f\le 2m$. Substitute $f=2-n+m$ from the Euler formula for a connected planar graph.^[ref]

The complete graph $K_4$ attains equality: $6=3\cdot 4-6$. The bound also holds for a disconnected finite simple graph that is a planar graph with $n\ge 3$: place its connected components of a graph in disjoint regions and join them with new bridges in a graph, then apply the connected bound.

Strong induction establishes a proposition $P(n)$ for all integers $n\ge n_0$. Prove $P(n_0)$; then, for each $n\ge n_0$, assume every $P(k)$ with $n_0\le k\le n$ and prove $P(n+1)$. The induction hypothesis includes all earlier cases, so the inductive step can use whichever smaller indices it needs. It cannot use $P(n+1)$ itself.^[ref]

This is useful when a construction breaks into smaller pieces of varying sizes. A prime factorization splits a composite number into smaller integer divisors, which need not be one less than the original integer. A recursive procedure can likewise call itself on several smaller inputs. The available cases should match those dependencies. Strong induction lets the mathematical proof follow the construction rather than forcing every step through the immediately preceding size.

Strong induction does not prove more than ordinary mathematical induction. Put $Q(n)={\bigwedge }_{k=n_0}^{n}P(k)$. Ordinary mathematical induction on $Q$ retains the earlier cases and obtains the next one, establishing the same conclusion. Conversely, an ordinary inductive step can ignore the extra cases allowed in a strong induction hypothesis.^[ref]

The well-ordering principle gives another view. A least counterexample would have all smaller cases true, so the strong induction step would establish that allegedly false case. The starting case excludes the beginning.

Every integer $n\ge 2$ has a prime factorization. The induction base case $2$ is prime. For a larger integer, either it is prime or it is $ab$, with $2\le a,b<n$. The induction hypothesis supplies prime factorizations of both smaller integer divisors; multiplying them gives one for $n$. This proves existence, while uniqueness requires the additional argument under the fundamental theorem of arithmetic.^[ref]

For the chosen number $84$, one possible recursive decomposition isThe step at $84$ needs results for $6$ and $14$, rather than a result for $83$. Each composite number in this factorization tree splits into strictly smaller positive integers, so recursive descent terminates at prime numbers. The tree describes one construction; another choice of integer divisors can give a differently shaped factorization tree.

The same pattern explains the usefulness of strong induction for algorithm correctness. When a recursive procedure returns answers for smaller inputs, the induction hypothesis can justify those answers; the remaining task is to justify how the current call combines them. Strictly decreasing a nonnegative integer size measure establishes termination. Correctness also requires each call's conditions and its combination rule; termination alone does not make an algorithm correct.

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]