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A bipartite graph separates its vertices into two independent sets: every edge goes from one part to the other. Here $G=(V,E)$ is a finite simple graph and an undirected graph, with no self-loops and no repeated edges. The two parts $X,Y\subseteq V$ satisfyThus neither part contains an edge. The parts need not have equal sizes, and some pairs across them may have no edge; requiring every cross pair gives a complete bipartite graph. The independent set and cross-edge definitions are equivalent.^[ref]

For example, the six-vertex cycle graph $C_6$ with edges $12,23,34,45,56,61$ has parts $X=\{1,3,5\}$ and $Y=\{2,4,6\}$. Drawing the same graph as a hexagon or in two columns changes neither its edges nor its bipartite property. An extra isolated vertex $z$ can go in either part.

We allow either part to be empty. Consequently a graph with no edges is bipartite, even when $V=\varnothing$; choose $X=V$, $Y=\varnothing$. Here "two colours" means two colours are available, with no requirement to use both. This convention explicitly includes the empty case, whereas definitions requiring two nonempty classes need an exception. For several connected components of a graph, choose the two parts separately in each graph component and take their unions. In a nonempty connected graph that is bipartite, fixing one vertex's colour forces all others along graph paths, so its two-colour assignment is unique up to swapping the colours. Each graph component, including an isolated vertex, can be swapped independently.

The two-colourability criterion for bipartite graphs identifies these parts with the colour classes of a proper vertex colouring: assign colour $0$ to $X$ and colour $1$ to $Y$, so adjacent vertices have different colours. More surprisingly, the odd-cycle characterization of bipartite graphs turns a global requirement into one precise obstruction:This statement is for the finite undirected graph and simple graph model above; a graph cycle has distinct vertices except for the repeated starting vertex. Along a graph cycle in a bipartite graph, colours alternate, so returning to the starting colour takes an even number of edges. Therefore any odd cycle, whether or not other edges join its vertices, prevents a bipartite split.^[ref]

Breadth-first search supplies the converse constructively. Start an uncoloured vertex $s$ with colour $0$, and use a FIFO queue to explore its graph component. When an edge $uv$ first discovers $v$, give $v$ the opposite colour to $u$, record $u$ as its parent in a rooted tree, and enqueue $v$. If both endpoints of any scanned edge already have the same colour, stop with a conflict. Restart from an uncoloured vertex until every graph component has been examined. If no conflict occurs, all edges cross the two colour classes, which are the required independent sets.^[ref]

A conflict also explains the failure. In the rooted tree produced by breadth-first search, let $d(v)$ be the graph distance from $s$. The colour of $v$ is its distance's integer parity. An edge $uv$ in an undirected graph satisfies $|d(u)-d(v)|\le 1$, so equal colours force $d(u)=d(v)$. Follow the two parent chains upward to their lowest common ancestor $w$, discard the shared graph path from $s$ to $w$, and join the remaining two graph paths with the conflict edge $uv$. The result is a genuine graph cycle withedges, an odd integer. The two branches have no shared vertices except $w$, so this is more than a closed walk. This odd cycle is a certificate that anyone can check directly, using only its listed vertices and edges.

In the example, breadth-first search from $s$ gives $d(w)=1$, $d(a)=d(b)=2$, and $d(u)=d(v)=3$. The conflict edge $uv$ closes the odd cycle $w,a,u,v,b,w$, with five edges. The common edge $sw$ is excluded. Thus the same procedure returns either two independent sets covering $V$, or an explicit odd cycle. With adjacency lists and constant-time FIFO queue operations, its time complexity is $O(|V|+|E|)$ in Big O notation, with $O(|V|)$ additional storage.^[ref]

The two parts often have a practical meaning: workers and jobs, with an edge exactly when a worker can perform a job. A matching in a graph chooses compatible worker–job pairs with no reused endpoint; merely being bipartite does not guarantee enough compatible pairs. Hall's marriage theorem characterizes when every worker can be matched, and the assignment problem adds costs to seek an optimal assignment. These connections make the bipartite structure useful after recognition, as the starting point for selecting edges under constraints.

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.

A chordal graph is a finite simple graph with no induced cycle of length at least four. Equivalently, every cycle in a graph with at least four vertices has a chord: an edge joining two nonconsecutive vertices of that graph cycle. The word "induced" matters: the induced subgraph on the chosen vertices must contain exactly the cycle's edges, with no chords. Triangles are allowed. This definition concerns adjacency, independently of how the graph is drawn.

For the square cycle graph $C_4$, take vertex set $\{a,b,c,d\}$ and set of edges $\{ab,bc,cd,da\}$, where $ab$ abbreviates $\{a,b\}$. Its graph neighbourhoods are $N(a)=N(c)=\{b,d\}$ and $N(b)=N(d)=\{a,c\}$. The square is an induced cycle, so $C_4$ is not a chordal graph. Add the single edge $ac$: now $N(a)=\{b,c,d\}$, $N(b)=\{a,c\}$, $N(c)=\{a,b,d\}$, and $N(d)=\{a,c\}$. The only possible four-vertex graph cycle has chord $ac$, so this new graph is a chordal graph. Its maximal cliques are exactly $\{a,b,c\}$ and $\{a,c,d\}$.

Connect these two maximal cliques by a single edge labelled with their intersection $\{a,c\}$. The result is a junction tree, also called a clique tree: its nodes are maximal cliques, and the nodes containing any fixed original vertex induce a connected graph inside the tree. Here $a,c$ occur in both nodes, while $b,d$ each occur in just one. This is the running intersection property. More generally, a nonempty connected graph is a chordal graph exactly when its maximal cliques admit such a tree; a disconnected chordal graph can be represented by one such tree per connected component of a graph.^[ref]

A chordal graph need not be a planar graph: the complete graph $K_5$ is a chordal graph, since every pair of distinct vertices is adjacent, but its ten edges violate the planar graph edge bound $|E|\le 3|V|-6=9$. Conversely, even a plane drawing whose bounded faces are triangles can hide an induced cycle: a wheel graph with a four-vertex rim and a central vertex joined to the whole rim still has the rim as an induced cycle. Adding chords to obtain a chordal graph is called triangulation of an undirected graph; this is an adjacency condition, rather than a requirement about planar faces.

Why is this structure useful? The maximal cliques collect local interactions, and the running intersection property makes their overlaps sufficient for consistent communication. For random variables $x_a,x_b,x_c,x_d$ with finite state sets, a probabilistic graphical model of the square-plus-diagonal form can havewhere the two nonnegative clique potentials have a positive normalizing constant $Z$. Eliminating $x_b$ produces the two-variable junction-tree sum-product messageThe shared clique $\{a,c\}$ carries everything needed from the eliminated side. For $q$ possible states per random variable, this example uses two sums over $q^3$ configurations after reusing the message, rather than directly summing over $q^4$ configurations. The example's message is an exact finite sum, with no simulated data. Large maximal cliques still make exact inference expensive; chordal graphs provide a decomposition, not a universal guarantee of cheap inference.^[ref]

The chromatic index $\chi \prime (G)$, also called the edge chromatic number, is the least number of labels in a proper edge colouring of a finite undirected graph without self-loops. Equivalently, it is the smallest number of matchings in a graph whose set partition covers all the edges. We take $\chi \prime (G)=0$ when there are no edges. For an arbitrary edge colouring, one colour always suffices whenever an edge exists, so properness is essential to this definition. The chromatic number $\chi (G)$ instead colours vertices.^[ref]

To establish an exact chromatic index, supply both a proper edge colouring and a reason that fewer colours cannot work. Every edge at a fixed vertex must have its own colour, giving the lower bound $\chi \prime (G)\ge \Delta (G)$, where $\Delta (G)$ is the maximum degree of a graph. The bound need not be attained: it counts a single resource's workload, while conflicts can also accumulate around a graph cycle.

For the triangle in a graph $K_3$ with edges $ab,bc,ca$, every pair of edges shares a vertex, so all three require different labels. The assignment $c(ab)=1,c(bc)=2,c(ca)=3$ works, proving $\chi \prime (K_3)=3$ although $\Delta (K_3)=2$. In its line graph, those three edges become a three-vertex clique. Treating this as three pair-resource jobs makes the obstruction tangible: each resource has only two jobs, but every pair of jobs conflicts, so no slot can contain two jobs.

For the square cycle graph $C_4$ with edges $ab,bc,cd,da$, use $c(ab)=c(cd)=1$ and $c(bc)=c(da)=2$. Each colour class is a matching in a graph, and each vertex has one edge of each colour. Since $\Delta (C_4)=2$, this proves $\chi \prime (C_4)=2$. More generally, alternation closes on any even cycle graph $C_{2m}$, $m\ge 2$. On an odd cycle $C_{2m+1}$, $m\ge 1$, two-colour alternation would make the last and first edges share a colour at their common vertex; a third colour on the last edge suffices. Thus even cycle graphs have chromatic index two and odd cycle graphs have chromatic index three. These exact examples show why maximum degree of a graph alone does not determine the answer.

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 circulant matrix applies the same rule at every position of a finite ring: moving both the input and output indices one step leaves its entries unchanged. Its first column determines the whole matrix. For a length-$N$ vector $c$, index rows and columns by $0,\dots ,N-1$ and defineThe indices use modular arithmetic. Thus each column is the previous column shifted down one place, with the last entry wrapping around to the top. Multiplication is exactly circular convolution:This makes a circulant matrix a natural model for a periodic finite difference rule or a filter acting on a periodic record. Strang's MIT lecture develops this connection between circulant matrices, circular convolution and the discrete Fourier transform.^[ref]

For example, the first column $c=(2,-1,0,-1{)}^T$ givesHere the indices wrap around modulo four. Each output compares a value with its two neighbors; this is the Graph Laplacian of a four-vertex cycle graph. A constant input gives zero, while an alternating input gives a large response. The Fourier diagonalization of a circulant matrix makes this distinction exact: its four eigenvalues are $0,2,4,2$.

The plotted discrete Fourier modes are complex vectors, shown through their real and imaginary parts. Each entry has modulus one; the Euclidean norm of each length-four vector is two. In particular, modes $k=1$ and $k=3$ have the same real part but opposite imaginary parts, so they are distinct eigenvectors with the same eigenvalue.

The useful surprise is that every length-$N$ circulant matrix shares the same basis of eigenvectors, whatever its first column. One change of basis therefore simplifies an entire family of linear maps, instead of requiring a separate eigenvalue computation for each one. Bamieh's tutorial develops the discrete Fourier transform from precisely this simultaneous diagonalization problem.^[ref]

The clique number $\omega (G)$ is the maximum number of vertices in a clique of a finite graph:We use $\omega (G)=0$ for the graph with no vertices. For a nonempty complete graph, $\omega (K_n)=n$; for the five-vertex example, $\omega (G)=3$.

If the graph contains $K_k$ and no $K_{k+1}$ as subgraphs, its clique number is exactly $k$, because every larger clique contains a clique with $k+1$ vertices. Also $\chi (G)\ge \omega (G)$: a proper vertex coloring must assign different colors to all members of any clique. This lower bound need not be an equality; the cycle graph $C_5$ has clique number $2$ and chromatic number $3$.

An edge contraction merges the two endpoints of an existing edge. For a finite simple graph $G$ and $e=uv$, replace $u,v$ by a new vertex $w$. Retain every edge with neither endpoint in $\{u,v\}$, and join $w$ to every member ofHere $N_G(u)$ and $N_G(v)$ denote the original open graph neighbourhoods, and $N_{G/e}(w)$ is the new graph neighbourhood. Remove the self-loop arising from $uv$ and merge duplicate edges, so the result $G/e$ is again a simple graph. This is not deletion of $uv$, which keeps both original vertices.^[ref]

The left example contracts $uv$ in the cycle graph $C_4$ to give $\{wa,ab,bw\}$. Its endpoints have no common neighbours, so four edges become three. The right example contracts $uv$ in the triangle in a graph with edges $uv,ua,va$. Its endpoints have the single common neighbour $a$, so the two edges $ua,va$ become one edge $wa$: three edges become one.

For any such edge contraction, the exact counts areThe contracted edge accounts for one loss. Each common neighbour $z$ contributes exactly one further loss, since $uz,vz$ become a single edge $wz$. All other edges correspond one-to-one to surviving edges. In particular the new degree of a vertex isThese formulas depend on simplifying to a simple graph; a multigraph convention that retains duplicate edges has different counts. In the six-vertex branch set of a graph minor diagram, contracting $12$ loses one edge, contracting the resulting $34$ loses two because of its common neighbour $A$, and contracting $56$ loses one. Hence its seven edges become the three edges of $K_3$: $7-1-2-1=3$.

An edge contraction can increase a degree of a vertex even while decreasing both total counts. For example, contracting $uv$ in the tree with edges $uv,ua,ub,vc,vd$ merges two vertices of degree of a vertex three into one of degree of a vertex four. This is the mechanism behind the distinction between a graph minor and a topological minor.

The Euler characteristic compresses a finite CW complex's homology into one integer: count its topological cells with alternating signs, or count its Betti numbers with the same signs. A filled triangle has Euler characteristic one; its cycle graph boundary has Euler characteristic zero. The missing face changes the answer even though the vertices and edges are identical.

For a finite CW complex $X$, including the geometric realization of a finite simplicial complex,Here $b_i={\dim }_{\mathbb{Q}}{H}_i(X;\mathbb{Q})$ is the rational Betti number. More generally this ordinary homology definition makes sense whenever every rational homology group is finite-dimensional and only finitely many are nonzero. An arbitrary infinite CW complex need not satisfy these assumptions; its formal infinite alternating cell count is not a definition of this invariant.

A filled triangle is a disk with three vertices, three edges and one two-dimensional topological cell, so $\chi =3-3+1=1$. Its simplicial homology has $(b_0,b_1,b_2)=(1,0,0)$: the filled triangle is contractible, and the three-edge chain cycle is the chain boundary of its face. Keeping only the cycle graph on those three vertices gives $\chi =3-3=0$ and $(b_0,b_1)=(1,1)$, since that chain cycle no longer bounds a face. The surrounding plane's exterior is not part of either simplicial complex; do not add it to $F$.

Build a closed annulus from two nested $n$-sided polygons, $n\ge 3$, with $n$ radial edges joining corresponding vertices. There are $2n$ vertices, $3n$ edges and $n$ quadrilateral topological cells, giving $\chi =2n-3n+n=0$. The central missing region and the exterior are not faces of the annulus. A deformation retraction onto a circle gives $(b_0,b_1,b_2)=(1,1,0)$, agreeing with the cell count. This is a useful reduction: the annulus is two-dimensional, yet its homology is that of a one-dimensional cycle graph.