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A common neighbour of vertices $u$ and $v$ is adjacent to both of them.

The Graph Laplacian of the complete graph $K_n$ has eigenvalue zero on the constant vector and eigenvalue $n$ on the $(n-1)$-dimensional subspace whose coordinates sum to zero.

For finite graphs $H,G$, the homomorphism density $t(H,G)$ is the probability that a uniformly random map $V(H)\to V(G)$ is a graph homomorphism. For bipartite graphs with prescribed vertex classes, the random map is chosen separately and uniformly into the corresponding classes.

The diameter of a connected finite graph is the maximum distance between two vertices.

Thus diameter is an extremal statistic of graph distance.

Given positive edge lengths $\ell (e)$, the path metric is the infimum of ${\sum }_{e\in \gamma }\ell (e)$ over paths $\gamma$ joining two vertices.

The chromatic polynomial turns a graph coloring problem into a counting problem: how many choices remain when adjacent vertices must have different colours? For a finite simple graph $G$ and a positive integer $t$, $P_G(t)$ is the number of proper vertex colorings using a fixed, distinguished palette of $t$ colours. A colouring need not use every available colour, and exchanging two palette colours generally gives a different colouring. Although $P_G$ is a polynomial, its values at arbitrary noninteger arguments are not counts of colourings.

Consider the three-vertex path graph $A$--$B$--$C$. Choose $A$ in $t$ ways, then $B$ in $t-1$ ways, then $C$ in $t-1$ ways: $A$ and $C$ may agree because they are not adjacent. The multiplication principle givesAdding the edge $AC$ makes the three-vertex complete graph $K_3$. Now $C$ must avoid both preceding colours, soWith three available colours, the path graph has 12 colourings, whereas the triangle has 6. With two colours the counts are $2$ and $0$: the extra constraint makes the triangle impossible to colour.

For a nonempty graph, the chromatic number records only the first positive integer $t$ for which $P_G(t)>0$. The chromatic polynomial also counts the choices for every larger palette. For the path the chromatic number is $2$, and for the triangle it is $3$. Counting assignments also gives a useful interpretation for scheduling: if adjacent vertices represent tasks that must use different slots, the chromatic polynomial counts the valid assignments to $t$ named slots. Adding a conflict edge can only remove assignments.

The chromatic polynomial also remembers some of the graph's structure. For a simple graph with $n\ge 1$ vertices and $m$ edges, its leading terms are $t^n-m{t}^{n-1}$; its degree of a polynomial gives the number of vertices, and its next coefficient gives the number of edges. It does not determine the graph completely. A four-vertex path graph and a four-vertex star graph are not related by a graph isomorphism, but both are trees and have chromatic polynomial $t(t-1{)}^3$.

The early connection to map colouring appears in George D. Birkhoff's article ^[ref]. For a derivation of the counting recurrence and its leading coefficients, see Paul Johnson's MAS 341 lecture on chromatic polynomials.

The edge chromatic number is the least number of colours required to colour edges so that incident edges receive different colours.

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 triangle is a set of three vertices joined by all three possible edges, equivalently a copy of $K_3$.

An undirected graph is connected when every pair $u,v$ of vertices is joined by a finite graph path. The length-zero graph path consisting of $u$ alone is allowed, so a single-vertex graph is connected. We use the term here for nonempty graphs; conventions for the graph with no vertices vary. Adjacent vertices are connected, but connected vertices need not be adjacent: in our six-vertex example, $A$ reaches $F$ along $A,C,D,F$, although $AF$ is absent.

Allowing walks instead of graph paths gives the same test for a connected graph. Whenever a finite walk revisits a vertex, delete the intervening closed portion; repeating this leaves a graph path with the same endpoints. This simple shortening argument explains why revisiting locations cannot help decide reachability.^[ref]

Connectivity can conceal fragile connections. Removing $CD$ from our graph produces two connected components of a graph, on $\{A,B,C\}$ and $\{D,E,F\}$: $CD$ is a bridge in a graph. Removing $AB$ instead preserves connectivity, because $A,C,B$ supplies an alternative graph path. If edges represent communication links, this calculation says precisely which single link failure separates the two groups. It is a conclusion about this model, independent of the lengths in its drawing.

A tree is a nonempty connected graph without a graph cycle; equivalently, each pair of vertices has exactly one graph path between them. Delete $BC$ and $EF$ from our example. The remaining edgesform a spanning tree: all six vertices remain, with $6-1=5$ edges. The graph path $A,C,D,F$ is now unique, and every remaining edge is a bridge in a graph. Adding $BC$ back creates exactly one graph cycle, $B,A,C,B$. Thus graph cycles express alternative routes, while a tree retains just enough connections for reachability. More generally, a finite connected graph on $n\ge 1$ vertices has at least $n-1$ edges, with equality exactly for a tree. One proof repeatedly removes an edge on a graph cycle, preserving connectivity, until a spanning tree remains.^[ref]

For a finite graph, breadth-first search tests connectivity by starting at one vertex and discovering neighbours in successive layers. Starting at $A$, the layers in our example are $\{A\}$, $\{B,C\}$, $\{D\}$, $\{E,F\}$. Every vertex is discovered, and the layer numbers give graph distances from $A$. Recording one discovery edge for each new vertex constructs a spanning tree. This turns an existence question into both a concrete route and a certificate of connectivity.

The square lattice has vertex set ${\mathbb{Z}}^2$ and joins vertices whose Euclidean distance is one.

The heavy-ball iteration adds momentum to gradient descent:

For a positive-definite matrix $A$, the conjugate gradient method chooses mutually $A$-conjugate search directions. Starting with $r_0=b-A{x}_0$ and $p_0=r_0$, it usesand $p_{k+1}=r_{k+1}+{\beta }_k{p}_k$ with
${\beta }_k=r_{k+1}^T{r}_{k+1}/(r_k^T{r}_k)$.

For $f(x)=x^{T}Ax/2-b^{T}x$ with $A$ positive definite and residual $r=b-Ax$, exact line search along $r$ usesThe objective error contracts by

A differentiable function has a $\beta$-Lipschitz gradient when

Stochastic gradient descent updates parameters using an unbiased or approximate gradient computed from one observation or a mini-batch rather than the complete data set.

For fixed prime $p$, a one-bounded function on ${\mathbb{F}}_p^n$ with $U^3$ norm at least $\delta$ has correlation bounded below in terms of $p$ and $\delta$ with a quadratic phase, with the usual nonclassical formulation in small characteristic.

A quadratic phase on ${\mathbb{F}}_p^n$ is a function $x\mapsto e_p(q(x))$, where $q$ is a quadratic form and $e_p(t)=e^{2\pi it/p}$. Its third multiplicative derivative is one, so its $U^3$ norm is one.

For functions indexed by $ϵ\in \{0,1{\}}^k$, the Gowers inner product iswhere $\mathcal{C}$ denotes complex conjugation.

The Fennoscandian Ice Sheet covered northern Europe during the last glacial period and produced ongoing uplift around the Baltic region after retreat.