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The 2-core is what remains of a finite simple graph after repeatedly removing every vertex with fewer than two surviving graph neighbours. It strips away dangling trees while keeping the part that can support itself under this local rule. Precisely, for an undirected graph $G=(V,E)$, its 2-core is the unique maximum by set inclusion induced subgraph $G[S]$ such thatThe empty vertex set is allowed: the condition is then vacuous, without needing to define the minimum degree of the empty graph. "Maximum by inclusion" means that $S$ contains every other vertex set satisfying the condition. It is stronger than merely saying that one chosen set cannot be enlarged: the 2-core simultaneously contains all possible qualifying induced subgraphs. In this particular family, every inclusion-maximal qualifying set does equal that unique maximum; the union argument in why the 2-core is unique explains why. ^[ref]

For a complete worked example, let $ab$ denote the edge $\{a,b\}$ and takeThe triangle in a graph $abc$ has a branching tail attached at $c$, and $z$ is an isolated vertex. In alphabetical order the initial degrees of vertices are $(2,2,3,3,2,1,1,0)$. Deleting $f$ turns $e$ into a leaf of a graph; deleting $g$ leaves $d$ with degree of a vertex two; deleting $z$ changes no other degree of a vertex. Now delete $e$, which makes $d$ a leaf of a graph, and finally delete $d$. Exactly the three edges of $abc$ remain: $C_2(G)=G[\{a,b,c\}]$.

A forest has empty 2-core: every nonempty finite forest has an isolated vertex or a leaf of a graph, so deletion cannot stop at a nonempty remainder. In particular, an isolated edge, an isolated vertex, a path graph, and the empty graph all have empty 2-core. Every cycle graph is its own 2-core. Two disjoint triangles in a graph also form their own 2-core, so a 2-core need not be a connected graph; attaching a tree to either triangle changes neither surviving triangle.

For $n\ge 2$ and $1\le k<n$, an $n$-cycle $c=(12\dots n)$ and $t=(11+k)$ generate $S_n$ exactly when $\gcd (n,k)=1$. Their conjugate transpositions connect labels differing by $k$ modulo $n$. This graph is connected exactly at coprime separation, so transpositions on a connected graph generate the symmetric group. Otherwise residue classes modulo $\gcd (n,k)$ form a nontrivial block system preserved by both generators.

An adjacency list represents a graph by keeping, for each vertex $v$, a collection $\mathrm{Adj}[v]$ of the vertices reached by its outgoing edges. An array indexed by vertex identifiers can locate each collection; the collection itself may use a linked list or an array. In an undirected graph, each nonloop edge supplies an entry at both endpoints.^[ref]

For a worked example, take the simple graphwhere $AB$ denotes the unordered pair $\{A,B\}$. Its adjacency lists, displayed in label order, areThus $C$ has three neighbours, although the graph has six vertices. Its list contains exactly the destinations available for a single step from $C$; no slots are reserved there for $C,E,F$. The fourteen neighbour entries represent seven edges, with $CD$ appearing as $D$ in $C$'s list and as $C$ in $D$'s list. The label order makes the example easy to read; an adjacency list need not be sorted.

The useful idea is to store existing connections explicitly. Neighbour enumeration can then follow exactly the possible next steps, which is what breadth-first search and depth-first search need. In a sparse graph, most pairs of vertices have no edge; adjacency lists avoid reserving a separate adjacency matrix entry for every such absence. Whether this saves work depends on the question: neighbour enumeration and testing one proposed edge have different costs.

Write $n=|V|$ and $m=|E|$. Suppose each vertex identifier, list descriptor, link and optional edge weight occupies a constant number of machine words, and indexed array access and identifier comparison take constant time. There are $n$ list descriptors and $m$ directed, or $2m$ undirected, neighbour entries. Hence adjacency lists use $O(n+m)$ machine words in Big O notation. A full adjacency matrix has $n^2$ entries, regardless of $m$. These are counts under a storage model, not fixed byte sizes.^[ref]

Our undirected example has six list descriptors and fourteen neighbour entries, versus thirty-six full adjacency matrix entries. The numbers $20$ and $36$ count different kinds of logical items: they do not prove a particular byte saving. A linked list also needs links; an array may have unused capacity. Storing only half of a symmetric matrix changes its exact count while leaving quadratic growth. If identifiers or weights require growing numbers of machine words, their size must be included in the space complexity.

An unsorted adjacency list answers "does the edge $u\to v$ exist?" by linear search through $\mathrm{Adj}[u]$. In the worst case it inspects all $d$ entries, giving $O(1+d)$ time, or $O(d)$ for $d\ge 1$. Even if $v$ is absent, an uninspected entry could still contain it, so arbitrary lists require the whole scan. On our example, looking for $CF$ in $C$'s list compares $F$ with $A,B,D$ and fails after three comparisons. A full adjacency matrix instead answers from its single $C,F$ entry in $O(1)$ time.^[ref]

Different promises change the query. A sorted array of neighbour identifiers permits binary search in $O(\log (d+1))$ time, but maintaining order after insertion may require shifting entries. A per-vertex hash table permits expected constant-time membership under the hashing and load assumptions of that data structure; it does not give that guarantee to an arbitrary adjacency list. A stored list length can likewise make a vertex degree query constant-time even though neighbour enumeration still has to visit its entries.

The advantage grows with scale. For a graph path on $n\ge 2$ vertices, $m=n-1$, so there are $2n-2$ undirected neighbour entries and $n$ descriptors, versus $n^2$ full adjacency matrix entries. Such a family is a sparse graph with $m=O(n)$. For a complete graph, $m=n(n-1)/2$, so the undirected neighbour entries already number $n(n-1)$: the adjacency list advantage in growth rate disappears. Choose by the workload: following existing edges favours adjacency lists; many arbitrary pair-membership queries can favour a full adjacency matrix. The representation organizes the same graph for different questions.

In the crystallographic convention, a Coxeter graph is admissible when it is realized by linearly independent unit vectors $v_i$ in a Euclidean space, with nonpositive mutual inner products and $4(v_i,v_j{)}^2=n_{ij}\in \{0,1,2,3\}$ for distinct vertices. The number $n_{ij}$ records a single, double or triple connection; equivalently, the displayed Gram matrix is positive-definite. In edge-label notation the nonzero connections have labels $3,4,6$. Forgetting multiplicity, the underlying graph is a forest: for any cycle of $k\ge 3$ distinct vertices, the sum of the corresponding unit vectors would have squared norm at most $k-k=0$, contradicting their linear independence. Multiple lines on one connection are labels and are not length-two cycles. This is the convention of Wehler's Lie algebra notes, section 6.3.

On a finite connected weighted graph, the interchange process and the single-label random walk with the same symmetric rates have equal spectral gaps. This is a substantial theorem, not just the elementary inclusion of the single-label spectrum in the full spectrum. A proof is given in Proof of Aldous' spectral gap conjecture, Theorem 1.1.

For $n\ge 3$, count the empty graph once, all one-edge graphs once, and each larger accepted graph under its unique center. The latter weighted contribution is $n(n-2)$, giving $1-\left(\genfrac{}{}{0ex}{}{n}{2}\right)+n(n-2)$. Its nonzero value proves evasiveness by the alternating-sum criterion for decision-tree evasiveness.

A bounded-degree infinite graph is amenable when there are finite nonempty vertex sets $W_n$ such that $|{\partial }^{+}{W}_n|/|W_n|\to 0$, where ${\partial }^{+}{W}_n$ is the set of vertices outside $W_n$ adjacent to it. These are graph Følner sets. Amenability makes surface contributions negligible compared with volume. The square lattice and any periodic finite decoration of it are examples.

A directed requirement for a binary constraint $R_{ij}$: every remaining $x\in D_i$ has at least one support $y\in D_j$ with $(x,y)\in R_{ij}$. Revising the arc $i\to j$ deletes unsupported values from $D_i$; it does not directly delete values from $D_j$. Full arc consistency requires the condition in both directions on every edge of the constraint graph. On general graphs, it need not imply that a complete solution exists.

The average vertex degree of a graph is the sum of its vertex degrees divided by its order. Double counting edge endpoints gives $\overline{d}(G)=2e(G)/|V(G)|$. It lies between the minimum degree of a graph and the maximum degree of a graph.

For a finite graph of maximum degree $\Delta \ge 1$, a random independent edge coloring and the Lovász local lemma prove existence of a coloring with no repetitively colored four-edge path when $k\ge 8{\Delta }^{3/2}$, and no repetitively colored six-edge path when $k\ge 5{\Delta }^{5/3}$.

A fixed edge lies in at most $\ell {\Delta }^{\ell -1}$ unoriented paths of length $\ell$. Hence a four-edge event has a closed dependency neighbourhood of size at most $16{\Delta }^3$ and probability $k^{-2}$; the symmetric local-lemma product is at most $16e{\Delta }^3/k^2$. The corresponding six-edge quantities are $36{\Delta }^5$, $k^{-3}$ and $36e{\Delta }^5/k^3$. The stated constants make both products less than one.

Both lengths can be excluded simultaneously with $k\ge C{\Delta }^{5/3}$ for an absolute constant $C$. Apply the asymmetric Lovász local lemma with parameters $x_4=2/k^2$, $x_6=2/k^3$. A path of length $\ell \in \{4,6\}$ meets at most $4\ell {\Delta }^3$ four-edge paths and $6\ell {\Delta }^5$ six-edge paths. The sums of neighbour parameters are therefore bounded by constants times $C^{-2}{\Delta }^{-1/3}+C^{-3}$. Choosing $C$ sufficiently large makes both products in the local-lemma inequalities at least one half, establishing the claim.

The balanced complete multipartite blow-up $K_r(t)$ replaces each vertex of the complete graph $K_r$ by an independent set of $t$ vertices, placing every possible edge between different sets and none inside a set. Every fixed graph of chromatic number at most $r$ embeds into $K_r(t)$ for sufficiently large $t$.

Choose an ideal absorbing chemical species with $\epsilon =8000\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}\mathrm{c}{\mathrm{m}}^{-1}$ at a fixed wavelength, and a cuvette with geometric length $l=1\mathrm{cm}$. These are stipulated model parameters. A micromolar molar concentration is $1\mu \mathrm{M}=10^{-6}\mathrm{mol}{\mathrm{L}}^{-1}$, so each additional $25\mu \mathrm{M}$ gives $\Delta A=8000\times 1\times 25\times 10^{-6}=0.2$. Five chosen standards therefore predictThe absorbance graph is a straight line of slope $0.008(\mu \mathrm{M}{)}^{-1}$, while the transmittance graph shows exponential decay. Equal increments of molar concentration multiply transmittance by $10^{-0.2}\approx 0.63096$; they do not subtract a fixed transmitted fraction.

Suppose a measurement returns the stipulated blank-normalized reading $T=0.40$ under those same conditions. The inferred absorbance and molar concentration areThis lies slightly below the $50\mu \mathrm{M}$ standard because $0.40$ is slightly above $10^{-0.4}\approx 0.39811$. A calibration determined experimentally by linear regression may instead have $A=mc+b$. Then use $c=(A-b)/m$, keeping the slope $m$ and intercept $b$ in their measured units. A nonzero intercept calls for checking the reference blank and sample preparation; it does not uniquely identify the cause.

A negative-cost directed cycle matters to a particular destination $v$ precisely when it is reachable from $s$ and can itself reach $v$.^[ref] Repeating it $m$ times adds $m$ times its negative total cost to a route. Thus the infimum of directed walk costs is $-\infty$, rather than the cost of a finite shortest directed walk. The distinction is about reachability, not simply the presence of a negative edge or a negative-cost directed cycle anywhere in the graph.

Add $C\to A$ of weight $-4$ to the worked example. Now $A\to B\to C\to A$ has total weight $-2+3-4=-3$. Add also an isolated vertex $U$ with a self-loop of weight $-1$. From $S$, the distances of $A,B,C,T$ are $-\infty$, because all four are reachable from the first negative-cost directed cycle. The source retains distance $0$: no directed edge leads back to $S$. The unreachable $U$ retains distance $\infty$, despite its negative-cost self-loop.

For example, $m$ extra laps before proceeding to $T$ give costs $7-3m$. This construction also shows why requiring a directed path with no repeated vertices would change the problem: only directed walks permit arbitrarily many laps. Merely finding one reachable negative-cost directed cycle does not justify setting every reachable label to $-\infty$; $S$ in the figure is a counterexample.

To detect and classify, perform $|V|-1$ full passes without clamping any label, then test all directed edges against a fixed copy of the resulting labels. Collect heads $v$ of edges $u\to v$ satisfying $d(u)<\infty$ and $d(u)+w_{uv}<d(v)$. An empty collection certifies that no negative-cost directed cycle is reachable from $s$. Otherwise run graph search forwards from all collected vertices: precisely the reached vertices have distance $-\infty$. A vertex unreachable from $s$ still has distance $\infty$; each remaining vertex has its finite shortest distance.

Why is this classification exact? Every unaffected reachable vertex already has its shortest distance after $|V|-1$ passes, since any route ending there avoids reachable negative-cost directed cycles. Thus it cannot be the head of a strict improvement, or be reached from such a head. Conversely, all vertices on a reachable negative-cost directed cycle have finite labels by that time. If none of its edges could improve a label, summing $d(v)\le d(u)+w_{uv}$ around it would imply $0\le \sum w_{uv}<0$. At least one of its heads is therefore collected, and the forward graph search reaches the entire directed cycle and every affected descendant. Detecting a directed cycle alone and marking its descendants are separate tasks; finite intermediate labels in the affected region are not answers.

The Bellman inequalities turn local directed edge checks into a global certificate for the shortest path problem. Here the convention is all-to-one: every journey ends at a fixed vertex $r$, and $c_{ij}$ is the cost of the original directed edge $i\to j$. Work in a finite directed graph with finite real costs. Write $L_i$ for candidate labels, normalized by $L_r=0$, on the vertices that can reach $r$. A feasible potential means precisely that these finite labels satisfyfor every directed edge within that set. These labels give lower bounds on costs to $r$. A discovered directed path of cost $U_i$ instead gives an upper bound. The following constructed graph shows both bounds and the exact distances between them; its weights are illustrative costs, not measured travel data.

The arrows in the picture point toward the destination, whereas edge relaxation propagates an improved label in the opposite direction: an improved $d_j$ can lower $d_i$. In the pictured graph, the directed paths from $a$ to $r$ cost $7$, $2+3=5$, and $2-2+4=4$. Thus $\delta (a,r)=4$. The candidate $L=(3,1,3,0)$ in vertex order $(a,b,c,r)$ is feasible, but underestimates the exact distances $(4,2,4,0)$. Merely satisfying the Bellman inequalities does not make a candidate exact.

Why do they give lower bounds? Along any directed path $i=v_0\to v_1\to \cdots \to v_k=r$, successive substitution givesThe same calculation works for a directed walk, with repeated vertices allowed. Taking the infimum of route costs gives $L_i\le \delta (i,r)$. A discovered route gives $\delta (i,r)\le U_i$. If the bounds meet, $L_i=U_i$, they certify an optimum without enumerating routes. Equivalently, check the Bellman inequalities for the proposed distances and exhibit directed paths realizing them. The authored Bellman-Ford algorithm implementation checks the corresponding single-source inequalities and equality on recorded route edges; reversing every directed edge gives this all-to-one convention.^[ref]

For the exact labels in the picture, the slack $c_{ij}+d_j-d_i$ is respectively $0,3,1,0,0$ on $a\to b$, $a\to r$, $b\to r$, $b\to c$, $c\to r$. Zero slack means equality; the zero-slack directed path $a\to b\to c\to r$ realizes the cost $4$. This is the useful beauty of the certificate: a few local inequalities rule out every cheaper journey, while one explicit journey attains the bound.

Keep unreachable vertices separate: $\delta (i,r)=\infty$ when no directed path to $r$ exists, and a finite lower bound alone cannot certify that fact. Extended labels can use $\infty +c=\infty$ for finite $c$, but never compute a slack as $\infty -\infty$. If $i$ can reach a negative-cost directed cycle and then reach $r$, repeatedly traversing the directed cycle makes route costs unbounded below: their infimum is $-\infty$, with no finite shortest directed walk. A finite feasible potential on such a directed cycle is impossible, since summing its Bellman inequalities would give $0\le \sum c_{ij}<0$. A negative-cost directed cycle that cannot reach $r$ does not affect distances to $r$. These are statements about unrestricted directed walks; requiring a directed path with no repeated vertices is a different optimization problem when negative-cost directed cycles exist.^[ref]

For the single-source convention, the same original directed edge $u\to v$ instead gives $L(v)\le L(u)+w_{uv}$ with $L(s)=0$, a lower bound on distance from $s$. The endpoint carrying the label on the left therefore changes with the direction of the problem.

For a finite simple graph $G$ and a matching in a graph $M\subseteq E(G)$, $M$ is a maximum cardinality matching if and only if $G$ has no augmenting path in a matching relative to $M$. The graph need not be a bipartite graph, and the assertion concerns cardinality, not edge benefits.^[ref]

The criterion turns a global comparison with every competing matching in a graph into a search for one repair. Starting from any matching in a graph, repeatedly find and toggle an augmenting path in a matching. Each step adds one selected edge, so finiteness gives termination after at most $\lfloor |V(G)|/2\rfloor -|M|$ steps; absence of any further augmenting path in a matching certifies maximum cardinality. This is an optimality criterion, not by itself a bound on the work needed to find each graph path.

For a short proof outline, an augmenting path in a matching immediately contradicts maximum cardinality. Conversely, if another matching in a graph $N$ has $|N|>|M|$, consider the subgraph whose set of edges is $M△N$. Its vertices have degree of a vertex at most two; its nontrivial connected components are graph paths and even cycles in a graph, alternating between the two matchings in a graph. A cycle in a graph contributes equal numbers from each. Hence some graph path contributes one more edge from $N$ than from $M$ and starts and ends with edges from $N$. Its endpoints are unmatched vertices in a matching relative to $M$, yielding the required augmenting path in a matching.

On a finite graph with symmetric edge-swap rates, the stirring representation of symmetric exclusion transports the occupations by a random permutation. An independent identically distributed Bernoulli distribution occupation field is invariant under every deterministic permutation, and hence under this random permutation when it is independent of the initial field. Therefore the displayed product measure is invariant, including densities zero and one. Equivalently, an edge swap preserves the number of particles and has the same reverse rate, giving detailed balance.

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.

Fix an odd cycle length. If the graph has edge density at least $1/2-ϵ$ and sufficiently few copies of that cycle, odd-cycle copies from positive triangle density forces few triangles in a graph. The triangle removal lemma deletes at most $ϵ\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ edges, after which the clique-free edit bound applies with two parts. Absorbing its linear rounding error into the density margin gives the displayed bound for large order.

A bipartition of a graph divides its vertex set into two independent sets, so every edge joins different parts. A graph admitting such a partition of a set is a bipartite graph. The two parts are not generally unique: separate connected components of a graph can have their sides exchanged independently, and isolated vertices can be assigned to either part.