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For a doubly stochastic matrix with a fractional entry, form the row-column bipartite graph of entries strictly between zero and one. Every incident vertex has degree at least two, so there is an even cycle. A matrix supported on that cycle with alternating entries $+1,-1$ has zero row and column sums. A sufficiently small positive multiple can be both added and subtracted while preserving entry bounds. The original matrix is then the midpoint of two distinct doubly stochastic matrices, so it is not an extreme point.

With the convention $H=J{\sum }_{⟨ij⟩}{\sigma }_i{\sigma }_j-h{\sum }_i{\sigma }_i$, positive $J$ favors opposite neighboring Ising spins. On a bipartite graph, the Néel order parameter measures opposite sublattice magnetizations; it is distinct from the uniform magnetization. Changing the signs of spins on one sublattice maps the zero-field model to the ferromagnetic Ising model.

The assignment problem chooses a one-to-one pairing of jobs and agents minimizing their total cost. Its minimum-cost flow formulation puts unit supplies on job vertices and unit demands on agent vertices of a bipartite graph. Integral feasible flows are perfect matchings.

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.

For a bipartite graph with ordered parts $X,Y$, its biadjacency matrix has rows indexed by $X$ and columns indexed by $Y$, with entry one for an edge and zero otherwise. It has size $|X|$ by $|Y|$, rather than the square full bipartite adjacency matrix on $X\bigsqcup Y$, which is the block matrix $\left(\begin{array}{cc}0& M\\ M^{\mathsf{T}}& 0\end{array}\right)$. Its singular values describe the normalized adjacency operator of a bipartite graph after division by $\sqrt{|X||Y|}$.

For a bipartite graph with bipartition $L,R$ and nonnegative integer bounds $b_v$, a bipartite b-matching is an edge subset $F$ whose degree of a vertex is at most $b_v$ at each $v$. Taking $b_v=1$ gives a matching in a graph. To maximize $|F|$, build a flow network with arcs $s\to u$ of flow network edge capacity $b_u$ for $u\in L$, arcs $u\to v$ of flow network edge capacity one for $uv\in E$, and arcs $v\to t$ of flow network edge capacity $b_v$ for $v\in R$. The integral max-flow theorem gives an integral optimum. Select precisely the middle arcs with flow one. Flow conservation identifies their endpoint degrees with the outside arc flows, and the flow value is $|F|$. Conversely, every such edge subset supplies a feasible flow of that value. This proves the reduction and its optimality.

If a bipartite graph with parts $X,Y$ has at least $\delta |X||Y|$ edges, then the number of ordered tuples $(x_1,x_2,y_1,y_2)\in X^2\times Y^2$ for which every $x_i{y}_j$ is an edge is at leastIndeed, if $d(x_1,x_2)$ is the number of common neighbours of $x_1,x_2$ in $Y$, two applications of the Cauchy-Schwarz inequality give

This is the part-preserving homomorphism density of a four-cycle in a bipartite graph with independently uniform labels $x,x^{\prime }$ in the first part and $y,y^{\prime }$ in the second. Repeated labels are included. It equals the fourth power of the four-cycle norm of the adjacency indicator function. It differs at finite sizes from a count requiring four distinct vertices.

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.

For every $\epsilon >0$ there is a positive integer $K(\epsilon )$ such that every finite bipartite graph has an $\epsilon$-regular pair of partitions with at most $K(\epsilon )$ cells on each side.

To prove this, give $X\times Y$ the uniform probability measure, let $f$ be the indicator function of the edge set, and for partitions $\mathcal{P},\mathcal{Q}$ define their energy byThis conditional expectation is an orthogonal projection, so $0\le \mathcal{E}\le \Vert f{\Vert }_2^2\le 1$. If the partitions are not $\epsilon$-regular, choose witnesses $A_{ij}\subseteq X_i$ and $B_{ij}\subseteq Y_j$ in every irregular cell pair and refine each $X_i$ by all its $A_{ij}$ and each $Y_j$ by all its $B_{ij}$. The refined energy exceeds the old energy byOn $A_{ij}\times B_{ij}$ the absolute value function of the mean of the difference is greater than $\epsilon$, while this rectangle occupies at least an ${\epsilon }^2$ fraction of $X_i\times Y_j$. The Cauchy-Schwarz inequality therefore contributes more than ${\epsilon }^4|X_i||Y_j|/(|X||Y|)$ from that cell pair. Irregular pairs have total weight greater than $\epsilon$, so every refinement raises the energy by more than ${\epsilon }^5$. The process stops after at most $\lceil {\epsilon }^{-5}\rceil$ refinements. If the current cell counts are $r,s$, the construction gives at most $r{2}^s,s{2}^r$ new cells, so a finite iterated bound depending only on $\epsilon$ supplies $K(\epsilon )$.

König's theorem equates the minimum vertex-cover size of a bipartite graph with its maximum matching size.

A bipartite vertex cover is a vertex cover restricted to a bipartite graph; Kőnig's theorem for bipartite matching computes its minimum size.

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.

A biregular graph is a bipartite graph whose vertices have constant degree on each part separately. The two degree constants can differ. If its parts are $X,Y$ and its edge density of a bipartite graph is $\gamma$, the two degrees are $\gamma |Y|$ and $\gamma |X|$. Thus its adjacency function has constant row and column averages $\gamma$.

For every finite simple graph, Vizing's theorem givesHere $\Delta (G)$ is the maximum degree of a graph, taken as zero when there are no vertices. A bipartite graph has the stronger equality $\chi \prime (G)=\Delta (G)$. The six-job complete bipartite graph above attains this lower bound, while the triangle in a graph attains the extra-colour alternative. The simple-graph restriction in the upper bound matters.^[ref]

The bipartite graph equality still holds for finite loopless multigraphs, with parallel edges counted separately in each vertex degree. For instance, $r$ parallel jobs between two resources need exactly $r$ slots: every pair conflicts, and $\Delta =\chi \prime =r$. For a general finite loopless multigraph, however, the simple graph upper bound can fail. Put two parallel edges on each side of a triangle in a graph. All six edges pairwise share a vertex, so each needs its own colour; every vertex has degree of a vertex four. Consequently $\chi \prime =6>\Delta +1=5$. A multiplicity-aware bound is $\chi \prime (G)\le \Delta (G)+\mu (G)$, where $\mu (G)$ is the largest number of parallel edges between a pair of vertices. The doubled triangle in a graph has $\mu =2$ and attains that bound. These statements exclude self-loops; the two-distinct-resource scheduling model excludes them too.^[ref]

For historical orientation, Graph Theory by Reinhard Diestel (Colouring preview, §5.3, p. 132) dates the bipartite equality to 1916 and Vizing's theorem to 1964. These are textbook attributions; the original papers have not been inspected here. The results state existence of optimal or nearly optimal proper edge colourings, rather than saying that arbitrary greedy choices will find them. The list-chromatic index asks the further question of success for every sufficiently large family of allowed-colour lists.

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.

If a bipartite graph with class sizes $m,N$ has at least $\beta mN$ edges, double count pairs consisting of an $s$-set in the first class and a common neighbour in the second. Convexity of binomial coefficients gives some $S$ with $|N(S)|\ge N\left(\genfrac{}{}{0ex}{}{\lfloor \beta m\rfloor }{s}\right)/\left(\genfrac{}{}{0ex}{}{m}{s}\right)$. When $s\le \beta m/2$, the quotient is at least $(\beta /2{)}^s$. This turns density into a large complete bipartite subgraph.

In a bipartite graph with parts of sizes $m,n$ and density $\delta >0$, a neighbourhood $V^{\prime }$ of size at least $\delta m/2$ can be chosen so that at least a proportion $1-2ϵ/{\delta }^2$ of its ordered pairs have at least $ϵn$ common neighbours. Choose a random vertex on the other side, use the second moment of its degree, and penalize pairs with few common neighbours. This is a simple dependent random choice tool for controlling restricted sumsets.

The maximum antichains in the Boolean lattice are exactly its full middle levels. For even $n$ there is one such level, and for odd $n$ either middle level works. The LYM inequality first confines a maximum antichain to those levels; connectedness of the regular inclusion bipartite graph rules out a mixture of the two odd middle levels.

In a bipartite graph, a self-avoiding walk ends in its starting class exactly when it has even length. If degrees are bounded by $\Delta$, deleting the last edge gives ${\sigma }_{2k+1}\le \Delta {\sigma }_{2k}$ and hence $Z_G^0(x)\le Z_G(x)\le (1+\Delta x)Z_G^0(x)$ for $x\ge 0$. Thus these generating functions have the same radius of convergence.