Part III · trunk B · combinatorial algorithms · spine

Matching

König and Hall, Hopcroft–Karp, the Hungarian method as primal–dual with its certificate in plain sight, Edmonds' blossoms, and stable matching for contrast.

needs 14~2 h slides + 3 h lab (+ blossom stretch)lab: uv run co lab 15

Matching is where the course's threads cross. Bipartite matching is a unit-capacity flow (unit 14). König's theorem is LP duality (unit 03) on a TU matrix (unit 06). The Hungarian method is the first algorithm in the course that maintains a dual solution explicitly and uses it to steer the primal. That pattern returns in Lagrangian relaxation (unit 11) and in primal–dual approximation (unit 24).

Non-bipartite matching is the historically decisive case. Its LP relaxation is not integral, but it is still solvable in polynomial time. Edmonds' 1965 paper used it to argue that "polynomial" is the right formal meaning of "efficient".

How to read this deck

Each section has a small worked example computed with the lab's reference solution: König's cover on a 3×3 bipartite graph, the Hungarian method on a 3×3 cost matrix with its potentials after every row, one blossom shrunk and expanded on eight vertices, and Gale–Shapley round by round on three proposers. The longer arguments (the Hopcroft–Karp phase bound, Edmonds' contraction lemma, König's construction) are in the notes.

15 · overview

The route

1 · bipartite cardinality

Berge's augmenting paths, König and Hall, Hopcroft–Karp's $O(E\sqrt V)$.

2 · assignment

The LP and its dual; the Hungarian method as primal–dual; the certificate.

3 · general graphs

Tutte's theorem, blossoms, the matching polytope and its odd-set inequalities.

4 · stable matching

Gale–Shapley: a different question with the same word.

getting around

Words with a dotted underline are glossary terms: click one for its definition. G lists every term in the unit, N shows the notes. Optional extra: interactive figures, linked from the slides they support by a small ▸ chip; skip them freely.

section 1

Bipartite cardinality

One augmenting path at a time, then many at once.

15 · bipartite

Augmenting paths: Berge's theorem

ℓ₂ and r₀ are free. The path r₀–ℓ₀–r₁–ℓ₁–r₂–ℓ₂ alternates free, matched, free, matched, free. Flip it and the matching grows from 2 to 3.

definitions

For a matching $M$, an alternating path alternates edges outside and inside $M$. It is augmenting if both its ends are free. Then the symmetric difference $M \mathbin{\triangle} P$ is a matching with one more edge.

Berge 1957

$M$ is maximum if and only if it has no augmenting path.

Why: overlay $M$ with a bigger matching. The overlay falls into paths and even cycles, and the extra edges of the bigger matching must sit on some path, which is augmenting for $M$.

full proof in the notes (N)

A maximal matching (nothing can be added) need not be maximum: the figure's thick edges are maximal. Berge's theorem holds in every graph. Only finding the path is harder when the graph is not bipartite (section 3).

This is unit 14's residual-graph argument again. Model bipartite matching as a flow with unit capacities (source to every left vertex, every edge left to right, every right vertex to sink). An augmenting path in the flow network is exactly an alternating path with free ends. The backward residual arcs are the matched edges traversed right to left.

Full proof

Proof of "if": let $M^*$ be larger. $M \mathbin{\triangle} M^*$ has maximum degree 2, so it splits into paths and even cycles. Cycles are balanced, so some path has more $M^*$ edges than $M$ edges: that path is augmenting for $M$. $\square$

The symmetric-difference proof is worth internalising: it's also why Hopcroft–Karp's phase bound works, and why the blossom algorithm is correct once contraction is shown to preserve augmenting paths.

15 · bipartite

König and Hall: min–max theorems

König 1931

In a bipartite graph, maximum matching size = minimum vertex cover size.

how to get the cover · lab step 2

König's construction: from the free left vertices, follow alternating paths. Take the left vertices you didn't reach and the right vertices you did. That set covers every edge and has exactly one vertex per matched edge. The next slide does it.

full proof in the notes (N)

Hall 1935

There is a matching that saturates $L$ iff $|N(S)| \ge |S|$ for every $S \subseteq L$, with $N(S)$ the neighbours of $S$.

From König: a cover $A \cup B$ with $A \subseteq L$, $B \subseteq R$ and $|A| + |B| < |L|$ would leave $S = L \setminus A$ with $N(S) \subseteq B$, so $|N(S)| \le |B| < |S|$.

through the course's lens

Matching LP: $\max \mathbf 1^\top x$, $x(\delta(v)) \le 1$. Its dual is the fractional vertex cover. The incidence matrix is TU (unit 06), so both LPs have integral optima, and LP strong duality is König.

Full proof of König

Take a maximum $M$. Let $Z$ be the vertices reachable from free left vertices by alternating paths. Cover $= (L \setminus Z) \cup (R \cap Z)$. Every edge is covered. Every cover vertex is matched (a free right vertex in $Z$ would end an augmenting path). No matched edge has both ends in the cover. So $|\text{cover}| \le |M|$. $\square$

Hall's condition is a certificate of impossibility, in the spirit of unit 00. If no perfect matching exists, a set $S$ with too few neighbours proves it, and König's construction finds one: $S$ = the reachable left vertices, $N(S)$ = the reachable right vertices, and $|N(S)| = |S| - (\text{free left vertices})$.

A triangle breaks König. The maximum matching has 1 edge, but the minimum cover needs 2 vertices, and $x_e = \tfrac12$ on all three edges is a fractional matching of value $\tfrac32$. The incidence matrix of an odd cycle is not TU (unit 06's odd_cycle). Section 3 repairs the LP.

15 · bipartite · example

König's cover, on a small graph

Maximum matching $\{\ell_0 r_0, \ell_2 r_1\}$, size 2. Filled: the cover $\{\ell_2, r_0\}$, size 2.

the construction, step by step

1. Free left vertices: $\ell_1$. Start there.

2. Unmatched edge $\ell_1 \to r_0$. Matched edge $r_0 \to \ell_0$. From $\ell_0$ there is no other edge. Reached: $Z = \{\ell_1, r_0, \ell_0\}$.

3. Cover $=$ left vertices not reached $\{\ell_2\}$ $\cup$ right vertices reached $\{r_0\}$.

Check: $r_0$ covers $\ell_0 r_0, \ell_1 r_0, \ell_2 r_0$; $\ell_2$ covers $\ell_2 r_1, \ell_2 r_2$. Size 2 = matching size, so both are optimal.

and a Hall violator

The reached left vertices $S = \{\ell_0, \ell_1\}$ have $N(S) = \{r_0\}$: two vertices, one neighbour. No matching saturates the left side, and this set proves it.

explore interactively

15 · bipartite

Hopcroft–Karp: many shortest paths per phase

BFS from all free left vertices at once, alternating edges, to get layers. Stop at the first layer that reaches a free right vertex.
DFS from each free left vertex along layer-increasing edges only, taking a maximal set of vertex-disjoint shortest augmenting paths. Remove dead ends.
Flip them all. Repeat.

Each phase costs $O(E)$. It is Dinic (unit 14) on the unit-capacity network.

Hopcroft & Karp 1973

$O(\sqrt V)$ phases, so $O(E\sqrt V)$ total.

Why: each phase makes the shortest augmenting path longer. After $\sqrt V$ phases the missing paths are all long, and long disjoint paths can't be numerous, so few phases remain.

full proof in the notes (N)

Lab then, degree 4: yours 0.070 s at 16 000 vertices a side, networkx 0.317 s.

Full proof

The shortest augmenting path length strictly increases every phase. After $\sqrt V$ phases every augmenting path has length $> \sqrt V$. $M \mathbin{\triangle} M^*$ holds $|M^*| - |M|$ vertex-disjoint augmenting paths, so at most $V / \sqrt V$ remain, and each later phase finds at least one. $\square$

The "length strictly increases" claim is the subtle part. After a phase has taken a maximal set of shortest paths, any remaining augmenting path of the same length would have to be vertex-disjoint from all of them. Maximality rules that out. A path sharing vertices with them must use at least one flipped edge in the other direction, which makes it longer. The full argument is again a symmetric-difference count.

In practice the number of phases on random graphs is usually far below the $\sqrt V$ bound, which is why the pure-Python lab code is fast at 16 000 vertices a side.

The functional reference augments a single shortest path per round. That gives the same answer with the Ford–Fulkerson bound $O(VE)$: the maximal-set-per-phase idea is precisely what the stateful DFS with dead-end removal buys.

section 2

Assignment

Where unit 03's duality first pays off outside LP.

15 · assignment

The assignment LP and its dual

primal: the assignment problem $$\begin{aligned} \min\ & \textstyle\sum_{i,j} c_{ij} x_{ij} \\ \text{s.t.}\ & \textstyle\sum_j x_{ij} = 1 \ \ \forall i \\ & \textstyle\sum_i x_{ij} = 1 \ \ \forall j, \quad x \ge 0 \end{aligned}$$

dual $$\max \sum_i u_i + \sum_j v_j \quad \text{s.t.}\ u_i + v_j \le c_{ij}$$

$u, v$: the dual potentials.

Integral vertices (Birkhoff–von Neumann: the doubly stochastic matrices are the convex hull of the permutation matrices), so the LP optimum is an assignment.

the certificate · lab step 4

If $\sigma$ is a permutation, $u_i + v_j \le c_{ij}$ everywhere, and $u_i + v_{\sigma(i)} = c_{i\sigma(i)}$ for every $i$, then $\sigma$ is optimal.

For any permutation $\tau$: $\ c(\tau) = \sum_i c_{i\tau(i)} \ge \sum_i (u_i + v_{\tau(i)}) = \sum u + \sum v = c(\sigma)$. $\square$

Checking takes $O(n^2)$ and needs no solver. The tests never call scipy for step 4.

Lab then: HiGHS's simplex vertex had zero fractional entries at every n up to 400.

15 · assignment

The Hungarian method is primal–dual

The Hungarian method, a primal–dual method. Keep dual-feasible $(u, v)$ and a partial assignment using only tight edges ($c_{ij} = u_i + v_j$). To add row $i$:

Grow an alternating tree from $i$ over tight edges.
If it reaches a free column, augment. Done.
Else let $\delta$ = the smallest reduced cost from a tree row to a non-tree column. Raise $u$ on tree rows by $\delta$ and lower $v$ on tree columns by $\delta$. Tree edges stay tight, all duals stay feasible, and at least one new edge becomes tight. Go to 1.

why this is cubic time

Keep $\text{minv}[j]$, the smallest reduced cost into column $j$ from the tree. Each dual step then costs $O(n)$, not $O(n^2)$. There are at most $n$ steps per row and $n$ rows.

the same thing, seen from unit 14

It is successive shortest paths for min-cost flow: Dijkstra on reduced costs, with $(u, v)$ as the potentials. $\text{minv}$ is Dijkstra's tentative distance, and step 3 is "settle the nearest node".

History. Kuhn named the method in 1955 after König and Egerváry, whose 1931 work contained its core. Munkres showed it polynomial in 1957 ($O(n^4)$), and Tomizawa (1971) and Edmonds–Karp (1972) gave the $O(n^3)$ shortest-path form the lab uses. Much later it emerged that Jacobi had described an equivalent method, published posthumously in 1890.

The lab's virtual column 0 is a trick from the widely copied e-maxx implementation. Setting p[0] = i makes the new row look like a row already matched to a column, so the first tree step needs no special case. Every row enters through the same loop.

Why $\sum u + \sum v = $ total at the end (a test in step 4): every row and every column is tight on its assigned edge, and summing those $n$ equalities gives exactly that.

15 · assignment · example

The Hungarian method, on a 3×3 matrix

$$c = \begin{pmatrix} 4 & 1 & 3 \\ 2 & 0 & 5 \\ 3 & 2 & 2 \end{pmatrix}$$

add row	what happens	$u$	$v$
1	cheapest entry 1, column 2: raise $u_1$ by 1, assign	$(1, 0, 0)$	$(0, 0, 0)$
2	wants column 2 too (reduced cost 0), which row 1 holds. Tree $\{2, 1\}$: next tight edge costs 2, to column 1. Shift by 2, reroute	$(3, 2, 0)$	$(0, -2, 0)$
3	best reduced cost 2, column 3 free: raise $u_3$ by 2, assign	$(3, 2, 2)$	$(0, -2, 0)$

Result: row 1 → column 2, row 2 → column 1, row 3 → column 3. Cost $1 + 2 + 2 = 5$.

the certificate, checked

Reduced costs $c_{ij} - u_i - v_j$ at the end:

$$\begin{pmatrix} 1 & \mathbf 0 & 0 \\ \mathbf 0 & 0 & 3 \\ 1 & 2 & \mathbf 0 \end{pmatrix}$$

All $\ge 0$: dual feasible. Zero on every assigned entry (bold): tight. And $\sum u + \sum v = 7 - 2 = 5$ = the cost. So 5 is optimal, without looking at the other 5 permutations.

Brute force over all 6 permutations agrees: the minimum is 5.

explore interactively

Row 2, in detail. With $u = (1, 0, 0)$ and $v = 0$, row 2's reduced costs are $(2, 0, 5)$. Column 2 is at 0 but taken by row 1, so the tree grows to row 1 (step with $\delta = 0$). Now the tree has rows 2 and 1, column 2. The cheapest way out: row 2 to column 1 costs 2, row 1 to column 1 costs $4 - 1 = 3$, row 1 to column 3 costs $3 - 1 = 2$, row 2 to column 3 costs 5. So $\delta = 2$: tree rows get $u \mathrel{+}= 2$, tree column 2 gets $v \mathrel{-}= 2$, and row 2 → column 1 becomes tight. Column 1 is free, so augment: row 2 takes column 1 and row 1 keeps column 2. (The column-3 edge from row 1 also became tight; the code stops at the first free column found, column 1.)

Every $(u, v)$ in the table is from a logged copy of the reference hungarian, which returns assignment $[1, 0, 2]$ (0-based), total 5, $u = (3, 2, 2)$, $v = (0, -2, 0)$, and certify_assignment accepts it.

15 · assignment

How much faster is production code?

Lab then, costs uniform in $0..10^6$, seconds. All four totals are equal.

n	yours	scipy	OR-Tools	HiGHS LP
50	0.002	0.0001	0.0014	0.051
100	0.012	0.0003	0.0056	0.046
200	0.085	0.0016	0.022	0.233
400	0.505	0.0063	0.089	1.332

scipy (linear_sum_assignment) is a C shortest-augmenting-path code, the same asymptotics as yours, about 80× faster.
OR-Tools uses cost-scaling push–relabel (Goldberg–Kennedy), designed for large sparse graphs rather than dense matrices.
The general LP is slowest, but is exact and integral here.

section 3

General graphs

Odd cycles, and why "polynomial" means "good".

15 · general graphs

Tutte's theorem, and why the bipartite search fails

From the free root, an alternating walk reaches the base, then goes round an odd cycle. Each cycle vertex is reachable at both parities, so "left/right" labels break.

Tutte 1947

$G$ has a perfect matching iff $\ \mathrm{odd}(G - U) \le |U|$ for every $U \subseteq V$.

$\mathrm{odd}(G - U)$ is the number of odd components after deleting $U$. Each needs a vertex of $U$ to match out. The Tutte–Berge formula generalises this: $\nu(G) = \min_U \tfrac12\bigl(|V| + |U| - \mathrm{odd}(G - U)\bigr)$.

In a bipartite graph every vertex has a fixed side, so the BFS labels it once. In the figure, a BFS that labels each vertex once can miss augmenting paths that go round the cycle.

15 · general graphs

Edmonds 1965: shrink the blossom

the key lemma

Let $B$ be a blossom (an odd alternating cycle reached at its base). The contraction $G / B$ is $G$ with $B$ shrunk to a single vertex. Then $M$ has an augmenting path in $G$ iff $M / B$ has one in $G / B$.

In words: shrinking a blossom never hides or invents an augmenting path. The next slide shrinks one.

The algorithm: grow alternating trees. When an edge joins two even vertices of the same tree, find their lowest common ancestor (the base) and contract the cycle. When you reach a free vertex, expand the blossoms back along the path and augment.

$O(V^3)$ as in the lab (contraction as base[v] relabelling). $O(E\sqrt V)$ by Micali–Vazirani (1980).

"Paths, trees, and flowers", §2

Edmonds argued that an algorithm whose running time is bounded by a polynomial in the input size is the right formal notion of a good algorithm. His evidence: perfect matching has exponentially many candidates, no totally unimodular LP, and yet a good algorithm.

Cobham (1965) made the same proposal independently. Their thesis is why P became the dividing line of unit 22.

How the lab's version represents contraction (the base[v] relabelling). It doesn't build $G / B$. Each vertex carries base[v], the base of the outermost blossom containing it. Edges inside the same blossom are skipped by comparing bases. mark_path rewrites parent pointers round the cycle in both directions, so a later augmentation can walk through the blossom on the correct side. Expansion is therefore implicit.

Two mutations of the reference implementation survive 800 random graphs. One drops the same-base skip test; the other marks the cycle from only one end. Neither changed a single answer, and whether either is truly redundant is not obvious. It's a reminder that "passes random tests" is weaker than "proved", and the blossom algorithm is the classic place where published implementations have had subtle bugs.

15 · general graphs · example

One blossom, shrunk and expanded

shrink

Matching $\{12, 34, 56\}$; free vertices 0 and 7. Search from 0: $0 - 1 = 2$, then $2 - 3 = 4$ and $2 - 6 = 5$ both reach the cycle, and edge $45$ joins two even vertices: an odd cycle $2,3,4,5,6$ with base 2. Shrink it to $B$.

In $G/B$: $0 - 1 = B - 7$ is an augmenting path (7 is adjacent to $B$ through 5).

expand, then flip

Inside $B$, enter at the base 2 and leave at 5 by the even side: $2 - 6 = 5$. The path in $G$ is $0 - 1 = 2 - 6 = 5 - 7$.

Flip it: matching $\{01, 26, 57, 34\}$, size 4, perfect on 8 vertices. Going round the other way, $2 - 3 = 4 - 5$, arrives at 5 on a free edge; the next edge would have to be matched ($5 = 6$), and 7 is out of reach.

Why the bipartite search fails here. Vertex 5 can be reached from 2 in two ways: $2 - 3 = 4 - 5$ (an odd number of steps, arriving on a free edge, so the next edge must be matched: $5 = 6$) or $2 - 6 = 5$ (arriving on the matched edge, so the next edge may be free: $5 - 7$). A search that labels each vertex once, as bipartite matching does, keeps whichever it meets first and can miss $5 - 7$. Shrinking treats the whole cycle as one even vertex, from which every exit is allowed.

The graph is edges $01, 12, 23, 34, 45, 56, 62, 57$. The reference blossom finds a perfect matching of size 4 on it ($[1, 0, 6, 4, 3, 7, 2, 5]$ as a mate list), which is exactly the flipped matching on the slide. It builds its matching from empty, so the starting matching $\{12, 34, 56\}$ is chosen by hand to show the blossom.

15 · general graphs

The matching polytope

Edmonds 1965

The convex hull of matchings of $G$, the matching polytope, is (with $E[S]$ the edges inside $S$)

$$\begin{aligned} x(\delta(v)) &\le 1 && \forall v \\ x(E[S]) &\le \tfrac{|S| - 1}{2} && \forall S \subseteq V,\ |S| \text{ odd} \\ x &\ge 0 \end{aligned}$$

The odd-set inequalities cut off $x_e = \tfrac12$ on a triangle: $x(E[S]) = \tfrac32 > 1$.

There are exponentially many of them, yet you can optimise over the polytope in polynomial time. Separation (Padberg–Rao 1982) is a minimum odd cut, via a Gomory–Hu tree. It's unit 09's pattern again: subtour elimination with an odd-parity twist.
No compact formulation exists. Rothvoss (2014) proved every extended formulation of the perfect matching polytope needs $2^{\Omega(n)}$ inequalities.
The weighted blossom algorithm keeps a dual variable on every odd set, and it's the primal–dual method on this LP.

section 4

Stable matching

A different question with the same word.

15 · stable matching

Gale–Shapley

$n$ proposers and $n$ receivers, each ranking the other side. A matching is stable if no pair $(p, r)$ both prefer each other to their partners (a blocking pair).

A free proposer proposes to the best receiver not yet tried.
The receiver keeps the better of the current holder and the proposer.

Gale & Shapley 1962

It ends after at most $n^2$ proposals with a stable perfect matching. Every proposer gets their best stable partner (proposer-optimal), and every receiver their worst.

what is different

No weights, nothing summed: stability is a local, strategic condition.
Who proposes matters: the algorithm is not symmetric.
Proposing truthfully is a dominant strategy for proposers (strategy-proof) (Dubins–Freedman, Roth), but not for receivers.

Used for the US medical residency match (NRMP, after Roth–Peranson 1999) and school choice. Roth and Shapley won the 2012 Nobel prize in economics.

Why it terminates stable: a receiver's holder only improves, and a proposer only moves down their list. If $(p, r)$ blocked the final matching, $p$ proposed to $r$ before reaching their final partner, and $r$ rejected $p$ for someone $r$ likes more. $r$ only improves after that, so $r$ can't prefer $p$ now.

Why proposer-optimal: suppose some proposer is rejected by a stable partner $r$. Look at the first such rejection in the run. It creates a blocking pair in the supposedly stable matching. Lab step 5 checks this by brute force over all stable matchings for $n \le 6$.

Despite the contrast drawn here, there is an LP after all. Vande Vate (1989) showed stable matchings are exactly the vertices of a polytope, so weighted stable matching is solvable by LP. The combinatorial lattice structure behind that is in Gusfield & Irving's book.

15 · stable matching · example

Gale–Shapley, round by round

	prefers, best first
A	X, Y, Z
B	X, Y, Z
C	Y, X, Z
X	B, A, C
Y	A, B, C
Z	A, B, C

A, B, C propose; X, Y, Z receive.

#	proposal	answer	held
1	A → X	accept	X:A
2	B → X	accept, drop A	X:B
3	C → Y	accept	X:B, Y:C
4	A → Y	accept, drop C	X:B, Y:A
5	C → X	reject (holds B)	X:B, Y:A
6	C → Z	accept	X:B, Y:A, Z:C

why nobody wants to break it

A would rather have X, but X holds B, whom X ranks higher. C would rather have Y or X, but each holds someone they rank above C. No blocking pair: stable.

explore interactively

15 · lab

Lab 15 — what you build

about 3 hours + a stretch session · uv run co lab 15

Hopcroft–Karp.
König's cover from your maximum matching.
The Hungarian method, returning its potentials.
The certificate: permutation + dual feasibility + tightness.
Gale–Shapley and a stability checker.
Stretch: Edmonds' blossom algorithm.

uv run co test 15
uv run co then 15     # vs scipy, OR-Tools, HiGHS · networkx for bipartite and general

15 · checkpoint

Done when

checkpoint

Lab steps 1–5 are ✓ (6 if you took the stretch).
You can verify your Hungarian algorithm's answer is optimal by checking complementary slackness against its dual potentials, without re-running the algorithm.
You can explain why König fails on a triangle and which inequality repairs the LP.

Reading

Korte & Vygen, ch. 10–11.
Schrijver, A Course in Combinatorial Optimization, ch. 3, 5.
Edmonds, "Paths, trees, and flowers" (1965).
Gale & Shapley, Amer. Math. Monthly 69 (1962).

16 dynamic programming · 18 Régin's alldifferent filter is a matching algorithm · 23 Christofides needs a minimum-weight perfect matching.

Matching

The route

Bipartite cardinality

Augmenting paths: Berge's theorem

König and Hall: min–max theorems

König's cover, on a small graph

Hopcroft–Karp: many shortest paths per phase

Assignment

The assignment LP and its dual

The Hungarian method is primal–dual

The Hungarian method, on a 3×3 matrix

How much faster is production code?

General graphs

Tutte's theorem, and why the bipartite search fails

Edmonds 1965: shrink the blossom

One blossom, shrunk and expanded

The matching polytope

Stable matching

Gale–Shapley

Gale–Shapley, round by round

Lab 15 — what you build

Done when

Reading

Next