Kruskal for maximum spanning forest

Chapters

1. [00:00] Einführung
2. [01:59] Der Algorithmus von Kruskal anhand eines Beispiels
3. [06:56] Und wie ist das bei mehreren Zusammenhangskomponenten?
4. [07:29] Welches Problem löst der Algorithmus von Kruskal genau?
5. [08:05] Wie lautet die Invariante?
6. [08:33] Warum ist der Algorithmus korrekt?
7. [08:59] Wie wird die Invariante sichergestellt?
8. [10:25] Was ist die asymptotische Komplexität des Algorithmus?

General information

Algorithmic problem: Maximum spanning forest

Prerequisites:

Type of algorithm: loop

Auxiliary data:

1. A sorted sequence $S$ whose items are the edges of $E$. The edges are sorted in descending order, that is, the edge $e\in E$ with the largest value of $\ell (e)$ is the first element of the sequence (ties arbitrarily broken). Below, $e_i$ denotes the $i$-th edge in $S$.
2. A union-find data structure $UF$ on $V$.

Abstract view

Invariant: After $i\ge 0$ iterations:

1. The undirected graph $F_i=(V,E_i)$, where $E_i$ is the set of all selected edges immediately after the $i$-th iteration, is cycle-free; in particular, it is maximally cycle-free in the sense that, for $j\in \{1,...,i\}$, adding an edge $e_J\notin F_i$ would close a cycle in $F_i$.
2. For the undirected graph $F_i=(V,E_i)$, there is a maximum spanning forest $F_i^{\prime }=(V,E_i^{\prime })$ of $G$ such that $E_i\subseteq E_i^{\prime }$.
3. The sets of $UF$ are the connected components of $F_i$.

Variant: $i$ increases by $1$.

Break condition: $i=|E|$.

Induction basis

Abstract view:

1. $E_0:=\mathrm{\varnothing }$
2. $UF$ is initialized by $V$, so each node $v\in V$ is a singleton in $UF$.

Implementation: Obvious.

Proof: Nothing to show.

Induction step

Abstract view: If inserting $e_i$ would close a cycle, reject $e_i$; otherwise, insert $e_i$.

Implementation:

1. Let $v,w\in V$ denote the endnodes of $e_i$, that is, $e_i=\{v,w\}$.
2. If $v$ and $w$ belong to different sets in $UF$:
   1. Insert $e_i$ in the result.
   2. Unite the sets of $v$ and $w$ in $UF$.

Correctness: The first and third invariants are obvious. So we focus on the second invariant. We may safely set $F_i^{\prime }:=F_{i-1}^{\prime }$ if

1. either $e_i\in E_{i-1}^{\prime }$ and $e_i$ is inserted in the $i$-th iteration
2. or $e_i\notin E_{i-1}^{\prime }$ and $e_i$ is rejected in the -th iteration.

Note that the case that $e_i\in E_{i-1}^{\prime }$ and $e_i$ is rejected cannot occur. In fact, $e_i\in E_{i-1}^{\prime }$ implies that $(V,E_{i-1}\cup \{e_i\})$ is cycle-free, but rejecting $e_i$ implies that $(V,E_{i-1}\cup \{e_i\})$ contains a cycle. So it remains to consider the case that $e_i$ is accepted but $e_i\notin E_{i-1}^{\prime }$.

As $\ell (e_i)>0$ and $F_{i-1}^{\prime }$ is maximal, $(V,E_{i-1}^{\prime }\cup \{e_i\})$ contains a cycle. Since $F_{i-1}$ is cycle-free, there must be an edge $e\in E_{i-1}^{\prime }\setminus E_{i-1}$ on that cycle. By assumption, $(V,E_{i-1}\cup \{e_i\})$ is cycle-free as well. Therefore, $e$ is none of the edges considered so far but one of $e_{i+1},...,e_{|E|}$, which implies $\ell (e)\le \ell (e_i)$ and, hence, maximality of the spanning forest $F_i^{\prime }:=(V,E_{i-1}^{\prime }\setminus \{e\}\cup \{e_i\})$. Clearly, this $F_i^{\prime }$ includes $F_i$, so the second invariant is maintained again.

Complexity

Statement: The asymptotic complexity is in $\mathcal{O}(m\cdot (T(n)+\log m)$ in the worst case, where $n=|V|$ and $m=|E|$, and $T(n)$ is the complexity of a single union-find operation.

Proof: The summand $\mathcal{O}(m\cdot log\cdot m)$ results from sorting all edges in the preprocessing. The main loop has $m$ iterations, and each iteration takes $\mathcal{O}(T(n))$ time.