随机算法 (Spring 2013)/Conditional Probability

This part of the lecture note is still under editting. This notice will be removed once the lecture note has been fully updated.

Conditional Probability

In probability theory, the word "condition" is a verb. "Conditioning on the event ..." means that it is assumed that the event occurs.

Definition (conditional probability)

The conditional probability that event [math]\displaystyle{ \mathcal{E}_1 }[/math] occurs given that event [math]\displaystyle{ \mathcal{E}_2 }[/math] occurs is [math]\displaystyle{ \Pr[\mathcal{E}_1\mid \mathcal{E}_2]=\frac{\Pr[\mathcal{E}_1\wedge \mathcal{E}_2]}{\Pr[\mathcal{E}_2]}. }[/math]

The conditional probability is well-defined only if [math]\displaystyle{ \Pr[\mathcal{E}_2]\neq0 }[/math].

For independent events [math]\displaystyle{ \mathcal{E}_1 }[/math] and [math]\displaystyle{ \mathcal{E}_2 }[/math], it holds that

[math]\displaystyle{ \Pr[\mathcal{E}_1\mid \mathcal{E}_2]=\frac{\Pr[\mathcal{E}_1\wedge \mathcal{E}_2]}{\Pr[\mathcal{E}_2]} =\frac{\Pr[\mathcal{E}_1]\cdot\Pr[\mathcal{E}_2]}{\Pr[\mathcal{E}_2]} =\Pr[\mathcal{E}_1]. }[/math]

It supports our intuition that for two independent events, whether one of them occurs will not affect the chance of the other.

Law of total probability

The following fact is known as the law of total probability. It computes the probability by averaging over all possible cases.

Theorem (law of total probability)

Let [math]\displaystyle{ \mathcal{E}_1,\mathcal{E}_2,\ldots,\mathcal{E}_n }[/math] be mutually disjoint events, and [math]\displaystyle{ \bigvee_{i=1}^n\mathcal{E}_i=\Omega }[/math] is the sample space. Then for any event [math]\displaystyle{ \mathcal{E} }[/math], [math]\displaystyle{ \Pr[\mathcal{E}]=\sum_{i=1}^n\Pr[\mathcal{E}\mid\mathcal{E}_i]\cdot\Pr[\mathcal{E}_i]. }[/math]

Proof. Since [math]\displaystyle{ \mathcal{E}_1,\mathcal{E}_2,\ldots,\mathcal{E}_n }[/math] are mutually disjoint and [math]\displaystyle{ \bigvee_{i=1}^n\mathcal{E}_i=\Omega }[/math], events [math]\displaystyle{ \mathcal{E}\wedge\mathcal{E}_1,\mathcal{E}\wedge\mathcal{E}_2,\ldots,\mathcal{E}\wedge\mathcal{E}_n }[/math] are also mutually disjoint, and [math]\displaystyle{ \mathcal{E}=\bigvee_{i=1}^n\left(\mathcal{E}\wedge\mathcal{E}_i\right) }[/math]. Then [math]\displaystyle{ \Pr[\mathcal{E}]=\sum_{i=1}^n\Pr[\mathcal{E}\wedge\mathcal{E}_i], }[/math] which according to the definition of conditional probability, is [math]\displaystyle{ \sum_{i=1}^n\Pr[\mathcal{E}\mid\mathcal{E}_i]\cdot\Pr[\mathcal{E}_i] }[/math]. [math]\displaystyle{ \square }[/math]

The law of total probability provides us a standard tool for breaking a probability into sub-cases. Sometimes, it helps the analysis.

A Chain of Conditioning

By the definition of conditional probability, [math]\displaystyle{ \Pr[A\mid B]=\frac{\Pr[A\wedge B]}{\Pr[B]} }[/math]. Thus, [math]\displaystyle{ \Pr[A\wedge B] =\Pr[B]\cdot\Pr[A\mid B] }[/math]. This hints us that we can compute the probability of the AND of events by conditional probabilities. Formally, we have the following theorem:

Theorem

Let [math]\displaystyle{ \mathcal{E}_1, \mathcal{E}_2, \ldots, \mathcal{E}_n }[/math] be any [math]\displaystyle{ n }[/math] events. Then [math]\displaystyle{ \begin{align} \Pr\left[\bigwedge_{i=1}^n\mathcal{E}_i\right] &= \prod_{k=1}^n\Pr\left[\mathcal{E}_k \mid \bigwedge_{i\lt k}\mathcal{E}_i\right]. \end{align} }[/math]

Proof. It holds that [math]\displaystyle{ \Pr[A\wedge B] =\Pr[B]\cdot\Pr[A\mid B] }[/math]. Thus, let [math]\displaystyle{ A=\mathcal{E}_n }[/math] and [math]\displaystyle{ B=\mathcal{E}_1\wedge\mathcal{E}_2\wedge\cdots\wedge\mathcal{E}_{n-1} }[/math], then [math]\displaystyle{ \begin{align} \Pr[\mathcal{E}_1\wedge\mathcal{E}_2\wedge\cdots\wedge\mathcal{E}_n] &= \Pr[\mathcal{E}_1\wedge\mathcal{E}_2\wedge\cdots\wedge\mathcal{E}_{n-1}]\cdot\Pr\left[\mathcal{E}_n\mid \bigwedge_{i\lt n}\mathcal{E}_i\right]. \end{align} }[/math] Recursively applying this equation to [math]\displaystyle{ \Pr[\mathcal{E}_1\wedge\mathcal{E}_2\wedge\cdots\wedge\mathcal{E}_{n-1}] }[/math] until there is only [math]\displaystyle{ \mathcal{E}_1 }[/math] left, the theorem is proved. [math]\displaystyle{ \square }[/math]

Polynomial Identity Testing (PIT)

Consider the following problem of Polynomial Identity Testing (PIT):

• Input: two [math]\displaystyle{ n }[/math]-variate polynomials [math]\displaystyle{ f, g\in\mathbb{F}[x_1,x_2,\ldots,x_n] }[/math] of degree [math]\displaystyle{ d }[/math].
• Output: "yes" if [math]\displaystyle{ f\equiv g }[/math], and "no" if otherwise.

The [math]\displaystyle{ \mathbb{F}[x_1,x_2,\ldots,x_n] }[/math] is the of multi-variate polynomials over field [math]\displaystyle{ \mathbb{F} }[/math]. The most natural way to represent an [math]\displaystyle{ n }[/math]-variate polynomial of degree [math]\displaystyle{ d }[/math] is to write it as a sum of monomials:

[math]\displaystyle{ f(x_1,x_2,\ldots,x_n)=\sum_{i_1,i_2,\ldots,i_n\ge 0\atop i_1+i_2+\cdots+i_n\le d}a_{i_1,i_2,\ldots,i_n}x_{1}^{i_1}x_2^{i_2}\cdots x_{n}^{i_n} }[/math].

The degree or total degree of a monomial [math]\displaystyle{ a_{i_1,i_2,\ldots,i_n}x_{1}^{i_1}x_2^{i_2}\cdots x_{n}^{i_n} }[/math] is given by [math]\displaystyle{ i_1+i_2+\cdots+i_n }[/math] and the degree of a polynomial [math]\displaystyle{ f }[/math] is the maximum degree of monomials of nonzero coefficients.

Alternatively, we can consider the following equivalent problem:

• Input: a polynomial [math]\displaystyle{ f\in\mathbb{F}[x_1,x_2,\ldots,x_n] }[/math] of degree [math]\displaystyle{ d }[/math].
• Output: "yes" if [math]\displaystyle{ f\equiv 0 }[/math], and "no" if otherwise.

If [math]\displaystyle{ f }[/math] is written explicitly as a sum of monomials, then the problem is trivial. Again we allow [math]\displaystyle{ f }[/math] to be represented in product form.

Example

The Vandermonde matrix [math]\displaystyle{ M=M(x_1,x_2,\ldots,x_n) }[/math] is defined as that [math]\displaystyle{ M_{ij}=x_i^{j-1} }[/math], that is [math]\displaystyle{ M=\begin{bmatrix} 1 & x_1 & x_1^2 & \dots & x_1^{n-1}\\ 1 & x_2 & x_2^2 & \dots & x_2^{n-1}\\ 1 & x_3 & x_3^2 & \dots & x_3^{n-1}\\ \vdots & \vdots & \vdots & \ddots &\vdots \\ 1 & x_n & x_n^2 & \dots & x_n^{n-1} \end{bmatrix} }[/math]. Let [math]\displaystyle{ f }[/math] be the polynomial defined as [math]\displaystyle{ f(x_1,\ldots,x_n)=\det(M)=\prod_{j\lt i}(x_i-x_j). }[/math] It is pretty easy to evaluate [math]\displaystyle{ f(x_1,x_2,\ldots,x_n) }[/math] on any particular [math]\displaystyle{ x_1,x_2,\ldots,x_n }[/math], however it is prohibitively expensive to symbolically expand [math]\displaystyle{ f(x_1,\ldots,x_n) }[/math] to its sum-of-monomial form.

Schwartz-Zippel Theorem

Here is a very simple randomized algorithm, due to Schwartz and Zippel.

Randomized algorithm for multi-variate PIT

fix an arbitrary set [math]\displaystyle{ S\subseteq \mathbb{F} }[/math] whose size to be fixed; pick [math]\displaystyle{ r_1,r_2,\ldots,r_n\in S }[/math] uniformly and independently at random; if [math]\displaystyle{ f(\vec{r})=f(r_1,r_2,\ldots,r_n) = 0 }[/math] then return “yes” else return “no”;

This algorithm requires only the evaluation of [math]\displaystyle{ f }[/math] at a single point [math]\displaystyle{ \vec{r} }[/math]. And if [math]\displaystyle{ f\equiv 0 }[/math] it is always correct.

In the Theorem below, we’ll see that if [math]\displaystyle{ f\not\equiv 0 }[/math] then the algorithm is incorrect with probability at most [math]\displaystyle{ \frac{d}{|S|} }[/math], where [math]\displaystyle{ d }[/math] is the degree of the polynomial [math]\displaystyle{ f }[/math].

Schwartz-Zippel Theorem

Let [math]\displaystyle{ f\in\mathbb{F}[x_1,x_2,\ldots,x_n] }[/math] be a multivariate polynomial of degree [math]\displaystyle{ d }[/math] over a field [math]\displaystyle{ \mathbb{F} }[/math] such that [math]\displaystyle{ f\not\equiv 0 }[/math]. Fix any finite set [math]\displaystyle{ S\subset\mathbb{F} }[/math], and let [math]\displaystyle{ r_1,r_2\ldots,r_n }[/math] be chosen uniformly and independently at random from [math]\displaystyle{ S }[/math]. Then [math]\displaystyle{ \Pr[f(r_1,r_2,\ldots,r_n)=0]\le\frac{d}{|S|}. }[/math]

Proof. We prove by induction on [math]\displaystyle{ n }[/math] the number of variables. For [math]\displaystyle{ n=1 }[/math], assuming that [math]\displaystyle{ f\not\equiv 0 }[/math], due to the fundamental theorem of algebra, the degree-[math]\displaystyle{ d }[/math] polynomial [math]\displaystyle{ f(x) }[/math] has at most [math]\displaystyle{ d }[/math] roots, thus [math]\displaystyle{ \Pr[f(r)=0]\le\frac{d}{|S|}. }[/math] Assume the induction hypothesis for a multi-variate polynomial up to [math]\displaystyle{ n-1 }[/math] variable. An [math]\displaystyle{ n }[/math]-variate polynomial [math]\displaystyle{ f(x_1,x_2,\ldots,x_n) }[/math] can be represented as [math]\displaystyle{ f(x_1,x_2,\ldots,x_n)=\sum_{i=0}^kx_n^{i}f_i(x_1,x_2,\ldots,x_{n-1}) }[/math], where [math]\displaystyle{ k }[/math] is the largest power of [math]\displaystyle{ x_n }[/math], which means that the degree of [math]\displaystyle{ f_k }[/math] is at most [math]\displaystyle{ d-k }[/math] and [math]\displaystyle{ f_k\not\equiv 0 }[/math]. In particular, we write [math]\displaystyle{ f }[/math] as a sum of two parts: [math]\displaystyle{ f(x_1,x_2,\ldots,x_n)=x_n^k f_k(x_1,x_2,\ldots,x_{n-1})+\bar{f}(x_1,x_2,\ldots,x_n) }[/math], where both [math]\displaystyle{ f_k }[/math] and [math]\displaystyle{ \bar{f} }[/math] are polynomials, such that as argued above, the degree of [math]\displaystyle{ f_k }[/math] is at most [math]\displaystyle{ d-k }[/math] and [math]\displaystyle{ f_k\not\equiv 0 }[/math]; [math]\displaystyle{ \bar{f}(x_1,x_2,\ldots,x_n)=\sum_{i=0}^{k-1}x_n^i f_i(x_1,x_2,\ldots,x_{n-1}) }[/math], thus [math]\displaystyle{ \bar{f}(x_1,x_2,\ldots,x_n) }[/math] has no [math]\displaystyle{ x_n^{k} }[/math] factor in any term. By the law of total probability, it holds that [math]\displaystyle{ \begin{align} &\Pr[f(r_1,r_2,\ldots,r_n)=0]\\ = &\Pr[f(\vec{r})=0\mid f_k(r_1,r_2,\ldots,r_{n-1})=0]\cdot\Pr[f_k(r_1,r_2,\ldots,r_{n-1})=0]\\ &+\Pr[f(\vec{r})=0\mid f_k(r_1,r_2,\ldots,r_{n-1})\neq0]\cdot\Pr[f_k(r_1,r_2,\ldots,r_{n-1})\neq0]. \end{align} }[/math] Note that [math]\displaystyle{ f_k(r_1,r_2,\ldots,r_{n-1}) }[/math] is a polynomial on [math]\displaystyle{ n-1 }[/math] variables of degree [math]\displaystyle{ d-k }[/math] such that [math]\displaystyle{ f_k\not\equiv 0 }[/math]. By the induction hypothesis, we have [math]\displaystyle{ \begin{align} (*) &\qquad &\Pr[f_k(r_1,r_2,\ldots,r_{n-1})=0]\le\frac{d-k}{|S|}. \end{align} }[/math] For the second case, recall that [math]\displaystyle{ \bar{f}(x_1,\ldots,x_n) }[/math] has no [math]\displaystyle{ x_n^k }[/math] factor in any term, thus the condition [math]\displaystyle{ f_k(r_1,r_2,\ldots,r_{n-1})\neq0 }[/math] guarantees that [math]\displaystyle{ f(r_1,\ldots,r_{n-1},x_n)=x_n^k f_k(r_1,r_2,\ldots,r_{n-1})+\bar{f}(r_1,r_2,\ldots,r_n)=g_{r_1,\ldots,r_{n-1}}(x_n) }[/math] is a single-variate polynomial such that the degree of [math]\displaystyle{ g_{r_1,\ldots,r_{n-1}}(x_n) }[/math] is [math]\displaystyle{ k }[/math] and [math]\displaystyle{ g_{r_1,\ldots,r_{n-1}}\not\equiv 0 }[/math], for which we already known that the probability [math]\displaystyle{ g_{r_1,\ldots,r_{n-1}}(r_n)=0 }[/math] is at most [math]\displaystyle{ \frac{k}{|S|} }[/math]. Therefore, [math]\displaystyle{ \begin{align} (**) &\qquad &\Pr[f(\vec{r})=0\mid f_k(r_1,r_2,\ldots,r_{n-1})\neq0]=\Pr[g_{r_1,\ldots,r_{n-1}}(r_n)=0\mid f_k(r_1,r_2,\ldots,r_{n-1})\neq0]\le\frac{k}{|S|} \end{align} }[/math]. Substituting both [math]\displaystyle{ (*) }[/math] and [math]\displaystyle{ (**) }[/math] back in the total probability, we have [math]\displaystyle{ \Pr[f(r_1,r_2,\ldots,r_n)=0] \le\frac{d-k}{|S|}+\frac{k}{|S|}=\frac{d}{|S|}, }[/math] which proves the theorem. In above proof, for the second case that [math]\displaystyle{ f_k(r_1,\ldots,r_{n-1})\neq 0 }[/math], we use an "probabilistic arguement" to deal with the random choices in the condition. Here we give a more rigorous proof by enumerating all elementary events in applying the law of total probability. You make your own judgement which proof is better. By the law of total probability, [math]\displaystyle{ \begin{align} &\Pr[f(\vec{r})=0]\\ = &\sum_{x_1,\ldots,x_{n-1}\in S}\Pr[f(\vec{r})=0\mid \forall i\lt n, r_i=x_i]\cdot\Pr[\forall i\lt n, r_i=x_i]\\ = &\sum_{x_1,\ldots,x_{n-1}\in S\atop f_k(x_1,\ldots,x_{n-1})=0}\Pr[f(\vec{r})=0\mid \forall i\lt n, r_i=x_i]\cdot\Pr[\forall i\lt n, r_i=x_i]\\ &+\sum_{x_1,\ldots,x_{n-1}\in S\atop f_k(x_1,\ldots,x_{n-1})\neq0}\Pr[f(\vec{r})=0\mid \forall i\lt n, r_i=x_i]\cdot\Pr[\forall i\lt n, r_i=x_i]\\ \le &\sum_{x_1,\ldots,x_{n-1}\in S\atop f_k(x_1,\ldots,x_{n-1})=0}\Pr[\forall i\lt n, r_i=x_i]\\ &+\sum_{x_1,\ldots,x_{n-1}\in S\atop f_k(x_1,\ldots,x_{n-1})\neq 0}\Pr[f(x_1,\ldots,x_{n-1},r_n)=0\mid \forall i\lt n, r_i=x_i]\cdot\Pr[\forall i\lt n, r_i=x_i]\\ = &\Pr[f_k(r_1,\ldots,r_{n-1})=0]+\sum_{x_1,\ldots,x_{n-1}\in S\atop f_k(x_1,\ldots,x_{n-1})\neq 0}\Pr[f(x_1,\ldots,x_{n-1},r_n)=0]\cdot\Pr[\forall i\lt n, r_i=x_i]. \end{align} }[/math] We have argued that [math]\displaystyle{ f_k\not\equiv 0 }[/math] and the degree of [math]\displaystyle{ f_k }[/math] is [math]\displaystyle{ d-k }[/math]. By the induction hypothesis, we have [math]\displaystyle{ \Pr[f_k(r_1,\ldots,r_{n-1})=0]\le\frac{d-k}{|S|}. }[/math] And for every fixed [math]\displaystyle{ x_1,\ldots,x_{n-1}\in S }[/math] such that [math]\displaystyle{ f_k(x_1,\ldots,x_{n-1})\neq 0 }[/math], we have argued that [math]\displaystyle{ f(x_1,\ldots,x_{n-1},x_n) }[/math] is a polynomial in [math]\displaystyle{ x_n }[/math] of degree [math]\displaystyle{ k }[/math], thus [math]\displaystyle{ \Pr[f(x_1,\ldots,x_{n-1},r_n)=0]\le\frac{k}{|S|}, }[/math] which holds for all [math]\displaystyle{ x_1,\ldots,x_{n-1}\in S }[/math] such that [math]\displaystyle{ f_k(x_1,\ldots,x_{n-1})\neq 0 }[/math], therefore the weighted average [math]\displaystyle{ \sum_{x_1,\ldots,x_{n-1}\in S\atop f_k(x_1,\ldots,x_{n-1})\neq 0}\Pr[f(x_1,\ldots,x_{n-1},r_n)=0]\cdot\Pr[\forall i\lt n, r_i=x_i] \le\frac{k}{|S|}. }[/math] Substituting these inequalities back to the total probability, we have [math]\displaystyle{ \Pr[f(\vec{r})=0] \le\frac{d-k}{|S|}+\frac{k}{|S|} =\frac{d}{|S|}. }[/math] [math]\displaystyle{ \square }[/math]

Min-Cut in a Graph

Let [math]\displaystyle{ G(V, E) }[/math] be a multi-graph, which allows parallel edges between two distinct vertices [math]\displaystyle{ u }[/math] and [math]\displaystyle{ v }[/math] but does not allow any self-loop, i.e. an edge connect a vertex to itself. Such a multi-graph can be represented as data structures like adjacency matrix [math]\displaystyle{ A }[/math], where [math]\displaystyle{ A }[/math] is symmetric (undirected graph) with zero diagonal, and each entry [math]\displaystyle{ A(u,v) }[/math] is a nonnegative integer giving the number of edges between vertices [math]\displaystyle{ u }[/math] and [math]\displaystyle{ v }[/math].

Mathematically, a multi-graph is a [math]\displaystyle{ G(V, E) }[/math] where [math]\displaystyle{ V }[/math] is the set of vertices and the set [math]\displaystyle{ E }[/math] of edges is a multiset on ground set [math]\displaystyle{ {V\choose 2} }[/math] (members of [math]\displaystyle{ E }[/math] are from [math]\displaystyle{ {V\choose 2} }[/math]).

A cut in a multi-graph [math]\displaystyle{ G(V,E) }[/math] is an edge set [math]\displaystyle{ C\subseteq E }[/math], which can be equivalently defined as

• there exists a nonempty [math]\displaystyle{ S\subset V }[/math], such that [math]\displaystyle{ C=\{uv\in E\mid u\in S,v\not\in S\} }[/math]; or
• removing of [math]\displaystyle{ C }[/math] disconnects [math]\displaystyle{ G }[/math], that is, [math]\displaystyle{ G'(V,E\setminus C) }[/math] disconnects.

The min-cut or minimum cut problem is defined as follows:

• Input: a multi-graph [math]\displaystyle{ G(V,E) }[/math];
• Output: a cut [math]\displaystyle{ C }[/math] in [math]\displaystyle{ G }[/math] with the minimum size [math]\displaystyle{ |C| }[/math].

The problem itself is well-defined on simple graph (without parallel edges), and our main goal is indeed solving the min-cut in simple graphs, however, as we shall see the algorithm creates parallel edges during its running, even though we start with a simple graph without parallel edges.

A canonical deterministic algorithm for this problem is through the max-flow min-cut theorem. A global minimum cut is the minimum [math]\displaystyle{ s }[/math]-[math]\displaystyle{ t }[/math] min-cut, which is equal to the minimum [math]\displaystyle{ s }[/math]-[math]\displaystyle{ t }[/math] max-flow.

Karger's Min-Cut Algorithm

We will introduce a very simple and elegant algorithm discovered by David Karger.

We define an operation on multi-graphs called contraction: For a multigraph [math]\displaystyle{ G(V, E) }[/math], for any edge [math]\displaystyle{ uv\in E }[/math], let [math]\displaystyle{ contract(G,uv) }[/math] be a new multigraph obtained by:

• replacing the vertices [math]\displaystyle{ u }[/math] and [math]\displaystyle{ v }[/math] by a new vertex [math]\displaystyle{ x\not\in V }[/math];
• for each [math]\displaystyle{ w\not\in\{u,v\} }[/math] replacing any edge [math]\displaystyle{ uw }[/math] or [math]\displaystyle{ vw }[/math] by the edge [math]\displaystyle{ xw }[/math];
• removing all parallel edges between [math]\displaystyle{ u }[/math] and [math]\displaystyle{ v }[/math] in [math]\displaystyle{ E }[/math];
• the rest of the graph remains unchanged.

To conclude, the [math]\displaystyle{ contract(G,uv) }[/math] operation merges the two vertices [math]\displaystyle{ u }[/math] and [math]\displaystyle{ v }[/math] into a new vertex which maintains the old neighborhoods of both [math]\displaystyle{ u }[/math] and [math]\displaystyle{ v }[/math] except for that all the parallel edges between [math]\displaystyle{ u }[/math] and [math]\displaystyle{ v }[/math] are removed.

Perhaps a better way to look at contraction is to interpret it as union of equivalent classes of vertices. Initially every vertex is in a dinstinct equivalent class. Upon call a [math]\displaystyle{ contract(G,uv) }[/math], the two equivalent classes corresponding to [math]\displaystyle{ u }[/math] and [math]\displaystyle{ v }[/math] are unioned together, and only those edges crossing between different equivalent classes are counted as valid edges in the graph.

Karger's min-cut algorithm

while [math]\displaystyle{ |V|\gt 2 }[/math] do choose an edge [math]\displaystyle{ uv\in E }[/math] uniformly at random; [math]\displaystyle{ G=contract(G,uv) }[/math]; return [math]\displaystyle{ C=E }[/math] (the parallel edges between the only remaining vertices in [math]\displaystyle{ V }[/math]);

A multi-graph can be maintained by appropriate data strucrtures such that each contraction takes [math]\displaystyle{ O(n) }[/math] time, where [math]\displaystyle{ n }[/math] is the number of vertices. We leave this as an exercise.

Analysis of accuracy

We first prove several lemma.

Lemma 1

If [math]\displaystyle{ C }[/math] is a cut in a multi-graph [math]\displaystyle{ G }[/math] and [math]\displaystyle{ e\not\in C }[/math], then [math]\displaystyle{ C }[/math] is still a cut in [math]\displaystyle{ G'=contract(G,e) }[/math].

Lemma 2

The size of min-cut in [math]\displaystyle{ G'=contract(G,e) }[/math] is at least as large as the size of min-cut in [math]\displaystyle{ G }[/math], i.e. contraction never reduces the size of min-cut.

Proof. Note that every cut in the contracted graph [math]\displaystyle{ G' }[/math] is also a cut in the original graph [math]\displaystyle{ G }[/math]. [math]\displaystyle{ \square }[/math]

Lemma 3

If [math]\displaystyle{ C }[/math] is a min-cut in a multi-graph [math]\displaystyle{ G(V,E) }[/math], then [math]\displaystyle{ |E|\ge \frac{|V||C|}{2} }[/math].

Proof. It must hold that the degree of each vertex [math]\displaystyle{ v\in V }[/math] is at least [math]\displaystyle{ |C| }[/math], or otherwise the set of adjacent edges of [math]\displaystyle{ v }[/math] forms a cut which separates [math]\displaystyle{ v }[/math] from the rest of [math]\displaystyle{ V }[/math] and has size less than [math]\displaystyle{ |C| }[/math], contradicting the assumption that [math]\displaystyle{ |C| }[/math] is a min-cut. And the bound [math]\displaystyle{ |E|\ge \frac{|V||C|}{2} }[/math] follows directly from the fact that every vertex in [math]\displaystyle{ G }[/math] has degree at least [math]\displaystyle{ |C| }[/math]. [math]\displaystyle{ \square }[/math]

For a multigraph [math]\displaystyle{ G(V, E) }[/math], fixed a minimum cut [math]\displaystyle{ C }[/math] (there might be more than one minimum cuts), we analyze the probability that [math]\displaystyle{ C }[/math] is returned by the above algorithm.

Initially [math]\displaystyle{ |V|=n }[/math]. After [math]\displaystyle{ (i-1) }[/math] contractions, denote the current multigraph as [math]\displaystyle{ G_i(V_i, E_i) }[/math]. Suppose that no edge in [math]\displaystyle{ C }[/math] has been chosen to be contracted yet. According to Lemma 1 and 2, [math]\displaystyle{ C }[/math] must be a minimum cut of the [math]\displaystyle{ G_i }[/math]. Then due to Lemma 3, the current edge number is [math]\displaystyle{ |E_i|\ge |V_i||C|/2 }[/math]. Uniformly choosing an edge [math]\displaystyle{ e\in E_i }[/math] to contract, the probability that the [math]\displaystyle{ i }[/math]th contraction contracts an edge in [math]\displaystyle{ C }[/math] is given by:

[math]\displaystyle{ \begin{align}\Pr_{e\in E_i}[e\in C] &= \frac{|C|}{|E_i|} &\le |C|\cdot\frac{2}{|V_i||C|} &= \frac{2}{|V_i|}.\end{align} }[/math]

Therefore, assuming that [math]\displaystyle{ C }[/math] is intact after [math]\displaystyle{ (i-1) }[/math] contractions, the probability that [math]\displaystyle{ C }[/math] survives the [math]\displaystyle{ i }[/math]th contraction is at least [math]\displaystyle{ 1-2/|V_i| }[/math]. Note that [math]\displaystyle{ |V_i|=n-i+1 }[/math], because each contraction decrease the vertex number by 1.

The probability that no edge in the minimum cut [math]\displaystyle{ C }[/math] is ever contracted is:

[math]\displaystyle{ \begin{align} &\quad\,\prod_{i=1}^{n-2}\Pr[\,C\mbox{ survives all }(n-2)\mbox{ contractions }]\\ &= \prod_{i=1}^{n-2}\Pr[\,C\mbox{ survives the }i\mbox{-th contraction}\mid C\mbox{ survives the first }(i-1)\mbox{-th contractions}]\\ &= \prod_{i=1}^{n-2}\left(1-\frac{2}{|V_i|}\right) \\ &= \prod_{i=1}^{n-2}\left(1-\frac{2}{n-i+1}\right)\\ &= \prod_{k=3}^{n}\frac{k-2}{k}\\ &= \frac{2}{n(n-1)}. \end{align} }[/math]

Therefore, we prove the following theorem,

Theorem

For any multigraph with [math]\displaystyle{ n }[/math] vertices, the MinCut algorithm returns a minimum cut with probability at least [math]\displaystyle{ \frac{2}{n(n-1)} }[/math].

Run MinCut independently for [math]\displaystyle{ n(n-1)/2 }[/math] times and return the smallest cut returned. The probability that this the minimum cut is found is:

[math]\displaystyle{ \begin{align} 1-\Pr[\mbox{failed every time}] &= 1-\Pr[\mbox{MinCut fails}]^{n(n-1)/2} \\ &\ge 1- \left(1-\frac{2}{n(n-1)}\right)^{n(n-1)/2} \\ &\ge 1-\frac{1}{e}. \end{align} }[/math]

A constant probability!