# Trading off speed against the probability of success in the Karger-Stein Algorithm

(Note: this is an analysis of one aspect of the Karger-Stein algorithm, it’s not meant to be a beginner-friendly introduction)

Karger’s algorithm randomly contracts a graph and surprisingly, this can be used to find a minimum cut with probability $$\mathcal{O}(n^{-2})$$, where $$n$$ is the number of vertices of the graph. This is a much, much higher probability than sampling a graph cut uniformly at random would give! But it still means we need to run the algorithm $$\mathcal{O}(n^2\log n)$$ times to get a high success probability. Karger’s algorithm can be implemented in $$\mathcal{O}(n^2)$$ time1, which gives an overall runtime of $$\mathcal{O}(n^4 \log n)$$ – not great, there are much faster deterministic algorithms.

To understand how the Karger-Stein algorithm improves upon that, we need the following key result that forms the foundation for Karger’s algorithm:

Theorem2: When Karger’s algorithm contracts a graph from $$n$$ to $$r$$ vertices, any given mincut survives with probability $$\geq {r \choose 2}/{n \choose 2} = \frac{r (r - 1)}{n (n - 1)}$$.

In particular, for $$r = 2$$, we get the $$\mathcal{O}(n^{-2})$$ probability mentioned above.

But note the following: if we make only a few contractions, $$r \lesssim n$$, then mincuts are almost guaranteed to survive! This is the key insight that allows us to improve the runtime of Karger’s algorithm, leading to the improved Karger-Stein algorithm.

The idea is the following: first we contract the graph down to roughly $$\frac{n}{b}$$ vertices, where $$b$$ is small enough that mincuts are very likely to survive. Then we branch: we again contract the graph down by a factor of $$b$$, but we do so $$a$$ times independently from one another. $$a$$ needs to be chosen high enough that mincuts are very likely to survive in at least one of the branches. We repeat this process until we have contracted the graph down to just 2 vertices3. If we chose $$a$$ and $$b$$ right, at least one of the final leaves of our computational tree will likely contain a mincut. So we return the best cut we’ve found among all the leaves.

The Karger-Stein algorithm as it was originally described and as it is usually presented uses $$a = 2$$ and $$b = \sqrt{2}$$. So we always split the computation into two branches and reduce the number of vertices by a factor of $$\sqrt{2}$$ before branching again. But in this post, I would like to motivate where these numbers come from, as well as show that they’re not the only ones that work. So in the following, we’re going to analyze the “generalized Karger-Stein algorithm” with arbitrary $$a$$ and $$b$$.

## Success probability

As mentioned above, any minimum cut survives a contraction from $$n$$ to $$r$$ vertices with probability $$\geq {r \choose 2}/{n \choose 2} = \frac{r (r - 1)}{n (n - 1)}$$. I said we first contract to “roughly” $$\frac{n}{b}$$ vertices – to be precise we contract until $$\lceil \frac{n}{b} + 1\rceil$$ vertices are left, this will give us a nice bound. The probability that a mincut survives this contraction is4:

$$\begin{split} p &\geq \frac{\lceil \frac{n}{b} + 1 \rceil \lceil \frac{n}{b} \rceil}{n (n - 1)}\\\ &\geq \frac{(\frac{n}{b} + 1) \cdot \frac{n}{b}}{n (n - 1)}\\\ &\geq \frac{1}{b^2} \end{split}$$

We can now apply this bound recursively: After we have contracted to $$\lceil \frac{n}{b} + 1 \rceil$$ vertices, we can forget that this is a partially contracted graph, and just treat this number as the “new $$n$$”.

We will write $$p_k$$ for the survival probability if there are $$k$$ levels of recursion left before we reach the leaves of the tree. So $$p_0$$ will be the probability in the leaves of the recursion tree. Depending on when precisely we stop the recursion and what method we use to finish the contraction, $$p_0$$ might take different values, but all that matters for us is that it is some constant.

Using the bound we found above, we get the following recurrence: $p_{k + 1} \geq 1 - \left( 1 - \frac{p_k}{b^2} \right)^a$ What’s going on here? $$\frac{p_k}{b^2}$$ is a lower bound on the probability that any given mincut survives in one particular branch. So $$\left(1 - \frac{p_k}{b^2}\right)^a$$ is an upper bound on the probability that the mincut survives in none of the $$a$$ branches, and consequently $$1 - \left( 1 - \frac{p_k}{b^2} \right)^a$$ is a lower bound on the probability that it survives in at least one.

This recurrence doesn’t have an obious solution we can just read off but with some rewriting, we can get something that’s good enough for our purposes. Substituting $$z_k := \frac{b^2}{p_k} - 1$$, we get

$$\begin{split} z_{k + 1} &= \frac{b^2}{p_{k + 1}} - 1 \\\ &\leq \frac{b^2}{1 - \left(1 - \frac{1}{z_k + 1}\right)^a} - 1\\\ &= \frac{b^2 \left(z_k + 1\right)^a}{\left(z_k + 1\right)^a - z_k^a} - 1\\\ &\leq \frac{b^2 \left(z_k + 1 \right)^a}{a z_k^{a - 1}} - 1\\\ &\leq \frac{b^2}{a} z_k + \text{const} \end{split}$$

where we used $$z_k \geq 1$$ in the last step. The constant term may depend on $$a$$ and $$b$$ but not on $$z_k$$.

If $$a \geq b^2$$, then $$z_k \in \mathcal{O}(k)$$ which means that $$p_k \in \Omega\left(\frac{1}{k}\right)$$. The depth of recursion for a graph with $$n$$ vertices is $$\Theta(\log n)$$, so the overall success probability is $$\Omega\left(\frac{1}{\log n}\right)$$.

What this means in words: if we create enough branches (at least $$b^2$$) compared to how long we contract before branching again, then we get quite a high success probability – $$\Omega\left(\frac{1}{\log n}\right)$$ means that $$\log^2 n$$ runs are enough to get an overall success probability that approaches 1 as $$n \to \infty$$.

But what if $$a < b^2$$, i.e. if we don’t have enough branches at each stage? Then the inequality derived above only yields $$z_k \in \mathcal{O}\left(\left(\frac{b^2}{a}\right)^k\right)$$ so the success probability $$p_k$$ can be exponentially low in $$k$$. This means we’d have to repeat the algorithm a potentially exponential number of times, which would make it useless.

That still leaves the question: why does the Karger-Stein algorithm use $$a = b^2$$ in particular, when $$a > b^2$$ would give a success probability at least as high? For that we need to turn to the runtime complexity.

## Runtime

The runtime of the Karger-Stein algorithm can be described with the following recurrence: $T(n) = aT\left(\frac{n}{b}\right) + \mathcal{O}(n^2)$ The $$\mathcal{O}(n^2)$$ term is for contracting down to roughly $$\frac{n}{b}$$ vertices. At that point, we solve $$a$$ smaller version of the original problem, each of size $$\frac{n}{b}$$. That leads to the $$aT\left(\frac{n}{b}\right)$$ term.

This kind of recurrence is exactly what the Master theorem is for. In this case, if we choose $$a = b^2$$, we get a runtime of $$\Theta(n^2 \log n)$$ for a single run of the Karger-Stein algorithm. We already saw that with $$a < b^2$$, we get an exponentially low success probability, so that choice isn’t interesting anyway. Finally, if $$a > b^2$$, we get a high success probability, but the runtime becomes $$\Theta(n^c)$$, where $$c := \frac{\log a}{\log b} > 2$$.

We can summarize all our results (and a few I didn’t mention) in one table:

Condition Time for single run Success probability Total runtime Comment
$$a < b^2$$ $$\Theta(n^2)$$ exponentially low in $$n$$ exponential in $$n$$ Too little branching
$$a = b^2$$ $$\Theta(n^2 \log n)$$ $$\Omega\left(\frac{1}{\log n}\right)$$ $$\mathcal{O}(n^2 \log^3 n)$$ Just right
$$a > b^2$$ $$\Theta(n^c)$$ with $$c > 2$$ $$\Omega(1)$$ $$\Theta(n^c)$$ with $$c > 2$$ Unnecessarily much branching

The “total runtime” column contains the runtime that is needed to achieve a high success probability by repeating the Karger-Stein algorithm often enough (at least if $$n$$ is large enough). This is the complexity that we want to minimize in practice.

We can now see that combining the analysis of the success probability with the runtime analysis explains why the Karger-Stein algorithm uses $$a = b^2$$: in the other cases, we are either very unlikely to succeed and therefore need too many runs, or we are taking unnecessarily long for a single run.

But also note that any choice of a “branching factor” $$a$$ works, as long as we then choose $$b = \sqrt{a}$$. So splitting the computation up into just two subproblems is a reasonable and simple choice, but from a purely asymptotic perspective it is arbitrary.

1. There is also an implementation in $$\mathcal{O}(m)$$, where $$m$$ is the number of edges, but for the Karger-Stein algorithm that won’t make a difference and we’ll ignore it. ↩︎

2. David Karger: Global min-cuts in RNC, and other ramifications of a simple min-cut algorithm, SODA 1993 ↩︎

3. Usually we stop a bit before 2 vertices and just compute the mincut from there using other methods but that doesn’t matter here ↩︎

4. This and the following analysis is based on the paper by Karger and Stein: A new approach to the minimum cut problem, Journal of the ACM 1996. The difference is just that I consider arbitrary $$a$$ and $$b$$, rather than just $$a = 2$$ and $$b = \sqrt{2}$$. ↩︎