Randomized Global Min-Cut (Karger’s Algorithm)

Let C be a specific global min-cut of size k. The algorithm succeeds if it never contracts any edge in C.

• Since the min-cut size is k, the degree of every vertex must be \ge k.

• Thus, the total number of edges m \ge \frac{nk}{2}.

• Probability of picking a min-cut edge in round 1: \Pr(e_1 \in C) = \frac{k}{m} \le \frac{k}{nk/2} = \frac{2}{n}

• Probability of survival: \Pr(e_1 \notin C) \ge 1 - \frac{2}{n} = \frac{n-2}{n}.

• Round i survival probability:

  • When n-i+1 vertices remain, the survival probability in round i is \ge 1 - \frac{2}{n-i+1} = \frac{n-i-1}{n-i+1}.

The probability that C survives all n-2 contraction rounds:

• By independence of conditional choices: \Pr(\text{Success}) \ge \prod_{i=1}^{n-2} \frac{n-i-1}{n-i+1} = \frac{n-2}{n} \cdot \frac{n-3}{n-1} \cdot \frac{n-4}{n-2} \cdot \frac{n-5}{n-3} \dots \frac{4}{6} \cdot \frac{3}{5} \cdot \frac{2}{4} \cdot \frac{1}{3}

• How terms cancel:

  • Every term j from 3 to n-2 appears exactly once in the numerator and once in the denominator (shifted by 2 positions).
  • Writing the remaining terms after cancellations: \Pr(\text{Success}) \ge \frac{(n-2)(n-3)\dots(3)(2)(1)}{(n)(n-1)(n-2)\dots(4)(3)} = \frac{2 \cdot 1}{n(n-1)} = \Omega\left(\frac{1}{n^2}\right)

• Note: This is small, but polynomially bounded!

We can use the success probability of Karger’s algorithm to prove a fundamental combinatorial theorem using the Probabilistic Method:

Theorem: Any undirected, connected graph G = (V,E) on n vertices has at most \binom{n}{2} = \frac{n(n-1)}{2} distinct global min-cuts.

• Let C_1, C_2, \dots, C_r be the distinct global min-cuts of G (each of size k).

• For any specific min-cut C_i, Karger’s algorithm outputs C_i with probability: \Pr(\text{Algorithm outputs } C_i) \ge \frac{2}{n(n-1)} = \frac{1}{\binom{n}{2}}

• Crucially, the algorithm outputs exactly one cut per run. The events of outputting different cuts C_i and C_j are mutually exclusive: \sum_{i=1}^r \Pr(\text{Algorithm outputs } C_i) \le 1

• Substituting our lower bound: \sum_{i=1}^r \frac{1}{\binom{n}{2}} \le 1 \implies r \cdot \frac{1}{\binom{n}{2}} \le 1 \implies r \le \binom{n}{2}

• Power of Randomization: A purely deterministic, structural combinatorial bound proved easily via a randomized algorithm!

We repeat the contraction algorithm N times independently and return the smallest cut found.

• The probability that we fail to find C in all N runs is: \Pr(\text{Failure}) \le \left(1 - \frac{2}{n(n-1)}\right)^N \approx \left(1 - \frac{2}{n^2}\right)^N

• Using the identity 1 - x \le e^{-x}, we set N = c \cdot \frac{n^2}{2} \ln n: \Pr(\text{Failure}) \le e^{-\left(\frac{2}{n^2}\right) \cdot c \frac{n^2}{2} \ln n} = e^{-c \ln n} = n^{-c}

• Conclusion: Repeating O(n^2 \log n) times finds the min-cut with high probability (1 - 1/n^c).

• Total Runtime: A single contraction run takes O(n^2) time. Repeating O(n^2 \log n) times yields a total runtime of O(n^4 \log n).

If deterministic Hao–Orlin runs in O(n^2 \log n) time, why study O(n^4 \log n) Karger?

A branching recursive contraction algorithm to speed up the process:

Why is the success probability \Omega(1/\log n) instead of \Omega(1/n^2)?

Answer each question based on Karger’s algorithm:

1. If a graph has a unique min-cut of size 2, what is the lower bound on the probability that a single run of Karger’s contraction finds it?
2. If we contract a graph down to 4 vertices instead of 2, how does the survival probability of the min-cut change?

• 1. \frac{2}{n(n-1)} (the size of the min-cut k=2 cancels out in the probability bound).
• 2. The telescoping product stops earlier, yielding a much higher survival probability: \ge \frac{\binom{4}{2}}{\binom{n}{2}} = \frac{12}{n(n-1)}.