The Karger-Stein Algorithm is Optimal for $k$-cut

Speaker: Jason Li , Carnegie Mellon University (CMU)

Date: Tuesday, February 11, 2020

Time: 4:00 PM to 5:00 PM Note: all times are in the Eastern Time Zone

Public: Yes

Location: 32-G575

Event Type: Seminar

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Host: Quanquan Liu, Sitan Chen, NIkhil Vyas, CSAIL MIT

Contact: Rebecca Yadegar,

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Reminder Subject: TALK: Jason Li: The Karger-Stein Algorithm is Optimal for $k$-cut

Abstract: In the $k$-cut problem, we are given an edge-weighted graph and want to find the least-weight set of edges whose deletion breaks the graph into $k$ connected components. Algorithms due to Karger-Stein and Thorup showed how to find such a minimum $k$-cut in time approximately $O(n^{2k-2})$. The best lower bounds come from conjectures about the solvability of the $k$-clique problem and a reduction from $k$-clique to $k$-cut, and show that solving $k$-cut is likely to require time $\Omega(n^k)$. Our recent results have given special-purpose algorithms that solve the problem in time $n^{1.98k + O(1)}$, and ones that have better performance for special classes of graphs (e.g., for small integer weights).

In this work, we resolve the problem for general graphs, by showing that for any fixed $k \geq 2$, the Karger-Stein algorithm outputs any fixed minimum $k$-cut with probability at least $\widehat{O}(n^{-k})$, where $\widehat{O}(\cdot)$ hides a $2^{O(\ln \ln n)^2}$ factor. This also gives an extremal bound of $\widehat{O}(n^k)$ on the number of minimum $k$-cuts in an $n$-vertex graph and an algorithm to compute a minimum $k$-cut in similar runtime. Both are tight up to $\widehat{O}(1)$ factors.

The first main ingredient in our result is a fine-grained analysis of how the graph shrinks---and how the average degree evolves---under the Karger-Stein process. The second ingredient is an extremal result bounding the number of cuts of size at most $(2-\delta) OPT/k$, using the Sunflower lemma.

Research Areas:
Algorithms & Theory

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Created by Rebecca Yadegar Email at Wednesday, January 22, 2020 at 1:30 PM.