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Advanced Algorithms (Winter 2026-27) - 04-08 Correctness for Closest String FPT

Sebastian Wild (Lectures) · 5:22 · Watch on YouTube

Part of University of Marburg CS 627. See all 14 lectures and search them

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Advanced Algorithms (Winter 2026-27) - 04-08 Correctness for Closest String FPT Watch on YouTube →

Overview

Sebastian Wild explains the correctness of a fixed-parameter algorithm for the p-closest string problem, which asks whether a consensus string lies within Hamming distance k of every input string. Starting from the first input string, the algorithm branches on k + 1 mismatching positions because every valid solution must fix at least one; with depth k, this yields a search space of (k + 1)^k, or roughly k^k.

Key takeaways

Chapters

0:00 p-Closest String Setup and the Dirty-Column Bound
1:19 Why Correct Branches Need Only k + 1 Mismatching Positions
4:12 Search-Tree Cost and Maintaining Distances

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Transcript excerpt

0:01 The problem we were looking at is the p-closest string problem in the parameterized setting. So we have strings of the same length and a threshold and the question is can you find some consensus string that's at most at Hamming distance k from all of them. And we had some supporting lemmas that if you have too many dirty columns where there's more than one letter then you're out of luck. Specifically if you have more than m * k then one string definitely has distance k + 1 to some um well any string you pick one of the strings in your input has distance k + 1 to uh your chosen consensus string. Um then I jumped straight to the the final solution for our FPT algorithm.

0:51 Um written down here maybe um just intuitively you start walking away from the first string as one potential guess of what this consensus string could be. And you give yourself a budget of k steps that you can okay mismatches to s 1 you're allowed to introduce. More than that would certainly not be a consensus string with respect to s 1 so that's no no point in trying. And the one step we left open correctness wise last time was uh why it's okay to just look at k + 1 positions here. So a simple argument but uh let's let's revisit that. Um first of all if we're even in this

1:38 position then we already ruled out that we're in a completely hopeless situation. If we've arrived somewhere where we have negative budget then we're in trouble. And also if we have the budget to still walk away uh to still walk and the distance we may at the end have if that is more than distance to any string, then we're also

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