Pergunta

I'm trying to show Amplification works for randomized algorithms, and for randomized approximation algorithms.


Amplification for randomized algorithms:
Given a randomized algorithm with time complexity $f(n)$ and error probability $\frac{1}{4}$, show there is an algorithm that solves the same problem, but with $O\left(f\left(n\right)log\left(\frac{1}{\varepsilon}\right)\right)$ run time, and error probability of $\varepsilon$, for a given $\varepsilon>0$.

I managed to show this running the original algorithm $k$ times, and choosing by majority vote the answer. Using Multiplicative Chernoff Bound (Which I've proven as well), I've shown $k\geq O\left(ln\left(\frac{1}{\varepsilon}\right)\right)$ gives us the required results.


Now I'm trying to do the same for approximation algorithms.

Amplification for randomized approximation algorithms:
Given a randomized approximation algorithm that returns an $\varepsilon$-additive-approximation with a probability of $\frac{3}{4}$, with run time complexity of $f\left(n\right)$, show there is an algorithm that returns an $\varepsilon$-additive-approximation with a probability of $1-\delta$, for the same problem, but with $O\left(f\left(n\right)log\left(\frac{1}{\delta}\right)\right)$ run time, for a given $\delta>0$.

The first thing that comes to my mind, is use a similar approach as before - run the original algorithm $k$ times, but this time instead of using majority vote, return the average.

I have to somewhow find a $k\geq O\left(ln\left(\frac{1}{\delta}\right)\right)$ such that $\mathbb{P}\left[\left|OPT-\frac{1}{k}\underset{i=1}{\overset{k}{\sum}}X_{i}\right|>\varepsilon\right]<\delta$.

I tried using the triangle inequality, then using union bound, but this doesn't lead me to being able to use Chernoff. In order to use Chernoff, I thought of defining an indicator random variable for each run of the original algorithm, stating if the result - $X_{i}$ is bad (that is, if $\left|X_{i}-OPT\right|>\varepsilon$), but this is where I get stuck.

Would appreciate any help. Thanks!

Nenhuma solução correta

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