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Ravi is an armchair futurist and an aspiring mad scientist. His mission is to create simplicity out of complexity and order out of chaos.

Friday, December 16, 2011

Randomized two point sampling

While reading "Randomized Algorithms" by Motwani and Raghavan, I felt that the authors' treatment of the topic of "two point sampling" was a little too short and summarized. In this article, I provide some details around this topic.

When using randomized algorithms, for a given input $x$, the usual strategy is to choose $n$ random numbers (a.k.a. "witnesses") and run a deterministic algorithm on the input using each of the random numbers. The intuition is that if the deterministic algorithm has a probability of error $\epsilon$ (e.g. $1/3$), $n$ independent runs reduce this probability to $\epsilon^n$ (e.g. $1/3^n$).

The above assumes that we have the option to choose $n$ independent random numbers. What happens if that option were constrained to choosing no more than a constant $c$ random numbers? We look at the simple case where we choose just 2 random numbers ($c=2$), giving it the name two-point.  So, the main question we are trying to answer is the following:

If our randomized algorithm reduced the probability of error to $\epsilon^n$ with $n$ random numbers, what can we expect about the probability of error with $2$ randoms? $\epsilon^2$ is an obvious bound, but can we do any better? It turns out that we can indeed do better and reduce the probability of error to $1/n$. This is significantly better than the constant, $\epsilon^2$.
The High-level Idea
The idea is to generate $n$ numbers based on the chosen 2 random numbers and use them in lieu of the $n$ purely independent random numbers. These generated numbers are dependent on the 2 chosen numbers. Hence they are not independent, but are pairwise independent. This loss of full independence reduces the accuracy of the algorithmic process, but it is still quite usable.

The Process

  1. Choose a large prime number, $\rho$ (e.g. Mersenne prime, $2^{31}-1$).
  2. Define $\mathbb{Z}_\rho$ as the ring of integers modulo $\rho$. (e.g. $0,1,\cdots,2^{31}-2$)
  3. Choose 2 random numbers, $a$ and $b$ from $\mathbb{Z}_\rho$. (e.g. $a=2^{20}+781$ and $b=2^{27}-44$). These are the only independent random numbers that are picked for two-point sampling.
  4. Generate $n$ "pseudo-random" numbers $y_i=(a*i+b)\mod{\rho}$. E.g.
    1. $y_0=2^{27}-44$
    2. $y_1=(2^{20}+781)*1+2^{27}-44$
    3. $y_2=(2^{20}+781)*2+2^{27}-44$ and so on.
  5. Use each of $y_i$s as witnesses in lieu of independent random witnesses, for algorithm $\mathbi{A}$.
  • Let $\mathbi{A}(x,y)$ denote the output of the execution of a deterministic algorithm $\mathbi{A}$ on input $x$ with witness $y$. We restrict this algorithm to return only $1$ or $0$. If the algorithm outputs $1$, we can be sure that $1$ is the right answer. If the algorithm outputs $0$, the right answer could be either $1$ or $0$.
  • $\mathbi{A}(x,y)$ produces correct output (which could be either $1$ or $0$) with probability $p$, meaning it produces a possibly erroneous output (i.e. a $0$ instead of $1$) with probability $1-p$.
  • Let $Y_i=\mathbi{A}(x,y_i)$ be a random variable, where $y_i$ is a psuedo-random number defined earlier. Note that the expected value of $Y_i$, $\mathbf{E}[Y_i]=p$, since $\mathbi{A}(x,y)$ produces a correct output with probability $p$.
  • Let $Y=\sum{Y_i}$. By linearity of expectations, $\mathbf{E}[Y]=\mathbf{E}[\sum{Y_i}]=\sum{\mathbf{E}[Y_i]}=np$. Note that linearity of expectations is agnostic to the dependence between the random variables $Y_i$.
  • Note that variance of $Y$, $\sigma^{2}_{Y}=np(1-p)$.

We are trying to determine the error for the randomized process after $n$ runs of the deterministic algorithm $\mathbi{A}$. This is at most the probability of the event where each of the $n$ runs produced an erroneous output of $0$, i.e. each of $Y_i=0$, or equivalently, $Y=0$. (Note that some outputs of $0$ are not erroneous, hence the use of the words "at most".) So we are trying to determine $\mathbf{P}[Y=0]$.

Since $Y_i$s are not all independent, we cannot say:

So, alternatively, we would like to know:
which, by Chebyshev's inequality is:

Intuitively, this means that for any given error bound $\epsilon$, instead of requiring $O(\log(1/\epsilon))$ runs of the deterministic algorithm $\mathbi{A}$ in the case of $n$ independent random witnesses, we will require $O(1/\epsilon)$ runs in the case $2$ independent random witnesses. This is exponentially larger. So, it is not as efficient, but it is not the end of the world either!

In Closing
The above derivation shows that two random numbers are sufficient to minimize the error of a randomized algorithm below any given bound, although this comes with loss of some efficiency.

  1. Motwani, R. and Raghavan, P., Randomized Algorithms, 2007.

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