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Let $s$ be an unknown bit string of length $n$. Let $p(i, b)$ be the probability that $i$-th bit of $s$ is equal to $b \in \{0,1\}$. What's the fastest method to find $s$, given the distribution $p()$? "Fastest" meaning using the least (in the expected value sense) number of queries to an oracle $O(x)$, such that $O(x)=1$ iff $x=SECRET$.

I'm interested in all observations about this problem, not only the optimal solution.

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Does $p(i,b) = p(j,b)$ for all $i$ and $j$? Are the events "$i$-th bit of $s$ is $1$" and "$j$-th bit of $s$ is $1$" independent for all $i \neq j$? – Dilip Sarwate Jun 27 '12 at 15:45
@DilipSarwate: The first equation does not hold (p can be anything). Events are independent. I think it's sufficient to start looking from the most probable bitstring to the least probable one, but I don't know how to show that. – zxc Jun 27 '12 at 15:54
@zxcv: The optimal search pattern is always in descending order of probability; the challenge is in finding an algorithm to enumerate the strings in this order. If the bits are independent, you may want to look at enumerating the strings in descending order of the log-likelihood $\lambda(s) = \log p(s) = \log \prod_{i=1}^n p(i,s_i) = \sum_{i=1}^n \log p(i,s_i)$, which is a linear map from $\{0,1\}^n$ to $\mathbb R^- \cup \{-\infty\}$; by the monotonicity of the logarithm, $p(a) < p(b)$ iff $\lambda(a) < \lambda(b)$. – Ilmari Karonen Jun 27 '12 at 16:42
up vote 2 down vote accepted

Since the metric is the number of queries to the oracle, how the words are represented doesn't matter so we can simply say that the secrets are words $w_1,\dots,w_n$, with probability $p_1\ge\dots\ge p_n$ (satisfying $\sum_i p_i=1$).

Now the probability of success of any algorithm after $k$ queries to the oracle is at most $\sum_{i=1}^k p_i$. Since the algorithm that simply queries $w_1,\dots,w_n$ in that order achieves this upper bound, that algorithm is optimal in every reasonable sense of the word (in particular, it has minimum expected number of queries, median number of queries, and so on).

So now, back to the original problem: to generate $w_1,\dots,w_n$ with decreasing probability, let's assume (up to flipping and sorting of the bits) that $p(i,0)\ge p(i,1)$, and $p(1,0)\ge p(2,0)\ge\dots\ge p(n,0)$.

Initially, let the candidate set be $\{0^n\}$. Then, generate a new word by selecting the most probable word $w$ from the candidate set (and removing it), and update the candidate set by adding all words $w'$ equal to $w$ except for one bit that is changed from 0 to 1. There are at most $np$ words in the candidate set (where $p$ is the number of words already generated) and computing the probability of a given word is $O(n)$, so the algorithm is polynomial in the number of requests to the oracle (and in $n$).

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By "secrets are words ..." do you mean $P(w_i = SECRET) = p_i$? – zxc Jun 27 '12 at 17:01
Exactly. Here "word" is synonymous with "bit string", since the alphabet is $\{0,1\}$. – Generic Human Jun 27 '12 at 17:05
There's only one secret, so "secrets" is a bit misleading. – zxc Jun 27 '12 at 17:08
How much does the skewed distribution help? For example, what's the expected number of queries for $p(.,.)$ such that $p(i,0)=1-\epsilon$ for all $i$? – zxc Jun 27 '12 at 17:12
Oh. I meant, the space of all possible secrets (in all possible instances of the problem), which is a subset of the space of words $\{0,1\}^*$. At that point I was just looking at the formulation of the problem, not fixing a particular instance of it. – Generic Human Jun 27 '12 at 17:12

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