Guessing a subset of $\{1,...,N\}$

I pick a random subset $S$ of $\{1,\ldots,N\}$, and you have to guess what it is. After each guess $G$, I tell you the number of elements in $G \cap S$. How many guesses do you need?

• see cs.mcgill.ca/~colt2009/papers/004.pdf if you know the cardinal of $S$ (which you can get in one guess) ( this could get a very good bound on average) Aug 25 '17 at 20:26
• Very nice! This answers my question, as it shows that the number of guesses required is $\Theta(N/\log(N))$. Aug 26 '17 at 1:48
• @DaveRadcliffe huh? didn't mjqxxxx show that $N/\log N$ guesses are required. The importance of the link is that $N/\log N$ guesses are sufficient. Apr 12 '20 at 19:31
• This is similar to brute force solving a true-false test, and the answer I give there applies here. Apr 2 '21 at 19:02
• My question was clear and complete, and it has been answered completely, for which I am greatly appreciative. Jul 16 '21 at 14:21

An obvious upper bound is $N$ queries, since you can test each element individually. On the other hand, it takes at least $\Omega(N/\log N)$ queries: $N$ bits of information are required to identify the target subset, and each query can yield at most $O(\log N)$ bits of information, since each query has only $O(N)$ possible answers. To see that the upper bound is not sharp, consider the following strategy for $N=5$, which takes at most $4$ queries:

1. Guess $\{1,2,3,4,5\}$. If the result is $0$ or $5$, we have the answer. If the result is $1$ or $4$, bisection search (for the single member or the single missing element) gives the answer in three more queries. Suppose the result is $2$ (the strategy for $3$ is the same by symmetry).
2. Guess $\{1,2\}$. If the result is $2$, we have the answer. If the result is $0$, then bisection search on $\{3,4,5\}$ (for the single missing element) gives the answer in two more queries. Suppose the result is $1$. Then we know the answer is $\{a,b\}$ for some $a \in \{1,2\}$ and $b \in \{3,4,5\}$.
3. Guess $\{1,3\}$. If the result is $2$, we have the answer. If the result is $0$, then the answer is $\{2,b\}$ for some $b\in\{4,5\}$, and one more query gives the answer. Suppose the result is $1$. Then we know the answer is $\{1,4\}$, $\{1,5\}$, or $\{2,3\}$.
4. Guess $\{1,4\}$. The answer is $\{1,4\}$ if the result is $2$, or $\{1,5\}$ if the result is $1$, or $\{2,3\}$ if the result is $0$.

This example gives an improved upper bound asymptotic to $4N/5$. It seems likely that the correct answer is strictly $o(N)$ (i.e., eventually less than $cN$ for any fixed $c$), but whether or not it's $\Theta(N/\log N)$, I can't say.

• "each query can yield at most O(logN) bits of information, since each query has only O(N) possible answers". However, since you get to choose what set to query, doesn't that change the amount of information you could theoretically gain?
– user2469
Mar 6 '11 at 17:19
• @barrycarter: Here's a longer explanation. Before each query, there are some number $X$ of possible subsets. The result of the query will be $i \in \{0,1,...,N\}$, so the $X$ subsets are partitioned into $N+1$ disjoint classes $X_i$ (where the subsets in $X_i$ are those consistent with the result $i$). The largest of these must have size at least $X / (N+1)$, and in the worst case, that is the result you'll get. So you've reduced the number of possible subsets from $X$ to no less than $X / (N+1)$; i.e., you've gained no more than $\log_{2}(N+1)$ bits of information about the answer. Mar 6 '11 at 18:20
• @mjqxxxx: we can say more. If your guess has $k$ elements, the central binomial coefficient is about $\frac {2^k}{\sqrt {\frac \pi 2 k}}$ so in the worst case you only get $\log_2 \sqrt {\frac \pi 2 k}$ bits (taking $\sqrt{\frac \pi 2}\approx 1)$. The first guess where $k=N$ will only get you about $\frac 12 \log_2N$ bits. I suspect after the first guess you want $k$ to be around $\frac N2$, so that gets you $\frac 12 \log_2 N -1$ bits Aug 25 '17 at 3:58
• @RossMillikan I must confess I don't know much about these things, but find them rather interesting nonetheless. Can you suggest any readings on the topic for me? Aug 25 '17 at 4:33
• @Fimpellizieri: I am not sure what you mean by "these things". The number of bits necessary to decide something is basic to information theory. A bit can give you the result of a yes/no question. Picking one item out of $N$ requires $\log_2 N$ bits because you can split the group in half and ask which half it is in $\log_2 N$ times and find it. One way to put a bound on the number of questions you need to ask is to count the bits needed and the number you get from each question, which is what we are doing here. Aug 25 '17 at 4:56

This can be solved in $$\Theta(N/\log N)$$ queries. First, here is a lemma:

Lemma: If you can solve $$N$$ in $$Q$$ queries, where one of the queries is the entire set $$\{1,\dots,N\}$$, then you can solve $$2N+Q-1$$ in $$2Q$$ queries, where one of the queries is the entire set.

Proof: Divide $$\{1,\dots,2N+Q-1\}$$ into three sets, $$A,B$$ and $$C$$, where $$|A|=|B|=N$$ and $$|C|=Q-1$$. By assumption, there exist subsets $$A_1,\dots,A_{Q-1}$$ such that you could find the unknown subset of $$A$$ alone by first guessing $$A$$, then guessing $$A_1,\dots,A_{Q-1}$$. Similarly, there exist subsets $$B_1,\dots,B_{Q-1}$$ for solving $$B$$. Finally, write $$C=\{c_1,\dots,c_{Q-1}\}$$.

The winning strategy is:

• Guess the entire set, $$\{1,\dots,2N+Q-1\}$$.

• Guess $$B$$.

• For each $$i\in \{1,\dots,Q-1\}$$, guess $$A_i\cup B_i$$.

• For each $$i\in \{1,\dots,Q-1\}$$, guess $$A_i\cup (B\setminus B_i)\cup \{c_i\}$$.

Using the parity of the the sum of the guesses $$A_i\cup B_i$$ and $$A_i\cup (B\setminus B_i)\cup \{c_i\}$$, you can determine whether or not $$c_i\in S$$. Then, using these same guesses, you get a system of equations which lets you solve for $$|A_i \cap S|$$ and $$|B_i\cap S|$$ for all $$i$$. This gives you enough info to determine $$A\cap S$$ and $$B\cap S$$, using the assumed strategy.$$\tag*{\square}$$

Let $$\def\Opt{\operatorname{Opt}}\Opt(N)$$ be the fewest number of guesses you need for $$\{1,\dots,N\}$$. Using the lemma and induction, you can show that $$\Opt(k2^{k-1}+1)\le 2^k\qquad \text{for all }k\in \{0,1,2,\dots\}$$ Note that when $$N=k2^{k-1}+1$$, we have $$\Opt(N)\le 2^k$$, and $$\frac N{\frac12\log_2 N}=\frac{k2^{k-1}+1}{\frac12\log_2(k2^{k-1}+1)}= 2^k(1+o(1))$$ It follows that $$\Opt(N)\in O(N/\log N)$$ when $$N$$ is of the form $$k2^{k-1}+1$$. Since $$\Opt(N+1)\le \Opt(N)+1$$, this extends to all $$N$$. Combined with the entropy argument, we get $$\Opt(N)\in \Theta(N/\log N)$$.

• very nice! ${}{}{}$ Jul 14 '21 at 21:30