TITLE: Puzzle on deleting k bits from binary vectors of length 3k
QUESTION [50 upvotes]: Consider all $2^n$ different binary vectors of length $n$ and assume $n$ is an integer multiple of $3$.  You are allowed to delete exactly $n/3$ bits from each of the binary vectors, leaving vectors of length $2n/3$ remaining. The number of distinct vectors remaining depends on which bits you delete. Assuming your aim is to leave as few remaining different vectors as possible, how few can you leave as a function of $n$?
Example, $n=3$.  You can leave only the two vectors $11$ and $00$.
Following comments at the math.se site (in particular by Jack D'Aurizio), in general for larger values of $n$ you can replace any block of three consecutive bits by either $00$ or $11$.  This gives an upper bound of $2^{n/3}$. Is this in fact the correct answer?
Now I have some code to solve small instances, we can start to fill in a table of optimal results. We use the notation $H(n,b)$ to indicate the smallest number of distinct vectors that results from starting with all vectors of length $n$ and removing $b$ from each.
$$12 \leq H(15,5) \leq17$$
$$H(12,4) = 10$$
For $n=10$ and $b = 1,2,3,4$ we have $\leq140,\leq 31,10, 4$
For $n=9$ and $b = 1,2,3,4$ we have $70,18,6,2$
For $n=8$ and $b = 1,2,3$ we have $40,10,4$
For $n=7$ and $b = 1,2,3$ we have $20,6,2$
For $n=6$ and $b = 1,2$ we have $12,4$
For $n=5$ and $b = 1,2$ we have $6,2$
$$H(4,1)  = 4, H(3,1) = 2, H(2,1) = 2$$
If we only allow symmetric solutions then $H(10,2)=32$. This implies (assumnig no error in the calculations) that for some instances there may be no symmetric optimal solutions as we already have $H(10,2) \leq 31$.

REPLY [4 votes]: I think it may be helpful and/or intersting to consider the problem for fractions other than $1/3$.
Specifically, if $n$ is a multiple of $q$, let $f(p/q,n)$ be the minimal size of a set of $n- (p/q)n$ bit strings such that deleting $(p/q)n$ bits from each $n$-bit string produces an element of the set. Then set
$$g(p/q) = \lim _{n\to \infty} \frac{\log f\left(\frac{p}{q},n\right)}{n}$$
The limit exists because, by the divide-into-independent-blocks argument, any particular value is a bound for the $\lim\sup$.
It is clear that $g$ is monotonic. It's not immediately obvious to me if we can prove that $g$ is continuous.
Clearly $g(x)=0$ for $x \geq 1/2$.
For a simple upper bound on $g$ for $x<1/2$, we can assume with the loss only of a constant that there are fewer $0$s than $1$s. Then divide the $n$ bits into $q$ equally sized parts, and delete all the $0$s in the $2p$ parts with the fewest ones. This takes $2^{ (1-2p/q)n }$, giving an upper bound $g(x) \leq (1-2x) \log 2$. 
Thanks to Brendan McKay and Yury,  we have a lower bound on $g$. If I understand this bound correctly, it is that $g(x) \geq \log 2 + x  \log x + (1-x) \log(1-x)$. (We can easily check that the maximal term in the $x=1/3$ case remains maximal for all $0<x<1/2$.)
Specific examples, like, Adam Goucher's, can give us tighter upper bounds for specific values of $x$.