# Approximating a partition

Let $$X = \{1,\ldots,2 \cdot n\}$$ be a set of numbers. For this question, an equal partition $$Y_1,Y_2$$ of $$X$$ is a partition of $$X$$ to two equal sizes. Let $$\varepsilon \in (0,1)$$ be an error parameter. Define an $$\varepsilon$$-cover of $$X$$ of size $$k$$ as equal partitions $$(Y^i_1,Y^i_2)_{i \in \{1,\ldots,k\}}$$ such that the following holds: For any equal partition $$Z_1, Z_2$$ of $$X$$, there is $$i \in \{1,\ldots,k\}$$ such that $$|Z_1 \cap Y^i_1| \geq (1-\varepsilon) \cdot n$$. That is, an $$\varepsilon$$-cover contains a good approximation for any equal partition.

My question is what is the smallest size of an $$\varepsilon$$-cover? Can it be of size independent of $$n$$? can it be polynomial in $$1/\varepsilon$$?

This seems like a fundamental combinatorial question and I assume variants of this problem have been studied. However, I am unfamiliar with the right keywords to find such results. Any form of help would be very appreciated.

Right keyword is "covering codes". If instead of a partition you consider two $$\{0,1\}$$-vectors consisting of $$n$$ 1's and $$n$$ 0's (and summing up to the all-1 vector), you want to cover the set of all such vectors by $$2k$$ Hamming balls of radius $$2\varepsilon n$$. The standard entropy bound shows that the volume of every ball grows as $$2^{H(2\varepsilon) n}$$, where $$H(t)=-t\log_2t-(1-t)\log_2(1-t)$$, that gives $$2^{(1-H(2\varepsilon))n}$$ lower bound for $$2k$$.