TITLE: Maths Puzzle: Partitioning a set into two disjoint sets
QUESTION [5 upvotes]: Le $X$ be the set of all non-empty subsets of $\{a,b,c,d,e,f\}$. So $X=\{a,b,c,d,e,f,ab,ac,ad,ae,af,bc,bd,be,bf,cd,ce,cf,de,df,ef,abc,\cdots,abcdef\}$; i.e., $|X|=63$. We want to partition $X$ into two disjoint sets $Y$ and $Z$ such that:
(i) $|Y|=13$;
(ii) Given any element $z \in Z$, $\exists ~ y_1,y_2 \in Y$ such that $y_1y_2=z$;
(iii) If $x_1,x_2 \in Y$, then $x_1x_2 \notin Y$.
Note the following:
(1.) $(X,\cdot)$ is commutative: Given $ x=ab,y=cd\in X$, then $x \cdot y=(ab)(cd)=abcd=adbc=dcba=\cdots=cdab=y \cdot x$;
(2.) Two same letters cancels themselves: $(ab)(ac)=aabc=bc$.
My target is to find $Y$. I have tried many things to no avail. For instance, $Y$ cannot consist of all words of lengths $1$, $5$ and $6$ since this cannot yield all the words of length $3$ or $4$.
Any help in solving for $Y$ will be highly appreciated.

REPLY [2 votes]: It has been pointed out in the comments that the problem is equivalent to the following: find a subset $Y$ of the vector space $V=\mathbb{F}_2^6$ such that $|Y|=13$, and $Y\oplus Y=V\setminus Y$.
Necessarily, the elements of $Y$ must be linearly independent (otherwise you could only get $32$ linear combinations, much less $50$ sums.)  The conditions on $Y$ are invariant under invertible linear transformations, so we can assume without loss of generality that the density-one vectors (i.e. singleton subsets) are in $Y$.
Thus, all density-two vectors must be in $V\setminus Y$.  We put two complementary density-three vectors into $Y$ -- we can assume these are $(1,1,1,0,0,0)$ and $(0,0,0,1,1,1)$ -- so that $(1,1,1,1,1,1)\in Y\oplus Y$. This leaves us $5$ vectors to choose in order to obtain all vectors of density $3$, $4$ and $5$.  Since the singletons are in $Y$, we pick all $5$ of the remaining vectors to have density $4$, such that (a) each density-$5$ vector contains one of them (i.e. the density $4$ vectors taken together have a zero in every position), and (b) each density-$3$ vector is contained in one of the density-$4$ vectors.  This is easy to do; one solution is:
$${a, b, c, d, e, f, a b c, d e f, b c d f, a c d e, a b d f, a c e f, 
 a c d f}$$
Incidentally, for $n=7$ letters, there is a solution with $|Y|=21$:
$${a, b, c, d, e, f, g, a c e, a b c e f g, a c d f g, b e f g, b c g, 
 a e f g, a d f, b c d f, a b f, a b d, a c d e f, d e f, c e f g, 
 a b d f g}$$
I suspect it is optimal.