Assume that you have a set S of having 2^2 elements first, let S={0,1,2,3}
Then the desired 2 partitions would be
1-{{0,1},{2,3}} 
2-{{0,2},{1,3}} 
3-{{0,3},{1,2}}


If S={0,1,2,3,4,5,6,7} having 2^3 elements then similarly the 2 partitions would be


1- {{0,1},{2,3},{4,5},{6,7}}
2- {{0,2},{1,3},{4,6},{5,7}}
3- {{0,3},{1,5},{2,6},{4,7}}
4- {{0,4},{1,6},{2,5},{3,7}}
5- {{0,5},{1,4},{2,7},{3,6}}
6- {{0,6},{1,7},{2,4},{3,5}}
7- {{0,7},{1,2},{3,4},{5,6}}


So it seems that for S having 2^n elements the initial table would like


1: {{0,1},{2,3},{4,5},{6,7},...,{(2^n)-2,(2^n)-1}}
2: {{0,2},... }
.
.
.
(2^n)-1: {{0,(2^n)-1},... }


I am just wondering whether there is an direct approach to generate the above table for a given n. 


Thanks



Is there any direct approach to generate discrete 2 partitions of a set of having 2^n elements for a given n ?
I presume you want $2^n-1$ partitions of ${0,\ldots,2^n-1\}$
into parts of size two so that each possible pair occurs
exactly once in a partition. You can do this as follows.
Write the numbers in question base $2$. Then each can be represented
by $n$ binary digits going from $00\cdots0$ to $11\cdots 1$. Define
an operation $\oplus$ on these as follows: we obtain $a\oplus b$
by add corresponding digits in $a$ and $b$ modulo $2$.
For example $11\oplus 14=5$ as in base $2$, $11$ and $14$ are
$1011$ and $1110$, and $1+1$ is $0$ mod $2$, $0+1$ is $1$ mod $2$,
$1+1$ is $0$ mod $2$ and $1+0$ is $1$ mod $2$ so in binary $a\oplus b$
is $0101$ which is $5$.
The operation $\oplus$ is sometimes known as "nim addition" as it is used
in the analysis of the game nim.


To get the $j$-th partition where $1\le j\le 2^n-1$ we just pair off
$a$ and $a\oplus j$. So for $n=4$, the partition for $j=5$
is $\{ \{0,5 \}, \{1,4 \}, \{2,7 \}, \{3,6 \}, \{8,13 \}, 
\{9,12 \}, \{10,15 \}, \{11,14 \} \}$
as $0\oplus 5=5$, $1\oplus 5=4$, $2\oplus 5=7$ etc.


