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