KGS math club/solution 11 2

Given n, k, and a k-member subset K of the n-member set N, we form the (n-k)-member subset it is paired with as follows.

Assign integers modulo n to the members of N, and cyclically order its members using them. Remove from N a member of K with a non-member of K that immediately follows it; and repeat until there are no members of K left. Append what we have left to K.

Solution by iceweasel.

He supplied this C code to compute it:

 main(int a, char **v) {   
   int n=atoi(v[1]), 
   ss=atoi(v[2]),
   x0=0,
   x1=0,
   m;  
   
   for ( m=1<<n ; m>>=1 ; (m&ss?(x1|=m):x1?(x1&=(x1-1)):(x0|=m)) );  
   for ( m=1<<n ; m>>=1 ; m&x0&&x1&&(x1&=(x0^=m,x1-1)) );   
   printf("%d\n",ss^x0^x1); 
 }

The arguments to the program are two integers: n, and one whose set bits correspond to the elements of K.

He also supplied this Perl code:

$_=shift;print $_;while(s/[A-Z][a-z]//){}while(s/^[a-z](.*)[A-Z]$/\1/){}print "^$_\n";

which requires presenting the subset as a word in mixed upper (members) and lower (non-member) case.