Soritng and inserting the rows of a bitset matrix....Brain teaser!!!

Question

Initially, i have two bitset matrices:
1. temp_matrix:Initially, it has all the rows,say n, in the sorted order based on their cradinality.
2.Final_matrix:Intially, empty.
Operation being performed on the bitset matrice:
1. Take each row of temp_matrix and perform "and" operation of it with all the other rows underneath it.Take the second row and perform and operation with all the orther rows underneath it and repeat the same with all the other rows till u reach the n-1the row which will be "Anded" with nth row.
2."Anded" result of each of two rows has to be stored in the Final_matrix in a sorted order based on its cardinality.
For example: say at any instant Final_matrix had the following state:
{1}
{3}
{1,2,3}
{3,4,5,6}
{1,2,3,4,5,6}
And suppose i have to insert a new row say: {4} in it at 3rd position
How should i do that..i mean which sorting method will be time efficent and how should i implement it in terms of inserting a new row in the list with the issues of shifting the rows etc.
The datastructure i am using is BitSet to store the bits.
Any suggestions !!!

Answer 1

Can u tell me how to wrap the BitSet entries in the Linked Lists.
I am pasting snap shot of my code,if that helps:
for( i=0;i<count_bit_vector_global-n;i=i+n)
long initial=0;
initial= System.currentTimeMillis();
count_j=0;
for(int j=i;j<i+n;j++)
if(temp_vector1.get(j)==true)
result_row_vector.set(count_j);
else
result_row_vector.clear(count_j);
count_j++;
}/*End for j:Result row created*/
if(result_row_vector.cardinality()>num_one)
for(int k=(i+n);k<count_bit_vector_global;k=k+n)
check=0;
if((temp_vector1.get(i,i+n)).intersects(temp_vector1.get(k,k+n))==true)/*All result elements are not zero*/
count_j=0;
for(int j=i;j<i+n;j++)
if(temp_vector1.get(j)==true)
result_row_vector.set(count_j);
else
result_row_vector.clear(count_j);
count_j++;
result_row_vector.and(temp_vector1.get(k,k+n));
while(check<count_bit_vector)
if((result_row_vector.get(0,count_j)).equals(bit_vector1.get(check,check+n)))
flag=true;
s2=track_vector.get(check/n).toString();
s1= temp_track_vector.get(i/n).toString().concat(b).concat(temp_track_vector.get(k/n).toString());
track_vector.set(check/n,concat_string(s2,s1));
break;
else
flag=false;
     check=check+n;
}/*End while*/
In this code n decides the length of one row.U can see i am simulating a 1D BitSet as 2D BitSet matrix.
Do pay any heed to the track_vector.
But i will summarize what i am doing here.
Temp_vector1 is the initial bitset storing the original matrix.
bit_vector1 is the bitset storing the new "Anded" rows.
"i" decides which row will be compared to all the other rows after that.
"k" decides the rows which are after the row determined by "i".
I am taking the row decided by "i" and then store that in result_vector and then "And" it with the other rows decided by "k".
"count_bit_vector" tells the number of bit elements in temp_vector
"check" tells the position of the row in the bit_vector1 which is similer to the row obtained after "And" operation.
If i find the row i don't add it in the bit_vector1 but if its not there then it is added.
Right now, before rejecting a row i have to compare all the rows which are there in the bit_vector1 and that is taking time.