Who owes who money optimization

Question

Say you have n people, each who owe each other money. In general it should be possible to reduce the amount of transactions that need to take place. i.e. if X owes Y £4 and

Answer 1

This problem may be tackled with the Warshall algorithm.

for(i=0;i owes[j][i] )
owes[i][j] -= (owes[i][j] - owes[j][i]), owes[j][i] = 0;

for(k=0;k owes[i][j] )
{
int diff = owes[j][k] - owes[i][j]; 
owes[i][j] = 0;
owes[i][k ] += diff;
owes[j][k] -= diff;
} 
}

After the algorithm finishes, the total number of transactions required would be the number of positive entries in the owes table.

I have not verified yet whether the algorithm will work, based on nature of the problem it may work. Solution is O(N^3).