I am writing a routine to store an $M$-by-$N$ sparse matrix in a balanced binary tree. The insertion routine calls a comparison function to determine where a new matrix entry $(i,j)$ should be inserted in the tree. I am defining the ordering of the matrix elements simliar to how C handles a row-major ordered dense matrix. In particular, a matrix element $(i_a,j_a)$ comes before $(i_b,j_b)$ if: $$ N i_a + j_a < N i_b + j_b $$ I need to write a routine which returns $-1$ if $(i_a,j_a)$ is less than $(i_b,j_b)$, $+1$ if $(i_a,j_a)$ is greater than $(i_b,j_b)$ and $0$ if they are equal. I want this routine to be as fast as possible since I will be calling it hundreds of millions of times.

I've tried two options so far, which I implement as macros to avoid the overhead of a function call.

Option 1: Comparing rows first and then compare columns

#define COMPARE(ia,ja,ib,jb) (ia < ib ? -1 : (ia > ib ? 1 : (ja < jb ? -1 : ja > jb)))

The worst case for this option is 4 comparisons between integers.

Option 2: Compare full linear index

#define COMPARE(ia,ja,ib,jb) (ia*N + ja < ib*N + jb ? -1 : ia*N + ja > ib*N + jb)

The worst case for this option is 2 multiplies, 2 adds, and 2 comparisons (assuming the compiler optimizes so that the quantities $N i_a + j_a$ and $N i_b + j_b$ are computed only once).

So far, Option 1 is a clear winner, and runs faster by around 20-30% on tests I have done.

My question is: can anyone suggest a way to optimize this even more, by somehow improving on Option 1, or perhaps suggesting a completely different method of sorting matrix elements in the tree?

Profiling has shown that a significant amount of time is spent in this comparison function, so it is worth it to me to make it as fast as possible.

Another idea I have is to use 2 levels of binary trees. The high-level tree uses only row indices for sorting. Then each node in that tree contains a pointer to another tree which uses the column indices for sorting. This way the comparison function only needs to compare two integers instead of 4. But having 2 levels of tree would make things more complicated. I have not implemented or benchmarked this approach yet.


1 Answer 1


Option 2: Compare full linear index
#define COMPARE(ia,ja,ib,jb) (ia*N + ja < ib*N + jb ? -1 : ia*N + ja > ib*N + jb)

Instead of N, use the minimal power of 2 that is not less than N. For example, if N = 10000, use $2^{14}=16348$. So we can have #define COMPARE(ia,ja,ib,jb) ((ia << 14) + ja < (ib << 14) + jb ? -1 : (ia << 14) + ja > (ib << 14) + jb).

Instead of using #define, we can use a function to make sure that long expression is computed only once. forceinline is defined to be the cross-platform forceinline modifier.

#define TWO_EXPONENT 14
forceinline int compare(int ia, int ja, int ib, int jb) {
    int t1 = ((ia - ib) << TWO_EXPONENT) + (ja - jb);
    return (t1 > 0) - (t1 < 0);  //branchless computation

You have specified that you need to return 3 values, 1, 0, and -1. That sounds more than enough. You may want to double check if it is enough to use 1 for the case of "not smaller than". That is enough, for example, when we just want to sort an array of numbers.

Just in case you have not noticed, there are a wealth of information on how to store sparse matrix, Google search or Wikipedia entry.


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