Fast indexing of symmetrical matrix - fortran

Question

In a large code written in Fortran08 for calculating thermodynamic equilibria and phase diagrams I use many symmetric matrices which I store as 1D arrays and index using a small function
integer function ixsym(i,j)
if(i.gt.j) then
ixsym=j+i*(i-1)/2
else
ixsym=i+j*(j-1)/2
endif
return
end
This works perfectly but after improving the speed of various other parts of the code this routine now takes 15-20% of the calculation time (it is used very often). I assume there are various ways of speeding this up but I do not know C or other way to replace this function so I am looking for help. I use gfortran but the replacement has to be portable.
Bo Sundman

Answer 1

The only thing you might consider is to get rid of the branching in that function:
The minimum and maximum of two numbers can be computed as:
max = (a+b + abs(a-b))/2
min = (a+b - abs(a-b))/2 = a+b - max
So you can use this as
integer function ixsym(i,j)
integer :: p, q
q = i+j; p = (q + abs(i-j))/2; q = q - p
ixsym = q + (p*(p-1))/2
return
end
which you can further reduce as
integer function ixsym(i,j)
integer :: p
ixsym = i+j; p = (ixsym + abs(i-j))/2;
ixsym = ixsym + (p*(p-3))/2
return
end

Answer 2

Fortran compilers used to have optimization on par or better than C compilers.
So I would not expect a gain just by switching language and rather focus on algorithmic improvements.
How about replacing the calculation of the index tranformation by a lookup table?
Do you have the memory to store the ixsym values for given i and j indices?
Yes it counters your memory for cpu trade-off, but if you have many matrices this extra one might help.
Is it really necessary to calculate the transformation at all times? E.g. if you iterate over elements: ixsym(i, j+1) = ixsym(i, j) + 1, if i < j.
Another idea, though hardware specific, might be to order your data differently, so that it stays within cache areas of the CPU. (Link)
About your index transformation:
I initially thought you used some variation of the Cantor pairing function to enumerate your symmetric 2D array. I asked my friend Ruby to plot a few pairs and she told me:
(0, 0) -> 0 (0, 1) -> 0 (0, 2) -> 1 (0, 3) -> 3 (0, 4) -> 6
(1, 0) -> 0 (1, 1) -> 1 (1, 2) -> 2 (1, 3) -> 4 (1, 4) -> 7
(2, 0) -> 1 (2, 1) -> 2 (2, 2) -> 3 (2, 3) -> 5 (2, 4) -> 8
(3, 0) -> 3 (3, 1) -> 4 (3, 2) -> 5 (3, 3) -> 6 (3, 4) -> 9
(4, 0) -> 6 (4, 1) -> 7 (4, 2) -> 8 (4, 3) -> 9 (4, 4) -> 10
I would have expected only two occurences of a calculated index, but I see three for some pairs. Is this intended?
Update:
It was the index start, as fellow user Jean-Claude Arbaut pointed out in his comment.
Here is Ruby's answer with indices starting at 1:
(1, 1) -> 1 (1, 2) -> 2 (1, 3) -> 4 (1, 4) -> 7 (1, 5) -> 11
(2, 1) -> 2 (2, 2) -> 3 (2, 3) -> 5 (2, 4) -> 8 (2, 5) -> 12
(3, 1) -> 4 (3, 2) -> 5 (3, 3) -> 6 (3, 4) -> 9 (3, 5) -> 13
(4, 1) -> 7 (4, 2) -> 8 (4, 3) -> 9 (4, 4) -> 10 (4, 5) -> 14
(5, 1) -> 11 (5, 2) -> 12 (5, 3) -> 13 (5, 4) -> 14 (5, 5) -> 15