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Editorial Team
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Asked: 2026-05-28T15:19:53+00:00

Recently, I read a post on Stack Overflow about finding integers that are perfect

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Recently, I read a post on Stack Overflow about finding integers that are perfect squares. As I wanted to play with this, I wrote the following small program:

PROGRAM PERFECT_SQUARE
IMPLICIT NONE
INTEGER*8 :: N, M, NTOT
LOGICAL :: IS_SQUARE

N=Z'D0B03602181'
WRITE(*,*) IS_SQUARE(N)

NTOT=0
DO N=1,1000000000
  IF (IS_SQUARE(N)) THEN
    NTOT=NTOT+1
  END IF
END DO
WRITE(*,*) NTOT ! should find 31622 squares
END PROGRAM

LOGICAL FUNCTION IS_SQUARE(N)
IMPLICIT NONE
INTEGER*8 :: N, M

! check if negative
IF (N.LT.0) THEN
  IS_SQUARE=.FALSE.
  RETURN
END IF

! check if ending 4 bits belong to (0,1,4,9)
M=IAND(N,15)
IF (.NOT.(M.EQ.0 .OR. M.EQ.1 .OR. M.EQ.4 .OR. M.EQ.9)) THEN
  IS_SQUARE=.FALSE.
  RETURN
END IF

! try to find the nearest integer to sqrt(n)
M=DINT(SQRT(DBLE(N)))
IF (M**2.NE.N) THEN
  IS_SQUARE=.FALSE.
  RETURN
END IF

IS_SQUARE=.TRUE.
RETURN
END FUNCTION

When compiling with gfortran -O2, running time is 4.437 seconds, with -O3 it is 2.657 seconds. Then I thought that compiling with ifort -O2 could be faster since it might have a faster SQRT function, but it turned out running time was now 9.026 seconds, and with ifort -O3 the same. I tried to analyze it using Valgrind, and the Intel compiled program indeed uses many more instructions.

My question is why? Is there a way to find out where exactly the difference comes from?

EDITS:

  • gfortran version 4.6.2 and ifort version 12.0.2
  • times are obtained from running time ./a.out and is the real/user time (sys was always almost 0)
  • this is on Linux x86_64, both gfortran and ifort are 64-bit builds
  • ifort inlines everything, gfortran only at -O3, but the latter assembly code is simpler than that of ifort, which uses xmm registers a lot
  • fixed line of code, added NTOT=0 before loop, should fix issue with other gfortran versions

When the complex IF statement is removed, gfortran takes about 4 times as much time (10-11 seconds). This is to be expected since the statement approximately throws out about 75% of the numbers, avoiding to do the SQRT on them. On the other hand, ifort only uses slightly more time. My guess is that something goes wrong when ifort tries to optimize the IF statement.

EDIT2:

I tried with ifort version 12.1.2.273 it’s much faster, so looks like they fixed that.

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1 Answer

  1. Editorial Team

    What compiler versions are you using?
    Interestingly, it looks like a case where there is a performance regression from 11.1 to 12.0 — e.g. for me, 11.1 (ifort -fast square.f90) takes 3.96s, and 12.0 (same options) took 13.3s.
    gfortran (4.6.1) (-O3) is still faster (3.35s).
    I have seen this kind of a regression before, although not quite as dramatic.
    BTW, replacing the if statement with 

    is_square = any(m == [0, 1, 4, 9])
if(.not. is_square) return

    makes it run twice as fast with ifort 12.0, but slower in gfortran and ifort 11.1.

    It looks like part of the problem is that 12.0 is overly aggressive in trying to vectorize things: adding 

    !DEC$ NOVECTOR

    right before the DO loop (without changing anything else in the code) cuts the run time down to 4.0 sec.

    Also, as a side benefit: if you have a multi-core CPU, try adding -parallel to the ifort command line 🙂

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