How to improve performance of this numerical computation in Haskell?

Question

I\'m in the middle of porting David Blei\'s original C implementation of Latent Dirichlet Allocation to Haskell, and I\'m trying to decide whether to leave some of the low-l

Answer 1

Use the same control and data structures, yielding:

{-# LANGUAGE BangPatterns #-}
{-# OPTIONS_GHC -fvia-C -optc-O3 -fexcess-precision -optc-march=native #-}

{-# INLINE trigamma #-}
trigamma :: Double -> Double
trigamma x = go 0 (x' - 1) p'
    where
        x' = x + 6
        p  = 1 / (x' * x')

        p' =(((((0.075757575757576*p-0.033333333333333)*p+0.0238095238095238)
                  *p-0.033333333333333)*p+0.166666666666667)*p+1)/x'+0.5*p

        go :: Int -> Double -> Double -> Double
        go !i !x !p
            | i >= 6    = p
            | otherwise = go (i+1) (x-1) (1 / (x*x) + p)

I don't have your testsuite, but this yields the following asm:

A_zdwgo_info:
        cmpq    $5, %r14
        jg      .L3
        movsd   .LC0(%rip), %xmm7
        movapd  %xmm5, %xmm8
        movapd  %xmm7, %xmm9
        mulsd   %xmm5, %xmm8
        leaq    1(%r14), %r14
        divsd   %xmm8, %xmm9
        subsd   %xmm7, %xmm5
        addsd   %xmm9, %xmm6
        jmp     A_zdwgo_info

Which looks ok. This is the kind of code the -fllvm backend does a good job.

GCC unrolls the loop though, and the only way to do that is either via Template Haskell or manual unrolling. You might consider that (a TH macro) if doing a lot of this.

Actually, the GHC LLVM backend does unroll the loop :-)

Finally, if you really like the original Haskell version, write it using stream fusion combinators, and GHC will convert it back into loops. (Exercise for the reader).

Answer 2

Before the optimization work, I wouldn't say that your original translation is the most idiomatic way to express in Haskell what the C code is doing.

How would the optimization process have proceeded if we started with the following instead:

trigamma :: Double -> Double
trigamma x = foldl' (+) p' . map invSq . take 6 . iterate (+ 1) $ x
where
  invSq y = 1 / (y * y)
  x' = x + 6
  p  = invSq x'
  p' =(((((0.075757575757576*p-0.033333333333333)*p+0.0238095238095238)
              *p-0.033333333333333)*p+0.166666666666667)*p+1)/x'+0.5*p