# Performance challenge: can you write a faster sum?

**URL:** <https://discourse.julialang.org/t/performance-challenge-can-you-write-a-faster-sum/130456>\
**Category:** Performance\
**Tags:** simd\
**Created:** [July 3, 2025, 3:47pm UTC](https://discourse.julialang.org/t/performance-challenge-can-you-write-a-faster-sum/130456 "2025-07-03T15:47:21Z")\
**Posts on this page:** 1\
**Showing post:** 33

<div class="post-metadata">

**Author:** ![epilliat](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/epilliat/32/219946_2.png) [@epilliat](https://discourse.julialang.org/u/epilliat)\
**Post date:** [December 11, 2025, 10:03pm UTC](https://discourse.julialang.org/t/performance-challenge-can-you-write-a-faster-sum/130456/33 "2025-12-11T22:03:55Z")

</div>

Hello, I just tried your sum from the PR, and I’ve generalized your ideas using `@generated` functions: implementing accumulators (for your `Base.jl` and `reduce_mb_4` versions), and incorporating tree reduction within the loop itself (`reduce_mb_3`).

Here is my attempt:

```julia
macro tree_reduce(op, prefix, N, M, stride)
    prefix_str = string(prefix)

    function build_tree(indices)
        count = length(indices)
        if count == 1
            return Symbol(prefix_str, :_, indices[1])
        elseif count == 2
            return :($op($(Symbol(prefix_str, :_, indices[1])),
                $(Symbol(prefix_str, :_, indices[2]))))
        else
            mid = count ÷ 2
            left = build_tree(indices[1:mid])
            right = build_tree(indices[mid+1:end])
            return :($op($left, $right))
        end
    end

    # Generate strided indices: N, N+stride, N+2*stride, ..., N+(M-1)*stride
    indices = [N + i * stride for i in 0:M-1]

    return esc(build_tree(indices))
end

@generated function reduce_abstracted(op, A, ::Val{Ntot}, ::Val{Nacc}) where {Ntot,Nacc}
    Pacc = trailing_zeros(Nacc)
    Ptot = trailing_zeros(Ntot)
    quote
        f = identity
        inds = eachindex(A)
        i1, iN = firstindex(inds), lastindex(inds)
        n = length(inds)
        @nexprs $Ntot N -> a_N = @inbounds A[inds[i1+(N-1)]]
        @nexprs $Nacc N -> v_N = @tree_reduce(op, a, N, $(cld(Ntot, Nacc)), $Nacc)

        for batch in 1:(n>>$Ptot)-1
            i = 1 + (batch << $Ptot)
            @nexprs $Ntot N -> begin
                a_N = @inbounds A[inds[i+(N-1)]]
            end
            @nexprs $Nacc N -> v_N = op(v_N, @tree_reduce(op, a, N, $(cld(Ntot, Nacc)), $Nacc))

        end
        v = @tree_reduce(op, v, 1, $Nacc, 1)
        i = i1 + (n >> $Ptot) * $Ntot - 1
        i == iN && return v
        for i in i+1:iN
            ai = @inbounds A[inds[i]]
            v = op(v, f(ai))
        end
        return v
    end
end

```

The Base.sum from your PR correspond to Nacc=8 (8 accumulators) and Ntot=8 (no intern tree reduction). You mb\_3 function correspond to Ntot=16 and Nacc=4.

Here are my benchmarks:

```julia
n = 2^15
a = rand(Float64, n)
for Ntot in (4, 8, 16, 32)
    for Nacc in (1, 2, 4, 8)
        if Nacc <= Ntot
            println("n=$n, a[1:n], view(a, 1:2:n) (above, below) Ntot=$Ntot, Nacc=$Nacc")
            @btime reduce_abstracted(+, $a, $(Val(Ntot)), $(Val(Nacc)))
            @btime reduce_abstracted(+, $(view(a, 1:2:n)), $(Val(Ntot)), $(Val(Nacc)))
        end
    end
end

```

```repl
n=32768, a[1:n], view(a, 1:2:n) (above, below) Ntot=4, Nacc=1
  3.294 μs (0 allocations: 0 bytes)
  2.475 μs (0 allocations: 0 bytes)
n=32768, a[1:n], view(a, 1:2:n) (above, below) Ntot=4, Nacc=2
  3.405 μs (0 allocations: 0 bytes)
  2.486 μs (0 allocations: 0 bytes)
n=32768, a[1:n], view(a, 1:2:n) (above, below) Ntot=4, Nacc=4
  3.649 μs (0 allocations: 0 bytes)
  2.449 μs (0 allocations: 0 bytes)
n=32768, a[1:n], view(a, 1:2:n) (above, below) Ntot=8, Nacc=1
  4.808 μs (0 allocations: 0 bytes)
  2.130 μs (0 allocations: 0 bytes)
n=32768, a[1:n], view(a, 1:2:n) (above, below) Ntot=8, Nacc=2
  2.396 μs (0 allocations: 0 bytes)
  2.181 μs (0 allocations: 0 bytes)
n=32768, a[1:n], view(a, 1:2:n) (above, below) Ntot=8, Nacc=4
  2.147 μs (0 allocations: 0 bytes)
  4.372 μs (0 allocations: 0 bytes)
n=32768, a[1:n], view(a, 1:2:n) (above, below) Ntot=8, Nacc=8
  3.427 μs (0 allocations: 0 bytes)
  2.225 μs (0 allocations: 0 bytes)
n=32768, a[1:n], view(a, 1:2:n) (above, below) Ntot=16, Nacc=1
  4.323 μs (0 allocations: 0 bytes)
  2.227 μs (0 allocations: 0 bytes)
n=32768, a[1:n], view(a, 1:2:n) (above, below) Ntot=16, Nacc=2
  2.061 μs (0 allocations: 0 bytes)
  2.123 μs (0 allocations: 0 bytes)
n=32768, a[1:n], view(a, 1:2:n) (above, below) Ntot=16, Nacc=4
  1.739 μs (0 allocations: 0 bytes)
  4.345 μs (0 allocations: 0 bytes)
n=32768, a[1:n], view(a, 1:2:n) (above, below) Ntot=16, Nacc=8
  2.468 μs (0 allocations: 0 bytes)
  4.318 μs (0 allocations: 0 bytes)
n=32768, a[1:n], view(a, 1:2:n) (above, below) Ntot=32, Nacc=1
  4.236 μs (0 allocations: 0 bytes)
  2.214 μs (0 allocations: 0 bytes)
n=32768, a[1:n], view(a, 1:2:n) (above, below) Ntot=32, Nacc=2
  2.270 μs (0 allocations: 0 bytes)
  2.369 μs (0 allocations: 0 bytes)
n=32768, a[1:n], view(a, 1:2:n) (above, below) Ntot=32, Nacc=4
  1.348 μs (0 allocations: 0 bytes)
  3.978 μs (0 allocations: 0 bytes)
n=32768, a[1:n], view(a, 1:2:n) (above, below) Ntot=32, Nacc=8
  1.651 μs (0 allocations: 0 bytes)
  3.959 μs (0 allocations: 0 bytes)

```

For comparison, the Base.sum from your PR on Float64:

```julia-auto
@btime Base.sum($a)
@btime Base.sum($(view(a, 1:2:n)))

```

```repl
3.459 μs (0 allocations: 0 bytes)
16295.798570780385

  4.040 μs (0 allocations: 0 bytes)
8128.148967556634

```

This is different from what I got because of the additionnal pairwise reduction I guess.  
Here is what the current Base.sum doing on Float64:

```julia-auto
2.488 μs (0 allocations: 0 bytes)
16410.273204764926

 6.539 μs (0 allocations: 0 bytes)
8170.225306474282

```

and just for completeness, the results from @simd:

```julia-auto
a=rand(Float32, 2^15)
function simdsum(a::AbstractArray{T}) where T
    s = T(0)
    @simd for x in a
        s += x
    end
    return s
end
@btime simdsum($a)
@btime simdsum($(view(a, 1:2:2^15)))

```

```repl
1.058 μs (0 allocations: 0 bytes)
16345.918f0

6.549 μs (0 allocations: 0 bytes)
8140.3384f0

```

On my Dell machine, the current `sum` implementation underperforms for non-view floating-point arrays. I’m curious whether you observe similar benchmark trends on your system:

- **Val(16), Val(2)**: Good all-around performance across array types
- **Val(32), Val(4)**: Optimal performance specifically for contiguous bitstype arrays

I’ve found that `@simd` performs poorly on views and larger element types, but on my machine it’s **unbeatable for contiguous bitstype arrays** —with one exception: very small arrays benefit more from vectorized loads followed by reduction.

For context, I started a discussion on this topic a few days ago (before discovering this thread):

> [@Imroving Base.mapreduce by 20% on vectors](https://discourse.julialang.org/t/imroving-base-mapreduce-by-20-on-vectors/134478):
>
> Hello everyone, I’ve been exploring reduction operations on GPUs recently, and this led me to investigate CPU performance as well. I discovered that we can improve mapreduce in Base while maintaining full precision for floating-point operations. My analysis suggests there may be an alignment issue with SIMD in the current Base implementation, which leads to suboptimal performance for common computational patterns. Additionally, I’ve implemented specialized functions for small arrays (size \< 32…

I also documented my initial attempt to outperform `mapreduce` on contiguous bitstype arrays using `@simd`:  
[https://epilliat.github.io/coding/notebooks/mapreduce\_cpu\_perf.html](https://epilliat.github.io/coding/notebooks/mapreduce_cpu_perf.html)

A key finding: **alignment is critical** for strong `@simd` performance, at least on my hardware. The current `Base.sum` implementation doesn’t properly initialize accumulators for 1024-element blocks, which significantly degrades performance.

---

_[View the full topic](https://discourse.julialang.org/t/performance-challenge-can-you-write-a-faster-sum/130456)._