Perhaps you were thinking of this?

At iteration number k, the value xk contains O(klog(k)) digits, thus the 
computation of xk+1 = kxk has cost O(klog(k)). Finally, the total cost with 
this basic approach is O(2log(2)+¼+n log(n)) = O(n2log(n)).
A better approach is the binary splitting : it just consists in recursively 
cutting the product of m consecutive integers in half. It leads to better 
results when products on large integers are performed with a fast method.

http://numbers.computation.free.fr/Constants/Algorithms/splitting.html

I think you can do recursive splitting without using function recursion by 
allocating N/2 array (where b = a+N-1) and iterating over it; each time the 
array "shrinks" by half. A "cleverer" algorithm would allocate an array of 
*words* of a bignum, as you know that the upper limit on size is N*64 (for 64 
bit numbers) so you can just reuse the same space for each outer iteration (N/2 
multiplie, N/4 ...) and apply Karatsuba 2nd outer iteration onwards. Not sure 
if this is easy in Go.

> On Jan 8, 2024, at 11:47 AM, Robert Griesemer <g...@golang.org> wrote:
> 
> Hello John;
> 
> Thanks for your interest in this code.
> 
> In a (long past) implementation of the factorial function, I noticed that 
> computing a * (a+1) * (a+2) * ... (b-1) * b was much faster when computed in 
> a recursive fashion than when computed iteratively: the reason (I believed) 
> was that the iterative approach seemed to produce a lot more "internal 
> fragmentation", that is medium-size intermediate results where the most 
> significant word (or "limb" as is the term in other implementations) is only 
> marginally used, resulting in more work than necessary if those words were 
> fully used.
> 
> I never fully investigated, it was enough at the time that the recursive 
> approach was much faster. In retrospect, I don't quite believe my own theory. 
> Also, that implementation didn't have Karatsuba multiplication, it just used 
> grade-school multiplication.
> 
> Since a, b are uint64 values (words), this could probably be implemented in 
> terms of mulAddVWW directly, with a suitable initial allocation for the 
> result - ideally this should just need one allocation (not sure how close we 
> can get to the right size). That would cut down the allocations massively.
> 
> In a next step, one should benchmark the implementation again.
> 
> But at the very least, the overflow bug should be fixed, thanks for finding 
> it! I will send out a CL to fix that today.
> 
> Thanks,
> - gri
> 
> 
> 
> On Sun, Jan 7, 2024 at 4:47 AM John Jannotti <janno...@gmail.com 
> <mailto:janno...@gmail.com>> wrote:
>> Actually, both implementations have bugs!
>> 
>> The recursive implementation ends with:
>> ```
>> m := (a + b) / 2
>> return z.mul(nat(nil).mulRange(a, m), nat(nil).mulRange(m+1, b))
>> ```
>> 
>> That's a bug whenever `(a+b)` overflows, making `m` small. 
>> FIX: `m := a + (b-a)/2`
>> 
>> My iterative implementation went into an infinite loop here:
>> `for m := a + 1; m <= b; m++ {`
>> if b is `math.MaxUint64`
>> FIX: add `&& m > a` to the exit condition is an easy fix, but pays a small 
>> penalty for the vast majority of calls that don't have b=MaxUint64
>> 
>> I would add these to `mulRangesN` of the unit test:
>> ```
>>  {math.MaxUint64 - 3, math.MaxUint64 - 1, 
>> "6277101735386680760773248120919220245411599323494568951784"},
>> {math.MaxUint64 - 3, math.MaxUint64, 
>> "115792089237316195360799967654821100226821973275796746098729803619699194331160"}
>> ```
>> 
>> On Sun, Jan 7, 2024 at 6:34 AM John Jannotti <janno...@gmail.com 
>> <mailto:janno...@gmail.com>> wrote:
>>> I'm equally curious.
>>> 
>>> FWIW, I realized the loop should perhaps be
>>> ```
>>> mb := nat(nil).setUint64(b) // ensure mb starts big enough for b, even on 
>>> 32-bit arch
>>> for m := a + 1; m <= b; m++ {
>>>   mb.setUint64(m)
>>>   z = z.mul(z, mb)
>>> }
>>> ```
>>> to avoid allocating repeatedly for `m`, which yields:
>>> BenchmarkIterativeMulRangeN-10      354685      3032 ns/op    2129 B/op     
>>>  48 allocs/op
>>> 
>>> On Sun, Jan 7, 2024 at 2:41 AM Rob Pike <r...@golang.org 
>>> <mailto:r...@golang.org>> wrote:
>>>> It seems reasonable but first I'd like to understand why the recursive 
>>>> method is used. I can't deduce why, but the CL that adds it, by gri, does 
>>>> Karatsuba multiplication, which implies something deep is going on. I'll 
>>>> add him to the conversation.
>>>> 
>>>> -rob
>>>> 
>>>> 
>>>> 
>>>> 
>>>> On Sun, Jan 7, 2024 at 5:46 PM John Jannotti <janno...@gmail.com 
>>>> <mailto:janno...@gmail.com>> wrote:
>>>>> I enjoy bignum implementations, so I was looking through nat.go and saw 
>>>>> that `mulRange` is implemented in a surprising, recursive way,.  In the 
>>>>> non-base case, `mulRange(a, b)` returns `mulrange(a, (a+b)/2) * 
>>>>> mulRange(1+(a+b)/2, b)` (lots of big.Int ceremony elided).
>>>>> 
>>>>> That's fine, but I didn't see any advantage over the straightforward (and 
>>>>> simpler?) for loop.
>>>>> 
>>>>> ```
>>>>> z = z.setUint64(a)
>>>>> for m := a + 1; m <= b; m++ {
>>>>>   z = z.mul(z, nat(nil).setUint64(m))
>>>>> }
>>>>> return z
>>>>> ```
>>>>> 
>>>>> In fact, I suspected the existing code was slower, and allocated a lot 
>>>>> more.  That seems true. A quick benchmark, using the existing unit test 
>>>>> as the benchmark, yields
>>>>> BenchmarkRecusiveMulRangeN-10       169417              6856 ns/op        
>>>>>     9452 B/op        338 allocs/op
>>>>> BenchmarkIterativeMulRangeN-10              265354              4269 
>>>>> ns/op            2505 B/op        196 allocs/op
>>>>> 
>>>>> I doubt `mulRange` is a performance bottleneck in anyone's code! But it 
>>>>> is exported as `int.MulRange` so I guess it's viewed with some value.  
>>>>> And seeing as how the for-loop seems even easier to understand that the 
>>>>> recursive version, maybe it's worth submitting a PR? (If so, should I 
>>>>> create an issue first?)
>>>>> 
>>>>> 
>>>>> 
>>>>> 
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