how do x^6 on sse or avx ?

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Roman

Joined: 21 Apr 2012
Posts: 2004

Roman 16 Oct 2025, 22:10

I search simple way do this x*x*x*x*x*x.
I not like do 5 times mulss xmm1,xmm1.
I searching couple sse or avx asm commands for this task.

I found this.

Code:

pow(base,power) = exp( power * log(base) )

In this case do 5 times mulss xmm1,xmm1 not bad solution.

AVX have vhaddps horizontal apply.
Maybe exist horizontal multiplication?
ymm0 could multiply 8 floats numbers.

16 Oct 2025, 22:10

revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 20742
Location: In your JS exploiting you and your system

revolution 16 Oct 2025, 23:42

A power of 6 can be done with three multiplies.

a = x*x ; x^2
b = a*x ; x^3
c = b*b ; x^6

16 Oct 2025, 23:42

Roman

Joined: 21 Apr 2012
Posts: 2004

Roman 17 Oct 2025, 00:08

Sorry my mistake.

Code:

;x^8
align 32
x dd 1f,2f,3f,4f,5f,6f,7f,8f

vmovaps ymm0,yword [x]
    vmulps ymm0, ymm0, ymm0    ; x^2
    vmulps ymm0, ymm0, ymm0    ; x^4
    vmulps ymm0, ymm0, ymm0    ; x^8  

;after  in ymm0 1f^8 ,2f^8 ,3f^8 ,4f^8 ,5f^8 ,6f^8 ,7f^8 ,8f^8

Code:

x dd 3.0

movss xmm1,[x]
movss xmm0,xmm1
mulss xmm0,xmm0 ;x^2
mulss xmm0,xmm0 ;x^4
mulss xmm1,xmm0 ;x^5 xmm1=3^5=243

17 Oct 2025, 00:08

revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 20742
Location: In your JS exploiting you and your system

revolution 17 Oct 2025, 06:30

In general x-to-power-of-n can be computed in no more than [ bsr(n) + popcnt(n) - 1 ] multiplies.

17 Oct 2025, 06:30

Roman

Joined: 21 Apr 2012
Posts: 2004

Roman 17 Oct 2025, 14:37

Quote:

[ bsr(n) + popcnt(n) - 1 ]

Very interesting see example how do this.

17 Oct 2025, 14:37

Tomasz Grysztar

Joined: 16 Jun 2003
Posts: 8463
Location: Kraków, Poland

Tomasz Grysztar 17 Oct 2025, 17:14

Roman wrote:

Quote:

[ bsr(n) + popcnt(n) - 1 ]

Very interesting see example how do this.

Take the binary representation of the number, for example 13 = 1101b:
$13 = 1101b = 1\cdot{2^3} + 1\cdot{2^2} + 0\cdot{2^1} + 1\cdot{2^0}$

$x^{13} = x^{2^3+2^2+2^0} = x^{2^3}\cdot{x^{2^2}}\cdot{x}$

So if we have the square powers of x ready, we only need to perform 2 more multiplications - this is the "popcnt(n) - 1" portion of revolution's formula.

Now, taking into consideration that $x^{2^i}=(x^{2^{i-1}})^2$ it follows that we can compute $x^{2^i}$ by squaring i times, and we get all the intermediate squares on the way. This is the "bsr(n)" part of the formula.

For n = 13 we get this sequence of multiplications:
a = x*x
b = a*a
c = b*b
x^13 = c*b*x
For final calculation you take the product of all the consecutive squares and remove the ones that correspond to zeros in the binary representation of n. Here, because 13 = 1101b, it meant removing "a" from "c*b*a*x" (as "a" corresponds to the only zero in 1101b).

17 Oct 2025, 17:14

Roman

Joined: 21 Apr 2012
Posts: 2004

Roman 18 Oct 2025, 08:13

Quote:

[ bsr(n) + popcnt(n) - 1 ]

I thinked this code help do two multiplication.
But if not, this code only confusing and complicating. But not help.

18 Oct 2025, 08:13

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