SQRT idea

[johnsa]

So I was extremely bored this morning, and although this is a rather pointless excercise it might be kind of interesting.
So I was thinking, how would I calculate SQRTs without using lookups, built-in SQRT or Newton Raphson. I came up with this little idea which is working perfectly for integer only operation. For floats it would be best to replace the linear interpolation with exponential for precision. It's based on the simple observation that the perfect squares go up according to 4+(2n+1) : n=0 or n>=2 IE: 4,9,16,25 etc.. and as such map back down to sqr roots that increase sequentially as 2,3,4,5,6.....

Code Select Expand


        public static int fastSQRT(int x)
        {
            uint i = 65536, n = 4, d = 5;
            uint first, second;
            while (n < x)
            {
                n += d;
                d += 2;
                i += 65536;
            }
            second = n;
            first = (n - (d - 2));
            //i = (uint)(65536 * (((x - 5) >> 1) + 2));
            //first = (uint)(4 + ((x - 5) >> 1) );
            //second = first + (uint)(4 + (((x - 5) >> 1) + 1));
            return ( (int)( i + (((x-first)<<16)/(second-first)) >> 16) );
        }


It's not exactly fast... although if we could take out the while loop to identify the perfect squares that are to the left/right of the number in question it would be roughly the same as the built in sqrt in C#.  In any event it was a random brain-exploration :)

[dedndave]

It's based on the simple observation that the perfect squares go up according to 4+(2n+1) : n=0 or n>=2 IE: 4,9,16,25 etc.. and as such map back down to sqr roots that increase sequentially as 2,3,4,5,6.....

that is an interesting observation, Johnsa

the squarerooter loops i have seen tend to work on bit pairs:
bit 0 of the squareroot is somewhat derived from input bits 0/1
bit 1 of the squareroot is somewhat derived from input bits 2/3
bit 2 of the squareroot is somewhat derived from input bits 4/5, and so on

[MichaelW]

Interesting, I wonder if the algorithm could be implemented as a macro that calculates a SQRT at assembly time.

[qWord]

I wonder if the algorithm could be implemented as a macro that calculates a SQRT at assembly time.

Code Select Expand

fastSQRT macro x:=<1>
fsqrt_i = 65536
fsqrt_n = 4
fsqrt_d = 5
WHILE fsqrt_n LT x
fsqrt_n = fsqrt_n + fsqrt_d
fsqrt_d = fsqrt_d + 2
fsqrt_i = fsqrt_i + 65536
ENDM
fsqrt_first = fsqrt_n-(fsqrt_d-2)
EXITM %((fsqrt_i + (((x-fsqrt_first) SHL 16)/(fsqrt_n-fsqrt_first))) SHR 16)
endm

[dedndave]

already putting the "%" thing to use, eh qWord   :U

[johnsa]

The interpolation will obviously lead to less precision the larger the number, but sofar I've tested 100,000,000 numbers in the range of 0 - 500,000,000 and it's been spot on using integers only.
The next trick would be to use non-linear interpolation between the sqrt results.

IE: sqrt(6.5) falls between 4 and 9 which yield perfect roots of 2 and 3... so the sqrt would be calculated with linear interpolation to be 2.5 (which is close.. its actually 2.51....).
This should be fairly easy to add, just need an exponential function in there. The second is how to get rid of the while loop... I was trying my luck with the commented out lines to see if i could replace it by using some sort of
inequality like 4+(2n+1) >= x... but I didn't get it right yet.

[jj2007]

Code Select Expand

fastSQRT macro x:=<1>


Code Select Expand

include \masm32\MasmBasic\MasmBasic.inc

fastSQRT macro x:=<1>
  fsqrt_i = 65536
  fsqrt_n = 4
  fsqrt_d = 5
  WHILE fsqrt_n LT x
fsqrt_n = fsqrt_n + fsqrt_d
fsqrt_d = fsqrt_d + 2
fsqrt_i = fsqrt_i + 65536
  ENDM
  fsqrt_first = fsqrt_n-(fsqrt_d-2)
  EXITM %((fsqrt_i + (((x-fsqrt_first) SHL 16)/(fsqrt_n-fsqrt_first))) SHR 16)
endm

Sqrt macro x:REQ
  ffree st(7)
  push x
  fild dword ptr [esp]
  fsqrt
  fstp MbDebugR8
  add esp, 4
  EXITM <MbDebugR8>
endm
Init
Print Str$("The fast square root of 30 is %9f\n", fastSQRT(30))
Inkey Str$("The slow square root of 30 is %9f", Sqrt(30))
Exit
end start


Code Select Expand

The fast square root of 30 is 5.00000000
The slow square root of 30 is 5.47722558
:green

[dedndave]

let's not start naming our procs starting with "Slow"   :lol
(or "Slower" or "Slowest" or "SnailSpeed")

[qWord]

 :bg

Code Select Expand

sqrti macro rm32:req
cvtsi2ss xmm0,rm32
sqrtss xmm0,xmm0
movd rm32,xmm0
endm

.data?
r4 REAL4 ?
.code
sqrti(ASM(mov r4,9))
print real4$(r4)

[johnsa]

I think we should drop the fast prefix off my idea (as it's not very fast at all.. .lol especially with the while in there). :)
Maybe call it... oddWayToGetaSqrt

[jj2007]

:bg

Code Select Expand

sqrti macro rm32:req
cvtsi2ss xmm0,rm32
sqrtss xmm0,xmm0
movd rm32,xmm0
endm

.data?
r4 REAL4 ?
.code
sqrti(ASM(mov r4,9))
print real4$(r4)

sqrtss is probably a tiny bit faster than fsqrt. Although I like the idea to have full Real10 precision...

What does ASM() do, by the way?
:bg

[qWord]

What does ASM() do, by the way?

from macros.inc:

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    ; --------------------------------------------------
    ; Macro for placing an assembler instruction either
    ; within another or within a procedure call
    ; --------------------------------------------------
work only with two operands instructions. The first operand is returned.
ASM(mov eax,9) ==> returns eax

[clive]

sqrtss is probably a tiny bit faster than fsqrt. Although I like the idea to have full Real10 precision...

RSQRTSS Is super fast (1 cycle on P3), low precision, table driven out of a ROM. Useful if you are dividing something by the square root.

[clive]

Probably in here, but I ran across it in one of his magazine articles.
http://www.amazon.com/dp/1929629095

Code Select Expand

// Jack Crenshaw's Integer Square Root

unsigned long integer_sqrt(unsigned long a)
{
  unsigned long rem = 0;
  unsigned long root = 0;
  unsigned long divisor = 0;
  int i;

  for(i=0; i<16; i++) // 32-bit
  {
    root <<= 1;

    rem = ((rem << 2) + (a >> 30));

    a <<= 2;

    divisor = (root << 1) + 1;

    if(divisor <= rem)
    {
      rem -= divisor;
      root++;
    }
  }

  return(root);
}


[dioxin]

Another square root, based on the method used to calculate square roots on hand calculating machines of the early 20th century
As written here it'll work for numbers up to about 1,000,000,000

Code Select Expand

!mov ecx,10000          ;Get SQRT of this number

!mov eax,40000000h      ;stepsize
!mov edx,40000000h      ;decrement
!mov esi,16             ;bits to do
     
lp1:
!sub ecx,edx
!jb x1
!add edx,eax
         
!shr eax,1
!shl ecx,1
!add edx,eax

!dec esi                ;next bit
!jne lp1

!jmp xit

x1:
!add ecx,edx
!sub edx,eax

!shr eax,1
!shl ecx,1
!add edx,eax

!dec esi                ;next bit
!jne lp1


XIT:
!shr edx,16             ;edx contains the answer