An algorithm to calculate square root??

merihevonen · Post by **merihevonen** » Fri Feb 02, 2007 6:44 pm

Hello!

Is there an algorithm to mathematically solve a square root expression??
For example, sqrt(25) equals 5, but what does the sqrt function do in the background of the executable?

Pseudo/Pure/ASM code is welcome.. I'm just making a Learn-O-Mathic program for 7+ kids to learn maths (Ofcourse, I'll sell it for 5000$ and that as a proprietary product only for Mac users

)

Thanks!

va!n · Post by **va!n** » Fri Feb 02, 2007 6:46 pm

http://www.purebasic.fr/english/viewtop ... =sqrt+fast

Trond · Post by **Trond** » Fri Feb 02, 2007 6:46 pm

but what does the sqrt function do in the background of the executable

It does !fsqrt, which is an fpu opcode.

Comtois · Post by **Comtois** » Fri Feb 02, 2007 9:25 pm

Code: Select all

Procedure.f Sqrt(n.f)
  ProcedureReturn Pow(10,Log10(n)/2.0)
EndProcedure

Procedure.f Racine(n.f, r)
  ProcedureReturn Pow(10,Log10(n)/r)
EndProcedure

a.f = 125.0
r = 3

Debug Sqrt(a)
Debug Sqr(a)

Debug Racine(a,r)
Debug Pow(a,1/r)

merihevonen · Post by **merihevonen** » Fri Feb 02, 2007 9:27 pm

Comtois wrote:Procedure.f Sqrt(n.f)
ProcedureReturn Pow(10,Log10(n)/2.0)
EndProcedure

a.f = 125.0

Debug Sqrt(a)
Debug Sqr(a)

I love you!

But I'm not gay.. I "love" you in the sense that I appreciate it a lot that you helped me.

EDIT: But how do I make a "pure" version of Log10?

Bonne_den_kule · Post by **Bonne_den_kule** » Sat Feb 03, 2007 12:04 am

Code: Select all

Procedure.f Sqrt(n.f) 
  ProcedureReturn Pow(n.f,0.5) 
EndProcedure 
Debug Sqrt(100)

Did I misunderstood something?
Can't it just be like this?
(Don't flame me thefool

)

Comtois · Post by **Comtois** » Sat Feb 03, 2007 12:11 am

Code: Select all

Procedure.f Sqrt(n.f)
  ProcedureReturn Pow(10,Log10(n)/2.0)
EndProcedure

Procedure.f Sqrt2(n.f)
  precision.f = 0.001
  s.f = 1.0
  r.f = 0.0
  Repeat 
    r=(s + n / s) / 2.0 
    s=r
    test.f = r*r - n
  Until Abs(test) < precision
  ProcedureReturn r
EndProcedure

Procedure.f InvSqrt(x.f)
   xhalf.f = 0.5*x;
   i = PeekL(@x); // get bits for floating value
   i = $5f3759df - (i >> 1); // gives initial guess y0
   x = PeekF(@i); // convert bits back to float
   x = x * (1.5 - xhalf * x * x); // Newton step, repeating increases accuracy
   ProcedureReturn x
EndProcedure

a.f = 671.0

Debug Sqrt(a)
Debug Sqrt2(a)
Debug a*InvSqrt(a) ; http://www.math.purdue.edu/~clomont/Math/Papers/2003/InvSqrt.pdf
Debug Sqr(a)
Debug Pow(a,1.0/2.0)

Comtois · Post by **Comtois** » Sat Feb 03, 2007 12:16 am

Bonne_den_kule wrote:Can't it just be like this?

Yes but it's too easy , not funny

PB · Post by PB » Sat Feb 03, 2007 12:17 am

> Can't it just be like this?

Yep. Any number to the power of 0.5 is the square root. To modify yours:

Code: Select all

Debug Pow(9,0.5) ; Returns square root of 9

merihevonen · Post by **merihevonen** » Sat Feb 03, 2007 12:55 am

Thanks everybody in MEGA advance!

Psychophanta · Post by **Psychophanta** » Sat Feb 03, 2007 12:11 pm

Comtois wrote:Yes but it's too easy , not funny

+1

thefool · Post by **thefool** » Sat Feb 03, 2007 12:53 pm

Bonne_den_kule wrote: (Don't flame me thefool )

*POW*

Well i wont flame you as your method is obviously the simplest (Hence its the smartest too!)

I've used it quite a lot dealing with dirty equations hehe

keep things simple!