square root?

BlitzMax Forums/BlitzMax Programming/square root?

Jeremy Paxman - Yeeeeeeeees

Im trying to make a 2d distance function
how do I do square root in blitzmax?

thanks

klepto2

sqr(x) ?

Czar Flavius

x^(1/n) where n is the root you wish to find.

GfK

sqr(x) is simpler.

x^(1/n) is faster (so use this if you need a lot of square roots).

fredborg

x^(1/n) is faster (so use this if you need a lot of square roots).

GfK

no, what?

using sqr(x) for 10 million iterations: 664ms
using x^(1/n) for 10 million iterations: 82ms.

OK, that's a lot of iterations but the bottom line is, x^(1/n) is eight times faster than Sqr(x).

Brucey

If you don't need the actual distance, but just need to know if A is beyond (or before) a certain distance, you don't even need to calculate the square root...

Gabriel

x^(1/n) where n is the root you wish to find.

Additionally, where N is a float or a double, otherwise you'll do an integer division and get one for the result no matter what you put in.

fredborg

SuperStrict

Local y:Float

Local a:Int = MilliSecs()
For Local i:Float = 0.0 To 1000000.0 Step 1.0
	y = Sqr(i)
Next
a = MilliSecs()-a

Local b:Int = MilliSecs()
For Local i:Float = 0.0 To 1000000.0 Step 1.0
	y = i^0.5
Next
b = MilliSecs()-b

Print a+" vs "+b

ziggy

#10

sqr is a lot faster than^1/n since version 1.24 of Max.

johnnyfreak

#11

why don't use lookup tables

SuperStrict

Local y:Float

Local a:Int = MilliSecs()
For Local i:Float = 0.0 To 1000000.0 Step 1.0
	y = Sqr(i)
Next
a = MilliSecs()-a

Local b:Int = MilliSecs()
For Local i:Float = 0.0 To 1000000.0 Step 1.0
	y = i^0.5
Next
b = MilliSecs()-b

Local arr:Float[1000000]

For Local i:Int = 0 To 1000000 Step 1
	arr[i] = Sqr(i)
Next

Local c:Int = MilliSecs()
For Local i:Float = 0.0 To 1000000.0 Step 1.0
	y = arr[i]
Next
c = MilliSecs() - c

Print a+" vs "+b + " vs " + c

GfK

#12

sqr is a lot faster than^1/n since version 1.24 of Max.

I'm using 1.28 and Sqr() is a lot slower. Unless there's something available via subversion that I don't have.

tonyg

#13

Fredborg's code came up with the same results on my system (e.g. sqr faster than ^1/n) and I am 1.28.
@GFK, what are your results with Fredborg's code and what code are you using?

ImaginaryHuman

#14

You don't actually have to do a square root at all, all it will do is scale the number. If you don't specifically need it to be in a certain range then just use the pre-square-rooted version of the result.

Brucey

#15

It's like an echo in here :-p

WendellM

#16

It's like an echo in here :-p

You gave the positive version of the answer while ImaginaryHuman gave the negative one. Both are needed since it's a square root question... :)

Czar Flavius

#17

I see I have sparked contraversy! It's my birthday today too. Yay.

Jim T

#18