This may be used in masking and some algos related to number systems (for ex. in crypting)
Any odd number may be represeted by difference of two squares like:
N = X^2 - Y^2
we can use X = (N+1)/2 and Y=X-1 ?Svin :)
so we can know those X and Y by three clocks in assembly
eax = N
lea ecx,
shr ecx,1 ;ecx = X
lea ebx, ;ebx = Y eax = ecx^2 - ebx^2
Now if we to know values of ecx^2 and edx^2 we can perform just one mul not two
;if edx < 10000h
mov eax,ebx
mul eax ;eax = ebx^2
lea edx,[1] ;edx = ecx^2
Posted on 2001-11-28 02:38:50 by The Svin
Maybe this can be used also:

N=X^3 - Y^3 what equals: (X-Y)(X^2 + XY + Y^2)

You need a buffer to store N, and of course X and Y.
The code would be:

mov eax, X
mov ebx, Y
push eax
sub eax, ebx ;X - Y
mov N1, eax ;store in N1
pop eax ;eax has lost its value
shl eax,1 ;X^2
mov N2, eax ;save in N2
shr eax,1 ;store X into eax
mul ebx ; XY
mov N3, eax ;save in N3
shl ebx,1 ; Y^2
mov N4, ebx ;save in N4
add N2, N3 ;X^2 + XY + Y^2)
add N2, N4 ;
mov ebx, N2 ;save in ebx
mov eax, N1 ;eax is (X - Y)
mul ebx ;(X-Y)(X^2 + XY + Y^2)
mov N, eax ;save into N
Posted on 2001-12-10 20:06:07 by CodeLover
I am having now a problem with vectors. Today a friend told me that he needed to calculate magnitude of vectors given angles and magnitude of the resultant. This is the example he showed me:

"There are two vectors. They represent strenghts. They each make a 18 grades angle with the axis of the resultant vector, that has a magnitude of 6000. I need to find the magnitude of the 18 grades vectors". It would be very good if you post some help, he needs the algorithm (i wont tell him i discovered it, i will give you the credits ;).

Thanks
Posted on 2001-12-12 16:18:32 by CodeLover
Maybe this illustrates the problem.
Posted on 2001-12-12 16:29:01 by CodeLover
The Svin, that is a nice piece of code. :)
I would love a class where you were the teacher!
Let me pull up a chair...

CodeLover, shl eax,2 does not equal (eax)^2.
Unless eax = 2.
Posted on 2001-12-12 22:42:50 by bitRAKE
CodeLover,

Each of the vectors can be broken down into an x
component and a y component. Using your diagram,
the x-axis is horizontal the y-axis is vertical,
and the two vectors appear to be 18 degrees above
and 18 degrees below the x-axis. And if the two
vectors are of equal magnitude then the problem
is just a matter of finding the magnitude so that
the x component of each vector is 3000, so that
when the x component of both vectors are added
together the resulting vector lies on the x-axis
with a magnitude of 6000; the sum of the two
vectors. Don't confuse my x with your
X(magnitude).

The x component of each vector would be:
vector.x = X( cos( 18 ))
and
6000 = 2( vector.x )= 2( X( cos( 18 ))) = 6000
solve for X

hope this helps and that I haven't made a stupid
error!

farrier
Posted on 2001-12-13 01:00:56 by farrier
bitRAKE, i don't see where i put "shl eax,2", i wrote always shl eax,1.

Thank you farrier, what i've done was seeing it as a triangle, with one side 18 and the other 6000. If you watch at the graph, the x axis works well as 6000, the magnitude of one of the vectors is the hypotenuse, and the angle is 18. So, as the cosine of an angle is Adj/hyp:

Cos (18)=6000/magnitude, so:

6000/cos (18)=magnitude, what is, approximately:

magnitude= 6308,77334543, what makes sense.

Thanks you farrier, your help enlightened my problem.
Posted on 2001-12-14 12:17:38 by CodeLover
Sorry, typo. Can you explain this line for me?

shl eax,1 ;X^2
Posted on 2001-12-14 13:03:26 by bitRAKE
Sorry bitRAKE, i made a stupid mistake :). Anyway, i have never written "shl eax,2" ;).

You were right friend, i should have written the piece of code like this:

mov eax, X
mov ebx, Y
push eax
sub eax, ebx ;X - Y
mov N1, eax ;store in N1
pop eax ;eax has lost its value
push eax
mul eax ;X^2
mov N2, eax ;save in N2
pop eax ;store X into eax
mul ebx ; XY
mov N3, eax ;save in N3
mov eax,ebx ;
mul eax ; Y^2
mov N4, eax ;save in N4
add N2, N3 ;X^2 + XY + Y^2)
add N2, N4 ;
mov ebx, N2 ;save in ebx
mov eax, N1 ;eax is (X - Y)
mul ebx ;(X-Y)(X^2 + XY + Y^2)
mov N, eax ;save into N
Posted on 2001-12-14 18:51:52 by CodeLover
N=X^3 - Y^3 what equals: (X-Y)(X^2 + XY + Y^2)

(X-Y)(X^2 + XY + Y^2) = (x-y)[(x+y)^2-xy]

mov ebx,x
mov ecx,y
mov eax,ebx
sub ebx,ecx ;x-y
push eax
mul ecx ;xy
mov edx,ecx ;save x
mov ecx,eax ; ecx=xy
pop eax ; get y
add eax,edx ;x+y
mul eax
sub eax,ecx ;(x+y)^2-xy
mul ebx
mov N,eax ;

bye
eko
Posted on 2001-12-14 20:11:51 by eko
This can be used also


mov eax,X
push eax
mul eax
pop ebx
mul ebx
mov N1, eax
mov eax, Y
push eax
mul eax
pop ebx
mul ebx
mov ebx, N1
sub ebx,eax
mov N, ebx
Posted on 2001-12-14 21:04:56 by CodeLover
How about this?:
	mov ebx,x

mov ecx,y
mov eax,ebx ; x
sub ebx,ecx ; ebx=x-y
mul ecx ; xy
add ecx,x ; x+y
xchg ecx,eax ; eax=(x+y) ; ecx=xy
mul eax
sub eax,ecx ; (x+y)^2-xy
mul ebx
mov N,eax
Still only three mul.
Posted on 2001-12-15 01:20:39 by bitRAKE
bitRake: xchg ecx,eax nice one . i havnt thought about it

codelover: your code is still with 4 mul


bye
eko:alright:
Posted on 2001-12-15 06:59:35 by eko
I would replace 2 commented instructions by these 3.
They take 2 clocks instead of 5 by xchg r,r and add r,m
Apart from this your code is very well organized and eko gave IMHO the best
possible formula transformation, though IMHO FPU fits better for long equations:
mov ebx,x
mov ecx,y
mov eax,ebx ; x
mul ecx ; xy
push eax ;!
sub ebx,ecx ; ebx=x-y

lea eax, ;!
; add ecx,x ; x+y
; xchg ecx,eax ; eax=(x+y) ; ecx=xy
pop ecx ;!
mul eax
sub eax,ecx ; (x+y)^2-xy
mul ebx
mov N,eax
Posted on 2001-12-18 00:24:04 by The Svin
; ebx-x
; ecx-y
1 lea eax, ; a=x+y
1 neg ecx ; c=-y
10 imul ebx ; a=(x+y)*x
1 lea edi, ; i=x-y
11 imul ecx,ecx ; s=y*y
1 add eax,ecx ; a=(x+y)*x+y*y
10 imul edi ; a=((x+y)*x+y*y)*(x-y)
=35
Posted on 2001-12-27 02:38:00 by The Svin
ebx=x
eax=y
 lea edx,[ebx+eax];d=x+y 

neg edx = d=-x-y
imul ebx;a=xy
lea ebx,[edx+2*ebx];b=-y-x+2x=x-y
imul edx,edx ; d =(x+y)^2
sub edx, eax;d=(x+y)^2-xy
imul edx,ebx ; (x-y)((x+y)^2-xy)


^ result in edx
or instead the last two lines

sub eax,edx (xy-(x+y)^2)
imul ebx,-1 ; result in eax (x-y)[-1(xy-(x+y)^2)]

3 registers
dont know how much clocks
Posted on 2001-12-27 07:09:29 by eko
How could you optimize not knowing clock table :)
But in the case it doesn't matter - use what I wrote as clock referrence.
The thing is that if you use imul reg,reg (additional parameter)
it result in one extra byte for prefix and one extra clock(11 instead of 10) so it will be 36 clocks instead of 35 in my version.
Nevertheless I enjoyed the formula transformation. It's good and nessesery training.
Need to be done on every day basics.
Most of uneffective code is written 'cause people don't care to calculate or don't use even basic (5th-6th form) arithmetics.

Unfortunatly I just don't have talent to show it :(
But with exchaging basic math solutions as we do now may be it would be a little bit better :)
Posted on 2001-12-27 07:24:55 by The Svin
hii ..

about the clock table .. i know the basic things . that lea is 1 clock the same for add,sub,and and etc..

but there are things i dont understand like :
you wrote imul with only one parameter 10 clock when i look at help files with clock table . imul reg 12-42 clocks on 486 ........
how do i know how much clocks mul takes ?



i realy like math . and I like to get into the little things ( thats way i code in asm ) but i have alot to learn (i'm in the 11 grade )


bye

eko
Posted on 2001-12-27 07:57:29 by eko
One of places where you can find clock tables is Anger.hlp which,
I belive is included in MASM32.
Despite of difference in clocking particular instructions the main
idea of my previous post was that
imul reg,reg
is always slower than
imul reg
and longer.
I'm glad that we are both agree on importance of math knoulidge and training in asm programming.
Couple thoughts:
1. Most of the beginners here know FPU badly.
And to know it better there are always three general methods:
- To see examples of particular equitions or other formulas realized in asm
- To read explonations and comments in those examples
- Try to make realization of math equtions yourselfe.

The most productive way is when complexity of tasks and examples
grows slow but at the number of exersizes to be done and examples to be learnt is quite big. (At least 15-20 a day)

It makes it easier even for very beginner to get used to the tasks and feel in making math as a fish in the water.

What do you think if we start giving ourself right here exersizes (starting with very basic arithmetics) and try to do exersizes of each other in both fpu and interger. And include comments and
explanations?
Posted on 2001-12-27 22:57:14 by The Svin
What do you think if we start giving ourself right here exersizes (starting with very basic arithmetics) and try to do exersizes of each other in both fpu and interger. And include comments and


yeah i realy like the idea ! and my fpu knowlege is bad :( so exercise will be good for me

here is a realy simple start(maybe continue) of integer math
(i dont have any idea for other equation maybe you can choose one)
n=x^2-y^2 finding n in the shortest way

(x-y)(X+Y)

eax -x
ebx -y

lea ecx, ;c=x+y
sub eax,ebx;a =x-y ;
mul ecx ; a =(x-y)(X+Y)

i guess this one too easy ,nothing comes into my mind

bye

eko
Posted on 2001-12-28 12:57:43 by eko