Question

Radial Blur Zoom

Asked by: ennixo

Hi, i programmed a function to radial blur (zoom) pictures in Vb6 (with a force from zero to 100% and a point of blur), it works well (better quality than photoshop) but it's a little slower.

my function is easy and i would like to know how to accelerate the process to get the same quality faster
i need a vb6 or C++ function.

here's my function :

'cx, cy = Center X and Center Y
'k = Force (0 to 100%)
'Quality = Quality : 1 = Best | 2 = Good | 4 = Medium | 16 = Bad
Public Sub RadialBlurZoom(ByVal cx As Long, ByVal cy As Long, ByVal k As Single, ByVal Quality as Long)

    Dim j               As Long
    Dim i               As Long
    Dim l               As Long
    Dim nX              As Long
    Dim nY              As Long
    Dim dX              As Long
    Dim dY              As Long
    Dim Alpha           As Single
    Dim Hypotenuse      As Long
    Dim vMax            As Long
    Dim TotalR          As Long
    Dim TotalG          As Long
    Dim TotalB          As Long
    Dim Dividor         As Long
    Dim NewBits()       As Byte

    'in the form, PicBits is a private array which contains bitmapbits (from GetBitmapBits)

    ReDim NewBits(LBound(PicBits) To UBound(PicBits))
   
    If K = 0 Then K = 1
    If K > 100 Then K = 100
    K = K / 400
   
    'shpSelection is a Shape which contains the rectangle area to be processed
    For j = shpSelection.Top + 1 To shpSelection.Top + shpSelection.Height
       
        dY = Abs(j - cY)
       
        For i = shpSelection.Left + 1 To shpSelection.Left + shpSelection.Width
           
            dX = Abs(i - cX)
            If dX = 0 Then dX = 1
           
            Alpha = Atn(dY / dX)
            Hypotenuse = Sqr(dX * dX + dY * dY)
           
            TotalR = 0
            TotalG = 0
            TotalB = 0
            Dividor = 0
           
            vMax = Hypotenuse * K
            
            For l = -vMax To vMax Step Quality
            If i < cX Then
                  nX = i + (Cos(Alpha) * l)
            ElseIf i >= cX Then
                  nX = i - (Cos(Alpha) * l)
            Else
                  nX = 0
            End If
            If j < cY Then
                  nY = j + (Sin(Alpha) * l)
            ElseIf j >= cY Then
                  nY = j - (Sin(Alpha) * l)
            Else
                  nY = 0
            End If
            
            'in this form, Matrix is a private array which contains indexes of
            'bits in PicBits to retrieve a point with X and Y values.
                If nX > 0 And nY > 0 And nX <= UBound(Matrix) And nY <= UBound(Matrix, 2) Then
                    TotalR = TotalR + PicBits(Matrix(nX, nY))
                    TotalG = TotalG + PicBits(Matrix(nX, nY) + 1)
                    TotalB = TotalB + PicBits(Matrix(nX, nY) + 2)
                    Dividor = Dividor + 1
                End If
            Next l
            If Dividor > 0 Then
                NewBits(Matrix(i, j)) = TotalR \ Dividor
                NewBits(Matrix(i, j) + 1) = TotalG \ Dividor
                NewBits(Matrix(i, j) + 2) = TotalB \ Dividor
            Else
                NewBits(Matrix(i, j)) = PicBits(Matrix(i, j))
                NewBits(Matrix(i, j) + 1) = PicBits(Matrix(i, j) + 1)
                NewBits(Matrix(i, j) + 2) = PicBits(Matrix(i, j) + 2)
            End If
        Next i
    Next j

    'P is the name of a pictureBox where the picture was loaded
    'sets the new bits to P.Image
    SetBitmapBits P.Image, UBound(NewBits), NewBits(1)
   
End Function

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Asked On
2004-04-22 at 05:02:57ID20963350
Tags

blur

,

radial

Topic

Miscellaneous Programming

Participating Experts
2
Points
500
Comments
12

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Answers

 

by: mokulePosted on 2004-04-23 at 01:25:54ID: 10897278

Maybe I can try. In case You would explain better in words what it should do.

 

by: ennixoPosted on 2004-04-23 at 02:30:43ID: 10897610

ok, for each pixel i compute the resulting color (function of its position, the center position and the force).
everything is calculated for one pixel, everything is calculated again for another pixel, etc...

what i would like is a function which won't recompute everything for each pixel but which will use what was already computed (when possible). this way, there will be less computes and it will be a lot faster (i think) !

but i don't really know how to implement this...

 

by: jamschadPosted on 2004-04-23 at 02:34:45ID: 10897632

Hi.

Depending on in what range the arguments (Alpha) for your sine and cosine functions vary, you can try to make your own sine and cosine functions, based on the Taylor series. As far as I know, VB (and other programming languages) actually does the same, but it usually uses very long series. You can choose to truncate the series earlier. For example write your own functions mCos and mSin which use the following truncated Taylor series:

mCos = 1 - (1/2)*x*x + (1/24)*x*x*x*x - (1/720)*x*x*x*x*x*x
mSin = x - (1/6)*x*x*x + (1/120)* x*x*x*x*x - (1/5040) *x*x*x*x*x*x*x

These two series give pretty good values for -pi<Alpha<pi. Try to make the series even shorter and compare the results. I am also interested to know if or how much this solution improves the speed of your code. So please keep me updated if you choose to give it a try.

There are other things you can do, for example bit-shifting instead of multiplication/division by 2, and so on. I know how it works in C but I am not sure how you can do it in VB, probably the same. Ask me later if you.

 

by: ennixoPosted on 2004-04-23 at 03:15:15ID: 10897883

jamchad:
you won't believe me because i really thought mCos and mSin would accelerate the radial blur... but for the same image in the same conditions with compiled code :
Cos & Sin : 27.189 sec
mCos & mSin : 30.564 sec
i did not compare the resulting images but mSin and mCos seems to give the same result effect.
perhaps it would be faster with simpler mCos & mSin but i don't really know how to "truncate the series" more.

for bitshifting i didn't found where i could do this (no multiplication or division by 2) and i don't know how to do this in vb.

 

by: mokulePosted on 2004-04-23 at 03:16:27ID: 10897889


1. But what function to calculate? I'm not sure what radial blur(zoom) should do.

2. You should not calculate Cos and Sin in this loop but once before entering loop.
        For l = -vMax To vMax Step Quality
          If i < cX Then
               nX = i + (Cos(Alpha) * l)
          ElseIf i >= cX Then
               nX = i - (Cos(Alpha) * l)

 

by: ennixoPosted on 2004-04-23 at 03:31:27ID: 10897967

mokule:
1.
original image :http://auchtre.free.fr/xpws/_flr_cmp_original.jpg
a radial blur (zoom) with photoshop : http://auchtre.free.fr/xpws/_flr_cmp_photoshop.jpg
a radial blur (zoom) with my algorithm : http://auchtre.free.fr/xpws/_flr_cmp_xpws.jpg

that's why i want to be faster, mine is better quality but slower.


2. GREAAAAAAT !!!!!!!
9.935 sec, same image, same conditions !!!

now it's better and faster than photoshop !!!

 

by: ennixoPosted on 2004-04-23 at 03:59:18ID: 10898148

even if you didn't do what i asked, i was so glad that i accepted your answer...
but if you want, you can still research to be faster & faster !

 

by: mokulePosted on 2004-04-23 at 04:07:25ID: 10898186


Well, take it into account.:)

 

by: jamschadPosted on 2004-04-23 at 08:02:04ID: 10900444

hi!

did the quality of the image deteriorate?

these are the more truncated versions (you simply remove the last term in the polynomials)

mCos = 1 - (1/2)*x*x + (1/24)*x*x*x*x
mSin = x - (1/6)*x*x*x + (1/120)* x*x*x*x*x

may be you won't need these now any more, but who knows? may be another time! cheers.


 

by: ennixoPosted on 2004-04-23 at 08:30:24ID: 10900796

jamschad:

with
mCos = 1 - (1/2)*x*x + (1/24)*x*x*x*x
mSin = x - (1/6)*x*x*x + (1/120)* x*x*x*x*x
image is a little deteriorated but it is still slower (!!!?)

with
mCos = 1 - (1/2)*x*x
mSin = x - (1/6)*x*x*x
image is deteriarated a lot more but it is faster
but the deterioration can not be a "quality" because the deformation is not the same so the constraint of the effect are not respected...

but thanx ;)

 

by: mokulePosted on 2004-04-23 at 09:18:04ID: 10901306

mCos = 1 - (1/2)*x*x + (1/24)*x*x*x*x           // 6 multiplications
mSin = x - (1/6)*x*x*x + (1/120)* x*x*x*x*x  // 8 multiplications

It can be optimized as follows. But I don't expect too mauch with it

y = x*x           // 1 multiplication

mCos = 1 - (1/2)*y + (1/24)*y*y
mSin = x - (1/6)*x*y + (1/120)* x*y*y

further

mCos = 1 - (1/2)*y* (1 + (1/12)*y)    // 3 multiplications
mSin = x - (1/6)*x*y*(1 + (1/20)*y)  // 4 multiplications

 

by: ennixoPosted on 2004-04-23 at 10:01:47ID: 10901734

i tried this kind of optimizations but the change is smaller than the loss.
Cos & Sin are definitively faster (i think it's because there are Cos & Sin instruction in asm so it is very fast because the processor can compute this, computing mCos & mSin needs a lot more instructions so it is slower)

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