Re: Evaluate ?
- To: mathgroup at smc.vnet.net
- Subject: [mg94551] Re: Evaluate ?
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Tue, 16 Dec 2008 02:33:45 -0500 (EST)
- Organization: Uni Leipzig
- References: <gi5je7$pi6$1@smc.vnet.net>
- Reply-to: kuska at informatik.uni-leipzig.de
Hi,
you explanation hal not much, because
one equation for two variables has never
a unique solution, it has infinite on and i your case
a -> -1 - b + i + j + pixelValuesOfImageA - PixelValuesOfImageB
will work for any b and give you an a.
Regards
Jens
rameshbog at gmail.com wrote:
> Hi,
> Very many thanks for your extremely prompt reply.
> Sorry; I was not quite clear about what I wanted - i was trying to be
> brief. Actually I was thinking of replacement of a pixel values in an
> image. It should replace only when it finds a solution to the equation
> [( i - a ) + ( j - b ) + (pixel values of ImageA - pixel values of
> ImageB ) <= 1].
> For example: In ImageA, if the position of a pixel is (2,5) and its
> intensity value is 100. substiting this information in the above
> equation [( 2 - a ) + ( 5 - b ) + (100 - pixel values of ImageB ) <=
> 1]. Then solve for ' a ' and ' b '. if there is any solution then the
> pixel(2,5) intensity should be changed to ' 0 ' otherwise ' 1'.
>
> My ultimate goal is 1). to check the each pixel of imageA whether it
> satisfies the equation with the imageB.
> 2). Count how many pixels (ImageA) satisfied the equation.
>
> thanks again for your precious time
>
> Ram
>
>
> On Dec 10, 11:12 am, Jens-Peer Kuska <ku... at informatik.uni-leipzig.de>
> wrote:
>
>> Hi,
>
> > NSolve[
> ( i - a ) + ( j - b ) + ( ImageA[[i,j]] - ImageFunction[x][y] ) <=
> 1,
> {a,b} ]
>
>> is *one* inequality for *two* variables {a,b}. There are either no
>> or an infinite number of solutions. x and y are not used in the whole
>> loop construction and it is unclear what you mean.
>
>> It would be better to describe what you wish to do and for what it is
>> good for. Because the most of us would give you an solution without
>> For[] loops and unsolvable equations.
>
>> Regards
>> Jens
>
> - Hide quoted text -
> - Show quoted text -
> ramesh... at gmail.com wrote:
>> Hi Mathematica Guru's,
>
>> I am stuck with a problem in evaluating a continous function. I have
>> two images (A and B), out of which ImageB is interpolated with a
>> continuous function (ListInterpolation). I want to replace the
>> intensity values of ImageA only when it finds a solution to the
>> equation. I have NO idea how to get a solution to this equation. So
>> please I ask for experts in the field to help me !!
>
>> Two Images:
>> ImageA = Import["image1.tiff];
>> ImageB = Import["image2.tiff];
>
>> Interpolation of image B:
>> ImageFunction = ListInterpolation[Transpose[ImageB]];
>
>> Length of the image:
>> Xvalues = Table[a, {a, 1, Length[ImageA]}];
>> Yvalues = Table[b, {b, 1, Length[ImageA[[2]]]}];
>
>> (* Here is the confusion should I use Nsolve or Evaluate for an
>> equation ?? so that when it is 'true' it should change the intensity
>> values of ImageA[[i,j]]=0 otherwise ImageA[[i,j]]=1.
>
>> Replacement and Evaluation:
>> For[i = 1, i = Length[Xvalues],
>> For[j = 1, j = Length[Yvalues],
>
>> (* Please somebody tell me how to Solve an equation in mathematica
>> with two different variables(x,y) with a continuous function )
>> Test[i,j] = NSolve[ ( i - a ) + ( j - b ) + ( ImageA[[i,j]] -
>> ImageFunction[x][y] ) <= 1, {a,b} ]
>> If there is any solution or if it finds any values of 'a' and 'b' for
>> the above equation then intenstiy values of ImageA[[i,j]] = 0
>> otherwise ImageA[[i,j]] = 1];
>> j++]; i++];
>
>> Finally display the image:
>> ListDensityPlot[ImageA[[All, All]], Mesh -> False];
>
>> Thanks for your help inadvance !!
>
>> Ram
>