how to convert to matlab from fortran

[hasan damaj]

continuation of the code is attached

Answer 1

[the cyclist]

You could try f2matlab. It might not get you all the way to the final code you need, but it will probably help.

[hasan damaj]

clc
clear
nx = 7;
ny = 13;
h = zeros(nx,ny);
h(:,:) = 100;
h;
Dx = 20;
amax = 1;

while amax >= 0.001
    amax = 0;
for j = 2 : ny-1
    h(1,j) = 0.02 * Dx * (j-2) + 100;
    for i = 1 : nx
        h(i,1)  = h(i,3);
        h(i,13) = h(i,11);
        for j = 2 : ny-1
            h(7,j) = h(i,11);
            for i = 2 : ny - 1
            for j = 2 : nx - 1
                oldval = h(i,j);
                h(i,j) = (h(i-1,j) + h(i+1,j) + h(i,j-1) + h(i,j+1))/4
                e = abs(h(i,j) - oldval);
                if e > amax
                    amax = e ;
                end
            end
            end
        end
    end
end
end
h

[hasan damaj]

i tried to run the above, it gave me an index exceeds matrix dimension,

can u help me sir and thank u.

attached is the answer

[the cyclist]

Is that code the output of f2matlab, or is it something you wrote?

The specific reason your code fails is that h(i,j) is a 7x13 matrix, and in the line

h(i,j) = (h(i-1,j) + h(i+1,j) + h(i,j-1) + h(i,j+1))/4

when i == 7, you try to access element h(8,2), which does not exist.

I really can't figure out what you are trying to do in this code. The fact that you have loops over the variables i and j, inside other loops over i and j, is very confusing to me.

[hasan damaj]

Thank you for your comment.

I'm trying to convert the Fortran code to MATLAB.

"""""this is a regional flow example using gauss-seidel iteration"""

[Image Analyst]

That is supposed to be a comment. In MATLAB, it would look like this:

% This is a regional flow example using Gauss-Seidel iteration.

[hasan damaj]

i know im not writing a code here :)

thank you

Answer 2

[Ben Barrowes]

OCR and some hand editing produced this fortran code which I compiled and ran:

      program flow
C     RE GIONAL FLOW SYST&M &XAMPLE 
      dimENSION H(13, 7) 
C INITIALIZE ALL H(I,.JJ VALUES TO BE l00. 
      DO 5 J=1,7
      do 5 I=1,l3
      H(I,J) = 100. 
 5    CONTINUE 
c WATER TABLE BOUNOARY 
      DX=28. 
      DO 10 i=2,12
      H(I,1) = 0.02*DX*(I-2)-100.
10    CONTINUE 
C KEEP TRACK OP NUMBER OF ITERATIONS AND OF larRGEST ERROR 
C NO-FLOW doundaries NEeD TO BE RESET WITHIN EACH ITERATION LOOP 
      NUMIT = 0 
35    AMAX =0. 
      NUmIT = NUMIT + 1
C     LeFT AND RIGHT NO-PLOW BOUNDARIES
      DO 20 J=1,7 
      H(1,J) = H(3,J) 
      H (13,J) = H(11,J) 
 20   COnTINUe

C BOT10M N!rtLCW SOUNDAFC:t 
      DO 30 i=2,12 
      H(i,7) =H(i,5) 
30    CONTINUE 
C SWEEP INTERIOR POINTS WITH 5-POltn' OPERATOR 
      DO 40 J=2,6
      DO 40 i=2,12 
      OLDVAL= H(I,J)
      H(i,J) = (H(I-1,J) + H(I+1,J) + H(I,J-1) + H(I,J+1))/4. 
      ERR= ABS(H(I,J) -OLdVAL)
      IF(ERR.GT.AKAX) AMAX=ERR 
40    CONTINUE 
C 00 ANOTHER ITERATION IF LARGEST ERROR AFFECTS JRD DECIMAL PLACE 
      IF (AMAX.GT.0.001) GO TO 35 
C WE ARE OONE. 
      PRINT 50,NUmIT, ((H(I,J),I=2,12),J=l,6) 
 50   formAT(///1X,'NUMBER OF ITERATIONS IS',I4,///6(11f8.2///)) 
      ENd

Using f2matlab and some other tools I have, this is the resulting matlab code. It seems to run, but I have not tested for accuracy of results. The fprintf statement could use some fixing for the results to look exactly the same.

      function flow(varargin)
 clear global;
 clear functions;
 global GlobInArgs nargs
 GlobInArgs={mfilename,varargin{:}};
 nargs=nargin+1;
 global unit2fid;
 if ~isempty(unit2fid), unit2fid={};
 end
 persistent akax amax dx err firstCall h i j l l3 numit oldval   format_50
 ; 
 if isempty(firstCall),firstCall=1;end; 

 if firstCall;
  format_50=[ '\n' , '\n' , '\n' ,blanks(1),'NUMBER OF ITERATIONS IS','%4d', '\n' , '\n' , '\n' ,'~',repmat([repmat('%8.2f',1,11), '\n' , '\n' , '\n' ] ,1,6)];
  akax=0;
  amax=0;
  dx=0;
  err=0;
  h=zeros(13,7);
  i=0;
  j=0;
  l=0;
  l3=0;
  numit=0;
  oldval=0;
 end

 firstCall=0;
 %     RE GIONAL FLOW SYST&M &XAMPLE
 % INITIALIZE ALL H(I,.JJ VALUES TO BE l00.
 for j = 1: 7
  for i = 1: l3
   h(i,j) = 100.;
  end
  i = fix(l3+1);
 end
 j = fix(7+1);
 % WATER TABLE BOUNOARY
 dx = 28.;
 for i = 2: 12
  h(i,1) = 0.02.*dx.*(i-2) - 100.;
 end
 i = fix(12+1);
 % KEEP TRACK OP NUMBER OF ITERATIONS AND OF larRGEST ERROR
 % NO-FLOW doundaries NEeD TO BE RESET WITHIN EACH ITERATION LOOP
 numit = 0;
 while (1);
  amax = 0.;
  numit = fix(numit + 1);
  %     LeFT AND RIGHT NO-PLOW BOUNDARIES
  for j = 1: 7
   h(1,j) = h(3,j);
   h(13,j) = h(11,j);
  end
  j = fix(7+1);

  % BOT10M N!rtLCW SOUNDAFC:t
  for i = 2: 12
   h(i,7) = h(i,5);
  end
  i = fix(12+1);
  % SWEEP INTERIOR POINTS WITH 5-POltn' OPERATOR
  for j = 2: 6
   for i = 2: 12
    oldval = h(i,j);
    h(i,j) =(h(i-1,j)+h(i+1,j)+h(i,j-1)+h(i,j+1))./4.;
    err = abs(h(i,j)-oldval);
    if(err > akax)
     amax = err;
    end
   end
   i = fix(12+1);
  end
  j = fix(6+1);
  % 00 ANOTHER ITERATION IF LARGEST ERROR AFFECTS JRD DECIMAL PLACE
  if(amax > 0.001)
   continue;
  end
  % WE ARE OONE.
  h(2:12,1:6)
  %fprintf(1,[format_50],numit,{{h(i,j),'i','2','1','12'},'j','l','1','6'});
  tempBreak=1;
  break;
 end
 clear all
end %program flow

[hasan damaj]

thank you mr. ben

but the results are not the same and they are negative and i think you got the rows and columns switched..

[Ben Barrowes]

Yes. As I wrote, I have not tested this for accuracy of the results. Please test and debug from here on. Hope this helps.

[hasan damaj]

Mr. Ben, i attached a possible code i figured can u just look at it,

clc
clear
nx = 7;
ny = 13;
h = zeros(nx,ny);
h(:,:) = 100;
h;
Dx = 20;
amax = 1;

% while amax >= 0.001
%     amax = 0;
for j = 2 : ny-1
    h(1,j) = 0.02 * Dx * (j-2) + 100;
    for i = 1 : nx
        h(i,1)  = h(i,3);
        h(i,13) = h(i,11);
        for j = 2 : ny-1
            h(7,j) = h(5,j);
            while amax >= 0.001
                amax = 0 ;
            for i = 2 : ny - 1
            for j = 2 : nx - 1
                oldval = h(i,j);
                h(i,j) = (h(i-1,j) + h(i+1,j) + h(i,j-1) + h(i,j+1))/4;
                e = abs(h(i,j) - oldval);
                if e > amax
                    amax = e ;
                end
            end
            end
        end
    end
end
end
h
is the location of while loop ok?? its giving me  index exceeds matrix dimensions.
the above comments shows you the result of table of numbers
thank you

[Ben Barrowes]

One thing I noticed looking at your code is that you have reused the same variable as the loop index variables. You have three loops that use J as the loop variable for example and two that use I as the loop variable. Use different variables for each loop index.

[hasan damaj]

i cant change the variable because its 1 equation where h is in terms of i and j