problem in recursive function

hi,

I have recurrent function , this function has two outputs and each output will be input next in time. I built divide function, this function divide the input into two parts , and each one of the two parts will be input to same function and get two outputs and so on

[a,b]=divede(c)

let divide whatever function

I need help to do that.

thanks

[Image Analyst]

I've never heard of a recurrent function. Do you mean recursive? If it outputs two outputs, how can they be an input into a function that takes only one input??? I think you need to think about this a lot more.

[Azzi Abdelmalek]

suppose you have to do it three times, what is the the expected result?

[Jan]

The question is not clear. What do you want to divide?

[Azzi Abdelmalek]

i think he means for two iterations

 it1) [a,b]=divede(c); 

 it2) [a1,a2]=divede(a)
      [a3,a4]=divede(b)

 ...

[Walter Roberson]

Sounds like a recursive function rather than a recurrant function.

Sounds like a factoring problem.

[huda nawaf]

thanks,

let this function

function [p o]=d(x)
%%x any vector of values
k=1;k1=1;

for i=1:length(x)

      if x(i)>median(x)

          s(k)=x(i);

          k=k+1;

      else

          s1(k1)=x(i);

          k1=k1+1;

      end

     end

   p=s;   o=s1;

end

what i need is,each of output will be the input to this function i.e [o p]=d(o) [o p]=d(p) each output will be input to this function and so on say do that for 5 times, in fact with my case i know the condition stop.

[huda nawaf]

yes, i mean recursive for image analyst I mean if I have

    while ()

    [a b]=d(p)

    [a1 a2]=d(a)

    [b1 b2]=d(b)

    [a11 a22]=d(a1)

    [b11 b22]=d(b1)

    and so on 

    end

i have stop condition.

Answer 1

[Image Analyst]

Here is some recursive code:

% Demo to rotate an ellipse over an image.
% By ImageAnalyst
function [aboveMedian belowMedian] = test2(v)
% Initialize
aboveMedian = v;
belowMedian = v;
% Stopping condition.
if length(v) <= 1
  return;
end
medianValue = median(v);
indexesAboveMedianVaue = v > medianValue;
aboveMedian = v(indexesAboveMedianVaue);
belowMedian = v(~indexesAboveMedianVaue);
% Stopping conditions.
if length(aboveMedian) <= 1 && ...
    length(belowMedian) <= 1
  return;
end
if length(belowMedian) == length(v)
  return;
end
fprintf('For v = ');
fprintf('%f, ', v);
fprintf('\n   Median = %f', medianValue);
fprintf('\n   Values above median = ');
fprintf('%.1f, ', aboveMedian);
fprintf('\n   Values at or below median = ');
fprintf('%.1f, ', belowMedian);
fprintf('\n');
% Now recurse in with the two outputs
test2(aboveMedian);
test2(belowMedian);

It sort of does what you were trying to do with the median. Try it with this code:

a=[2 3 4 56 7 85 3 5 7 7];
[am bm] = test2(a)

But it will quit early because I'm not really sure when you want to stop. What are your stopping conditions when you won't recurse in anymore? I don't have the right stopping conditions in there. Anyway, it's a start and I'll leave it up to you to finish. Good luck.

[huda nawaf]

thanks Aziz,

The problem in second itr

*??? Attempted to access a(1); index out of bounds because numel(a)=0.

Error in ==> test2 at 4 a(1)

Error in ==> d at 22 [a,b]=test2(c{i1}) *

This test code: %%%this ish function%%%

function [c,d]=h(a)
 b1=magic(7);
 b=b1(1,:);
 a(1)
if a(1)>1
   c=a*b';
 d=a+b;
 else
    c=a*b'
    d=[];
end
end

%%%%this is your code

n=5;c={rand(1,7)};
 for k=1:n
     k
     c1=[];
     for i1=1:length(c)
     [a,b]=d(c{i1})
     c1=[c1 ;{a};{ b}];
     end
     c=c1
 end

[huda nawaf]

thanks image analyst, I will try your code

[huda nawaf]

image analyst,

I do not understand this condition what it is working

if length(belowMedian) == length(v)
  return;
end

thanks

[Image Analyst]

Just run it and see. Sometimes, like with my example, you'll get [7 7 7], and if you don't have that check, it will just do that over and over again because it tries to keep going until there's only one value left. But with 7 7 7 it will never stop unless you check for that. You might also use unique().

[huda nawaf]

Yes, I know now what is meaning. I did what u did regarding stop condition, why when ran this function , the code is closed and exit from matlab.

function [G1,G2]=Divide(B)
        G1=B;
        G2=B;
        [p o]=size(B);

%%%compute eignvalue and eignvector
        [U Beta]=eig(B);
%%convert the Beta matrix(eignvalues) into vector%%convert the Beta matrix(eignvalues) into vector
        k=1;
        for i=1:p
            for j=1:o
                if i==j
                    Beta1(k)=Beta(i,j);
                    k=k+1;
                end
            end
        end
%%%sort the Beta1 vector 
  [Beta1 ind]=sort(Beta1,'descend');
  if Beta1(1)>0
        for j=1:o
               if U(ind(1),j)>0
                s1(j)=1;
               else
                s1(j)=-1;
               end
        end
     v=s1*B*s1';
       if v>=0
            %%%divide the eignvector into two groups

               k=1;k1=1;
             for j=1:length(s1)
                 if s1(j)>0
                     for j1=1:o
                     GGrp_1(k,j1)=B(j,j1);
                     track(k)=j;
                     end
                     k=k+1;
                 else
                     for j2=1:o
                     GGrp_2(k1,j2)=B(j,j2);
                     track1(k1)=j;
                     end
                     k1=k1+1;
                 end
             end
                   d=0;d1=0;
                  for i=1:length(track1)
                  GGrp_1(:,track1(i)-d)=[];
                  d=d+1;
                  end
                  [m n]=size(GGrp_1);
                  for i=1:m
                      f(i)=sum(GGrp_1(i,:));
                  end
                  for i=1:m
                      for j=1:n
                          if i~=j
                           B_updat(i,j)=GGrp_1(i,j);
                          else
                           B_updat(i,j)=GGrp_1(i,j)-f(i);
                          end
                      end
                  end
      %     
                 for i=1:length(track)

                     GGrp_2(:,track(i)-d1)=[];
                     d1=d1+1;
                 end
                  [m1 n1]=size(GGrp_2);
                 for i=1:m1
                      f1(i)=sum(GGrp_2(i,:));
                 end
                 for i=1:m1
                      for j=1:n1
                          if i~=j
                           B1_updat(i,j)=GGrp_2(i,j);
                          else
                           B1_updat(i,j)=GGrp_2(i,j)-f1(i);
                          end
                      end
                 end

                   G1=B_updat;

                   G2=B1_updat;

                   [m n]=size(G1);
                   [m1 n1]=size(G2);
                  if m==p | n==o
                   return;
                  end

                  if m1==p | n1==o
                   return;
                  end
         end
   else
            G1=B;
            G2=0;
  end
           Divide(G1)
           Divide(G2)
  end

thanks in advance

[huda nawaf]

this matrix what i try

sim=[0 70 40 -50 10;
    70 0 20  -20 -20;
    40 20 0  -10 -10;
    -50 -20 -10 0 50;
    10 -20 -10 50 0];

[Image Analyst]

Well keep working at it. I'm sure with the debugger, you'll figure it out.

[huda nawaf]

thanks image analyst, your code is very very help me to solve some problems in my code.

Now, I have another query related recursive function. I did modify for stop condition in above code:

   Q_updat=(s1*B*s1')/2
                   if Q_updat<=0
                       return;
                   end

now when Divide(G1) is running and test stop condition , and when met it will exit from code. How I can make the control of program go to Divide(G2),where stop condition stop the programm before raunning the secoind function. how solve this problem?

thanks

[Image Analyst]

I don't know. Maybe just check the stop condition again or output a variable called "keepGoing" that you set to true or false and can check.

[farah fadel]

Image Analyst, Is there a way to keep storing all th outputs of this recursion?

[Image Analyst]

Yes, just accept the variable and stuff it into some array, like a global variable

global allResults
theseResults = func(); % Call your recursive function.
allResults = [allResults, theseResults];

or something like that. If you want, you can attach the array to the handles structure if you're using GUIDE.

handles.allResults = [handles.allResults, theseResults];

The handles structure (or app if using App Designer) is pretty much like a global variable.

[farah fadel]

These methods didn't work(It just stored one result) because the two outputs of the function is matrix with different dimensions, so I was thinking to use cell structure, but the implemntation is not working

[Walter Roberson]

global saved
if isempty(saved); saved = cell(0);
[out1, out2] = yourfunction(whateve)
saved(end+1) = {out1, out2};

[farah fadel]

It still stores just the first two outputs

[Walter Roberson]

How many outputs does the function have? You said it has two outputs, and now you are complaining that it only saves the first two outputs.

You would use the above code everywhere that you call your function, if you want to save all the outputs of the recursion.

[farah fadel]

thank you, it worked

Answer 2

[Azzi Abdelmalek]

 n=5;c={rand(1,100)};
 for k=1:5
     c1=[];
     for i1=1:length(c)
     [a,b]=divede(c{i1})
     c1=[c1 ;{a};{ b}];
     end
     c=c1
 end

% your results are

c{1},c{2} ,c{3}  % ...

[huda nawaf]

thanks, i will try what u wrote to see if work with my case.

[huda nawaf]

but u collect a and b in c1. and in fact a, and b are independent . if c is vector, the out put will be two vectors a, and b

So, each one of outputs will be input to same function and will get new pair a, and b and so on like binary tree, each child will be parted into two parts.

[Azzi Abdelmalek]

no, i 'am using one at time, c1(1), c1(2),...

 c=[a b];
 [a,b]=divede(c(1))
 c1=[a b];
 [a,b]=divede(c(2))
 c1=[c1 a b]
 % for the next iteration c=c1 contains 4 ellements
 % at the end the vector c contain all your 2^n rsults

[Azzi Abdelmalek]

can you post your first value c? you dd'nt mention that c is a vector

[huda nawaf]

can modify code to make first c as vector not one value?

[Azzi Abdelmalek]

ok, check the updated code

[huda nawaf]

my code is complex , i tried write code of modularity and community structure algorithm of newman.

the important part of algorithm is recurrent. In fact the input is array and the output are two arrays, but I simplfied the task . So, supposed the input is vector and the output two vectors.

like this code

function [p o]=d(x)

%%x any vector of values

k=1;k1=1;

for i=1:length(x)

        if x(i)>median(x) 

           s(k)=x(i);

       k=k+1;

        else      

      s1(k1)=x(i);   

       k1=k1+1;

    end  

   end 

   p=s;   o=s1;

end

what i need is,each of output will be the input to this function i.e [o p]=d(o) and [o p]=d(p) each output will be input to this function and so on say do that for 5 times, in fact with my case i know the condition stop.

[Azzi Abdelmalek]

there is a problem in your funcion, when the if is skipped, p=s causes an error because s is not calculated, maybe yo need to initialize it

[huda nawaf]

so just place p=[]; and o=[].

right?

[Azzi Abdelmalek]

no , at the begening

s=[] and  s1=[]

[huda nawaf]

ok , what if place s=[] and s1=[]?

how my problem of recurrent will be resolved? is not there way to make each output as input and give me each time two outputs and repeat the process?

[Azzi Abdelmalek]

that 's what the code i've posted i guess is doing. did you try it? if yes what is the problem?

[Azzi Abdelmalek]

%maby we are not using the same function dived; s and s1 should be initialized

function [p o]=dived(x)
s=[];s1=[]
k=1;k1=1;
for i=1:length(x)
    if x(i)>median(x)
       s(k)=x(i);
       k=k+1;
    else
        s1(k1)=x(i);
        k1=k1+1;
    end
end
 p=s;   o=s1;

the code using dived

 n=5;c={rand(1,100)};
 for k=1:5
     c1=[];
     for i1=1:length(c)
     [a,b]=dived(c{i1})
     c1=[c1 ;{a};{ b}];
     end
     c=c1
 end

% your results are

c{1},c{2} ,c{3}  % ...