Recursive loops in MATLAB

I have a question regarding recursive loops in MATLAB.

Suppose I have a variable called "dimension" with value 3. The script that I need to run has the following structure:

dimension=3;
order=5;

for i1=1:order
    for i2=1:order
        for i3=1:order
            if i1+i2+i3<=order
                disp([i1,i2,i3])
            end
        end
    end
end

That is, if "dimension=3", 3 nested loops will be required to run this script.

If "dimension=4", then the script shown above becomes:

dimension=4;
order=5;

for i1=1:order
    for i2=1:order
        for i3=1:order
            for i4=1:order
                if i1+i2+i3+i4<=order
                    disp([i1,i2,i3,i4])
                end
            end
        end
    end
end

That is, 4 nested loops in this case. And so on.

To solve this problem, I tried the following first (without the if condition i1+i2...<=order shown above, because I didn't know how to incorporate it into the script):

function result=recursiveLoop(dimension,order,level,iter,counter,expression)
if nargin==2 % initial call
    level=1;
    iter=zeros(dimension,1);
    counter=0;
    expression=zeros(order^dimension,dimension);
end

for i=1:order
    iter(level)=i;

      counter=counter+1;

      expression(counter,:)=iter;

      if level==dimension && i==order
          result=expression;
      end

      if level~=dimension
          result=recursiveLoop(dimension,order,level+1,iter,counter,expression);
      end
  end

end

But the output is not the one that one would expect from the above definition. Any ideas about how to solve this problem using, say, a recursive function? Is there a better way to solve this problem?

Thank you,

Ricardo

[David Goodmanson]

Hi Ricardo, do you care what order the results are displayed in, i.e. right now the last index changes most quickly and the first index changes least quickly.

[Ricardo Prada]

Hi David. I don't. I just want to execute some code inside the if condition. Thanks for your prompt reply.

Answer 1

[David Goodmanson]

Hi Ricardo, here is a method using a function I created recently that is related to the adjustable-dimension issue. This is the first method that came to mind and other contributors here may well come up with more compact code. This creates a matrix whose rows are what you display in the loop.

dim = 3;  ord = 5;
a = ord*(ones(1,dim));
z = ind2subb(a,1:ord^dim);
ind = (sum(z,2)>ord);
z(ind,:) = [];          % knock out rows that don't meet the condition

% -----------------------------

function subs = ind2subb(siz,ind)

% matrix of sub indices from a linear index, for an array
% of size siz = [s1,s2 ...].
% each row in subs gives an index for each dimension of siz. 
% same as ind2sub except subs is a matrix rather than a set of vectors.
%
% subs = ind2subb(siz,ind)

ndims = length(siz);
subs = zeros(length(ind),ndims)
ind = ind-1;
for j = 1:ndims
  r = rem(ind,siz(j))
  subs(:,j) = r;
  ind = (ind -r)/siz(j);
end
subs = subs+1;

[Ricardo Prada]

Hi David,

Your script solves very nicely the problem shown above. But unfortunately, my problem is more complicated than that. What I want is to execute a "sub-script" inside of the if condition, which is a function of i1, i2, ...

For say, if "dimension=4", then

dimension=4;
order=5;

for i1=1:order
    for i2=1:order
        for i3=1:order
            for i4=1:order
                if i1+i2+i3+i4<=order
                    % Do some stuff... (I have about 1500 lines of code here)
                end
            end
        end
    end
end

Thanks! and sorry for not having been clear from the very beginning.

Best,

Ricardo

[Ricardo Prada]

Hi David,

Never mind. Your approach will solve my actual problem too. What I need to do now is to use the resulting "z" matrix, and put it in another loop.

Thanks a lot for your time and help. I greatly appreciate it.

Best,

Ricardo

[David Goodmanson]

Hi Ricardo, you're welcome. Looking at this again I think better coding in the last two script lines would be

ind = (sum(z,2)<=ord);
z = z(ind,:);           % keep rows that meet the condition