How do I modify this code so it counts how many times the recursive function is called upon?

I'm working on a project to evaluate different sudoku, but I would like to be able to count how many times this recursive function is called upon for each sudoku. wondering if you could help me modify it so it uses a global variable to do so. Here the function I'm currently using?

function X = sudoku(X) 
% SUDOKU  Solve Sudoku using recursive backtracking. 
%   sudoku(X), expects a 9-by-9 array X. 
 % Fill in all “singletons”. 
 % C is a cell array of candidate vectors for each cell. 
 % s is the first cell, if any, with one candidate. 
 % e is the first cell, if any, with no candidates. 
 [C,s,e] = candidates(X); 
 while ~isempty(s) && isempty(e) 
    X(s) = C{s}; 
    [C,s,e] = candidates(X); 
 end 
 % Return for impossible puzzles. 
 if ~isempty(e) 
    return 
 end 
 % Recursive backtracking. 
 if any(X(:) == 0) 
    Y = X; 
    z = find(X(:) == 0,1);    % The first unfilled cell. 
    for r = [C{z}]            % Iterate over candidates. 
       X = Y; 
       X(z) = r;              % Insert a tentative value. 
       X = sudoku(X);         % Recursive call. 
       if all(X(:) > 0)       % Found a solution. 
          return 
       end 
     end 
   end 
% ------------------------------ 
   function [C,s,e] = candidates(X) 
      C = cell(9,9); 
      tri = @(k) 3*ceil(k/3-1) + (1:3); 
      for j = 1:9 
         for i = 1:9 
            if X(i,j)==0 
               z = 1:9; 
               z(nonzeros(X(i,:))) = 0; 
               z(nonzeros(X(:,j))) = 0; 
               z(nonzeros(X(tri(i),tri(j)))) = 0; 
               C{i,j} = nonzeros(z)’; 
            end 
         end 
      end 
 L = cellfun(@length,C);   % Number of candidates. 
 s = find(X==0 & L==1,1); 
 e = find(X==0 & L==0,1); 
 end % candidates 
end % sudoku

Answer 1

[Guillaume]

You could also not bother with the persistent variable and just pass the count as input/output:

function [X, count] = sudoku(X, count)
    if nargin < 2
        count = 0;
    else
        count = count + 1;
    end
    %... rest of the code that leads to the recursive call
    [X, count] = sudoku(X, count);
end

Answer 2

[Matt J]

Use a persistent counter variable:

function [X,count] = sudoku(X, restart) 

 persistent count
 
 if nargin<2, restart=false; end
 
 if isempty(count) || restart, 
     count=1;
 else
     count=count+1;
 end

 [C,s,e] = candidates(X); 
 
 .... %the rest of your code

[Peter Takach]

unfortunately working on a bit older version and aparently this is pulling up a bug...

[Walter Roberson]

What error are you seeing ?

[Guillaume]

It's much better if you just copy/paste the full text of the error rather than giving us a (misspelled) interpretation of it.

Assuming that the error is what I think it is, it is still true in the latest version of matlab. The persistent variable must be declared before it's used, and it can't be an output variable. So, while Matt's answer explained the correct concept, the actual execution is a bit off.

function [X, count] = sudoku(X, restart)
    persistent internalcount;  %can't be a return variable
    if isempty(internalcount) || restart
        internalcount = 1;
    else
        internalcount = internalcount + 1;        
    end
    %... rest of the code
    count = internalcount;
end

[Peter Takach]

yes i quite agree unfortuantely i was sending it from my phone at work. thanks so much for the help though. appreciate you working through my spelling errors.