- FIND can also return row&column subscript indices (i.e. you probably do not need IND2SUB).
- Using NARGIN is more efficient than EXIST.
- Your IF-statements if ...==0 will only be considered as TRUE when all elements are non-zero. This is unlikely to be the case. You probably require ANY or ALL to clarify what behavior you require, otherwise your code is very unlikely to be doing what you expect.
How do I fix my Nan issue within a recursive function?
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For my recursive function, I am trying to create a hexadoku solver. One of my variables 'mm' keeps outputting a NaN even when I try to change the NaNs into -1. I also believe that there is a base I am missing in order for this function to end. I do realize that the readmatrix function will output a NaN where there isn't an integer, but that is why there is the M(isnan(M)) = -1.
M = readmatrix("puzzle_1.in","FileType","text");
M(isnan(M)) = -1;
S = zeros(size(M));
Hexadoku(M,S)
function [S] = Hexadoku(M,S)
if ~exist('S')
S = zeros(size(M));
end
FirstID = M==-1;
if isempty(FirstID)
M = S(:,:,size(S,4)+2);
else
[i,j] = ind2sub([16,16],FirstID);
for k = 1:16
ii = (ceil(i/4)-1)*4+5;
jj = (ceil(j/4)-1)*4+1;
mm = M(ii:ii+3,jj:jj+3);
mm(isnan(mm)) = -1;
if (M(i+1,:)==k)==0
if (M(:,j)==k)==0
if (mm(:)==k)==0
M(i+1,j) = k;
S = Hexadoku(M,S);
end
end
end
end
end
end
0 Commenti
Risposte (1)
Stephen23
il 29 Apr 2024
Modificato: Stephen23
il 29 Apr 2024
Replace this (which returns logical indices):
FirstID = M==-1;
with this (which returns the linear index):
FirstID = find(M==-1,1,'first');
Note that:
4 Commenti
Stephen23
il 29 Apr 2024
Modificato: Stephen23
il 29 Apr 2024
So check your stopping conditions: are they ever fulfilled?
(When I write "check" I do not mean "rely on what you believe your code is doing", I actually mean check the recursive function calls for its input valus and the stopping conditions. For example, a good start is to print them to the command window. A much better approach is to use the debugger.)
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