could anyone help me how to reduce the circshift to half of it.

3 vues (au cours des 30 derniers jours)
Prabha Kumaresan
Prabha Kumaresan le 16 Août 2018
Commenté : Prabha Kumaresan le 16 Août 2018
By using circshift i am getting 12 times what can be done inorder to reduce 12 to 6(which means half of it). Could anyone please help me on this.
  4 commentaires
Prabha Kumaresan
Prabha Kumaresan le 16 Août 2018
Modifié(e) : Rik le 16 Août 2018
yes i can understand circshift does not reduce the length.
this was the code which i am working on
A_master=1:12;
for k=1:numel(A_master)
fprintf('\nPossibilities %02d:\n',k);
A=circshift(A_master,1-k,2);
B=0;
while ~isempty(A)
B=B+1;
if B>numel(A)
ind=1;
B=1;%reset B
else
ind=[1 (numel(A)-B+2):numel(A) ];
end
C=A(ind)
A(ind)=[];
end
end
If i run the code it executes and gives me the result.But I dont want to have all the possibilities. Instead I want to get the possibilities 1,3,5,7,9,11. Could you help me on this.
Stephen23
Stephen23 le 16 Août 2018
Modifié(e) : Stephen23 le 16 Août 2018
@Prabha Kumaresan: this is clearly related to your earlier questions of the last two days. Rather than showing us broken code with no comments or help or explanation, it would actually make things a lots easier if you explained what you are trying to achieve. Forget about your code attempts, they don't help us and are possibly not a very good solution anyway. Show us a simple example, complete with input and output arrays, so that we can understand what you want to achieve.

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Rik
Rik le 16 Août 2018
Just change the for-loop header:
A_master=1:12;
for k=1:2:numel(A_master)%[1 3 5 7 9 11] for 12, [1 3 5] for 6, etc
fprintf('\nPossibilities %02d:\n',k);
A=circshift(A_master,1-k,2);
B=0;
while ~isempty(A)
B=B+1;
if B>numel(A)
ind=1;
B=1;%reset B
else
ind=[1 (numel(A)-B+2):numel(A) ];
end
C=A(ind)
A(ind)=[];
end
end

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