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Create number of for loops depending on size of N

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Hi, i have a question regarding number of nested loops:
In this case N would be 4 and hence there are 4 for loops
But if N = 2 in need 2 for loops and the formula also changes to i_1+1_2/N where N=2
is it possible to create code that creates the correct amount of for loops (corresponding to the value of N)
and also changes the formula for i_value in a correct way.
i_Max = 8
i_value = [];
i_real = [];
i_first = [];
i_second = [];
i_third = [];
i_forth = [];
tol = 0.01;
i_good = false;
while i_good == false
% generate new ratio's
for i_1 = 1:0.1:i_Max
for i_2 = 1:0.1:i_Max
for i_3 = 1:0.1:i_Max
for i_4 = 1:0.1:i_Max
i_value(end+1) = (i_1+i_2+i_3+i_4)/N;
if (i_average-tol <i_value(end)) && (i_value(end)<i_average+tol)...
&& (i_1>i_2) && (i_2>i_3) && (i_3>i_4)
i_good =true;
i_real(end+1) = i_value(end);
i_first(end+1) = i_1;
i_second(end+1) = i_2;
i_third(end+1) = i_3;
i_forth(end+1) = i_4;
end
end
end
end
end
end
  4 Comments
Torsten
Torsten on 26 May 2024 at 19:08
Edited: Torsten on 26 May 2024 at 19:08
I think you changed the loops in the meantime. That's not fair :-)
Victor
Victor on 27 May 2024 at 8:49
Thank you Torsten for providing code for my problem!

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Accepted Answer

Image Analyst
Image Analyst on 26 May 2024 at 17:50
OK, a not-clever but brainless and verbose approach is to just make a set of "if" blocks
if N == 2
% Code for N=2
elseif N == 3
% Code for N=3
elseif N == 4
% Code for N=4
end
Hopefully you have a small, known and limited number of possibilities for N, like 2, 3, or 4. If you have hundreds of possibilities then you should re-think your algorithm.
  1 Comment
Victor
Victor on 26 May 2024 at 17:53
Thank you, yes N_max = 6 so then it would be possible. Again thank you for your time answering my question

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More Answers (1)

Torsten
Torsten on 26 May 2024 at 18:23
Edited: Torsten on 26 May 2024 at 18:56
The N columns of the resulting C-matrix contain i_first, i_second,...
imax = 8;
N = 4;
i_average = (sum(0:N-1)/N+sum(imax:-1:imax-N+1)/N)/2;
tol = 0.5;
C = nchoosek(0:imax,N)
C = 126x4
0 1 2 3 0 1 2 4 0 1 2 5 0 1 2 6 0 1 2 7 0 1 2 8 0 1 3 4 0 1 3 5 0 1 3 6 0 1 3 7
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
C = sort(C,2,'descend')
C = 126x4
3 2 1 0 4 2 1 0 5 2 1 0 6 2 1 0 7 2 1 0 8 2 1 0 4 3 1 0 5 3 1 0 6 3 1 0 7 3 1 0
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
i_value = sum(C,2)/N
i_value = 126x1
1.5000 1.7500 2.0000 2.2500 2.5000 2.7500 2.0000 2.2500 2.5000 2.7500
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
idx = abs(i_value-i_average)<tol
idx = 126x1 logical array
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 1 1 0 0 0 1 1 1 1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
C = C(idx,:)
C = 34x4
8 6 1 0 8 7 1 0 8 5 2 0 7 6 2 0 8 6 2 0 8 7 2 0 8 4 3 0 7 5 3 0 8 5 3 0 7 6 3 0
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

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