Create number of for loops depending on size of N

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

[Stephen23]

There are essentially two approaches:

• the complex approach is to write a recursive function,
• the simple approach is to use exactly one FOR-loop:

https://www.mathworks.com/matlabcentral/answers/79986-how-can-i-condition-the-number-of-nested-for-loops-in-a-function-on-an-input-argument

https://www.mathworks.com/matlabcentral/answers/345551-function-with-varying-number-of-for-loops

https://www.mathworks.com/matlabcentral/answers/381505-creating-a-variable-number-of-nested-for-loops

https://www.mathworks.com/matlabcentral/answers/602395-different-number-of-for-loops#answer_502861

[Victor]

Thank you for commenting on the problem!

I have looked into the links you have send me but i dont know how to start. In my case the iterations remain the same for all index values. Can you please give me some advice on how to tackle this problem?

[Torsten]

@Victor

I think you changed the loops in the meantime. That's not fair :-)

[Victor]

Thank you Torsten for providing code for my problem!

Answer 1

[Image Analyst]

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.

[Victor]

Thank you, yes N_max = 6 so then it would be possible. Again thank you for your time answering my question

Answer 2

[Torsten]

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

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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

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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

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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

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