Order of iterations through a 5 dimensional nested for loop

Order = NaN([5 5 5 5 5]); %Creates a 5 dimensional matrix, where 
                          % the length of each dimension is 5
n=1

for Loop5 = [1:5] 
for Loop4 = [1 5];
for Loop3 = [1 5];
for Loop2 = [1:5];
for Loop1 = [1:5] 
Order(Loop1,Loop2,Loop3,Loop4,Loop5) = n;         %Line10
n = n + 1
end
end
end
end
 end

To my understanding, this code should fill the "Order" matrix with sequential numbers from 1 through 3125, as line 10 should run 5^5 times total.

I expect loop 1 to be the first to iterate, through all 5 of it's steps,as below:

Order(1,1,1,1,1) = 1 ,

Order(2,1,1,1,1) = 2 ....

...and on to Order(3,1,1,1,1) = 5,

Then, Loop2 should tick forward one step, to 2, and again, loop 1 iterates 5 steps, while the loop2 variable stays at a value of 2 by this logic, Order(5,5,1,1,1) should be 25, and it is.

Up to here, everything works, but the third dimension, Order(1,1,2,1,1,)Never gets filled. Instead, number 26 ends up at Order(1,1,5,1,1).

How can I change the code so the 26th element to be filled is Order(1,1,2,1,1), and every element in the "Order" matrix has a value?

[Stephen23]

Getting rid of the superfluous square brackets makes the error very easy to spot:

for Loop5 = 1:5;
for Loop4 = 1 5; % <- missing colon
for Loop3 = 1 5; % <- missing colon
for Loop2 = 1:5;
for Loop1 = 1:5;

https://www.mathworks.com/matlabcentral/answers/35676-why-not-use-square-brackets

Answer 1

[Walter Roberson]

Order = NaN([5 5 5 5 5]); %Creates a 5 dimensional matrix, where 
                          % the length of each dimension is 5
n=1

 for Loop5 = 1:5 
    for Loop4 = 1:5
        for Loop3 = 1:5
             for Loop2 = 1:5
                for Loop1 = 1:5
                    Order(Loop1,Loop2,Loop3,Loop4,Loop5) = n;
                    n = n + 1;
                end
            end
        end
    end
 end

[Alejandro Navarro]

Hello, that works perfectly, thank you!

Can you please explain why that works? If you don't indent the for loops, how does the algorithm change? Also, if I were to add functions between lines 6-8, would I need to indent those as well?

[Walter Roberson]

Reread your code more carefully. You had

for Loop4 = [1 5];

That is the specific values 1 and 5, not all the values in the range 1 to 5.

[Alejandro Navarro]

lol whoops, totally missed that. thanks everyone!