MATLAB: Simple question in matlab

general matlab usage

Hello all,

I have the following for loop: for n = 3:32 r(n) = rank(magic(n)); end r

Shouldn't this loop start from n=3 and the size of r be 1×30? but why size of r is 1×30 and it puts r(1)=r(2)=0? how can I make the loop start from 3 not 1?

Best Answer

You need to start your index value in r with 1 or else it populates up to the index value specified with zeros.

r = zeros(1,30);
for n = 3:32 
    r(n-2) = rank(magic(n)); 
end
r

Related Solutions

MATLAB: Does r have these many number of exact elements

Here is your code.

r = [];
a = 0;
n = 3;
while a<15
  a = rank(magic(n));
  r(n) = a;
  n=n+1;
end

I initialized a as 0, and r as an empty vector to allow the loop to operate properly.

I've assumed you started out n as at least 3, because magic(1) and magic(2) are not useful things. That is magic(2) is not in fact a magic square at all.

magic(2)
ans =
     1     3
     4     2

And magic(1) just returns the scalar 1.

So what is r after that loop? I'll append it to the numbers 1:n so we can look at what happens.

[1:numel(r);r]
ans =
   1  2  3  4  5  6  7  8  9 10 11 12 13 14 15
   0  0  3  3  5  5  7  3  9  7 11  3 13  9 15

That is, only for an index of [3,5,7,9,11,13,15] is the rank of the resulting magic suare full. In all even cases, magic does not generate a full rank matrix. (In fact, I often use this property of magoic to generate nice integer matrices that are not full rank in some of my answers.)

That is, if we look at the matrix magic(4), we see that it is indeed a magic matrix, with the necessary properties.

magic(4)
ans =
    16     2     3    13
     5    11    10     8
     9     7     6    12
     4    14    15     1
     
>> sum(magic(4),1)
ans =
    34    34    34    34
    
>> sum(magic(4),2)
ans =
    34
    34
    34
    34
    
>> sum(diag(magic(4)))
ans =
    34
    
sum(diag(flip(magic(4))))
ans =
    34
    
>> rank(magic(4))
ans =
     3

So all rows and columns, and the two main diagonals sum to 34.

A quick search online finds the paper:

R. Bruce Mattingly (2000), Even Order Regular Magic Squares Are Singular, The American Mathematical Monthly, 107(9): 777-782.

The title tends to suggest that even order magic squares seem to have a problem.

But I also find the claim that some even order magic squares can be non-singular, see here:

https://math.stackexchange.com/questions/1136380/why-the-determinant-of-even-sized-magic-matrices-are-zero-anything-special

In fact, in the next link, we find the statement that magic squares can be classified into three types, where the order is odd, singly even (divisible by 2, but not by 4), and doubly even (divisible by 4).

https://answers.yahoo.com/question/index;_ylt=AwrC1CnZZjldXjcAFS1PmolQ;_ylu=X3oDMTByOHZyb21tBGNvbG8DYmYxBHBvcwMxBHZ0aWQDBHNlYwNzcg--?qid=20080629073644AANAcIt

In fact, if you look at the code for magic,

type magic

you will see it does different things depending on the value of n.

To answer why it is that even order magic squares produced by magic are not full rank will be lengthy. But if we accept that even order magic squares are not full rank, then the simple answer to your question is why those numbers are repeated in the vector r. To answer that, we first look at the ranks of all doubly even magic squares as produced by magic. See that the ranks of magic(n), for n in the set [4 8 12...] seem to be always 3.

rank(magic(16))
ans =
     3
>> rank(magic(20))
ans =
     3
>> rank(magic(24))
ans =
     3
>> rank(magic(32))
ans =
     3
>> rank(magic(104))
ans =
     3

You should see the pattern. The scheme used in magic to produce doubly even magic squares for doubly even values of n always produces squares of rank 3. So the number 3 will appear very regularly in the list of ranks in r. To prove that scheme always produces rank 3 matrices for doubly even n will take some minor effort, and I've already managed to make this a lengthy response.

MATLAB: How to add for loop in the code to get a table of result and plot it

Yes, there are some problems with your for loop. But problems with density of states are fun!

Remember, that when you say "for R=2:7; ... end", matlab automatically increments R on every loop. You don't have to include R+1.

Also, you want to collect a table of R and N_states values, but you've only made scalar values which are getting over written on each iteration of the loop. You need to allocate an array for the R and N_states values before the loop. In this case since you're ultimately using a table, you can allocate a table.

R = (2:7)'; % allocate column vector R, integers 2,3,4..7
N_states = zeros(size(R)); % allocate N_states as zeros. This vector is the same shape as R
T = table(R,N_states); % create a table with these values
for m = 1:height(T) % loop through all the rows of T with index m
    for nx = 1:T.R(m) % nx and ny range from 1 to R, the R specified by THIS row m of the table
    for ny = 1:T.R(m) % are you sure you don't want to go from 0?
       if ((nx.*nx + ny.*ny) <=  T.R(m).*T.R(m))
          T.N_states(m)= T.N_states(m) + 1; % increment this N_states in table
       end
    end
    end
    % No incrementing of R or m or nx, ny here.
end
% tables don't have an explicit plot function. Extract the vectors and plot like this:
plot(T.R,T.N_states)

Computationally there are much faster ways to solve this problem, but let me know how this feels to you.

Best Answer

Related Solutions

MATLAB: Does r have these many number of exact elements

MATLAB: How to add for loop in the code to get a table of result and plot it

Related Question