is this code correct?

1 view (last 30 days)
redcavalry
redcavalry on 28 Sep 2011
Write a MATLAB code that creates Magic Square of user defined odd order(for example, 5x5, 7x7, etc.) using Terrace method
1. Calculates elements of the Magic Square. 2. Displays the Magic Square.
Note : you can use the special fuction magic to test your own fuction only
M=zeros(n,n);
row=1;
column=(n+1)/2;
for k=1:n^2
M(row,column)=k;
row=row-1;
column=column+1;
if(row==0 && column<=n)
row=n;
elseif(row>0 && column>n)
column=1;
elseif((row==0 && column>n) || M(row,column)>0 )
row=row+2;
column=column-1;
end
end
  1 Comment
Jan
Jan on 28 Sep 2011
@redcavalry: Please delete your former thread, which contains a Pascal version of the code.

Sign in to comment.

Answers (4)

the cyclist
the cyclist on 28 Sep 2011
I see that you have calculated the elements of the magic square (which look correct to me), but I do not see that you have displayed the magic square.

Daniel Shub
Daniel Shub on 28 Sep 2011
you might want to try
type magic
TMW do it in 4 lines.
  2 Comments
the cyclist
the cyclist on 28 Sep 2011
Guessing that TMW doesn't use the "terrace method", but have to admit that I don't know for sure.
Walter Roberson
Walter Roberson on 28 Sep 2011
TMW needs 11 lines for the "singly even" case, at least in R2008b.
(Singly even is a size which is of the form 4*K+2 for non-negative integers K.)

Sign in to comment.


Image Analyst
Image Analyst on 28 Sep 2011
You can determine if your code is correct by subtracting your M from the correct answer. If everything is zero then you matched the correct magic square.
M % Display M
result = M - magic(n) % Display difference

Andrei Bobrov
Andrei Bobrov on 29 Sep 2011
testing your code (edited)
all(diff([sum(M),sum(M,2)',sum(diag(M)),sum(diag(M(:,end:-1:1)))])==0) &&...
isequal(sort(M(:)'),1:numel(M))
ADDED
Hi redcavalry! You used siamese method. Terrace method is variant of siamese method. About of Terrace method in russian Wikipedia (see "Метод террас").
Cod: 1 variant (eg,for n = 5)
n = 5;
k = n*2-1;
B = zeros(k);
B(1:2:end) = 1;
a = hankel(1:k,k:-1:1);
B = (min(a,a(:,end:-1:1))>=n&B)+0;
A = zeros(n,k);
A(:,1:2:end) = 1;
A(~~A)=1:n^2;
A = reshape(A,k,[])';
B(~~B)=A(~~A);
i1 = [(n+1)/2:n-1,n+1:n+(n-1)/2];
i2 = [i1(end)+1:k,1:i1(1)-1];
B(:,i1) = B(:,i1) + B(:,i2);
B(i1,:) = B(i1,:) + B(i2,:);
M = B(i1(1):i1(end),i1(1):i1(end))
2 variant with use siamese method
i1 = (n+1)/2;
j1 = i1+1;
M = zeros(n);
for k = 1:n^2
M(i1,j1) = k;
i1 = i1 - 1;
j1 = j1 +1;
if i1 == 0 && j1 <= n
i1 = n;
elseif i1 > 0 && j1 > n
j1 = 1;
elseif i1 == 0 && j1 > n || M(i1,j1) > 0
j1 = rem(j1,n)+1;
i1 = i1 + 1;
end
end
  3 Comments
Daniel Shub
Daniel Shub on 29 Sep 2011
I think to be a magic square the sorted elements have to equal 1:M
Andrei Bobrov
Andrei Bobrov on 29 Sep 2011
Yes, Daniel. Added.

Sign in to comment.

Categories

Find more on Particle & Nuclear Physics in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!