is this code correct?
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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 个评论
Jan
2011-9-28
@redcavalry: Please delete your former thread, which contains a Pascal version of the code.
回答（4 个）
the cyclist
2011-9-28
0 个投票
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
2011-9-28
you might want to try
type magic
TMW do it in 4 lines.
2 个评论
the cyclist
2011-9-28
Guessing that TMW doesn't use the "terrace method", but have to admit that I don't know for sure.
Walter Roberson
2011-9-28
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.)
Image Analyst
2011-9-28
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
2011-9-29
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 个评论
Walter Roberson
2011-9-29
length(unique([sum(M),sum(M,2)',sum(diag(M)),sum(flipud(M))]))==1
Daniel Shub
2011-9-29
I think to be a magic square the sorted elements have to equal 1:M
Andrei Bobrov
2011-9-29
Yes, Daniel. Added.
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2011-9-28
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