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 Kommentar
Jan
am 28 Sep. 2011
@redcavalry: Please delete your former thread, which contains a Pascal version of the code.
Antworten (4)
the cyclist
am 28 Sep. 2011
0 Stimmen
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
am 28 Sep. 2011
you might want to try
type magic
TMW do it in 4 lines.
2 Kommentare
the cyclist
am 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
am 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.)
Image Analyst
am 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
am 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 Kommentare
Walter Roberson
am 29 Sep. 2011
length(unique([sum(M),sum(M,2)',sum(diag(M)),sum(flipud(M))]))==1
Daniel Shub
am 29 Sep. 2011
I think to be a magic square the sorted elements have to equal 1:M
Andrei Bobrov
am 29 Sep. 2011
Yes, Daniel. Added.
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