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# is this code correct?

Asked by 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 Simon on 28 Sep 2011

@redcavalry: Please delete your former thread, which contains a Pascal version of the code.

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## 4 Answers

Answer by 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.

## 0 Comments

Answer by Daniel on 28 Sep 2011

you might want to try

`   type magic`

TMW do it in 4 lines.

## 2 Comments

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 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.)

Answer by 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
```

## 0 Comments

Answer by 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

Walter Roberson on 29 Sep 2011

length(unique([sum(M),sum(M,2)',sum(diag(M)),sum(flipud(M))]))==1

Daniel on 29 Sep 2011

I think to be a magic square the sorted elements have to equal 1:M

Andrei Bobrov on 29 Sep 2011

Yes, Daniel. Added.

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