How to scan matrix in spiral way?

I enter an array from me or user defined like A= [1 2 3 4 5 6

7 8 9 10 11 12

13 14 15 16 17 18]

and a spiral display is made for it like:

A=[1 2 3 4 5 6

14 15 16 17 18 7

13 12 11 10 9 8]

, then I want another code in which I can retrieve my original array agin in A' matrix. I need code for two processes and I would be grateful for your help.

[Image Analyst]

Why do you want to do this quirky thing? What is your "use case"?

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[Arshub]

@Image Analyst I've seen previous answers for the spiral scan, but I couldn't find the reverse process because I needed it for an image encryption process.

This is not my homework, but I am a master's student and I build my algorithm using Matlab, and I need this code in my work.

[Arshub]

@Image AnalysI asked before, but I did not get the desired response and the right answer, so I decided to ask the question again in a short way.

Answer 1

[DGM]

There is a demo function that can generate NxN arrays in an outward spiral. That could be used (with some arithmetic and flips/transposes) to generate NxN spirals in other directions/orientations.

n = 5;
sp = spiral(n)

sp = 5×5

21 22 23 24 25 20 7 8 9 10 19 6 1 2 11 18 5 4 3 12 17 16 15 14 13

The result could also be used as an indices into another array.

mg = magic(n)

mg = 5×5

17 24 1 8 15 23 5 7 14 16 4 6 13 20 22 10 12 19 21 3 11 18 25 2 9

mg(sp)

ans = 5×5

15 16 22 3 9 2 5 6 12 18 21 24 17 23 1 20 11 10 4 7 14 8 25 19 13

However, that's strictly a square array. If you want something different, you'll have to do something else.

As one possibility, MIMT has a tool for reshaping images using spiral devectorizing. It could be used for this.

A = reshape(1:18,6,3).'

A =

1 2 3 4 5 6

7 8 9 10 11 12

13 14 15 16 17 18

B = jellyroll(A.').'

B =

1 2 3 4 5 6

14 15 16 17 18 7

13 12 11 10 9 8

Arecovered = jellyroll(B.','reverse').'

Arecovered =

1 2 3 4 5 6

7 8 9 10 11 12

13 14 15 16 17 18

Obviously, I can't run that here without MIMT, but this should show that jellyroll() does work on nonsquare matrices. The direction and orientation can be specified with the appropriate options. It just happens that this one case works fine with the defaults.

I should point out though that jellyroll() is intended to be a novelty for use in manual image processing workflows where computational efficiency is unimportant. Keep that in mind if you stick it in a short loop with tiny arrays.

[Arshub]

@DGM

can you take alook for this cod, I hope that you can help me in modifying this code so that I enter an array from me like A= [2 3 5

12 13 14

19 0 4]

and a spiral display is made for it, then I want another code in which I can retrieve my original array. I would be grateful for your help.

I try to modifiey this cod but i got errors .

A = zeros(3,6);%You can change this mxn to whatever dims you like, be it square(m=n) or rectangular matrix.
[m,n] = size(A);
%m = no of rows
%n = no of columns
%A = matrix to be filled with spiral values
 %This code works fine. I commented an eg also below
top = 1;
bottom = m;
left = 1;
right = n; %To keep track of the four directions
value = 1;%Let us strat from 1
while true
    %Top Row First
    if left>right
        break;
    end
    for i = left:right
        A(top,i) = value;
        value = value + 1;
    end
    top = top + 1;
    %Then The RightMost Column
    if top>bottom
        break;
    end
    for i = top:bottom
        A(i,right) = value;
        value = value + 1;
    end
    right = right - 1;
    %Then Bottom Row
    if left>right
        break;
    end
    for i = right:-1:left
        A(bottom,i) = value;
        value = value + 1;
    end
    bottom = bottom - 1;
    %Then The Left Column
    if top>bottom
        break;
    end
    for i = bottom:-1:top
        A(i,left) = value;
        value = value + 1;
    end
    left = left + 1;
end
A

[DGM]

You can reverse it the same way that jellyroll() does internally, though changing the spiral orientation will complicate things.

A = zeros(3,6);%You can change this mxn to whatever dims you like, be it square(m=n) or rectangular matrix.
[m,n] = size(A);
%m = no of rows
%n = no of columns
%A = matrix to be filled with spiral values
%This code works fine. I commented an eg also below
top = 1;
bottom = m;
left = 1;
right = n; %To keep track of the four directions
value = 1;%Let us strat from 1
while true
    %Top Row First
    if left>right
        break;
    end
    for i = left:right
        A(top,i) = value;
        value = value + 1;
    end
    top = top + 1;
    %Then The RightMost Column
    if top>bottom
        break;
    end
    for i = top:bottom
        A(i,right) = value;
        value = value + 1;
    end
    right = right - 1;
    %Then Bottom Row
    if left>right
        break;
    end
    for i = right:-1:left
        A(bottom,i) = value;
        value = value + 1;
    end
    bottom = bottom - 1;
    %Then The Left Column
    if top>bottom
        break;
    end
    for i = bottom:-1:top
        A(i,left) = value;
        value = value + 1;
    end
    left = left + 1;
end
A

A = 3×6

1 2 3 4 5 6 14 15 16 17 18 7 13 12 11 10 9 8

% revert the transformation
B = A.';
B(B) = 1:(n*m);
B = B(A)

B = 3×6

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

[Arshub]

@DGM can you edit the code to enter array wirh any onathers values, I mean that enter an array from me like A= [2 3 5

12 13 14

19 0 4]

i need implement this code with several matices to get spiral and inverse spiral.

[DGM]

If you want to use the example you're using, you'll have to use the output as indices into the given matrix.

A = [2 3 5; 12 13 14; 19 0 4]

[m,n] = size(A);
Aidx = zeros(m,n);

top = 1;
bottom = m;
left = 1;
right = n; %To keep track of the four directions
value = 1;%Let us strat from 1
while true
    %Top Row First
    if left>right
        break;
    end
    for i = left:right
        Aidx(top,i) = value;
        value = value + 1;
    end
    top = top + 1;
    %Then The RightMost Column
    if top>bottom
        break;
    end
    for i = top:bottom
        Aidx(i,right) = value;
        value = value + 1;
    end
    right = right - 1;
    %Then Bottom Row
    if left>right
        break;
    end
    for i = right:-1:left
        Aidx(bottom,i) = value;
        value = value + 1;
    end
    bottom = bottom - 1;
    %Then The Left Column
    if top>bottom
        break;
    end
    for i = bottom:-1:top
        Aidx(i,left) = value;
        value = value + 1;
    end
    left = left + 1;
end

% do forward transformation
B = A.';
B = B(Aidx)

% do reverse transformation
Bidx = Aidx;
Bidx(Bidx) = 1:(n*m);
Arecovered = reshape(B(Bidx),n,m).'

Or something like that. Again, it'll probably need tweaking if the orientation changes.

If you're using jellyroll(), the usage is way simpler. You just give it the matrix itself. Everything else is the same as before.

A = [2 3 5; 12 13 14; 19 0 4]
B = jellyroll(A.').'
Arecovered = jellyroll(B.','reverse').'

[Arshub]

@DGM output of edited code is not spiral matrix for matrix A = [2 3 5; 12 13 14; 19 0 4], all output are wrong.

Aidx =

     1     2     3
     8     9     4
     7     6     5
     
Bidx =

     1     8     3
     4     9     2
     7     6     5
     
Arecovered =

     1     9    13
     2    15     7
     3    14     8

[DGM]

When I run the above code, I get this:

A =
     2     3     5
    12    13    14
    19     0     4
    
B =
     2     3     5
     0     4    12
    19    14    13
    
Arecovered =
     2     3     5
    12    13    14
    19     0     4

I get the same using jellyroll().

Like I said, jellyroll() might not be the fastest approach (i don't know that Piyush Aggarwal's example is much faster), but it has some overhead due to the other things for which it's intended to be used. The advantage to using jellyroll() is that all these little issues of orientation and mapping have already been ironed out and they don't have to be reinvented again. If it seems confusing, it's because it can be. At least it is for me.

[Arshub]

@DGM Thanks alot for your help.

[DGM]

If and when you find that the answer satisfies your needs, you can click "Accept". That way the post is labeled as having an accepted answer and may be more likely to be found by the next person with a similar question.