How to scan matrix in spiral way?

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Arshub 2022-5-16

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1720060-how-to-scan-matrix-in-spiral-way

评论： DGM 2022-5-17

采纳的回答： DGM

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.

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Arshub 2022-5-16

编辑：Arshub 2022-5-16

@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 2022-5-16

@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.

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Answer 1

DGM 2022-5-16

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https://ww2.mathworks.cn/matlabcentral/answers/1720060-how-to-scan-matrix-in-spiral-way#answer_964625

编辑：DGM 2022-5-16

在 MATLAB Online 中打开

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
  24     1     8    15
   5     7    14    16
   6    13    20    22
  12    19    21     3
  18    25     2     9
mg(sp)
ans = 5×5
  16    22     3     9
   5     6    12    18
  24    17    23     1
  11    10     4     7
   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.

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Arshub 2022-5-16

在 MATLAB Online 中打开

@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 2022-5-16

在 MATLAB Online 中打开

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

DGM 2022-5-16

在 MATLAB Online 中打开

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 2022-5-17

@DGM Thanks alot for your help.

DGM 2022-5-17

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.

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How to scan matrix in spiral way?

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