Alternating up and down in my script to find path through 'maze'

Scrip folder is attached! (.zip)

The point of this script is to find if a path exists of 1's between the bottom row and top row. A path consisting of straight lines only, no diagonals.

Concept of the script

The maze:

0 1 1 0 1 1 0 (path is made bold)

0 1 0 0 0 0 0

0 1 1 1 1 0 0

0 0 0 0 1 0 0

1 1 1 0 1 1 0

1 0 1 1 1 0 0

1 1 0 0 0 0 0

Initiating step:

Search the bottom row for values equal to 1, then change these into 2's.

Looping step:

Now scan the row above for values equal to 1, which are attached to values of 2. (Like a virus)

Now scan left and right of these new 2's for other 1's that are attached to these, and turn these into 2's aswell.

0 1 1 0 1 1 0

0 1 0 0 0 0 0

0 1 1 1 1 0 0

0 0 0 0 1 0 0

2 2 2 0 1 1 0

2 0 1 1 1 0 0

2 2 0 0 0 0 0

As can be seen above we now encounter a point where there's only 0s above and so the script should go down.

I am unable to find an efficient way for alternating up and down within my script, so that it keeps searching for new 1's connected to 2's.

I have been struggling with this specific problem for over 5 hours already and am getting really upset.

Any help is greatly aprecciated!

Answer 1

[Meg Noah]

Here's a solution to finding paths that connect the top of the grid to the bottom using 4-neighbors

clc
close all
clear all

maze = [ ...
0	1	1	0	1	1	0; ...
0	1	0	0	0	0	0; ...
0	1	1	1	1	0	0; ...
0	0	0	0	1	0	0; ...
1	1	1	0	1	1	0; ...
1	0	1	1	1	0	0; ...
1	1	0	0	0	0	0];

amazing = bwlabel(maze,4);
[ny,nx] = size(amazing);
stats = regionprops(amazing,'PixelList');
for iblob = 1:length(stats)
    % check if the path has a pixel in row 1 and a pixel in the last row
    if ((min(stats(iblob).PixelList(:,2)) == 1) && ...
       (max(stats(iblob).PixelList(:,2)) == ny))
        disp(['Success: path ' num2str(iblob) ' connects top and bottom.']);
        amazing(amazing == iblob) = 10*iblob;
    else
        % path fails
        amazing(amazing == iblob) = 0;
    end
end
npaths = sum(unique(amazing))-1;
disp(['[Found ' num2str(npaths) ' paths in the maze.']);
amazing

The solution prints out

Success: path 1 connects top and bottom.
[Found 9 paths in the maze.

amazing =

     0    10    10     0     0     0     0
     0    10     0     0     0     0     0
     0    10    10    10    10     0     0
     0     0     0     0    10     0     0
    10    10    10     0    10    10     0
    10     0    10    10    10     0     0
    10    10     0     0     0     0     0

[The Legend]

'amazing' hahaha, nice one.

Thank you alot for the help!