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

3 次查看(过去 30 天)
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!

采纳的回答

Meg Noah
Meg Noah 2020-1-14
Try this algorithm:
Goal:to assign a label to each connected track of 1's on a background of 0's.
Procedure: Scan image left to right and top to bottom.
Mask/Template:
- U
L C
for evaluating the C center pixel compared to the U up pixel and the L left pixel.
Algorithm:
let the initial label be k = 1
scan image left to right then top to bottom evaluating each pixel:
if f( C ) == 0
do nothing
else
if f(U)==1 & f(L) ==0
label( C) = label(U);
elseif f(U)==0 & f(L) ==1
label( C) = label(L)
elseif f(U) == 1 & f(L) ==1
label ( C) = label(L)
label(L) <-> label(U)
elseif f(U) == 0 and f(L) == 0
label( C) = k
k = k + 1 (increment label value)
end
end

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Labyrinth problems 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by