How do I write a function that checks whether there is a horizontal, vertical or diagonal connect N?

Input:

M = [0,0,0,2,0,0,0;0,0,0,2,0,1,0;0,2,0,1,1,2,0;0,1,2,1,1,1,0;1,2,1,2,2,2,0;2,2,1,2,2,1,2]

Visualization of input:

---------------        x = 1
| | | |o| | | |        o = 2
| | | |o| |x| |        
| | | |x|x|o| |
| |x|o|x|x|x| |
|x|o|x|o|o|o| |
|o|o|x|o|o|x|o|
---------------

A connect 4 is present:

0 0 0 2 0 0 0

0 0 0 2 0 1 0

0 0 0 1 1 2 0

0 1 2 1 1 1 0

1 2 1 2 2 2 0

2 2 1 2 2 1 2

My question:

How do I write a function that checks whether there is a horizontal, vertical or diagonal connect N (connect 4/5/6 etc. (program is scalable)) in this matrix?

[Stephen23]

Duplicate:

https://www.mathworks.com/matlabcentral/answers/496768-check-if-connect-4-occurs-in-matrix

[The Legend]

Not really, the answers I received almost worked, but not entirely.

I really do appreciate the help, but it didn't work.

[Stephen23]

" the answers I received almost worked, but not entirely."

Based on your comment I updated my answer to work with all cases. So far the corrected code works with everything that you have shown, but you apparently have not tried it.

Is there are specific reason why you continue to ignore my edited answer?

"I really do appreciate the help but it didn't work."

In my earlier comment I asked you to provide any case that did not work with my edited answer. So far you have not provided any examples, or even replied to my comment.

[The Legend]

Your code does work, but it doesnt work for diagonals that go like this // for some reason

[Stephen23]

"Your code does work, but it doesnt work for diagonals that go like this // for some reason"

Please show an example, together with the exact code that you are using.

It works:

>> F = @(x)conv2(x,eye(4),'same')>=4 | conv2(x,flipud(eye(4)),'same')>=4 | conv2(x,ones(4,1),'same')>=4 | conv2(x,ones(1,4),'sa
me')>=4;
>> M = [0,1,0,0,0;0,0,1,0,0;0,0,0,1,0;0,0,0,0,1;0,0,0,0,0]
M =
   0   1   0   0   0
   0   0   1   0   0
   0   0   0   1   0
   0   0   0   0   1
   0   0   0   0   0
>> X1 = F(M==1);
>> any(X1(:))
ans = 1
>> M = [0,0,0,0,0;0,0,0,1,0;0,0,1,0,0;0,1,0,0,0;1,0,0,0,0]
M =
   0   0   0   0   0
   0   0   0   1   0
   0   0   1   0   0
   0   1   0   0   0
   1   0   0   0   0
>> X1 = F(M==1);
>> any(X1(:))
ans = 1

Notice how I provide complete working examples, complete with the matrices and the function, which makes it easy for anyone reading this forum to check if they work or not.

Do you see a problem with my examples?

(notice also that the code you are using is NOT that from my updated answer).

[Stephen23]

I have no idea why you are asking me about someone else's code: if you have a problem with their answer, you should be asking them in a comment to their answer.

In any case, my edited answer detects both of those diagonals perfectly:

>> F = @(x)conv2(x,eye(4),'same')>=4 | conv2(x,flipud(eye(4)),'same')>=4 | conv2(x,ones(4,1),'same')>=4 | conv2(x,ones(1,4),'same')>=4;
>> M = [0,0,0,2,0,0,0;0,0,0,2,0,1,0;0,2,0,1,1,2,0;0,1,2,1,1,1,0;1,2,1,2,2,2,0;2,2,1,2,2,1,2]
M =
   0   0   0   2   0   0   0
   0   0   0   2   0   1   0
   0   2   0   1   1   2   0
   0   1   2   1   1   1   0
   1   2   1   2   2   2   0
   2   2   1   2   2   1   2
>> X1 = F(M==1);
>> X2 = F(M==2);
>> any(X1(:)) % from (2,6) to (5,2)
ans = 1
>> any(X2(:)) % from (3,2) to (6,5)
ans = 1

Once again, I will repeat my request: please provide one example which does not work using my edited answer. I look forward to your reply.

[The Legend]

Thank you it is working now! accepted your answer in the other post, will accept ur answer too if you post it here. thanks!

[Stephen23]

"This function shouldnt return a 2 value, but a 0 value, idk why that's happning... there's no connect N."

Actually there is. Let me highlight it for you:

   0   0   0   2   0   0   0
   0   0   0   2   0   0   0
   0  *2*  0   1   1   2   0
   0   1  *2*  1   1   1   0
   1   2   1  *2*  2   2   0
   2   2   1   2  *2*  1   2   

Why are you still using the outdated answer? Even though I have written around ten times that I updated/fixed my answer, you are still using the version that does not detect anti-diagonals.

[The Legend]

I had commented right before you posted this haha, my bad, sorry. Really appreciate the help!