Help checking a grid

I did a hexagon problem like this earlier . . . it helped me to make the matrix 12x6 as well, but when checking fromt op to bottom sometimes you have to say:

check(r+2, c);
check(r+1, c+1);
check(r+1, c-1);

and I can't remember, but I also thing I ended up trying:

check(r-1, c-1); and check(r-1, c+1); as well . . ., making sure it didn't go backwards to the spot it just came from, but I'm not sure about those.

That got somewhat confusing at times, but it worked for me. Good luck man, those hexagon problems are a beast...lol.

When the function calls itself, how do you make sure it's not going back to the same place? What if you had a circle of them? That would infinite loop...

*thinks* I guess I could get an array of grid locations and use it like a stack.

Start at 0,1: Store 0,1 in 'stack'
Find next at 1,1: Store 1,1 in 'stack'
Find next at 1,2: Store 1,2 in 'stack'
Find next at 2,2: Store 2,2 in 'stack'
Find next at 2,1: Store 2,1 in 'stack'
Find next at 1,1: 1,1 already exists in 'stack' do not take.

And then when I fall out of an iteration, I can set the most recent location in the 'stack' back to -1,-1. That may be hard to code correctly...

I suppose if I sat down and worked on it, I might find out... but meh. I'll do it later.

When the function calls itself, how do you make sure it's not going back to the same place? What if you had a circle of them? That would infinite loop... *thinks* I guess I could get an array of grid locations and use it like a stack. Start at 0,1: Store 0,1 in 'stack' Find next at 1,1: Store 1,1 in 'stack' Find next at 1,2: Store 1,2 in 'stack' Find next at 2,2: Store 2,2 in 'stack' Find next at 2,1: Store 2,1 in 'stack' Find next at 1,1: 1,1 already exists in 'stack' do not take. And then when I fall out of an iteration, I can set the most recent location in the 'stack' back to -1,-1. That may be hard to code correctly... I suppose if I sat down and worked on it, I might find out... but meh. I'll do it later.

Yea, I ran across a problem with it bouncing between two spots. It wasn't a very elegant method to solve it, but I passed two more arguments to the function: the old r and the old c values, and made sure it didn't go back to that location. That made it work for me.

I guess you could have a parallell boolean grid and mark true false for spots you've been also . . . I haven't tried that though.

Ya, that would work even better I think.

That way, I'd just have to set the current (i,j) location to false when the iteration fell back a step.