Need guidance on how to solve line follower problem.

Projects General Guidance
1

So my line follower path is basically a path with ONLY right angle turns, lots of 'fake' turns with dead ends and lots of 'T' joints and junction joints. So you can probably think of it as a maze. Except that my goal isn't to reach the end of it (there is no end), but it is to traverse every single section of the maze (or as many as I can), and then STOP my bot.

I coded the maze 'solver' succesfuly, using the priority order of RIGHT>FORWARD>LEFT.

I have 7 sensors, arranged as "o o o o o o o"
------------------------------6 5 4 3 2 1 0
I've stored their's values (digital) into an array a[7].

I have functions for Fwd, for Right, for Left, and I call them with the following logic:

if(a[0]==0)
    {right();}
    
    else if(a[0]==1 && a[1]==1 && a[2]==1 && a[3]==1 && a[4]==1 && a[5]==1 && a[6]==1)
    {left();}
    
    else
    {fwd();}

So my bot moves right whenever it encounters a right turn, OR a T joint. If it ever encounters a sole left turn, it keeps going forward till it leaves the path, and THEN turns left. This solves the U Turn at a dead end problem simoultaneously.

There are obviously other ways to code this but this seemed the shortest and it works well.

Now back to the problem. I need idea/advice/guidance on how to recognize when it has completely traversed the maze (every part of it).
I'm using Atmega168 with 512 bytes of EEPROM.

2

cyanide911:
I need idea/advice/guidance on how to recognize when it has completely traversed the maze (every part of it).

You need a way to build a model of the maze and work out which parts you have visited. This would be simplified if you are able to assume that your starting position is at an end, but it will still need somethinking to work out how to model the maze.

cyanide911:
I'm using Atmega168 with 512 bytes of EEPROM.

That's going to severely limit the size of maze you can model. Suggest you measure how much of that memory is actually free once you have got your basic sketch running - I suspect it will be close to zero.