How to avoid the local minima problem in mobile robot control and execute it in matlab

I am working on mobile robot control using potential field and wanted to implement the local minima issue and how to overcome that issue. I have written code for a normal mobile robot control. I have two obstacles very close to each other and when executing the code, the robot gets stuck there and not able to each the final point. Can someone help me with it. I can send you the code over the email address. Please let me know. Thank you in advance.

[Walter Roberson]

Yes, this kind of problem can happen any time you use a "greedy" algorithm that does not have the ability to back up. The two destinations do not even need to be close to each other as long as the path towards the first of them is cheaper until you get past a choke point.

For example,

     1113333333111
     1A13s222221B1
     1113333333111

if you are starting from s with a destination of A, then a greedy algorithm will always immediately keep heading towards B, and when it gets there, it will be unable to get back, because circling around B is lower cost per step than heading back towards A.

A is not "far" from s, but greedy algorithms never look back.

[Abhibrata Adhikary]

I didn't get it sir.

[Walter Roberson]

Suppose you start at the location marked s and you need to get to A, and the 1's, 2's, and 3's are the cost of movement with 3 being more expensive. Then if you use an algorithm that always chooses lowest cost, then you would immediately move to the right from the s because the 2 there is cheaper than the 3's that are otherwise around s. And once you have moved to the first 2, then a greedy algorithm would keep moving further right along the 2's, because at each point, doing so would be less expensive than "climbing the hill" of 3's. Eventually you get to the ring of 1's around B. Following the 1's around B is less expensive on any one step than "climbing the hill" of 2 or 3's to head back towards A, so a greedy algorithm would never be able to get back to A.

You might perhaps have an algorithm that is not permitted to re-visit any location it has already visited. If so, then let * indicate a wall that cannot be passed:

     11133333*3111
     1A13s222221B1
     11133333*3111

the greedy algorithm would get to B, follow the ring around once, and get to one of the two *3's. But once there, they would not be able to go back to the 2 that is

*
2
*

because it had already been there and there are walls that do not permit passing through them.

You need to use a different algorithm for deciding where to go.