Saturday, May 9, 2009

Re: Regarding graph planning in Homework 5

First of all, since I discussed mutual exclusion propagation only briefly in the class, you are not responsible for that part.

Since you asked however, just because two actions are mutex doesn't mean that  their effects are mutex--since after all the effects may also have been given by other actions.

Consider for example a situation where p is given by m different actions, and q is given by n different actions.

In order for p and q to be mutex, it must be the case that every pair of actions in the cartesian product must be
mutex--ie there must be m*n mutexes.  If there exists even one pair--say ai giving p and bk giving q such that
ai and bk are not mutex, then p and q are not mutex.

[Suppose you started with the belief that you shouldn't hit people because god might punish you. If you then went on to become an atheist, it doesn't necessarily mean that you should now believe that hitting people is fine. You may have found other reasons why hitting people is not reasonable.]

[In contrast, if *any* pair of preconditions of two actions are mutex, then the actions themselves are mutex.
I.e. if action a has m precods and action b has n preconds, if any of the m*n precondition pairs are mutex, then the action a and b are mutex. ]


On Sat, May 9, 2009 at 3:05 PM, Sidharth Gupta <> wrote:
The mutexes shown between variables at level 2  in the problem in planning graph for homework 5 solutions dont quite seem right.... Shouldnt there be a mutexes between all the pairs of variables whose actions were also having a mutex...? Many seem to be missing like between R and S whose actions o1 and o2 are also in mutex?

Sidharth Gupta

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