Planning

1
Planning

• Points
• Elements of a planning problem
• Planning as resolution
• Conditional plans
• Actions as preconditions and effects
• Goal regression

2
Elements of a planning problem

• When is a problem a good application for
  planning?
• It can be decomposed into subproblems in such a
  way that solving each subproblem (in some order)
  solves the whole problem.
• It is conjunctive, that is, it can be expressed
  as a conjunction of conditions, each representing
  an elementary aspect if the world, the initial
  situation, the final situation.
• Each subproblem can be reduced to facts about
  domain objects, and relations among such objects.

3
Elements of a planning problem (2)

• States, state spaces.
• Actions with preconditions and postconditions.
• An action is a function from states to states. It
  makes local changes that affect few objects.
• Actions are generic, facts are specific.
• Actions are elementary one-step plans.
• A design decision granularity of actions.
• Conditional actions If( C, A1, A2 ).
• Frame axioms and the Closed World Assumption.
• The timing of actions is a real-time plan
  needed, or do we only need a sequence of actions
  with relative time?

4
Elements of a planning problem (3)

• A plan is a sequence of actions that lead from an
  initial state to a goal state.
• In a linear plan, the actions required to solve a
  subproblem precede the actions required to solve
  the next subproblem.
• In a non-linear plan, actions for subproblems may
  be (or even may have to be) interleaved.
• Optimality of plans a simple criterion is the
  number of actions.
• A plan may be executed either in batch mode
  (after the whole plan has been created), or with
  every action performed immediately (with feedback
  for more planning).

5
Elements of a planning problem (4)

• A formal representation of plans may look, for
  example, like this
• Do( Action3, Do( Action2, Do( Action1, init ) ) )
• where init denotes an initial state.
• There are two ways of implementing state changes
  (we must be able to undo such changes, because
  planning algorithms usually backtrack).

• Explicit changes in the knowledge base.
• Implicit changes, not in the knowledge base but
  only in the plan representation, recomputed as
  needed (this is costly, but undoing is easier).

6
Planning as resolution

• The blocks world (find its description in any
  textbook ?)
• T( p, s ) means "predicate p is true in state s"
• Elementary conditions On, OnTbl and Clear.
• Examples of actions Stack, Unstack and NoOp.
• T( OnTbl( x ), s ) ? T( Clear( x ), s ) ? T(
  Clear( y ), s ) ? x ? y ? T( On( x, y ), Do(
  Stack( x, y ), s ) )
• T( On( x, y ), s ) ? T( Clear( x ), s ) ? T(
  OnTbl( x ), Do( Unstack( x, y ), s ) ) ? T(
  Clear( y ), Do( Unstack( x, y ), s ) )
• T( p, s ) ? T( p, Do( NoOp, s ) )

7
Planning as resolution (2)

• On means sitting on another block.
• OnTbl means sitting on the table.
• Some facts about this worlds are always true.For
  example
• T( OnTbl( x ), s ) ? ? ? y T( On( x, y ), s)
• Frame axioms two examples
• T( Clear( u ), s ) ? u ? y ? T( Clear( u ), Do(
  Stack( x, y ), s ) )
• T( On( u, w ), s ) ? u ? x ? T( On( u, w ), Do(
  Unstack( x, y ), s ) )

8
Planning as resolution (3)

• An initial state
• T( Clear( C ), S0 ) T( On( C, A ), S0 )
• T( Clear( B ), S0 )
• T( OnTbl( A ), S0 ) T( OnTbl( B ), S0 )
• Another (possibly final) state
• T( Clear( A ), S1 ) T( On( A, B ), S1 )
• T( On( B, C ), S1 ) T( OnTbl( C ), S1 )

9
Planning as resolution (4)
Plan construction by resolution(a very simple
example)

• We want to prove that T( OnTbl( A ), t ) given
  S1.
• We assume that we can do it in one action ?, so
  that state t should look like Do( ?, S1 ).
• If this is not possible, we will try Do( ?, Do(
  ?, S1 ) ),Do( ?, Do( ?, Do( ?, S1 ) ) ), and so
  on.
• Lets negate the fact that we want proven.
• ? T( OnTbl( A ), Do( ?, S1 ) )
• Action ? could be Unstack. That is because one of
  the clauses in its clausal-form definition is
  this
• ? T( On( x, y ), s ), ? T( Clear( x ), s ),T(
  OnTbl( x ), Do( Unstack( x, y ), s ) )

10
Planning as resolution (5)
Plan construction by resolution(continued)

• ? T( OnTbl( A ), Do( ?, S1 ) )
• T( OnTbl( x ), Do( Unstack( x, y ), s ) ),?
  T( On( x, y ), s ), ? T( Clear( x ), s )
• This gives
• ? T( On( A, y ), S1 ), ? T( Clear( A ), S1 )
• after assigning x ? A, s ? S1, ? ? Unstack( A,
  y ).
• This is given T( On( A, B ), S1 ).
• We resolve and get ? T( Clear( A ), S1 ) with y ?
  B.
• This too is given T( Clear( A ), S1 ).
• so we produce the empty resolvent done.
• t ? Do( ?, S1 ) Do( Unstack( A, B ), S1 )

11
Conditional plans

• Let T( Clear( A ), S2 ) be the only given fact.
• We can find a plan for the goal
• T( OnTbl( A ), Do( ?, S2 ) )
• even though the problem is under-constrained.
• Axioms for conditional actions
• T( p, s ) ? T( q, Do( ?, s ) ) ? T( q, Do(
  If( p, ?, ? ), s ) )
• T( p, s ) ? T( q, Do( ?, s ) ) ? T( q, Do( If(
  p, ?, ? ), s ) )
• Again, we begin by assuming a one-step plan, but
  now it will be a conditional plan.

12
Conditional plans (2)

• T( OnTbl( A ), Do( ?, S2 ) )
• Assume ? ? If( p, ?, ? ).
• T( OnTbl( A ), Do( If( p, ?, ? ), S2 ) )
• The first axiom as a clause
• T( q, Do( If( p, ?, ? ), s ) ), T( p, s ),
  T( q, Do( ?, s ) )
• Resolve with q ? OnTbl( A ), s ? S2 and get
• T( p, S2 ), T( OnTbl( A ), Do( ?, S2 ) )
• We need again that axiom for Unstack
• T( OnTbl( x ), Do( Unstack( x, y ), s ) ),? T(
  On( x, y ), s ), ? T( Clear( x ), s )

13
Conditional plans (3)

• T( p, S2 ), T( OnTbl( A ), Do( ?, S2 ) )
• T( OnTbl( x ), Do( Unstack( x, y ), s ) ),? T(
  On( x, y ), s ), ? T( Clear( x ), s )
• With x ? A, ? ? Unstack( A, y ), s ? S2 we get
• T( p, S2 ), ? T( On( A, y ), S2 ), ? T(
  Clear( A ), S2 )
• We resolve this with our only fact
• T( p, S2 ), ? T( On( A, y ), S2 )
• ... and this is it. We need something new to move
  on.
• One possibility is factoring look for matching
  literals on the resolvent. Here we can assign p ?
  On( A, y ) and reduce the resolvent to ? T( On(
  A, y ), S2 ).

14
Conditional plans (4)

• We can conclude the proof if we assume
• T( On( A, y ), S2 )
• Here, the conditional nature of the plan
  helps.We combine everything we have found so
  far.Our one-action plan looks like this
• Do( If( On( A, y ), Unstack( A, y ), ? ), S2 )
• To also determine ?, we return to the point
  marked and look at the second axiom for
  conditional actions
• T( q, Do( If( p, ?, ? ), s ) ), T( p, s ), T(
  q, Do( ?, s ) )
• T( OnTbl( A ), Do( ?, S2 ) )
• Resolve with q ? OnTbl( A ), s ? S2 and get
• T( p, S2 ), T( OnTbl( A ), Do( ?, S2 ) )

15
Conditional plans (5)

• T( p, S2 ), T( OnTbl( A ), Do( ?, S2 ) )
• The clausal form of action NoOp
• T( p, s ), T( p, Do( NoOp, s ) )
• Assign p ? OnTbl( A ), ? ? NoOp and get this
• T( p, S2 ), T( OnTbl( A ), S2 )
• It is time to recall that useful fact about
  OnTbl
• T( OnTbl( x ), s ) ? ? ? y T( On( x, y ), s)
• So, we can replace one side with the other
• T( p, S2 ), ? y T( On( A, y ), S2)
• Skolemize (? is the Skolem constant)
• T( p, S2 ), T( On( A, ? ), S2)

16
Conditional plans (6)

• T( p, S2 ), T( On( A, ? ), S2)
• We perform factorization, assigning p ? On( A, ?
  )
• T( On( A, ? ), S2)
• To conclude the second part of the proof, we
  assume
• ? T( On( A, ? ), S2 )
• Now, our one-action plan is either of these
• Do( If( On( A, y ), Unstack( A, y ), ? ), S2 )
• Do( If( On( A, ? ), ?, NoOp ), S2 )
• We can finally put these partial plans together
• Do( If( On( A, ? ), Unstack( A, ? ), NoOp ), S2 )

17
Another representation of actions

• Preconditions of an action
• facts added (made true) by this action
• facts deleted (made false) by this action.
• Pre( Unstack( x, y ) ) On( x, y ), Clear( x
  )
• Add( Unstack( x, y ) ) OnTbl( x ), Clear( y
  )
• Del( Unstack( x, y ) ) On( x, y )
• The Closed World Assumption serves as the frame
  axiom
• facts not on the Add list and on the Del listare
  by default preserved.
• A state is represented as a set of facts, for
  example
• OnTbl( A ), OnTbl( B )

18
Goal regression

• Reduce one of the goals to its subgoal or
  subgoals.
• This means working backward from a state to a
  state by looking at the preconditions and effects
  of some action.
• Let Regr( q, ? ) be a state achieved by
  regression from q through ? (undoing the
  effects of ?).
• ? cannot delete a fact that must be true in q
• q ? Del( ? ) Ø
• If ? has this property, we can define
• Regr( q, ? ) Pre( ? ) ? (q Add( ? ) )
• So keep the preconditions of ?, undo the effects
  of ?.

19
Goal regression (2)

• Example
• If we begin with
• OnTbl( A ), OnTbl( B )
• we can recreate
• Clear( A ), On( A, B ), OnTbl( B )
• by regression through Unstack( A, B ).
• Protection violation
• Actions for a subproblem must not undo the
  effects of earlier actions for already solved
  subproblems.

20
A planning strategy for conjunctive problems

• Going left to right, make a plan for each
  conjunct and freeze the effects (put the added or
  preserved goals on a list of protected goals).
• This simple method is weak. It will not work even
  for the following trivial problem.
• Sinitial On( C, A ) ? OnTbl( A ) ? OnTbl( B )
• Sfinal On( C, A ) ? On( A, B )
• On( C, A ) is true in Sinitial and must be
  protected. But it must be temporarily destroyed
  to achieve On( A, B ). Goal protection makes
  planning impossible.
• Here, rearrangement helps On( A, B ) ? On( C, A
  ) works. In general non-linear planning is
  necessary.
• One such a smarter planner is Warplan.