04 Delete Relaxation

Introduction to Relaxation

Delete relaxation is a method to relax planning tasks and automatically compute a heuristic function \(h\)

Delete relaxation is a widely used heuristic, highly successful for satisficing planning
Note that every \(h\) function typically yields good performance in only in some domains (balacing search reduction versus computational overhead).
- Therefore we aim for multiple alternative relaxation methods, depending upon domain

Relaxation Types

Four known types of relxation that generate heuristics include

Critical path heuristics
Delete relaxation (covered in this Module)
Abstractions
Landmarks

Relaxing the World by Ignoring Delete Lists

“What was once true remains true forever”

The Delete Relaxation

Definition: The Delete Relaxation

We denote a STRIPS action \(a^+\), corresponding \(a\), as the delete relaxed action, defined by \(pre_{a^+} := pre_a\), \(add_{a^+} := add_a\), and \({\color{red}del_{a^+} =}\) \(\emptyset\).

A set of actions \(A^+\) corresponds to a set of relaxed actions \(A^+ := \{a^+ \mid a \in A\}\); and for a sequence of actions \(\vec{a} = \langle a_1, \dots, a_n\rangle\) denoted by \(\vec{a}^+\), we denote the corresponding sequence of relaxed actions as \(\vec{a}^+ := \langle a_1^+, \dots, a_n^+\rangle\).
For STRIPS planning task \(\Pi=(F,A,c,I,G)\) we denote the delete relaxed planning task as \(\Pi^+ := (F,A^+,c,I,G)\).

The “\(^+\)” super-script denotes delete relaxed states encountered in the relaxed version of the problem, e.g. \(s^+\) is a fact set just like \(s\).
The “\(^+\)” symbol is chosen to capture the additive (monotone) aspect of delete relaxed planning problems.

The Relaxed Plan

Definition: The Relaxed Plan

Let \(\Pi=(F,A,c,I,G)\) be a STRIPS planning task, and let \(s\) be a state.

An optimal relaxed plan for \(s\) is an optimal plan for \(\Pi_s^+\).
A relaxed plan for \(I\) is also called a relaxed plan for \(\Pi\).

Question: Do you recall what \(\Pi_s\) is?:

\(\Pi_s = (F,A,c,s,G)\)

A Relaxed Plan for “TSP” in Australia

Initial state: \(\{\mathit{at}(\mathit{Sy}), \mathit{v}(\mathit{Sy})\}\)

Apply \(\mathit{drive}(\mathit{Sy},\mathit{Br})^+\): \(\{\mathit{at}(\mathit{Br}), \mathit{v}(\mathit{Br}),\) \(\mathit{at}(\mathit{Sy}), \mathit{v}(\mathit{Sy})\}\)
Apply \(\mathit{drive}(\mathit{Sy},\mathit{Ad})^+\): \(\{\mathit{at}(\mathit{Ad}), \mathit{v}(\mathit{Ad}),\) \(\mathit{at}(\mathit{Br}), \mathit{v}(\mathit{Br}),\) \(\mathit{at}(\mathit{Sy}), \mathit{v}(\mathit{Sy})\}\)
Apply \(\mathit{drive}(\mathit{Ad},\mathit{Pe})^+\): \(\{\mathit{at}(\mathit{Pe}), \mathit{v}(\mathit{Pe}),\) \(\mathit{at}(\mathit{Ad}), \mathit{v}(\mathit{Ad}),\) \(\mathit{at}(\mathit{Br}), \mathit{v}(\mathit{Br}),\) \(\mathit{at}(\mathit{Sy}), \mathit{v}(\mathit{Sy})\}\)
Apply \(\mathit{drive}(\mathit{Ad},\mathit{Da})^+\): \(\{\mathit{at}(\mathit{Da}), \mathit{v}(\mathit{Da}),\) \(\mathit{at}(\mathit{Pe}), \mathit{v}(\mathit{Pe}),\) \(\mathit{at}(\mathit{Ad}), \mathit{v}(\mathit{Ad}),\) \(\mathit{at}(\mathit{Br}), \mathit{v}(\mathit{Br}),\) \(\mathit{at}(\mathit{Sy}), \mathit{v}(\mathit{Sy})\}\)

Definition of State Dominance

Definition: Dominance

Let \(\Pi^+=(F,A^+,c,I,G)\) be a STRIPS planning task, and let \(s^+ and s^{\prime +}\) be states. We say that state \(s^{\prime +}\) dominates state \(s^+\) if \(s^{\prime +} \supseteq s^+\), meaning every fact true in \(s^+\) is also true in \(s^{\prime +}\).

For example, on the previous slide, what state dominates what?

Each state along the relaxed plan dominates the previous one, because the actions don’t delete any facts.

State Dominance Properties

State Dominance Properties

Let \(\Pi^+=(F,A^+,c,I,G)\) be a delete relaxed STRIPS planning task, and let \(s^+\) and \(s^{\prime +}\) be states such that \(s^{\prime +}\) dominates \(s^+\).

If \(s^+\) is a goal state, then \(s^{\prime +}\) is also a goal state.
If an action sequence \(\vec a^+\) is applicable in \(s^+\), then it is also applicable in \(s^{\prime +}\), and the state resulting from applying \(\vec a^+\) in \(s^{\prime +}\) dominates the state resulting from applying \(\vec a^+\) in \(s^+\).

Proof: (i) is trivial. (ii) by induction over the length \(n\) of \(\vec{a}^+\). The base case \(n=0\) is trivial. The inductive case \(n \rightarrow n+1\) follows directly from the induction hypothesis and the definition of \(appl(.,.)\). Note that it is always better to have more facts true.

The Delete Relaxation & Admissibility

Proposition: Action Dominance

Let \(\Pi=(F,A,c,I,G)\) be a STRIPS planning task, let \(s\) be a state, and let \(a \in A\). Then \(appl(s,a^+)\) dominates both (i) \(s\), and (ii) \(appl(s,a)\).

Proof. Trivial from the definitions of \(appl(s,a)\) and \(a^+\).

Proposition: Relaxed plan

Let \(\Pi=(F,A,c,I,G)\) be a STRIPS planning task, let \(s\) be a state, and let \(\vec{a}\) be a plan for \(\Pi_s\). Then \(\vec{a}^+\) is a relaxed plan for \(\Pi_s^+\).

Proof: Prove by induction over the length of \(\vec{a}\) that \(appl(s,\vec{a}^+)\) dominates \(appl(s,\vec{a})\). Base case is trivial, inductive case follows from (ii) above.

The Delete Relaxation & Admissibility (continued)

Proposition: Delete Relaxation is Admissible

Let \(\Pi=(F,A,c,I,G)\) be a STRIPS planning task, let \(s\) be a state, and let \(\vec{a}\) be a plan for \(\Pi_s\). Then \(\vec{a}^+\) is a relaxed plan for \(s\).

Therefore optimal relaxed plans admissibly estimate the cost of optimal plans

Applying a relaxed action can only ever make more facts true ((i) above).
That can only be good, i.e., cannot render the task unsolvable (dominance proposition). [

So how do we find a relaxed plan?:

We keep applying relaxed actions, and only stop if the goal is true.

Greedy Relaxed Planning

Greedy Relaxed Planning for \(\Pi^+_s\)

\(s^+ := s\); \(\vec{a}^+ := \langle \rangle\)
while \(G \not \subseteq s^+\) do:
     if \(\exists a \in A\) such that \(pre_a \subseteq s^+\) and \(appl(s^+,a^+) \neq s^+\) then
         select one such \(a\)
         \(s^+ := appl(s^+,a^+)\); \(\vec{a}^+ := \vec{a}^+ \circ \langle a^+ \rangle\)
     else return “\(\Pi_s^+\) is unsolvable” endif
endwhile
return \(\vec{a}^+\)

Proposition Greedy relaxed planning is sound, complete, and terminates in time polynomial in the size of \(\Pi\).

Proof: Soundness: If \(\vec{a}^+\) is returned then, by construction, \(G \subseteq appl(s,\vec{a}^+)\). Completeness: If “\(\Pi_s^+\) is unsolvable” is returned, then no relaxed plan exists for \(s^+\) at that point; since \(s^+\) dominates \(s\), by the dominance proposition this implies that no relaxed plan can exist for \(s\). Termination: Every \(a \in A\) can be selected at most once because afterwards \(appl(s^+,a^+) = s^+\).

It is easy to decide whether a relaxed plan exists

Greedy Relaxed Planning to Generate a Heuristic Function?

Using greedy relaxed planning to generate \(h\)

In search state \(s\) during forward search, run greedy relaxed planning on \(\Pi_s^+\).
Set \(h(s)\) to the cost of \(\vec{a}^+\), or \(\infty\) if “\(\Pi_s^+\) is unsolvable” is returned.

Is this heuristic safe?:

Yes: \(h(s)=\infty\) only if no relaxed plan for \(s\) exists, which by admissibility of delete relaxation implies that no plan for \(s\) exists.

Is this heuristic goal-aware?

Yes, we’ll have \(G \subseteq s^+\) right at the start.

Is this heuristic admissible?

Would be if relaxed plans were optimal; but clearly aren’t. So \(h\) isn’t consistent either.
To be informed (to be an accurate estimate of \(h^*\)), a heuristic must approximate the needed to reach the goal. Greedy relaxed planning therefore doesn’t do this because it may select arbitrary actions that aren’t relevant at all.

\(h^+\) in Travelling Saleman Problem (TSP)

\(h^+\) in Towers of Hanoi

\(h^+\)(Hanoi) \(= O(n)\), as opposed to problem complexity of \(O(2^n)\)