Planning and Optimization B3. General Regression Gabriele R¨oger and Thomas Keller

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Gabriele R¨oger and Thomas Keller

Universit¨at Basel

October 10, 2018

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Planning

Classical

Tasks Progression/

Regression Complexity Heuristics

Probabilistic

MDPs Uninformed Search

Heuristic Search Monte-Carlo

Methods

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With disjunctions and conditional effects, things become more tricky. How to regressa∨(b∧c) with respect tohq,d Bbi?

In this chapter, we show how to regressgeneral sets of states through general operators.

We extensively use the idea of representing sets of states as formulas.

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Regressing State Variables

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Key question for general regression:

Assume we are applying an operator with effecte. What must be true in the predecessor state

for state variable v to be true in the successor state?

If we can answer this question, a general definition of regression is only a small additional step.

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Assume we are in states and apply effecte to obtain successor states⁰.

State variablev is true ins⁰ if effect e makes it true, or

it remains true, i.e., it is true ins and not made false bye.

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Definition (Regressing a State Variable Through an Effect) Lete be an effect of a propositional planning task,

and letv be a state variable.

Theregression ofv throughe, writtenregr(v,e), is defined as the following logical formula:

regr(v,e) =effcond(v,e)∨(v∧ ¬effcond(¬v,e)).

Question: Does this capture add-after-delete semantics correctly?

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Example

Lete = (bBa)∧(c B¬a)∧b∧ ¬d. v regr(v,e) a b∨(a∧ ¬c) b > ∨(b∧ ¬⊥)≡ >

c ⊥ ∨(c∧ ¬⊥)≡c d ⊥ ∨(d ∧ ¬>)≡ ⊥

Reminder: regr(v,e) =effcond(v,e)∨(v∧ ¬effcond(¬v,e))

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Lemma (Correctness ofregr(v,e))

In a propositional planning task, let s be a state, e be an effect and v be a state variable.

Then s|=regr(v,e) iff sJeK|=v .

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Proof.

(⇒): We know s |=regr(v,e), and hence s |=effcond(v,e)∨(v∧ ¬effcond(¬v,e)).

Do a case analysis on the two disjuncts.

Case 1: s |=effcond(v,e).

ThensJeK|=v by the first case in the definition ofsJeK. Case 2: s |= (v∧ ¬effcond(¬v,e)).

Thens |=v ands 6|=effcond(¬v,e).

We may additionally assumes 6|=effcond(v,e) because otherwise we can apply Case 1 of this proof.

ThensJeK|=v by the third case in the definition ofsJeK. . . .

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Proof.

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Proof.

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Proof (continued).

(⇐): Proof by contraposition.

We show that ifregr(v,e) is falsein s, thenv is false insJeK. By prerequisite, s 6|=effcond(v,e)∨(v∧ ¬effcond(¬v,e)).

Hence s |=¬effcond(v,e)∧(¬v∨effcond(¬v,e)).

From the first conjunct, we get s |=¬effcond(v,e) and hence s 6|=effcond(v,e).

From the second conjunct, we gets |=¬v∨effcond(¬v,e).

Case 1: s |=¬v. Then v is false before applying e and remains false, so sJeK6|=v.

Case 2: s |=effcond(¬v,e). Then v is deleted by e and not simultaneously added, so sJeK6|=v.

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Proof (continued).

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Proof (continued).

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Proof (continued).

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Proof (continued).

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Proof (continued).

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Proof (continued).

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Regressing Formulas Through Effects

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We can now generalize regression from state variables to general formulas over state variables.

The basic idea is to replace every occurrenceof every state variable v byregr(v,e) as defined in the previous section.

The following definition makes this more formal.

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Definition (Regressing a Formula Through an Effect) In a propositional planning task, lete be an effect, and letϕbe a formula over state variables.

Theregression ofϕthrough e, written regr(ϕ,e), is defined as the following logical formula:

regr(>,e) =>

regr(⊥,e) =⊥

regr(v,e) =effcond(v,e)∨(v∧ ¬effcond(¬v,e)) regr(¬ψ,e) =¬regr(ψ,e)

regr(ψ∨χ,e) =regr(ψ,e)∨regr(χ,e) regr(ψ∧χ,e) =regr(ψ,e)∧regr(χ,e).

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Example

Lete = (bBa)∧(c B¬a)∧b∧ ¬d. Recall:

regr(a,e)≡b∨(a∧ ¬c) regr(b,e)≡ >

regr(c,e)≡c regr(d,e)≡ ⊥ We get:

regr((a∨d)∧(c∨d),e)≡((b∨(a∧ ¬c))∨ ⊥)∧(c∨ ⊥)

≡(b∨(a∧ ¬c))∧c

≡b∧c

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Lemma (Correctness ofregr(ϕ,e))

Letϕbe a logical formula, e an effect and s a state.

Then s|=regr(ϕ,e) iff sJeK|=ϕ.

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Proof.

The proof is by structural induction onϕ.

Induction hypothesis: s |=regr(ψ,e) iffsJeK|=ψ for all proper subformulasψ ofϕ.

Base caseϕ=>:

We haveregr(>,e) =>, ands |=>iff sJeK|=> is correct.

Base caseϕ=⊥:

We haveregr(⊥,e) =⊥, ands |=⊥iff sJeK|=⊥ is correct.

Base caseϕ=v:

We haves |=regr(v,e) iff sJeK|=v from the previous lemma. . . .

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Proof.

Base caseϕ=>:

Base caseϕ=⊥:

Base caseϕ=v:

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Proof.

Base caseϕ=>:

Base caseϕ=⊥:

Base caseϕ=v:

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Proof.

Base caseϕ=>:

Base caseϕ=⊥:

Base caseϕ=v:

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Proof.

Base caseϕ=>:

Base caseϕ=⊥:

Base caseϕ=v:

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Proof (continued).

Inductive caseϕ=¬ψ:

iffsJeK|=ψ∨χ Inductive caseϕ=ψ∧χ:

Like previous case, replacing “∨” by “∧”

and replacing “or” by “and”.

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Proof (continued).

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Proof (continued).

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Regressing Formulas Through

Operators

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We can now regress arbitrary formulas through arbitrary effects.

The last missing piece is a definition of regression through operators, describing exactly in which statess applying a given operator o leads to a state satisfying a given formula ϕ.

There are two requirements:

The operatoro must beapplicablein the states.

Theresulting state sJoKmustsatisfyϕ.

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Definition (Regressing a Formula Through an Operator) In a propositional planning task, leto be an operator, and letϕbe a formula over state variables.

Theregression ofϕthrough o, written regr(ϕ,o), is defined as the following logical formula:

regr(ϕ,o) =pre(o)∧regr(ϕ,eff(o)).

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Theorem (Correctness ofregr(ϕ,o))

Letϕbe a logical formula, o an operator and s a state.

Then s|=regr(ϕ,o) iff o is applicable in s and sJoK|=ϕ.

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Reminder: regr(ϕ,o) =pre(o)∧regr(ϕ,eff(o))

Proof.

Case 1: s |=pre(o).

Theno is applicable ins and the statement we must prove simplifies to: s |=regr(ϕ,eff(o)) iffsJeK|=ϕ.

This was proved in the previous lemma.

Case 2: s 6|=pre(o).

Thens 6|=regr(ϕ,o) ando is not applicable in s.

Hence both statements are false and therefore equivalent.

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Reminder: regr(ϕ,o) =pre(o)∧regr(ϕ,eff(o))

Proof.

Case 1: s |=pre(o).

Theno is applicable ins and the statement we must prove simplifies to: s |=regr(ϕ,eff(o)) iffsJeK|=ϕ.

This was proved in the previous lemma.

Case 2: s 6|=pre(o).

Thens 6|=regr(ϕ,o) ando is not applicable in s.

Hence both statements are false and therefore equivalent.

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Examples: compute regression and simplify to DNF regr(b,ha,bi)

≡a∧(> ∨(b∧ ¬⊥))

≡a

regr(b∧c ∧d,ha,bi)

≡a∧(> ∨(b∧ ¬⊥))∧(⊥ ∨(c∧ ¬⊥))∧(⊥ ∨(d∧ ¬⊥))

≡a∧c∧d

regr(b∧ ¬c,ha,b∧ci)

≡a∧(> ∨(b∧ ¬⊥))∧ ¬(> ∨(c∧ ¬⊥))

≡a∧ > ∧ ⊥

≡ ⊥

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Examples: compute regression and simplify to DNF regr(b,ha,c Bbi)

≡a∧(c∨(b∧ ¬⊥))

≡a∧(c∨b)

≡(a∧c)∨(a∧b)

regr(b,ha,(c Bb)∧((d ∧ ¬c)B¬b)i)

≡a∧(c∨(b∧ ¬(d∧ ¬c)))

≡a∧(c∨(b∧(¬d∨c)))

≡a∧(c∨(b∧ ¬d)∨(b∧c))

≡a∧(c∨(b∧ ¬d))

≡(a∧c)∨(a∧b∧ ¬d)

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Summary

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Regressing a state variablethrough an (arbitrary) operator must consider two cases:

state variablesmade true (by add effects)

state variablesremaining true(by absence of delete effects) Regression of state variables can be generalized

to arbitrary formulas ϕby replacing each occurrence of a state variable in ϕby its regression.

Regressing a formula ϕthrough an operatorinvolves

regressingϕthrough the effect and enforcing the precondition.