Lecture 14 Situation Calculus 13 th February 2020 Outline 2 / 24 - - PowerPoint PPT Presentation

Informatics 2D ⋅ Agents and Reasoning ⋅ 2019/2020

Lecture 14 ⋅ Situation Calculus

Claudia Chirita

School of Informatics, University of Edinburgh

13th February 2020

Based on slides by: Jacques Fleuriot, Michael Rovatsos, Michael Herrmann, Vaishak Belle

Outline

• Planning
• Situations
• Frame problem

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Using Logic to Plan

We need ways of representing

• the world
• the goal
• how actions change the world

We haven’t said much about changing the world. Diffjculty After an action, new things are true, and some previously true facts are no longer true.

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Situations

Situations extend the concept of a state by additional logical terms – consist of initial situation (usually called 𝑇) and all situations generated by applying an action to a situation Providing facts about situations – by relating predicates to situations e.g. instead of saying just 𝑃𝑜(𝐵, 𝐶), say (somehow)

𝑃𝑜(𝐵, 𝐶) in situation 𝑇

Actions are thus – performed in a situation, and – produce new situations with new facts – e.g. Forward and Turn(Right)

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Representing Predicates Relative to a Situation

• 1. Can add an argument for a situation to each predicate

that can change. – e.g. instead of On(𝐵, 𝐶), write On(𝐵, 𝐶, 𝑇)

• 2. Alternatively, introduce a predicate Holds

– On etc. become functions – e.g. Holds(On(𝐵, 𝐶), 𝑇) – What do things like On(𝐵, 𝐶) mean now? A set of situations in which 𝐵 is on 𝐶.

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How This Will Work

Before some action, we might have in our KB

• On(𝐵, 𝐶, 𝑇)
• On(𝐶, Table, 𝑇)

A B After an action that moves A to the table, say, we add

• Clear(𝐶, 𝑇)
• On(𝐵, Table, 𝑇)

A B All these propositions are true. We have dealt with the issue

• f change, by keeping track of what is true when.

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Same Thing, Slightly Difgerent Notation

Before

• Holds(On(𝐵, 𝐶), 𝑇)
• Holds(On(𝐶, Table), 𝑇)

A B After, add

• Holds(Clear(𝐶), 𝑇)
• Holds(On(𝐵, Table), 𝑇)

A B

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Representing Actions

We need to represent – results of doing an action – conditions that need to be in place to perform an action For convenience, we will defjne functions to abbreviate actions – e.g. Move(𝐵, 𝐶) denotes the action type of moving 𝐵 onto 𝐶 – these are action types, because actions themselves are specifjc to time, etc. Now, introduce a function Result, designating “the situation resulting from doing an action type in some situation”. – e.g. Result(Move(𝐵, 𝐶), 𝑇) means “the situation resulting from doing an action of type Move(𝐵, 𝐶) in situation 𝑇”.

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How This Works

Keep in mind that things like

Result(Move(𝐵, 𝐶), 𝑇)

are terms and denote situations. They can appear anywhere we would expect a situation. So we can say things like

𝑇 = Result(Move(𝐵, 𝐶), 𝑇) On(𝐵, 𝐶, Result(Move(𝐵, 𝐶), 𝑇)) ≡ On(𝐵, 𝐶, 𝑇)

Alternatively,

Holds(On(𝐵, 𝐶), Result(Move(𝐵, 𝐶), 𝑇))

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Axiomatising Actions

We can describe the results of actions, together with their preconditions. e.g. “If nothing is on 𝑦 and 𝑧, then one can move 𝑦 to on top

• f 𝑧, in which case 𝑦 will then be on 𝑧.”

∀𝑦, 𝑧, 𝑡.Clear(𝑦, 𝑡) ∧ Clear(𝑧, 𝑡) → On(𝑦, 𝑧, Result(Move(𝑦, 𝑧), 𝑡))

Alternatively,

∀𝑦, 𝑧, 𝑡.Holds(Clear(𝑦), 𝑡) ∧ Holds(Clear(𝑧), 𝑡) → Holds(On(𝑦, 𝑧), Result(Move(𝑦, 𝑧), 𝑡))

This is an efgect axiom. It includes a precondition as well.

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Situation Calculus

This approach is called the situation calculus. We axiomatise all our actions, then use a general theorem prover to prove that a situation exists in which our goal is true. The actions in the proof would comprise our plan.

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Example

KB

On(𝐵, Table, 𝑇) On(𝐶, 𝐷, 𝑇) On(𝐷, Table, 𝑇) Clear(𝐵, 𝑇) Clear(𝐶, 𝑇)

+ axioms about actions Goal

∃𝑡.On(𝐵, 𝐶, 𝑡) 𝑇

B A C Table

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What happens?

• We want to prove On(𝐵, 𝐶, 𝑡) for some 𝑡.

– Find axiom

∀𝑦, 𝑧, 𝑡.Clear(𝑦, 𝑡) ∧ Clear(𝑧, 𝑡) → On(𝑦, 𝑧, Result(Move(𝑦, 𝑧), 𝑡))

– Goal would be true if we could prove

Clear(𝐵, 𝑡) ∧ Clear(𝐶, 𝑡) by backward chaining.

– But both are true in 𝑇, so we can conclude

On(𝐵, 𝐶, Result(Move(𝐵, 𝐶), 𝑇))

• We are done!

We look at the proof and see only one action, Move(𝐵, 𝐶), which is executed in situation 𝑇, so this is our plan.

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Example ⋅ Same Situation, Harder Goal1

KB

On(𝐵, Table, 𝑇) On(𝐶, 𝐷, 𝑇) On(𝐷, Table, 𝑇) Clear(𝐵, 𝑇) Clear(𝐶, 𝑇)

+ axioms about actions Goal

∃𝑡.On(𝐵, 𝐶, 𝑡) ∧ On(𝐶, 𝐷, 𝑡) 𝑇

B A C Table

It’s not really harder, is already on , and we just showed how to put on . 14 / 24

With Goal On(𝐵, 𝐶, 𝑡′) ∧ On(𝐶, 𝐷, 𝑡′)

• Suppose we try to prove the fjrst subgoal, On(𝐵, 𝐶, 𝑡).

– Use same axiom

∀𝑦, 𝑧, 𝑡.Clear(𝑦, 𝑡) ∧ Clear(𝑧, 𝑡) → On(𝑦, 𝑧, Result(Move(𝑦, 𝑧), 𝑡))

– Again, by chaining, we can conclude

On(𝐵, 𝐶, Result(Move(𝐵, 𝐶), 𝑇))

– Abbreviating Result(Move(𝐵, 𝐶), 𝑇) as 𝑇, we have

On(𝐵, 𝐶, 𝑇).

• Substituting 𝑇 for 𝑡 in our other subgoal makes that On(𝐶, 𝐷, 𝑇).

If this were true, we are done.

• But we have no reason to believe this is true!
• Sure, On(𝐶, 𝐷, 𝑇), but how does the planner know this is still

true, i.e., On(𝐶, 𝐷, 𝑇)?

• In fact, it doesn’t, so it fails to fjnd an answer!

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The Frame Problem

We have failed to express the fact that everything that isn’t changed by an action should really stay the same. Can fjx by adding frame axioms.

∀𝑦, 𝑡.Clear(𝑦, 𝑡) → Clear(𝑦, Result(Paint(𝑦), 𝑡)) …

There are lots of these! Is this a big problem?

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Better Frame Axioms

• Can fjx with neater formulation:

∀𝑦, 𝑧, 𝑡, 𝑏.On(𝑦, 𝑧, 𝑡) ∧ (∀𝑨.𝑏 = Move(𝑦, 𝑨) → 𝑧 = 𝑨) → On(𝑦, 𝑧, Result(𝑏, 𝑡))

• Can combine with efgect axioms to get successor-state axioms:

∀𝑦, 𝑧, 𝑡, 𝑏.On(𝑦, 𝑧, Result(𝑏, 𝑡)) ↔ On(𝑦, 𝑧, 𝑡) ∧ (∀𝑨.𝑏 = Move(𝑦, 𝑨) → 𝑧 = 𝑨) ∨(Clear(𝑦, 𝑡) ∧ Clear(𝑧, 𝑡) ∧ 𝑏 = Move(𝑦, 𝑧))

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How Does This Help Our Example?

• We want to prove On(𝐶, 𝐷, Result(Move(𝐵, 𝐶), 𝑇)) given that

On(𝐶, 𝐷, 𝑇).

• Axiom says

∀𝑦, 𝑧, 𝑡, 𝑏.On(𝑦, 𝑧, Result(𝑏, 𝑡)) ↔ On(𝑦, 𝑧, 𝑡) ∧ (∀𝑨.𝑏 = Move(𝑦, 𝑨) → 𝑧 = 𝑨) ∨(Clear(𝑦, 𝑡) ∧ Clear(𝑧, 𝑡) → 𝑏 = Move(𝑦, 𝑧))

• So we need to show

On(𝐶, 𝐷, 𝑇) ∧ (∀𝑨.Move(𝐵, 𝐶) = Move(𝐶, 𝑨) → 𝐷 = 𝑨) is true: – The fjrst conjunct is in the KB. – The second one is true since actions are the same ifg they have the same name and involve the exact same objects∗:

𝐵(𝑦, … , 𝑦) = 𝐵(𝑧, … , 𝑧) ifg 𝑦 = 𝑧, … , 𝑦 = 𝑧.

So Move(𝐵, 𝐶) = Move(𝐶, 𝑨) is false.

∗Another assumption in KB: (, … , ) (, … , ).

These are known as Unique Action Axioms.

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Refutation Theorem Proving (Dual) Skolemisation

Suppose ∀𝑦.∃𝑧.𝐻(𝑦, 𝑧) is the goal in resolution refutation. We need to negate the goal:

¬∀𝑦.∃𝑧.𝐻(𝑦, 𝑧) ≡ ∃𝑦.∀𝑧.¬𝐻(𝑦, 𝑧)

Then skolemise (i.e drop the existential quantifjer):

¬𝐻(𝑌, 𝑧)

Intuition

𝑧 is to be unifjed to construct witness. 𝑌 must not be instantiated.

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KB and Axioms as Clauses

Variables

𝑏, 𝑦, 𝑧, 𝑨, 𝑡

Constants

𝐵, 𝐶, 𝐷, 𝑇

Initial State

On(𝐵, Table, 𝑇) On(𝐶, 𝐷, 𝑇) On(𝐷, Tabl𝑓, 𝑇) Clear(𝐵, 𝑇) Clear(𝐶, 𝑇)

(neg.) Goal

¬On(𝐵, 𝐶, 𝑡) ∨ ¬On(𝐶, 𝐷, 𝑡)

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KB and Axioms as Clauses

Efgect Axiom

¬Clear(𝑦, 𝑡) ∨ ¬Clear(𝑧, 𝑡) ∨ On(𝑦, 𝑧, Result(Move(𝑦, 𝑧), 𝑡))

Skolem function

Frame Axioms

↙ ¬On(𝑦, 𝑧, 𝑡) ∨ 𝑏 = Move(𝑦, 𝑎(𝑦, 𝑧, 𝑡, 𝑏)) ∨ On(𝑦, 𝑧, Result(𝑏, 𝑡)) ¬On(𝑦, 𝑧, 𝑡) ∨ ¬𝑧 = 𝑎(𝑦, 𝑧, 𝑨, 𝑡, 𝑏) ∨ On(𝑦, 𝑧, Result(𝑏, 𝑡))

Unique Action Axioms

¬Move(𝐵, 𝐶) = Move(𝐶, 𝑨)

Unique Name Axiom

¬𝐷 = 𝐷 for every pair of distinct constants 𝐷 and 𝐷 in KB.

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Resolution Refutation

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Frame problem partially solved

• This solves the representational part of the frame problem.
• Still have to compute that everything that was true and

wasn’t changed is still true.

• Ineffjcient (as is general theorem proving).
• Solution: Special purpose representations, special purpose

algorithms, called planners.

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Summary

• Planning
• Situations
• Frame problem

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