Knowledge and Application Flashcards
Declarative approach
Tell the agent what it needs to know and let it use reasoning to deduce consequences
Knowledge base
Contains facts, rules and general knowledge about the domain in some formal language
Reasoning engine
Produces relevant consequences of the knowledge base
Unary rule-based
Let Ka (assertions) contain:
CompetesInPremierLeague(LiverpoolFC)
CompetesInPremierLeague(EvertonFC)
Kr (rules):
CompetesInPremierLeague(x) -> CompetesInFACup(x)
Atomic assertions follow from K:
CompetesInFACup(Liverpool FC)
FootballCLub(Everton)
class(individual variable) -> individual is part of class
K be knowledge base and A(b) an atomic assertion
K |= A(b)
if whenever K is true, then A(b) is true since Kr allows us to prove this (acts like a bridge).
Ka - {A1(a)}
Kr = {A1(x) -> A2(x), A2(x) -> A3(x)} if x is in a1, then it is in a3
K |= {A1(a), A2(a), A3(a)}
Time Complexity
IndividualName*ClassName
Relation Name
Peter is the son of John
John is the son of Joseph
Relation name R denotes a set of pairs of individual objects, also called binary predicates:
onOf
grandsonOf
sonOf(Peter, John) - R(a,b) is atomic assertion
Propositional Logic
A statement that can be true or false.
Proof
a and q = m therefore I(m) = 1 (1 being true, 0 being false)*
f = q
!p = a
I(p) exists {0,1} -> I is the interpretation
=>
If P is false, then P=>Q is true.
If the first initial value is false.
Satisfiable
Propositional formula is satisfiable if there exists an interpretation under which it is true.
A . !A is a contradiction, because everything in the truth table adds up to false.
2^n interpretations
N propositional atoms creates combinatorial explosion.
Propositional Knowledge Base
Is a finite set of propositional formulas.
∧
And
P=>Q
If P is false, Q is true.
{(p1 and p2)} |= (p1 or p2)
True, because all elements of AND that are true are true in OR.
{p1, p1 => p2} |= p2
True, because when p1 AND p1=>p2 is equal true, p2 is equal to true.
∨
Or
¬
Not
And
P1 … Pn is in X
Q = p1 … pn AND ¬P
If Q is unsatisfiable then X |= P
Q is not satisfiable then P is true because
p1…pn implies P stands true
Q = X AND ¬P
if Q was satisfiable, then at one point ¬P would stand true
therefore Q needs to be non-satisfiable
Rule-based approach
They cannot contain ors, ands or nots since if it only an if something holds then something else holds.
Only propositional logic can.
Propositional logic, however, cannot express ‘any’ but rule-based can.
Propositional Atoms
They are just propositions.
Interpretation
Is the rule.
I(P) = true under interpretation I
Combinatorial Explosion
2^n, with n being propositions and 2 being either they are true (1) or false (0).
=>
Implies
whatever is on the left side, which is true, implies the right side is true
the sky is overcast |= the sun is not visible
