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CSE 415 -- (c) S. Tanimoto, 2005 Knowledge Representation 1 Knowledge Representation (Introduction) Knowledge Information Data Knowledge representation:

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Presentation on theme: "CSE 415 -- (c) S. Tanimoto, 2005 Knowledge Representation 1 Knowledge Representation (Introduction) Knowledge Information Data Knowledge representation:"— Presentation transcript:

1 CSE 415 -- (c) S. Tanimoto, 2005 Knowledge Representation 1 Knowledge Representation (Introduction) Knowledge Information Data Knowledge representation: Information representation with support for inference

2 CSE 415 -- (c) S. Tanimoto, 2005 Knowledge Representation 2 Procedural vs Declarative Knowledge can be operational: “When you see a grizzly bear, run away fast!” if Condition is true then do Action e.g., production rules. Knowledge can be declarative: “A grizzly bear is a dangerous animal.” predicate(object) e.g., logical statements or relations

3 CSE 415 -- (c) S. Tanimoto, 2005 Knowledge Representation 3 Example of Declarative Representation: An “Isa” Hierarchy Living-thing PlantAnimal Bird Fish Aquatic-birdLand-bird Great-blue-heron

4 CSE 415 -- (c) S. Tanimoto, 2005 Knowledge Representation 4 Inference with Isa Hierarchies “A great-blue-heron is an aquatic-bird.” Isa(great-blue-heron, aquatic-bird). “An aquatic-bird is a bird.” Isa(acquatic-bird, bird). “Is a great-blue-heron a bird?” Isa(great-blue-heron, bird)? “Isa” represents a relation which has certain properties that support types of inference.

5 CSE 415 -- (c) S. Tanimoto, 2005 Knowledge Representation 5 Binary Relations A binary relation R over a domain D is a set of ordered pairs (x,y) where x and y are in D. For example, we could have, D = {a,b,c,d} R = {(a, b), (c, a), (d, a)} If (x, y) is in R, then we write R(x, y) or x R y.

6 CSE 415 -- (c) S. Tanimoto, 2005 Knowledge Representation 6 Partial Orders If for each x in D we have x R x, then R is reflexive. If for each x in D and y in D we have x R y and y R x imply x = y, then R is antisymmetric. If for each x in D, y in D, and z in D we have x R y and y R z imply x R z, then R is transitive. If R has all 3 properties, R is a partial order.

7 CSE 415 -- (c) S. Tanimoto, 2005 Knowledge Representation 7 Examples of the Partial Order Properties Isa is reflexive: A bear is a bear. -> Isa(bear, bear) Isa is antisymmetric: A bear is a wowie, and a wowie is a bear. Therefore, bear and wowie represent the same things. Isa(bear, wowie) & Isa(wowie, bear) -> bear = wowie Isa is transitive: A grizzly is a bear, and a bear is a mammal. Therefore, a grizzly is a mammal. Isa(grizzly, bear) & Isa(bear, mammal) -> Isa(grizzly, mammal)

8 CSE 415 -- (c) S. Tanimoto, 2005 Knowledge Representation 8 Redundant Facts Any fact implied by others via the reflexive, antisymmetric, or transitive properties of a partial order can be considered redundant. Suppose a <= b b <= c c <= b Then a <= c is redundant. a <= a is redundant. b and c could be represented by one name.

9 CSE 415 -- (c) S. Tanimoto, 2005 Knowledge Representation 9 Inheritance of Properties via Isa A fox is a mammal. A mammal bears live young Therefore a fox bears live young.

10 CSE 415 -- (c) S. Tanimoto, 2005 Knowledge Representation 10 Non-inheritance of Certain Properties A neutron is an atomic particle. There are three types of atomic particles. Therefore there are three types of neutrons (??)

11 CSE 415 -- (c) S. Tanimoto, 2005 Knowledge Representation 11 Propositional Logic The propositional calculus is a mathematical system for reasoning about the truth or falsity of statements. The statements are made up from atomic statements using AND, OR, NOT, and their variations. A more powerful representation system is the predicate calculus. It’s an extension of the propositional calculus.

12 CSE 415 -- (c) S. Tanimoto, 2005 Knowledge Representation 12 Propositional Logic Examples P: a grizzly is a bear. Q: a bear is a mammal. R: a grizzly is a mammal. P & Q -> R Each proposition symbol represents an atomic statement. Atomic statements can be combined with logical connectives to form compound statements. ~P, P & Q, P v Q, P = Q, P -> Q


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