For the test (next Monday)
- Search algorithms
- Best First
- Depth First
- Breadth First
- Hill Climbing
- A*
- El Paso
- GPS (go over proofs of theorems)
- Expert Systems
- Semantic Networks
- Frames
Predicate Calculus will not be included on the test.
Frames — A frame is a structure that represents a priori knowledge about something. Compare to a semantic network, which is an association of nodes. The example used in the book is a hotel form.
Another example is an office. There are different kinds of offices, and there are subclasses of those kinds of offices, and on and on.
You can also have a frame of 'restaurant' and within that frame there are different sub-frames.
Default reasoning is the knowledge we have about things that can be considered common knowledge. It can be considered true unless there is something you know that goes against it.
Procedure attachment (consult the book on this). Procedures are functions are stored in the frame entries and are executed as needed. They may ask for information from the user, etc.
Constraints are conditions that need to be satisfied before a frame is modified. Database systems are not intelligent; frames address the question of "how can you have an intelligent database system that won't let you make mistakes?"
(FLORIDA (POPULATION (16 MILLION) (PROCEDURE)) (CAPITAL (TALLAHASSEE)) (MAIN.CITIES (MIAMI, TAMPA, ORLANDO, JACKSONVILLE)) )
The superframe of Florida would be "American States." There may be another frame in between like "Southeastern State."
Predicate Calculus
Example: All Floridians are mortal. This is equivalent to: ∀x(F(x) ⇒ M(x)). The capital letters represent the predicates.
Example: Some Floridians are nice. Equivalent to ∃x(F(x) ∧ nice)
Example: All men are mortal. All Greeks are men. == All Greeks are mortal. Equivalent to:
∀x(M(x) → Mo(x))
∀x(G(x) → M(x))
C: ∀x(G(x) → Mo(x))
Proof by contradiction is also known as an indirect proof. You negate what you need to prove.
Axiom 1:
∃xA(x)
C: A(a)
Axiom 2:
∀x A(x)
A(x)
⌉∀x(A(x)⇔∃x⌉A(x)
⌉∃x(A(x)⇔∀x⌉A(x)
∃x⌉(G(x)→Mo(x))
⌉(G(a)→Mo(a))
⌉(⌉(G(a)∨Mo(a))
G(a) ∧⌉Mo(a)
Some assertions:
1. ⌉M(x) ∨ Mo(x)
2. ⌉G(y) ∨ M(y)
3. G(x)
4. ⌉Mo(a)
Unify 2 and 3 and you get: (replace y with a, or {y/a}) M(a) which contradicts 1. You can replace the x with a.
Suppose you have G(a)∨⌉G(x)∨... This violates modus ponens.
Modus ponens:
p → q
p
--
q


