Download Inference and resolution for problem solving

Inference and Resolution
for Problem Solving
Fall 2016
Comp3710 Artificial Intelligence
Computing Science
Thompson Rivers University
Course Outline
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Part I – Introduction to Artificial Intelligence
Part II – Classical Artificial Intelligence
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Knowledge Representation
Searching
Knowledge Represenation and Automated Reasoning
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Propositinoal and Predicate Logic
Inference and Resolution for Problem Solving
Rules and Expert Systems
Part III – Machine Learning
Part IV – Advanced Topics
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Chapter Objectives
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How to convert BNF to CNF
Use of the resolution rule for some problems
How to construct Horn clauses from given rules and facts
Use of forward chaining
Use of backward chaining
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Chapter Outline
1.
2.
3.
4.
Introduction
Resolution in propositional logic
An application of resolution – graph coloring
Inference methods – forward and backward chaining
1.
2.
3.
Horn clauses
Forward chaining
Backward chaining
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1. Introduction
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Topics
Problem solving using propositional calculus:
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Based on rules, knowledge, and facts,
Decide if a given query is valid.
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Example
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Rules and facts (or observations)
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Question:
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If it is raining, Jenny does not go camping.
It is raining.
It is a hot day.
It is afternoon.
Does Jenny go fishing?
Can you convert them to a logical form, i.e., wff (or sentence)?
(Raining  ~Camping)  Raining  HotDay  Afternoon  ~Fishing ???
We need to see the above sentence is contradictory to answer “Is Jenny
coming?”.
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2. Resolution in Propositional Logic
1.
2.
3.
4.
5.
6.
CNF(Conjunctive Normal Form)
How to convert propositions to CNFs
Clauses instead of CNFs for convenience
The Resolution rule
Resolution refutation
Proof by refutation
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Conjunctive Normal Forms (CNF)
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BNF (Backus-Naur Form) is not convenient to use.
A wff is in Conjunctive Normal Form (CNF) if it is a conjunction of
disjunctions:
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2.
conjunction of disjunctions of literals
clauses
E.g., (A  B)  (B  C  D)
E.g., A  (B  C)  (¬A  ¬B  ¬C  D)
Every sentence of propositional logic is logically equivalent to a
conjunction of disjunctions of literals.
[Q] Really? How to convert?
[Q] Why do we need to use CNF?
Similarly, a wff is in Disjunctive Normal Form (DNF) if it is a
disjunction of conjunctions.

E.g., A  (B  C)  (¬A  ¬B  ¬C  D)
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Converting to CNF
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Any wff can be converted to CNF by using the following
equivalences:
1.
2.
3.
4.
5.
6.
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A ↔ B  (A → B)  (B → A)
A → B  ¬A  B
¬(A  B)  ¬A  ¬B
¬(A  B)  ¬A  ¬B
¬¬AA
A  (B  C)  (A  B)  (A  C)
biconditional elimination
implication elimination
de Morgan’s law
de Morgan’s law
double-negation elimination
distributivity
Importantly, this can be converted into an algorithm – this will be
useful when we come to automating resolution.
Examples – convert the following propositions to CNFs.
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(¬A → B) → ¬C
A  (B  C)
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2.
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Examples – convert the following propositions to CNFs.
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(~A  B)  ~C
~(~A  B)  ~C
(~~A  B)  ~C
(A  B  ~C)
A  (B  C)
(A  B)  (B  A)
(~A  B)  (~B  A)
(A → B) → C
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Clauses
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2.
Having converted a wff to CNF, it is usual to write it as a set of
clauses.
E.g.:
(A → B) → C
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In CNF is:
(A  C)  (¬B  C)
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[Q] How?
In clause form, we write:
{(A, C), (~B, C)}
Disjunction
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Conjunction
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The Resolution Rule
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2.
The resolution rule is written as follows:
A  B, ¬B  C
AC
A  B, ¬B
A
(A  B)  (¬B  C) ╞ (A  C)
(A  B)  ¬B ╞ A
If the upper part is valid, then the lower part should be valid.
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This tells us that if we have two clauses that have a literal and its
negation, we can combine them by removing that literal.
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[Q] Can you prove it using the truth table?
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E.g.: if we have {(A, C), (~A, D)}, we would apply resolution to get
{(C, D)}
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Resolution Refutation
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2.
Let us resolve: {(~A, B), (~A, ~B, C), A, ~C}
[Q] We begin by resolving the first clause with the second clause, thus
eliminating B and ~B:
{(~A, B), (~A, ~B, C), A, ~C}
=> {(~A, C), A, ~C}
=> {C, ~C}
=> falsum
Why falsum? Because the initial assumption, ~C, is not valid. Therefore C
is valid.

Now we can resolve both remaining literals, which gives falsum:
^
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If we reach falsum, we have proved that our initial set of clauses were
inconsistent.
This can be written as:
{(~A, B), (~A, ~B, C), A, ~C} ╞ ^
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Proof by Refutation
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If we want to prove that a logical argument is valid, we negate its
conclusion, convert it to clause form, and then try to derive falsum
using resolution.
If we derive falsum, then our clauses were inconsistent, meaning the
original argument was valid, since we negated its conclusion.
Proof by refutation is also called proof by contradiction.
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
[Q] Is the following argument valid?
(A ˄ ¬B) → C
A ˄ ¬B
C
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Start with the negation of the conclusion and convert to clauses:
{(~A, B, C), A, ~B, ~C}
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Let’s resolve:
{(~A, B, C), A, ~B, ~C}
=> {(B, C), ~B, ~C}
=> {C, ~C}
=> ^
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[Q] How?
How?
How?
We have reached falsum.
[Q] Why?
Because of the assumption of ~C. Therefore C. Hence, our original
argument is valid.
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Another example:
AB
BC
CD
DEF
AF
Is this argument valid?
~(A  F) 
~(~A  F)  A  ~F
Clause form?
{(~A, B), (~B, C), (~C, D), (~D, E, F), A, ~F }
Resolution?
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2.
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It is often sensible to represent a refutation proof in tree form:
Topics
How to
automate the
resolution?
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In this case, the
proof has failed, as we
are left with E instead of falsum.
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3. Applications of Resolution
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Resolution refutation can be used to determine if a solution exists for a
particular combinatorial problem.
For example, for graph coloring, we represent the assignment of
colors to the nodes and the constraints regarding edges as propositions,
and attempt to prove that the complete set of clauses is satisfiable (i.e.,
to be deduced to non-false sum).
This does not tell us how to color the graph, simply that it is possible.
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Topics
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3-coloring problem: Available colors are r (red), g (green) and b (blue).
Ar means that node A has been coloured red.
Each node must have exactly one color:
Ar  Ag  Ab
¬Ar  ¬Ag
( ¬(Ar  Ag))
¬Ag  ¬Ab
¬Ab  ¬Ar
If (A, B) is an edge, then:
¬Ar  ¬Br
( ¬(Ar  Br))
¬Ab  ¬Bb
¬Ag  ¬Bg
(B, C), (A, C), (C, D) and (A, D) are also edges.
Now we construct the complete set of clauses for our graph, and try to
see if they are satisfiable, using the resolution rule.
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4. Forward and backward chaining
1.
2.
3.
Horn clauses – a simpler form
Forward chaining
Backward chaining
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Horn Clauses
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Horn Form (a more restricted version of propositional language) –
conjunction of Horn clauses
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Horn clause =
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E.g., C  (B  A)  (C  D  B)
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proposition symbol; or
(conjunction of symbols)  symbol
Fact: C
Rules: if B then A; if C and D then B
Modus Ponens (for Horn Form): complete for Horn KBs
α1[, … ,αn],
α1 [ …  αn]  β
β
Meaning: Facts and the rule with the facts  then  is a fact.
E.g., Rule: if red light, then stop; Fact: red light; then ???
Can be used with forward chaining or backward chaining. These
algorithms are very natural and run in linear time.
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How to model Horn clauses?
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Example:
Rules
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IF P THEN Q
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IF L and M THEN P
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IF B and L THEN M
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IF A and P THEN L
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IF A and B THEN L
Facts
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A
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B
[Q] Horn clauses?
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and-or graph
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
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Example:
Rules
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If X croaks and X eats flies, then X is a frog.
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If X chirps and X sings, then X is a canary.
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If X is a frog, then X is green.
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If X is a canary, then X is yellow.
Facts
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Question
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Fritz croaks.
Fritz eats flies.
The color of Fritz.
How to solve?
Can you convert the problem to Horn clauses?
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
Another more complex example – ZooKeeper:
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If X has hair, then X is a mammal.
If X gives milk, then X is a mammal.
If X has feathers, then X is a bird.
If X flies and lays eggs, then X is a bird
If X is a mammal and X eats meat, then X is a carnivore.
If X is a mammal and X has pointed teeth and X has claws and X has forward
pointing eyes, then X is a carnivore.
If X is a mammal and X has hoofs, then X is an ungulate.
If X is a mammal and X chews cud, then X is an ungulate.
If X is a carnivore and X has tawny color and X has dark spots, then X is a cheetah.
If X is a carnivore and X has tawny color and X has black strips, then X is a tiger.
If X is an ungulate and X has long legs and X has a long neck and X has tawny color
and X has dark spots, then X is a giraffe.
If X is an ungulate and X has white color and X has black strips, then X is a zebra.
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4.
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Facts/observations
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Stretch has hair.
Stretch chew cud.
Stretch has long legs.
Stretch has a long neck.
Stretch has tawny color.
Stretch has dark spots.
Question
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If X is a bird and X does not fly and X has long neck and X has long legs and X is
black and white, then X is an ostrich.
If X is a bird and X does not fly and X swims and X is black and white, then X is a
penguin.
If X is a bird and X is a good flier, then X is an albatross.
What animal is Stretch?
How to solve?
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Forward chaining
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[Q] The query Q is valid?
Knowledge base
Facts
Query = Q
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Idea:
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Start with facts, and fire any rule whose premises are
satisfied in the KB; Keep adding its conclusion to the KB, until
query is found.
Knowledge base
OR
AND
Facts
Query = Q
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AND-OR graph
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Query
Facts
Rules
triggered
Rule fired
A, B
OR
# of premises
AND
Not inferred
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Facts
Rules
triggered
Rule fired
A, B
Reduced by 1
Set true
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Facts
A, B
Rules
triggered
5
Rule fired
5
OR
AND
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Facts
A, B
A, B, L
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Inference and Resolution
Rules
triggered
5
3
Rule fired
5
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Facts
A, B
A, B, L
A, B, L, M
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Inference and Resolution
Rules
triggered
5
3
2
Rule fired
5
3
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Facts
A, B
A, B, L
A, B, L, M
A, B, L, M, P
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Rules
triggered
5
3
2
4, 1
Rule fired
5
3
2
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Facts
A, B
A, B, L
A, B, L, M
A, B, L, M, P
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Inference and Resolution
Rules
triggered
5
3
2
4, 1
Rule fired
5
3
2
4
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Facts
A, B
A, B, L
A, B, L, M
A, B, L, M, P
A, B, L, M, P, Q
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Rules
triggered
5
3
2
4, 1
1
STOP
Rule fired
5
3
2
4
1
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4.
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Example – Fritz:
Rules
1.
Croak  EatFly  Frog
2.
Chirp  Sing  Canary
3.
Frog  Green
4.
Canary  Yellow


Croak
EatFly
Frog
Canary
Facts
Croak, EatFly
Question
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Yellow
Croak
Facts
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Green
EatFly
Chirp
Rules triggered
Sing
Rule fired
The color of Fritz.
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Backward chaining

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[Q] Problem of forward chaining (FC)?
FC starts with all the asserted symbols, i.e., facts -> longer execution
time
(We do not have to use all the facts to conclude the given query.)
Idea of backward chaining (BC): work backwards from the query q:
to prove q by BC,
check if q is known already, or
prove by BC – only the premises of some rules concluding q

Avoid loops: check if new subgoal is already on the goal stack
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Avoid repeated work: check if new subgoal
1.
has already been proved true, or
2.
has already failed
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Unknown
Facts
A, B
Goals
Q
Matching rules
?
TRUE
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Unknown
Facts
(subgoal)
A, B
A, B
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Goals
Q
P
Matching rules
1
?
38
Facts
A, B
A, B
A, B
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Goals
Q
P
L, M
Matching rules
1
2
L: ?
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Already in the stack of subgoals
Facts
A, B
A, B
A, B
Goals
Q
P
L, M
Matching rules
1
2
L: 4, 5
Not fired
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Facts
A, B
A, B
A, B
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Inference and Resolution
Goals
Q
P
L, M
Matching rules
1
2
L: 4, 5
41
Facts
A, B
A, B
A, B
A, B, L
Goals
Q
P
L, M
M
Matching rules
1
2
L: 4, 5
?
Became TRUE
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Facts
A, B
A, B
A, B
A, B, L
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Inference and Resolution
Goals
Q
P
L, M
M
Matching rules
1
2
L: 4, 5
3
43
Facts
A, B
A, B
A, B
A, B, L
A, B, L, M
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Goals
Q
P
L, M
M

STOP
Matching rules
1
2
L: 4, 5
3
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

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Example – Fritz:
Rules
1.
Croak  EatFly  Frog
2.
Chirp  Sing  Canary
3.
Frog  Green
4.
Canary  Yellow


Croak
EatFly
Question

The color of Fritz.
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Yellow
Frog
Canary
Croak
Facts

Green
EatFly
Chirp
Facts
Croak, EatFly
Croak, EatFly
Croak, EatFly
Goals
Yellow
Canary
Chirp, Sing
Croak, EatFly
Green
Inference and Resolution
Sing
Matching rules
4
2

Failure
47

How to use BC to solve ZooKeeper?
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Animals – Cheetah, Tiger, Giraffe, Zebra, Ostrich, Penguin, Albatross
For each of the above animals,
Run BC to see if the animal can be deduced.
Can you implement BC for ZooKeeper?
Recursion with the next representation of rules and facts.
var rule = {};
rule['Carnivore'] = [
[Mammal, EatMeat],
[Mammal, PointedTooth, Claw, ForwardPointingEye]
]; // (Mammal  EatMeat) 
// (Mammal  PointedTooth  Claw  ForwardPointingEye)
...
var fact = [];
fact[1] = 'Hair';
...
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Forward vs. backward chaining
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FC is data-driven, automatic, unconscious processing,
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May do lots of work that is irrelevant to the goal
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BC is goal-driven, appropriate for problem-solving,
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Topics
E.g., object recognition, routine decisions
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4.
E.g., Where are my keys?
E.g., How do I get into a PhD program?
E.g., My car won’t start. What is the problem?
Complexity of BC can be much less than linear in size of KB
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Summary
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