9
Machine Learning: Symbol-based
9.0
Introduction
9.5
Knowledge and Learning
9.1
A Framework for
Symbol-based Learning
9.6
Unsupervised Learning
9.7
Reinforcement Learning
9.8
Epilogue and
References
9.9
Exercises
9.2
Version Space Search
9.3
The ID3 Decision Tree
Induction Algorithm
9.4
Inductive Bias and
Learnability
Additional source used in preparing the slides:
Jean-Claude Latombe’s CS121 (Introduction to Artificial Intelligence) lecture
notes, http://robotics.stanford.edu/~latombe/cs121/2003/home.htm (version
spaces, decision trees)
Tom Mitchell’s machine learning notes (explanation based learning)
1
Chapter Objectives
• Learn about several “paradigms” of symbolbased learning
• Learn about the issues in implementing and
using learning algorithms
•The agent model: can learn, i.e., can use prior
experience to perform better in the future
2
A learning agent
Critic
Learning
element
KB
sensors
environment
actuators
3
A general model of the learning process
4
A learning game with playing cards
I would like to show what a full house is. I give
you examples which are/are not full houses:
6 6 6 9 9
is a full house
6 6 6 6 9
house
is not a full
3 3 3 6 6
is a full house
1 1 1 6 6
is a full house
Q Q Q 6 6
is a full house
1 2 3 4 5
house
is not a full
1 1 3 4 5
house
is not a full
1 1 1 4 5
is not a full
5
A learning game with playing cards
If you haven’t guessed already, a full house is
three of a kind and a pair of another kind.
6 6 6 9 9
is a full house
6 6 6 6 9
house
is not a full
3 3 3 6 6
is a full house
1 1 1 6 6
is a full house
Q Q Q 6 6
is a full house
1 2 3 4 5
house
is not a full
1 1 3 4 5
house
is not a full
1 1 1 4 5
is not a full
6
Intuitively,
I’m asking you to describe a set. This set is the
concept I want you to learn.
This is called inductive learning, i.e., learning a
generalization from a set of examples.
Concept learning is a typical inductive learning
problem: given examples of some concept,
such as “cat,” “soybean disease,” or “good
stock investment,” we attempt to infer a
definition that will allow the learner to correctly
recognize future instances of that concept.
7
Supervised learning
This is called supervised learning because we
assume that there is a teacher who classified
the training data: the learner is told whether an
instance is a positive or negative example of a
target concept.
8
Supervised learning?
This definition might seem counter intuitive. If
the teacher knows the concept, why doesn’t
s/he tell us directly and save us all the work?
The teacher only knows the classification, the
learner has to find out what the classification is.
Imagine an online store: there is a lot of data
concerning whether a customer returns to the
store. The information is there in terms of
attributes and whether they come back or not.
However, it is up to the learning system to
characterize the concept, e.g,
If a customer bought more than 4 books, s/he
will return.
If a customer spent more than $50, s/he will
return.
9
Rewarded card example
• Deck of cards, with each card designated by [r,s],
its rank and suit, and some cards “rewarded”
• Background knowledge in the KB:
((r=1)  …  (r=10))  NUM (r)
((r=J)  (r=Q)  (r=K))  FACE (r)
((s=S)  (s=C))  BLACK (s)
((s=D)  (s=H))  RED (s)
• Training set:
REWARD([4,C])  REWARD([7,C]) 
REWARD([2,S])  REWARD([5,H]) 
REWARD([J,S])
10
Rewarded card example
Training set:
REWARD([4,C])  REWARD([7,C]) 
REWARD([2,S])  REWARD([5,H]) 
REWARD([J,S])
Card
4
7
2
5
J
In the target set?
yes
yes
yes
no
no
Possible inductive hypothesis, h,:
h = (NUM (r)  BLACK (s)  REWARD([r,s])
11
Learning a predicate
• Set E of objects (e.g., cards, drinking cups,
writing instruments)
• Goal predicate CONCEPT (X), where X is an
object in E, that takes the value True or False
(e.g., REWARD, MUG, PENCIL, BALL)
• Observable predicates A(X), B(X), … (e.g.,
NUM, RED, HAS-HANDLE, HAS-ERASER)
• Training set: values of CONCEPT for some
combinations of values of the observable
predicates
• Find a representation of CONCEPT of the form
CONCEPT(X)  A(X)  ( B(X) C(X) )
12
How can we do this?
• Go with the most general hypothesis possible:
“any card is a rewarded card”
This will cover all the positive examples, but will
not be able to eliminate any negative examples.
• Go with the most specific hypothesis possible:
“the rewarded cards are 4, 7, 2”
This will correctly sort all the examples in the
training set, but it is overly specific, will not be
sort any new examples.
• But the above two are good starting points.
13
Version space algorithm
• What we want to do is start with the most
general and specific hypotheses, and
when we see a positive example, we
minimally generalize the most specific
hypotheses
when we see a negative example, we
minimally specialize the most general
hypothesis
• When the most general hypothesis and the
most specific hypothesis are the same, the
algorithm has converged, this is the target
concept
14
Pictorially
+
-
-
-
+
+ -
+ - ++
+
-
+
+
?
?
?
?
+
+
?
?
?
-
-
?
+
boundary of G
-
-
?
?
-
boundary of S
+
+ +
?
-
?
-
-
-
potential target concepts
+
-
?
-
-
+
+ +
+
+
-
-
-
-
15
Hypothesis space
• When we shrink G, or enlarge S, we are
essentially conducting a search in the
hypothesis space
• A hypothesis is any sentence h of the form
CONCEPT(X)  A(X)  ( B(X) C(X) )
where, the right hand side is built with
observable predicates
• The set of all hypotheses is called the
hypothesis space, or H
• A hypothesis h agrees with an example if it
gives the correct value of CONCEPT
16
Size of the hypothesis space
• n observable predicates
• 2^n entries in the truth table
• A hypothesis is any subset of observable
predicates with the associated truth tables: so
there are 2^(2^n) hypotheses to choose from:
BIG!
2n
2
• n=6  2 ^ 64 = 1.8 x 10 ^ 19
BIG!
• Generate-and-test won’t work.
17
Simplified Representation for the card
problem
For simplicity, we represent a concept by rs, with:
• r = a, n, f, 1, …, 10, j, q, k
• s = a, b, r, , , , 
For example:
• n represents:
NUM(r)  (s=)  REWARD([r,s])
• aa represents:
ANY-RANK(r)  ANY-SUIT(s)  REWARD([r,s])
18
Extension of an hypothesis
The extension of an hypothesis h is the set of
objects that verifies h.
For instance,
the extension of f is: {j, q, k}, and
the extension of aa is the set of all cards.
19
More general/specific relation
Let h1 and h2 be two hypotheses in H
h1 is more general than h2 iff the extension of
h1 is a proper superset of the extension of h2
For instance,
aa is more general than f,
f is more general than q,
fr and nr are not comparable
20
More general/specific relation (cont’d)
The inverse of the “more general” relation is the
“more specific” relation
The “more general” relation defines a partial
ordering on the hypotheses in H
21
A subset of the partial order for cards
aa
na
4a
ab
a
nb
n
4b
4
22
G-Boundary / S-Boundary of V
An hypothesis in V is most general iff no
hypothesis in V is more general
G-boundary G of V: Set of most general
hypotheses in V
An hypothesis in V is most specific iff no
hypothesis in V is more general
S-boundary S of V: Set of most specific
hypotheses in V
23
Example: The starting hypothesis space
G
aa
na
4a
ab
n
4b
S
1
a
nb
…
4
…
k
24
4 is a positive example
We replace every
hypothesis in S whose
extension does not
contain 4 by its
generalization set
The generalization set
of a hypothesis h is
the set of the
hypotheses that are
immediately more
general than h
aa
na
4a
ab
a
nb
n
4b
4
Generalization
set of 4 25
7 is the next positive example
Minimally generalize
the most specific
hypothesis set
We replace every
hypothesis in S whose
extension does not
contain 7 by its
generalization set
aa
na
4a
ab
a
nb
n
4b
4
26
7 is positive(cont’d)
Minimally generalize
the most specific
hypothesis set
aa
na
4a
ab
a
nb
n
4b
4
27
7 is positive (cont’d)
Minimally generalize
the most specific
hypothesis set
aa
na
4a
ab
a
nb
n
4b
4
28
5 is a negative example
Minimally specialize
the most general
hypothesis set
aa
na
4a
Specialization
set of aa
ab
a
nb
n
4b
4
29
5 is negative(cont’d)
Minimally specialize
the most general
hypothesis set
aa
na
4a
ab
a
nb
n
4b
4
30
After 3 examples (2 positive,1 negative)
G and S, and all hypotheses in between
form exactly the version space
ab
a
nb
1. If an hypothesis between
G and S disagreed with an
example x, then an hypothesis
G or S would also disagree with x,
hence would have been removed
n
31
After 3 examples (2 positive,1 negative)
G and S, and all hypotheses in between
form exactly the version space
ab
nb
a
2. If there were an hypothesis
n
not in this set which agreed
with all examples, then it would
have to be either no more specific than any member
of G – but then it would be in G – or no more general
than some member of S – but then it would be in S
32
At this stage
ab
a
nb
No
Yes
n
Maybe
Do 8, 6, j
satisfy CONCEPT?
33
2 is the next positive example
Minimally generalize
the most specific
hypothesis set
ab
a
nb
n
34
j is the next negative example
Minimally specialize
the most general
hypothesis set
ab
nb
35
Result
+ 4 7 2
– 5 j
nb
NUM(r)  BLACK(s)  REWARD([r,s])
36
The version space algorithm
Begin
Initialize S to the first positive training instance
N is the set of all negative instances seen so far;
For each example x
If x is positive, then
(G,S)  POSITIVE-UPDATE(G,S,x)
else
(G,S)  NEGATIVE-UPDATE(G,S,x)
If G = S and both are singletons, then the algorithm has found
a single concept that is consistent with all the data and the
algorithm halts
If G and S become empty, then there is no concept that covers
all the positive instances and none of the negative instances
End
37
The version space algorithm (cont’d)
POSITIVE-UPDATE(G,S,x)
Begin
Delete all members of G that fail to match x
For every s  S, if s does not match x, replace s with its most
specific generalizations that match x;
Delete from S any hypothesis that is more general than some
other hypothesis in S;
Delete from S any hypothesis that is neither more specific
than nor equal to a hypothesis in G; (different than the
textbook)
End;
38
The version space algorithm (cont’d)
NEGATIVE-UPDATE(G,S,x)
Begin
Delete all members of S that match x
For every g  G, that matches x, replace g with its most
general specializations that do not match x;
Delete from G any hypothesis that is more specific than some
other hypothesis in G;
Delete from G any hypothesis that is neither more general nor
equal to hypothesis in S; (different than the textbook)
End;
39
Comments on Version Space Learning (VSL)
• It is a bi-directional search. One direction is
specific to general and is driven by positive
instances. The other direction is general to
specific and is driven by negative instances.
• It is an incremental learning algorithm. The
examples do not have to be given all at once (as
opposed to learning decision trees.) The
version space is meaningful even before it
converges.
• The order of examples matters for the speed
of convergence
• As is, cannot tolerate noise (misclassified
examples), the version space might collapse
40
Examples and near misses for the
concept “arch”
41
More on generalization operators
• Replacing constants with variables. For example,
color (ball,red)
generalizes to
color (X,red)
• Dropping conditions from a conjunctive
expression. For example,
shape (X, round)  size (X, small)  color (X, red)
generalizes to
shape (X, round)  color (X, red)
42
More on generalization operators (cont’d)
• Adding a disjunct to an expression. For example,
shape (X, round)  size (X, small)  color (X, red)
generalizes to
shape (X, round)  size (X, small) 
( color (X, red)  (color (X, blue) )
• Replacing a property with its parent in a class
hierarchy. If we know that primary_color is a
superclass of red, then
color (X, red)
generalizes to
color (X, primary_color)
43
Another example
• sizes = {large, small}
• colors = {red, white, blue}
• shapes = {sphere, brick, cube}
• object (size, color, shape)
• If the target concept is a “red ball,” then size
should not matter, color should be red, and
shape should be sphere
• If the target concept is “ball,” then size or
color should not matter, shape should be
sphere.
44
A portion of the concept space
45
Learning the concept of a “red ball”
G : { obj (X, Y, Z)}
S:{}
positive: obj (small, red, sphere)
G: { obj (X, Y, Z)}
S : { obj (small, red, sphere) }
negative: obj (small, blue, sphere)
G: { obj (large, Y, Z), obj (X, red, Z), obj (X, white, sphere)
obj (X,Y, brick), obj (X, Y, cube) }
S: { obj (small, red, sphere) }
delete from G every hypothesis that is neither more
general than nor equal to a hypothesis in S
G: {obj (X, red, Z) }
S: { obj (small, red, sphere) }
46
Learning the concept of a “red ball”
(cont’d)
G: { obj (X, red, Z) }
S: { obj (small, red, sphere) }
positive: obj (large, red, sphere)
G: { obj (X, red, Z)}
S : { obj (X, red, sphere) }
negative: obj (large, red, cube)
G: { obj (small, red, Z), obj (X, red, sphere),
obj (X, red, brick)}
S: { obj (X, red, sphere) }
delete from G every hypothesis that is neither more general
than nor equal to a hypothesis in S
G: {obj (X, red, sphere) }
S: { obj (X, red, sphere) } converged to a single concept
47
LEX: a program that learns heuristics
• Learns heuristics for symbolic integration problems
• Typical transformations used in performing integration
include
OP1:  r f(x) dx  r  f(x) dx
OP2:  u dv  uv -  v du
OP3: 1 * f(x)  f(x)
OP4:  (f1(x) + f2(x)) dx   f1(x) dx +  f2(x) dx
• A heuristic tells when an operator is particularly useful:
If a problem state matches  x transcendental(x) dx
then apply OP2 with bindings
u=x
dv = transcendental (x) dx
48
A portion of LEX’s hierarchy of symbols
49
The overall architecture
• A generalizer that uses candidate elimination
to find heuristics
• A problem solver that produces positive and
negative heuristics from a problem trace
• A critic that produces positive and negative
instances from a problem traces (the credit
assignment problem)
• A problem generator that produces new
candidate problems
50
A version space for OP2 (Mitchell et al.,1983)
51
Comments on LEX
• The evolving heuristics are not guaranteed to
be admissible. The solution path found by the
problem solver may not actually be a shortest
path solution.
• The problem generator is the least developed
part of the program.
• Empirical studies:
before: 5 problems solved in an average
of 200 steps
train with 12 problems
after: 5 problems solved in an average of
20 steps
52
More comments on VSL
• Still lots of research going on
• Uses breadth-first search which might be
inefficient:
 might need to use beam-search to prune hypotheses
from G and S if they grow excessively
 another alternative is to use inductive-bias and restrict
the concept language
• How to address the noise problem?
Maintain several G and S sets.
53
Decision Trees
• A decision tree allows a classification of an
object by testing its values for certain
properties
• check out the example at:
www.aiinc.ca/demos/whale.html
• The learning problem is similar to concept
learning using version spaces in the sense that
we are trying to identify a class using the
observable properties.
• It is different in the sense that we are trying to
learn a structure that determines class
membership after a sequence of questions.
This structure is a decision tree.
54
Reverse engineered decision tree of the
whale watcher expert system
see flukes?
no
yes
see dorsal fin?
yes
no
size?
vlg
med
blue
blow
whale forward?
yes
no
sperm
whale
humpback
whale
(see next page)
size med?
yes
no
blows?
1
gray
whale
Size?
2
lg
right bowhead
whale
whale
vsm
narwhal
whale
55
Reverse engineered decision tree of the
whale watcher expert system (cont’d)
see flukes?
no
yes
(see previous page)
yes
see dorsal fin?no
blow?
no
yes
size?
lg
sm
dorsal fin and
blow visible
at the same time?
yes
no
sei
whale
fin
whale
dorsal fin
tall and pointed?
yes
no
killer
whale
northern
bottlenose
whale
56
What does the original data look like?
Place
Time
Group Fluke
Kaikora 17:00
Yes
Yes
Kaikora
7:00
No
Yes
Kaikora
8:00
Yes
Yes
Kaikora
9:00
Yes
Yes
Cape
Cod
Cape
Cod
Newb.
Port
Cape
Cod
…
18:00
Yes
Yes
20:00
No
Yes
18:00
No
No
Dorsal Dorsal
fin
shape
Yes
small
triang.
Yes
small
triang.
Yes
small
triang.
Yes
squat
triang.
Yes
Irregular
Yes
Irregular
No
Curved
6:00
Yes
Yes
No
None
Size
Blow …
Very
large
Very
large
Very
large
Medium
Yes
Blow Type
fwd
No
Blue whale
Yes
No
Blue whale
Yes
No
Blue whale
Yes
Yes
Sperm
whale
Hump-back
whale
Hump-back
whale
Fin
whale
Right
whale
Medium Yes
No
Medium Yes
No
Large
Yes
No
Medium Yes
No
57
The search problem
Given a table of observable properties, search
for a decision tree that
• correctly represents the data (assuming that
the data is noise-free), and
• is as small as possible.
What does the search tree look like?
58
Comparing VSL and learning DTs
A hypothesis learned in
VSL can be represented
as a decision tree.
Consider the predicate
that we used as a VSL
example:
NUM(r)  BLACK(s) 
REWARD([r,s])
NUM?
True
False
BLACK?
True
True
False
False
False
The decision tree on the
right represents it:
59
Predicate as a Decision Tree
The predicate CONCEPT(x)  A(x)  (B(x) v C(x)) can
be represented by the following decision tree:
Example:
A mushroom is poisonous iff
it is yellow and small, or yellow,
big and spotted
• x is a mushroom
• CONCEPT = POISONOUS
• A = YELLOW
• B = BIG
• C = SPOTTED
True
• D = FUNNEL-CAP
• E = BULKY
True
A?
True
B?
True
C?
False
False
False
True
False
False
60
Training Set
Ex. #
A
B
C
D
E
CONCEPT
1
False
False
True
False
True
False
2
False
True
False
False
False
False
3
False
True
True
True
True
False
4
False
False
True
False
False
False
5
False
False
False
True
True
False
6
True
False
True
False
False
True
7
True
False
False
True
False
True
8
True
False
True
False
True
True
9
True
True
True
False
True
True
10
True
True
True
True
True
True
11
True
True
False
False
False
False
12
True
True
False
False
True
False
13
True
False
True
True
True
True
61
Possible Decision Tree
D
T
E
Ex. #
A
B
C
D
E
CONCEPT
1
False
False
True
False
True
False
2
False
True
False
False
False
False
3
False
True
True
True
True
False
4
False
False
True
False
False
False
5
False
False
False
True
True
False
6
True
False
True
False
False
True
7
True
False
False
True
False
True
8
True
False
True
False
True
True
9
True
True
True
False
True
True
10
True
True
True
True
True
True
11
True
True
False
False
False
False
12
True
True
False
False
True
False
13
True
False
True
True
True
True
T
C
T
A
F
F
B
F
T
E
A
F
A
T
T
F
62
Possible Decision Tree
CONCEPT 
(D  (E v A)) v
(C  (B v ((E  A) v A)))
CONCEPT  A  (B v C)
True
T
E
A
A?
D
F
C
T
B
F
False
T decision
F
T
KIS bias  Build smallest
tree
E
False
B?
True
False
True
C? Computationally
A
intractable problem
False
True
True
False
 greedy algorithm
F
T
A
T
F
63
Getting Started
The distribution of the training set is:
True: 6, 7, 8, 9, 10,13
False: 1, 2, 3, 4, 5, 11, 12
64
Getting Started
The distribution of training set is:
True: 6, 7, 8, 9, 10,13
False: 1, 2, 3, 4, 5, 11, 12
Without testing any observable predicate, we
could report that CONCEPT is False (majority rule)
with an estimated probability of error P(E) = 6/13
65
Getting Started
The distribution of training set is:
True: 6, 7, 8, 9, 10,13
False: 1, 2, 3, 4, 5, 11, 12
Without testing any observable predicate, we
could report that CONCEPT is False (majority rule)
with an estimated probability of error P(E) = 6/13
Assuming that we will only include one observable
predicate in the decision tree, which predicate
should we test to minimize the probability of error?
66
Assume It’s A
A
T
True:
False:
6, 7, 8, 9, 10, 13
11, 12
F
1, 2, 3, 4, 5
If we test only A, we will report that CONCEPT is True
if A is True (majority rule) and False otherwise
The estimated probability of error is:
Pr(E) = (8/13)x(2/8) + (5/8)x0 = 2/13
67
Assume It’s B
B
T
True:
False:
9, 10
2, 3, 11, 12
F
6, 7, 8, 13
1, 4, 5
If we test only B, we will report that CONCEPT is False
if B is True and True otherwise
The estimated probability of error is:
Pr(E) = (6/13)x(2/6) + (7/13)x(3/7) = 5/13
68
Assume It’s C
C
T
True:
False:
6, 8, 9, 10, 13
1, 3, 4
F
7
1, 5, 11, 12
If we test only C, we will report that CONCEPT is True
if C is True and False otherwise
The estimated probability of error is:
Pr(E) = (8/13)x(3/8) + (5/13)x(1/5) = 4/13
69
Assume It’s D
D
T
True:
False:
7, 10, 13
3, 5
F
6, 8, 9
1, 2, 4, 11, 12
If we test only D, we will report that CONCEPT is True
if D is True and False otherwise
The estimated probability of error is:
Pr(E) = (5/13)x(2/5) + (8/13)x(3/8) = 5/13
70
Assume It’s E
E
T
True:
False:
8, 9, 10, 13
1, 3, 5, 12
F
6, 7
2, 4, 11
to testis False,
is A
If weSo,
test the
only Ebest
we willpredicate
report that CONCEPT
independent of the outcome
The estimated probability of error is unchanged:
Pr(E) = (8/13)x(4/8) + (5/13)x(2/5) = 6/13
71
Choice of Second Predicate
A
F
T
C
T
True:
False:
6, 8, 9, 10, 13
False
F
7
11, 12
The majority rule gives the probability of error Pr(E|A) = 1/8
and Pr(E) = 1/13
72
Choice of Third Predicate
A
F
T
False
C
F
T
True
T
True:
False:
11,12
B
F
7
73
Final Tree
True
True
True
True
A
True
C
C?
B?
A?
False
False
False
False
False
True
False
False
False
True
B
True
True
False
False
True
L  CONCEPT  A  (C v B)
74
Learning a decision tree
Function induce_tree (example_set, properties)
begin
if all entries in example_set are in the same class
then return a leaf node labeled with that class
else if properties is empty
then return leaf node labeled with disjunction of all
classes in example_set
else begin
select a property, P, and make it the root of the current tree;
delete P from properties;
for each value, V, of P
begin
create a branch of the tree labeled with V;
let partitionv be elements of example_set with values V
for property P;
call induce_tree (partitionv, properties), attach result to
branch V
end
If property V is Boolean: the partition will contain two 75
end
sets, one with property V true and one with false
end
What happens if there is noise in the
training set?
The part of the algorithm shown below handles this:
if properties is empty
then return leaf node labeled with disjunction of all
classes in example_set
Consider a very small (but inconsistent) training set:
A
T
F
F
classification
T
F
T
A?
True
True
False
False

True
76
Using Information Theory
Rather than minimizing the probability of error,
most existing learning procedures try to
minimize the expected number of questions
needed to decide if an object x satisfies
CONCEPT.
This minimization is based on a measure of the
“quantity of information” that is contained in
the truth value of an observable predicate and
is explained in Section 9.3.2. We will skip the
technique given there and use the “probability
of error” approach.
77
% correct on test set
Assessing performance
100
size of training set
Typical learning curve
78
The evaluation of ID3 in chess endgame
79
Other issues in learning decision trees
• If data for some attribute is missing and is
hard to obtain, it might be possible to
extrapolate or use “unknown.”
• If some attributes have continuous values,
groupings might be used.
• If the data set is too large, one might use
bagging to select a sample from the training
set. Or, one can use boosting to assign a weight
showing importance to each instance. Or, one
can divide the sample set into subsets and train
on one, and test on others.
80
Explanation based learning
• Idea: can learn better when the background
theory is known
• Use the domain theory to explain the
instances taught
• Generalize the explanation to come up with a
“learned rule”
81
Example
• We would like the system to learn what a cup
is, i.e., we would like it to learn a rule of the
form: premise(X)  cup(X)
• Assume that we have a domain theory:
liftable(X)  holds_liquid(X)  cup(X)
part (Z,W)  concave(W)  points_up  holds_liquid (Z)
light(Y)  part(Y,handle)  liftable (Y)
small(A)  light(A)
made_of(A,feathers)  light(A)
• The training example is the following:
cup (obj1)
small(obj1)
owns(bob,obj1)
part(obj1, bowl)
concave(bowl)
small(obj1)
part(obj1,handle)
part(obj1,bottom)
points_up(bowl)
color(obj1,red)
82
First, form a specific proof that obj1 is a cup
cup (obj1)
liftable (obj1)
light (obj1)
part (obj1, handle)
holds_liquid (obj1)
part (obj1, bowl)
points_up(bowl)
concave(bowl)
small (obj1)
83
Second, analyze the explanation
structure to generalize it
84
Third, adopt the generalized the proof
cup (X)
liftable (X)
light (X)
part (X, handle)
holds_liquid (X)
part (X, W)
points_up(W)
concave(W)
small (X)
85
The EBL algorithm
Initialize hypothesis = { }
For each positive training example not covered
by hypothesis:
1. Explain how training example satisfies
target concept, in terms of domain theory
2. Analyze the explanation to determine the
most general conditions under which this
explanation (proof) holds
3. Refine the hypothesis by adding a new rule,
whose premises are the above conditions, and
whose consequent asserts the target concept
86
Wait a minute!
• Isn’t this “just a restatement of what the
learner already knows?”
• Not really
 a theory-guided generalization from examples
 an example-guided operationalization of theories
• Even if you know all the rules of chess you get
better if you play more
• Even if you know the basic axioms of
probability, you get better as you solve more
probability problems
87
Comments on EBL
• Note that the “irrelevant” properties of obj1 were
disregarded (e.g., color is red, it has a bottom)
• Also note that “irrelevant” generalizations were
sorted out due to its goal-directed nature
• Allows justified generalization from a single
example
• Generality of result depends on domain theory
• Still requires multiple examples
• Assumes that the domain theory is correct (errorfree)---as opposed to approximate domain theories
which we will not cover.
 This assumption holds in chess and other search problems.
 It allows us to assume explanation = proof.
88
Two formulations for learning
Inductive
Analytical
Given:
Given:
• Instances
• Instances
• Hypotheses
• Hypotheses
• Target concept
• Target concept
• Training examples of the
target concept
• Training examples of the
target concept
• Domain theory for
explaining examples
Determine:
Determine:
• Hypotheses consistent
with the training examples
• Hypotheses consistent
with the training examples
and the domain theory
89
Two formulations for learning (cont’d)
Inductive
Analytical
Hypothesis fits data
Hypothesis fits domain theory
Statistical inference
Deductive inference
Requires little prior
knowledge
Learns from scarce data
Syntactic inductive bias
Bias is domain theory
DT and VS learners are “similarity-based”
Prior knowledge is important. It might be one of the
reasons for humans’ ability to generalize from as few as
a single training instance.
Prior knowledge can guide in a space of an unlimited
number of generalizations that can be produced by
training examples.
90
An example: META-DENDRAL
• Learns rules for DENDRAL
• Remember that DENDRAL infers structure of
organic molecules from their chemical formula
and mass spectrographic data.
• Meta-DENDRAL constructs an explanation of
the site of a cleavage using
 structure of a known compound
 mass and relative abundance of the fragments produced
by spectrography
 a “half-order” theory (e.g., double and triple bonds do not
break; only fragments larger than two carbon atoms
show up in the data)
• These explanations are used as examples for
constructing general rules
91
Analogical reasoning
•Idea: if two situations are similar in some
respects, then they will probably be in others
• Define the source of an analogy to be a
problem solution. It is a theory that is relatively
well understood.
• The target of an analogy is a theory that is not
completely understood.
• Analogy constructs a mapping between
corresponding elements of the target and the
source.
92
Example: atom/solar system analogy
• The source domain contains:
yellow(sun)
blue(earth)
hotter-than(sun,earth)
causes(more-massive(sun,earth), attract(sun,earth))
causes(attract(sun,earth), revolves-around(earth,sun))
• The target domain that the analogy is intended to
explain includes:
more-massive(nucleus, electron)
revolves-around(electron, nucleus)
• The mapping is: sun  nucleus and earth  electron
• The extension of the mapping leads to the inference:
causes(more-massive(nucleus,electron), attract(nucleus,electron))
causes(attract(nucleus,electron), revolves94
around(electron,nucleus))
A typical framework
• Retrieval: Given a target problem, select a
potential source analog.
• Elaboration: Derive additional features and
relations of the source.
• Mapping and inference: Mapping of source
attributes into the target domain.
• Justification: Show that the mapping is valid.
95