CMPS 441 Artificial Intelligence Notes
0.0 Course Information
Primary textbook: Artificial Intelligence A guide to Intelligent Systems; Second Edition;
Michael Negnevitsky
Secondary textbook: Machine Learning; Tom M. Mitchell
1.1 Quizzes, Assignments, Exams, Projects, Etc.
There will be 3 quizzes on the following material:
1. Overview and Classic AI
2. Statistical and Fuzzy Systems
3. Neural Networks and Evolutionary Systems
The final exam will have two parts. If you have a 93% average on your work prior to the
final, your final will have one part, otherwise 2 parts.
There will several homework assignments, and 3 or more programming projects.
If you do all your work on time and study for the quizzes you should do well.
1.2 Grading
The grade will be based on a weighted average. Quizzes and the final will comprise 60%
of the grade, homework, programs and projects fill out the remaining points.
A university grade scale will be used:
A
B
C
D
F
90 – 100
80 – 89
70 – 79
60 – 69
< 60
There will little to no curving done. Only in certain cases, for example if a student has a
79.5 average, and did all their work on time, they are likely to be bumped from a C to a
B. If on the other hand a student has a 79.8 average, smoked the tests, but did not turn in
all of their assignments, they will receive a C.
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On another note, quality of work will also be taken into consideration. An A will be
awarded for work that it of excellent quality and fulfills all requirements. If work is
turned in that is of excellent quality and goes beyond the requirements this will be noted
and will definitely be taken into account in the grading process.
1.3 Miscellaneous
Learning is like many other endeavors in that practice and repetition aid in the process.
This class will present several topics. It is important to be clear on the concepts that are
being taught and it is equally important to put those concepts to practical use through
various homework assignments and projects. The homework assignments and projects
help by tying together the concepts learned in the classroom in a practical manner.
For an average student, an easy rule of thumb to ensure an “A” is that for every hour
spent in class, 3 hours should be spent outside of class. So for this class, spend about 9
hours a week outside of class and you should get an “A”. If it takes time, that is almost
better. Difficult material will help you learn strategies to deal with complex material.
Many times gifted students have real trouble when they finally run into material that is
hard for them. This is because they have never experienced cases in which they don’t
what to do, so they have a difficult time developing a strategy to find answers.
1.0 Introduction
Read chapter 1
Objective:
 Gain an overview of knowledge about the field.
Assignments:
 Page 21, Questions 2, 4, 5, 6, 7, 8, 10, 11
2.1 Definitions:



Intelligence – the ability to understand and learn things.
o Understand – analyze
o Learn – create a behavior in order to cope with a new situation.
Intelligence – the ability to think and understand instead of doing things by
instinct or automatically.
o The second of these two definitions imply that there is local adaptation
and learning occurring rather than preprogrammed or generational
learning.
o Many behaviors that bugs, animals, and people exhibit are inherited. For
example, deer and horses begin to walk within a short time from being
born.
Intelligence – The ability to learn and understand, to solve problems and to make
decisions.
2

Thinking is the activity of using your brain to consider a problem or to create an
idea. (page 1)
o Can something without a brain think?
o Does something have to be alive to think?
o If something thinks, is it alive?
2.2 History:


Alan Turing – Alan Turing, “Computing Machinery and Intelligence”, 1950.
o Proposed the concept of a universal machine (Turing Machine)
o Helped break codes in WWII.
o Designed “Automatic Computing Engine”
o Wrote the first program that could play chess.
o Some key questions that he asked – still relevant today
 Is there thought without experience?
 Is there intelligence with life?
o Turing game – Can a computer communicate with a person so well that
the person cannot tell whom they are talking with, computer or person?
 Turing thought that by the year 2000 computers would pass this
test.
Early History (1943 – 1956)
o Warren McCulloch and Walter Pitts designed a model of human nerves in
which each neuron was either in an on state or off state. They showed that
there model was equivalent to a Turing machine and showed some basic
structures.
o Claude Shannon showed the need for heuristics in order to solve complex
problems.
 Heuristic – (book definition) A strategy that can be applied to
complex problems; it usually - but not always – yields a correct
solution. Heuristics, which are developed from years of
experience, are often used to reduce complex problem solving to
more simple operations based on judgement. Heuristics are often
expressed as rules of thumb.
 What does this definition imply in terms of the solution
space?
 What does it imply in terms of solution optimality.
 How does this compare with a search algorithm or greedy
algorithm?
 Heuristic search – A technique that applies heuristics to guide the
reasoning and thus reduce the search space for a solution.
 Algorithm – (American Heritage Dictionary) A rule or procedure
for solving a problem.
 What does this definition imply?
o Under defined circumstances, an algorithm will find
a correct solution.
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





Optimal Solution – The best solution to a problem. Many
algorithms that find optimal solutions have to check every solution
in some way. When the solution space is large, this can be
prohibitive.
 Claude Shannon’s 1950 paper on chess playing programs pointed
out the number that a typical chess game had about 10120 possible
moves.
 How long would it take a computer to pick the first move if
it could evaluate 1 move in every time cycle and the
computer were a 10 gigahertz (109 evaluation cycles per
second) machine?
o Other notable scientists:
 John McCarthy
 John Von Neumann
 Marvin Minsky
Middle History (Great expectations)
o John McCarthy develops LISP
o Marvin Minsky focuses on formal logic
o McCulloch and Pitts neural networks further developed by Rosenblatt
(perceptrons)
o Approach to solving problems:
 General methods (Weak methods)
Reality sets in (late 60’s and early 70’s)
Expert systems show success – contrast with weak methods.
Neural networks rebirth
Evolutionary Computation
2.3 Summary
1. What is Artificial Intelligence – AI is the study of making machines think.
2. Two main branches of AI
a. Classic AI techniques - rooted in heuristics, symbolic computing and
expert systems.
i. Chess playing – Alan Turing
ii. Mycin, Prospector – expert systems developed at Stanford
iii. Symbolic processing - Deterministic searching of solution spaces.
b. Machine Learning techniques – numerically based, neural networks, nondeterministic solution space searching (genetic algorithm, simulated
annealing.
i. McCulloch/Pitts neurons
ii. Non-deterministic searching – genetic algorithm, simulated
annealing, evolutionary programming.
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2.0 LISP
1. LISP
o Second high level language developed. FORTRAN was the first.
o Program and data take the same form.
Functions:











(car <list>)
o (car (1 2 3)) => 1
(cdr <list>)
o (cdr (1 2 3)) => (2 3)
o (cdr (1)) => NIL
(list <element1 element2 element3 ….. elementN>)
o (setf fullhouse (list 2 2 3 3 3)))
(setf <variable [list or atom]> <value>)
o (setf eat 8)
o (setf eatthis (list pie cake))
(length <list>)
o (length fullhouse) => 5
(defun <function name> <arg list> (exp1) (exp2) (exp3) ….(expN) )
(cond ( (condition 1) (exp1)..(expN)) ( (condition 2) (exp1)..(expN)) (t
(exp1)..(expN)) )
(append (list1) (list2))
o (append (list 1 2 3) (list 4 5)) => (1 2 3 4 5)
(cons (list1) (list2))
o (cons (list 1 2 3) (list 4 5)) => ((1 2 3) 4 5)
(eval (list))
o (eval (car (1 2 3)) => 1
(mapcar <function name> <list 1> <list 2> ….<list n>)
o (mapcar 'car '((1 a) (2 b) (3 c))) => (1 2 3)
o (mapcar 'abs '(3 -4 2 -5 -6)) => (3 4 2 5 6)
o (mapcar 'cons '(a b c) '(1 2 3)) => ((A . 1) (B . 2) (C . 3))
Examples:
(defun init ()
(setf big (list 1 2 3 4 5 6 7 8 9 10))
(setf fullhouse (list 2 2 2 3 3))
)
(defun lastone (list)
(cond
(( = (length list) 1)
(first list)
)
( t
5
(print list)
(lastone (cdr list))
)
)
)
(defun sumList(addlist)
(cond
((= (length addlist) 1)
(car addlist)
)
(t
(+ (car addlist)(sumList(cdr addlist)))
)
)
)
(defun average (list)
(/ (sumList list) (length list))
)
(defun middleNum (nums)
(setf mid (/ (length nums) 2))
(getNum mid nums)
)
(defun getNum (mid nums)
(cond
((= 1 mid)
(car nums)
)
(t
(getNum (- mid 1) (cdr nums))
)
)
)
(defun insultMe (message)
(append (list 'Yo 'Mamma) message)
)
Link to LISP Help
http://www.lisp.org/HyperSpec/FrontMatter/Chapter-Index.html
6
; Homework assigment 2
; Lisp review
; Fall 2007
(setf big '(1 2 3))
(defun revMe (rlist)
(cond
((= (length rlist) 1)
rlist
)
(t
(append (revMe (cdr rlist)) (list (car rlist)))
)
)
)
; (1 2 3)
; (2 3)
; (3)
; (append (3) (list 2)) => (3 2)
; (append (3 2) (list 1)) = > (3 2 1)
(defun countMe (rlist)
(cond
((equal NIL (cdr rlist))
1
)
(t
(+ 1 (countMe (cdr rlist)))
)
)
)
(defun fact (num)
(cond
((= 1 num)
num
)
(t
(* (fact (- num 1)) num)
)
)
)
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(defun funTheList (op nums)
(setf comList (append (list op) nums))
(print comList)
(eval comList)
)
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3.0 Classic AI Systems (Rule-Based Systems)
While important to our study, we don’t want to spend the whole semester on this topic.
We will concentrate on rule-based reasoning and do some LISP.
Here is a link to some LISP literature from Wikipedia:
http://en.wikipedia.org/wiki/Lisp_programming_language#Syntax_and_semantics
Early methods were based on providing the computer general methods of problems
solving broad classes of problems. The computer searched for solutions using these
general methods. This approach is referred to as a “weak method”. They applied “weak”
as in non-problem specific information to a task domain. The results were not
satisfactory.
Domain specific systems showed success so we will talk about those.
Summary – Weak methods relied upon general problem solving techniques and little to
no a-priori domain specific information. This approach has shown merit for more trivial
problems, but has not been shown to be scaleable to larger more complex problems.
Expert systems operate with an abundance of domain specific knowledge and are very
narrow in their abilities and are quite inflexible. They have had good success in several
areas, such as medicine and prospecting.
Key points of interest:
2. Knowledge (page 25)
o What is it?
o Where does it come from
o How do we represent it?
 Production rules (if – then statements)
3. Structure of AI program (page 31)
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The production rules are rules of thumb that we obtained through research; reading books
and papers; experimentation; asking experts what they would do in a variety of situations.
The short-term memory consists of a set of facts that describes the current problem to be
solved.
The reasoning function acts as an inference engine, user interface, and a tracing
mechanism that can be used to explain the programs line of reasoning.
 Inference engine – matches facts in the short-term memory against the rules in the
long-term memory.
 User interface – communicates to the user. Typically its job would be collect
facts from the user and store them in the short-term memory. When it draws
conclusions from its rules, these may also be stored as facts. If the inference
engine comes to impasse before drawing a conclusion, it may direct the user
interface to ask the user for more information so it can continue.
 Explanation facility – tracks the rule/fact matching process so that the system can
show its line of reasoning.
Other modules may include:
 External databases
 Models/programs
 Developer interface
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The key to the functionality of the system is the inference engine. The diagram below
shows the inference engine operational procedure.
4. Forward chaining – The accumulation of facts to support a conclusion. We gather
information and then infer whatever we can from it. Data driven reasoning.
Database Facts:
{a, b, c, d, e}
Rules:
/* Rule 1. */
If (y == TRUE) && (d == TRUE) {
Z = TRUE;
}
/* Rule 2. */
If (x == TRUE) && (b == TRUE) && (e == TRUE) {
Y = TRUE;
}
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/* Rule 3. */
If (a == TRUE) {
X = TRUE;
}
/* Rule 4. */
If (c == TRUE) {
L = TRUE;
}
/* Rule 5. */
If (l == TRUE) && (M == TRUE) {
N = TRUE;
}
Step 1:
Facts = {a, b, c, d, e}
Rule 3 fires => Facts = {a, b, c, d, e, x}
Step 2:
Facts = {a, b, c, d, e, x}
Rule 4 fires => Facts = {a, b, c, d, e, x, l}
Step 3:
Facts = {a, b, c, d, e, x, l}
Rule 2 fires => Facts = {a, b, c, d, e, x, l, y}
Step 4:
Facts = {a, b, c, d, e, x, l, y}
Rule 1 fires => Facts = {a, b, c, d, e, x, l, y, z}

Here our rule search pattern always picked up from the last rule that fired and
continued down the list.

When hit the bottom of the list we started at the top again.

What would happen if we always started at rule 1 and searched downward.
o Would the results be different?
o Would the path of reasoning be different?
o Does it matter?
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2. Backward chaining – Given a likely conclusion, can we find the facts to
support it. Goal driven reasoning.
Database Facts:
{a, b, c, d, e}
Rules:
/* Rule 1. */
If (y == TRUE) && (d == TRUE) {
Z = TRUE;
}
/* Rule 2. */
If (x == TRUE) && (b == TRUE) && (e == TRUE) {
Y = TRUE;
}
/* Rule 3. */
If (a == TRUE) {
X = TRUE;
}
/* Rule 4. */
If (c == TRUE) {
L = TRUE;
}
/* Rule 5. */
If (l == TRUE) && (M == TRUE) {
N = TRUE;
}
Step 1:
Facts = {a, b, c, d, e}
Goal = z; Rule 1 => z ; sub-goal required => y
Step 2:
Facts = {a, b, c, d, e}
Goal = y; Rule 2 => y ; sub-goals required => x, b, e; b and e exist, x is required
Step 3:
Facts = {a, b, c, d, e}
Goal = x; Rule 3 => x ; sub-goals required => none
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Now we start forward chaining process to confirm that the goal z is supportable with the
existing fact set.
Step 4:
Facts = {a, b, c, d, e}
Rule 3 fires => Facts = {a, b, c, d, e, x}
Step 5:
Facts = {a, b, c, d, e, x}
Rule 2 fires => Facts = {a, b, c, d, e, x, y}
Step 6:
Facts = {a, b, c, d, e, x, y}
Rule 1 fires => Facts = {a, b, c, d, e, x, y, z}
4.0 Statistical Based Systems
Here we will introduce probability, conditional probability, and Bayesian reasoning. We
will relate it back to section 3.
4.1 Introduction to Probability
The science of odds and counting.
Flip a coin. What is the chance of it landing heads side up? There are 2 sides, heads and
tails. The heads side is one of them, thus
P(Heads) =
Heads side
Heads side + Tails side
P(Heads) = 1/(1+1) = ½ = .50 = 50%
Throw a die. What is the probability of getting a 3.
P(die = 3) =
1 side with a 3
6 sides total
P(die = 3) = 1/6 = .167 = 16.7%
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4.2 Conditional Probability (page 59)
Events A and B are events that are not mutually exclusive, but occur conditionally on the
occurrence of one another.
The probability that event A will occur:
p(A)
The probability that event B will occur:
p(B)
The number of times that both A and B occur or the probability that both events A and B
will occur is called the joint probability.
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Mathematically joint probability is defined as:
p(A  B)
i.e. the probability that both A and B will occur
The probability that event A will occur if event B occurs is called conditional
probability.
P(A|B) = the number of times A and B can occur
the number of times B can occur
or
P(A|B) = P(A and B)
P(B)
or
P(A|B) = P(A  B)
P(B)
Lets take an example.
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Suppose we have 2 dice and we want to know what the probability of getting an 8 is.
Normally if we roll both at the same time the probability is 5/36.
1,1
2,1
3,1
4,1
5,1
6,1
1,2
2,2
3,2
4,2
5,2
6,2
1,3
2,3
3,3
4,3
5,3
6,3
1,4
2,4
3,4
4,4
5,4
6,4
1,5
2,5
3,5
4,5
5,5
6,5
1,6
2,6
3,6
4,6
5,6
6,6
But what happens if we roll the first die and get a 5, now what is the probability of
getting an 8?
There is only one way to get an 8 after a 5 has been rolled. You have to roll a 3.
Looking at the formula:
P(A|B) = P(A  B)
P(B)
Lets rephrase it for our problem:
P(getting an 8 using 2 die | given that we roll a 5 with first dice) = P(rolling 5 and 3)
P(rolling a 5)
P(A|B) = (1/36) / (1/6) = (1/36) * (6/1) = 6/36 = 1/6
So the chances improve slightly 1/6 > 5/36 by only 1/36.
Why do they improve, but only slightly?
An intuitive explanation is that the only way to not be able to get an 8 using two dice
rolled in sequence (one then the other) is if the first die is a 1. The chance of that is only
1/6. So it would not really matter what the first die is, as long as is not a 1, you can still
get an 8 with the combination of the two.
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4.3 Bayes Rule
Bayes theorem is an algebraic rewriting of conditional probability. The explanation from
Negnevitsky follows (page 59) :
Starting with conditional probability:
P(A|B) = P(A  B)
P(B)
P(A  B) = P(A|B) * P(B)
Recognizing that the intersection of A and B is communitve:
P(A  B) = P(B  A)
And that the following is true:
P(B|A) = P(B  A)
P(A)
P(B  A) = P(B|A) * P(A)
Remembering that the intersection is communitve we can form the following equation:
P(A  B) = P(B|A) * P(A)
Now this can be substituted into the conditional probability equation:
P(A|B) = P(A  B)
P(B)
Yielding Bayes theorem:
P(A|B) = P(B|A) * P(A)
P(B)
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We like to rewrite it using h and D, h being a hypotheses that we are testing and D being
data that we want to support our hypotheses.
Bayes theorem:
p(h|D) = p(D|h) * p(h)
p(D)
Machine Learning, Mitchell, page 156 gives some intuitive meaning to the equations on
pages 57 – 60 of Negnevitsky.
P(h|D) - the probability that hypotheses h is true given the data D.
 Often referred to as the posterior probability. It reflects our confidence that h is
true after we have seen the data D.
 The posterior probability reflects the influence of the data D while the prior
probability P(h) does not, it is independent of D.
P(D|h) – The probability that data D exists in when hypotheses h is true.
 Think of h as the answer to what happened in a murder case. One way to think of
this is that if h is true, what is the probability that the data (evidence) D will exist.
 D is the evidence that backs up the case, it is what proves that h is true.
 In general P(x|y) denotes the probability of x given y. It is the conditional
probability as discussed above.
P(h) – initial probability that hypotheses h holds, before we have observed the data.


Often called the prior probability of h. It will reflect any background knowledge
that we have about the correctness of hypothesis h.
If we have no initial knowledge about the hypotheses (h0…..hn), we would divide
the probability equally among the set of available hypotheses (in which h is a
member).
P(D) – The prior probability that the data D will be observed (this the probability of D
given no prior knowledge that h will hold). Remember that it is completely independent
of h.
Looking at the equation we can make some observations:


P(h|D), the probability of h being true given the presence of D increases with
commonness of h being true independently.
P(h|D), the probability of h being true given the presence of D increases with the
likelihood of data D being associated with hypotheses h. That is the higher our
confidence is in saying that data D is present only when h is true, the more we can
say for our hypothesis depends on D.
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
When p(D) is high that means our evidence is likely to exist independently of h,
so it weakens the link between h and D.
4.4 Bayes Rule in AI
Brute Force Bayes Algorithm
Here is a basic Bayes rule decision algorithm:


Initialize H a set of hypothesis such that h0……hn  H
For each hi in H calculate the posterior probability

p(hi|D) = p(D|hi) * p(hi)
p(D)
Output MAX(h0……hn)
Naïve Bayes Classifier
This classifier applies to tasks in which each example is described by a conjunction of
attributes and the target value f(x) can take any value from the set of v.
A set of training examples for f(x) is provided.
In this example we want to use Bayes theorem to find out the likelihood of playing tennis
for a given set weather attributes.
f(x)  v = (yes, no) i.e. v = (yes we will play tennis, no we will not play tennis)
The attribute values are a0…a3 = (Outlook, Temperature, Humidity, and Wind).
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To determine our answer (if we are going to play tennis given a certain set of conditions)
we make an expression that determines the probability based on our training examples
from the table.
Day
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Outlook
Sunny
Sunny
Overcast
Rain
Rain
Rain
Overcast
Sunny
Sunny
Rain
Sunny
Overcast
Overcast
Rain
Temperature
Hot
Hot
Hot
Mild
Cool
Cool
Cool
Mild
Cool
Mild
Mild
Mild
Hot
Mild
Humidity
High
High
High
High
Normal
Normal
Normal
High
Normal
Normal
Normal
High
Normal
High
Wind
Weak
Strong
Weak
Weak
Weak
Strong
Strong
Weak
Weak
Weak
Strong
Strong
Weak
Strong
Play Tennis
No
No
Yes
Yes
Yes
No
Yes
No
Yes
Yes
Yes
Yes
Yes
No
Remembering Bayes Rule:
p(h|D) = p(D|h) * p(h)
p(D)
We write our f(x) in that form:
P(Play Tennis | Attributes) = P(Attributes | Play Tennis) * P(Play Tennis)
P(Attributes)
Or
P(v|a) = P(a|v) * P(v)
P(a)
Lets look closely at P(a|v)
P(a|v) = P(a0…a3 | v0,1)
Or
P(a|v) = P(Outlook, Temperature, Humidity, Wind | Play tennis, Don’t Play tennis)
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In order to get a table with reliable measurements every combination of each attribute
a0…a3 for each hypotheses v0,1 our table would have be of size 3*3*2*2*2 = 72 and each
combination would have to be observed multiple times to ensure its reliability. Why,
because we are assuming an inter-dependence of the attributes (probably a good
assumption). The Naïve Bayes classifier is based on simplifying this assumption. That
is to say, cool temperature is completely independent of it being sunny and so on.
So :
P(a0…a3 | vj=0,1)  P(a0|v0) * P(a1|v0) * P(an|v0)
 P(a0|v1) * P(a1|v1) * P(an|v1)
or
P(a0…an | vj)  i P(ai | vj)
A more concrete example:
P(outlook = sunny, temperature = cool, humidity = normal, wind = strong | Play tennis) 
P(outlook = sunny | Play tennis) * P(temperature = cool | Play tennis) *
P(humidity = normal | Play tennis) * P(wind = strong | Play tennis)
The probability of observing P(a0…an | vj) is equal the product of probabilities of
observing the individual attributes. Quite an assumption.
Using the table of 14 examples we can calculate our overall probabilities and conditional
probabilities.
First we estimate the probability of playing tennis:
P(Play Tennis = Yes) = 9/14 = .64
P(Play Tennis = No) = 5/14 = .36
Then we estimate the conditional probabilities of the individual attributes. Remember
this is the step in which we are assuming that the attributes are independent of each other:
Outlook:
P(Outlook = Sunny | Play Tennis = Yes) = 2/9 = .22
P(Outlook = Sunny | Play Tennis = No) = 3/5 = .6
P(Outlook = Overcast | Play Tennis = Yes) = 4/9 = .44
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P(Outlook = Overcast | Play Tennis = No) = 0/5 = 0
P(Outlook = Rain | Play Tennis = Yes) = 3/9 = .33
P(Outlook = Rain | Play Tennis = No) = 2/5 = .4
Temperature
P(Temperature = Hot | Play Tennis = Yes) = 2/9 = .22
P(Temperature = Hot | Play Tennis = No) = 2/5 = .40
P(Temperature = Mild | Play Tennis = Yes) = 4/9 = .44
P(Temperature = Mild | Play Tennis = No) = 2/5 = .40
P(Temperature = Cool | Play Tennis = Yes) = 3/9 = .33
P(Temperature = Cool | Play Tennis = No) = 1/5 = .20
Humidity
P(Humidity = Hi | Play Tennis = Yes) = 3/9 = .33
P(Humidity = Hi | Play Tennis = No) = 4/5 = .80
P(Humidity = Normal | Play Tennis = Yes) = 6/9 = .66
P(Humidity = Normal | Play Tennis = No) = 1/5 = .20
Wind
P(Wind = Weak | Play Tennis = Yes) = 6/9 = .66
P(Wind = Weak | Play Tennis = No) = 2/5 = .40
P(Wind = Strong | Play Tennis = Yes) = 3/9 = .33
P(Wind = Strong | Play Tennis = No) = 3/5 = .60
Suppose the day is described by :
a = (Outlook = sunny, Temperature = cool, Humidity = high, Wind = strong)
What would our Naïve Bayes classifier predict in terms of playing tennis on a day like
this?
Which ever equation has the higher probability (greater numerical value)
P(Playtennis = Yes | (Outlook = sunny, Temperature = cool, Humidity = high, Wind = strong))
Or
P(Playtennis = No | (Outlook = sunny, Temperature = cool, Humidity = high, Wind = strong))
23
Is the prediction of the Naïve Bayes Classifier. (for brevity we have omitted the attribute names)
Working through the first equation….
P(Yes|(sunny, cool, high, strong)) = P((sunny, cool, high, strong) | Yes) * P(Yes)
P(sunny, cool, high, strong)
Now we do the independent substitution for:
P((sunny…..)|Yes)
And noting that the denominator, P(sunny, cool, high, strong), includes both:
P((sunny, cool, high, strong) | Yes) and
P((sunny, cool, high, strong) | No)
cases, our equation expands to:
=
P(sunny|Yes)*P(cool|Yes)*P(high|Yes)*P(strong|Yes) * P(yes)
P((sunny, cool, high, strong) | Yes) + P((sunny, cool, high, strong) | No)
Remember the quantities in the denominator are expanded using the independent
assumption in a similar way that the first term in the numerator.
=
(.22 * .33 * .33 * .33) * .64
(.22 * .33 * .33 * .33)*64 + (.6 * .2 * .8 * .6)*.36
=
.0051
.0051 + .0207
= .1977
Working through the second equation in a similar fashion….
P(No|(sunny, cool, high, strong)) = P((sunny, cool, high, strong) | No) * P(No)
P(sunny, cool, high, strong)
=
.0207
.0051 + .0207
= .8023
24
As we can see, the Bayes Naïve classifier gives a value of just about 20% for playing
tennis in the described conditions, and value of 80% for not playing tennis in these
conditions, therefore the prediction is that no tennis will be played if the day is like these
conditions.
5.0 Fuzzy Systems
Chapter 4 in the book – page 87.
Statements and Definitions:




Father of fuzzy logic is Lotfi Zadeh
The word fuzzy is the opposite of crisp. In our conventional programming logic
we use Boolean logic. Boolean logic uses sharp distinctions. (Yes/no, true/false,
etc.)
o Question with crisp answer: Q: Do you want a donut? A: Yes
o Fuzzy answer to same question: Q: Do you want a donut? A: Sort of.
Fuzzy logic is not logic that is fuzzy, but logic that is used to describe fuzziness.
It is the theory of fuzzy sets, sets that describe fuzziness.
Fuzzy logic is useful to us because it helps us quantify and describe the words that
people use when describing and thinking about problems and situations.
Examples:
o The car is really quick.
o Tom’s back is quite hairy.
o The motor is running really hot.
Considering the examples above, using Boolean logic we would need to define a point at
which a car is really quick, but where is that.
25
A Camaro is considered really quick to many people. If it runs a 14.1 second quarter
mile are all cars that are slower than Camaros really slow?
How many hairs does someone need to have on their back for it to be quite hairy? If
Tom’s back has 2500 hairs, and Mary’s has 2200, is Mary lucky because she does not
have a hairy back?
If we define really hot as 245 degree’s or higher, is it normal for my car to run at 243
degrees?
As can be seen from the example above, a car is either quick, or slow.
26
5.1 Fundamentals of Fuzzy Logic
The basic idea of fuzzy set theory is that an element of a fuzzy set belongs with a certain
degree of membership. Thus a proposition may be partly true or partly false. This degree
is a real number between 0 to 1.
The universe of discourse is the range of all possible values applicable to a chosen
variable.
The classic example is the set of tall men. The elements of the set are all men, but they
have varying degrees of membership based on their heights. The universe of discourse is
their heights.
Name
Chris
Mark
John
Tom
David
Mike
Bob
Steven
Bill
Peter
Degree of membership
Crisp
Fuzzy
1
1.0
1
1.0
1
.98
1
.82
0
.78
0
.24
0
.15
0
.06
0
.01
0
.00
Height (cm)
208
205
198
181
179
172
167
158
155
152
Below we graph the degree of membership versus height for both Crisp and Fuzzy sets.
27
5.2 Set definitions
Crisp



Let X be the universe of discourse. x is a member of X
Let A be a crisp set.
The membership function fA(x) determines if x is a member of A.
fA(x) : X → 0, 1
where
fA(x) = 1, if x is a member of A, 0 if x is not a member of A.
Fuzzy



Let X be the universe of discourse. x is a member of X
Let A be a fuzzy set.
The membership function µA(x) determines the degree of membership element x
has in fuzzy set A.
µA(x) : X → [ 0, 1]
µA(x) = 1 if x is totally in A;
28
µA(x) = 1 if x is not in A;
0 < µA(x) < 1 if x is partly in A.
5.3 Defining membership functions
So the key here is how to define the membership function µA(x)
Lets take another example:
Universe of discourse X – men’s heights
X = {x1, x2, x3, x4, x5}
A is a crisp subset of X such that A = { x2, x3 }
Membership to A can be described by :
A = {(x1, 0), (x2, 1), (x3, 1), (x4, 0), (x5, 0)}
That is A is a set of pairs {(xi, µA(xi))} where µA(xi) is the membership function of
element xi in the subset of A.
29
Representation of Fuzzy sets:
A = { (x1, µA(x1), (x2, µA(x2), ………(xn, µA(xn)}
Or
A = { µA(x1)/ x1, µA(x2)/ x2,………… µA(xn)/ xn}
The function µA is the function that determines an elements degree of membership in the
set. Typical functions are Gaussian, Sigmoid, etc. Linear functions are used commonly,
they take less computation time, thus the linear fit function. A fit vector can be used.
Below are the fit vectors for the ‘short, average, tall men’ graph.
For the above example the fit vectors are:
Tall men = { 0/180, .5/185, 1/190} = {0/180, 1/190}
Short men = {1/160, .5/165, 0/170} = {1/160, 0/170}
Average men = {0/165, 1/175, 0/185}
30
5.4 Linguistic Variables and Hedges
Example –
John is tall.
The linguistic variable John takes the linguistic value tall.
We can use these variables in fuzzy expert systems.
IF wind is strong
THEN sailing is good
IF John is tall
THEN John runs fast
We use hedges (very, somewhat, quite, more or less, etc) to modify our linguistic
variables.
tall
height
170
180
181
182
183
184
185
186
187
188
189
190
200
linear
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1
a little
(µA(x)^1.3)
Very
(µA(x)^2)
more or
less
sqrt(µA(x))
0
0
0.050119
0.123407
0.209054
0.303863
0.406126
0.51475
0.628966
0.748199
0.871998
1
1
0
0
0.01
0.04
0.09
0.16
0.25
0.36
0.49
0.64
0.81
1
1
0
0
0.316228
0.447214
0.547723
0.632456
0.707107
0.774597
0.83666
0.894427
0.948683
1
1
31
Linguistic varialbles and hedges
1.2
1
membership
0.8
tall (linear)
0.6
tall (a little)
tall (very)
0.4
tall (more or less)
0.2
0
170
175
180
185
190
195
200
-0.2
height
To calculate these variable values use the line equation to find the linear function:
y = m*x + b
m = (y1 – y0)/(x1 – x0)
b = y0 – m*x0
Then do the operation that the hedge calls for.
5.45
Operations on Fuzzy sets – see page 97
5.5 Putting it all together
Fuzzy Rules:
IF x is A
THEN y is B
x, y : linguistic variables (crisp)
A, B : linguistic values determined by fuzzy sets on the universe of discourses X, and Y.
32
Example:
IF speed over speed limit is somewhat fast
THEN ticket is very expensive
speed over speed limit (somewhat fast)
1.2
1
0.8
0.6
speed
0.4
0.2
0
0
10
20
30
40
50
60
70
80
ticket expense (very expensive)
1.2
1
0.8
0.6
ticket expense
0.4
0.2
0
0
20
40
60
80
100
120
140
Let speed over speed limit = 30mph
That gives a degree of membership of .77 in the somewhat fast set.
33
160
Then using the .77 degree of membership in the somewhat fast set maps to a ticket
costing between 80 and 100 dollars in the ticket very expensive set.
speed over
0
5
10
15
20
25
30
35
40
45
50
55
60
65
70



fast 0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1
1
1
1
ticket
cost
fast 1
0
0.316228
0.447214
0.547723
0.632456
0.707107
0.774597
0.83666
0.894427
0.948683
1
1
1
1
1
20
40
60
80
100
120
140
expensive
0.2
0.4
0.6
0.8
1
1
1
expensive
2
0.04
0.16
0.36
0.64
1
1
1
To find the exact value of the ticket, you would need to interpolate between 80 and
100 dollars.
You need to make sure that you can do this.
What are the fit vectors for these two sets?
How does it all work?
Rule 1:
IF
project_funding is adequate
OR
project_staffing is small
THEN risk is low
Rule 2:
IF
project_funding is marginal
AND project_staffing is large
THEN risk is normal
Rule 3:
IF
project_funding is inadequate
THEN risk is high
Rule 1:
IF
x is A3
34
OR
y is B1
THEN z is C1
Rule 2:
IF
x is A2
AND y is B2
THEN z is C2
Rule 3
IF
x is A1
THEN z is C3
adequate
0
0
0
0
0.33
0.67
1
0.67
0.33
0
0
low
1
1
1
0.8
0.6
0.4
0.2
0
0
0
0
high
0
0
0
0
0
0
0
0.25
0.5
0.75
1
staffing
percent
0
10
20
30
40
50
60
70
80
90
100
small
1
1
1
0.8
0.6
0.4
0.2
0
0
0
0
large
0
0
0
0
0.2
0.4
0.6
0.8
1
1
1
risk
percent
0
10
20
30
40
50
60
70
80
90
100
low
1
1
0.66
0.33
0
0
0
0
0
0
0
high
0
0
0
0
0
0
0
0.33
0.66
1
1
normal
0
0
0
0.33
0.67
1
0.67
0.33
0
0
0
funding
1.2
1
degree of mebership
funding
percent
0
10
20
30
40
50
60
70
80
90
100
0.8
adequate
0.6
low
high
0.4
0.2
0
0
20
40
60
80
percent of funding
35
100
120
staffing
degree of membership
1.2
1
0.8
small
0.6
large
0.4
0.2
0
0
20
40
60
80
100
120
percent staffing
risk
degree of membership
1.2
1
0.8
low
0.6
high
normal
0.4
0.2
0
0
20
40
60
80
percent risk
5.5 Mamdani-style inference
1.
2.
3.
4.
Fuzzification
Rule Evaluation
Aggregation of rule outputs
Defuzzificaiton
36
100
120
Example:
Step : Fuzzify inputs
Let project funding = 35%
µ(inadequate_funding) = µ(A1) = .5;
µ(marginal_funding) = µ(A2) = .2;
µ(adequate_funding) = µ(A3) = 0.0;
Let project staffing = 60%
µ(staffing_small) = µ(B1) = .1;
µ(staffing_large) = µ(B2) = .7;
Step 2: Evaluate the rules
When we evaluate the rules, we must evaluate each all if parts (antecedents) and find
values for the then parts (consequents). For conjunctions like AND and OR we do
operations like min and max.
Rule 1:
(A3 OR B1) ≈ max(0, .1) = .1
Mapping .1 to the risk_low set we get:
C1 = .1
Rule 2:
(A2 AND B2) ≈ min(.2, .7) = .2
Mapping .2 to the risk_normal set we get:
C2 = .2
Rule 3:
A1 = .5
C3 = .5
37
Step 3: Aggregation of consequents
In this step we unify the outputs of all the rules.
In our example we clipped the top off each of our 3 risk sets, so now we sum these areas
into one resultant area. See the bottom of figure 4.10 on page 108.
Step 4: Defuzzification
In order to turn our resultant fuzzy sum (we summed the clipped fuzzy sets from each of
our rules in 1 big fuzzy resultant set) we find the center of gravity (COG) of the set.
What we mean by center of gravity is that we want to locate the point at where we can
slice the set into two equal pieces.
We can use the following formula to get an estimate of the COG.
COG ≈ ∑ab µA(x)x / ∑ab µA(x)
µA(x)
x
b
a
the consequent of each of our rules
the x axis value of the set
the upper range of the x domain of the set
the lower range of the x domain of the set
x
y1
9.99
10
20
30
30.001
x
0
0.1
0.1
0.1
0
y2
30.001
30.001
40
50
60
60.001
0
0.3
0.3
0.3
0.3
0
38
For our example problem:
COG = (0 + 10 + 20)*.1 + (30 + 40 + 50 + 60)*.2 + (70 + 80 + 90 + 100)*.5
.1 + .1 + .1 + .2 + .2 + .2 + .2 + .5 + .5 + .5 +.5
= 3 + 36 + 170
3.1
= 209/3.1 = 67.4%
So the aggregate risk is 67.4%
Assignment questions – page 126
1, 2, 3, 4, 5, 6, 7, 8, 11
6.0 Evolutionary Systems/Non Deterministic Searching
Chapter 7, page 219
Now we change our tact from use of a-priori knowledge to one of adaptation.
Evolutionary Computation consists of
 Genetic Algorithm – Computer optimization based on Darwinian processes in
biology.
 Evolutionary strategies – Similar to Genetic Algorithm but uses statistically
generated offsets to vary solution.
 Genetic Programming – Use of Genetic algorithm to generate code.
Based on:
 Darwin’s classical theory of evolution
 Weismanns’s theory of natural selection
 Mendals concept of genetics
Process
 Reproduction
 Mutation
 Competition
 Selection
Species undergo various pressures:
 Environmental
39


Competition for resources among:
o Themselves
o Other species
Dangers from other species
Topics include:



Genetic algorithm
Genetic programming
Simulated annealing
6.1 Basic Genetic algorithm
40
Pseudo Code
#define popSize 100
#define chromoSize 20
#define numGens 100
#define MU 10
char population[popSize][chromoSize]
char newPop[popSize][chromoSize]
int scores[popSize]
void main()
{
int j, k
int mom, dad;
/* Set up random number generator. */
srand(time);
Generate Population(popSize)
for j = 1 to numGens
rankPop(popSize, chromoSize, "hello tv land!)
sortLo2Hi(popSize)
for k = 1 to popSize
kid = k
mom = getMom(popSize)
dad = getDad(popSize)
crossOver(mom, dad, kid, chromoSize)
mutate(MU, chromoSize, kid)
}
population = newPop
}
}
void Create Chromosome (int chromoIndex, int chromoSize)
{
range = high limit - lower limit
for j = 1 to chromoSize
chromo[chromoIndex][j] = rand()%range + lower limit
41
end for
}
void Generate Population (int popSize)
{
for j = 1, popSize
population [j] = Create a chromosome
end for
}
int Get fitness (int chromoIndex, int chromoSize, char fitString)
{
int j;
int sum;
sum = 0;
for j = 1 to chromoSize
sum = sum + abs(fitString[j] - population[chromoIndex][j])
end for
return sum;
}
void rankPop(int popSize, int chromoSize, char *string)
{
int j;
for j = 1 to popSize
scores[j] = Get fitness (j, chromoSize, string);
end for
}
void sortLo2Hi(int popSize)
{
int j,k
for j = 1 to popSize
for k = j + 1 to popSize
if (scores[j] < scores[k]) {
exchange(scores[j], scores[k])
exchange(population[j], population[k])
42
end if
end for k
end for j
}
int getMom(int popSize)
{
top = .3 * popSize
mom = rand() % top
return mom;
}
int getPop(int popSize)
{
getMom(popSize)
}
void crossOver(int mom, int dad, int kid, int chromoSize)
{
int cross, j;
cross = rand() % chromoSize
for j = 1 to cross
newPop[kid][j] =population[mom][j]
end j
for j = cross to chromoSize
newPop[kid][j] = population[dad][j]
end j
}
void mutate(int chance, int chromosize, int kid)
{
int muGene;
if (rand() % 100 < chance)
muGene = rand() % chromoSize
.......Alter newPop[kid][muGene] ......
end if
}
43
Example – Spell “hello tv land”
Generations: 100
Population: 100.
Mutation rate: 1
hello tv land
D≤N{╫WnX(╬q?I generation
0; best score 608.000000
'M[ª}ëïï‼→Çg generation
1; best score 460.000000
ÇÇsÇÇÇÇÇÇÇÇg generation
2; best score 431.000000
ÇÇsÇÇ►ïì‼▬Çj generation
3; best score 355.000000
fÇÇÇÇ►ïì‼▬Çj generation
4; best score 342.000000
fÇÇÇÇ►|Ç~|Çgk► generation
5; best score 320.000000
`ÇÇÇÇ►ïì‼|~gk► generation 6; best score 259.000000
fÇÇÇê►ïÇ‼|sgh► generation
7; best score 246.000000
fÇÇÇÇ►|ì‼|}gh► generation
8; best score 246.000000
fÇÇÇÇ►|Ç‼|~gk► generation 9; best score 237.000000
SÇÇÇÇ►ïÇ‼u~gh► generation
10; best score 223.000000
SpÇÇê►ïÇ▬|sgh☼ generation
11; best score 209.000000
SÇ⌂ÇÇ►|Ç‼|sgh► generation
12; best score 203.000000
SpÇyÇ►ïÇ‼usgh► generation
13; best score 189.000000
SpÇyÇ►|Ç‼ufgh► generation
14; best score 161.000000
RiÇÇÇ►|Ç‼ufgh► generation 15; best score 160.000000
SpÇyy►|Ç‼ufgh► generation 16; best score 154.000000
SpÇyÇ►|x‼ufgh► generation
17; best score 153.000000
SiÇwÇ►|z‼ufgh► generation
18; best score 146.000000
SiÇwÇ►|x‼ufgh► generation 19; best score 144.000000
RiÇyÇ►|x‼ufgh↔ generation
20; best score 132.000000
RhÇly►|Ç ufgh► generation
21; best score 119.000000
ShÇjy►|v‼ofgh↔ generation
22; best score 106.000000
ShÇjy►|v‼ofgh↔ generation
23; best score 106.000000
RhÇly►|z ufgh↔ generation
24; best score 100.000000
Oisy|∟yz ufgh↔ generation
25; best score 86.000000
Oisjy∟yz ufgh↔ generation
26; best score 72.000000
Oisjy∟yz ofgh↔ generation
27; best score 66.000000
Oisjy∟yv ufjh▲ generation
28; best score 64.000000
Oisjy∟yv ufjh▲ generation
29; best score 64.000000
Oiljy∟yv ufjh▲ generation
30; best score 57.000000
Khljy∟|v u`gh▲ generation
31; best score 54.000000
Khljy∟|v u`jh▲ generation
32; best score 51.000000
Ohljy∟yv u`kh generation
33; best score 49.000000
Khlky∟zv ufmh generation
34; best score 47.000000
Khljy∟yv s`kh generation
35; best score 43.000000
Kills∟yv u`kh▲ generation
36; best score 40.000000
Khljl∟yv u`lh generation
37; best score 37.000000
Kilks∟yv qclh generation
38; best score 35.000000
Kdlls∟yv s`mh generation
39; best score 31.000000
44
Kdlls∟yv h`mh generation
40; best score 28.000000
Kdlls▼yv g`mh generation
41; best score 26.000000
Kdllo▼yt q`mh generation
42; best score 24.000000
Kdllo▼yv q`mh generation
43; best score 22.000000
Hdllo▼yt h`mh generation
44; best score 20.000000
Hdllo▼yv g`me generation
45; best score 16.000000
Hdllo▼yv i`me generation
46; best score 14.000000
Hdllo▼yv j`me generation
47; best score 13.000000
Heljo!uv j`me generation
48; best score 10.000000
Hello!uv j`me generation
49; best score 8.000000
Hello!uv j`me generation
50; best score 8.000000
Hello▼sv l`me generation
51; best score 6.000000
Hello▼uv m`me generation
52; best score 7.000000
Hello▼uv m`me generation
53; best score 7.000000
Hello!tv m`me generation
54; best score 6.000000
Hello!uv m`md generation
55; best score 6.000000
Hello!tv m`md generation
56; best score 5.000000
Hello!tv mamd generation
57; best score 4.000000
Hello tv m`md generation
58; best score 4.000000
Hello!tv lamd generation
59; best score 3.000000
Hello tv lamd generation
60; best score 2.000000
Hello tv lamd generation
61; best score 2.000000
Hello tv lamd generation
62; best score 2.000000
Hello tv lamd generation
63; best score 2.000000
Hello tv lamd generation
64; best score 2.000000
Hello tv lamd generation
65; best score 2.000000
Hello tv land generation
66; best score 1.000000
Hello tv land generation
67; best score 1.000000
Hello tv lamd generation
68; best score 2.000000
Hello tv lamd generation
69; best score 2.000000
Hello tv lamd generation
70; best score 2.000000
Hello tv lamd generation
71; best score 2.000000
Hello tv lamd generation
72; best score 2.000000
Hello tv lamd generation
73; best score 2.000000
Hello tv lamd generation
74; best score 2.000000
Hello tv lamd generation
75; best score 2.000000
Hello tv land generation
76; best score 1.000000
Hello tv land generation
77; best score 1.000000
Hello tv land generation
78; best score 1.000000
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Hello tv land generation
Hello tv land generation
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6.2 Discussion of Parent Selection Functions
We talk about several methods of parent selection:
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Elite
Roulette
Polygamy
Dog and Cat
Alien
6.3 Evolutionary Strategies (page 242)
Similar to GA.
Major differences:
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Uses only mutation
Population consists of 1 individual, many parameters.
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6.4 Simulated Annealing
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6.5 Genetic Programming
6.6 Fitness Functions
7.0 Neural Networks
Book Sections: 6.1 to 6.3; pages 163 to175

Brief intro to neural networks
o Nerve cells transmit information to and from various parts of the body.
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o As creatures get more complex and thing called cephalization occurs. This is
when nerves tend to lead towards a central point.
o The more advanced creatures are, the higher the degree of cephalization.
o A little biology
 Nerve cell
o Artificial Neural Network
o Structure of ANN
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o Supervised learning
 PERCEPTRON
 y = x1* w1 + x2* w2….. xn* wn
 e(p) = y(d) – y(p); p = 1,2, 3, 4,….
 wi(p+1) = wi(p) + α* xi(p) * e(p)
 PERCEPTRON training algorithm
1. Create training set, each example needs to have a desired
output matched to it.
2. Initialization (set weights to random numbers (-1 to 1)
3. For each training example
a. Present to perceptron
b. Calculate y
c. Execute transfer function
d. Adjust weights.
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