6.5. Here are the corresponding sentences in propositional logic.
Mythical => Immortal
~Mythical => ~Immortal ^ Mammal
Immortal v Mammal => Horned
Horned => Magical
Proof whether unicorn is mythical, magical, horned.
First convert the sentences in KB into some standard from
CNF or Horn Clauses.
Doing inference in propositional logic, we can use resolution,
forward chaining or backward
chaining.
In order to proof Mythical we use backward chaining:
We have to look for a clause which implies mythical
and then check what needs
to be true in order for mythical to follow.
Since none of the clauses implies mythical
hence mythical is NIL - is not true.
in almost CNF:
1. (~Mythical v Immortal ) ^
2. (Mythical v (~Immortal ^ Mammal)^
3. (~(Immortal v Mammal) v Horned) ^
4. (~Horned v Magical)
to proof Magical :
resolving 1 and 2 yields (Immortal)
v (~Immortal ^ Mammal) which is 5. (Immortal v Mammal)
resolving 5 and 3 yields 6. Horned
resolving 6 and 4 yields 7. Mythical
Hence both Horned and Magical are True.
6.12.
You have to represent following facts using propositional
logic.
First you need a symbol for each person/job assignment
e.g. JM, JP, JE stands for Jones manager,
Jones Programmer and Jones Enginneer
SM, SP, SE and CM, CP, CE similarly. Then
you need to specify that each person
can have one job and each job can be held by only one
person.
(JM ^ ~JP ^ ~JE) v (~JM ^ JP ^ ~JE) v (~JM
^ ~JP ^ JE) states that Jones has one job.
etc for the others. From the statements you can
deduce that Smith is not married, hence he cannot
be the manager ~SM and from the other statements
you also ~JP and ~JM.
The thorough mechanical proof is rather lengthy. You can
conclude immediatelly
that JE is true and since you know that ~SM and also
~SE then SP has to be true
and then CM must be true, since that's the only job which
is left.
7.3. Here are few examples :
a) while naturally you would write Outburst => Contempt
this is not correct meaning.
you should have Outburst
^ Contempt
b) correct AnnieHallOn ^ (Interested => CanWatch)
intended Interested = >
AnnieHallOn
etc.
9.5. a. If you try to prove ThereExist h Horse (h) you get into an infinite loop
Horse (h)
/
\
Offspring(h,y)
Horse(Bluebeard)
|
|
Parent(y,h)
Offspring(Bluebeard, y)
Yes y/Bluebeard
Parent(y,Bluebeard)
h/Charlie
Offspring(Bluebeard, y)
....
b. particular rule above generates infinite loops
c. Both Charlie and Bluebeard cane proven to be horses
9.8. ForAll x Horse(x)
= > Animal(x)
ForAll x,h Horse(x) ^ HeadOf(h,x) => ThereExist y Animal(y) ^ HeadOf(h,y)
after converting to CNF , negating the conclusion, skolemization,
removing quantifiers
we get following (where H, G are Skolem constants)
1. ~Horse(x) v Animal(x)
2. Horse(G),
3. HeadOf(H,G)
4. ~Animal(y) v ~HeadOf(H,y) where H and G are Skolem constants
Resolving 4 and 3 yields 5. ~Animal(G)
Resolving 5. and 1. yields ~Horse(G)
This is a contradiction with 2.
14.3. We are given following information:
P(test | disease) = 0.99
P(~test | ~disease) = 0.99
P(disease) = 0.0001
test which means that test is positive
What needs to be computed is P(disease | test) from the available information
P(test | disease) P(disease)
P(disease | test) = ------------------------------------
P(test | disease) P(disease) + P(test | ~disease) P(~disease)
14.4. Rational beliefs are those who obey axioms of probability hence:
We
know that P(A) = 0.4 regardless B
B ~B is the probability table
and P(B) = 0.3 regardless A
A a b
~A c d
P(A)
= a + b = 0.4
P(B)
= a + c = 0.3
P(A
v B) = a + b + c = 0.5
From
there we can compute a = 0.2 , b = 0.2, c = 0.1 and d = 0.5 and any
other query.
hence
it is rational to believe that P(A ^ B) = 0.2;
if P(A v
B) = 0.7 then a = 0, but it is still rational.
if P(A
v B) = 0.8 as in the example in the text the axioms of probability are
not obeyed.
14.6. In oder to do the excercise you have to use apply the definition of conditional probablity
P(X | Y) = P( X, Y) / P(Y)
first express the left hand side of both statements in terms of their joint distribution
P(A,B,C)
P(B ,C)
P(A, B, C)
------- = P(A | C) ------
and -------
= P(A | C)
P(C)
P(C)
P(B, C)
with little more algebra you can show that they are the same.
14. 7. I did in the class.
14.11. We need two random variables LB - looked
blue and B - was blue
The statement in the excercise says that :
P(LB |
B) = 0.75 and also P(~LB | ~B) = 0.75
We
need to know probability P(B | LB) = using Bayes Rule ~ P(LB |
B) P(B) ~ 0.75 *P(B)
P(~B |
LB) ~ P(LB | ~B) P(~B) ~ (1 - P(LB | B)) P(~B) ~ 0.25*(1-P(B))
where we use
~ insted of = to denote equiaility within constant factor (scale)
which is
in this case
P(LB) . We are still missing probability of P(B),
so
if we assume that the
possibilities are
equally likely P(B) = 0.5 we can compute the probability that
taxi was blue.
19.1. First check whether XOR is linearly separable => it is not, hence we need a hidden layer.
w = 0.3
w = 0.3 ------------
\ w =-0.6
\
t = 0.5 --- -- -- t = 0.2
/
/
w = 0.3 -------- ----
w = 0.3
XOR
is like OR with the AND case ruled out , so we need just one hidden
node the which
implements the AND
. There are different choices of weights you can make as long as the
function is preserved.
19.2. This depends on the number of
examples and inputs. Initially the data may be linearly
separable,
and the hyperplane will successfully separate them. But they won't be implementing
XOR.
Once the data are inseperable the hyperplane plane will start moving around
is different direct
with
XOR (or parity data), if we have the same number of positive and negative
examples then
the
hyperplane will converge exactly in the middle such that weight configurations
will be
such that output will be 0.5. But when there is more examples of
one kind the hyperplane will
move around a lot.
14.5. Some of you asked me about the derivation
of the equation on page 429. The idea is related to
the excercise
14.5 and the story goes as follows. We know that Bayesian rule is good
for computing
the probabilities
of some hypothesis given
evidence
P(H
| E) It lets us do it by having some idea how
often when the hypothesis was observed with a
particular
evidence .etc. .. P(H|E) = P(E|H) P(H)/ P(E)
.
If we have multiple pieces of evidence E1 and E2 , then
we eventhough
we can still use Bayes rule, we would end up on the left hand side
with expression
P(E1
^ E2 | H), which would have to be represented using some conditional
probaility table by
n^2
entries . So the point of the excersise is to figure out how we can
do it by understanding
(and hence
needing to store) only individual conditional probablities P(E1|H) and
P(E2| H) .
The derivation
in formula on page 429, is using the relationship derived in excercise
14.5. b)
where you
are asked to prove that following statement holds
P(A|B,C) = P(B|A, C) P(A|C) / P(B|C)
Everytime when
you have to show some relationship regarding the conditional probabilities,
you
have to related them to joing probability, since that is what they
have in common.
So LHS can
be written in terms of joints as P(A,B,C)/P(B,C)
RHS can be
written in terms of joints as
P(A,B,C)
P(A,C) P(C)
------
. ----- . ------
P(A,C)
P(C) P(B,C)
which is the same as LHS.
Now is you
then apply the result from 14b to the derivation of P(Cavity
| Tootache ^ Catch) and
also substitute
the expression from P(Cavity|Tootach) from page 428 , you
will get the desired result.