# Instructions:
12 FRQ, written answers
3 MCQ, multiple choice
1 Matching question, algorithms (role of algorithms)
15 total questions (?)

1 A4 size double sided hand written notes allowed (Important!!!)
2 hour exam
Partial marks allowed for partially correct answers 
Bring a calculator (Important!!!)
# Part 1 Important Questions
Horn form for logic?
Why are these conditions not solvable without a truth table?

# Part 2 Important Questions 
## 1
Arithmetic assertions can be written in first order logic with the predicate symbol <, the function symbols + and x, and the constant symbols 0 and 1. Additional predicates can also be defined with bi-conditionals
a) Represent the property "x is and even number"
Ax Even(x) <=> Ey x=y+y
b) Represent the property "x is prime"
Ax Prime(x) <=> Ey,z x=y * z => y = 1 V z = 1  
c) Goldbach's conjecture is the conjecture (unproven as of yet) that "every even number is equal to the sum of two primes". Represent this conjecture as a logical sentence.
Ax Even(x)=> Ey,z Prime(y) /\ Prime(z) /\ x=y+z

# 2
Find the values for the probabilities a and b in joint probability table below so that the binary variables X and Y are independent

| X   | Y   | P(X, Y) |
| --- | --- | ------- |
| t   | t   | 3/5     |
| t   | f   | 1/5     |
| f   | t   | a       |
| f   | f   | b       |
Due to probability being max 1, we know that a + b must be 1/5
P(Yt)/P(Yf) = a/b = 3
b = 1/20
a = 3/20

# 3
idk where R comes from, look into slides about bayes theorem
Show the three forms of independence in Equation (12.11) are equivalent
P(a|b) = P(a) or P(b|a) = P(b) or P(a /\ b) = P(a) * P(b) / R(?)

First two are logically the same, just inverted

From bayes theorem
P(a | b) * P(b) = P(a) * P(b) / R(?)

P(a /\ b) = P(a | b) * P(b)

# 4
Consider the following propability distrobutions:

| A   | P(A) |
| --- | ---- |
| t   | 0.8  |
| f   | 0.2  |

| A   | B   | P(B\|A) |
| --- | --- | ------- |
| t   | t   | 0.9     |
| t   | f   | 0.1     |
| f   | t   | 0.6     |
| f   | f   | 0.4     |

| B   | C   | P(C\|B) |
| --- | --- | ------- |
| t   | t   | 0.8     |
| t   | f   | 0.2     |
| f   | t   | 0.8     |
| f   | f   | 0.2     |
Given these tables and no other assumptions, calculate the following probabilities.
a. P(a, ~b)
 = P(a) * P(~b|a)
 = 0.8 * 0.1
 = 0.08 
b. P(b)
 = P(bt|a) * P(a) + P(bt | ~a) * P(~a)
 = 0.9 * 0.8 + 0.6 * 0.2 
 = 0.84
# 5
Let A and B be Boolean Random variables. You are given the following probabilities 
P(A=true) = 0.5
P(B=true |A=true) = 1
P(B=true) = 0.75

What is P(B=true|A=false)?

# 6
Consider the XOR function of three binary input attributes, which produces the value 1 if and only if an odd number of the three input attributes has value 1.

Draw a minimal sized decision tree for the three input XOR function.

Three layer decision three, A > B > C. Output of tree would be
0 1 1 0 1 0 0 1 if on the left of the decision is always 0 and 1 is right

# 7
Consider the problem of separating N data points into +ve and -ve examples using a linear separator. Clearly this can always be done for N=2 points on a line of dimension d=1, regardless of how many points are labeled or where they are located (unless the points are in the same place)

a) Show that it can always be done for N=3 points on a plane of dimension d=2 unless they are co-linear.

b) Show that it cannot (or can we?) always be done for N=4 points on a plane of dimension d=2