2025-04-09 13:22:21

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UNB/Year 5/Semester 2/CS4725/Final Review.md

@@ -0,0 +1,102 @@
+# 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