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"state": {
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"type": "markdown",
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"state": {
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"file": "UNB/Year 5/Semester 2/HIST3925/Lecture Notes.md",
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"file": "UNB/Year 5/Semester 2/CS4725/Final Review.md",
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"icon": "lucide-file",
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"title": "Lecture Notes"
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"title": "Final Review"
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"UNB/Year 5/Semester 2/HIST3925/Breakout Questions.md",
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"UNB/Year 5/Semester 2/CS4725/Final Review.md",
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"UNB/Year 5/Semester 2/HIST3925/Lecture Notes.md",
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"UNB/Year 5/Semester 2/HIST3925/Breakout Questions.md",
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102
UNB/Year 5/Semester 2/CS4725/Final Review.md
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UNB/Year 5/Semester 2/CS4725/Final Review.md
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# Instructions:
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12 FRQ, written answers
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3 MCQ, multiple choice
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1 Matching question, algorithms (role of algorithms)
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15 total questions (?)
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1 A4 size double sided hand written notes allowed (Important!!!)
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2 hour exam
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Partial marks allowed for partially correct answers
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Bring a calculator (Important!!!)
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# Part 1 Important Questions
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Horn form for logic?
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Why are these conditions not solvable without a truth table?
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# Part 2 Important Questions
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## 1
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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
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a) Represent the property "x is and even number"
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Ax Even(x) <=> Ey x=y+y
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b) Represent the property "x is prime"
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Ax Prime(x) <=> Ey,z x=y * z => y = 1 V z = 1
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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.
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Ax Even(x)=> Ey,z Prime(y) /\ Prime(z) /\ x=y+z
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# 2
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Find the values for the probabilities a and b in joint probability table below so that the binary variables X and Y are independent
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| X | Y | P(X, Y) |
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| --- | --- | ------- |
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| t | t | 3/5 |
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| t | f | 1/5 |
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| f | t | a |
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| f | f | b |
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Due to probability being max 1, we know that a + b must be 1/5
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P(Yt)/P(Yf) = a/b = 3
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b = 1/20
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a = 3/20
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# 3
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idk where R comes from, look into slides about bayes theorem
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Show the three forms of independence in Equation (12.11) are equivalent
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P(a|b) = P(a) or P(b|a) = P(b) or P(a /\ b) = P(a) * P(b) / R(?)
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First two are logically the same, just inverted
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From bayes theorem
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P(a | b) * P(b) = P(a) * P(b) / R(?)
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P(a /\ b) = P(a | b) * P(b)
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# 4
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Consider the following propability distrobutions:
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| A | P(A) |
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| --- | ---- |
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| t | 0.8 |
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| f | 0.2 |
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| A | B | P(B\|A) |
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| --- | --- | ------- |
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| t | t | 0.9 |
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| t | f | 0.1 |
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| f | t | 0.6 |
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| f | f | 0.4 |
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| B | C | P(C\|B) |
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| --- | --- | ------- |
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| t | t | 0.8 |
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| t | f | 0.2 |
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| f | t | 0.8 |
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| f | f | 0.2 |
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Given these tables and no other assumptions, calculate the following probabilities.
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a. P(a, ~b)
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= P(a) * P(~b|a)
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= 0.8 * 0.1
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= 0.08
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b. P(b)
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= P(bt|a) * P(a) + P(bt | ~a) * P(~a)
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= 0.9 * 0.8 + 0.6 * 0.2
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= 0.84
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# 5
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Let A and B be Boolean Random variables. You are given the following probabilities
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P(A=true) = 0.5
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P(B=true |A=true) = 1
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P(B=true) = 0.75
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What is P(B=true|A=false)?
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# 6
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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.
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Draw a minimal sized decision tree for the three input XOR function.
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Three layer decision three, A > B > C. Output of tree would be
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0 1 1 0 1 0 0 1 if on the left of the decision is always 0 and 1 is right
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# 7
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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)
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a) Show that it can always be done for N=3 points on a plane of dimension d=2 unless they are co-linear.
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b) Show that it cannot (or can we?) always be done for N=4 points on a plane of dimension d=2
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