1. There are two webcams, X and Y that can automatically switch On or Off at any given time to record a live video streaming session. To ensure the smooth recording session, the live feed input system is programmed with certain criteria. The conditions set are; if webcam X and webcam Y are switch on, the input system can enable the recording input and the live session will start. If either webcam X or webcam Y is switched on, the input system cannot enable the recording and the live session will stop. In all other conditions, the input system will enable the recording and the live session streaming will commence. a. Based on the scenario above, generate a simple pseudocode with a function that has x and y as inputs and livestream and nolivestream as outputs. a. On a particular day, Prof. Z is intending to have a live session discussion with his students. He informs his students before the session that “If webcam X works while the other doesn't or the other way round, I’m afraid that the live discussion session would not take place”. Represent it in a simple pseudocode the scenario of what was said by Prof. Z. b. Translate the pseudocodes in part (a.) and (b.) into propositional logic statements. c. Using a Truth table, determine if what Prof. Z had stated is consistent with what the input system was programmed to do. 2. There is an island with two mythical creatures, the Komos and the Dragos. Komos loves eating visitors to the island while Dragos hates to eat visitors. One fine day, you mysteriously end up on the island and encounter two creatures, Blue and Red. The Blue creature smiles and tell you, “Don’t be afraid! We both are not human eaters”. Determine, if possible, whether the Blue and Red creatures is a Komos or a Dragos. 3. Let A, B and C denotes the sets with the universal set, U. Prove the following equivalence: π΄∩(π΅∪πΆ)=(π΄∩π΅)∪(π΄∩πΆ). Let A and B denote the sets with the universal set, U. The symmetric difference of two sets, A & B is denoted by A Δ B and is defined as such: A Δ B = (A-B) ∪ (B-A) 4. Prove using Set Builder notation that an equivalent definition of the symmetric difference of two sets A & B is A Δ B = (A∪B) − (A∩B) 5. In a set of integers between the numbers 100 and 100,000, how many of these numbers are divisible by 2, 5, 9 and 13? 6. Find the result for the following double summation. Question1 1a. Pseudocode Start x ο webcam x y ο webcam y function checkLivestream(x,y) if (x=1 and y=1) return livestream if (x=1 and y=0) return nolivestream if (x=0 and y=1) return nolivestream if (x=0 and y=0) return livestream end 1b. Live discussion pseudocode Start x ο webcam x y ο webcam y function check live discussion(x,y) if x=1 and y=1 return live discussion if x=1 and y=0 return no live discussion if x=0 and y=1 return no live discussion if x=0 and y=0 return live discussion end 1c. Let: X = webcam x is turned on Y = webcam y is turned on L = the livestream will occur D = the live discussion will occur For part 1a: π∧π =πΏ ¬π ∧ π = ¬πΏ π ∧ ¬π = ¬πΏ ¬π ∧ ¬π = πΏ For part 1b: ¬π ∧ π = ¬π· π ∧ ¬π = ¬π· π∧π =π· ¬π ∧ ¬π = π· 1d. X Y ¬π ¬π T T F F T F T F F F T T F T F T π∧π T F F T ¬π ∧ ¬π T F F T π ∧ ¬π F T T F ¬π ∧ π F T T F Based on the truth table above, the truth values of X ∧ Y = ¬X ∧ ¬Y and X ∧ ¬Y= ¬X ∧ Y, therefore what Prof. Z had stated is consistent with what the input system was programmed to do. 2. 3. πππ‘ π₯ ππ π΄ ∩ (π΅ ∪ πΆ), πβππ π₯ ππ ππ π΄ πππ (π΅ ∪ πΆ), πππ‘ π₯ ππ (π΄ ∩ π΅) ∪ (π΅ ∩ πΆ), πβππ π₯ ππ ππ π΄ ∩ (π΅ ∪ πΆ), πΌπ π₯ ππ ππ π΅, π‘βππ π₯ ππ ππ (π΄ ∩ π΅) πΌπ π₯ ππ ππ πΆ, π‘βππ π₯ ππ ππ (π΄ ∩ πΆ) ∴ (π΄ ∩ π΅) ∪ (π΄ ∩ πΆ) ⊆ π΄ ∩ (π΅ ∪ πΆ) ∴ π₯ ππ ππ (π΄ ∩ π΅) ∪ (π΄ ∩ πΆ) π΄ ∩ (π΅ ∪ πΆ) ⊆ (π΄ ∩ π΅) ∪ (π΄ ∩ πΆ) Therefore, π΄ ∩ (π΅ ∪ πΆ) ⊆ (π΄ ∩ π΅) ∪ (π΄ ∩ πΆ) 4. π΄βπ΅ = (π΄ − π΅) ∪ (π΅ − π΄) πΏππ‘ π₯ ∈ (π΄ − π΅) ∪ (π΅ − π΄) πβππ π₯ ∈ (π΄ ∪ π΅) − (π΄ ∩ π΅) π₯ ∈ (π΄ ∩ π΅)πππ π₯ ∉ (π΄ ∩ π΅) (π₯ ∈ π΄ πππ π₯ ∉ π΅) ππ (π₯ ∈ π΅ πππ π₯ ∉ π΄) (π₯ ∈ π΄ ππ π₯ ∈ π΅) πππ (π₯ ∉ π΅ ππ π₯ ∉ π΄) π₯ ∈ (π΄ ∪ π΅) − (π΄ ∩ π΅) (π₯ ∈ π΄ ππ π₯ ∈ π΅) πππ (π₯ ∉ π΅ ππ π₯ ∉ π΄) (π₯ ∈ π΄ πππ π₯ ∉ π΅) ππ (π₯ ∈ π΅ πππ π₯ ∉ π΄) π₯ ∈ (π΄ − π΅) ∪ (π΅ − π΄) ∴ (π΄ − π΅) ∪ (π΅ − π΄) ⊆ (π΄ ∪ π΅) − (π΄ ∩ π΅) (π΄ ∪ π΅) − (π΄ ∩ π΅) ⊆ (π΄ − π΅) ∪ (π΅ − π΄) Therefore, (π΄ − π΅) ∪ (π΅ − π΄) = (π΄ ∪ π΅) − (π΄ ∩ π΅) 5. 100ο100,000 a. For numbers divisible by 2 less than 100,000 100,000 ⌋ 2 = 50,000 =⌊ Numbers divisible by 2 that is less than 100 100 ⌋−1 2 = 50 − 1 = 49 =⌊ = 50,000 − 49 = 49,951 Therefore, there are 49,951 numbers that are divisible by 2 whose values are less than 100,000 and more than or equal to 100. b. For numbers divisible by 5 that is less than 100,000 100,000 =⌊ ⌋ 5 = 20,000 Numbers divisible by 5 that is less than 100 100 =⌊ ⌋−1 5 = 20 − 1 = 19 = 20,000 − 49 = 19,981 Therefore, there are 19,981 numbers that are divisible by 5 whose values are less than 100,000 and more than or equal to 100. c. For numbers divisible by 9 that is less than 100,000 100,000 =⌊ ⌋ 9 = 11,111 Numbers divisible by 9 that is less than 100 100 =⌊ ⌋ 9 = 11 = 11,111 − 11 = 11,100 Therefore, there are 11,100 numbers that are divisible by 9 whose values are less than 100,000 and more than or equal to 100. d. For numbers divisible by 13 that is less than 100,000 100,000 =⌊ ⌋ 13 = 7692 Numbers divisible by 13 that is less than 100 100 =⌊ ⌋ 13 =7 = 7692 − 7 = 7685 Therefore, there are 7625 numbers that are divisible by 13 whose values are less than 100,000 and more than or equal to 100. e. For numbers divisible by 2, 5, 9, and 13 that is less than 100,000 Prime factors: 2=2 5=5 9 =3×3 13 = 13 LCM of 2, 5, 9, and 13 = 2 × 32 × 5 × 13 =1170 ⌊ 100,000 ⌋ = 85 1170 Therefore, there are 85 numbers that are divisible by 2, 5, 9, and 13 whose values are less than 100,000 but more than or equal to 100. 6. π 5 ∑ ∑ 3ππ π=1 π=1 π = ∑[3π + 3(2)π + 3(3)π + 3(4)π + 3(5)π] π=1 π = ∑[3π + 6π + 9π + 12π + 15π] π=1 π = ∑[45π] π=1 π = 45 ∑ π π=1 Using Sum Formula π 1 ∑ π = π(π + 1) 2 π=1 1 = 45 ( π(π + 1)) 2 45π(π + 1) = 2