Let L₁ = {aa, bb}, L₂ = {A, a, aa}, L3 = {x | x = {a,b}* ^ |x| ≤ 3} = {A, a, b, aa, ab, ba, bb, aaa, aab, aba, abb, baa, bab, bba, bbb}. Evaluate: (L₁n L3) · L2 = •
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- 1. Big-O Notation Let f and g be functions from the set of integers or the set of real numbers to the set of real numbers. We say that f ( x ) is O ( g ( x ) ), read as "f ( x ) is big-oh of g ( x )", if there are constants C and k such that | f ( x ) | ≤ C | g ( x ) | whenever x > k. (a) Show that f(x) = x2 + 2x + 1 is O(x2) Solution: When x>1; ? 2 (1 + 2 ? + 1 ? 2 ) < ? 2 (1 + 2 1 + 1 1 2 ) = 4? 2 So, ??? ? > 1, ? 2 + 2? + 1 < 4? 2 From the definition 0 ≤ f(x) ≤ cg(x) for x≥1 Hence, for N0 = 1; c=4; and g(x)=x2 for N0 = 2; c=3; and g(x)=x2 for N0 = 3; c=2; and g(x)=x2 … Therefore, ? ? + ?? + ? = ?(? ? ) O(g(x)) = {f(x)|there exist positive constant c and N0 such that 0 ≤ f(x) ≤ cg(x) for all x≥N0} 2. Show that 7x2 is O(x3). 3. Suppose there are x number of boxes to be delivered to x number of household that is 2km apart, what is the distance travelled by the transport delivery service? 4. In number 3, suppose that each boxes…Q. No. 2: Let Σ = {a, b, c}. a. Draw a DFSA that rejects all words for which the last two letters match. b. Draw a DFSA that rejects all words for which the first two letters match.Assume the following scoring matrix. A T C G - A 1 -1 -1 -2 -1 T 1 -1 -1 -1 C 2 -1 -1 G 1 -1 Fill out the dynamic programming table for the global alignment between sequences v = TAGCTCCG and w = GCATCCA using the scoring matrix above. Use the following recursive scoring formula. w= G C A T C C A 0 1 2 3 4 5 6 7 v= 0 T 1 A 2 G 3 C 4 T 5 C 6 C 7 G 8 What is the score and global alignment between v and w?
- QUESTION 4 State whether the following statement is true or false : there exists x element of R comma space there exists y space element of R space left parenthesis x plus y equals 2 right parenthesis space logical and left parenthesis x minus y equals 3 right parenthesis True False"Equatable and Comparable" in the Swift Programming: The Big Nerd Ranch Guide (2nd Ed.) e-book: Point’s current conformance to Comparable yields some confusing results. let c = Point(x: 3, y: 4) let d = Point(x: 2, y: 5) let cGreaterThanD = (c > d) // false let cLessThanD = (c < d) // false let cEqualToD = (c == d) // false As the above example demonstrates, the trouble arises in comparing two points when one point’s x and y properties are not both larger than the other point’s. In actuality, it is not reasonable to compare two points in this manner. Fix this problem by changing Point’s conformance to Comparable. Calculate each point’s Euclidean distance from the origin instead of comparing x and y values. This implementation should return true for a < b when a is closer to the origin than b. Use the formula shown in below Figure 25.1 to calculate a point’s Euclidean distance. Figure 25.1 Euclidean distance Euclidean distanceHi! I get this error message with this code. Can you help me? # Write your solution here def who_won(game_board): # Initialize counters for each player's encircled area player1_area = 0 player2_area = 0 # Iterate through each square in the game board for i in range(len(game_board)): for j in range(len(game_board[i])): # Check if the square is encircled by player 1's game pieces if game_board[i][j] == 1: # Check if all four surrounding squares are also player 1's game pieces if (i > 0 and game_board[i-1][j] == 1) and (i < len(game_board)-1 and game_board[i+1][j] == 1) and (j > 0 and game_board[i][j-1] == 1) and (j < len(game_board[i])-1 and game_board[i][j+1] == 1): player1_area += 1 # Check if the square is encircled by player 2's game pieces elif game_board[i][j] == 2: # Check if all four surrounding squares are also player…
- Not pseudocode Suppose the economies of the world use a set of currencies C1, . . . , Cn; think of these as dollars, pounds, Bitcoin, etc. Your bank allows you to trade each currency Ci for any other currency Cj, and finds some way to charge you for this service. Suppose that for each ordered pair of currencies (Ci, Cj ), the bank charges a flat fee of fij > 0 dollars to exchange Ci for Cj (regardless of the quantity of currency being exchanged). Describe an algorithm which, given a starting currency Cs, a target currency Ct, and a list of fees fij for all i, j ∈ {1, . . . , n}, computes the cheapest way (that is, incurring the least in fees) to exchange all of our currency in Cs into currency Ct. Also, justify the its runtime. [We are expecting a description of the algorithm, as well as a brief justification of its runtime.]Write a classifier algorithm for p(Y |X,α)4. Let N(x) be the statement “x has visited North Dakota,” where the domain consists of the students in your school. Express each of these quantifications in English. a) ∃x N(x) b) ∀x N(x) c) ˺∃x N(x) d) ∃x ˺N(x) e) ˺∀x N(x) f ) ∀x ˺N(x)
- PYTHON DATASET given x = np.array([i*np.pi/180 for i in range(60,300,4)]) np.random.seed(10) #Setting seed for reproducibility y = 4*x + 7 + np.random.normal(0,3,len(x)) Write a function inspired by sklearn’s polynomial preprocessing: (https://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.PolynomialFeatures.html) your function should have: degree and include bias parameters only. For this assignment, assume that input is a 1-dimensional numpy array. For example, if an input sample is np.array([a, b]), the degree-2 polynomial features with "include_bias=True" are [1, a, b, a2, b2].Given g = {(1,c),(2,a),(3,d)}, a function from X = {1,2,3} to Y = {a,b,c,d}, and f = {(a,r),(b,p),(c,δ),(d,r)}, a function from Y to Z = {p, β, r, δ}, write f o g as a set of ordered pairs.Simplify the following assertions (so that ¬ does not appear). a) ¬((∃a ∈ A,((∀b ∈ B, a × b = 2) ⇒ (∃b ∈ B, a + b ≠ 3))) & (∀a ∈ A, ∃b ∈ B, ((a+b = 5)∨(a−b = 5)))) b) ¬((∃a ∈ A, ∀b ∈ B, ((a + b = 3) ∨ (a - b = 3))) & (∀a ∈ A, ((∃b ∈ B, a × b = 7) ⇒ (∀b ∈ B, a + b ≠ 4)))) c) ¬((∃a ∈ A, ∀b ∈ B, ((a + b = 2) & (a - b = 2))) ∨ (∀a ∈ A, ((∃b ∈ B, a + b = 7) ⇒ (∀b ∈ B, a + b = 5)))) d) ¬((∀a ∈ A,((∃b ∈ B, a - b = 4) ∨ (∀b ∈ B, a + b ≠ 2))) & (∃a ∈ A, ∀b ∈ B, ((a+b = 5)∨(a−b = 5))))