For the code below, draw a picture of the program stack when the function findZero() is called the 2nd time.
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- 2. code locations Consider the following C code and the corresponding assembly code: void baz(int a, int * p) { pushq %rbp movq %rsp,%rbp subq $16,%rsp int c; C = a-*p; movl %ecx,16(%rbp) movq %rdx, 24(%rbp) movq 24(%rbp), %rax movl (%rax),%eax movl 16(%rbp),%edx subl %eax,%edx movl %edx,%eax movl %eax,-4(%rbp) cmpl $0, -4(%rbp) je L2 if (c) { int d = c*2; movl -4(%rbp),%eax addl %eax,%eax movl %eax,-8(%rbp) *p -= d; movl 24(%rbp),%rax movl (%rax),%eax subl -8(%rbp),%eax movl %eax,%edx movq 24(%rbp), %rax movl %edx, (%rax) jmp L4: (continued on next page)Q2: Simplify F(A, B, C, D) = E0,4,5,6,7,8,12)? (= %3DLet L10 = (a^bmcki k = n + m and n,m,k2 0}. Design a %3D CFG to generate L10.Then design a PDA to generate L10. This can be based on your CFG using the method described in the book, but does not have to be.
- Execute the following codes: import numpy as np mean = [0,0] cov = [[1,-0.6],[-0.6,7]] x,y = np.random.multivariate_normal(mean,cov,500).T Save x and y using save method to your computer. Read them and calculate the correlation coefficient matrix. Make a 2-dimensional scatter plot of x and y. Do the straight line fit to y vs x.Q4/Full the following blanks (1101 1010)BCD(5211) = ( )BCD(3321)=()EX-3=()16 * (1101 1010)BCD(5211) = (1110 1100)BCD(3321)=(1010110)EX-3= (A6)16 (1101 1010)BCD(5211) = (1110 0111)BCD(3321)=(1011001)EX-3= (56)16 (1101 1010)BCD(5211) = (1101 0111)BCD(3321)=(1011001)EX-3= (2B)16 (1101 1010)BCD(5211) = (1011 1110)BCD(3321)=(1000111)EX-3= (44)16 None of them IIn 0, T(n - а) + 1 п>а. T(n} =
- function result = result = ''; 1 tokens_to_str_code(token_mat, time_unit) 3 for i =1:size(token_mat,1) 4 5 if(token_mat (i,1) if(token_mat(i,2) > 4*time_unit) result = strcat(result,"-"); elseif((token_mat(i,2) > time_unit) && (token_mat(i,2) 8*time_unit) result = strcat(result,"/"); elseif((token_mat(i,2) 11 == 0) 12 13 14 4*time_unit) && (token_mat(i,2) < 8*time_unit)) 15 result strcat(result," "); %3D 16 end 17 end 18 end 19 end 20 21 Check Test Expected Got tokens %3D [1 4; е 3; 1 3; е 1; 1 3; е 1; 1 1; ө 1; 1 1 ]; time_unit = .5; disp( tokens_to_str_code( tokens, time_unit )) tokens = [ 0 4; 1 1457; 0 463; 1 1457; 0 463; 1 497; 0 1423; 1 1457; ---/..--.. --.---/..--.. о 463; 1 1457; 0 463; 1 1457; ө 3343; 1 497; ө 463; 1 497; 0 463; 1 1457; 0 1423; 1 1457; 0 463; 1 497; 0 463; 1 497; e 379 ]; time_unit = 240; disp( tokens_to_str_code( tokens, time_unit ))Check this code and make a code to plot the curve def maxProfitBrute(arr): # O(n^2) arr_len = len(arr) profit = -1 for i in range(arr_len-1): for j in range(i+1, arr_len): diff = arr[j]-arr[i] if (profit <= diff): profit = diff start = i end = j return start, end, arr[end]-arr[start] def maxProfit(prices): # O(n) arr = prices if len(arr) < 2: return 0 elif len(arr) == 2: return arr[1]-arr[0] if arr[1] > arr[0] else 0 mini = arr[0] profit = arr[1]-arr[0] for idx, item in enumerate(arr): mini = min(mini, item) profit = max(profit, item-mini) return profit arr = [5, 5] profit = maxProfit(arr) print(f'\nProfit = {profit}')HW7_2 This problem uses an interpolating polynomial to estimate the area under a curve. Fit the interpolating polynomial to the following set of points. These points are the actual values of f(x) = sin (e* – 2) 0.4 0.8 1.2 1.6 y -0.8415 |-0.4866 0.2236 0.9687 0.1874 a) Plot the function f(x) and the interpolating polynomial, using different colors. Use polyfit and polyval. Also include the data points using discrete point plotting. b) We wish to estimate the area under the curve, but this function is difficult to integrate. Hence, instead 1.6 of finding ° sin(e* – 2) dx (which is the same as finding the area under the curve sin (e* – 2) ), we will compute the area under the interpolating polynomial over the domain 0SEE MORE QUESTIONS