T(n) = aT(^) + f(n) 1. F(n) = F(n-1) + F(n-2) donde: F(0) = 0 F(1) = 1
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Use the master Theorem to solve the problem
Step by step
Solved in 2 steps
- Construct a DFA A so that L(A) = L(N) where N is the following NFA:▸ Part 1 ▾ Part 2 Construct a truth table to prove if the argument is valid or invalid. The argument is Ovalid Ⓒinvalid P 9 pvq (pvq) ^p q ((pv q) ^p) →9Function Name: odd_even_diagParameters: a 2D list (list of lists)Returns: list of lists Description: Given a 2-dimensional matrix (list of lists) with any size (n*n), modify it according to the following rules: Find the sum of the main diagonal. If the sum is an odd number, change all the values of the given matrix (except the main diagonal) to 0. If the sum is an even number, change all the values of the given matrix (except the main diagonal) to 1. Return the resulting matrix. Example 1:If argument is: [[1, 2], [4, 3]] odd_even_diag should return: [[1, 1], [1, 3]] because the sum 1 + 3 is even. Example 2:If argument is: [[1, 2, 3], [4, 5, 6], [7, 8, 9]] odd_even_diag should return: [[1, 0, 0], [0, 5, 0], [0, 0, 9]] because the sum 1 + 5 + 9 is odd.
- if A=[1 10 30 ; 5 -1 3] choose the correct code to get the matrix B =[10 ;-1]: None of above B=A(1,:) B=A(:,2) O B=A(2,:)Define, A = Find x. Hint: use numpy.linalg.inv and the @operator for matrix multiplication. 3 6 2 b = 10 5 2 1 7 5 Y = 3 Ax + b = Y Nt(1) = 5 , t(n) = t(n/2) substitution recurrence relation
- if A=[1 10 30 ; 5 -13] choose the correct code to get the matrix B = [110 30 ; 5 -13] B=A(2,:) B=A' None of above B=A(;,2)105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 40 M ㅏ void sequence::insert(const value_type& entry) if(used capacity) { resize((1.5* capacity) + 1); } if (current_index >= used) { current_index = 0; } for (int i = used; i > current_index; i--) { data[i] = data[i-1];The formula C=numpy.add(A, B) adds the two matrices A and B and stores the result in C.• To subtract matrix B from matrix A, enter C=numpy.subtract(A, B). The result is stored in C.• C=numpy.divide(A, B): Split matrix A into two equal parts and store the result in C.• C stands for numpy.multiply(A, B): Multiply matrices A and B, then store the result in matrices C.• C=numpy.sum(A): Calculate the sum of each element in the matrix A, then store the result in c R.• C=numpy.sum(A, axis = 0): Create a vector C by adding the columns of the matrix A.• C=numpy.sum(A, axis = 1): Summarize matrix A row-by-row and store the result in the vector C. Create Python code that demonstrates the use of these techniques in a sample matrix.