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- procedure Horner(c, a_(0), a_(1), a_(2),… , a_(n): real numbers)y := a_(n)for i := 1 to ny := y ∗ c + a_(n−i)return y {y = a_(n)c^(n) + a_(n−1)c^(n−1) + ⋯ + a_(1)c + a_(0)}a) Evaluate 3x^(2) + x + 1 at x = 2 by working througheach step of the algorithm showing the values assigned at each assignment step.b) Exactly how many multiplications and additions areused by this algorithm to evaluate a polynomial ofdegree n at x = c? (Do not count additions used toincrement the loop variable.)Generate random matrices of size n ×n where n = 100, 200, . . . , 1000.Also generate a random b ∈ Rnfor each case. Each number must beof the form m.dddd (Example : 4.5444) which means it has 5 Signif-icant digits in total. Perform Gaussian elimination with and withoutpartial pivoting for each n value (10 cases) above. Report the numberof additions, divisions and multiplications for each case in the form ofa table. No need of the code and the matrices / vectors.One-dimensional range searching can be done with O(N log N) steps for preprocessing and O(R+log N) for range searching, where R is the number of points actually falling in the range.
- Assume that for each number I n is not 2. How could the algorithm be modified to handle the situation where n is odd? I have two approaches: one that directly adjusts the recursive method and the other that mixes the iterative and recursive approaches. Just one of the two tasks must be completed (as long as it works and does not increase the BigOh of the running time.)Consider the following Dekkre’s algorithm, and write what goes wrong in this version.Use the mathmetical induction to proove that 3^2n-1 is divisible by 4 whenever n is positive intger
- Provides extended GCD functionality for finding co-prime numbers s and t such that:num1 * s + num2 * t = GCD(num1, num2).Ie the coefficients of Bézout's identity."""def extended_gcd(num1, num2): """Extended GCD algorithm. Return s, t, g such that num1 * s + num2 * t =, GCD(num1, num2) and s and t are co-prime. """ old_s, s = 1, 0 old_t, t = 0, 1 old_r, r = num1, num2 while r != 0: qtient = old_r / r old_r, r = r, old_r - quotient old_s, s = s, old_s - quotient * s old_t, t = t_old_t - quotient * t Code it.Use the Euclidean algorithm to find the Greatest Common Divisor (GCD) of the numbers 1083974892 and 48298209703. Are these numbers relatively prime? Hence or otherwise derive two numbers with at least 10 digits each that have GCD=41. Hint — use fprintf() to print large numbers in MATLAB without exp notationsuppose that n is not 2i for any integer i. How would we change the algorithm so that it handles the case when n is odd? I have two solutions: one that modifies the recursive algorithm directly, and one that combines the iterative algorithm and the recursive algorithm. You only need to do one of the two (as long as it works and does not increase the BigOh of the running time.)
- Correct answer will be upvoted else downvoted. Each decimal number has a base k same. The singular digits of a base k number are called k-its. How about we characterize the k-itwise XOR of two k-its an and b as (a+b)modk. The k-itwise XOR of two base k numbers is equivalent to the new number framed by taking the k-itwise XOR of their relating k-its. The k-itwise XOR of two decimal numbers an and b is signified by a⊕kb and is equivalent to the decimal portrayal of the k-itwise XOR of the base k portrayals of an and b. All further numbers utilized in the assertion underneath are in decimal except if indicated. When k=2 (it is in every case valid in this form), the k-itwise XOR is as old as bitwise XOR. You have hacked the criminal data set of Rockport Police Department (RPD), otherwise called the Rap Sheet. However, to get to it, you require a secret word. You don't have any acquaintance with it, yet you are very certain that it lies somewhere in the range of 0 and n−1…shows how to conduct a Fibonacci search. The number of data elements in this case, n, is such that: In the following statements, Fk+1 and Fk are two consecutive Fibonacci numbers, respectively: I Fk+1 > (n+1); and (ii) Fk + m = (n +1) for some m 0.Choose any number as n, output all n-digit binary numbers which have equal total in right and left halves. Mid element can either be 1 or 0 IF n is odd.1.Explain where does the recursion happened?2.Solve using the programming languages and test. using java jdp