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- Correct answer will be upvoted else Multiple Downvoted. Don't submit random answer. Computer science. Ridbit begins with an integer n. In one action, he can perform one of the accompanying tasks: partition n by one of its appropriate divisors, or take away 1 from n in case n is more prominent than 1. An appropriate divisor is a divisor of a number, barring itself. For instance, 1, 2, 4, 5, and 10 are appropriate divisors of 20, however 20 itself isn't. What is the base number of moves Ridbit is needed to make to decrease n to 1? Input The principal line contains a solitary integer t (1≤t≤1000) — the number of experiments. The main line of each experiment contains a solitary integer n (1≤n≤109). Output For each experiment, output the base number of moves needed to lessen n to 1.Given the following function, what happens if a[] contains just one element that doesn't match val? int binarySearch(int a[], int first, int last, int val){ if (first > last) return -1; int middle = (first + last) / 2; if (a[middle] == val) return middle; if (a[middle] < val) return binarySearch(a, middle+1, last, val); else return binarySearch(a, first, middle-1, val);} Group of answer choices binarySearch never calls itself again and terminates (recursion never happens) binarySearch calls itself once then terminates (recursion happens once) binarySearch calls itself twice then terminates (recursion happens twice) binarySearch calls itself 3 times then terminates (recursion happens 3 times)5. Let F(n) denote the Fibonacci numbers. Prove the following property for all n ≥ 1 (Hint: Induction)F(1) + F(2) + · · · + F(n) = F(n + 2) − 1 7. The binary search algorithm is used with the following sorted list, with parameter x having the value“Chicago”. Name the elements against which x is compared.Boston, Charlotte, Indianapolis, New Orleans, Philadelphia, San Antonio, Yakima 8. The binary search algorithm is used with the following sorted list, with parameter x having the value“flour”. Name the elements against which x is compared.butter, chocolate, eggs, flour, shortening, sugar
- Correct answer will be upvoted else downvoted. Computer science. Presently Nezzar has a beatmap of n particular focuses A1,A2,… ,An. Nezzar might want to reorder these n focuses so the subsequent beatmap is great. Officially, you are needed to find a change p1,p2,… ,pn of integers from 1 to n, to such an extent that beatmap Ap1,Ap2,… ,Apn is great. In case it is unthinkable, you ought to decide it. Input The primary line contains a solitary integer n (3≤n≤5000). Then, at that point, n lines follow, I-th of them contains two integers xi, yi (−109≤xi,yi≤109) — directions of point Ai. It is ensured that all focuses are unmistakable. Output In case there is no arrangement, print −1. In any case, print n integers, addressing a legitimate change p. In case there are numerous potential replies, you can print any.Count consecutive summers def count_consecutive_summers(n): Like a majestic wild horse waiting for the rugged hero to tame it, positive integers can be broken down as sums of consecutive positive integers in various ways. For example, the integer 42 often used as placeholder in this kind of discussions can be broken down into such a sum in four different ways: (a) 3 + 4 + 5 + 6 + 7 + 8 + 9, (b) 9 + 10 + 11 + 12, (c) 13 + 14 + 15 and (d) 42. As the last solution (d) shows, any positive integer can always be trivially expressed as a singleton sum that consists of that integer alone. Given a positive integer n, determine how many different ways it can be expressed as a sum of consecutive positive integers, and return that count. The number of ways that a positive integer n can be represented as a sum of consecutive integers is called its politeness, and can also be computed by tallying up the number of odd divisors of that number. However, note that the linked Wikipedia de0inition…Binary Search? When Jojo was a little kid, his teacher asked him a question. What was the answer of 1^2+2^2+3^2+...+N^2? Of course that problem was hard enough for a primary student. Buts ince he is a university student, and he knows that the answer is 1/6∗N∗(N +1)∗(2N +1), such problem is to easy for him. Now Jojo came up with a new problem. If he has an integerM, what is the smallest integer N such that 1^2 + 2^2 + 3^2+ ...+N^2 is greater than or equal to M. After thinking for some time, he came up with a solution, but his solution takes a lot of time. Since you are his nemesis, he challenges you to solve the problem quicker than him. Of course you accept his challenge! Format InputThe first line is an integer T representing the number of test cases.For each test case there will be 1 line consisting of an integerM. Format OutputFor each test case output “Case #X: N”. X is the test case number andN is the smallest integer such that 1^2 + 2^2 + 3^2+ ...+N^2is greater than or equal to…
- 1. def f(x):"""Evaluates function `f(x) = x^2 - 15 \sin(x * \pi/3)`Parameters----------x : array_likeInput(s) to the functionReturns-------out : ndarrayFunction `f`, evaluated at point(s) x"""# your code here 2. def grad_f(x):"""Evaluates gradient of a function `f(x) = x^2 - 15 \sin(x * \pi/3)`Parameters----------x : array_likePoint(s) at which gradient should be avalueatedReturns-------out : ndarrayGradient of the function `f` evaluaed at point(s) x"""# your code here # TEST 1. f(x)assert f(0.) == 0.assert np.allclose(f(np.array([2.5, 7.5])), np.array([-1.25, 41.25])) assert np.allclose(f(np.arange(-10, 10, 1)), np.arange(-10, 10, 1)**2 - 15*np.sin(np.pi/3*np.arange(-10, 10, 1)))x_min = 1.33668375assert np.allclose(f(x_min), -12.9944407) # TEST 2. grad_f(x)assert np.allclose(grad_f(0.), -15.7079)assert isinstance(grad_f(0.), float) assert np.allclose(grad_f(0), - 5*np.pi*np.cos(0)), 'Gradient at point 0 is wrong'assert np.allclose(grad_f(np.arange(-10, 10, 1)), 2*np.arange(-10,…Can you please help update this? //A6Utilities.java import java.io.IOException;import java.nio.file.Files;import java.nio.file.Paths;import java.security.SecureRandom;import java.util.List;import java.util.TreeMap;import java.util.function.Function;import java.util.stream.Collectors;public class A6Utilities {/*** Returns the corresponding letter grade. Don't modify this method, simply use it when converting single* grades to their equivalent letter grade** @param grade the integer grade to convert* @return the letter grade equivalent to the integer grade*/private static char letterGrade(final int grade) {if (grade < 0) return '?';else if (grade < 60) return 'F';else if (grade < 70) return 'D';else if (grade < 80) return 'C';else if (grade < 90) return 'B';else if (grade <= 100) return 'A';else return '?';}/*** The method uses Java Streams only. Don't use any looping structures, conditional statements are OK (IF, TERNARY, SWITCH)* <p>* Generates a list of secure…Count consecutive summers def count_consecutive_summers(n): Like a majestic wild horse waiting for someone to come and tame it, positive integers can be broken down as sums of consecutive positive integers in various ways. For example, the integer 42 often used as placeholder in this kind of discussions can be broken down into such a sum in four different ways: (a) 3 + 4 + 5 + 6 + 7 + 8 + 9, (b) 9 + 10 + 11 + 12, (c) 13 + 14 + 15 and (d) 42. As the last solution (d) shows, any positive integer can always be trivially expressed as a singleton sum that consists of that integer alone. Given a positive integer n, determine how many different ways it can be expressed as a sum of consecutive positive integers, and return that count. The count of how many different ways a positive integer n can be represented as a sum of consecutive integers is also called its politeness, and can be alternatively computed by counting how many odd divisors that number has. However, note that the linked…
- Euclid’s algorithm (or the Euclidean algorithm) is an algorithm that computes thegreatest common divisor, denoted by gcd, of two integers. Below are the original versions ofEuclid’s algorithm that uses repeated subtraction and another one that uses the remainder.int gcd_sub(int a, int b){ if (!a) return b; while (b) if (a > b) a = a – b; else b = b – a; return a;}int gcd_rem(int a, int b){ int t; while (b) { t = b; b = a % b; a = t; } return a;}1. Trace each of the above algorithm using specific values for a and b.2. Compare both algorithms.Implement the sieve of Eratosthenes: a function for computing prime numbers, known to the ancient Greeks. Choose an integer n. This function will compute all prime numbers up to, and including, n. First insert all numbers from 1 to n into a set. Then erase all multiples of 2 (except 2); that is, 4, 6, 8, 10, 12, ... . Erase all multiples of 3, that is, 6, 9, 12, 15, ... . Go up to ?⎯⎯√n. The remaining numbers are all primes. Make sure that you use functions in your solution, including a main function. Define def sieveOfEratosthenes(n) that receives the maximum value and returns a list of all prime numbers less than or equal to this value in main() print the list of prime numbers.1.Given a positive integer, N, the ’3N+1’ sequence starting from N is defined as follows:If N is an even number, then divide N by two to get a new value for NIf N is an odd number, then multiply N by 3 and add 1 to get a new value for N.Continue to generate numbers in this way until N becomes equal to 1For example, starting from N = 3 the complete ’3N+1’ sequence would be:3, 10, 5, 16, 8, 4, 2, 1Write c++ code to ask the user to enter a positive integer (N) in the main() function. Write a function sequence()that receives the integer value N and display the ‘3N+1’ sequence starting from the integer value that wasreceived (entered by the user). The function must also count and return the numbers that the sequenceconsists of. The returned value must be displayed from the main() function.
