Given an integer n, you need to find all non-negative even numbers that are no greater thann. Please write a function named get_even_numbers(n) which returns a list of the required even numbers. You should use recursion to solve this problem. Please put your code in Q1_get_even_numbers_student.py. The following figure shows the expected outputs. n = 0, get_even_numbers(0) returns [0] n = 1, get_even_numbers(1) returns [0] n = 20, get_even_numbers(20) returns [20, 18, 16, 14, 12, 10, 8, 6, 4, 2, ®] 21 get even numbers(21) return s (20 18 16 14 12 10 4 2
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- (Program) Write a program that tests the effectiveness of the rand() library function. Start by initializing 10 counters, such as zerocount, onecount, twocount, and so forth, to 0. Then generate a large number of pseudorandom integers between 0 and 9. Each time 0 occurs, increment zerocount; when 1 occurs, increment onecount; and so on. Finally, display the number of 0s, 1s, 2s, and so on that occurred and the percentage of time they occurred.In C programming Mathematically, given a function f, we recursively define fk(n) as follows: if k = 1, f1(n) = f(n). Otherwise, for k > 1, fk(n) = f(fk-1(n)). Assume that there is an existing function f, which takes in a single integer and returns an integer. Write a recursive function fcomp, which takes in both n and k (k > 0), and returns fk(n). int f(int n);int fcomp(int n, int k){c++ Implement a function that recursively calculates the nth number in the Fibonacci sequence of numbers. The Fibonacci sequence starts at 0 and 1. Each number thereafter is the sum of the two preceding numbers. This gives us the following first ten numbers: 0,1,1,3,5,8,13,21,34, .... Input Expected output0 01 17 139 34
- Write a recursive function that, given a sequence of comparable values, returns the count of elements where the current element is less than the following ( next ) element in the given sequence. See the examples given below. def count_ordered ( seq ) : """ Input : A sequence of comparable elements Output : The number of elements that are less than the following element in the sequence Example : >>> count_ordered ( [ 1 , 2 , 3 , 4 , 5 , 6 ] ) 5 >>> count_ordered ( ( 1 , 12, 7.3 , -2,4 ) ) 2 >>> count_ordered ( 'Python' ) 2 >>> count_ordered ( [ 6 ] ) 0 >>> count_ordered ( [ ] ) 0 """ In the first example above , count_ordered ( [ 1,2,3,4,5,6 ] )the returned answer is 5 because for all the first 5 numbers the current number is less than the next number. In the second example above, count_ordered ( ( 1,12,7.3 , -2,4 ) )the…The Polish mathematician Wacław Sierpiński described the pattern in 1915, but it has appeared in Italian art since the 13th century. Though the Sierpinski triangle looks complex, it can be generated with a short recursive function. Your main task is to write a recursive function sierpinski() that plots a Sierpinski triangle of order n to standard drawing. Think recursively: sierpinski() should draw one filled equilateral triangle (pointed downwards) and then call itself recursively three times (with an appropriate stopping condition). It should draw 1 filled triangle for n = 1; 4 filled triangles for n = 2; and 13 filled triangles for n = 3; and so forth. API specification. When writing your program, exercise modular design by organizing it into four functions, as specified in the following API: public class Sierpinski { // Height of an equilateral triangle whose sides are of the specified length. public static double height(double length) // Draws a filled equilateral…Need some help with this c++ recursive function: Write the function int countEights(int n). Inside the function, compute recursively (no loops) the count of the occurrences of 8 as a digit inside the parameter n. Except that an 8 with another 8 immediately to its left counts double, so 8818 yields 4. Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6), while divide (/) by 10 removes the rightmost digit (126 / 10 is 12). Here are some examples: countEights(8) returns 1 countEights(818 returns 2 countEights(8818) returns 4
- Write a recursive function for Euclid's algorithm to find the greatest common divisor (gcd) of two positive integers. gcd is the largest integer that divides evenly into both of them. For example, the gcd(102, 68) = 34. You may recall learning about the greatest common divisor when you learned to reduce fractions. For example, we can simplify 68/102 to 2/3 by dividing both numerator and denominator by 34, their gcd. Finding the gcd of huge numbers is an important problem that arises in many commercial applications. We can efficiently compute the gcd using the following property, which holds for positive integers p and q: If p > q, the gcd of p and q is the same as the gcd of q and p % q.The following function f uses recursion: def f(n): if n <= 1 return n else return f(n-1) + f(n-2) Let n be a valid input, i.e., a natural number. Which of the following functions returns the same result but without recursion? a) def f(n): a <- 0 b <- 1 if n = 0 return a elsif n = 1 return b else for i in 1..n c <- a + b a <- b b <- c return b b) def f(n): a <- 0 i <- n while i > 0 a <- a + i + (i-1) return a c) def f(n): arr[0] <- 0 arr[1] <- 1 if n <= 1 return arr[n] else for i in 2..n arr[i] <- arr[i-1] + arr[i-2] return arr[n] d) def f(n): arr[0..n] <- [0, ..., n] if n <= 1 return arr[n] else a <- 0 for i in 0..n a <- a + arr[i] return aPlease answer fast Question 1 Answer following questions and write corresponding algorithms. Algorithm: sqrt(n) [main] 1. return sqrtHelper(Arg 1, Arg 2) Algorithm: sqrtHelper(n,m) 1. if m*m <= n 2. return m 3. else 4. return sqrtHelper(Arg 3, Arg 4) 5. endif (a) Write an recursive algorithm called isPrime(n) that takes a positive integer and returns True if it is prime and False otherwise. You must call the function sqrt(n) as a sub- algorithm. Trace your algorithm for isPrime(41). You only need to show the detail in the main algorithm and its helper function (if there is any).
- The following function f uses recursion: def f(n): if n <= 1 return n else return f(n-1) + f(n-2) 5 Let n be a valid input, i.e., a natural number. Which of the following functions returns the same result but without recursion? a) def f(n): a <- 0 b <- 1 if n = 0 return a elsif n = 1 return b else for i in 1..n c <- a + b a <- b b <- c return b f(n): a <- 0 i <- n while i > 0 a <- a + i + (i-1) return a f(n): arr[0] <- 0 arr[1] <- 1 if n <= 1 return arr[n] else for i in 2..n arr[i] <- arr[i-1] + arr[i-2] return arr[n] f(n): arr[0..n] <- [0, ..., n] if n <= 1 return arr[n] else a <- 0 for i in 0..n a <- a + arr[i] return aa) Write a non-recursive function in C++/ to multiply all even numbers from 2 to n, where n is an input to the function, and n>=2. (reminder: An even number is divisible by 2 and generates a remainder of 0. for example 2,4,6,... are even numbers). b) Analyze your algorithm in part (a) in the worst-case. Show all your work. Then express the time as Big-O().The goal is to rewrite the function, below, such that passes in a different list of parameters, particularly eliminating the need to pass low and high for each recursive call to binary_search. defbinary_search(nums,low,high,item): mid=(low+high)//2iflow>high:returnFalse #The item doesn't exist in the list!elifnums[mid]==item:returnTrue# The item exists in the list!elifitem<nums[mid]:returnbinary_search(nums,low,mid-1,item)else:returnbinary_search(nums,mid+1,high,item) The new function should be prototyped below. The number of changes between the given version, and the one requested is not significant. defbinary_search(nums,item):pass# Remove this and fill in with your code Tip: If you consider that high and low are used to create a smaller version of our problem to be processed recursively, the version requested will do the same thing, just through a different, more Pythonic technique.