8. Know how to do these, to trace functions like these and to debug functions like these: // recursive power , compute xn int exp(int x, int n){ if(n== return return *exp } void main(){ int a,b; cin >>a>>b; cout<>a; rev_print(a); }
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- 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…Write a recursive function named reverseWithinBounds that has an argument that is an array of characters and two arguments that are bounds on array indices. The function should reverse the order of those entries in the array whose indices are between the two bounds (including the bounds). For example, if the array is: a[0] == 'A' a[1] == 'B' a[2] == 'C' a[3] == 'D' a[4] == 'E' and the bounds are 1 and 4, then after the function is run the array elements should be: a[0] == 'A' a[1] == 'E' a[2] == 'D' a[3] == 'C' a[4] == 'B' Embed the function in a program and test it. After you have fully debugged this function, define another function named reverseCstring that takes a single argument that is a C string and modifies the argument so that it is reversed. This function will include a call to the recursive definition you did for the first part of this project, and need not be recursive. Embed this second function in a program and test it.Write a recursive function named reverseWithinBounds that has an argument that is an array of characters and two arguments that are bounds on array indices. The function should reverse the order of those entries in the array whose indices are between the two bounds (including the bounds). For example, if the array is: a[0] == 'A' a[1] == 'B' a[2] == 'C' a[3] == 'D' a[4] == 'E' and the bounds are 1 and 4, then after the function is run the array elements should be: a[0] == 'A' a[1] == 'E' a[2] == 'D' a[3] == 'C' a[4] == 'B' Embed the function in a program and test it. After you have fully debugged this function, define another function named reverseCstring that takes a single argument that is a C string and modifies the argument so that it is reversed. This function will include a call to the recursive definition you did for the first part of this project, and need not be recursive. Embed this second function in a program and test it. Turn in only this final result (with output,…
- Write a recursive function named reverseWithinBounds that has an argument that is an array of characters and two arguments that are bounds on array indices. The function should reverse the order of those entries in the array whose indices are between the two bounds (including the bounds). For example, if the array is: a[0] == 'A' a[1] == 'B' a[2] == 'C' a[3] == 'D' a[4] == 'E' and the bounds are 1 and 4, then after the function is run the array elements should be: a[0] == 'A' a[1] == 'E' a[2] == 'D' a[3] == 'C' a[4] == 'B'In C++ Consider the following recursive function (Chapter 17, #9, modified) void recFun(int x) { if (x > 10) { recFun(x / 10); cout<< x % 10; } else cout<< x; } How can the code be modified so that the sum of all digits is printed?write each of the functions LENGTH, WRITESTR, READSTR and CONCAT so that they operate iteratively without using recursion.
- Question: First write a function mulByDigit :: Int -> BigInt -> BigInt which takes an integer digit and a big integer, and returns the big integer list which is the result of multiplying the big integer with the digit. You should get the following behavior: ghci> mulByDigit 9 [9,9,9,9] [8,9,9,9,1] Your implementation should not be recursive. Now, using mulByDigit, fill in the implementation of bigMul :: BigInt -> BigInt -> BigInt Again, you have to fill in implementations for f , base , args only. Once you are done, you should get the following behavior at the prompt: ghci> bigMul [9,9,9,9] [9,9,9,9] [9,9,9,8,0,0,0,1] ghci> bigMul [9,9,9,9,9] [9,9,9,9,9] [9,9,9,9,8,0,0,0,0,1] ghci> bigMul [4,3,7,2] [1,6,3,2,9] [7,1,3,9,0,3,8,8] ghci> bigMul [9,9,9,9] [0] [] Your implementation should not be recursive. Code: import Prelude hiding (replicate, sum)import Data.List (foldl') foldLeft :: (a -> b -> a) -> a -> [b] -> afoldLeft =…Write a recursive function definition for the following function: int squares(int n); //Precondition: n >= 1 //Returns the sum of the squares of numbers 1 through n. For example, squares(3) returns 14 because 12 + 22 + 32 is 14.IN C Programing Let us define the weighted sum of an integer array a[0], a[1], a[2], …, a[n-1] be a[0]*1 + a[1]*2 + a[2]*3 + …+a[n-1]*n. For example, the weighted sum of the array [5,2,6] would be 5*1+2*2+6*3 = 27. Write a recursive function that takes in an array numbers and its length n, and returns its weighter sum. You can assume n is non-negative integer. int weightedSum(int numbers[], int n) {
- Implement a recursive function called evens that returns an integer with only theeven numbers. Note this function is returning an integer, not printing. There shouldbe no use of cout within your function. The function declaration should look asfollows:int evens(int n);(ex. evens(234567); returns 246)(ex. cout << evens(56032); prints 602)Write a recursive function recursiveMinimum that takes an integer array, a starting subscript, and an ending subscript as arguments, and returns the smallest element of the array. The function should stop processing and return when the starting sub- script equals the ending subscript. c++in C programing Write a recursive function that returns 1 if an array of size n is in sorted order and 0 otherwise. Note: If array a stores 3, 6, 7, 7, 12, then isSorted(a, 5) should return 1 . If array b stores 3, 4, 9, 8, then isSorted(b,4) should return 0.int isSorted(int *array, int n){