Below,enter code to complete implementation of a recursive function sum_all_integers(), which takes an input n and adds all intergers preceding it, up to n: add_all_integers(n):
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Below,enter code to complete implementation of a recursive function sum_all_integers(), which takes an input n and adds all intergers preceding it, up to n:
add_all_integers(n):
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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…Below is a recursive version of binarySearch:int binarySearch(int nums[], int low, int high, int target){ if (low > high) return -1; int mid = (low + high)/2; if (target == nums[mid]) return mid; else if (target < nums[mid]) return binarySearch(nums, low, mid - 1, target); else return binarySearch(nums, mid + 1, high, target);} Code Analysis ()If we change the first line of code in the function to be “if (low >= high) return -1;”, and we have an array defined as “int nums[] = {2, 3, 5, 7, 8, 10};”, answer the questions below if we called this function with “int index = binarySearch(nums, 0, nums.length - 1, 3);”For each iteration of binarySearch:1) What are the values of low, high, mid, and target variables?Below is a recursive version of binarySearch:int binarySearch(int nums[], int low, int high, int target){ if (low > high) return -1; int mid = (low + high)/2; if (target == nums[mid]) return mid; else if (target < nums[mid]) return binarySearch(nums, low, mid - 1, target); else return binarySearch(nums, mid + 1, high, target);} Code Analysis ()If we change the first line of code in the function to be “if (low >= high) return -1;”, and we have an array defined as “int nums[] = {2, 3, 5, 7, 8, 10};”, answer the questions below if we called this function with “int index = binarySearch(nums, 0, nums.length - 1, 3);”For each iteration of binarySearch:3) If the base case(s) are false, which recursive function call will be made?
- Below is a recursive version of binarySearch:int binarySearch(int nums[], int low, int high, int target){ if (low > high) return -1; int mid = (low + high)/2; if (target == nums[mid]) return mid; else if (target < nums[mid]) return binarySearch(nums, low, mid - 1, target); else return binarySearch(nums, mid + 1, high, target);} Code Analysis ()If we change the first line of code in the function to be “if (low >= high) return -1;”, and we have an array defined as “int nums[] = {2, 3, 5, 7, 8, 10};”, answer the questions below if we called this function with “int index = binarySearch(nums, 0, nums.length - 1, 3);”For each iteration of binarySearch:2) Will a base case be true? If so, what value is being returnedWrite 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'Write a RECURSIVE function, without using any loops, that prints the contents of a matrix with 3 columns. The function should take the matrix and the number of rows as arguments. void print_matrix(int arr[][3], int num_rows);For function decToBinary, write the missing parts of the recursion case. This function should return a string that stores the binary equivalent for int variable num. Example: The binary equivalent of 13 may be found by repeatedly dividing 13 by 2. So, 13 in base 2 is represented by the string "1101". Examples: decToBinary(13) -> "1101" public String decToBinary (int num) { if (num < 2) return Integer.toString(num); else return <<Missing recursive call>> + <<Missing calculation>>;}
- : Write a recursive function to multiply two positive integers without usingthe * operator (or / operator). You can use addition, subtraction, and bit shifting, but you shouldminimize the number of those operations.Write a function that takes in an integer n and computes n!. Do this without recursion. In [ ]: deffactorial_iter(n):"""Takes in an integer n>0 and returns the product of all integers from 1 to n."""# YOUR CODE HEREraiseNotImplementedError() In [ ]: In [ ]: assert factorial_iter(6) == 720 assert factorial_iter(7) == 5040 assert factorial_iter(10) == 3628800Write a recursive function, reverseDigits, that takes an integer as a parameter and returns the number with the digits reversed. Also, write a program to test your function.