Define an iterative function named append_ordered; it is passed two arguments: a linked list (11) whose values are ordered from smallest to biggest (they can be ints, strs, anything that can be compared), and another value (v). It returns a reference to the front of a linked list that includes all the values of the original linked list, and v, all in order. We call it like x = append_ordered (x, v). You may create exactly one new LN, for storing the value of v. For example, if we defined x = list_to_11 ([1, 3, 8, 12]). and wrote x = append_ordered (x, 10), then str_11(x) returns the string "1->3->8->10->12->No ne". Your code should work correctly regardless of whether the other value is added at the front, middle, or rear of the linked list. You may not use any other data structures (e. g., you may not put all the values into a list, sort the list, and then put all the values into a linked list), nor call any other helper functions: write the iterative code. 3b. (3 pts)
Define a recursive function named append_ordered _r that is given the same arguments and produces the same result as the iterative version specified above and abiding by the same restrictions). Also, use no looping, no local variables, etc. Hint: use the 3 proof rules to help synthesis your code. You might first write code that always appends the value at the end of the linked list.