Summary: Although most of our prior experiments with recursion have emphasized recursion over lists, it is also possible to use other values as the basis of recursion. In this laboratory, you will explore the use of natural numbers (non-negative integers) as the basis of recursion.

Here is a template for the simplest kind of numeric recursive procedures. (define recursive-proc (lambda (n) (if (zero? n) base-case (combine n (recursive-proc (- n 1)))))) If you choose to use a local kernel, here is one common form. (define recursive-proc (letrec ((kernel (lambda (so-far n) (if (zero? n) so-far (kernel (update so-far n) (- n 1)))))) (lambda (n) (kernel starting-value n)))) We can also use a kernel to count up from 1 to n. (A slight change lets you count up from 0 to n.) (define recursive-proc (letrec ((kernel (lambda (so-far i n) (if (> i n) so-far (kernel (update so-far i) (+ i 1) n))))) (lambda (n) (kernel starting-value 1 n))))

Make a copy of numeric-recursion-lab.scm, which contains important code from the reading.

Rewrite termial it so that the kernel is local.

Define and test a recursive Scheme procedure, (count-down val), that takes a natural number as argument and returns a list of all the natural numbers less than or equal to that number, in descending order. > (count-down 5) (5 4 3 2 1 0) > (count-down 0) (0) Note that you should use cons to build up the list. Note also that you are better off writing this with direct recursion (the first pattern above), rather than using a helper procedure. When you are finished, you may want to read the notes on this exercise.

Define and test a Scheme procedure, (my-make-list count value), that takes two arguments, the first of which is a natural number, and returns a list consisting of the specified number of repetitions of the second argument. > (my-make-list 5 "sample") ("sample" "sample" "sample" "sample" "sample") > (my-make-list 3 10) (10 10 10) > (my-make-list 1 null) (()) > (my-make-list 2 null) (() ()) > (my-make-list 0 "hello") () You should not call the built-in make-list procedure. Instead, implement my-make-list recursively. When you are finished writing this procedure, compare it to the notes on this exercise.

As you may recall, iota is a procedure that takes a natural number as argument and returns a list of all the natural numbers that are strictly less than the argument, in ascending order. As you've seen, the iota procedure is quite useful. Unfortunately, it does not come as part of standard Scheme. (We include it in MediaScheme because it is so useful.) Implement and test your own version of iota. (You should not call the MediaScheme iota.) For example, > (my-iota 3) (0 1 2) > (my-iota 5) (0 1 2 3 4) > (my-iota 1) (0) Note that you will probably need to use a helper of some sort to write my-iota. You might use the traditional form of helper, which adds an extra parameter. You might also use a helper that simply computes the list in the reverse order. (Most students write a backwards version in the first attempt; instead of throwing it away, rename it and call it from my-iota.)

You may recall the count-from procedure from the reading on recursion over natural numbers, which is available in the code for this lab. a. What is the value of the call (count-from -10 10)? b. Check your answer experimentally. c. It is possible to implement count-from in terms of iota. The implementation looks something like the following. (define count-from (lambda (lower upper) (map (lambda (n) (+ _____ n)) (iota _____)))) Finish this definition. d. What do you see as the advantages and disadvantages of each way of defining count-from?

Write a procedure, (my-list-ref lst n), that extracts element n of a list. For example, > (my-list-ref (list "red" "orange" "yellow" "green" "blue" "indigo" "violet") 5) "indigo" > (my-list-ref (list "red" "orange" "yellow" "green" "blue" "indigo" "violet") 0) "red" Even though this procedure does the same thing as list-ref, you should not use list-ref to implement it. Instead, your goal is to figure out how list-ref works, which means that you will need to implement this procedure using direct recursion. Hint: When recursing, you will need to simplify the numeric parameter (probably by subtracting 1) and the list parameter (probably by taking its cdr).

a. Define and test a recursive procedure, (list-prefix lst n), that returns a list consisting of the first n elements of the list, lst, in their original order. You might also think of list-prefix as returning all the values that appear before index n. For example, > (list-prefix (list "a" "b" "c" "d" "e") 3) ("a" "b" "c") > (list-prefix (list 2 3 5 7 9 11 13 17) 2) (2 3) > (list-prefix (list "here" "are" "some" "words") 0) () > (list-prefix (list null null) 2) (() ()) > (map rgb->color-name (list-prefix (map color-name->rgb (list "black" "white" "green") 1)) ("black") While your procedure has much the same purpose as list-take, you should not call list-take. Rather, your goal is to implement the procedure recursively. b. Update your procedure to check appropriate preconditions. In particular, your procedure should signal an error if lst is not a list, if n is not an exact integer, if n is negative, or if n is greater than the length of lst. Note that in order to signal such errors, you may want to take advantage of the husk-and-kernel programming style. You can use named let or letrec. c. Compare your error messages for various invalid parameters with those given by list-take.

Rewrite list-prefix to use whichever of named let and letrec you didn't use in the previous exercise.

Here's a possible solution to the problem. The base case is easy. If the number is zero, then the list of all non-negative numbers less than or equal to zero is the list that contains only zero. (if (zero? n) (list 0) In the recursive case, we assume that we can compute the list of all numbers less than or equal to n-1. To get the complete list, we simply add n to the front. (cons n (count-down (- n 1))) Putting it all together, we get Return to the problem.

We begin by considering the base case. The last example gives a hint: If you want zero copies, you end up with the empty list. (if (zero? n) null Now, on to the recursive case. If we can create a list of n-1 copies, we can then create a list of n copies by prepending one more copy. (cons val (value-replicate val (- n 1))) Putting it all together, we get Return to the problem.