Assignment 7 List Recursion

Write and document a procedure (safe-sum values) that, given a list of values as a parameter, computes the sum of the numeric values in the list. That is, safe-sum should ignore all non-numeric values.

> (safe-sum (list 1 2 3))
6
> (safe-sum (list 3 'a 'b 5))
8
> (safe-sum (list 'a 'b))
0

a. Write a procedure, (closest-to-zero values), that, given a list of real numbers (including both positive and negative numbers), returns the value closest to zero in the list. Your solution should use basic recursion.

Hint: Think about how, given two numbers, you determine which is closer to zero.

b. Write a second version of closest-to-zero that uses helper recursion. That is, you should have an additional helper procedure that takes closest-so-far and remaining as parameters.

c. Explain which version of closest-to-zero you prefer and why.

Averaging two colors is a fairly simple task: simply average each of the respective red, green, and blue components, as in the following procedure.

;;; Procedure:
;;;   rgb-average
;;; Parameters:
;;;   rgb1, an RGB color
;;;   rgb2, an RGB color
;;; Purpose:
;;;   Compute the "average" of two RGB colors.  
;;; Produces:
;;;   rgb-avg, an RGB color
;;; Preconditions:
;;;   [No additional]
;;; Postconditions:
;;;   (rgb-red rgb-avg) is the average of (rgb-red rgb1) and (rgb-red rgb2)
;;;   (rgb-green rgb-avg) is the average of (rgb-green rgb1) and (rgb-green rgb2)
;;;   (rgb-blue rgb-avg) is the average of (rgb-blue rgb1) and (rgb-blue rgb2)
(define rgb-average
  (lambda (rgb1 rgb2)
    (rgb-new (quotient (+ (rgb-red rgb1) (rgb-red rgb2)) 2)
             (quotient (+ (rgb-green rgb1) (rgb-green rgb2)) 2)
             (quotient (+ (rgb-blue rgb1) (rgb-blue rgb2)) 2))))

We also saw how to average a list of colors in the lab on Recursion Basics. But what if we want to do something different: Given a list of colors, we want averages, but only of neighboring elements in the list.

Write a procedure, (rgb-averages colors), that, given a list of colors, computes a new list of colors, by averaging subsequent pairs of colors. For example, if the input list is the standard seven rainbow colors (red, orange, yellow, green, blue, indigo, and violet), the output list will consist of a red-orange average, an orange-yellow average, a yellow-green average, a green-blue average, a blue-indigo average, and an indigo-violet average.

The length of the resulting list will be one less than the length of the input list.

a. Using the 6 P's, document the reverse procedure.

b. Suppose the reverse procedure were not included in Scheme. Could you write it yourself? Certainly! It should be possible to implement reverse recursively.

Using the helper recursion pattern, implement your own version of reverse called my-reverse.

c. Modify my-reverse so that it verifies its preconditions.

Write and document a function (riffle first second) that produces a new list containing alternating elements from the lists first and second. If one list runs out before the other, then the remaining elements should appear at the end of the new list.

> (riffle (list 'a 'b 'c) (list 'x 'y 'z))
(a x b y c z)
> (riffle (list 'a 'b 'c) (iota 10))
(a 0 b 1 c 2 3 4 5 6 7 8 9)

Define and test a Scheme procedure, (unriffle lst), that takes a list as argument and returns a list of two lists, one comprising the elements in even-numbered positions in the given list, the other comprising the elements in odd-numbered positions.

> (unriffle (list 'a 'b 'c 'd 'e 'f 'g 'h 'i))
((a c e g i) (b d f h))
> (unriffle (list))
(() ())
> (unriffle (list 'a))
((a) ())
> (unriffle (list 'b))
((b) ())
> (unriffle (list 'a 'b))
((a) (b))

Hint: There are many ways to solve this problem. Before writing code, try solving it by hand to develop your algorithm.