Lab: Pairs and pair structures

Held

Wednesday, 10 April 2019

Writeup due

Friday, 12 April 2019

Summary

In this laboratory, you will further ground your understanding of what happens “behind the scenes” when Scheme deals with lists and other pair-based structures.

Preparation

a. Make sure you have some blank pieces of paper (lined is okay) and something with which to write.

b. Open DrRacket and add the following procedure.

;;; Procedure:
;;;   tree->code
;;; Parameters:
;;;   tree, a tree (or tree-like structure)
;;; Purpose:
;;;   Build a list that resembles the code to construct the tree.
;;; Produces:
;;;   code, a list
;;; Preconditions:
;;;   [No additional]
;;; Postconditions:
;;;   If you evaluate code, you get tree
;;; Practica:
;;;   > (tree->code (cons 'a 'b))
;;;   '(cons a b)
;;;   > (tree->code '(a . b))
;;;   '(cons a b)
;;;   > (tree->code '((a . b) c . d))
;;;   '(cons (cons a b) (cons c d))
;;;   > (tree->code (list 'a 'b 'c 'd))
;;;   '(cons a (cons b (cons c (cons d null))))
(define tree->code
  (lambda (tree)
    (cond
      [(null? tree)
       'null]
      [(pair? tree)
       (list 'cons (tree->code (car tree)) (tree->code (cdr tree)))]
      [else
       tree])))

d. If you have not already, add an appropriate citation for the procedure.

Exercises

Exercise 1: Some Pictures

Draw box-and-pointer diagrams for each of the following lists:

• '((x) y z)
• '(x (y z))
• '((a) b (c ()))

If you’re not sure about how any of these are built with cons, you can use (tree->code VALUE), e.g., (tree->code '((x) y z)).

Be prepared to share your pictures with me.

Exercise 2: Some pairs

Enter each of the following expressions into Scheme. In each case, explain why Scheme does or does not use the dot notation when displaying the value.

• (cons 'a "Walker")
• (cons 'a null)
• (cons 'a "null")
• (cons 'a "()")
• (cons null 'a)
• (cons null (cons null null))

Exercise 3: More pictures

Draw a box-and-pointer representation of the value of the last two expressions in the previous exercise.

Exercise 4: Are they pairs?

What do you think that pair? will return for each of the following? How about list?. Attempt to confirm each answer experimentally and explain any that you found particularly tricky.

• (cons 'a 'b)
• (cons 'a (cons 'b 'c))
• (cons 'a null)
• null
• (list 'a 'b 'c)
• (list 'a)
• (list)

Exercise 5: Is it a list?

You may recall that we told you that many kinds of data are defined recursively. For example, a list is either (1) null or (2) cons of anything and a list.

Using that recursive definition of lists, write a procedure, (listp? val), that determines whether or not val is a list.

You may not use list? in your definition of listp?.

For those with extra time

If you were able to complete the primary exercises with time to spare, you might want to consider the following problems:

Extra 1: Rewriting listp?

Write listp? without using if or cond.

Hint: You will need to use and and or.

Extra 2: Finding the last element

Write a procedure, (last pairthing) that finds the “last” value in a list-like pairs structure. If the pair structure is actually a list, return the last element of the list. Otherwise, follow the cdrs until you find the last pair, and return the cdr of that pair.

In solving this problem you should only step through the structure once.

Extra 3: Lists and list-likee things to strings

a Write a procedure, (intlist->string ints) that takes as input a list of integers and returns a string that represents the list.

> (intlist->string null)
"()"
> (intlist->string '(1))
"(1)"
> (intlist->string '(1 2 3))
"(1 2 3)"
> (intlist->string (cons 1 2))
BOOM!

b. Extend your procedure to print a period and the cdr when the cdr is not null.

> (intlist->string null)
"()"
> (intlist->string (cons 1 2))
"(1 . 2)"
> (intlist->string (cons 1 (cons 2 3)))
"(1 2 . 3)"

c. Extend your procedure to handle nested pair structure.

> (intpairs->string null)
"()"
> (intpairs->string (list (list 1)))
"((1))"
> (intpairs->string (list 1 (list 2 3) 4))
"(1 (2 3) 4)"
> (intpairs->string (list 1 (cons 2 3) 4))
"(1 (2 . 3) 4)"