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.

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

Draw box-and-pointer diagrams for each of the following lists: ((x) y z) (x (y z)) ((a) b (c ())) Be prepared to share your pictures with me.

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))

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

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)

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?.

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

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 list once.

Write listp? without using if or cond.