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2 81 > (fun > (exit It evaluates the two expressions containing square, and the one using fun and then quits. It is important that the program ends with (exit so that the Scheme interpreter will exit and return you to the Unix shell. Now try the same thing by running these within Emacs using the directions given in the document Using emacs with Scheme available via the course web page. In general, you should find this to be a better experience in debugging your programs. Please remember that the dialect of LISP we use, Scheme, is similar to what is described in the book, but many of the key words are different. Use the ideas in the book, but look up precise key words in the Quick reference guide and tutorial linked to via the course web pages. (a What is the value of the following expressions? Try to work them out yourself, and verify your answers on the computer: i. (car (inky blinky pinky clyde ii. (cons inky (cdr (clyde blinky pinky iii. (car (car (cdr (inky (blinky pinky clyde iv. (cons (+ 1 2 (cdr (+ 1 2 v. (map (lambda (x (/ x 2 ( vi. (map (lambda (x (car x ((WOMBATS 3 (WALLAROOS 1 (NUMBATS 33 vii. (map (lambda (x (cdr x ((WOMBATS 3 (WALLAROOS 1 (NUMBATS 33 Write a function called list-length that returns the length of a list. Do not use the built-in length function in your solution. > (list-length (cons 1 (cons 2 (cons 3 (cons 4 nil 4 > (list-length (A B (C D 3 Write a function double that doubles every element in a list of numbers. Write this two different ways first use recursion over lists and then use map. > (double (1 2 3 (2 4 6 (b Recursive Definitions Not all recursive programs take the same amount of time to run. Consider, for instance, the following function that raises a number to a power: (define (power base exp (if (eq? exp 0 1 (* base (power base (- exp 1 There are more efficient means of exponentiation. First write a Lisp function that squares an integer, and then using this function, design a Lisp function fastexp which calculates b e for any e 0 by the rule: b 0 = 1 b e = (b (e/2 2 if e is even b e = b (b e 1 if power is odd 2

3 It is easy to blow this optimization and not save any multiplications. Show that the program you implemented is indeed faster than the original by comparing the number of multiplications that must be done for some exponent e in the first and second algorithms. (Hint it is easiest to calculate the number of multiplications if the exponent, e, is of the form 2 k for the second algorithm. Give the answer for exponents of this form and then try to give an upper bound for the others. (c Recursive list manipulation Write a function merge-list that takes two lists and joins them together into one large list by alternating elements from the original lists. If one list is longer, the extra part is appended onto the end of the merged list. The following examples demonstrate how to merge the lists together: > (merge-list (1 2 3 (A B C (1 A 2 B 3 C > (merge-list (1 2 (A B C D (1 A 2 B C D > (merge-list (1 2 3 nil (1 2 3 > (merge-list ((1 2 (3 4 (A B ((1 2 A (3 4 B (d Reverse Write a function rev that takes one argument. If the argument is an atom it remains unchanged. Otherwise, the function returns the elements of the list in reverse order: > (rev nil nil > (rev A A > (rev (A (B C D (D (B C A > (rev ((A B (C D ((C D (A B (Yes, I know that Scheme already includes a predefined function reverse that does exactly this. I want you to write your own without using the built-in function! (e Mapping functions Write a function censor-word that takes a word as an argument and returns either the word or XXXX if the word is a bad word: > (censor-word lisp lisp > (censor-word midterm XXXX Recall the function ismember? that we defined in class. Using that function definition, (member word (extension algorithms graphics AI midterm evaluates to#t if word is in the given list of bad words, i.e., the list (extension algorithms graphics AI midterm is the list of bad words. Use this function to write a censor function that replaces all the bad words in a sentence: 3

5 2. (10 points Lisp Programming The following code is a slight rewrite of code that appears in Mitchell, Chapter 3 (page 31. ;; substitute exp1 for all occurrences of var in exp2 (define (subst exp1 var exp2 (cond ((null? exp2 ( ((not (list? exp2 (if (eq? var exp2 exp1 exp2 (else (cons (subst exp1 var (car exp2 (subst exp1 var (cdr exp2 (define (subst-and-eval exp1 var exp2 (eval (subst exp1 var exp2 The subst function replaces all occurences of atom var with exp1 in exp2. The function subst-and-eval performs such a substitution and then evaluates the resulting list. For example, (subst-and-eval * + ( evaluates to 60. Your job is to write a function multi-subst-and-eval that performs multiple substitutions on a list before evaluating it. For example, (multi-subst-and-eval ((x 10 (y 10 (* + (* y (- 12 x evaluates to 12. First write a function multi-subst that performs the substitutions and verify that it is constructing a valid expression to evaluate. The expression (multi-subst ((x 10 (y 10 (* + (* y (- 12 x evaluates to the list (+ 10 ( You may use the above subst function as part of your solution. Writing multi-subst-and-eval is straight-forward once you have done that. Here is some additional sample input: > (multi-subst-and-eval ((x 10 (y 10 (* y (- 12 x 20 > (multi-subst-and-eval ((x 10 (y 20 (+ * (* y (+ y (* y y x > (multi-subst-and-eval ((x y (y 20 (+ * (* y (+ y (* y y x > (multi-subst-and-eval ((y 20 (* cons (* y (* y (* y nil ( Substitutions should be performed from left to right in the list. For example, the substitutions ((x y (y 20 will change any x in a term to y, and then replace every y with 20. Thus, 5

6 (multi-subst ((x y (y 20 (* x x evaluates to (* 20 20, and (multi-subst-and-eval ((x y (y 20 (* x x returns (10 points Cons Cell Representations Mitchell, Problem (20 points Conditional Expressions in Lisp Mitchell, Problem 3.2 6

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