# Announcements. The current topic: Scheme. Review: BST functions. Review: Representing trees in Scheme. Reminder: Lab 2 is due on Monday at 10:30 am.

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1 The current topic: Scheme! Introduction! Object-oriented programming: Python Functional programming: Scheme! Introduction! Numeric operators, REPL, quotes, functions, conditionals! Function examples, helper functions, let, let*! More function examples, higher-order functions! More higher-order functions, trees Next up: More trees, lambda reductions, mutual recursion, examples, letrec Python GUI programming (Tkinter) Types and values Logic programming: Prolog Syntax and semantics Exceptions Announcements Reminder: Lab 2 is due on Monday at 10:30 am. Term Test 2 is on November 3rd in GB05. Aids allowed: Same as Term Test 1. Reminder: Deadline for Term Test 1 re-mark requests is today. 1 2 Review: Representing trees in Scheme Review: BST functions Trees are represented as lists. Each node contains its data value followed by all its children. If the "child" is a "null pointer" (that is, there is no child), it is represented by the empty list. Example: Binary trees. / \ / 5 5 ( (2 () ()) (6 (5 () ()) ()) ) Getting the data value in a given node: > (define (key node) (car node)) > (key '( (2 () ()) (6 (5 () ()) ()) ) ) Getting the left subtree of a given node: > (define (left node) (cadr node)) > (left '( (2 () ()) (6 (5 () ()) ()) ) ) (2 () ()) Getting the right subtree of a given node: > (define (right node) (caddr node)) > (right '( (2 () ()) (6 (5 () ()) ()) ) ) (6 (5 () ()) ()) 3

2 Printing binary trees We want a function print-tree to do something like this: > (print-tree '( (2 () (3 () ())) (6 (5 () ()) (7 () ())))) > (print-tree '( (2 () ()) (6 () ()))) 2 6 Another printing example > (print-tree (list2tree '( ))) (In all examples, we've omitted the #t that print-tree likes to finish off with.) 5 6 Printing a binary tree Environments and local variables in lambda expressions > (define (print-tree tree) (print-tree-help tree 0)) > (define (print-tree-help tree D) (cond ((null? tree)) (else (print-spaces D) (display (key tree)) (newline) (print-tree-help (left tree) (+ D 1)) (print-tree-help (right tree) (+ D 1))))) > (define (print-spaces N) (cond ((= N 0)) (else (display #\space) (display #\space) (print-spaces (- N 1))))) Recall that function definitions are equivalent to lambda expressions: > (define (mult x y) (* x y)) is equivalent to > (define mult (lambda (x y) (* x y))) Lambda expressions are a formal notation for establishing an environment (a local context) in which the lambda variables (the parameters to the function) are defined. Note that the above code doesn't completely follow functional programming style. 7 8

3 Environments and local variables in lambda expressions Lambda reduction Analogy: Logic expressions also establish an environment within which variables are defined. For example:!!x (P(x) Q(x)) The variable x in the above expression is a "bound" variable it has meaning only within the expression. The expression establishes the environment in which x has meaning. But the analogy isn't perfect. In a lambda expression, the variables have individual values, assigned by the caller, when the expression is evaluated. Lambda expressions get values by the process of lambda reduction: when a lambda expression is followed by a sequence of expressions, the values of those expressions are substituted for the lambda variables. ( (lambda (x y) (* x y)) 3 (+ 5))! [by lambda reduction] (* 3 9)! [by evaluation] Function calls and lambda reduction let and let* are not primitive A function call in Scheme is really a lambda reduction: > (mult 3 (+ 5))! [by evaluation] ( (lambda (x y) (* x y)) 3 9)! [by lambda reduction] (* 3 9)! [by evaluation] 27 That is, we can define them in terms of other forms. (let ((v1 e1) (vn en)) expr) is equivalent to: ((lambda (v1 vn) expr) e1 en) For example: (let ((x 5) (y 3)) (* x y)) is equivalent to: ((lambda (x y) (* x y)) 5 3) 11 12

4 let and let* are not primitive let and let* are not primitive Similarly: (let* ((v1 e1) (v2 e2)) expr) is equivalent to: ((lambda (v1) ((lambda (v2) expr) e2)) e1) For example: (let* ((x 5) (y (* x 2))) (+ x y)) is equivalent to: ((lambda (x) ((lambda (y) (+ x y)) (* x 2))) 5) Tracing the previous example: ((lambda (x) ((lambda (y) (+ x y)) (* x 2))) 5)! [by lambda reduction] ((lambda (y) (+ 5 y)) (* 5 2))! [by evaluation] ((lambda (y) (+ 5 y)) 10)! [by lambda reduction] (+ 5 10)! [by evaluation] 15 All binding of values to variables in let and let* is by parameter passing (that is, lambda reduction), not by assignment! 13 1 What cons really does What cons really does We've been treating cons as a function that "appends" to the beginning of a "list". This is the right general idea. But it leaves out details about what's happening "behind the scenes". Some of you have (accidentally?) noticed what happens when the second argument given to cons is not a list: > (cons 'a 'b) (a. b) > (cons 1 2) (1. 2) List are implemented as linked lists. Each node has two parts. A part storing (pointing to) data. This is the node's car. A part that's meant to store a pointer to the next node. This is the node's cdr. Think of this as the node's "next" pointer. cons creates a linked list node (also known as a pair). The first argument to cons is stored in the new node's car part. The second argument to cons is stored in the new node's cdr part. e.g. (cons 'a '()) produces: Notice that the return values include a dot. a 15 16

5 Displaying lists To display (output) a list, Scheme traverses the list's linked list, printing each node's car part. For example, the following list is displayed as (a b c). (cons 'a 'b) produces: Displaying lists a b a b c Observe that a properly-formed list ends with a null pointer. Another example: the nested list (a (b d) c) is stored as: When traversing this list, Scheme finds that the node's cdr part doesn't point to the next node, but instead points to a symbol. This list is displayed as (a. b) This list's cdr is the atom b, not the list (b). The list (a b) produced by (cons 'a '(b)) is stored as: a c b d a b Mutual recursion map-even and map-odd Mutual recursion is a form of recursion where two functions call each other (rather than themselves). Functions f1 and f2 are mutually recursive if f1 calls f2 and f2 calls f1. Let's define variants of map that only apply the given function to certain parts of the given list (and leave other parts unchanged). map-even takes a function f and a list L, and returns a new list in which each even-positioned element is the result of applying f to the corresponding element in L, and each odd-positioned element is simply the corresponding element in L unchanged. map-odd takes a function f and a list L, and returns a new list in which each odd-positioned element is the result of applying f to the corresponding element in L, and each even-positioned element is simply the corresponding element in L unchanged. : > (map-even car '((1 2 3) ( 5 6) (7 8) (a b c))) ((1 2 3) (7 8) a) > (map-odd car '((1 2 3) ( 5 6) (7 8) (a b c))) (1 ( 5 6) 7 (a b c)) > (map-even (lambda (x) (* 2 x)) '( )) ( ) > (map-odd (lambda (x) (* 2 x)) '( )) ( ) 19 20

6 map-even and map-odd We'll define map-even and map-odd so that they're mutually recursive: > (define (map-odd f L) (cond ((null? L) ()) (else (cons (f (car L)) (map-even f (cdr L)))) )) > (define (map-even f L) (cond ((null? L) ()) (else (cons (car L) (map-odd f (cdr L)))) )) map-even and map-odd Call: (map-even car '((a b) (c d) (1 2) (3 ))) Trace: (map-even car '((a b) (c d) (1 2) (3 ))) (map-odd car '((c d) (1 2) (3 ))) (map-even car '((1 2) (3 ))) (map-odd car '((3 ))) (map-even car '()) () (3) ((1 2) 3) (c (1 2) 3) ((a b) c (1 2) 3) Write a function maketester that takes two unary predicates f1 and f2 (a predicate is a function that returns true or false), and returns a function that takes a list and returns true iff all odd-positioned elements satisfy f1 and all even-positioned elements satisfy f2. For example: > ((maketester list? symbol?) '((a b) a (c) d)) #t > ((maketester symbol? number?) '(a 1 2 a)) #f Defining maketester, first solution: > (define (maketester f1 f2) (lambda (L) ((f1 (car L)) ((maketester f2 f1) (cdr L))) ))) This works, but notice that the function that's returned by maketester calls maketester each time it's called. Let's modify maketester so the returned function does not call maketester. 23 2

7 Defining maketester, second solution (using map-odd and mapeven): > (define (maketester f1 f2) (lambda (L) (eval (cons 'and (map-even f2 (map-odd f1 L)))) )) Observe we use map-odd to check if the odd-positioned elements satisfy f1, we use map-even to check if the even-positioned elements satisfy f2, and we use and to combine all the results. Now let's try to define maketester using mutually recursive lambda expressions. > (define (maketester f1 f2) (let ((test-odd (lambda (L) ((f1 (car L)) (test-even (cdr L))) ))) (test-even (lambda (L) ((f2 (car L)) (test-odd (cdr L))) )))) test-odd)) letrec General idea: test-odd checks if odd-positioned elements satisfy f1. test-even checks if even-positioned elements satisfy f2. test-odd and test-even take turns doing the checking. This is accomplished using mutual recursion. But the code doesn't work: > ((maketester symbol? number?) '(a 1 2 a)) reference to undefined identifier: test-even What's going on? The definition of test-odd refers to test-even, but we're using let, so the name test-even doesn't "exist" within the definition of test-odd. Using let* instead of let won't solve the problem, since then test-odd exists within the definition of test-even, but test-even still doesn't exist within the definition of test-odd. Solution: Use letrec, which allows lambda expressions to refer to each other (which allows for mutual recursion). > (define (maketester f1 f2) (letrec ((test-odd (lambda (L) ((f1 (car L)) (test-even (cdr L))) ))) (test-even (lambda (L) ((f2 (car L)) (test-odd (cdr L))) )))) test-odd)) 27 28

8 Write a function findsequence that takes two unary predicates f1 and f2, and returns a function that takes a list and returns the leftmost pair of adjacent elements in the list such that the first element of the pair satisfies f1 and the second element satisfies f2, if such a pair exists, and returns #f otherwise. For example: > ((findsequence list? symbol?) '(1 (a b) a (c) d)) ((a b) a) > ((findsequence symbol? number?) '((z) 1 a 3 2 a)) (a 3) > ((findsequence symbol? number?) '((z) 1 a (d) 2 3)) #f Defining findsequence: > (define (findsequence f1 f2) (letrec ((g (lambda (L) (cond ((null? L) #f) ((null? (cdr L)) #f) ((and (f1 (car L)) (f2 (cadr L))) (list (car L) (cadr L))) (else (g (cdr L))) )))) g)) Observe that letrec is needed here, since otherwise function g won't be able to call itself (since the name g won't exist within its own definition) Exercises More exercises Fix print-tree (defined in this lecture) so that it's clear from the output whether an only child is a right-child or a left-child. Hint: Do something special when the left child is null but the right child is not null. Write a function map-odd-even that takes functions f1 and f2, and a list L, and returns a new list in which each odd-positioned element is the result of applying f1 to the corresponding element in L, and each evenpositioned element is the result of applying f2 to the corresponding element in L. Do not define any helper functions, and do not use map-odd or map-even. : > (map-odd-even car cdr '((a b) (1 2) (#t #f) (3) ( 5))) (a (2) #t () ) Write a function make-odd-even that takes functions f1 and f2, and returns a function that takes a list returns a new list in which each oddpositioned element is the result of applying f1 to the corresponding element in L, and each even-positioned element is the result of applying f2 to the corresponding element in L. Do not use any helper functions. Instead, use letrec and mutually recursive lambda expressions. : > ((make-odd-even car cdr) '((a b) (1 2) (#t #f) (3) ( 5))) (a (2) #t () ) > ((make-odd-even list? symbol?) '((a b) (a b) c d (e) f))) (#t #f #f #t #t #t) > (map-odd-even list? symbol? '((a b) (a b) c d (e) f))) (#t #f #f #t #t #t) 31 32

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