Reflection Consider the member? function, which has the following - PowerPoint PPT Presentation

Reflection Consider the member? function, which has the following algebraic laws: (member? m ’()) == #f (member? m (cons m ks)) == #t (member? m (cons k ks)) == (member? m ks), m != k This algorithm searches for an element in a list. List other algorithms you are familiar with that operate over a list (or linked list, or an array) and explain briefly what they do.

Predefined list algorithms Some classics: • exists? (Example: Is there a number?) • all? (Example: Is everything a number?) • filter (Example: Select only the numbers) • map (Example: Add 1 to every element) • foldr ( Visit every element; also called reduce , accum , a “catamorphism”)

Defining exists? ; (exists? p? ’()) = #f ; (exists? p? (cons y ys)) = #t, if (p? y) ; (exists? p? (cons y ys)) = (exists? p? ys), otherwise -> (define exists? (p? xs) (if (null? xs) #f (if (p? (car xs)) #t (exists? p? (cdr xs))))) -> (exists? number? ’(1 2 zoo)) #t -> (exists? number? ’(apple orange)) #f

Coding Interlude: exists? Define a list ys and two predicates p1? and p2? such that exists? p1? ys evaluates to #t and exists? p2? ys evalutes to #f . Run the code to confirm your answers are correct. An alternative formulation of the algebraic laws for the exists? function is: ;; (exists? p? ’() = #f ;; (exists? p? (cons x xs)) = (|| (p? x) ;; (exists? p? xs)) Modify the definition of exists? following this algebraic formulation instead, and re-run your

examples.

Defining filter ; (filter p? ’()) == ’() ; (filter p? (cons y ys)) == ; (cons y (filter p? ys)), when (p? y) ; (filter p? (cons y ys)) == ; (filter p? ys), when (not (p? y)) -> (define filter (p? xs) (if (null? xs) ’() (if (p? (car xs)) (cons (car xs) (filter p? (cdr xs))) (filter p? (cdr xs)))))

Running filter -> (filter (lambda (n) (> n 0)) ’(1 2 -3 -4 5 6)) (1 2 5 6) -> (filter (lambda (n) (<= n 0)) ’(1 2 -3 -4 5 6)) (-3 -4)

Coding Interlude: filter Use lambda to define a predicate that returns true iff the argument is a number larger than 9 . Define a list ys that contans a mix of one- and two-digit numbers. Use your lambda function and the filter function defined in the video to produce a new list zs that contains only the elements in ys with at least two-digits. Run your code. How would you modify the code to produce a list with only one-digit numbers?

Your turn: map Complete the algebraic laws for the map function: -> (map add3 ’(1 2 3 4 5)) (4 5 6 7 8) ;; (map f ’()) = ;; (map f (cons y ys)) =

Answers: map -> (map add3 ’(1 2 3 4 5)) (4 5 6 7 8) ; (map f ’()) == ’() ; (map f (cons y ys)) == (cons (f y) (map f ys))

Defining and running map ; (map f ’()) == ’() ; (map f (cons y ys)) == (cons (f y) (map f ys)) -> (define map (f xs) (if (null? xs) ’() (cons (f (car xs)) (map f (cdr xs))))) -> (map number? ’(3 a b (5 6))) (#t #f #f #f) -> (map (lambda(x)(* x x)) ’(5 6 7)) ((25 36 49)

Foldr

Algebraic laws for foldr Idea: � 0 + 0 : x 1 + x n � + : + � � � (foldr (plus zero ’())) = zero (foldr (plus zero (cons y ys))) = (plus y (foldr plus zero ys)) Note: Binary operator + associates to the right. Note: zero might be identity of plus .

Code for foldr Idea: � 0 + 0 : x 1 + x n � + : + � � � -> (define foldr (plus zero xs) (if (null? xs) zero (plus (car xs) (foldr plus zero (cdr xs))))) -> (val sum (lambda (xs) (foldr + 0 xs))) -> (sum ’(1 2 3 4)) 10 -> (val prod (lambda (xs) (foldr * 1 xs))) -> (prod ’(1 2 3 4)) 24

Another view of operator folding ’(1 2 3 4) (cons 1 (cons 2 (cons 3 (cons 4 ’())))) = (foldr + 0 ’(1 2 3 4)) (+ 1 (+ 2 (+ 3 (+ 4 0 )))) = (foldr f z ’(1 2 3 4)) (f 1 (f 2 (f 3 (f 4 z )))) =

Check your understanding: foldr What does the following code evaluate to? -> (define foldr (plus zero xs) (if (null? xs) zero (plus (car xs) (foldr plus zero (cdr xs))))) -> (define combine (x a) (+ 1 a)) -> (foldr combine 0 ’(2 3 4 1)) • 0 • 10 • 24 • 4 Answer: 4 . The plus function for this invocation of

foldr is the function combine , which adds one to an accumulator a for each element x in the list. The zero function for this invocation of foldr is the literal 0 . Hence this invocation of foldr starts with 0 and adds 1 for each element in the list, returning 4 when applied to the list '(2 3 4 1) because that list has four elements. This use of foldr is one way of writing the list length function.

Your turn: Explain the design 1. Functions like exists? , map , filter are subsumed by 2. Function foldr , which is subsumed by 3. Recursive functions Seems redundant: Why?

Currying: The motivation Remember me? q-with-y? = (lambda (z) (q? y z)) Happens so often, there is a function for it: q-with-y? = ((curry q?) y) Called a partial application (one now, one later)

Map/search/filter love curried functions -> (map ((curry +) 3) ’(1 2 3 4 5)) ; add 3 to each element -> (exists? ((curry =) 3) ’(1 2 3 4 5)) ; is there an element equal to 3? -> (filter ((curry >) 3) ’(1 2 3 4 5)) ; keep elements that 3 is greater then

The idea of currying The function curry • Input: a binary function f(x,y) • Output: a function f' – Input: argument x – Output: a function f'' * Input: argument y * Output: f(x,y) What is the benefit? • Functions like exists? , all? , map , and filter expect a function of one argument. To get there, we use currying and partial application. Slogan: Curried functions take their arguments “one-at-a-time.”

What’s the algebraic law for curry ? ... (curry f) ... = ... f ... Keep in mind: All you can do with a function is apply it! (((curry f) x) y) = (f x y) Three applications: so implementation will have three lambda s

From law to code ;; curry : binary function -> value -> function ;; (((curry f) x) y) = (f x y) -> (val curry (lambda (f) (lambda (x) (lambda (y) (f x y))))) -> (val positive? ((curry <) 0))

Check your understanding: curry What does the following code evaluate to? -> (map ((curry +) 3) ’(1 2 3 4 5)) • 15 • 18 • '(4 5 6 7 8) • '(1 2 3 3 4 5) • '(3 4 5 6 7 8) • 4 Answer: '(4 5 6 7 8) . The map function applies its argument function ((curry +) 3) to each element in the list '(1 2 3 4 5) , to produce the list '(4 5 6 7 8) .

The function ((curry +) 3) is the “plus-three” function obtained by partially applying the curried + function to the constant 3 . The + function has to be curried before being applied to 3 because it otherwise takes two arguments at a time, not just one. An invariant of the map function is that it will always return a list with the same length as its argument. -> (exists? ((curry =) 3) ’(1 2 3 4 5)) • '() • #t • #f • 0

Answer: #t . The exists? function applies its argument predicate ((curry =) 3) to each list element until it finds one where the predicate evalutes to true ( #t ). If it reaches the end of the list without finding such an element, it returns false ( #f ). The predicate is the “equal-three?” function, obtained by partially applying the curried equality function to the constant 3 . The equality function has to be curried before being applied to 3 because it otherwise takes two arguments at a time, not just one. -> (filter ((curry >) 3) ’(1 2 3 4 5))

; tricky • '(1 2) • '(1 2 3) • ’() • '(3 4 5) • '(4 5) Answer: '(1 2) . The filter function applies its argument predicate ((curry > ) 3) to each list element, including in the resulting list only those that evaluate to true ( #t ). The predicate is the “three-is-greater-than?” function, obtained by partially applying the curried greater-than function to the constant 3 . This

question requires careful thought because a quick glance can make it seem like the code is testing whether something is greater than 3 rather than the opposite. The greater-than function has to be curried before being applied to 3 because it otherwise takes two arguments at a time, not just one.

Composing Functions In math, what is the following equal to? (f o g)(x) == ??? Another algebraic law, another function: (f o g) (x) = f(g(x)) (f o g) = \x. (f (g (x)))

One-argument functions compose -> (define o (f g) (lambda (x) (f (g x)))) -> (define even? (n) (= 0 (mod n 2))) -> (val odd? (o not even?)) -> (odd? 3) #t -> (odd? 4) #f

Coding Interlude: compose The predicate null? returns #t iff the argument list is empty, while the function not inverts a boolean. Use the compose function ( o ) to write a predicate that tests whether a list is non-empty. Run your code on both empty and non-empty lists.

Proofs about functions Function consuming A is related to proof about A • Q: How to prove two lists are equal? A: Prove they are both ’() or that they are both cons cells cons-ing equal car’s to equal cdr’s • Q: How to prove two functions equal? A: Prove that when applied to equal arguments they produce equal results.