1 Append: process  (append list1 list2) (cons 1 (append ‘(2) list2)) (cons 1 (cons 2 (append ‘() list2))) (cons 1 (cons 2 list2)) (define (append list1 list2) (cond ((null? list1) list2) ; base (else (cons (car list1) ; recursion (append (cdr list1) list2))))) list2 list1 12 ( )  (define list1 (list 1 2))  (define list2 (list 3 4)) Quote: what you see is what you get.

10/22/20152 Reverse of a list (reverse (list )) (define (reverse lst) (cond ((null? lst) lst) (else ( (reverse (cdr lst)) ))))) append (list (car lst)) (append ) (reverse (cdr lst))(list (car lst)) Append: T(n) = c*n =  (n) Reverse: T(n) = c*(n-1) + c*(n-2)+ … + c*1 =  (n 2 )

3 Reverse (reverse (list 1 2 3)) (append (reverse (2 3)) (1)) (append (append (reverse (3)) (2)) (1)) (append (append (append (reverse ()) (3)) (2)) (1)) (append (append (append null (3)) (2)) (1)) (append (append (3) (2)) (1)) (append (3 2) (1)) (3 2 1) (define (reverse lst) (cond ((null? lst) lst) (else (append (reverse (cdr lst)) (list (car lst))))))) Append: T(n1) = c*n1 =  (n1) (n1 is length of list1) Reverse: T(n) = c*(n-1) + c*(n-2) … c*1 =  (n 2 )  (reverse (list 1 2 3)) (3 2 1)

4 Trees Abstract tree: a leaf (a node that has no children, and contains data) - is a tree an internal node (a node whose children are trees) – is a tree leaf internal node Implementation of tree: a leaf - will be the data itself an internal node – will be a list of its children

5 Count Leaves of a Tree Strategy –base case: count of an empty tree is 0 –base case: count of a leaf is 1 –recursive strategy: the count of a tree is the sum of the countleaves of each child in the tree. Implementation: (define (leaf? x) (atom? x))

6 Count Leaves (define (countleaves tree) (cond ((null? tree) 0) ;base case ((leaf? tree) 1) ;base case (else ;recursive case (+ (countleaves (car tree)) (countleaves (cdr tree)))))) (define my-tree (list 4 (list 5 7) 2)) 42 57

7 Countleaves (countleaves my-tree ) ==> 4 (cl (4 (5 7) 2)) + (cl 4)(cl ((5 7) 2) ) + (cl (5 7))(cl (2)) + (cl 2) (cl null) + (cl 5) (cl (7)) + (cl 7) (cl null) my-tree

8 Enumerate-Leaves Goal: given a tree, produce a list of all the leaves Strategy –base case: list of empty tree is empty list –base case: list of a leaf is one element list –otherwise, recursive strategy: build a new list from a list of the leaves of the first child and a list of the leaves of the rest of the children

9 Enumerate-Leaves (define (enumerate-leaves tree) (cond ((null? tree) null) ;base case ((leaf? tree) ) ;base case (else ;recursive case ( (enumerate-leaves (car tree)) (enumerate-leaves (cdr tree))))))

10 Enumerate-Leaves (define (enumerate-leaves tree) (cond ((null? tree) null) ;base case ((leaf? tree) (list tree)) ;base case (else ;recursive case (append (enumerate-leaves (car tree)) (enumerate-leaves (cdr tree))))))

מבוא מורחב - שיעור 811 Enumerate-leaves (el (4 (5 7) 2)) ap (el 4)(el ((5 7) 2) ) ap (cl (5 7))(el (2)) ap (el 2) (el nil) ap (el 5) (el (7)) ap (el 7) (el nil) (4) (5) (7)() (2) () (7) (5 7)(2) (5 7 2) ( )

מבוא מורחב - שיעור 812 Your Turn: Scale-tree Goal: given a tree, produce a new tree with all the leaves scaled Strategy –base case: scale of empty tree is empty tree –base case: scale of a leaf is product –otherwise, recursive strategy: build a new tree from a scaled version of the first child and a scaled version of the rest of children

13 Scale-tree (define (scale-tree tree factor) (cond ((null? tree) ) ;base case ((leaf? tree) ) (else ;recursive case (cons )))) (scale-tree (car tree) factor) null (* tree factor) (scale-tree (cdr tree) factor)

14 Enumeration (integers-between 2 4) (cons 2 (integers-between 3 4))) (cons 2 (cons 3 (integers-between 4 4))) (cons 2 (cons 3 (cons 4 (integers-between 5 4)))) (cons 2 (cons 3 (cons 4 null))) (2 3 4) (define (integers-between lo hi) (cond ((> lo hi) null) (else (cons lo (integers-between (+ 1 lo) hi))))) 234

15 Count Leaves (define (countleaves tree) (cond ((null? tree) 0) ;base case ((leaf? tree) 1) ;base case (else ;recursive case (+ (countleaves (car tree)) (countleaves (cdr tree)))))) (define my-tree (list 4 (list 5 7) 2)) 42 57

16 List abstraction (sequence operations) Find common high order patterns Distill them into high order procedures Use these procedures to simplify list operations Mapping Filtering Accumulating Patterns:

17 Mapping (define (map proc lst) (if (null? lst) null (cons (proc (car lst)) (map proc (cdr lst))))) (define (square-list lst) (map square lst)) (define (scale-list lst c) (map (lambda (x) (* c x)) lst)) (scale-list (integers-between 1 5) 10) ==> ( )

18 Mapping: process (define (map proc lst) (if (null? lst) null (cons (proc (car lst)) (map proc (cdr lst))))) (map square (list 1 2 3)) (cons (square 1) (map square (list 2 3))) (cons 1 (map square (list 2 3))) (cons 1 (cons (square 2) (map square (list 3)))) (cons 1 (cons 4 (map square (list 3)))) (cons 1 (cons 4 (cons (square 3) (map square null)))) (cons 1 (cons 4 (cons 9 (map square null)))) (cons 1 (cons 4 (cons 9 null))) (1 4 9)

19 Generalized Mapping (map + (list 1 2 3) (list ) (list )) ==> ( ) (map (lambda (x y) (+ x (* 2 y))) (list 1 2 3) (list 4 5 6)) ==> ( ) (map … ) Returns a list in which proc is applied to the i-th elements of the lists respectively. We will see how to write such a procedure later!

20 Filtering (define (filter pred lst) (cond ((null? lst) null) ((pred (car lst)) (cons (car lst) (filter pred (cdr lst)))) (else (filter pred (cdr lst))))) (filter odd? (integers-between 1 10)) ( )

21 Filtering: process (define (filter pred lst) (cond ((null? lst) null) ((pred (car lst)) (cons (car lst) (filter pred (cdr lst)))) (else (filter pred (cdr lst))))) (filter odd? (list )) (cons 1 (filter odd? (list 2 3 4))) (cons 1 (filter odd? (list 3 4))) (cons 1 (cons 3 (filter odd? (list 4)))) (cons 1 (cons 3 (filter odd? null))) (cons 1 (cons 3 null)) (1 3)

22 Finding all the Primes: Sieve of Eratosthenes (a.k.a. Beta)

23.. And here’s how to do it! (define (sieve lst) (if (null? lst) ‘() (cons (car lst) (sieve (filter (lambda (x) (not (divisible? x (car lst)))) (cdr lst)))))) (cons 2 (sieve (filter (lambda (x) (not (divisible? X 2) (list …100 )))) ==> (sieve (list … 100))

24 How sieve works Sieve takes as argument a list of numbers L and returns a list M. Take x, the first element of L and make it the first element in M. Drop all numbers divisible by x from (cdr L). Call sieve on the resulting list, to generate the rest of M. (define (sieve lst) (if (null? lst) ‘() (cons (car lst) (sieve (filter (lambda (x) (not (divisible? x (car lst)))) (cdr lst))))))

25 Another example Find the number of integers x in the range [1…100] s.t.: x * (x + 1) is divisible by 6. (length (filter _____________________________________ (map _________________________________ _________________________________)))) (lambda(n) (* n (+ n 1))) (integers-between 1 100) (lambda(n) (= 0 (remainder n 6))) Any bets on the result???? 66 (about two thirds)

26 A new pattern: Accumulation

27 Accumulating (define (add-up lst) (if (null? lst) 0 (+ (car lst) (add-up (cdr lst))))) (define (mult-all lst) (if (null? lst) 1 (* (car lst) (mult-all (cdr lst))))) (define (accumulate op init lst) (if (null? lst) init (op (car lst) (accumulate op init (cdr lst))))) Add up the elements of a list Multiply all the elements of a list

28 Accumulating (cont.) (define (accumulate op init lst) (if (null? lst) init (op (car lst) (accumulate op init (cdr lst))))) (define (add-up lst) (accumulate + 0 lst)) (define (mult-all lst) (accumulate * 1 lst)) el n init op el n-1 el 1 …….. op...

29 Accumulate: process (define (accumulate op init lst) (if (null? lst) init (op (car lst) (accumulate op init (cdr lst)))))  (accumulate append ‘()(list (list 1) (list 2) (list 3))) (append ‘(1) (accumulate append ‘() ‘((2)(3)))) (append ‘(1) (append ‘(2) (accumulate append ‘()‘((3))))) (append ‘(1) (append ‘(2) (append ‘(3) (accumulate append ‘()‘())))) (append ‘(1) (append ‘(2) (append ‘(3) ‘()))) (append ‘(1) (append ‘(2) ‘(3))) (append ‘(1) ‘(2 3)) (1 2 3)

30 Length and append as accumulation (define (length lst) (accumulate (lambda (x y) (+ 1 y)) 0 lst)) (define (append lst1 lst2) (accumulate cons lst2 lst1) 0 el n +1 el n-1 el 1 …… el n lst2 el n-1 el 1 …….....

31 Implementing map and filter using accumulate (define (map proc lst) (accumulate (lambda (x y) (cons (proc x) y)) ‘() lst)) (define (filter pred lst) (accumulate (lambda (x y) (if (pred x) (cons x y) y) ‘() lst))

32 Another accumulate example We need to implement (more-evens lst), that receives a list of integers and returns #t iff lst contains more even integers than odd integers. (define (more-evens? l) (positive? (accumulate _________________________ _________________________ ___ l))) (lambda(x y) (+ y (if (even? x) 1 –1))) 0

33 Yet another example: dot product (dot-product x y) returns  i x i y i (define (dot-product x y) (accumulate ) + 0 (map * x y) The generalized map!

10/22/ Lists as interfaces

10/22/ (define (sum-odd-squares tree) (cond ((null? tree) 0) ((leaf? tree) (if (odd? tree) (square tree) 0)) (else (+ (sum-odd-squares (car tree)) (sum-odd-squares (cdr tree)))))) (define (even-fibs n) (define (next k) (if (< n k) nil (let ((f (fib k))) (if (even? f) (cons f (next (+ k 1))) (next (+ k 1)))))) (next 0)) Common Structure

10/22/ Even-fibs Enumerates the integers from 0 to n Computes the Fibonacci number for each integer Filters them selecting the even ones Accumulates the results using cons Sum-odd-squares Enumerates the leaves of a tree Filters them, selecting the odd ones Squares each of the selected ones Accumulates the results using +

10/22/ enumerate leaves filter map accumulate tree odd? square + 0 integers between map filter accumulate 0, n fib even? consnil even-fibs sum-odd-squares in a tree Interface: Along horizontal arrows flow lists of numbers

10/22/ even-fibs - Implementation (define (even-fibs n) (accumulate cons nil (filter even? (map fib (integers-between 0 n))))) integers between map filter accumulate 0, n fib even? consnil

sum-odd-squares - Implementation (define (sum-odd-squares tree) (accumulate + 0 (map square (filter odd? (enumerate-leaves tree))))) enumerate leaves filter map accumulate tree odd? square + 0

40 Example : Even-fibs-prod (product of all fib numbers with even index between lo and hi) Enumerate the integers from lo to hi Filter them selecting the even ones Compute the Fibonacci number for each integer using map Accumulates the results using * and 1 Lists as interfaces An algorithm as a series of steps performed on sequences (lists). Each step performs a simple operation. Start with a given list, or by creating a list. Other steps use filter or map. Finally, generate final result by accumulation.

41 (define (even-fibs-prod lo hi) (accumulate * 1 (map fib (filter even? (integers-between lo hi))))) Compare with: (define (even-fibs-prod lo hi) (cond ((> lo hi) 1) ((even? lo) (* (fib lo) (even-fibs-prod (+ lo 1) hi))) (else (even-fibs-prod (+ lo 1) hi)))) even-fibs-prod

42 Even-fibs-prod Interface: Along horizontal arrows flow lists of numbers generate filter map accumulate lo, hi even? fib *0 integers between generate filter/map accumulate.... (optional) A common way to structure a computation:

43 What did we gain A conventional way to partition a problem into small tasks. Resulting code is clearer, easier to design and understand. But sometimes less efficient.