Scheme Exercises - Mutable List, RPN Calculator, Vectors and Sorting

1. מבוא מורחב למדעי המחשב בשפת Scheme תרגול 11

2. Outline • Mutable list structure • RPN calculator • Vectors and sorting

3. (define z2  (cons (list 'a 'b)  (list 'a 'b))) (set-to-wow! z2) ((wow b) a b) (define (set-to-wow! x)  (set-car! (car x) 'wow)  x) (define x (list 'a 'b)) (define z1 (cons x x)) (set-to-wow! z1) ((wow b) wow b) eq? point to the same object, equal? same content (eq? (car z1) (cdr z1)) is true (eq? (car z2) (cdr z2)) is false

4. Map without list copying (define (map! f s) (if (null? s) 'done (begin (set-car! s (f (car s))) (map! f (cdr s))))) (define s '(1 2 3 4)) (map! square s) => done s => (1 4 9 16)

5. append! (define (append! x y) (set-cdr! (last-pair x) y) x) where: (define (last-pair x) (if (null? (cdr x)) x (last-pair (cdr x))))

6. append vs. append! (define x ‘(a b)) (define y ‘(c d)) (define z (append x y)) z ==> (a b c d) (cdr x) ==> ? (define w (append! x y)) w ==> (a b c d) (cdr x) ==> ?

7. Cycle (define (last-pair x) (if (null? (cdr x)) x (last-pair (cdr x)))) (define (make-cycle x) (set-cdr! (last-pair x) x) x) (define z (make-cycle (list 'a 'b 'c)))

8. What happens? (last-pair z) => z b a c

9. is-cycle? Examine the list and determine whether it contains a cycle, whether a program that tried to find the end of the list by taking successive cdrs would go into an infinite loop (define (is-cycle? c) (define (loop fast slow) (cond ((null? fast) #f) ((null? (cdr fast)) #f) ((eq? (cdr fast) slow) #t) (else (loop (cddr fast) (cdr slow))))) (loop c c))

10. Bounded Counter A bounded counter can be incremented or decremented in steps of 1, until upper and lower bounds are reached. syntax: (make-bounded-counter init bottom top) example: (define c (make-bounded-counter 3 1 5)) (counter-inc! c)  4 (counter-inc! c)  5 (counter-inc! c)  5 (counter-dec! c)  4

11. List Implementation Constructor: (define (make-bounded-counter init bottom top) (list init bottom top)) Selectors: (define (counter-value c) (car c)) (define (counter-bottom c) (cadr c)) (define (counter-top c) (caddr c))

12. List Implementation – cont. Mutators: (define (counter-inc! c) (if (< (counter-value c) (counter-top c)) (set-car! c (+ 1 (counter-value c)))) (counter-value c)) (define (counter-dec! c) (if (> (counter-value c) (counter-bottom c)) (set-car! c (- (counter-value c) 1))) (counter-value c))

13. Polish Notation • PostFix Notation: operands before operators • Expressions with binary operators can be written without Parentheses • Apply operators, from left to right, on the last two numbers • 5 7 4 - 2 * + • 5 3 2 * + • 5 6 + • 11 • Stack implementation: • Init: Empty stack • Number: insert! Into stack • Operator: apply on 2 top stack elements and insert! result into stack

14. Simulation (calc) 5 7 4 - TOP= 3 2 * TOP= 6 + TOP= 11 exit ok

15. Read & Eval • (read) - returns an expression from the user • (eval exp) - evaluates an expression • We will use these functions in: • (perform m) - receives a symbol of an operation, applies it on the two top numbers in the stack, and returns the result to the stack • (iter) - reads an input from the user. If it is a number it is pushed to the stack, if it is an operator we call perform

16. perform (define (perform m) (let ((arg2 ((stk 'top)))) ((stk 'delete!)) (let ((arg1 ((stk 'top)))) ((stk 'delete!)) ((stk 'insert!) ((eval m) arg1 arg2)))))

17. iter (define (iter) (let ((in (read))) (if (eq? in 'exit) 'ok (begin (cond ((number? in) ((stk 'insert!) in)) (else (perform in) (display "TOP= ") (display ((stk 'top))) (newline))) (iter)))))

18. Calc (define (calc) (let ((stk (make-stack))) (define (perform m) ...) (define (iter) ... ) (iter)))

19. Vectors Constructors: (vector v1 v2 v3 . . .) (make-vector size init) Selector: (vector-ref vec place) Mutator: (vector-set! vec place value) Other functions: (vector-length vec)

20. Example - accumulating (define (accumulate-vec op base vec) (define (helper from to) (if (> from to) base (op (vector-ref vec from) (helper (+ from 1) to)))) (helper 0 (- (vector-length vec) 1)))

21. Bucket Sort • Problem: Sorting numbers that distribute “uniformly” across a given interval (for example, [0,1) ). • Observation: The number of elements that fall within a sub-interval is proportional to the sub-interval’s size • Idea: • Divide into sub-intervals (buckets) • Throw each number into appropriate bucket • Sort within each bucket (any sorting method) • Adjoin all sorted buckets

22. Example Sorting: 0.31 0.44 0.72 0.89 0.10 0.05 0.97 0.23 0.56 0.68 If we want ~3 numbers in each bucket, we need 3-4 buckets. Using 3 buckets we have: Bucket 0.000 - 0.333: 0.31 0.10 0.05 0.23 Bucket 0.333 – 0.667: 0.44 0.56 Bucket 0.667 – 1 : 0.72 0.89 0.97 0.68 Now sort! Bucket 0.000 - 0.333:0.05 0.10 0.23 0.31 Bucket 0.333 – 0.667:0.44 0.56 Bucket 0.667 – 1 : 0.68 0.72 0.89 0.97

23. Analysis • Observation: If number of buckets is proportional to the number of elements, then the number of elements in each bucket is more or less the same, and bounded by a constant. • Dividing into buckets: O(n) • Sorting one bucket: O(1) • Sorting all buckets: O(n) • Adjoining sorted sequences: O(n)

24. Implementation • A bucket is a list • The set of buckets is a vector • The elements are sorted while inserted into the buckets(very similar so insertion sort): (define (insert x s) (cond ((null? s) (list x)) ((< x (car s)) (cons x s)) (else (cons (car s) (insert x (cdr s))))))

25. Implementation – for-each • Scheme primitive • Similar to map, without returning a value • Useful for side-effects (define (for-each proc lst) (if (null? lst) ‘done (begin (proc (car lst)) (for-each proc (cdr lst)))))

26. Implementation – cont. (define (bucket-sort s) (let* ((size 5) (n (ceiling (/ (length s) size))) (buckets (make-vector n null))) (define (insert! x) (let ((bucket (inexact->exact (floor (* n x))))) (vector-set! buckets bucket (insert x (vector-ref buckets bucket))))) (for-each insert! s) (accumulate-vec append null buckets)))

27. Streams 3.5, pages 316-352 definitions file on web 27

28. cons, car, cdr (define s (cons 9 (begin (display 7) 5))) -> prints 7 The display command is evaluated while evaluating the cons. (car s) -> 9 (cdr s) -> 5 28

29. cons-stream, stream-car, stream-cdr (define s (cons-stream 9 (begin (display 7) 5))) Due to the delay of the second argument,cons-stream does not activate the displaycommand (stream-car s) -> 9 (stream-cdr s) -> prints 7 and returns 5 stream-cdr activates the display which prints 7, and then returns 5. 29

30. List enumerate (define (enumerate-interval low high) (if (> low high) nil (cons low (enumerate-interval (+ low 1) high)))) (enumerate-interval 2 8) -> (2 3 4 5 6 7 8) (car (enumerate-interval 2 8)) -> 2 (cdr (enumerate-interval 2 8)) -> (3 4 5 6 7 8) 30

31. Stream enumerate (define (stream-enumerate-interval low high) (if (> low high) the-empty-stream (cons-stream low (stream-enumerate-interval (+ low 1) high)))) (stream-enumerate-interval 2 8) -> (2 . #<promise>) (stream-car (stream-enumerate-interval 2 8)) -> 2 (stream-cdr (stream-enumerate-interval 2 8)) -> (3 . #<promise>) 31

32. List map (map <proc> <list>) (define (map proc s) (if (null? s) nil (cons (proc (car s)) (map proc (cdr s))))) (map square (enumerate-interval 2 8)) -> (4 9 16 25 36 49 64) 32

33. Stream map (map <proc> <stream>) (define (stream-map proc s) (if (stream-null? s) the-empty-stream (cons-stream (proc (stream-car s)) (stream-map proc (stream-cdr s)) ))) (stream-map square (stream-enumerate-interval 2 8)) -> (4 . #<promise>) 33

34. List of squares (define squares (map square (enumerate-interval 2 8))) squares -> (4 9 16 25 36 49 64) (car squares) -> 4 (cdr squares) -> (9 16 25 36 49 64) 34

35. Stream of squares (define stream-squares (stream-map square (stream-enumerate-interval 2 8))) stream-squares -> (4 . #<promise>) (stream-car stream-squares) -> 4 (stream-cdr stream-squares) -> (9 . #<promise>) 35

36. List reference (define (list-ref s n) (if (= n 0) (car s) (list-ref (cdr s) (- n 1)))) (define squares (map square (enumerate-interval 2 8))) (list-ref squares 3) -> 25 36

37. Stream reference (define (stream-ref s n) (if (= n 0) (stream-car s) (stream-ref (stream-cdr s) (- n 1)))) (define stream-squares (stream-map square (stream-enumerate-interval 2 8))) (stream-ref stream-squares 3) -> 25 37

38. List filter (filter <predicate> <list>) (define (filter pred s) (cond ((null? s) nil) ((pred (car s)) (cons (car s) (filter pred (cdr s)))) (else (filter pred (cdr s))))) (filter even? (enumerate-interval 1 20)) -> (2 4 6 8 10 12 14 16 18 20) 38

39. Stream filter (stream-filter <predicate> <stream>) (define (stream-filter pred s) (cond ((stream-null? s) the-empty-stream) ((pred (stream-car s)) (cons-stream (stream-car s) (stream-filter pred (stream-cdr s)))) (else (stream-filter pred (stream-cdr s))) ))) (stream-filter even? (stream-enumerate-interval 1 20)) -> (2 . #<promise>) 39

40. Generalized list map (generalized-map <proc> <list1> … <listn>) (define (generalized-map proc . arglists) (if (null? (car arglists)) nil (cons (apply proc (map car arglists)) (apply generalized-map (cons proc (map cdr arglists)))))) (generalized-map + squares squares squares) -> (12 27 48 75 108 147 192) 40

41. Generalized stream map (generalized-stream-map <proc> <stream1> … <streamn>) (define (generalized-stream-map proc . argstreams) (if (stream-null? (car argstreams)) the-empty-stream (cons-stream (apply proc (map stream-car argstreams)) (apply generalized-stream-map (cons proc (map stream-cdr argstreams)))))) (generalized-stream-map + stream-squares stream-squares stream-squares) -> (12 . #<promise>) 41

42. List for each (define (for-each proc s) (if (null? s) 'done (begin (proc (car s)) (for-each proc (cdr s))))) 42

43. Stream for each (define (stream-for-each proc s) (if (stream-null? s) 'done (begin (proc (stream-car s)) (stream-for-each proc (stream-cdr s))))) useful for viewing (finite!) streams (define (display-stream s) (stream-for-each display s)) (display-stream (stream-enumerate-interval 1 20)) -> prints 1 … 20 done 43

44. Lists (define sum 0) (define (acc x) (set! sum (+ x sum)) sum) (define s (map acc (enumerate-interval 1 20))) s -> (1 3 6 10 15 21 28 36 45 55 66 78 91 105 120 136 153 171 190 210) sum -> 210 (define y (filter even? s)) y -> (6 10 28 36 66 78 120 136 190 210) sum -> 210 (define z (filter (lambda (x) (= (remainder x 5) 0)) s)) z -> (10 15 45 55 105 120 190 210)sum -> 210 44

45. (list-ref y 7) -> 136 sum -> 210 (display z) -> prints (10 15 45 55 105 120 190 210) sum -> 210 45

46. Streams (define sum 0) (define (acc x) (set! sum (+ x sum)) sum) (define s (stream-map acc (stream-enumerate-interval 1 20))) s -> (1 . #<promise>) sum -> 1 (define y (stream-filter even? s)) y -> (6 . #<promise>) sum -> 6 (define z (stream-filter (lambda (x) (= (remainder x 5) 0)) s)) z -> (10 . #<promise>) sum -> 10 46

47. (stream-ref y 7) -> 136 sum -> 136 (display-stream z) -> prints 10 15 45 55 105 120 190 210 done sum -> 210 47

48. Defining streams implicitlyby delayed evaluation Suppose we needed an infinite list of Dollars. We can (define bill-gates (cons-stream ‘dollar bill-gates)) If we need a Dollar we can take the car (stream-car bill-gates) -> dollar The cdr would still be an infinite list of Dollars. (stream-cdr bill-gates)->(dollar . #<promise>) 48

49. Infinite Streams Formulate rules defining infinite series wishful thinking is key 49

50. 1,1,1,… = ones = 1,ones (define ones (cons-stream 1 ones)) 50