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# Class 17: Golden Sneezewort

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Generalizing Loops
Fibonacci
Measuring Cost
Golden Ratio

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### Class 17: Golden Sneezewort

1. 1. Class 17:GoldenSneezewort cs1120 Fall 2011 David Evans 30 September 2011
2. 2. PlanGeneralizing LoopsIntroducing Cost Book: Chapter 7 and 8 (okay to read after Exam 1) Not covered on Exam 1 2
3. 3. Recap/Wake-up: A General Loop(define (loop index result test update proc) 3
4. 4. General Loop with Result(define (loop index result test update proc) (if (test index) (loop (update index) (proc index result) test update proc) result)) 4
5. 5. Using Loop (define (loop index result test update proc) (if (test index) (loop (update index) (proc index result) test update proc) result))(define (gauss-sum n) (loop 1 0 (lambda (i) (<= i n)) (lambda (i) (+ i 1)) (lambda (i res) (+ res i)))) 5
6. 6. (define (loop index result test update proc) (if (test index) (loop (update index) (proc index result) test update proc) result))(define (factorial n) 6
7. 7. (define (loop index result test update proc) (if (test index) (loop (update index) (proc index result) test update proc) result))(define (list-length p) 7
8. 8. (define (loop index result test update proc) (if (test index) (loop (update index) (proc index result) test update proc) result))(define (list-accumulate f base p) (if (null? p) base (f (car p) (list-accumulate f base (cdr p)))) 8
9. 9. (define (loop index result test update proc) (if (test index) (loop (update index) (proc index result) test update proc) result))(define (list-accumulate f base p) (if (null? p) base (f (car p) (list-accumulate f base (cdr p))))(define (list-accumulate f base p) (loop p base not-null? cdr (lambda (p res) (f (car p) res)))) 9
10. 10. Challenge ProblemDefine an efficient edit-distanceprocedure (as defined in PS2) usingloop (without using memoization). 10
11. 11. Sneezewort! 11
12. 12. Sneezewort GrowthFirst Time Unit Second Time Unit Offshoot Could we model Sneezewort with PS3 code? 12
13. 13. Sneezewort Numbers 13 8? 5 3 2 1 1 13
14. 14. Page fromLiber abbaci,Leonardo da Pisa(1202) 14
15. 15. Fibonacci’s ProblemSuppose a newly-born pair of rabbits, onemale, one female, are put in a field.Rabbits mate at the age of one month sothat at the end of its second month afemale can produce another pair ofrabbits.Suppose that our rabbits never die andthat the female always produces one newpair (one male, one female) every monthfrom the second month on.How many pairs will there be in one year? Filius Bonacci, 1202 (Pisa) Liber Abaci (“Book of Calculation”) 15
16. 16. RabbitsFrom http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibnat.html 16
17. 17. Fibonacci NumbersGEB p. 136: These numbers are best defined recursively by the pair of formulas FIBO (n) = FIBO (n – 1) + FIBO (n – 2) for n > 2 FIBO (1) = FIBO (2) = 1 Can we turn this into a Scheme procedure? 17
18. 18. Defining FIBO These numbers defined recursively by: FIBO (n) = FIBO (n – 1) + FIBO (n – 2) for n > 1 FIBO (1) = 1 FIBO (0) = 0 18
19. 19. Defining fibo(define (fibo n) > (fibo 1) 1 (if (= n 0) 0 > (fibo 2) (if (= n 1) 1 2 (+ (fibo (- n 1)) > (fibo 3) 3 (fibo (- n 2)))))) > (fibo 10) 55 19
20. 20. Defining fibo with loop? (define (loop index result test update proc) (if (test index) (loop (update index) (proc index result) test update proc) result)) 20
21. 21. (define (fibo-loop n) ; doesnt work for n=0 (cdr (loop 1 (cons 0 1) (lambda (i) (< i n)) inc (lambda (i v) (cons (cdr v) (+ (car v) (cdr v))))))) 21
22. 22. Which is better? (define (fibo-rec n) (if (= n 0) 0 (if (= n 1) 1 (+ (fibo-rec (- n 1)) (fibo-rec (- n 2))))))(define (fibo-loop n) ; doesnt work for n=0 (cdr (loop 1 (cons 0 1) (lambda (i) (< i n)) inc (lambda (i v) (cons (cdr v) (+ (car v) (cdr v))))))) 22
23. 23. Fibo Results> (fibo-rec 2)1> (fibo-rec 3)2> (fibo-rec 4)3> (fibo-rec 10)55> (fibo-rec 60)Still working… 23
24. 24. Tracing Fibo> (require racket/trace)> (trace fibo-rec)> (fibo-rec 3) > (fibo-loop 3)|(fibo-rec 3) >(fibo-loop 3)| (fibo-rec 2) > (loop 1 (0 . 1) #<procedure> #<procedure:inc> #<procedure> )| 1 > (loop 2 (1 . 1) #<procedure> #<procedure:inc> #<procedure> )| (fibo-rec 1) > (loop 3 (1 . 2) #<procedure> #<procedure:inc> #<procedure> )| 1 < (1 . 2)|2 <22 2 24
25. 25. > (fibo-rec 4)>(fibo-rec 4)> (fibo-rec 3)> >(fibo-rec 2)> > (fibo-rec 1)<<1 To compute (fibo 4) we calculated:> > (fibo-rec 0)<<0 (fibo 3) 1 times< <1 (fibo 2) 2 times> >(fibo-rec 1) (fibo 1) 3 times< <1<2 (fibo 0) 3 times> (fibo-rec 2) = 3+6 calls to fibo> >(fibo-rec 1)< <1 How many calls to calculate (fibo 60)?> >(fibo-rec 0)< <0<1<33 25
26. 26. > (fibo-rec 5)>(fibo-rec 5)> (fibo-rec 4)> >(fibo-rec 3)> > (fibo-rec 2)> > >(fibo-rec 1) To calculate (fibo 5) we calculated:< < <1> > >(fibo-rec 0) (fibo 4) 1 time< < <0<<1 (fibo 3) 2 times> > (fibo-rec 1)<<1 (fibo 2) 3 times< <2> >(fibo-rec 2) (fibo 1/0) 8 times> > (fibo-rec 1)<<1> > (fibo-rec 0)<<0< <1<3> (fibo-rec 3) How many calls to calculate (fibo 60)?> >(fibo-rec 2)> > (fibo-rec 1)<<1> > (fibo-rec 0)<<0< <1> >(fibo-rec 1)< <1<2<55 26
27. 27. Fast-Fibo Results > (time (fibo-loop 61)) cpu time: 0 real time: 0 gc time: 0 25047307819612.5 Trillion applications2.5 GHz computer does 2.5 Billion simple operations per second,so 2.5 Trillion applications operations take ~1000 seconds.Each application of fibo involves hundreds of simple operations… 27
28. 28. ;;; The Earths mass is 6.0 x 10^24 kg> (define mass-of-earth (* 6 (expt 10 24))) ;;; A typical rabbits mass is 2.5 kilograms> (define mass-of-rabbit 2.5)> (/ (* mass-of-rabbit (fibo-loop 60)) mass-of-earth)6.450036483e-013> (/ (* mass-of-rabbit (fibo-loop 120)) mass-of-earth)2.2326496895795693According to Bonacci’s model, after less than 10 years, rabbitswould out-weigh the Earth! Beware the Bunnies!! Beware the Sneezewort!! 28
29. 29. The Verge of Survival by Filip and Colton 29
30. 30. Evaluation CostActual running times How does timevary according to:– How fast a processor scale with the you have size of the input?– How much memory you have If the input size increases by– Where data is located in one, how much longer will it take? memory– How hot it is If the input size doubles, how– What else is running much longer will it take?– etc...
31. 31. Running Time of fibo-recFibonacci (n) = Fibonacci(n-2) + Fibonacci(n-1) = X Fibonacci(n-1) = Xn What is X? > (exact->inexact (/ (fibo-loop 2) (fibo-loop 1))) 1.0 > (exact->inexact (/ (fibo-loop 3) (fibo-loop 2))) 2.0 > (exact->inexact (/ (fibo-loop 4) (fibo-loop 3))) 1.5
32. 32. > (for 1 50 (lambda (i) (printf "~a~n" (exact->inexact (/ (fibo-loop (+ i 1)) (fibo-loop i))))))1.02.01.51.66666666666666671.61.6251.61538461538461541.6190476190476191.61764705882352941.61818181818181821.61797752808988761.61805555555555561.61802575107296141.61803713527851461.6180327868852461.6180344478216821.6180338134001253…1.618033988749895
33. 33. Running Time of fibo-rec Fibonacci (n) = Fibonacci(n-2) + Fibonacci(n-1) = ? Fibonacci(n-1) = ΦnΦ = (/ (+ 1 (sqrt 5)) 2)“The Golden Ratio” 1.618033988749895...> (exact->inexact (/ (fibo-loop 41) (fibo-loop 40)))1.618033988749895
34. 34. “Golden” Rotunda?
35. 35. The Golden Ratio Euclid: “extreme and mean ratio” Parthenonhttp://www.mathsisfun.com/numbers/golden-ratio.html Nautilus Shell 35
36. 36. ChargePS4 due Monday 3 OctoberExam 1: out October 7, due October 12 Covers: Problem Sets 1-4 including PS Comments Course Book Chapters 1-6 Classes 1-18Reading: no reading assignment until after Exam 1, but I will start covering things in Chapter 7-8 soon and encourage you to read The Information, Chapters 5-7 36