1 / 34

Lecture 3

Lecture 3. Material in the textbook Sections 1.1.5 to 1.2.1. Today. Continue Lecture 2 (sqrt) Refine our model Applicative vs. Normal Order Understand how it captures the nature of processes Recursive vs. Iterative processes Orders of growth. מבוא מורחב. 2. X = 2. G = 1. X/G = 2.

robbin
Download Presentation

Lecture 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Lecture 3 Material in the textbook Sections 1.1.5 to 1.2.1 מבוא מורחב

  2. Today Continue Lecture 2 (sqrt) Refine our model Applicative vs. Normal Order Understand how it captures the nature of processes Recursive vs. Iterative processes Orders of growth מבוא מורחב 2

  3. X = 2 G = 1 X/G = 2 G = ½ (1+ 2) = 1.5 X/G = 4/3 G = ½ (3/2 + 4/3) = 17/12 = 1.416666 X/G = 24/17 G = ½ (17/12 + 24/17) = 577/408 = 1.4142156 Computing SQRT: A Numeric Algorithm • To find an approximation of square root of x, use the following recipe: • Make a guess G • Improve the guess by averaging G and x/G • Keep improving the guess until it is good enough

  4. (define initial-guess 1.0) (define precision 0.0001) (define (sqrt-iter guess x) (if (good-enough? guess x) guess (sqrt-iter (improve guess x) x))) (define (good-enough? guess x) (< (abs (- (square guess) x)) precision)) (define (improve guess x) (average guess (/ x guess))) (define (sqrt x) (sqrt-iter initial-guess x))

  5. Good programming Style • 1. Divide the task to well-defined, natural, and • simple sub-tasks. • E.g: good-enough? and improve. • Rule of thumb : If you can easily name it, it does a well-defined task. 2. Use parameters. E.g.: precision, initial-guess. 3. Use meaningful names.

  6. Procedural abstraction • It is better to: • Export only what is needed • Hide internal details. The procedure SQRT is of interest for the user. The procedure improve-guessis an internal detail. • Exporting only what is needed leads to: • A clear interface • Avoids confusion

  7. Rewriting SQRT (Block structure) (define (sqrt x) (define (sqrt-iter guess x) (if (good-enough? guess x) guess (sqrt-iter (improve guess x) x))) (define (good-enough? guess x) (< (abs (- (square guess) x)) precision)) (define (improve guess x) (average guess (/ x guess))) (define initial-guess 1.0) (define precision 0.00001) (sqrt-iter initial-guess x))

  8. Further improving sqrt Note that in every application of sqrt we substitute for x the same value in all subsequent applications of compound procedures ! Therefore wedo not have to explicitly pass x as a formal variable to all procedures. Instead, can leave it unbounded (“free”).

  9. SQRT again, taking advantage of the refined substitution model (define (sqrt x) (define (good-enough? guess) (< (abs (- (square guess) x)) precision)) (define (improve guess) (average guess (/ x guess))) (define (sqrt-iter guess) (if (good-enough? guess) guess (sqrt-iter (improve guess)))) (define initial-guess 1.0) (define precision 0.00001) (sqrt-iter initial-guess))

  10. SQRT (cont.) ==>(sqrt 2) (define (good-enough? guess) (< (abs (- (square guess) 2)) precision)) (define (improve guess) (average guess (/ 2 guess))) (define (sqrt-iter guess) (if (good-enough? guess) guess (sqrt-iter (improve guess)))) (define initial-guess 1.0) (define precision 0.00001) (sqrt-iter initial-guess))

  11. Lexical Scoping - again The lexical scoping rules means that the value of a variable which is unbounded (free) in a procedure f is taken from the procedure in which f was defined. It is also called static scoping

  12. Proc1.x Proc3.x Another example for lexical scope (define (proc1 x) (define (proc2 y) (+ x y)) (define (proc3 x) (proc2 x)) (proc3 (* 2 x))) (proc1 4) proc1.x = 4 (proc3 8) proc3.x = 8 (proc2 8) proc2.y = 8 proc2.x=proc1.x=4 12

  13. Applicative order vs. Normal OrderEvaluation To Evaluate a combination: (other than special form) Evaluate all of the sub-expressions in any order Applythe procedure that is the value of the leftmost sub-expression to the arguments (the values of the other sub-expressions) מבוא מורחב

  14. Applicative order evaluation rules Combination... (<operator> <operand1> …… <operand n>) • Evaluate <operator> to get the procedure and evaluate <operands> to get the arguments • If <operator> is primitive: do whatever magic it does • If <operator> is compound: evaluate body with formal parameters replaced by arguments מבוא מורחב

  15. Normal order evaluation Combination … (<operator> <operand1> …… <operand n>) • Evaluate <operator> to get the procedure and evaluate <operands> to get the arguments • If <operator> is primitive: do whatever magic it does • If <operator> is compound: evaluate body with formal parameters replaced by arguments מבוא מורחב

  16. The Difference Normal ((lambda (x) (+ x x)) (* 3 4)) (+ (* 3 4) (* 3 4)) (+ 12 12) 24 Applicative ((lambda (x) (+ x x)) (* 3 4)) (+ 12 12) 24 This may matter in some cases: ((lambda (x y) (+ x 2)) 3 (/ 1 0)) Scheme is an Applicative Order Language! מבוא מורחב

  17. Compute ab (Recursive Approach) • wishful thinking : • base case: ab=a * a(b-1) a0 = 1 • (define exp-1 • (lambda (a b) • (if (= b 0) • 1 • (* a (exp-1 a (- b 1)))))) מבוא מורחב

  18. ab=a * a * a*…*a • Which is: b ab = a2 *a*…*a= a3 *…*a • Operationally: • Halting condition: product  product * a counter  counter - 1 counter = 0 Compute ab (Iterative Approach) • Another approach: מבוא מורחב

  19. Syntactic Recursion Compute ab (Iterative Approach) (define (exp-2 a b) (define (exp-iter a counter product) (if (= counter 0) product (exp-iter a (- counter 1) (* a product))) (exp-iter a b 1)) How then, do the two procedures differ? They give rise to different processes – lets use our model to understand how. מבוא מורחב

  20. Recursive Process • (define exp-1 • (lambda (a b) • (if (= b 0) • 1 • (* a (exp-1 a (- b 1)))))) • (exp-1 3 4) • (* 3 (exp-1 3 3)) • (* 3 (* 3 (exp-1 3 2))) • (* 3 (* 3 (* 3 (exp-1 3 1)))) • (* 3 (* 3 (* 3 (* 3 (exp-1 3 0))))) • (* 3 (* 3 (* 3 (* 3 1)))) • (* 3 (* 3 (* 3 3))) • (* 3 (* 3 9)) • (* 3 27) • 81 מבוא מורחב

  21. Iterative Process (define (exp-2 a b) (define (exp-iter a counter product) (if (= counter 0) product (exp-iter a (- counter 1) (* a product))) (exp-iter a b 1)) (exp-2 3 4) (exp-iter 3 4 1) (exp-iter 3 3 3) (exp-iter 3 2 9) (exp-iter 3 1 27) (exp-iter 3 0 81) 81 מבוא מורחב

  22. (exp-1 3 4) • (* 3 (exp-1 3 3)) • (* 3 (* 3 (exp-1 3 2))) • (* 3 (* 3 (* 3 (exp-1 3 1)))) • (* 3 (* 3 (* 3 (* 3 (exp-1 3 0))))) • (* 3 (* 3 (* 3 (* 3 1)))) • (* 3 (* 3 (* 3 3))) • (* 3 (* 3 9)) • (* 3 27) • 81 (exp-2 3 4) Growing amount of space (exp-iter 3 4 1) (exp-iter 3 3 3) (exp-iter 3 2 9) Constant amount of space (exp-iter 3 1 27) (exp-iter 3 0 81) 81 The Difference מבוא מורחב

  23. operation pending no pending operations Why More Space? • Recursive exponentiation: (define exp-1 (lambda (a b) (if (= b 0) 1 (* a (exp-1 a (- b 1))))) • Iterative exponentiation: • (define (exp-2 a b) • (define (exp-iter a counter product) • (if (= counter 0) • product • (exp-iter a (- counter 1) (* a product)))) • (exp-iter a b 1))

  24. Example: Factorial wishful thinking : base case: n! =n * (n-1)! n = 1 (define fact (lambda (n) (if (= n 1) 1 (* n (fact (- n 1)))))) Iterative or Recursive? מבוא מורחב 24

  25. Summary • Recursive process num of deferred operations “grows proportional to b” • Iterative process num of deferred operations stays “constant” (actually it’s zero) Can we better quantify these observations? Orders of growth… מבוא מורחב

  26. Order of Growth: Recursive Process • (exp-1 3 4) • (* 3 (exp-1 3 3)) • (* 3 (* 3 (exp-1 3 2))) • (* 3 (* 3 (* 3 (exp-1 3 1)))) • (* 3 (* 3 (* 3 (* 3 (exp-1 3 0))))) • (* 3 (* 3 (* 3 (* 3 1)))) • (* 3 (* 3 (* 3 3))) • (* 3 (* 3 9)) • (* 3 27) • 81  4 • (exp-1 3 5) • (* 3 (exp-1 3 4)) • (* 3 (* 3 (exp-1 3 3))) • (* 3 (* 3 (* 3 (exp-1 3 2)))) Dependent on b • (* 3 (* 3 (* 3 (* 3 (exp-1 3 1))))) • (* 3 (* 3 (* 3 (* 3 (* 3 (exp-1 3 0))))) • (* 3 (* 3 (* 3 (* 3 (* 3 1)))) • (* 3 (* 3 (* 3 (* 3 3)))) • (* 3 (* 3 (* 3 9))) • (* 3 (* 3 27)) • (* 3 81)  5 • 243

  27. Iterative Process (define (exp-2 a b) (define (exp-iter a b product) (if (= b 0) product (exp-iter a (- b 1) (* a product)))) (exp-iter a b 1) (exp-2 3 4) (exp-2 3 5) (exp-iter 3 4 1) (exp-iter 3 4 1) (exp-iter 3 3 3) (exp-iter 3 3 3) (exp-iter 3 2 9) (exp-iter 3 2 9) (exp-iter 3 1 27) (exp-iter 3 2 27) (exp-iter 3 0 81) (exp-iter 3 0 81) (exp-iter 3 0 243) 81 243 Some constant, independent of b

  28. The worst-case over all inputs of size n Orders of Growth • Suppose n is a parameter that measures the size of a problem (the size of its input) • R(n)measures the amount of resources needed to compute a solution procedure of size n. • Two common resources are space, measured by the number of deferred operations, and time, measured by the number of primitive steps. מבוא מורחב

  29. Orders of Growth • Want to estimate the “order of growth” of R(n): R1(n)=100n2 R2(n)=2n2+10n+2 R3(n) = n2 Are all the same in the sense that if we multiply the input by a factor of 2, the resource consumption increases by a factor of 4 Order of growth is proportional to n2 מבוא מורחב

  30. Orders of Growth • We say R(n)has order of growth Q(f(n))if there are constants c1 0and c2 0 such that for all n c1f(n)<= R(n)<= c2f(n) • R(n)O(f(n)) if there is a constant c  0such that for all n • R(n) <= cf(n) • R(n)(f(n)) if there is a constant c  0such that for all n • cf(n)<= R(n) מבוא מורחב

  31. Orders of Growth True or False? f 100n2 O(n) t 100n2 (n) 100n2 Q(n) f 100n2 Q(n2) t True or False? f 2100 (n) t 2100n  O(n2) 2n  Q(n) f 210 Q(1) t מבוא מורחב

  32. (exp-1 3 4) “n”=b=4 • (* 3 (exp-1 3 3)) • (* 3 (* 3 (exp-1 3 2))) • (* 3 (* 3 (* 3 (exp-1 3 1)))) • (* 3 (* 3 (* 3 (* 3 (exp-1 3 0))))) • (* 3 (* 3 (* 3 (* 3 1)))) • (* 3 (* 3 (* 3 3))) • (* 3 (* 3 9)) • (* 3 27) • 81 Resources Consumed by EXP-1 • Space b <= R(b) <= b which is Q(b) • Time b <= R(b) <= 2b which is Q(b) Linear Recursive Process מבוא מורחב

  33. (exp-2 3 4) “n”=b=4 (exp-iter 3 4 1) (exp-iter 3 3 3) (exp-iter 3 2 9) (exp-iter 3 1 27) (exp-iter 3 0 81) 81 Resources Consumed by EXP-2 • Space Q(1) • Time Q(b) Linear Iterative Process מבוא מורחב

  34. Summary • Trying to capture the nature of processes • Quantify various properties of processes: • Number of steps a process takes (Time Complexity) • Amount of Space a process uses (Space Complexity) מבוא מורחב

More Related