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Your turn (Review)

Your turn (Review). What does a lambda expression return when it is evaluated? the value of a lambda expression is a procedure What three things are in the code of every recursive algorithm? a recursive case a base case a test

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Your turn (Review)

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  1. Your turn (Review) • What does a lambda expression return when it is evaluated? • the value of a lambda expression is a procedure • What three things are in the code of every recursive algorithm? • a recursive case a base case a test • Which of the following is a use of lambda?1. (define fred +) 2. (define (fred x) (+ x x)) • only 2: (define fred (lambda (x) (+ x x))) מבוא מורחב

  2. (define (sqrt x) (sqrt-iter 1.0 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)) 0.001)) (define (improve guess x) (average guess (/ x guess))) מבוא מורחב

  3. Block Structure (define (sqrt x) (define (good-enough? guess x) (< (abs (- (square guess) x)) 0.001)) (define (improve guess x) (average guess (/ x guess))) (define (sqrt-iter guess x) (if (good-enough? guess x) guess (sqrt-iter (improve guess x) x))) (sqrt-iter 1.0 x)) מבוא מורחב

  4. Binding names קשירת שמות • A procedure F binds a name x if either • 1. x is a formal parameter of F • 2. x is a name defined internally in F (define (F x y z) (define (h y) (* y y) (+ (h x) (h y) (h z))) • h, x, y, z are bound in F • y is bound in h

  5. עידון מודל ההצבה An occurrence of a name is bound by the innermost procedure that binds the name and contains the occurrence in its body. When replacing a formal parameter by the corresponding argument, do not substitute for occurences that are bound by an internal definition. מבוא מורחב

  6. refining the substitution model (define (sqrt x) (define (good-enough? guess x) (< (abs (- (square guess) x)) 0.001)) (define (improve guess x) (average guess (/ x guess))) (define (sqrt-iter guess x) (if (good-enough? guess x) guess (sqrt-iter (improve guess x) x))) (sqrt-iter 1.0 x)) (sqrt 2) מבוא מורחב

  7. (sqrt 2) (define (good-enough? guess x) (< (abs (- (square guess) x)) 0.001)) (define (improve guess x) (average guess (/ x guess))) (define (sqrt-iter guess x) (if (good-enough? guess x) guess (sqrt-iter (improve guess x) x))) (sqrt-iter 1.0 2) מבוא מורחב

  8. Scheme uses Lexical Scoping • ערך משתנה חופשי בפונקציהF נקבע על פי מקום הגדרתF. (ולא על פימקןםהפעלתF) • פונקציה המכילה את הגדרתF תקבע את ערכו. מבוא מורחב

  9. (define (sqrt x) (define (good-enough? guess x) (< (abs (- (square guess) x)) 0.001)) (define (improve guess x) (average guess (/ x guess))) (define (sqrt-iter guess x) (if (good-enough? guess x) guess (sqrt-iter (improve guess x) x))) (sqrt-iter 1.0 x)) מבוא מורחב

  10. Taking advantage of Lexical Scoping (define (sqrt x) (define (good-enough? guess) (< (abs (- (square guess) x)) 0.001)) (define (improve guess) (average guess (/ x guess))) (define (sqrt-iter guess) (if (good-enough? guess) guess (sqrt-iter (improve guess)))) (sqrt-iter 1.0)) (sqrt 2) מבוא מורחב

  11. (sqrt 2) (define (good-enough? guess) (< (abs (- (square guess) 2)) 0.001)) (define (improve guess) (average guess (/ 2 guess))) (define (sqrt-iter guess) (if (good-enough? guess) guess (sqrt-iter (improve guess)))) (sqrt-iter 1.0) (if (good-enough? 1.0) 1.0 (sqrt-iter (improve 1.0))) מבוא מורחב

  12. (define (h x) (define (f y) (+ x y)) (define (g x) (f x)) (g (* 2 x))) => (h 1) (define (f y) (+ 1 y)) (define (g x) (f x)) (g (* 2 1)) (f 2) (+ 1 2) 3 מבוא מורחב

  13. Applicative 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

  14. 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

  15. Normal ((lambda (x) (+ x x)) (* 3 4)) (+ (* 3 4) (* 3 4)) (+ 12 12) 24 Applicative ((lambda (x) (+ x x)) (* 3 4)) ((lambda (x) (+ x x)) 12) (+ 12 12) 24 This may matter in some cases: ((lambda (x y) (+ x 2)) 3 (/ 1 0)) מבוא מורחב

  16. Compute ab • (exp-1 2 2) ==> 4 • wishful thinking : • base case : a * a(b-1) a0 = 1 • (define exp-1 • (lambda (a b) • (if (= b 0) • 1 • (* a (exp-1 a (- b 1)))))) מבוא מורחב

  17. exp-1 is a recursive algorithm • In a recursive algorithm, bigger operands => more space • (define exp-1 • (lambda (a b) • (if (= b 0) 1 • (* a (exp-1 a (- b 1))))))(fact 4) • (exp-1 2 3) • (if (= 3 0) 1 (* 2 (exp-1 2 (- 3 1)))) • (* 2 (exp-1 2 2)) • .. • (* 2 (* 2 (exp-1 2 1))) • .. • (* 2 (* 2 (* 2 (exp-1 2 0)))) • An iterative algorithm uses constant space מבוא מורחב

  18. Intuition for iterative ab • same as you would do if calculating 24 by hand: 1. start with 1 (4 more to go) 1. multiply 1 by 2 gives 2 (3 more to go) 2. multiply 2 by 2 gives 4 (2 more to go) 3. multiply 4 by 2 gives 8 (1 more to go) 4. multiply 8 by 2 gives 16 (0 more to go) • At each step, only need to remember: current product, how many times has yet to multiply • Therefore, constant space מבוא מורחב

  19. In scheme: (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) מבוא מורחב

  20. A trace (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-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

  21. operation pending Recursive = pending operations when procedure calls itself • Recursive exponentiation: (define exp-1 (lambda (a b) (if (= b 0) 1 (* a (exp-1 a (- b 1)))))) (exp-1 2 3) (if (= 3 0) 1 (* 2 (exp-1 2 (- 3 1)))) (* 2 (exp-1 2 2)) .. (* 2 (* 2 (exp-1 2 1))) .. (* 2 (* 2 (* 2 (exp-1 2 0))))(fact 4) • Pending ops make the expression grow continuously

  22. Iterative = no pending operations • Iterative exponentiation: (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-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 • Fixed size because no pending operations no pending operations

  23. Summary • Iterative algorithms have constant space • Using substitution the expression doesn’t grow. • Iterative algorithms have no pending operations when the procedure calls itself • How to develop an iterative algorithm • figure out a way to accumulate partial answers • translate rules into scheme code מבוא מורחב

  24. Orders of growth of processes • Suppose n is a parameter that measures the size of a problem • Let R(n) be the amount of resources needed to compute a 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. • 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 increase by a factor of 4 מבוא מורחב

  25. Orders of growth of processes • We say R(n) has order of growth Q(f(n)) if there are constants k1 and k2 such that k1f(n)<= R(n)<= k2f(n)for all n • R(n)  O(f(n)) if there is a constant k such that R(n) <= kf(n) for all n • R(n)  (f(n)) if there is a constant k such that kf(n)<= R(n) for all n 100n2 O(n) x 100n2  (n)  k1, k2, k  0 x 100n2 Q (n)  100n2 Q (n2) מבוא מורחב

  26. Resources consumed by exp-1 • In a recursive algorithm, bigger operands => more space • (define exp-1 • (lambda (a b) • (if (= b 0) 1 • (* a (exp-1 a (- b 1))))))(fact 4) • (exp-1 2 3) • (if (= 3 0) 1 (* 2 (exp-1 2 (- 3 1)))) • (* 2 (exp-1 2 2)) • .. • (* 2 (* 2 (exp-1 2 1))) • .. • (* 2 (* 2 (* 2 (exp-1 2 0)))) • Space Q(b) • Time Q(b) מבוא מורחב

  27. Resources consumed by exp-2 (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-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 • Space Q(1) • Time Q(b) מבוא מורחב

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