Lecture 2

1. Lecture 2 מבוא מורחב

2. Review: syntax vs. semantics Syntax Semantics מבוא מורחב

3. ==> (define score 23) ==> (* (+ 5 6 ) (- score (* 2 3 2 ))) * + 5 6 - 23 * 2 3 2 11 12 11 121 An example מבוא מורחב

4. Substitution model • To Apply a compound procedure to a list of arguments: • Replace the formal parameters with the corresponding actual values. • Evaluate the body of the procedure with these values. • (define square (lambda (x) (* x x))) • 1. (square 4) • 2. (* 4 4) • 3. 16 מבוא מורחב

5. Review: evaluation of an expression • To Evaluate a combination (other than special form) or a compound procedure • Evaluate all of the sub-expressions in any order • Apply the procedure that is the value of the leftmost sub-expression to the arguments (the values of the other sub-expressions) • The value of a numeral: number • The value of any name: the associated object in the environment • The value of a built-in operator: machine instructions to execute מבוא מורחב

6. IF special form (if <predicate> <consequent> <alternative>) (if (< 2 3) 2 3) ==> 2 (if (< 2 3) 2 (/ 1 0)) ==> ERROR 2 • If the value of <predicate> is #t, • Evaluate <consequent> and return it • Otherwise • Evaluate <alternative> and return it מבוא מורחב

7. IF is a special form • In a general form, we first evaluate all arguments and then apply the function • (if <predicate> <consequent> <alternative>)is different: <predicate> determines whether we evaluate <consequent> or <alternative>. We evaluate only one of them Can you see why it has to be that way? מבוא מורחב

8. Lambda special form • lambda syntax (lambda (x y) (+ x y x 2)) • 1st operand position: the parameter list(x y) • a list of names (perhaps empty) • determines the number of operands required • 2nd operand position: the body(+ x y x 2) • may be any expression • not evaluated when the lambda is evaluated • evaluated when the procedure is applied מבוא מורחב

9. Naming procedures • An application of an unnamed procedure: • ((lambda (x) (* x x)) 4) ==> 16 • 2. Naming the procedure: • (define square (lambda (x) (* x x))) • (square 3) 3. Syntactic Sugar: (define (square x) (* x x))

10. Some examples: • (define twice ) • (twice 2) ==> 4 (twice 3) ==> 6 • (define second ) • (second 2 15 3) ==> 15(second 34 -5 16) ==> -5 (lambda (x) (* 2 x)) (lambda (x y z) y) מבוא מורחב

11. An example for the substitution model (define square (lambda (x) (* x x)))(define average (lambda (x y) (/ (+ x y) 2))) (average 5 (square 3))(average 5 (* 3 3))(average 5 9) first evaluate operands,then substitute (applicative order) (/ (+ 5 9) 2)(/ 14 2) if operator is a primitive procedure, 7replace by result of operation מבוא מורחב

12. S(n-1) Sum of squares • S(n) = 02 + 12 + 22 ………. …… (n-1)2 + n2 • Notice that: • S(n) = S(n-1) + n2 • S(0) = 0 • These two properties completely define the function

13. An algorithm for sum of squares (define sum-squares (lambda (n) (if (= n 0) 0 (+ (sum-squares (- n 1)) (square n))))

14. Evaluating (sum-squares 3) (define (sum-squares n) (if (= n 0) 0 (+ (sum-squares (- n 1)) (square n)))) (sum-squares 3) (if (= 3 0) 0 (+ (sum-squares (- 3 1)) (square 3))) (+ (sum-squares (- 3 1)) (square 3)) (+ (sum-squares (- 3 1)) (* 3 3)) (+ (sum-squares(- 3 1)) 9) (+ (sum-squares 2) 9) (+ (if (= 2 0) 0 (+ (sum-squares (- 2 1)) (square 2))) 9) … (+ (+ (sum-squares 1) 4) 9) … (+ (+ (+ (sum-squares 0) 1) 4) 9) (+ (+ (+ (if (= 0 0) 0(+ (sum-squares (- 0 1)) (square 0))) 1) 4) 9) (+ (+ (+ 0 1) 4) 9) … 14 מבוא מורחב

15. How to design recursive algorithms • Show how to solve big instances, if you know the solution to • smaller instances. • Wishful thinking: if I could only solve the smaller instance … • Solve the “base cases”. For example, for sum of squares: S(n) = S(n-1) + n2 (induction rule) S(0)=0 (base case) מבוא מורחב

16. General form of recursive algorithms • test, base case, recursive case (define sum-sq (lambda (n) (if (= n 0) ; test for base case 0 ; base case (+ (sum-sq (- n 1)) (square n)) ; recursive case ))) • base case: non-decomposable problem • recursive case: larger (decomposable) problem מבוא מורחב

17. Evaluating (sum-squares 3) in the substitution model,with IF a regular form…. (sum-squares 3) (if (= 3 0) 0 (+ (sum-squares (- 3 1)) (square 3))) (if #f 0 (+ (sum-squares 2) 9)) (if #f 0 (+ (if #f 0 (+(sum-squares 1)4))) 9)) .. calling (sum-squares 0) …. calling (sum-squares -1) (define (sum-squares n) (if (= n 0) 0 (+ (sum-squares (- n 1)) (square n)))) We evaluate all operands. We always call (sum-squares) again. We get an infinite loop…….. OOPS מבוא מורחב

18. Another example of a recursive algorithm • even? (define even? (lambda (n) (not (even? (- n 1))) ; recursive case ))) • (if (= n 0) ; test for base case • #t ; base case מבוא מורחב

19. Short summary • Design a recursive algorithm by 1. Solving big instances using the solution to smaller instances. 2. Solving directly the base cases. • Recursive algorithms have 1. test 2. recursive case 3. base case מבוא מורחב

20. 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 SQRT • To find an approximation of square root of x: • Make a guess G • Improve the guess by averaging G and x/G • Keep improving the guess until it is good enough מבוא מורחב

21. (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)) מבוא מורחב

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

23. 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 מבוא מורחב

24. 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)) מבוא מורחב

25. Lexical Scoping • Every variable is: • Recognized within the procedure where it is defined (its scope). • Not recognized outside it. • Allows different procedures to use the same variable name. E.g., different procedures can use the variable name i as a counter. • A variable can be used globally within its scope. No need to pass it from one procedure to another. מבוא מורחב

26. SQRT again, x is used globally. (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)) מבוא מורחב

27. Variables scope (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)) מבוא מורחב

28. Proc1.x Proc3.x An example (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 מבוא מורחב