מבוא מורחב - שיעור 2 1 Lecture 2 - Substitution Model (continued) - Recursion - Block structure and scope (if time permits)

מבוא מורחב - שיעור Evaluation of An Expression To Apply a compound procedure: (to a list of arguments) Evaluate the body of the procedure with the formal parameters replaced by the corresponding actual values To Evaluate a combination: (other than special form) a.Evaluate all of the sub-expressions in any order b.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 a built-in operator: machine instructions to execute The value of any name: the associated object in the environment

מבוא מורחב - שיעור 2 3 lambda: (lambda (x y) (+ x y x 2)) 1st operand position: the parameter list (x y)  a list of names (perhaps empty) 2nd operand position: the body (+ x y x 2)  may be any sequence of expressions The value of a lambda expression is a compound procedure. Reminder: Lambda special form

מבוא מורחב - שיעור 2 4 Evaluating expressions To Apply a compound procedure: (to a list of arguments) Evaluate the body of the procedure with the formal parameters replaced by the corresponding actual values ==> ((lambda(x)(* x x)) 5) Proc(x)(* x x) 5 (* 5 5) 25

מבוא מורחב - שיעור Using Abstractions ==> (square 3) 9 ==> (+ (square 3) (square 4)) ==> (define square (lambda(x)(* x x))) (* 3 3) (* 4 4) Environment Table NameValue squareProc (x)(* x x)

מבוא מורחב - שיעור Yet More Abstractions ==> (define f (lambda(a) (sum-of-two-squares (+ a 3) (* a 3)))) ==> (sum-of-two-squares 3 4) 25 ==> (define sum-of-two-squares (lambda(x y)(+ (square x) (square y)))) Try it out…compute (f 3) on your own

מבוא מורחב - שיעור Syntactic Sugar for naming procedures (define square (lambda (x) (* x x)) (define (square x) (* x x)) Instead of writing: We can write:

מבוא מורחב - שיעור (define second ) (second ) ==> 15 (second ) ==> -5 Some examples: (lambda (x) (* 2 x)) (lambda (x y z) y) Using “syntactic sugar”: (define (twice x) (* 2 x)) Using “syntactic sugar”: (define (second x y z) y) (define twice ) (twice 2) ==> 4 (twice 3) ==> 6

מבוא מורחב - שיעור Lets not forget The Environment ==> (define x 8) ==> (+ x 1) 9 ==> (define x 5) ==> (+ x 1) 6 The value of (+ x 1) depends on the environment!

מבוא מורחב - שיעור 2 10 Using 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 (/ (+ 5 9) 2) (/ 14 2)if operator is a primitive procedure, 7replace by result of operation

מבוא מורחב - שיעור 2 11 Booleans Two distinguished values denoted by the constants #t and #f The type of these values is boolean ==> (< 2 3) #t ==> (< 4 3) #f

מבוא מורחב - שיעור 2 12 Values and types Values have types. For example: In scheme almost every expression has a value Examples: 1)The value of 23 is 23 2)The value of + is a primitive procedure for addition 3)The value of (lambda (x) (* x x)) is the compound procedure proc (x) (* x x) 1)The type of 23 is numeral 2)The type of + is a primitive procedure 3)The type of proc (x) (* x x) is a compound procedure 4)The type of (> x 1) is a boolean (or logical)

מבוא מורחב - שיעור 2 13 No Value? In scheme almost every expression has a value Why almost? Example : what is the value of the expression (define x 8) In scheme, the value of a define expression is “undefined”. This means “implementation-dependent” Dr. Scheme does not return (print) any value for a define expression. Other interpreters may act differently.

מבוא מורחב - שיעור 2 14 More examples ==> (define x 8) Name Value Environment Table 8x ==> (define x (* x 2)) ==> x 16 ==> (define x y) reference to undefined identifier: y ==> (define + -) # + ==> (+ 2 2) 0

מבוא מורחב - שיעור 2 15 The IF special form ERROR2 (if ) If the value of is #t, Evaluate and return it Otherwise Evaluate and return it (if ( 2 (if (

מבוא מורחב - שיעור 2 16 IF is a special form In a general form, we first evaluate all arguments and then apply the function (if ) is different: determines whether we evaluate or. We evaluate only one of them !

מבוא מורחב - שיעור 2 17 Using 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 (/ (+ 5 9) 2) (/ 14 2)if operator is a primitive procedure, 7replace by result of operation

מבוא מורחב - שיעור 2 18 Recursive Procedures How to create a process of unbounded length? Needed to solve more complicated problems. Start with a simple example.

מבוא מורחב - שיעור 2 19 Example: Sum of squares S(n) = ………. …… (n-1) 2 + n 2 Notice that: S(n) = S(n-1) + n 2 S(0) = 0 These two properties completely define the function S(n-1) Wishful thinking: if I could only solve the smaller instance …

מבוא מורחב - שיעור 2 20 An algorithm for computing sum of squares (define sum-squares (lambda (n) (if (= n 0) 0 (+ (sum-squares (- n 1)) (square n))))

מבוא מורחב - שיעור 2 21 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 What would have happened if ‘ if ’ was a function ?

מבוא מורחב - שיעור 2 22 (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)) (if #f 0 (+ (if #f 0 (+ (if #f 0 (+ (sum-squares 0) 1)) 4)) 9)) (if #f 0 (+ (if #f 0 (+ (if #f 0 (+ (if #t 0 (+ (sum-squares -1) 0)) 1)) 4)) 9)).. Evaluating (sum-squares 3) with IF as regular form (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 - שיעור 2 23 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: small (non-decomposable) problem recursive case: larger (decomposable) problem at least one base case, and at least one recursive case.

מבוא מורחב - שיעור 2 24 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 - שיעור 2 25 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(s) 3. base case(s)

מבוא מורחב - שיעור 2 26 Block Structure Lets write a procedure that given x, y, and z computes f(x,y,z) = (x+y) 2 + (x+z) 2 (define (sum-and-square x y) (square (+ x y))) (define (f x y z) (+ (sum-and-square x y) (sum-and-square x z)))

מבוא מורחב - שיעור 2 27 Block structure (cont.) (define (f x y z) (define (sum-and-square x y) (square (+ x y))) (+ (sum-and-square x y) (sum-and-square x z))) Lets write a procedure that given inputs x, y, and z, computes f(x,y,z) = (x+y) 2 + (x+z) 2 while keeping sum-and-square private to f (hidden from the outside world):

מבוא מורחב - שיעור 2 28 We still need to clarify the substitution model.. (define (f x y z) (define (sum-and-square x y) (square (+ x y))) (+ (sum-and-square x y) (sum-and-square x z))) ==> (f 1 2 3) (define (sum-and-square 1 2) (square (+ 1 2))) (+ (sum-and-square 1 2) (sum-and-square 1 3)))

מבוא מורחב - שיעור 2 29 Bounded variables and scope A procedure definition binds its formal parameters The scope of the formal parameter is the body of the procedure. This is called lexical scoping (define (f x y z) (define (sum-and-square x y) (square (+ x y))) (+ (sum-and-square x y) (sum-and-square x z))) x,y,z x,y

מבוא מורחב - שיעור 2 30 Evaluation of An Expression (refined) To Apply a compound procedure: (to a list of arguments) Evaluate the body of the procedure with the formal parameters replaced by the corresponding actual values. Do not substitute for occurrences that are bound by an internal definition. To Evaluate a combination: (other than special form) a.Evaluate all of the sub-expressions in any order b.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 a built-in operator: machine instructions to execute The value of any name: the associated object in the environment

מבוא מורחב - שיעור 2 31 The refined substitution model (define (f x y z) (define (sum-and-square x y) (square (+ x y))) (+ (sum-and-square x y) (sum-and-square x z))) ==> (f 1 2 3) (define (sum-and-square x y) (square (+ x y))) (+ (sum-and-square 1 2) (sum-and-square 1 3)))

מבוא מורחב - שיעור 2 32 The refined substitution model ==> (f 1 2 3) (define (sum-and-square x y) (square (+ x y))) (+ (sum-and-square 1 2) (sum-and-square 1 3))) (+ (sum-and-square 1 2) (sum-and-square 1 3))) Sum-and-square Proc (x y) (square (+ x y))

מבוא מורחב - שיעור 2 33 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 G = 1X = 2 X/G = 2G = ½ (1+ 2) = 1.5 X/G = 4/3G = ½ (3/2 + 4/3) = 17/12 = X/G = 24/17G = ½ (17/ /17) = 577/408 =

מבוא מורחב - שיעור 2 34 (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)) (define initial-guess 1.0) (define precision )

מבוא מורחב - שיעור 2 35 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.

מבוא מורחב - שיעור 2 36 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-guess is an internal detail. Exporting only what is needed leads to: A clear interface Avoids confusion

מבוא מורחב - שיעור 2 37 Rewriting SQRT (Block structure) (define (sqrt x) (define (good-enough? guess x) (< (abs (- (square guess) x)) precision)) (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))) (define initial-guess 1.0) (define precision ) (sqrt-iter initial-guess x))

מבוא מורחב - שיעור 2 38 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 we do not have to explicitly pass x as a formal variable to all procedures. Instead, can leave it unbounded (“free”).

מבוא מורחב - שיעור 2 39 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 ) (sqrt-iter initial-guess))

מבוא מורחב - שיעור 2 40 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 ) (sqrt-iter initial-guess))

מבוא מורחב - שיעור 2 41 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

מבוא מורחב - שיעור 2 42 Another example for lexical scope (define (proc1 x) (define (proc2 y) (+ x y)) (define (proc3 x) (proc2 x)) (proc3 (* 2 x))) Proc3.x Proc1.x (proc1 4) proc1.x = 4 (proc3 8) proc3.x = 8 (proc2 8) proc2.y = 8 proc2.x=proc1.x=4 12