מבוא מורחב - שיעור 2 1 Lecture 2 - Substitution Model (continued) - Recursion - Block...

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Lecture 2 - Substitution Model (continued)

- Recursion

- Block structure and scope (if time permits)

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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 orderb. 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: numberThe value of a built-in operator: machine instructions to executeThe value of any name: the associated object in the environment

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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

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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

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Using Abstractions

==> (square 3)

9

==> (+ (square 3) (square 4))

==> (define square (lambda(x)(* x x)))

(* 3 3) (* 4 4)

9 16+

25

Environment Table

Name Value

square Proc (x)(* x x)

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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

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Syntactic Sugar for naming procedures

(define square (lambda (x) (* x x))

(define (square x) (* x x))

Instead of writing:

We can write:

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(define second )

(second 2 15 3) ==> 15

(second 34 -5 16) ==> -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

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Lets not forget The Environment

==> (define x 8)

==> (+ x 1)

9

==> (define x 5)

==> (+ x 1)

6The value of (+ x 1) depends on the environment!

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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, 7 replace by result of operation

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Booleans

Two distinguished values denoted by the constants

#t and #f

The type of these values is boolean

==> (< 2 3)

#t

==> (< 4 3)

#f

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Values and types

Values have types. For example:

In scheme almost every expression has a value

Examples:

1) The value of 23 is 232) The value of + is a primitive procedure for addition3) 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 procedure3) The type of proc (x) (* x x) is a compound procedure4) The type of (> x 1) is a boolean (or logical)

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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.

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More examples==> (define x 8)

Name Value

Environment Table

8x

==> (define x (* x 2))

==> x

16 16

==> (define x y)

reference to undefined identifier: y

==> (define + -)

>-<#+

==> (+ 2 2)

0

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The IF special form

ERROR2

(if <predicate> <consequent> <alternative>)If the value of <predicate> is #t,

Evaluate <consequent> and return it

Otherwise

Evaluate <alternative> and return it

(if (< 2 3) 2 3) ==> 2

(if (< 2 3) 2 (/ 1 0)) ==>

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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 !

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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, 7 replace by result of operation

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Recursive Procedures• How to create a process of unbounded length?• Needed to solve more complicated problems.• Start with a simple example.

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Example: 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

S(n-1)

Wishful thinking: if I could only solve the smaller instance …

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An algorithm for computing sum of squares

(define sum-squares (lambda (n) (if (= n 0)

0 (+ (sum-squares (- n 1)) (square n))))

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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 ?

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(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

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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.

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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

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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)

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Block StructureLets 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)))

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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):

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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)))

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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,zx,y

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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 orderb. 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: numberThe value of a built-in operator: machine instructions to executeThe value of any name: the associated object in the environment

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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)))

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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))

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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 = 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

.2for :Example xx

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(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 0.0001)

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Good programming Style1. 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.

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Procedural abstractionIt 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

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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 0.00001)

(sqrt-iter initial-guess x))

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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”).

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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))

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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))

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Lexical Scoping - again

The lexical scoping rules means thatthe 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

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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=412