Lecture #9

מבוא מורחב1

Lecture #9

142-151, sections 2.3.1, 2.3.2

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Symbol: a primitive type

constructors: (quote alpha)

Selectors: none

Methods: symbol? : anytype -> boolean

eq? ; discuss in a minute

(symbol? (quote x)) ==> #t

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eq?

A primitive procedure that returns #t if its two arguments are the same object

(eq? (quote eps) (quote eps)) ==> #t

(eq? (quote delta) (quote eps)) ==> #f

(eq? (cons 1 2) (cons 1 2)) #f

Eq? Can be used for any type except numbers or strings. One should use = for equality of numbers.

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Symbols are ordinary values

In particular we can form list of symbols.

(list 1 2) ==> (1 2)

(list (quote delta) (quote gamma)) ==> (delta gamma)

symbolgamma

symboldelta

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Syntactic Sugar.

Expression: Rewrites to:

1. (quote (a b)) (list (quote a) (quote b))

2. (quote ()) nil

3. (quote (1 (b c)))

(list (quote 1) (quote (b c)))

(list 1 (quote (b c)))

(list 1 (list (quote b) (quote c)))

(1 (b c))

4. 'a is shorthand for (quote a) '(1 (b c)) (quote (1 (b c)))

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Your turn:

(define a 2) (define b 5)

(+ a b) ==>

'(+ a b) ==>

(+ 'a b) ==>

(list '+ a b) ==>

(list + ‘a b) ==>

(eq? '(cons 1 2) '(cons 1 2)) #f

7

(+ a b)

([proc#..] a 3) - error

(+ 2 5)

([procedure #…] a 5)

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Summary so far

• symbol is a primitive type• the special form quote is its constructor• quoting the same name always returns the same object• the primitive procedure eq? tests for sameness

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Manipulating lists and trees of symbols

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Example 1: memq

(define (memq item x) (cond ((null? x) #f) ((eq? item (car x)) x) (else (memq item (cdr x)))))

(memq 'a '(b a b c)) =>

(memq 'a '((a b) b c)) =>

(memq 'a ‘((a b) b a (c a) d)) =>

(a b c)

#f

(a (c a) d)

If item does not appear in x, returns falseOtherwise – returns the sublist beginning with item

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Example 2: rember

(define (rember item x) (cond ((null? x) ‘()) ((eq? item (car x))(cdr x)) (else (cons (car x) (rember item (cdr x))))))

(rember 'a '(d a b c a)) =>

(rember 'b '(a b c)) =>

(rember 'a '(b c d)) =>

(d b c a)

(a c)

(b c d)

Removes the first occurrence of item in x.

11

Example 3: rember*

(define (rember* a x) (cond ((null? x) ‘()) ((atom? (car x)) (cond ((eq? (car x) a)(rember* a (cdr x))) (else (cons (car x) (rember* a (cdr x)))))) (else (cons (rember* a (car x)) (rember* a (cdr x))))))

(rember* 'a '(a b)) =>

(rember* 'a '(a (b a) c (a)) =>

(b)

((b) c ())

Recursively removes all occurances of a

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An example: symbolic differentiation

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Symbolic differentiation

(deriv <expr> <with-respect-to-var>) ==> <new-expr>

(deriv '(+ x 3) 'x) ==> 1

(deriv '(+ (* x y) 4) 'x) ==> y

(deriv '(* x x) 'x) ==> (+ x x)

The expressions we deal with: x+y x+5 2x (x+5) + 2 x y

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Step 1: Legal expressions.

First we have to choose what expressions we work with,And how we represent them.

Suppose we allow addition and multiplication.

Do we accept x+y or (+ x y) ?What about x+y+z and (+ x y z) ?Do we interpret x+ y * z as (x+y)*z or as x+(y*z) ?

How do we represent the output? Are we trying to simplify the output.

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Step 1: Legal Expressions- formal

Expr = SimpleExpr | CompoundExpr

SimpleExpr = number | symbol

CompoundExpr = LIST( OPERATOR, Expr, Expr )

Operator = +|*

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Step 2: Breaking the problem to smaller subproblems

• Base case: (corresponds to SimpleExpr)• deriv constant dx = 0• deriv variable dx = 1 if variable is the same as x

= 0 otherwise

• Induction step: (corresponds to CompoundExpr)

• (f+g)’ = f’ + g’ • (f*g)’ = f * g’ + g * f’

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Step 3: An algorithm assuming an abstract interface.(define deriv (lambda (expr var)

(cond

((number? expr) 0)

((variable? expr) (if (eq? expr var) 1 0))

((sum-expr? expr)

(make-sum (deriv (addend expr) var)

(deriv (augend expr) var)))

((product-expr? expr)

(make-sum

(make-product (augend expr)

(deriv (addend expr) var))

(make-product (addend expr)

(deriv (augend expr) var)))

(else (error "unknown expression" expr))))

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Step 4: The underlying interface

Constructors:

make-product

make-sum

Methods:

number?

variable?

sum-expr?

product-expr?

Selectors:

addend

augend

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Step 5: The underlying interface – constructors and selectors.

We represent constants with integers.

We represent symbols with (quote ..)

And the selectors:

(define (addend expr) (cadr expr))

(define (augend expr) (caddr expr))

We define the constructors:

(define (make-sum e1 e2) (list '+ e1 e2))

(define (make-product e1 e2) (list '* e1 e2))

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Step 5: The underlying interface – methods.

(define (sum-expr? expr) (and (pair? expr)

(eq? (car expr) '+)))

(define (variable? expr) (and (not (pair? expr))

(symbol? expr)))

(define (product-expr? expr) (and (pair? expr)

(eq? (car expr) ‘*)))

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Writing, Debuging, Testing…

(deriv 5 ‘x) 0(deriv ‘x ‘x) 1 (deriv ‘x ‘y) 0…

(deriv ‘(+ x y) ‘x) (+ 1 0)(deriv ‘(+ x y) ‘z) (+ 0 0)

(deriv '(* (+ x y) (- x y)) 'x) unknown expression (- x y)

(deriv '(* (+ x y) (+ x (* -1 y))) 'x) (+ (* (+ x (* -1 y)) (+ 1 0)) (* (+ x y) (+ 1 (+ (* y 0) (* -1 0)))))

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We would like to simplify expressions.

Instead of the expression (+ (* (+ x (* -1 y)) (+ 1 0)) (* (+ x y) (+ 1 (+ (* y 0) (* -1 0)))))

We would like to have 2x

What should we change?

How easy is it to make the change?

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We change make-sum and make-product

It’s easier to change a well-written code.

(define (make-sum a1 a2) (cond ((and (number? a1) (= a1 0)) a2) ((and (number? a2) (= a2 0)) a1) ((and (number? a1) (number? a2)) (+ a1 a2)) (else (list '+ a1 a2))))

(define (make-product a1 a2) (cond ((and (number? a1) (= a1 0)) 0) ((and (number? a2) (= a2 0)) 0) ((and (number? a1) (= a1 1)) a2) ((and (number? a2) (= a2 1)) a1) ((and (number? a1) (number? a2)) (* a1 a2)) (else (list ‘* a1 a2))))

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Testing again..

(deriv '(* (+ x y) (+ x (* -1 y))) 'x) (+ (+ x (* -1 y)) (+ x y))

This is an improvement.

There is still a long way to go….

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Allowing (- x y)

(define (make-minus a1 a2) (cond ((and (number? a1) (= a1 0)) a2) ((and (number? a2) (= a2 0)) a1) ((and (number? a1) (number? a2)) (+ a1 a2)) (else (list ‘- a1 a2))))

(define (minus-expr? expr) (and (pair? expr)

(eq? (car expr) ‘-)))

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Allowing x^k, …. ,1/x, (ln x)

(define (ln-expr? expr) (and (pair? expr)

(eq? (car expr) ‘ln)))

(define deriv (lambda (expr var)

(cond …

((minus-expr? expr)

…

((ln-expr? expr)

…

)..)

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Allowing (+ x y z … w) (* x y z … w)

How about the following change:

(define (addend expr) (cadr expr))

(define (augend expr)

(let ((first (car expr))

(second (cadr expr))

(rest (cddr expr)))

(cond ((> (length rest) 1) (cons first rest))

((= (length rest) 1) (car rest)))))

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Testing..

> (deriv '(* x y z w x t y ) 't)(* x (* y (* z (* w (* x y)))))> (deriv '(* x y z w x t y) 'x)(+ (* y z w x t y) (* x (* y (* z (* w (* t y))))))

> (deriv '(+ x y z w a b c x y z e r t ) 'y)2> (deriv '(+ x y z w a b c x y z e r t ) 'f)0

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Guidelines

• Carefully think about • the problem • type of data• The algorithm.

• Identify • Constructors • Selectors, • And the methods you need from your data.

• Use data abstractions.• Respect abstraction barriers.

• Avoid nested if expressions, complicated procedures and always use good names.