מבוא מורחב - שיעור 101

Symbols

Manipulating lists and trees of symbols:symbolic differentiation

Lecture 10

מבוא מורחב - שיעור 102

Symbols

• We want to deal with data that is not just numeric.• For example:

(a b c d) ((rami 5743346) (dina 7584345))

• We have used symbols already as variable names.• With symbols, we can write application like

Symbolic differentiation Pattern searching in text An interpreter for scheme

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The special form quote

• quote is a special form• (quote <argument>) return the argument

without evaluating it.• syntactic sugar:

‘ alpha is the same as (quote alpha)

‘(a b c) is the same as (quote (a b c))

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Symbols are ordinary values

In particular we can form lists of symbols.

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

(list 'delta 'gamma) ==> (delta gamma)

symbolgamma

symboldelta

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The special form quote

(quote a)a 'aa (define a 1) (define bcd 2) (list a bcd)(1 2) (list 'a 'bcd);; same as (list (quote a) (quote bcd)(a bcd) (list a 'bcd)(1 bcd) '(a bcd) ;; same as (quote (a bcd))(a bcd)

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(define a 2) (define b 5)

(+ a b) ==>

'(+ a b) ==>

(+ 'a b) ==>

(list '+ a b) ==>

(list + 'a b) ==>

7

(+ a b)

error:+: expects type number…

(+ 2 5)

(#<primitive:+> a 5)

Examples using quote

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Syntactic Sugar

Expression:

Rewrites to: evaluates to:

'2 (quote 2) 2

'() (quote ()) ()

'(a b)

(quote (a b))

(a b)

'(1 (b c)) (quote (1(b c))) (1 (b c))

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The predicate symbol?symbol? : anytype -> boolean

Examples:(symbol? 'x)#t(symbol? x)Reference to undefined identifier: x(symbol? 3)#f(symbol? '3)#f(define b 2)(symbol? b)#f(define b 'a)(symbol? b)#t

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The predicate eq?A primitive procedure that tests if the pointers representing the objects point to the same place.Based on two important facts:

• A symbol with a given name exists only once.

• Each application of cons creates a new pair, different from any other previously created.

Therefore

(eq? ‘a ‘a) #t

But

(eq? ‘(a b) ‘(a b)) #f

מבוא מורחב - שיעור 1010

x y

ba

z

eq? example

(define x ‘(a b))

(define y ‘(a b))

(define z y)

(eq? x y) (eq? y z) (eq? y ‘(a b))

#f#t

#f (a new list is created)

מבוא מורחב - שיעור 1011

The predicate equal?A primitive procedure that tests if the pointers

represent identical objects

1. For symbols, eq? and equal? are equivalent

2. If two pointers are eq?, they are surely equal?

3. Two pointers may be equal? but not eq?

• (equal? ‘(a b) ‘(a b)) #t

• (equal? ‘((a b) c) ‘((a b) c)) #t

• (equal? ‘((a d) c) ‘((a b) c)) #f

מבוא מורחב - שיעור 1012

Symbolic differentiation

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

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

deriv of (x+y) with respect to x = 1

deriv of (2x(x+5) + 2xy ) w.r.t x = 2(x+5) + 2x + 2y

For now we allow only addition and multiplication

מבוא מורחב - שיעור 1013

Step 1: Representing expressions.

Use list of symbols.

There are still few possible lists one can think of

For example, we can represent 2x+5 as'(2 * x + 5) or '(2 x + 5)

We shall use a list structure corresponding to the same prefix notation used by scheme in combinations.

'(+ (* 2 x) 5)

מבוא מורחב - שיעור 1014

Step 1: Legal Expressions- formal

Expr ::= SimpleExpr | CompoundExpr

SimpleExpr ::= number | symbol

CompoundExpr ::= ( OPERATOR Expr Expr )

OPERATOR ::= +|*

Using Backus-Naur-Form BNF, a formalism fordefining the syntax of programming languages.

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Example

(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)

(deriv 7 'x) ==> 0

(deriv 'x 'x) ==> 1

Want deriv to be such that

מבוא מורחב - שיעור 1016

Step 2: Abstraction: The underlying interface

Constructors: make-product make-sumPredicates: number? variable? sum-expr? product-expr?Selectors: addend augend multiplier multiplicand

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Step 3: Breaking the problem to smaller subproblems

• Base case: (corresponds to SimpleExpr) deriv number w.r.t. x = 0 deriv variable w.r.t. x = 1 if variable is the same as x

= 0 otherwise

• Recursive case: (corresponds to CompoundExpr)

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

מבוא מורחב - שיעור 1018

Step 4: 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 (multiplier expr)

(deriv (multiplicand expr) var))

(make-product (multiplicand expr)

(deriv (multiplier expr) var)))

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

מבוא מורחב - שיעור 1019

Step 5: The underlying interface – constructors and selectors.

And the selectors:

(define (addend expr) (cadr expr)) (define (augend expr) (caddr expr))

(define (multiplier expr) (cadr expr)) (define (multiplicand expr) (caddr expr))

We define the constructors:

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

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

מבוא מורחב - שיעור 1020

Step 5: The underlying interface – predicates.

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

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

(define (variable? expr) (symbol? expr))

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

(eq? (car expr) '*)))

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

Writing, Debugging, 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)

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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))))) [= (x-y)+ (x+y)]We would like to have 2x or actually (* 2 x)

What should we change?

How easy is it to make the change?

מבוא מורחב - שיעור 1023

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))))

מבוא מורחב - שיעור 1024

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 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)))

((minus-expr? expr)

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

(deriv (augend expr) var)))

((product-expr? expr)

. . . .

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

מבוא מורחב - שיעור 1026

Allowing (- x y)

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

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

(eq? (car expr) ‘-)))

• We use the same selectors as in sum-expr. • Could do the same for product-expr• Is it a good idea?

מבוא מורחב - שיעור 1027

Allowing (- x y)

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

Now instead of:

We can write:

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

There’s still room for improvement!

מבוא מורחב - שיעור 1028

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)

…

)..)

מבוא מורחב - שיעור 1029

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)))))Similarly for multiplier and multiplicand

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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 Other operations

• Use data abstractions.• Respect abstraction barriers.• Avoid nested if expressions, complicated

procedures and always use good names.