מבוא מורחב למדעי המחשב Scheme בשפת

2תרגול

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Outline

• Scoping and block structure

• Recursive and iterative processes

• Orders of growth

Scoping• We have seen two methods to introduce names into a program:

– Define form introduces a name associated with a procedure or an expression:

(define x 5)

(define (f x) (* x x))

– Lambda expression introduces names for formal parameters(define (g a b) (+ a b))

((lambda(a b) (+ a b)) 2 3)

• Need rules to interpret names.

(define x 5)

((lambda(x)(+ x 1)) x)3

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Reminder: Bounded variables

• A procedure definition binds its formal parameters (their names)

• These names are called bounded variables. • Other variables are free variables

(define y 7)

(define (foo x) (+ x y))

x is bounded to foo, while y is free

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Reminder: Scope

• The set of expressions for which binding defines a name is called a scope of this name

• A bound variable has its binding function’s body as a scope, and is unknown outside its scope

• Free variables are known anywhere in the program

• For variables with the same name and overlapping scopes, the more internal binding overrides the other

Example(define y 7)

(define z 5)

(define (foo x z) (+ x z y))

• Foo binds x and z. x and z are bounded to foo

• Scope of x and z is the body of foo

• y is known everywhere, z=5 is only outside foo

• The meaning of foo will not change if we’ll change the names of its formal parameters.

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Internal Definitions

• Can define a name inside the lambda expression (local to this procedure/lambda expression)

• It’s scope is limited to the rest of the body of the procedure

• (define (g x)

(define a 5)

(* x a))

• Note: Body of a procedure(lambda expression) can consist of multiple sub-expressions.

– The value of the last one is defined to be the value of the procedure

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Reminder: Block structure

• Binding/Internal definitions isolate a variable from the rest of the program (limit its scope)

• Can we isolate procedure from the rest of the

program (to limit its scope)?

• Block structure: defining procedures inside other procedure

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Example

• (define (f x)

(define (g z) (* z z))

(+ (g x) (g x))

)

• (define (h x)

(define (g x) (+ x x))

(- (g x) (g x)))

)9

Why do wee need block structure and scoping?

• It makes our life easier:

– Don’t have to remember which names are already used

– Hide away helper procedures to see the overall structure of the program

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Example 2

(define x 3)

(define (foo y z)

(define (g z x)

(+ x y (* 2 z)))

(g (+ x z) y))

(foo x 5)

z,x y,z

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Example 2: Cont’d

(define x 3)

(define (foo y z)

(define (g z x)

(+ x y (* 2 z)))

(g (+ x z) y))

(foo x 5)3

3 5

3 5 3

8 3

83 3

z,x y,z

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Reminder: Recursive algorithm

• Reduce problem to one or more sub-problems of smaller sizes (linear or tree recursion) and solve them recursively

• Solve the very small sized problems directly

• Usually some computations are required in order to split problem into sub-problems or /and to combine the results

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Example 1

• Compute f(n)=n!

• Notice that f(n)=n*f(n-1) if n>1 and f(1)=1;

• Algorithm:1. If n=1 return 1

2. Computing factorial for n-1 if n>1

3. Multiply result by n and return

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Example 2a

• Compute the average of set of n=2^k numbers

– Avg({x1 .. xn})=(x1 +.. +xn)/n

• Notice that:

– Avg({x1 .. xn})=[Avg({x1 .. Xn/2})+Avg({xn/2+1 .. xn})]/2

• Algorithm

1. If input set size is 1, return the input as its average

2. Divide set into 2 subsets of n/2 if n>1

3. Find an average of each subset

4. Return average of two results

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Example 2b

• Compute the average of set of numbers

– Avg({x1 .. xn})=(x1 +.. xn)/n

• Notice that:

– Avg({x1 .. xn})=[xn + (n-1)*Avg({x1 .. Xn-1})]/n

• Algorithm

1. If input set size is 1, return the input as its average

2. Find average of last n-1 numbers

3. Multiply result by (n-1) add xn and the divide by n

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Example 3• Find if the number x is a power of 2

F(x)=#t if x=2^n for some n or #f otherwise

• Notice– F(x)=F(x/2) if x is even, F(x)=#t if x=1 and #f

otherwise

• Algorithms

1. If x is odd return #f , if x=1 return #t

2.Compute and return for x/2

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Recursive calls

• Function/procedure that calls to itself called recursive procedure

• Can’t call to itself every time as have to stop somewhere

• Recursive algorithms are usually implemented using recursive calls

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The conditional form(cond (<test-1> <consequent-1>)

(<test-2> <consequent-2>)

. . .

(<test-n> <consequent-n>)

(else <consequent-else>))

(define (abs x) (cond ((> x 0) x)

((= x 0) 0) (else (- x))))

Scheme Examples

• (define (f n) (if (= n 1) 1 (* n f (- n 1)))))

• (define (is_pow2 x)

(cond (

((= x 1) #t)

((odd? x) #f)

(else (is_pow2 (/ x 2))))

))

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Special case of recursive processes

• No need to do computations after return of recursive call

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call call call

return valreturn valreturn valcalccalc calccalc

call

call call call

return valreturn same val

return same val

calcno calc

no calc

no calc

call

return same val

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converting a tail recursive call to an iterative process

call call call

calc

call

return val

don’t wait

don’t wait

don’t wait

In Scheme, a tail recursive procedure results in an iterative process!

Different Meanings of Recursions

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Recursive Algorithms Iterative Algorithms

Recursive Scheme Function

Recursive Process Iterative Process

Tail Recursion

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Fibonacci Series

Recursive algorithmfib(0) = 0

fib(1) = 1

fib(n) = fib(n-1) + fib(n-2) :n > 1

Tree recursion

(define (fib n)

(cond ((= n 0) 0)

((= n 1) 1)

(else (+ (fib (- n 1))

(fib (- n 2))))))

Iterative algorithm Initialize: a = 1, b = 0

count = n

Step: anew = aold + bold

bnew = aold

count = count - 1

Tail Recursion

(define (fib n)

(define (fib-iter a b count)

(if (= count 0) b

(fib-iter (+ a b) a (- count 1))))

(fib-iter 1 0 n))

Every element is the sum of its predecessors: 1, 1, 2, 3, 5, 8, 13, 21…

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Tree Recursion

Implementing loops using tail recursion

• Usually using helper local function (iterator)

• Pass loop local variables as an argument (usually includes some sort of counter)

• Check stopping condition

• Make initial call with initial state arguments

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Example(Recursion)

• Input: Amount of money : $1.50

• Output: Number of different ways to construct the amount using coins of 1,2,5,10,15,20,50 cents

• Assume there is unlimited number of coins of each value

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Counting Change

= 2x

= 3x + 4x

= + + 93x

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Counting Change

CC( , { } ) =

CC( , { } ) +

CC( - , { } )

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Counting Change

(define (count-change amount)  (cc amount 6))

(define (cc amount kinds-of-coins)  (cond ((= amount 0)  )        ((or (< amount 0) (= kinds-of-coins 0))  )        (else (+ (cc                        )                 (cc 

                                               )))))

(define (first-denomination kinds-of-coins)  (cond ((= kinds-of-coins 1) 1)

        ((= kinds-of-coins 2) 2)       ((= kinds-of-coins 3) 5) ((= kinds-of-coins 4) 10)         ((= kinds-of-coins 5) 20)         ((= kinds-of-coins 6) 50))))

1 0amount (- kinds-of-coins 1) (- amount (first-denomination kinds-of-coins)) kinds-of-coins

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Greatest Common Divisor

Given integers a>b0

gcd(a,0) = a

gcd(a,b) = gcd(b, a mod b)

By Substitution

(gcd 30 21)

(gcd 21 9)

(gcd 9 3)

(gcd 3 0)

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(define (gcd a b)

(if (= b 0) a

(gcd b (remainder a b))))