Outline Scoping and block structure Recursive and iterative

Outline • Scoping and block structure • Recursive and iterative processes 2

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 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

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 4

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 5

Example (define y 7) (define z 5) (define (foo x z) (+ x z y)) • Foo binds 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. 6

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 7

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 8

Example • (define (f x) (define (g z) (* z z)) (+ (g x)) ) • (define (h x) (define (g x) (+ x x)) (- (g x))) ) 9

Why do we 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 10

Example 2 (define x 3) (define (foo y z) (define (g z x) (+ x y (* 2 z))) (g (+ x z) y)) y, z z, x (foo x 5) 11

Example 2: Cont’d (define x 3) 3 5 (define (foo y z) 8 3 (define (g z x) 3 3 8 (+ x y (* 2 z))) 3 5 3 (g (+ x z) y)) y, z z, x 3 (foo x 5) 22 12

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 13

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 14

Example 2 a • Compute the average of set of n=2^k numbers – Avg({x 1. . xn})=(x 1 +. . + xn)/n • Notice that: – Avg({x 1. . xn})=[Avg({x 1. . 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 15

Example 2 b • Compute the average of set of numbers – Avg({x 1. . xn})=(x 1 +. . + xn)/n • Notice that: – Avg({x 1. . xn})=[xn + (n-1)*Avg({x 1. . 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 16

Example 3 • Decide 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=1 return #t; If x is odd return #f; else 2. Compute and return for x/2 17

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 18

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)))) 19

Scheme Examples • n! (define (f n) (if (= n 1) 1 (* n f (- n 1))))) • Decide whether x is a power of 2 (define (is_pow 2 x) (cond ( ((= x 1) #t) ((odd? x) #f) (else (is_pow 2 (/ x 2)))) )) 20

Special case of recursive processes • No need to do computations after return of recursive call calc return val call return same val calc return val call no calc call return val call no calc return same val calc call no calc return val 21

converting a tail recursive call to an iterative process call don’t wait calc return val In Scheme, a tail recursive procedure results in an iterative process! 22

Different Meanings of Recursions Recursive Algorithms Iterative Algorithms Recursive Scheme Function Tail Recursion Recursive Process Iterative Process 23

Fibonacci Series Every element is the sum of its predecessors: 1, 1, 2, 3, 5, 8, 13, 21… Recursive algorithm fib(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)) 24

Tree Recursion 25

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 26

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 27

Counting Change = 2 x = 3 x = + 4 x + + 93 x 28

Counting Change CC( , { CC( - })= })+ , { }) 29

Counting Change (define (count-change amount) (cc amount 6)) (define (cc amount kinds-of-coins) (cond ((= amount 0) 1 ) ((or (< amount 0) (= kinds-of-coins 0)) 0 ) (else (+ (cc amount (- kinds-of-coins 1) ) (cc (- amount (first-denomination kinds-of-coins)) kinds-of-coins ))))) (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)))) 30

Greatest Common Divisor Given integers a>b 0 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) 3 (define (gcd a b) (if (= b 0) a (gcd b (remainder a b)))) 31