Recursion and Iteration Let the entertainment begin…

Recursion and IterationRecursion and IterationLet the entertainment begin…Let the entertainment begin…

CS 480/680 – Comparative LanguagesCS 480/680 – Comparative Languages

Recursion and Iteration 2

No Iteration??No Iteration?? Yes, it’s true, there is no iteration in Scheme.

So, how do we achieve simple looping constructs? Here’s a simple example from Cal-Tech:


Summing integersSumming integers How do I compute the sum of the first N

integers?


Decomposing the sumDecomposing the sum

Sum of first N integers = N + N-1 + N-2 + … 1 N + ( N-1 + N-2 + … 1 ) N + [sum of first N-1 integers]


to Schemeto Scheme Sum of first N integers =

N + [sum of first N-1 integers]

Convert to Scheme (first attempt):

(define (sum-integers n)

(+ n (sum-integers (- n 1))))


Recursion in SchemeRecursion in Scheme



sum-integers defined in terms of itself This is a recursively defined procedure

... which is incorrect Can you spot the error?


Almost…Almost… What’s wrong?



Gives us:

N + N-1 + …+ 1 + 0 + -1 + -2 + …


DebuggingDebugging

Fixing the problem:• it doesn’t stop at zero!

Revised:


(if (= n 0)

0

(+ n (sum-integers (- n 1))))) How does this evaluate?


WarningWarning

The substitution evaluation you are about to see is methodical, mechanistic, • and extremely tedious.

It will not all fit on one slide. Don't take notes, but instead try to grasp the

overall theme of the evaluation.


EvaluatingEvaluating

Evaluate: (sum-integers 3)• evaluate 3 3• evaluate sum-integers (lambda (n) (if …))• apply (lambda (n)

(if (= n 0)

0

(+ n (sum-integers (- n 1))))) to 3

(if (= 3 0) 0 (+ 3 (sum-integers (- 3 1))))


Evaluating…Evaluating…

Evaluate: (if (= 3 0) 0 (+ 3 (sum-integers (- 3 1))))• evaluate (= 3 0)

evaluate 3 3 evaluate 0 0 evaluate = =

apply = to 3, 0 #f• since expression is false,

replace with false clause: (+ 3 (sum-integers (- 3 1)))

• evaluate: (+ 3 (sum-integers (- 3 1)))



evaluate (+ 3 (sum-integers (- 3 1)))• evaluate 3 3• evaluate (sum-integers (- 3 1))

evaluate (- 3 1) ...[skip steps] … 2 evaluate sum-integers(lambda (n) (if …))




evaluate (- 3 1) ...[skip steps] … 2 evaluate sum-integers(lambda (n) (if …))

Note: now pending (+ 3 …)



apply (lambda (n) (if (= n 0)

0

(+ n (sum-integers (- n 1))))) to 2

(if (= 2 0) 0 (+ 2 (sum-integers (- 2 1)))

Pending (+ 3 …)



evaluate: (if (= 2 0) 0 (+ 2 (sum-integers (- 2 1)))

• evaluate (= 2 0) … [skip steps] … #f• since expression is false,

replace with false clause• (+ 2 (sum-integers (- 2 1)))• evaluate: (+ 2 (sum-integers (- 2 1)))

Pending (+ 3 …)




evaluate (- 2 1) ...[skip steps] … 1 evaluate sum-integers(lambda (n) (if …)) apply (lambda (n) …) to 1 (if (= 1 0) 0 (+ 1 (sum-integers (- 1 1))))

– evaluate (= 1 0) ...[skip steps] … #f

Pending (+ 3 …)

Note: pending (+ 3 (+ 2 …))



Evaluate (+ 1 (sum-integers (- 1 1)))• evaluate 1 1• evaluate (sum-integers (- 1 1))

evaluate (- 1 1) ...[skip steps] … 0 evaluate sum-integers(lambda (n) (if …)) apply (lambda (n) …) to 0 (if (= 0 0) 0 (+ 1 (sum-integers (- 0 1))))

– evaluate (= 0 0) #t

– result: 0

Pending (+ 3 (+ 2 …))

Note: pending (+ 3 (+ 2 (+ 1 0)))



Back to pending: Know (sum-integers (- 1 1)) 0 [prev slide]

Evaluate (+ 1 (sum-integers (- 1 1)))• evaluate 1 1• evaluate (sum-integers (- 1 1)) 0• evaluate + +• apply + to 1, 0 1

Pending (+ 3 (+ 2 (+ 1 0)))

Note: pending (+ 3 (+ 2 1))



Back to pending: Know (sum-integers (- 2 1)) 1 [prev slide]

Evaluate (+ 2 (sum-integers (- 2 1)))• evaluate 2 2• evaluate (sum-integer (- 2 1)) 1• evaluate + +• apply + to 2, 1 3

Pending (+ 3 (+ 2 1))

Note: pending (+ 3 3)


EvaluatingEvaluatingFinish pending: (+ 3 3) … final result: 6


……Yeah!!! …Yeah!!! …Substitution model… …works fine for recursion. Recursive calls are well-defined. Careful application of model shows us what

they mean and how they work.


Recursive FunctionsRecursive Functions Since there are no iterative constructs in

Scheme, most of the power comes from writing recursive functions

This is fairly straightforward if you want the procedure to be global:

(define factorial (lambda (n) (if (= n 0) 1 (* n (factorial (- n 1))))))

(factorial 5) » 120


The trickThe trick Must find the structure of the problem:

• How can I break it down into sub-problems?• What are the base cases?


List ProcessingList Processing Suppose I want to search a list for the max

value? A standard list:

What is the base case? What state do I need to maintain?

3 7 4 1

lst


More list processingMore list processing How about a list of X-Y pairs?

• What would the list look like?

Given X, find Y:• Base case• State• Helper procedures?

3 7 4 1

lst

5 2 3 9


Using RecursionUsing Recursion Write avg.scheme See list-position.scheme See count_atoms.scheme See printtype.scheme


Mutual RecursionMutual Recursion

Note: Scheme has built-in primitives even? and odd?

(define is-even? (lambda (n) (if (= n 0) #t (is-odd? (- n 1)))))

(define is-odd? (lambda (n) (if (= n 0) #f (is-even? (- n 1)))))


Recursion and Local DefinitionsRecursion and Local Definitions Recursion gets more difficult when you want

the functions to be local in scope:

(let ((local-even? (lambda (n) (if (= n 0) #t (local-odd? (- n 1))))) (local-odd? (lambda (n) (if (= n 0) #f (local-even? (- n 1)))))) (list (local-even? 23) (local-odd? 23)))

local-odd? is not yet definedlocal-odd? is not yet defined

Can we fix this with let* ?Can we fix this with let* ?


letrecletrec letrec – the variables introduced by letrec are

available in both the body and the initializations• Custom made for local recursive definitions

(letrec ((local-even? (lambda (n) (if (= n 0) #t (local-odd? (- n 1))))) (local-odd? (lambda (n) (if (= n 0) #f (local-even? (- n 1)))))) (list (local-even? 23) (local-odd? 23)))


Recursion for loopsRecursion for loops

Instead of a for loop, this letrec uses a recursive procedure on the variable i, which starts at 10 in the procedure call.

(letrec ((countdown (lambda (i) (if (= i 0) 'liftoff (begin (display i) (newline) (countdown (- i 1))))))) (countdown 10))


Named Named letlet

This is the same as the previous definition, but written more compactly

Named let is useful for defining and running loops

((let countdown ((i 10)) (if (= i 0) 'liftoff (begin (display i) (newline) (countdown (- i 1)))))


Recursion and Stack FramesRecursion and Stack Frames What happens when we

call a recursive function? Consider the following

Fibonacci function:

int fib(int N) { int prev, pprev;

if (N == 1) { return 0; } else if (N == 2) {

return 1; } else { prev = fib(N-1); pprev = fib(N-2); return prev + pprev; }}


Stack FramesStack Frames Each call to Fib

produces a new stack frame

1000 recursive calls = 1000 stack frames!

Inefficient relative to a for loop

Nprev pprev

Nprev pprev


How can we fix this inefficiency?

Here’s an answer from the Cal-Tech slides:


ExampleExample

Recall from last time:


(if (= n 0)

0

(+ n (sum-integers (- n 1)))))


Revisited (summarizing steps)Revisited (summarizing steps)

Evaluate: (sum-integers 3) (if (= 3 0) 0 (+ 3 (sum-integers (- 3 1)))) (if #f 0 (+ 3 (sum-integers (- 3 1)))) (+ 3 (sum-integers (- 3 1))) (+ 3 (sum-integers 2)) (+ 3 (if (= 2 0) 0 (+ 2 (sum-integers (- 2 1))))) (+ 3 (+ 2 (sum-integers 1))) (+ 3 (+ 2 (if (= 1 0) 0 (+ 1 (sum-integer (- 1 1))))))


Revisited (summarizing steps)Revisited (summarizing steps) (+ 3 (+ 2 (+ 1 (sum-integers 0)))) (+ 3 (+ 2 (+ 1 (if (= 0 0) 0 …)))) (+ 3 (+ 2 (+ 1 (if #t 0 …)))) (+ 3 (+ 2 ( + 1 0))) … 6


Evolution of computationEvolution of computation (sum-integers 3) (+ 3 (sum-integers 2)) (+ 3 (+ 2 (sum-integers 1))) (+ 3 (+ 2 (+ 1 (sum-integers 0)))) (+ 3 (+ 2 (+ 1 0)))


What can we say about the What can we say about the computation?computation?

On input N: How many calls to sum-

integers?N+1

How much work per call?

(sum-integers 3)

(+ 3 (sum-integers 2))

(+ 3 (+ 2 (sum-integers 1)))

(+ 3 (+ 2 (+ 1 (sum-integers 0))))

(+ 3 (+ 2 (+ 1 0)))


Linear recursive processesLinear recursive processes

This is what we call a linear recursive process

Makes linear # of calls• i.e. proportional to N

Keeps a chain of deferred operations linear in size w.r.t. input• i.e. proportional to N

(sum-integers 3)


(+ 3 (+ 2 (sum-integers 1)))

(+ 3 (+ 2 (+ 1 (sum-integers 0))))

(+ 3 (+ 2 (+ 1 0)))

time = C1 + N*C2


Another strategyAnother strategyTo sum integers…

add up as we go along: 1 2 3 4 5 6 7 … 1 1+2 1+2+3 1+2+3+4 ... 1 3 6 10 15 21 28 …


Alternate definitionAlternate definition(define (sum-int n)

(sum-iter 0 n 0)) ;; start at 0, sum is 0

(define (sum-iter current max sum)

(if (> current max)

sum

(sum-iter (+ 1 current)

max

(+ current sum))))


Evaluation of Evaluation of sum-intsum-int

(define (sum-int n)

(sum-iter 0 n 0))

(define (sum-iter current

max

sum)

(if (> current max) sum


max

(+ current

sum))))

(sum-int 3) (sum-iter 0 3 0)



(define (sum-int n)

(sum-iter 0 n 0))


max

sum)



max

(+ current

sum))))

(sum-int 3) (sum-iter 0 3 0) (if (> 0 3) … ) (sum-iter 1 3 0)



(define (sum-int n)

(sum-iter 0 n 0))


max

sum)



max

(+ current

sum))))

(sum-int 3) (sum-iter 0 3 0) (if (> 0 3) … ) (sum-iter 1 3 0) (if (> 1 3) … ) (sum-iter 2 3 1)



(define (sum-int n)

(sum-iter 0 n 0))


max

sum)



max

(+ current

sum))))

(sum-int 3) (sum-iter 0 3 0) (if (> 0 3) … ) (sum-iter 1 3 0) (if (> 1 3) … ) (sum-iter 2 3 1) (if (> 2 3) … ) (sum-iter 3 3 3)



(define (sum-int n)

(sum-iter 0 n 0))


max

sum)



max

(+ current

sum))))

(sum-int 3) (sum-iter 0 3 0) (if (> 0 3) … ) (sum-iter 1 3 0) (if (> 1 3) … ) (sum-iter 2 3 1) (if (> 2 3) … ) (sum-iter 3 3 3) (if (> 3 3) … ) (sum-iter 4 3 6)



(define (sum-int n)

(sum-iter 0 n 0))


max

sum)



max

(+ current

sum))))

(sum-int 3) (sum-iter 0 3 0) (if (> 0 3) … ) (sum-iter 1 3 0) (if (> 1 3) … ) (sum-iter 2 3 1) (if (> 2 3) … ) (sum-iter 3 3 3) (if (> 3 3) … ) (sum-iter 4 3 6) (if (> 4 3) … ) 6



on input N: How many calls to sum-

iter?N+2

How much work per call?

(sum-int 3) (sum-iter 0 3 0) (sum-iter 1 3 0) (sum-iter 2 3 1) (sum-iter 3 3 3) (sum-iter 4 3 6) 6



on input N: How many calls to sum-

iter?N+2

How much work per call?constant

one comparison, one if, two additions, one call

How many deferred operations?none

(sum-int 3) (sum-iter 0 3 0) (sum-iter 1 3 0) (sum-iter 2 3 1) (sum-iter 3 3 3) (sum-iter 4 3 6) 6


What’s different about these What’s different about these computations?computations?

(sum-integers 3)


(+ 3 (+ 2 (sum-integers 1)))

(+ 3 (+ 2 (+ 1 (sum-integers 0))))

(+ 3 (+ 2 (+ 1 0)))

(sum-int 3)

(sum-iter 0 3 0)

(sum-iter 1 3 0)

(sum-iter 2 3 1)

(sum-iter 3 3 3)

(sum-iter 4 3 6)

Old New


Linear iterative processesLinear iterative processes

This is what we call a linear iterative process

Makes linear # calls State of computation kept

in a constant # of state variables• state does not grow with

problem size

• requires a constant amount of storage space

• (sum-int 3)• (sum-iter 0 3 0)• (sum-iter 1 3 0)• (sum-iter 2 3 1)• (sum-iter 3 3 3)• (sum-iter 4 3 6)• 6

time = C1 + N*C2


Linear recursive vs linear iterativeLinear recursive vs linear iterative

Both require computational time which is linear in size of input

L.R. process requires space that is also linear in size of input• why?

L.I. process requires constant space• not proportional to size of input• more space-efficient than L.R. process


Linear recursive vs linear iterativeLinear recursive vs linear iterative

Why not always use linear iterative instead of linear recursive algorithm?• often the case in practice

Recursive algorithm sometimes much easier to write• and to prove correct

Sometimes space efficiency not the limiting factor


Tail-call eliminationTail-call elimination

Countdown uses only tail-recursion• Each call to countdown either does not call itself

again, or does it as the very last thing done

Scheme recognizes tail recursion and converts it to an iterative (loop) construct internally

((let countdown ((i 10)) (if (= i 0) 'liftoff (begin (display i) (newline) (countdown (- i 1)))))


Mapping a procedure across a listMapping a procedure across a list

(map add2 '(1 2 3))» (3 4 5)

(map cons '(1 2 3) '(10 20 30))» ((1 . 10) (2 . 20) (3 . 30))

(map + '(1 2 3) '(10 20 30))» (11 22 33)


ExercisesExercises Include error checking into avg.scheme

• Divide by zero error• Non-numeric list items

Write a scheme function that accepts a single numeric argument and returns all numbers less than 10 that the argument is divisible by

Factor.scheme – described in class Write a scheme function that returns the first n

Fibonacci numbers as a list (n is an argument)

Documents

Recursion and Iteration Let the entertainment begin…