CS790 – Introduction to Bioinformatics

Recursion and Iteration

Let the entertainment begin…

CS 480/680
–

Comparative Languages

Recursion and Iteration

2

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:

Recursion and Iteration

3

Summing integers


How do I compute the sum of the first N
integers?

Recursion and Iteration

4

Decomposing 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
]

Recursion and Iteration

5

to 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 and Iteration

6

Recursion in Scheme


(define (
sum
-
integers

n)


(+ n (
sum
-
integers

(
-

n 1))))



sum
-
integers

defined in terms of itself


This is a
recursively defined

procedure

... which is incorrect


Can you spot the error?

Recursion and Iteration

7

Almost…


What’s wrong?

(define (sum
-
integers n)

(+ n (sum
-
integers (
-

n 1))))



Gives us:



N + N
-
1 + …+ 1 + 0 +
-
1 +
-
2 + …


Recursion and Iteration

8

Debugging


Fixing the problem:

•
it doesn’t stop at zero!


Revised:

(define (sum
-
integers n)


(if

(= n 0)


0


(+ n (sum
-
integers (
-

n 1)))
)
)


How does this evaluate?

Recursion and Iteration

9

Warning


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.

Recursion and Iteration

10

Evaluating


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))))

Recursion and Iteration

11

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)))

Recursion and Iteration

12

Evaluating…


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 …))

Recursion and Iteration

13

Evaluating…


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 …))

Note: now pending (+ 3 …)

Recursion and Iteration

14

Evaluating…


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 …)

Recursion and Iteration

15

Evaluating…


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 …)

Recursion and Iteration

16

Evaluating…


evaluate (+ 2 (sum
-
integers (
-

2 1)))

•
evaluate 2


2

•
evaluate (sum
-
integers (
-

2 1))


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 …))

Recursion and Iteration

17

Evaluating…


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)))

Recursion and Iteration

18

Evaluating

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
))

Recursion and Iteration

19

Evaluating

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
)

Recursion and Iteration

20

Evaluating

Finish pending:


(+ 3 3)


… final result:
6

Recursion and Iteration

21

…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.

Recursion and Iteration

22

Recursive 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

Recursion and Iteration

23

The trick


Must find the structure of the problem:

•
How can I break it down into sub
-
problems?

•
What are the base cases?

Recursion and Iteration

24

List 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

Recursion and Iteration

25

More 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

Recursion and Iteration

26

Using Recursion


Write avg.scheme


See list
-
position.scheme


See count_atoms.scheme


See printtype.scheme


Recursion and Iteration

27

Mutual 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 Iteration

28

Recursion 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 defined

Can we fix this with let* ?

Recursion and Iteration

29

letrec


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 and Iteration

30

Recursion 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))

Recursion and Iteration

31

Named
let


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 Iteration

32

Recursion 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;


}

}

Recursion and Iteration

33

Stack Frames


Each call to Fib
produces a new
stack frame


1000 recursive calls
= 1000 stack frames!


Inefficient

relative to
a
for

loop

N

prev

pprev

N

prev

pprev

Recursion and Iteration

34


How can we fix this inefficiency?


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

Recursion and Iteration

35

Example


Recall from last time:

(define (sum
-
integers n)


(if (= n 0)


0


(+ n (sum
-
integers (
-

n 1)))))


Recursion and Iteration

36

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))))))

Recursion and Iteration

37

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


Recursion and Iteration

38

Evolution of computation


(sum
-
integers 3)


(+ 3 (sum
-
integers 2))


(+ 3 (+ 2 (sum
-
integers 1)))

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


(+ 3 (+ 2 (+ 1 0)))

Recursion and Iteration

39

What can we say about the
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)))


Recursion and Iteration

40

Linear 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 (sum
-
integers 2))

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

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

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



time = C
1

+ N*C
2

Recursion and Iteration

41

Another strategy

To 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 …


Recursion and Iteration

42

Alternate 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))))

Recursion and Iteration

43

Evaluation of
sum
-
int

(define (sum
-
int n)


(sum
-
iter 0 n 0))

(define (sum
-
iter current


max


sum)


(if (> current max) sum


(sum
-
iter (+ 1 current)


max


(+ current


sum))))


(sum
-
int 3)


(sum
-
iter 0 3
0
)

Recursion and Iteration

44

Evaluation of
sum
-
int

(define (sum
-
int n)


(sum
-
iter 0 n 0))

(define (sum
-
iter current


max


sum)


(if (> current max) sum


(sum
-
iter (+ 1 current)


max


(+ current


sum))))


(sum
-
int 3)


(sum
-
iter
0

3
0
)


(if (> 0 3) … )


(sum
-
iter 1 3
0
)

Recursion and Iteration

45

Evaluation of
sum
-
int

(define (sum
-
int n)


(sum
-
iter 0 n 0))

(define (sum
-
iter current


max


sum)


(if (> current max) sum


(sum
-
iter (+ 1 current)


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
)

Recursion and Iteration

46

Evaluation of
sum
-
int

(define (sum
-
int n)


(sum
-
iter 0 n 0))

(define (sum
-
iter current


max


sum)


(if (> current max) sum


(sum
-
iter (+ 1 current)


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
)

Recursion and Iteration

47

Evaluation of
sum
-
int

(define (sum
-
int n)


(sum
-
iter 0 n 0))

(define (sum
-
iter current


max


sum)


(if (> current max) sum


(sum
-
iter (+ 1 current)


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
)

Recursion and Iteration

48

Evaluation of
sum
-
int

(define (sum
-
int n)


(sum
-
iter 0 n 0))

(define (sum
-
iter current


max


sum)


(if (> current max) sum


(sum
-
iter (+ 1 current)


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

Recursion and Iteration

49

What can we say about the
computation?

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


Recursion and Iteration

50

What can we say about the
computation?

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


Recursion and Iteration

51

What’s different about these
computations?

(sum
-
integers 3)

(+ 3 (sum
-
integers 2))

(+ 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

Ne
w

Recursion and Iteration

52

Linear 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 = C
1

+ N*C
2

Recursion and Iteration

53

Linear 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

Recursion and Iteration

54

Linear 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

Recursion and Iteration

55

Tail
-
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)))))

Recursion and Iteration

56

Mapping 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)


Recursion and Iteration

57

Exercises


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)