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Instructions for students & labbies:

Students use DrScheme, following the design recipe,
on the exercises at their own pace,
while labbies wander among the students, answering questions,
bringing the more important ones to the lab's attention.

Generative Recursion Design Recipe Review

In class, we concluded that we now have a new methodology for use with
generative recursion. If a function is generative recursive, we modify our
design recipe with the following changes:

  • When making function examples, we need even more examples, as they help us
    to find the algorithm we will use.

  • When making a template, we don't follow the structure of the data.
    Instead, we ask ourselves six questions:

    • What is (are) the trivial case(s)?

    • How do we solve the trivial case(s)?

    • For the non-trivial case(s), how many subproblems do we use?

    • How do we generate these subproblems?

    • Do we combine the results of these subproblems?

    • How do we combine the results of these subproblems?

    We use the answers to fill in the following very general template:

    (define (f args …)
    (cond [(trivial? args …)
    (solve-trivial args …)]
    [else
    (combine (f (generate-subproblem1 args …) …)

    (f (generate-subproblemn args …) …))]))

  • We add a step to reason about why our algorithm terminates,
    since this is no longer provided by simply following the structure
    of the data.

Converting a token stream to a list input stream

The Scheme reader reads list inputs rather than symbols or characters where a list input is defined by the following set of mutually recursive data definitions:

An input is either:

  • any built-in Scheme value that is not a list ( empty? or =cons?=)
  • a general list.

A general list is either:

  • empty ,
  • (cons i l) where i is an input and l is a general list.

Note that the second clause above could be written as the following two clauses:

  • (cons p l) where p is a primitive Scheme value that is not a list, or
  • (cons el l) where el and l are general lists.

Your task is to write the core function listify used in an idealized Scheme reader
stream. listify converts a list of tokens, where a token is defined below, to
a list of inputs as defined above.

A token is either

  • any built-in Scheme value (including symbol, boolean, number, string, empty
    (the empty list), etc.; or
  • (make-open), or
  • (make-close)
    given the structure definitions
(define-struct open ())
(define-struct close ())

The open and close parenthesis objects (not the symbols '|(| and '|)| ) clearly play a critical role in token streams because they delimit general lists.

Examples:

  • (listify empty) {{ =empty}} .
  • (listify (list (make-open) (make-close) ) {{ =(list empty)}} .
  • (listify (list (make-open) 'a (make-close) ) {{ =(list '(a))}}
  • (listify (list (make-open) 'a 'b (make-close) ) {{ =(list '(a b))}}
  • (listify (list (make-open) (make-open) 'a (make-close) 'b (make-close)) {{ =(list '((a) b))}}

You will probably want to define the following predicate

(define (input? t) (and (not (open? t)) (not (close? t))))

in your program. As a rough guideline follow the parse.ss (https:wikiriceedudisplayCSWIKI211lab04parsess) program that we wrote in class on Monday, Feb 2. Note that finding the first input is more involved than finding the first line. You will probably want to explicitly check for errors in the input because they correspond to natural tests on the form of the input.

Insertion Sort vs. Quicksort

In class, we've described two different ways of sorting numbers,
insertion sort and quicksort.

      (define (isort l)
         (cond
            [(empty? l) empty]
            [(cons? l)  (insert (first l) (isort (rest l)))]))

      (define (insert new sortedl)
         (cond
            [(empty? sortedl) (list new)]
            [else
             (cond
                 [(< new (first sortedl))
                  (cons new sortedl)]
                 [else
                  (cons (first sortedl)
                        (insert new (rest sortedl)))])]))

      ;;;;;;;;;;;;;;;;;

      (define (qsort l)
         (cond
            [(empty? l) empty]
            [(cons? l)
             (local [(define pivot (first l))
                     (define lt (filter (lambda (n) (< n pivot)) l))
                     (define eq (filter (lambda (n) (= n pivot)) l))
                     (define gt (filter (lambda (n) (> n pivot)) l))]
                (append (qsort lt)
                        eq
                        (qsort gt)))]))

So we have two fundamentally different approaches to sorting a list
(and there are lots of others, too).
It seems unsurprising that each might behave differently.
Can we observe this difference? Can we provide explanations?
We'll do some timing experiments,
outline the theoretical analysis,
and see if the two jive.

If your programs only sort small lists a few times, it doesn't matter;
any sort that is easy to write and have correct is fine.
However, for longer lists, the difference really is huge.
In Real Life (tm), sorting is an operation often done
on very long lists (repeatedly).

There is an expression

(time <em>expr</em>)

, which
can be used to time how long it takes the computer to
evaluate something. It is used like

      (time (isort big-list))

where

big-list

would be previously defined.
This expression prints out the time taken during evaluation,
along with the evaluation result. Since we're only interested in
the time, we can avoid seeing the long sorted list by using
something like

      (time (empty? (isort big-list)))
time

prints three times, each in milliseconds.
The first is the amount of computing time elapsed -- thus, it doesn't
count whatever else your computer was doing at the same time.
This is what we're interested in.

Timing Exercises

Partner with someone else in lab to split the following work.

First, we need some data to use for our timing experiments.

  1. Write a function

    nums-up

    which creates
    a list of naturals in ascending order.
    E.g.,

    (nums-up 7)

    returns

    (list 0 1 2 3 4 5 6)

    .
    Hint: Use

    build-list

    or dig out your solution
    from a previous lab.

  2. Similarly, write

    nums-down

    , where

    (nums-down 7)

    returns

    (list 6 5 4 3 2 1 0)

    .

  3. Now write a function

    nums-rand

    , which similarly creates
    a list of random numbers in the range from 0 to 32767.
    To create a single random numbers in the range from 0 to 9,
    you could use

    (random 10)

    .
    For larger numbers, try:

    
      (define max-rand 32768)
      (random max-rand)         ; random is built-in
      

    (Note that

    random

    is certainly a function in the Scheme
    sense, but not in the mathematical sense, since the output is
    not determined by the input.)

Now, make sure you know how to use

time

.

  1. Time

    isort

    on a list of 200 elements.

  2. Run the exact same expression again a couple times.
    Note that the time varies some.
    We typically deal with such variation by taking the
    average of several runs.
    The variation is caused by a number of low-level hardware
    reasons that are beyond the scope of this course (see
    ELEC 220 and
    COMP 320).

Continue timing, filling in the following chart.
When you're done, compare results with other pairs in the lab.

input size

200

1000

5000

isort

up

 

up

 

up

 

down

 

down

 

down

 

rand

 

rand

 

rand

 

qsort

up

 

up

 

up

 

down

 

down

 

down

 

rand

 

rand

 

rand

 

COMP 280
introduces concepts of how these algorithms behave in general.
The punchline is that we can say that both insertion sort and
quicksort on lists are "_O(n

2

)_"
in the worst case.
I.e., for a list of n elements, they each take about
_c × n

2

_ evaluation steps to sort
the list, for some constant

c

.
Furthermore, in the "average" case, Quicksort does better:

O(n log n).
COMP 280, and later COMP 482, show precisely what these statements
mean and how to derive them.

Ackermann's Function Example

If you have time…

Here's the definition of a strange numerical function on natural numbers,
called Ackermann's function:

A(m,n) = 0, if n=0
= 2×n, if n>0 and m=0
= 2, if n=1 and m>0
= A(m-1,A(m,n-1)), if n>1 and m>0

Note that this definition is not structurally recursive.
In fact, it cannot be defined in a structurally recursive manner.
(In technical jargon, the function is not

primitive recursive

.
See

COMP 481

for what that means.)

Here's the equivalent Scheme code:

(define (ack m n)
(cond
[(= n 0) 0]
[(and (> n 0) (= m 0)) (* 2 n)]
[(and (= n 1) (> m 0)) 2]
[(and (> n 1) (> m 0)) (ack (sub1 m) (ack m (sub1 n)))]))

Ackermann's function exercises

  1. Try it out on some inputs.

    Warning:

    Try very very small inputs, like 0, 1, 2, 3, and maybe 4, because the result gets very big
    very fast.

  2. Argue why this always terminates (for natural numbers).

This function isn't very useful, except as a theoretical example of a
function that grows faster than probably any one you've seen before.
In

COMP 482

,
you'll learn one use for this function.

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