# Lab: Analyzing procedures

Held
Friday, 19 April 2019
Writeup due
Monday, 22 April 2019
Summary
In this laboratory, you will explore the running time for a few algorithm variants.

## Preparation

a. Make a copy of analysis-lab.rkt, which contains most of the procedures you will need for this lab.

b. Review the file to see what procedures are included. You may find it easiest to look at the list provided by the (define …) menu.

c. If you have not done so already, go over the self checks from the reading with your partner.

## Exercises

### Exercise 1: Analyzing the reverse procedure.

a. Add a set of counters for expermenting with reverse. For example,

(define RC (make-counter "experiments with reverse"))


b. Add a line of code to each of the procedures involved in reversing so that it increments the appropriate counter. You should plan to count list-reverse-1, list-reverse-2, and list-append. For example,

(define list-append
(lambda (front back)
(counter-increment! RC 'list-append)
...))


c. Find out how many times list-append is called in reversing a list of seven elements by entering the following commands in the interactions pane.

> (counter-reset-all! RC)
> (list-reverse-1 (range 7)
> (counter-print RC)


d. Did you get the same answer as in self-check 2(c)? If not, why do you think you got a different result?

e. Find out how many times kernel is called in reversing a list of seven elements.

f. Did you get the same answer as in self-check 2(e)? If not, what difference do you see?

What if we also care about calls to car, cdr, and such? We’ll need to create our own versions of those procedures, along with counters. As a start, we might write:

(define LC (make-counter "core list procedures")

(define $car (lambda (lst) (counter-increment! LC 'car) (car lst))) (define$cdr
(lambda (lst)
(counter-increment! LC 'cdr)
(cdr lst)))
(define $cons (lambda (val lst) (counter-increment! LC 'cons) (cons val lst))) (define$null?
(lambda (val)
(counter-increment! LC 'null?)
(null? val)))


We then have to update all of our calls to car to use $car instead. For example, here’s an updated definition of list-reverse-1. (define list-reverse-1 (lambda (lst) (if ($null? lst)
null
(list-append (list-reverse-1 ($cdr lst)) (list ($car lst))))))


Unfortunately, we can’t quite do that for list, because list takes a variable number of parameters. So, we might rewrite the list-reverse-1 procedure to more explicitly use cons.

(define list-reverse-1
(lambda (lst)
(if ($null? lst) null (list-append (list-reverse-1 ($cdr lst))
($cons ($car lst) null)))))


a. Update the definition of list-append so that it uses $car, $cons, $cdr, and $null?.

b. Find out how many procedure calls are done for each procedure in reversing a list of length seven, using list-reverse-1, with the following.

> (counter-reset-all! LC)
> (list-reverse-1 (iota 7))
> (counter-print LC)


b. How does the number of calls seem to relate to the number of calls to list-append?

c. Update list-reverse-2 and find out how many procedure calls (including calls to kernel) are made when we use that procedure to reverse a list of length seven.

d. How does that number of calls seem to relate to the number of calls to kernel?

### Exercise 3: Predicting calls

a. Fill in the following chart to the best of your ability. Note that the total function calls should include calls to car, cdr, and the rest.

List Length rev1: Calls to list-append rev1: Total function calls rev2: Calls to kernel rev2: Total function calls
2
4
8
16
;               rev-1   rev-1   rev-2   rev-2
;       length  l-a     total   kernel  total
;        2
;        4
;        8
;       16


b. Predict what the entries will be for a list size of 32.

d. Write a formula for the columns, to the best of your ability.

### Exercise 4: The alphabetically first string, revisited

Here is a third version of alphabetically-first, which should already be in your definitions.

(define alphabetically-first-3
(lambda (strings)
(let kernel ([first-so-far (car strings)]
[remaining (cdr strings)])
(if (null? remaining)
first-so-far
(kernel (first-of-two first-so-far (car remaining))
(cdr remaining))))))


a. Find out how many steps this procedure takes on lists of length 2, 4, 8, and 16 in which the elements are arranged from alphabetically first to alphabetically last. As in the previous exercises, you’ll need to build a counter (or use AF) and add a line to increment the counter. In this case, you should make sure to count kernel, car, cdr, and null?.

b. Find out how many steps this procedure takes on lists of length 2, 4, 8, and 16 in which the elements are arranged from alphabetically last to alphabetically first. (You can reverse the lists from the previous step to create these lists.)

c. Find out how many steps this procedure takes on lists of length 2, 4, 8, and 16 in which the elements are in no particular order.

d. Predict the number of steps this procedure will take on each kind of list, where the length is 32.

### Exercise 5: The effects of precondition checking

Consider alphabetically-first-4, a variant of an efficient version of alphabetically-first that has additional error checking added.

(define alphabetically-first-4
(lambda (strings)
(when (not (all-string? strings))
(error "alphabetically-first: expects a list of strings; received" strings))
(if (null? (cdr strings))
(car strings)
(first-of-two (car strings)
(alphabetically-first-4 (cdr strings))))))


a. Predict the number of calls to alphabetically-first-4 in finding the alphabetically first in a list of eight strings. Once again, you will need to make sure to add lines to count various calls.

c. Predict the number of calls to all-string? in finding the alphabetically first in a list of eight strings. You will, of course, need to add a line to count each call to all-string?.

### Exercise 6: Precondition checking, revisited

Rewrite alphabetically-first-4 so that it continues to check preconditions, but precondition checking does not exact such a heavy penalty.

## For those with extra time

### Extra 1: Extracting successors

Consider the problem of extracting successors of a particular element in a list of elements.

> (successors 'a (list 'b 'a 'c 'd 'a 'b 'd 'a 'b))
'(c b b)


There are a variety of approaches that students tend to use.

a. Some use direct recursion.

(define successors-1
(lambda (val lst)
(cond
[(or (null? lst) (null? (cdr lst)))
null]
[(equal? val (car lst))
(cons (car (cdr lst)) (successors-1 val (cdr lst)))]
[else
(successors-1 val (cdr lst))])))


b. Some make a list of pairs of element and successor, filter the list for those whose first element matches, and then take the cadr of the matching elements

(define successors-2
(lambda (val lst)
(let* ([pairs (map make-pair lst (append (cdr lst) (list "<end>")))]
[matches (filter (o (l-s equal? val) car) pairs)]
[result (map (o car cdr) matches)])
result)))

(define make-pair
(lambda (a b)
(cons a (cons b null))))


c. Some make a list of pairs of element and position, filter the list for those whose element matches, extract the indices, increment the indices, and then use list-ref to find the elements.

(define successors-3
(lambda (val lst)
(let* ([pairs (map make-pair lst (iota (length lst)))]
[matches (filter (o (l-s equal? val) car) pairs)]
[indices (map (o car cdr) matches)]
[result (map (o (l-s my-list-ref lst) increment) indices)])
result)))

(define my-list-ref
(lambda (lst index)
(if (zero? index)
(car lst)
(my-list-ref (cdr lst) (decrement index)))))


Determine experimentally how many calls to the core list procedures each of these procedure does. You need only count the explicit calls to car, cdr, and cons, in each of the above.