This lab is due at the beginning of your next lab period.
It should be submitted as "lab02" and contain a single file called
lab.scm". See the "submit" link on the left menu for
details. You are highly encouraged to submit whatever
you have done by the end of this lab. Remember that duplicate
submissions are fine; I will just grade the latest one.
Recall that a symbol is just a quoted identifier, such as
We can use the function
symbol? to find out whether
something is a symbol, and
symbol=? to compare two
1 US dollar = .6932 Euros (euro) 1 US dollar = 76.733 Japanese Yen (yen) 1 US Dollar = .9787 Canadian dollars (cad)
(to-usdollar amt cur)that takes an amount of money
amtin some foreign currency
curand returns that amount in US dollars. The parameter
curwill be one of the symbols
(to-usdollar 500 'yen)produces 500/76.733, which comes out to
(from-usdollar amt cur)that does the opposite: takes an amount in US Dollars and converts it to the named currency.
(convert amt fromCur toCur)that takes an amount in currency
fromCurand converts it to crrency
toCur. Use your functions from parts 1 and 2 and life will be easy!
Recall "big 4" of list processing:
null: the name for an empty list.
(cons a L): Take an item
aand a list
Land return the new list with
ainserted in the front of
(car L): Returns the first item in
(cdr L): Returns the list containing all the items in
Lafter the first item.
If you're faced with nested lists, you sometimes need cars of
cdrs, and cdrs of cars, and so forth. The shortcut for
a bunch of these in a row is
X is either
> (define L '((4 a) (3 x) (3 11 4))) > (car (cdr L)) (3 x) > (cadr L)) (3 x) > (car (cdr (car (cdr L)))) x > (cadadr L) x
There are two more extremely useful shortcuts:
list: Takes an arbitrary number of items and makes a single list containing them.
append: Takes an arbitrary number of lists and makes a single list containing all their items.
(define (my-list a b) (cons a (cons b null))) (define (my-append L1 L2) (if (null? L1) L2 (cons (car L1) (my-append (cdr L1) L2))))
Make sure you understand what is going on here. In particular,
my-append is recursively processing
and producing a new list.
The "recursive processing" part means that
we check when
L1 is empty, and when it is, we have the
"base case" and just return
not empty, we break it down with
and make a recursive call on
The "producing a new list" part means that in the base case we
return a list (in this case
L2; often the list returned in
the base case is just
null). And in the recursive case,
we add on to the list returned by the recursive call, using
All the exercises below either recursively process a list, or create a new list, or both. Keep in mind what is going, and you should see some of the ingredients above cropping up over and over again.
squaresthat takes integers i and j and returns list of the squares i^2, (i+1)^2, ..., (j-1)^2, j^2.
> (squares 2 12) (4 9 16 25 36 49 64 81 100 121 144)
Linto your definitions window:
(define L '((a x 2) (('the) b c) (z 12)))Now, write expressions to pick apart L to get the following values (save each of these in your definition file -- put a comment next to each one explaining what it is):
sum-cashthat returns the value in US dollars of a collection of amounts of different currencies (same currencies as above). The amounts will be given in a list L, such that each element of L is itself a list, whose first element is an amount and whose second element is a currency name. So, for example,
( (12.20 usdollar) (340 yen) (8.50 euro) )as an arguement to
sum-cashwould mean adding 12.20 dollars, 340 yen and 8.50 euros, and giving the total in dollars. Here's an example:
> (sum-cash '((12.20 usdollar) (340 yen) (8.50 euro))) 28.892922331103236
std-devthat takes a list of numbers and returns their standard deviation. (Recall: std dev of x1, x2, ..., xn is
__________________________________________ / /(x1 - u)^2 + (x2 - u)^2 + ... + (xn - u)^2 / ------------------------------------------ \/ n - 1... where u is the average of x1, x2, ..., xn. When you write this function, use top-down design. In scheme this means writing the std-dev function using functions you wish were already defined, then going back and defining them later. It can be useful to quote these function calls before you implement them, so you can even do top-down testing!
> (std-dev '(34 18 25 23 29 11 28 24 27 29)) 6.460134157533676
let construct in Scheme allows you to give a
name to a common subexpression. For example, consider the
(33 * (501 - 33)) / ( 1 - (33 * (501 - 33)))The natural way to code this in Scheme is probably
(/ (- (* 33 501) 33) (- 1 (- (* 33 501) 33)))But you could say "let a = 33 * (501 - 33), and return a / (1 - a)". That's essentially what
letallows you to do.
(let ((a (- (* 32 501) 33))) (/ a (- 1 a)))
let is the Scheme way of getting local
variable functionality. What you've got is the
let keyword, followed by a list of
variable-name/value pairs, followed by an expression
(presumably using the new names) that provides the value of
let expression. For example:
> (let ((a 2) (b 4) (c 6)) (+ a b c) ) 12
There is power in using
let in functions. Suppose I want to define a function called
shifted-cube, which computes (x + 1)^3 for a
> (define (shifted-cube x) (let ((a (+ x 1))) (* a a a))) > (shifted-cube 2) 27
test-sinthat computes 1/(sin(x)^2 + 1) + sqrt(sin(x)^2 + 1) - (sin(x)^2 + 1)^2
distthat computes the difference (in inches) between two lengths (given in feet and inches). Example:
> (dist 3 7 2 11) ;;; difference between 3'7'' and 2'11'' 8Write this function using a
letexpression to create values
L2for the lengths in inches of the input feet-and-inches lengths.
As you have perhaps noticed, you never actually tell scheme what the type of a function parameter is. Thus, if I define some function (f x y), there's nothing to stop me from calling it like this: (f sqrt 16). In other words, there's nothing to stop me from passing f a function as one of its arguments. Does this make sense? It depends what f does with the parameter x. If it uses it like a function, then it makes sense:
> (define (f x y) (* y (x y))) > (f sqrt 4) 8What happened here? Well since x is sqrt and y is 4, f evaluates to
(* 4 (sqrt 4))which is 8. So f is the "apply function x to argument y and then multiply by y" function. Passing functions to functions like this is very powerful. This is part of what we mean when we say that "functions are first class objects" in a functional language.
(fd-at g n)that takes a function g and a value n and returns the finite difference of g at n. For example:
> (define (f x) (* x x)) > (fd-at f 3) 7
The function map is a really useful function that takes functions as arguments.
(map f L) applies the function f to each element of the list L
and puts the results together in a list. For example:
> (map abs '(-4 12 -3 -8 11 0)) (4 12 3 8 11 0)
If you have a function with k arguments, then you give map k lists, and it will take the first argument from list1, the second from list2, etc.
> (map * '(2 3 4) '(6 5 4)) (12 15 16)
Another useful function of this type is apply.
(apply f L)
calls the function
f with arguments the elements of
are some examples:
> (apply max '(4 6 2)) ; same as (max 4 6 2) 6 > (apply - '(3 7)) ; same as (- 3 7) -4
Combining map and apply can be very interesting.
> (define (sqr x) (* x x)) > (map sqr '(1 2 3 4 5 6 7 8 9 10)) (1 4 9 16 25 36 49 64 81 100) > (apply + (map sqr '(1 2 3 4 5 6 7 8 9 10))) 385
Yet another useful function-that-takes-a-function is filter.
(filter f? L) applies the predicate function
f? to each element in the list
L, and returns the
list containing only the elements for which the predicate returns true.
> (filter number? '(a b 2 #t + 4)) (2 4)
For these, you will probably want to use this function:
;;Returns a list containing integers a, a+1, ..., b. (define (range a b) (if (> a b) null (cons a (range (+ a 1) b))))
sqrmight also be useful.
(define P1 '((3 5) (9 2) (11 6) (8 8) (4 6)))P1 represents a polygon whose vertex coordinates are given by the pairs in P1. Write a function
(translate p d)that takes a polygon (in the sense of P1, i.e., a list of coordinate pairs) and an offset d (represented by a pair, the x offset and the y offset) and translates the given polygon by the given offset.
> (define P1 '((3 5) (9 2) (11 6) (8 8) (4 6))) > (translate P1 '(1 2)) ((4 7) (10 4) (12 8) (9 10) (5 8)) >Depending on your approach, the following function might be useful to you:
(define (repeat x k) ;Creates a list of k copies of x (if (= k 0) '() (cons x (repeat x (- k 1)))))