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
'si413, or 'wombats.
We can use the function symbol? to find out whether
something is a symbol, and symbol=? to compare two
symbols.
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 amt in some foreign currency cur
and returns that amount in US dollars.
The parameter cur will be one of
the symbols
'euro,
'yen,
'cad, or
'usdollar.(to-usdollar 500 'yen)
produces 500/76.733, which comes out to
6.516101286278394.
(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 fromCur
and 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 a and a list
L and return the new list with a inserted
in the front of L.
(car L): Returns the first item in L.
(cdr L): Returns the list containing all the
items in L after 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 cXXXr, where
each X is either a or d,
corresponding to car and cdr:
> (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,
notice that my-append is recursively processing L1
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 L2. When L1 is
not empty, we break it down with car and cdr
and make a recursive call on (cdr L1).
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
cons.
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.
squares that 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)
L
into 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):
a in L'the in L12 in Lsum-cash that 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-cash would 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-dev that 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
The let construct in Scheme allows you to give a
name to a common subexpression. For example, consider the
expression
(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
let allows you to do.
(let ((a (- (* 32 501) 33))) (/ a (- 1 a)))
Essentially, 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
the whole 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
value x.
> (define (shifted-cube x) (let ((a (+ x 1))) (* a a a))) > (shifted-cube 2) 27
test-sin that computes
1/(sin(x)^2 + 1) + sqrt(sin(x)^2 + 1) - (sin(x)^2 + 1)^2
dist that 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''
8
Write this function using a let expression
to create values L1 and L2 for
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.
The expression
(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.
The expression (apply f L)
calls the function f with arguments the elements of
L. Here
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.
The expression (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.
For example:
> (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))))
sqr might 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)))))