Fundamentals of Computer Science I: Media Computing (CS151.02 2007F)
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Reference: [Scheme Report (R5RS)] [Scheme Reference] [DrScheme Manual]
Related Courses: [CSC151.01 2007F (Davis)] [CSC151 2007S (Rebelsky)] [CSCS151 2005S (Stone)]
Summary: In this laboratory, you will explore various aspects of the Vector data type that Scheme provides as an alternative to lists.
a. In the interaction window, type in a vector literal that denotes a vector containing just the two elements 3.14159 and 2.71828. How does DrFu display the value of this vector?
b. Create a vector that contains the same two values by using the
vector
procedure.
c. Create a vector that contains the same two values by using the
make-vector
and vector-set!
procedures.
Consider the following code. (That is, read it, don't enter it.)
>
(define aardvark (list 1 2 3 4))
>
(define baboon aardvark)
>
(define aardvark (cons 5 (cdr aardvark)))
>
(define chinchilla (vector 1 2 3 4))
>
(define dingo chinchilla)
>
(vector-set! chinchilla 0 5)
a. What do you expect the output of the following commands to be?
>
(list-ref aardvark 0)
_____>
(list-ref baboon 0)
_____
b. Verify your answer experimentally. (That is, you can type in the commands now.)
c. What do your results suggest about Scheme?
d. What do you expect the output of the following commands to be? (That is, think about the answer; don't just type it in.)
>
(vector-ref chinchilla 0) _____>
(vector-ref dingo 0) _____
e. Verify your answer experimentally. (You can type the commands now.)
f. What do your results suggest about Scheme?
Write a procedure, (
, which takes one argument,
a vector of numbers, and returns the sum of the elements of that vector.
vector-sum
numbers
)
You can use vector->list
from the reading as
a pattern for vector-sum
;
only a few judicious changes are needed. However, you should
not use vector->list
as a helper.
;;; Procedure: ;;; vector->list ;;; Parameters: ;;; vec, a vector ;;; Purpose: ;;; Builds a list that contains the elements of vec in the same order. ;;; Produces: ;;; lst, a new list. ;;; Preconditions: ;;; [None] ;;; Postconditions: ;;; The element at position i of lst is the element at position ;;; i of vec for all reasonable i. ;;; Does not change vec. (define vector->list (lambda (vec) (vector->list-kernel vec 0 (vector-length vec)))) ;;; Procedure: ;;; vector->list-kernel ;;; Parameters: ;;; vec, a vector ;;; start, an integer ;;; finish, an integer ;;; Purpose: ;;; Builds a list that contains elements start ... finish-1 of vec ;;; Produces: ;;; lst, a new list. ;;; Preconditions: ;;; start >= 0 ;;; finish >= start ;;; finish <= length of vec ;;; Postconditions: ;;; The element at position i of lst is the element at position ;;; i+start of vec for all reasonable i. ;;; Does not change vec. (define vector->list-kernel (lambda (vec start finish) ; We stop at the finish. (if (= start finish) ; There are no more elements to put into a list, so use ; the empty list. null ; Otherwise, add the element at position start to a list of ; the remaining elements. (cons (vector-ref vec start) (vector->list-kernel vec (+ 1 start) finish)))))
In the reading on vectors, we saw that it was possible
to implement list->vector
and
vector->list
by using more primitive
operations (particularly vector-set!
and
vector-length
).
Here's the definition of list->vector
and its kernel.
;;; Procedure: ;;; list->vector ;;; Parameters: ;;; lst, a list. ;;; Purpose: ;;; Convert the list to a vector. ;;; Produces: ;;; vec, a vector ;;; Preconditions: ;;; lst is a list ;;; Postconditions: ;;; vec is a vector. ;;; The length of vec equals the length of lst. ;;; The ith element of vec equals the ith element of lst for ;;; all "reasonable" i. (define list->vector (lambda (lst) (list->vector-kernel! lst ; Copy the whole list 0 ; Starting at position 0 (make-vector (length lst) null) ; Into a new vector of the appropriate length ))) ;;; Procedure: ;;; list->vector-kernel! ;;; Parameters: ;;; lst, a list to copy ;;; pos, a position in the vector ;;; vec, a vector ;;; Purpose ;;; Copy values from lst into positions pos, pos+1, ... ;;; Produces: ;;; vec, the same vector but with different contents. ;;; Preconditions: ;;; The length of vec = pos + the length of list. [Unverified] ;;; pos >= 0. [Unverified] ;;; Postconditions: ;;; Element pos of vec now contains element 0 of list. ;;; Element pos+1 of vec now contains element 1 of lst. ;;; Element pos+2 of vec now contains element 2 of lst. ;;; ... ;;; The last element of vec now contains the last element of lst. (define list->vector-kernel! (lambda (lst pos vec) ; We don't bother to verify the preconditions because this ; procedure should only be called by something that has already ; verified the preconditions (explicitly or implicitly). (cond ; If there's nothing left to copy, stop. ((null? lst) vec) ; Otherwise, copy the initial element of the list and ; then copy the remaining elements (else (vector-set! vec pos (car lst)) (list->vector-kernel! (cdr lst) (+ pos 1) vec)))))
a. The design of list->vector
is a bit
subtle. Explain what the kernel does. If you're not completely
sure, you may want to add a call to display
to the kernel, as in
(display (list 'l2v-kernel! lst pos vec)) (newline)
b. Write your own version of vector-fill!
. Remember that
vector-fill!
takes two parameters, a vector and a value,
and puts that value in every position of the vector.
Note: You may find that you want to do two things for a particular
position: fill the value at that position and recurse. Remember that
when you want to sequence actions if a test succeeds, you should use
a cond
rather than a if
.
It may be that vectors provide a better mechanism for storing palettes than lists. How can we tell? We can rewrite some of the procedures to check.
Write a procedure, (
, that
given a vector of colors and a color as parameters, finds the index of
the color in the vector closest to color.
vpalette.index-of
vpalette
color
)
Write a procedure, (
that rotates the elements in
rotate
!
vec
)vec
. That is, rotate!
puts the
initial element of vec
at the end, the element
at position 1 in position 0, the element at position 2 in position 1,
and so on and so forth.
a. Write a procedure, (
, that creates a new vector
whose elements appear in the reverse order of the elements in
vector.reverse
vec
)vec
.
b. Write a procedure,
(vector.reverse!
,
that reverses vec
)vec
“in place”.
That is, instead of producing a new vector, it rearranges the elements
within vec
.
Write a procedure, (
,
that rotates the values in rotate!
vec
amt
)vec
by
amt
positions. That is, the first
amt
values in vec
move
to the end, the value in position amt
moves to
position 0, the value in position amt
+1 moves
to position 1, and so on and so forth.
list->vector
, you will probably
want to define a helper procedure that fills only part of the vector.
Your termination condition will certainly be different and should probably
involve the length of the vector.
Primary: [Front Door] [Glance] - [Academic Honesty] [Instructions]
Current: [Outline] [EBoard] [Reading] [Lab] [Assignment]
Groupings: [Assignments] [EBoards] [Examples] [Exams] [Handouts] [Labs] [Outlines] [Projects] [Readings] [Reference]
Reference: [Scheme Report (R5RS)] [Scheme Reference] [DrScheme Manual]
Related Courses: [CSC151.01 2007F (Davis)] [CSC151 2007S (Rebelsky)] [CSCS151 2005S (Stone)]
Copyright © 2007 Janet Davis, Matthew Kluber, and Samuel A. Rebelsky. (Selected materials copyright by John David Stone and Henry Walker and used by permission.)
This material is based upon work partially supported by the National Science Foundation under Grant No. CCLI-0633090. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
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