Summary: In this laboratory, you will begin to explore some of the key built-in higher-order procedures. You will also write your first higher-order procedures.
best.scm, our example from class.
(map proc lst) builds a list by applying
proc to each element of
lst in succession.
map to compute the successors to the squares of
the integers between 1 and 10. Your result should be
(2 5 10 17 26 37 50 65 82 101).
map to take the last element of each list in a
list of lists. The result should be a list of the last elements.
For example, given
((1 2 3) (4 5 6) (7 8 9 10) (11 12))
as input, you should produce the list
(3 6 10 12).
map to sum the last elements
of each list in a list of lists of numbers. The result should be a
Note that you may have written a similar expression for another lab. Your goal here is to see whether you can solve the problem more concisely.
Although we often use the
map procedure with only two
parameters (a procedure and a list), it can take more than two
parameters, as long as the first parameter is a procedure and the
remaining parameters are lists.
a. What do you think the value of the following expression will be?
(map (lambda (x y) (+ x y)) (list 1 2 3) (list 4 5 6))
b. Verify your answer through experimentation.
c. What do you think the value of the following expression will be?
(map list (list 1 2 3) (list 4 5 6) (list 7 8 9))
d. Verify your answer through experimentation.
e. What do you think Scheme will do when evaluating the following expression?
(map list (list 1 2 3) (list 4 5))
f. Verify your answer through experimentation.
g. What do you think Scheme will do when evaluating the following expression?
(map (lambda (x y) (+ x y)) (list 1 2) (list 3 4) (list 5 6))
h. Verify your answer through experimentation.
map to concisely define a
(dot-product list1 list2), that
takes as arguments two lists of numbers, equal in length, and returns
the sum of the products of corresponding elements of the arguments:
> (dot-product (list 1 2 4 8) (list 11 5 7 3)) 73 ; ... because (1 x 11) + (2 x 5) + (4 x 7) + (8 x 3) = 11 + 10 + 28 + 24 = 73 > (dot-product null null) 0 ; ... because in this case there are no products to add
Sarah and Steven Schemer suggest that
apply is irrelevant. After all,
when you write
(apply proc (arg1 ... argn))
you're just doing the same thing as
(proc arg1 arg2 ... argn).
Given your experience in the previous exercise, are they correct? Why or why not?
a. Document and write a procedure,
that counts the number of values in list for which
b. Demonstrate the procedure by tallying the number of odd values in the list of the first twenty integers.
c. Demonstrate the procedure by tallying the number of multiples of three in the list of the first twenty integers.
Document and write a procedure,
that builds a procedure that takes a list as a parameter and tallies
the values in the list for which the predicate holds. For example
> (define count-odds (make-tallier odd?)) > (count-odds (list 1 2 3 4 5)) 3
You can assume that
tally already exists for the purpose
of this problem.
Write a procedure,
that creates a procedure that takes a list as a parameter and
removes all elements for which predicate holds.
> (define remove-whitespace (make-remover char-whitespace?))
> (remove-whitespace (list #\a #\space #\b #\c))
(#\a #\b #\c)
> ((make-remover odd?) (list 1 2 3 4 5))
make-tallier so that it doesn't call
and so that it doesn't call itself. (You'll have to create a local
recursive procedure and then return it.)
Write a procedure,
(map! proc vec), that
replaces each element of vec with the result of applying proc
to the original element.
Thursday, 2 November 2000 [Samuel A. Rebelsky]
Wednesday, 14 February 2001 [Samuel A. Rebelsky]
Sunday, 8 April 2001 [Samuel A. Rebelsky]
Tuesday, 15 October 2002 [Samuel A. Rebelsky]
Wednesday, 16 October 2002 [Samuel A. Rebelsky]
Sunday, 16 February 2003 [Samuel A. Rebelsky]
if you have extra timesection.
Monday, 16 February 2004 [Samuel A. Rebelsky]
I usually create these pages
on the fly, which means that I rarely
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It also means that I tend to update them regularly (see the history for
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