Summary: In this laboratory, we consider merge sort, a more efficient technique for sorting lists of values.
Make a copy of
mergesort-lab.scm, the code for this lab.
a. Write an expression to merge the lists
(1 2 3) and
(1 1.5 2.3).
b. Write an expression to merge two identical lists of numbers.
For example, you might merge the lists
(1 2 3 5 8 13 21)
(1 2 3 5 8 13 21)
c. Write an expression to merge two lists of strings. (You may choose the strings yourself. Each list should have at least three elements.)
a. What will happen if you call
merge with unsorted
lists as the two list parameters?
b. Check your answer by experimentation. To help you understand what is
happening, you may wish to modify
merge so that it
displays the values of
c. What will happen if you call
merge with sorted
lists of very different lengths as the first two parameters?
d. Check your answer by experimentation.
split to split:
a. A list of numbers of length 6
b. A list of numbers of length 5
c. A list of strings of length 6
merge-sort on a list you design of fifteen integers.
new-merge-sort on a list you design of ten strings.
c. Add the following lines to the
new-merge-sort. (Add them directly after
the line that contains
(display list-of-lists) (newline)
d. What output do you expect to get if you rerun
new-merge-sort on the list from step b?
e. Check your answer experimentally.
new-merge-sort on a list of twenty integers.
As we've seen, in exploring any algorithm, it's a good idea to check a few special cases that might cause the algorithm difficulty. Here are some to start with.
a. Run both versions of merge sort on the empty list.
b. Run both versions of merge sort on a one-element list.
c. Run both versions of merge sort on a list with duplicate elements.
We've claimed that merge sort takes approximately
steps. Let's explore that claim experimentally. We'll focus on the
number of calls to
a. Update the definitions of
define$ rather than
analyze, count the number of calls to
may-precede? in sorting a few lists of size
8, 16, 32, and 64. (Try a few lists of each size. You may find
it helpful to use
random-numbers to generate the lists.)
c. Are the number of calls to
or different for the same size list? What explains the similarities
d. Does the running time seem to grow slower than n2? (In such functions, when you double the input size, you should quadruple the number of steps.)
As you've probably noticed, there are two key postconditions of a procedure that sorts lists: The result is a permutation of the original list and the result is sorted.
that the unit test framework lets us test permutations (with
test-permutation!). Hence, if we wanted to test
merge sort in the unit test framework, we might write
(define some-list ...) (test-permutation! (merge-sort some-list pred?) some-list)
We need a way to make sure that the result is sorted, particularly if the result is very long.
Write a procedure,
( that checks whether or not
is sorted by
(sorted? (list 1 3 5 7 9) <=)
(sorted? (list 1 3 5 4 7 9) <=)
(sorted? (list "alpha" "beta" "gamma") string-ci<=?)
Note that we can use that procedure in a test suite for merge sort with
(test! (sorted? (merge-sort some-list may-precede?) may-precede?) #t)
Assume that we represent names as lists of the form
Write an expression to merge the following two lists
(define mathstats-faculty (list (list "Chamberland" "Marc") (list "French" "Chris") (list "Jager" "Leah") (list "Kuiper" "Shonda") (list "Moore" "Emily") (list "Moore" "Tom") (list "Mosley" "Holly") (list "Romano" "David") (list "Shuman" "Karen") (list "Wolf" "Royce"))) (define more-faculty (list (list "Moore" "Chuck") (list "Moore" "Ed") (list "Moore" "Gordon") (list "Moore" "Roger")))
Some computer scientists prefer to define
like the following.
(define split (lambda (ls) (let kernel ((rest ls) (left null) (right null)) (if (null? rest) (list left right) (kernel (cdr rest) (cons (car rest) right) left)))))
a. How does this procedure split the list?
b. Why might you prefer one version of split over the other?
We've written a procedure that checks whether a list is sorted, which is one of the postconditions of a list-based sorting routine. However, we have not yet written a procedure to check whether two lists are permutations of each other.
Write a procedure,
that determines if
lst2 is a permutation
You might find the following strategy, which involves iterating through the first list, an appropriate strategy for testing for permutations.
Base case: If both lists are empty, the two lists are permutations of each other.
Base case: If the first list is empty and the second is not (i.e., it's a pair), the two lists are not permutations of each other.
Base case: If the first list is nonempty and car of the first list is not contained in the second, the two lists are not permutations of each other.
Recursive case: If the first list is not empty and the first element of the first list is in the second list, then the two lists are permutations of each other only if the cdr of the first list is a permutation of what you get by removing the car of the first list from the second list.
You may also find the following two procedures useful.
;;; Procedure: ;;; list-contains? ;;; Parameters: ;;; lst, a list ;;; val, a value ;;; Purpose: ;;; Determines if lst contains val. ;;; Produces: ;;; contained?, a Boolean ;;; Preconditions: ;;; [No additional] ;;; Postconditions: ;;; If there is an i such that (list-ref lst i) equals val, ;;; then contained? is true (#t). ;;; Otherwise, ;;; contained? is false. (define list-contains? (lambda (lst val) (and (not (null? lst)) (or (equal? (car lst) val) (list-contains? (cdr lst) val))))) ;;; Procedure: ;;; list-remove-one ;;; Parameters: ;;; lst, a list ;;; val, a value ;;; Purpose: ;;; Remove one copy of val from lst. ;;; Produces: ;;; newlst ;;; Preconditions: ;;; lst contains at least one value equal to val. ;;; Postconditions: ;;; (length newlst) = (length lst) - 1 ;;; lst is a permutation of (cons val newlst). ;;; Ordering is preserved. That is, if val1 and val2 appear in ;;; both lst and newlst, and val1 precedes val2 in lst, then ;;; val1 precedes val2 in newlst. (define list-remove-one (lambda (lst val) (cond ((null? lst) null) ((equal? val (car lst)) (cdr lst)) (else (cons (car lst) (list-remove-one (cdr lst) val))))))
Copyright (c) 2007-9 Janet Davis, Matthew Kluber, Samuel A. Rebelsky, and Jerod Weinman. (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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