Summary: In this laboratory, we consider merge sort, a more efficient technique for sorting lists of values.
Make a copy of
mergesort-lab.rkt, the code for this lab.
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.
a. Uncomment the following line in the
(write list-of-lists) (newline)
b. What output do you expect to get if you run your updated
new-merge-sort on the list from the reading's
self-check 3, step b?
c. 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. Briefly review the definitions of
counter-print! in the definitions pane.
b. Define a counter named
a counter called
c. Add the following function.
(define int-may-precede? (lambda (left right) (counter-count! may-precede-counter) (<= left right)))
d. Add the following line to the beginning of the
e. Add the following helpful procedure to your definitions.
(define experiment (lambda (lst) (counter-reset! may-precede-counter) (counter-reset! merge-counter) (let ([result (merge-sort lst int-may-precede?)]) (counter-print! may-precede-counter) (counter-print! merge-counter) result)))
f. Using these counters, count the number of calls to
in sorting a few lists of size 8, 16, 32, and 64. (Try a
few lists of each size. You should use
random-numbers to generate the lists.)
Is the number of calls to
merge similar or
different for different lists of the same size? Is the number of calls
may-precede? similar or different for different
lists of the same size? What explains the similarities or differences?
h. 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.)
Assume that we represent names as lists of the form
Write an expression to merge the following two lists:
(define mathstats-faculty (list (list "Blanchard" "Jeff") (list "Chamberland" "Marc") (list "Fellers" "Pamela") (list "French" "Chris") (list "Jonkman" "Jeff") (list "Kuiper" "Shonda") (list "Mileti" "Joseph") (list "Moore" "Emily") (list "Moore" "Tom") (list "Olsen" "Chris") (list "Paulhus" "Jennifer") (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))))))