Summary: In this laboratory, you will reflect on the appropriate and inappropriate uses of tail recursion.
a. Review the reading on tail recursion.
b. Start DrScheme.
Identify three (or more) tail-recursive procedures you've already written.
In DrScheme, you can find out how long it takes to evaluate an
expression by using
which prints out the time it takes to evaluate
and then returns the value computed.
Please scan the reading on using
time before continuing this exercise.
a. Try the two versions of
factorial on some large numbers.
Does one seem to be faster than the other? You can find the two versions
of factorial in the notes on this problem.
b. Try the three versions of
add-to-all on some lists
of varying sizes (you'll probably need at least a hundred values in
each list, and possibly more) until you can determine a difference
in running times. Note that you should make sure to build the list
before you start timing.
Note that you can easily create a long list of using
Here's an attempt to write a tail-recursive version of the
select procedure that selects all the values in a list for
which a predicate holds.
;;; Procedure: ;;; select ;;; Parameters: ;;; pred?, a unary predicate ;;; lst, a list of values ;;; Purpose: ;;; Selects all values in lst for which pred? holds. ;;; Produces: ;;; selected, a list ;;; Preconditions: ;;; pred? can be applied to any element of lst. ;;; Postconditions: ;;; every value in selected appears in lst. ;;; (pred? val) holds for every val in selected. ;;; (pred? val) holds for every val in lst not in selected. ;;; values in selected appear in the same order in which they appear in lst. (define select (lambda (pred? lst) (let kernel ((lst lst) (selected null)) (cond ; If nothing's left in the original list, use selected ((null? lst) selected) ; If the predicate holds for the first value in lst, ; select it and continue ((pred? (car lst)) (kernel (cdr lst) (cons (car lst) selected))) ; Otherwise, skip it and continue (else (kernel (cdr lst) selected))))))
a. What do you expect the results of the following two expression to be?
(select odd? (list 1 2 3 4 5 6 7)) (select even? (list 1 2 3 4 5 6 7))
b. Verify your answer through experimentation.
c. Correct any errors in
select that you observed in a or b.
Write a tail-recursive
longest-string-on-list procedure which,
given a list of strings as a parameter, returns the longest string in
Note that you can use
string-length to find the length of
Define a tail-recursive procedure
lst) that returns the index of val
in lst. That is, it should return zero-based location of
lst. If the item is not in the list,
the procedure should return -1. Test your procedure on:
(index 3 (list 1 2 3 4 5 6)) => 2 (index 'so (list 'do 're 'mi 'fa 'so 'la 'ti 'do)) => 4 (index "a" (list "b" "c" "d")) => -1 (index 'cat null) => -1
(iota n) is to be the list of all
nonnegative integers less than n in increasing order.
Define and test a tail-recursive version of
Write your own tail-recursive version of
Here's a non-tail-recursive version of
(define append (lambda (list-one list-two) (if (null? list-one) list-two (cons (car list-one) (append (cdr list-one) list-two)))))
a. Write your own tail-recursive version.
b. Determine experimentally which of the three versions (built-in, given above, tail-recursive) is fastest.
Here are the two versions of
;;; Procedures: ;;; factorial ;;; factorial-tr ;;; Parameters: ;;; n, a non-negative integer ;;; Purpose: ;;; Computes n! = 1*2*3*...*n ;;; Produces: ;;; result, a positive integer ;;; Preconditions: ;;; n is a non-negative integer [unchecked] ;;; Postconditions: ;;; result = 1*2*3*...*n (define factorial (lambda (n) (if (<= n 0) 1 (* n (factorial (- n 1)))))) (define factorial-tr (lambda (n) (letrec ((kernel (lambda (m acc) (if (<= m 0) acc (kernel (- m 1) (* acc m)))))) (kernel n 1))))
Here are the three versions of
;;; Procedures: ;;; add-to-all-1 ;;; add-to-all-2 ;;; add-to-all-3 ;;; Parameters: ;;; value, a number ;;; values, a list of numbers ;;; Purpose: ;;; Creates a new list by adding value to each member of values. ;;; Produces: ;;; new-values, a list of numbers ;;; Preconditiosn: ;;; value is a number [unchecked] ;;; values is a list of numbers [unchecked] ;;; Postconditions: ;;; (list-ref new-values i) equals (+ value (list-ref values i)) ;;; for all reasonable values of i. (define add-to-all-1 (lambda (value values) (if (null? values) null (cons (+ value (car values)) (add-to-all-1 value (cdr values)))))) (define add-to-all-2 (lambda (value values) (let kernel ((values-remaining values) (values-processed null)) (if (null? values-remaining) values-processed (kernel (cdr values-remaining) (append values-processed (list (+ value (car values-remaining))))))))) (define add-to-all-3 (lambda (value values) (let kernel ((values-remaining values) (values-processed null)) ; If we've run out of values to process, (if (null? values-remaining) (reverse values-processed) (kernel (cdr values-remaining) (cons (+ value (car values-remaining)) values-processed))))))
Monday, 30 October 2000 [Samuel A. Rebelsky]
Sunday, 8 April 2001 [Samuel A. Rebelsky]
add-to-endafter adding the answer to the reading.
Thursday, 7 November 2002 [Samuel A. Rebelsky]
if you have time.
Monday, 11 November 2002 [Samuel A. Rebelsky]
factorialand the three versions of
Sunday, 11 April 2003 [Samuel A. Rebelsky]
Thursday, 13 November 2003 [Samuel A. Rebelsky]
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