CMPS 10 Winter 2011- Homework Assignment 2
Problems:
Chapter 2 (p.75): 1abc, 2, 4, 9, 12, 13, 15, 16abc, 17ab, 18ab
1. Write pseudocode instructions to carry out each of the following computational operations.
a.) Determine the area of a triangle given values for the base b and the height h.
b.) Compute the interest earned in 1 year given the starting account balance B and the annual
interest rate I and assuming simple interest, that is, no compounding. Also determine the
final balance at the end of the year.
c.) Determine the flying time between two cities given the mileage M between them and the
average speed of the airplane.
Solutions:
(a): Set the value of area to ½ * b * h.
(b): Set the value of FinalBalance to (1 + I) * B, assuming I is represented as a decimal fraction.
Interest only is equal to I * B.
(c): Set the value of FlyingTime to M/AvgSpeed.
2. Using only the sequential operations described in Section 2.2.2, write an algorithm that gets
values for the starting account balance B, annual interest rate I, and annual service charge S.
Your algorithm should output the amount of interest earned during the year and the final account
balance at the end of the year. Assume that the interest is compounded monthly and the service
charge is deducted once, at the end of the year.
Solution:
Here is one possible solution
Step
Operation
1.) Input values for B, I, and S
2.) Set the value of FinalBalance to B * (1 + I/12)12
3.) Set the value of TotalInterest to FinalBalance – B
4.) Set the value of FinalBalance to FinalBalance – S
5.) Print the message “Total Interest Earned: “, and print the value of TotalInterest
6.) Print the message “Final Balance: “, and print the value of FinalBalance
4. Write an algorithm that gets the price for item A plus the quantity purchased. The algorithm
prints out the total cost, including a 6% sales tax.
Solution:
One possible solution is:
Step
Operation
1.) Input values for A and N (the quantity purchased).
2.) Set the value of TotalCost to 1.06 * (A * N)
3.) Print out the value of TotalCost.
9. Modify the sales algorithm of Exercise 4 so that after finishing the computation for one item, it
starts on the computation for the next. This iterative process is repeated until the total cost
exceeds $1000.
Solution:
Following directly from the previous algorithm of problem 4, we obtain:
Step
1.)
2.)
3.)
4.)
5.)
Operation
Set TotalCost to 0.
While TotalCost is less than or equal to $1000, repeat steps 3 and 4.
Input values for A and N (the quantity purchased).
Set the value of TotalCost to TotalCost + (1.06 * (A * N)).
Print out the value of TotalCost.
12. Develop a formal argument that “proves” that the sequential search algorithm shown
in figure 2.13 cannot have an infinite loop; that is, prove that it will always stop after a
finite number of operations.
Solution:
Steps 1, 2, 5, 6, 7, 9, and 10 are sequential operations and thus cannot go into an infinite loop.
Steps 4 and 8 are conditionals, and also cannot go into an infinite loop. Step 3 is the only one
left to consider, and since it is a repeating loop, it could potentially become an infinite loop.
The looping conditions is “if Found = no and i  10,000.” If Found ever becomes yes (the
algorithm finds the name), then the condition is no longer satisfied and the loop will end. If
NAME is never found, then line 7 adds 1 to i per iteration. Since i starts at 1 (from line 2), it will
advance to 10,001 after 10,000 iterations of the loop. At this point, the second portion of the
condition will no longer be true and the loop will halt. Then lines 8, 9, and 10 will be executed,
and the program will stop.
13. Modify the sequential search algorithm of Figure 2.13 so that it works correctly even if
the names in the directory are not unique, that is, if the desired name occurs more than
once. Your modified algorithm should find every occurrence of NAME in the directory
and print out the telephone numbers corresponding to every match. In addition, after all
the numbers have been displayed, your algorithm should print out how many occurrences
of NAME were located. For example, if NAME occurred three times, the output of the
algorithm might look something like this:
528-5638
922-7874
488-2020
A total of three occurrences were located.
Solution:
Here is one possible solution to the problem.
numbers from the directory.
Let N denote names and T denote telephone
Step
Operation
1.) Get values for NAME, N1, … , N10000 and T1, … , T10000
2.) Set the value of i to 1, and set the value of NumberFound to 0.
3.) Repeat steps 4 through 7 while i  10,000.
4.)
If Ni = NAME, then
5.)
Print Ti
6.)
Set the value of NumberFound to NumberFound + 1.
7.)
Add 1 to the value of i.
8.) Print “A total of <NumberFound> occurrences were located.”
9.) STOP.
15. With regards to the Find Largest algorithm of Figure 2.14, if the numbers in our list
were not unique and therefore the largest number could occur more than once, would the
algorithm find the first occurrence? The last occurrence? Every occurrence? Explain
precisely how this algorithm would behave when presented with this new condition.
Solution:
This algorithm will find the first occurrence of the largest element in the collection. This first
occurrence will become LargestSoFar, and from then on, Ai will be tested to see if it is greater
than LargestSoFar. Some of the other elements are equal to LargestSoFar but since none are
greater, the first element will be retained.
16. On the sixth line of the Find the Largest algorithm of Figure 2.14 there is an
instruction that reads:
While (i  n) do
Explain exactly what would happen if we changed that instruction to read as follows:
a.) While (i  n) do
b.) While (i < n) do
c.) While (i = n) do
Solution:
a.) If n < 2, then the test would be true, so the loop would be executed. In fact, the algorithm
would enter an infinite loop because the condition (i  n) would never become false, or the
algorithm would generate an error when attempting to refer to an invalid Ai value. On the other
hand, if n > 2, then the test condition would be false at the first check, and the loop would be
skipped entirely. Thus A1 would be reported as the largest value, probably erroneously.
b.) In the case of (i < n), the algorithm would work almost correctly. Instead of finding the
largest value of the 1st through nth element, it would find the largest value of the 1st through nth –
1 element because the loop would exit when n = i.
c.) If n = 2, then the loop would execute once, comparing the A1 and A2 values to determine
which was larger. For any other value of n, the loop would never execute, and the value of A1
would be reported.
17. On the seventh line of the Find the Largest algorithm of Figure 2.14 is an instruction
that reads:
If Ai > largest so far then…
Explain exactly what would happen if we changed that instruction to read as follows:
a.) If Ai  largest so far then …
b.) If Ai < largest so far then …
Solution:
a.) The algorithm would still correctly find the largest element on the list. However, if the
largest element was duplicated, this algorithm would find the last occurrence of the largest
element instead of the first.
b.) The algorithm would find the smallest element on the list instead of the largest element,
completely reversing the goal of the algorithm.
18.
a.) Refer to the pattern-matching algorithm in Figure 2.16. What is the output of the
algorithm as it currently stands if our text is:
Text: We must band together and handle adversity
and we search for the pattern “and”?
b.) How could we modify the algorithm so that it finds only the complete word and rather
than the occurrence of the character sequence a, n, and d that are contained within another
word such as band?
Solution:
a.) The algorithm will find three occurrences of the pattern “and.” The first occurrence is the
last three characters of “band.” The second occurrence is the word “and.” The third
occurrence is the “and” portion of “handle.” Remembering that the counter starts at 1 for “W”
and spaces count as part of the text, the output of the program will be:
There is a match at position 10
There is a match as position 23
There is a match as position 28
b.) In order to find only complete words, the algorithm must also examine the characters
immediately before and after the word. Searching for spaces on either side would catch all
normal stand-alone instances of the word. In other words, search for “ and ”. However, this
does not account for possibilities when the word is inside quotation marks or has other
punctuation surrounding it. For example, the string “and?” would not match with the above
method. Therefore, in addition to searching for spaces, also search for punctuation surrounding
the word. Lastly, even searching for punctuation may not be enough. If you wanted to search
for the word “an” and were scanning a text with the word “Qur’an” would produce a positive
match because it would find the string “’an “. All in all, one needs to be very careful in
construction rules of this nature.