CMPS 10 Winter 2011- Homework Assignment 3 Chapter 2 (p.76 ...

CMPS 10 Winter 2011- Homework Assignment 3
Problems:
Chapter 2 (p.76): 20, 21, 22
Chapter 3 (p.120): 3abc, 6, 17abc, 18, 19, 20, 21, 22
Chapter 2:
20. Design an algorithm that is given a positive integer N and determines whether N is a
prime number, that is, not evenly divisible by any value other than 1 and itself. The
output of your algorithm is either the message “not prime”, along with a factor of N, or
the message “prime”.
Solution:
Step Operation
1.) Input N
2.) set i to 2
3.) while ( i

6.) code_dist = dist + k
7.) if ( code_dist < 26 )
8.) Codei = ‘a’ + code_dist
9.) else
10.) Codei = ‘a’ + code_dist – 26
11.) set i to i+1
12.) print Code
13.) STOP
Explanation: This algorithm assumes that alphabets ‘a’ –‘z’ have consecutive values
assigned to them, for example, as in ASCII code. (see page 141 in Text, 5 th ed). Thus ‘b’
=’a’+1, ‘c’ = ‘a’ + 2 so on. Each character has a fixed distance from ‘a’. In step 5, we
compute the distance of ith letter in text from ‘a’. In step 6, we add k to this value to get
distance of coded letter. If this code_dist is less then 26, we can get the character by adding it
to ‘a’ as in step 8. If code_dist is greater than 26, we wrap around. A code_dist of 27 would
actually be distance 1.
22. Design and implement an algorithm that is given as input an integer value k 0
and a list of k numbers N1, N2, … , Nk. Your algorithm should reverse the order of the
numbers in the list. That is, if the original list contained:
N1 = 5, N2 = 13, N3 = 8, N4 = 27, N5 = 10 (k = 5)
Then when your algorithm has completed, the values stored in the list will be:
N1 = 10, N2 = 27, N3 = 8, N4 = 13, N5 = 5.
Solution:
There are several possible solutions to this problem. Here is one possible algorithm:
Step Operation
1.) Input k and N1 … Nk
2.) Create and set i to 1, and create Temp
3.) Repeat steps 4, 5, 6, and 7 while (i k/2) (See note in explanation)
4.) Set Temp to Ni
5.) Set Ni to N(k + 1 – i)
6.) Set N(k + 1 – i) to Temp
7.) Set i to i + 1
8.) STOP
Explanation: This algorithm reverses the list in place by utilizing a temporary storage
variable. One could replace lines 4-6 with a line saying “Swap Ni and N(k + 1 – i)” if an
appropriate “swap” algorithm were available. In step 3, we assume integer division for
“k/2,” which truncates the answer. For example, if k = 5, we get 5/2 = 2. Therefore, again
using k = 5, we swap N1 with N(5 + 1 – 1) (or N5) and we swap N2 with N(5 + 1 – 2) (or N4). Due
to integer division, we stop at this point and leave N3 alone, which is correct. If we are using
real number division, the algorithm will attempt to swap N3 with N3. This does not change

the answer or break the algorithm, but it is a wasted operation. If k is an even number, then
the type of division does not matter since there is no “middle element” that would be swapped
with itself.
Chapter 3:
3. A tennis tournament has 342 players. A single match involves 2 players. The winners
of a match will play the winner of a match in the next round, whereas losers are
eliminated from the tournament. The 2 players who have won all previous rounds play
in the final game, and the winner wins the tournament. What is the total number of
matches needed to determine the winner?
a. Here is one algorithm to answer this question. Compute 342 / 2 = 171 to get the
number of pairs (matches) in the first round, which results in 171 winners to go on to the
second round. Compute 171 / 2 = 85 with 1 left over, which results in 85 matches in the
second round and 85 winners, plus the 1 left over, to go to the third round. For the third
round compute 86 / 2 = 43, so the third round has 43 matches, and so on. The total
number of matches is 171 + 85 + 43 + …. Finish this process to find the total number of
matches.
Solution:
The number of matches in the fourth round is 43 / 2 = 21, with 1 left over
The number of matches in the fifth round is 22 / 2 = 11
The number of matches in the sixth round is 11 / 2 = 5, with 1 left over
The number of matches in the seventh round is 6 / 2 = 3
The number of matches in the eighth round is 3 / 2 = 1, with 1 left over
The number of matches in the fifth round is 2 / 2 = 1
So, the total number of matches is
171 + 85 + 43 + 21 + 11 + 5 + 3 + 1 + 1 = 341
b. Here is another algorithm to solve the problem. Each match results in exactly one
loser, so there must be the same number of matches as losers in the tournament.
Compute total number of losers in the entire tournament. (Hint: This isn’t really a
computation; it is a one-sentence argument.)
Solution:
At the end of the tournament, there is only one winner, 341 losers. Since each match results in
exactly one loser, the total number of matches is 341 for the tournament.
c. What are your opinions on the relative clarity, elegance, and efficiency of the two
algorithms?
Solution:
Both algorithms are clear and easy to understand, but the second algorithm is more elegant
and much efficient. Especially when the total number of players in the tournament is large,
there is a big difference in term of efficiency.

6. Perform a selection sort on the list 7, 4, 2, 9, 6. Show the list after each exchange
that has an effect on the list ordering.
Solution:
Initial List: 7, 4, 2, 9, 6
Exchange 6 and 9: 7, 4, 2, 6, 9
Exchange 7 and 6: 6, 4, 2, 7, 9
Exchange 6 and 2: 2, 4, 6, 7, 9
List sorted: 2, 4, 6, 7, 9
17. Consider the following list of names.
Arturo, Elsa, JoAnn, John, Jose, Lee, Snyder, Tracy
1 2 3 4 5 6 7 8
a. Use binary search to decide whether Elsa is in this list. What names will be
compared to Elsa?
Solution:
(The names being compared are bold and italic.)
There are totally 8 names in the list. So the midpoint position is location 4, which is
John. So we first compare the target name Elsa with John. Elsa precedes John, so we
go to the front half of the list and compare Elsa with the midpoint of this half
segment at location 2, which is Elsa – the target name is found!
b. Use binary search to decide whether Tracy is in this list. What names will be
compared to Tracy?
Solution:
(The names being compared are bold and italic.)
1) Compare Tracy with the midpoint of the whole list, which is John. Since Tracy
follows John, go to the back half segment [5,6,7,8].
2) Compare Tracy with the midpoint of the segment [5,6,7,8], which is Lee. Since
Tracy follows Lee, go the back half segment [7,8].
3) Compare Tracy with midpoint of the segment [7,8], which is Snyder. Since Tracy
follows Snyder, go to the back half segment [8].
4) Compare Tracy with Tracy at location 8 – the target name is found!
c. Use binary search to decide whether Emile is in this list. What names will be
compared to Emile?
Solution:
(The names being compared are bold and italic.)
1) Compare Emile with the midpoint of the whole list, which is John. Since Emile
precedes John, go to the front half segment [1,2,3].

2) Compare Emile with the midpoint of the segment [1,2,3], which is Elsa. Since
Emile follows Elsa, go to the back half segment [3].
3) Compare Emile with JoAnn at location 3 – not match! So Emile is not in the list.
18. Use the binary search algorithm to decide whether 35 is in the following list:
3, 6, 7, 9, 12, 14, 18, 21, 22, 31, 43
What numbers will be compared to 35?
Solution:
The binary search algorithm compares the target number (35) to the middle number of
each segment it searches. Thus it compares 35 to 14, then 22, then 31 (rounding down
for the middle value), and then 43. Thus it compares four numbers to 35: 14, 22, 31, 43.
19. If a list is already sorted in ascending order, a modified sequential search algorithm
can be used that compares against each element in turn, stopping if a list element
exceeds the target value. Write a pseudo code version of this short sequential search.
One possible solution:
1. Get value for n, the number of elements in the list
2. Get value for Target, the target value is being searched
3. Set the value of i to 1
4. While i is less than or equal to n, do steps 5 through 13
5. If Target is equal to the value of the ith element in the list
6. Print the message “The target value is found: ”
7. Print the value of the ith element
8. Go to Step 14 (Exit from the “While” loop)
9. Else if Target is greater than the value of the ith element
10. Print the message “The target value is not in the list”
11. Go to Step 14
12. Else (Target is less than the value of the ith element)
13. Increment i by 1
14. Stop
20. This exercise refers to short sequential search (see Exercise 19).
a. What is the worst case number of comparisons of short sequential search on sorted
n-element list?
b. What is the approximate average number of comparisons to find an element that is in
a sorted list using short sequential search?
c. Is short sequential search ever more efficient than regular sequential search? Explain.
Solution:
a. Worst case comparisons is n.
b. Average number of comparisons is (n+1)/2.

c. The algorithms have similar efficiency.
21. Draw the tree structure that describes binary search on the 8-element list in Excise
17. What is the number of comparisons in the worst case? Give an example of a
name to search for that requires that many comparisons.
Solution:
The worst case scenario requires logN+1 = 4 comparisons. See Excise 17(b) for the
example.
22. Draw the tree structure that describes binary search on a list with 16 elements.
What is the number of comparisons in the worst case?
Solution:
John
Elsa Lee
Arturo JoAnn Jose Snyder
Tracy
Because there are five levels to this tree, the worst case scenario occurs with element 16
and requires 5 comparisons.