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ser222_exam_practice_deprecated.lyx
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#LyX 2.3 created this file. For more info see http://www.lyx.org/
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SER 222
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Deprecated Questions
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(SOLN)
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Updated 3/19/2022
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Last Name:
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Last 4 digits of ASU ID:
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$
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Exam Instructions
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The exam is open textbook (Algorithms 4e by Sedgewick and Wayne), as well
as open note.
\bar under
No electronic items are allowed.
Write legibly.
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Please use a pen (instead of a pencil) if you have one.
There are 44 points available and the exam must be completed in 37.5 minutes.
This exam has three types of questions:
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Programming questions:
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The programming questions are given near the end of the paper.
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There are 24 points of write-in programming questions.
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Topic
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Earned
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Possible
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MC/SA: Recursion
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20
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Prog: Recursion
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24
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Total:
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44
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\begin_inset Formula $ $
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\begin_inset Formula $ $
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Short Answer: Recursion
\end_layout
\begin_layout Enumerate
Consider the problem of reversing a string.
For example: "ser222" becomes "222res".
Should this problem be solved with recursion? [5 points]
\begin_inset Note Comment
status open
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from FA19 exam
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\begin_inset Note Comment
status collapsed
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LO2
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Yes - using recursion is necessary for the structure of this problem.
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Yes - using recursion is ideal to develop a fast algorithm.
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\begin_layout Enumerate
No - using recursion will require many more function calls than a loop.
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ANS
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\begin_layout Enumerate
No - using recursion will require writing much more complex code.
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\end_deeper
\begin_layout Enumerate
Consider the following method.
Is this a correct recursive implementation for factorial? [5 points]
\end_layout
\begin_deeper
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lstparams "language=Java"
inline false
status open
\begin_layout Plain Layout
int factorial(int n) {
\end_layout
\begin_layout Plain Layout
if(n>1)
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return n * factorial(n-1);
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else
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return 0;
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}
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Yes, it calls itself on the n-1 sub-problem.
\end_layout
\begin_layout Enumerate
Yes, it includes both a base case and recursive step.
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\begin_layout Enumerate
No, the n>1 condition is not correct for factorial.
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\begin_layout Enumerate
No, the return value is not correct for factorial.
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ANS
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\begin_layout Enumerate
Consider an algorithm for determining the size of some nested folders/files
recursively.
(Imagine right clicking an item on your desktop and viewing it's size.)
What would be the base case(s) for this problem? [10 points]
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Ans:
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An empty folder, or a folder containing only files.
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Programming: Recursion
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\end_deeper
\begin_layout Enumerate
Implement a recursive method called
\series bold
countOccurrences
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that takes the first node in a singly linked list and returns the number
of times a particular value occurs in the list.
For example, if countOccurrences("A") is called on a list with nodes containing
: "A", "B", "C", "B", "E", it would return 1.
If countOccurrences("B") was called on the same list, it would return 2.
For reference, a standard implementation for the nodes of a singly linked
list is given below.
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Source: MT Fall 2017.
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public class LinearNode<T> {
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private T element;
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public LinearNode(T elem) { next = null; element = elem; }
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public LinearNode<T> getNext() { return next; }
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public void setNext(LinearNode<T> node) { next = node; }
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public T getElement() { return element; }
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public void setElement(T elem) { element = elem; }
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}
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\begin_layout Enumerate
Using the fantastic four approach, determine the size n problem for the
method countOccurrences.
[2 point]
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\series bold
Ans:
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Finding the total number of times a value occurs in the whole list.
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\end_layout
\end_inset
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Identify the stopping condition(s) and the return value, if any, for the
problem.
[2 point]
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Ans:
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An empty list, for which zero will be returned.
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\end_inset
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\end_deeper
\begin_layout Enumerate
Determine the size m problem(i.e.
the “sub-problem”) for the problem.
[2 point]
\end_layout
\begin_deeper
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\begin_inset Branch blank
inverted 0
status collapsed
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\begin_inset Branch soln
inverted 0
status collapsed
\begin_layout Standard
\series bold
Ans:
\end_layout
\begin_layout Standard
Finding the total number of times a value occurs in the tail of the list
(i.e., search everything after the first node).
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\end_layout
\end_inset
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\begin_layout Enumerate
How is the size-n problem constructed from the size m problem? [2 point]
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\begin_deeper
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\begin_inset Branch blank
inverted 0
status collapsed
\begin_layout Standard
\begin_inset VSpace 1.5cm
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\begin_layout Standard
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Ans:
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Add the sub-problem count to 1 if the head has a match, otherwise add 0.
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\begin_layout Standard
\begin_inset VSpace 0.75cm
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\end_deeper
\begin_layout Enumerate
Implement the recursive method
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public static int countOccurrences<T>(LinearNode node, T target)
\series default
(Hint: use .equals to compare the target with the contents of each node.)
[16 points]
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\begin_inset VSpace 4cm
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Ans:
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lstparams "language=Java"
inline false
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\begin_layout Plain Layout
public static int countOccurrences<T>(LinearNode node, T target) {
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if(node == null)
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return 0;
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\begin_layout Plain Layout
if(node.element.equals(target))
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return 1 + countOccurrence(node.getNext(), target);
\end_layout
\begin_layout Plain Layout
else
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\begin_layout Plain Layout
return countOccurrence(node.getNext(), target);
\end_layout
\begin_layout Plain Layout