The University of Melbourne

School of Computing and Information Systems

COMP90038 Algorithms and Complexity

Assignment 1, Semester 2 2019

Released: Friday August 23 2019. Deadline: Sunday September 8 2019 23:59

This assignment is marked out of 30 and is worth 15% of your grade for COMP90038.

Objectives

To improve your understanding of the time complexity of algorithms and recurrence relations. To develop

problem-solving and design skills. To improve written communication skills; in particular the ability to

present algorithms clearly, precisely and unambiguously.

Problems

Let’s play a game. Or, more precisely, design and analyse some algorithms that might be used in a simple

two-dimensional (2D) game.

Figure 1: A tiny example game scenario. Enemy (AI-controlled) players are shown in blue. The fixed

location of the human player is marked by the red cross.

Consider a game played on an two-dimensional grid, whose Cartesian coordinates range from

(-M; M) : : : (M; M). Figure 1 depicts a game board for which M = 4.

The game contains a fixed number N of enemy players, who are AI-controlled. Each player (including

the human player and each enemy AI) is located at some arbitrary but fixed position (x; y) on the board,

i.e. for which -M ≤ x ≤ M and -M ≤ y ≤ M.

The human player can perform certain attacks, which damage all enemy players within a certain

range of the human player. To avoid the need to calculate with expensive multiplication and square root

operations, we will approximate the range by a square situated about the human player.

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Specifically, given some fixed bound b, for which 0 ≤ b ≤ M, if the human player is located at position

(px; py) then all enemy players that lie within or on the borders of the box whose bottom-left corner is

(px – b; py – b) and whose top-right corner is (px + b; py + b) are affected by the attack. Figure 1 depicts

an example box for the bound b = 1:5.

1. [1 mark] Implement an Θ(1) function to determine whether an enemy at position (x; y) is affected

by the player’s attack, given the player’s location (px; py):

function IsAffected((px; py); (x; y); b)

. . .

2. [2 marks] Suppose (for now) that the positions of the enemy players are stored in an unsorted array

A[0] : : : A[N – 1].

The following function uses the divide and conquer approach to mark all enemy players affected by

a player attack. Marking is implemented by the Mark function whose details are not important.

(Note: the division below is integer division, i.e. it rounds down to the nearest integer.)

function MarkAffected((px; py); A; b)

MarkAffectedDC((px; py); A; 0; N – 1; b)

function MarkAffectedDC((px; py); A; lo; hi; b) if lo = hi then | . assume lo ≤ hi |

if IsAffected((px; py); A[lo]; b) then

Mark(A[lo])

else

mid lo + (hi – lo)=2

MarkAffectedDC((px; py); A; lo; mid; b)

MarkAffectedDC((px; py); A; mid + 1; hi; b)

Explain the purpose of the outermost \if/else” test. In particular, suppose we removed the line

\if lo = hi” and the \else” line. In no more than a paragraph explain how this would affect the

algorithm.

3. [2 marks] Consider the following recurrence relations. Which one best describes the worst-base

complexity of the MarkAffectedDC function, whose input size is n = hi -lo +1 and whose basic

operations are calling IsAffected and Mark? Justify your answer in no more than a paragraph

of text.

T (1) = 1

T (n) = 2T (n=2)

T (1) = 2

T (n) = 2T (n=2) + 2

T (1) = 0

T (n) = 2T (n=2)

T (1) = 2

T (n) = 2T (n=2)

4. [2 marks] Use telescoping (aka back substitution) to derive a closed form for your chosen recurrence

relation and thereby prove that the worst case complexity of the MarkAffectedDC algorithm is

in Θ(n).

5. [2 marks] Can we do better than this? Recall that the human player is at some fixed location (px; py).

Your task is to work out how you would sort the array A so that those enemy AIs that need to be

marked can be identified in log(N) time.

Specifically, complete the following comparison function that would be used while sorting the array A. Here, (x1; y1) and (x2; y2) are two points from the array A. The function should return

true if it considers the first point to be less than or equal to the second, and should return false

otherwise. Your function can use the player’s coordinate (px; py) as global variables, i.e. you are

allowed to refer to px and py in your function.

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function LessOrEqualTo((x1; y1); (x2; y2))

. . .

6. [3 marks] Now, supposing the array has been sorted using your comparison function, implement

an algorithm whose worst case complexity is in Θ(log(N)) that determines which array elements

should be marked. Your function should take the bound b as an argument, and may also take the

player’s coorinates (px; py).

7. [2 marks] How many elements will need to be marked in the worst case and what, therefore, would

be the worst-case complexity of an algorithm that, given an array A sorted according to your

comparison function, implemented the behaviour of MarkAffected above? Express your answer

in Big-Theta notation.

8. [2 marks] The worst case is quite pessimistic. On average, we might expect that enemy AIs are

uniformly distributed throughout the game board.

Let d be the expected number of enemy AIs contained in or on the borders of a box of bound b (i.e.

whose width and height are each 2b) on the board. For simplicity, we limit our attention to boxes

that are contained entirely within the game baord.

Consider an algorithm that, given an array A sorted according to your comparison function, implemented the behaviour of MarkAffected above by first using your Θ(log(N)) function to determine

the elements that need to be marked (of which we expect there to be d), and then marking those d

elements. We might characterise its expected complexity as:

log(N) + d

Derive a formula for d in terms of b, M and N.

9. [2 marks] As a point of comparison, what is the expected complexity of the divide-and-conquer

algorithm above in this average case, expressed in Big-Theta notation? Justify your answer in no

more than a paragraph of text.

10. [2 marks] Now compare the best case complexity of the divide-and-conquer algorithm and the one

that uses a pre-sorted array A described above. Express the best-case complexity of each in BigTheta notation as a function of N. Justify both in no more than a paragraph of text.

11. [4 marks] Suppose now that the enemy AIs can send each other messages. However, two AIs,

whose positions are (x1; y1) and (x2; y2) respectively, can directly communicate only if the line

(x1; y1) | (x2; y2) that connects them does not pass through the rectangle (including its borders)

surrounding the human player whose bottom left corner is (px-b; py -b) and whose top-right corner

is (px + b; py + b), where (px; py) is the human player’s position.

Implement a function that determines whether two enemy AIs can directly communicate. Include

a brief description (no more than a single paragraph) of how your function works, i.e. the rationale

behind its design.

function CanDirectlyCommunicate((x1; y1); (x2; y2); (px; py); b)

. . .

12. [6 marks] Imagine that your CanDirectlyCommunicate function has been used to construct a

graph whose nodes are the enemy AIs, such that there exists an edge between node n1 and n2 if and

only if those two AIs can directly communicate. This graph is stored as an adjacency matrix M.

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Here the nodes ni are simply the indices of enemy AIs in the array A, so each of them is simply an

integer in the range 0 ≤ ni ≤ N -1. Therefore the adjacency matrix M is simply a two-dimensional

array, M[0][0] : : : M[N – 1][N – 1].

Implement an O(N2) algorithm that, given two enemy AIs n1 and n2 determines whether there exists a path by which they might communicate via other enemy AIs. If a path exists, your algorithm

should return the shortest path by which communication is possible. Otherwise, your algorithm

should return the empty path. Your algorithm also takes as its input the adjacency matrix representation of the graph described above.

You should represent a path as a linked list of enemy AI indices ni. The empty path is therefore

the empty linked list without any nodes.

Submission and Evaluation

• You must submit a PDF document via the LMS. Note: handwritten, scanned images, and/or Microsoft Word submissions are not acceptable | if you use Word, create a PDF version for submission.

• Marks are primarily allocated for correctness, but elegance of algorithms and how clearly you communicate your thinking will also be taken into account. Where indicated, the complexity of algorithms also matters.

• We expect your work to be neat { parts of your submission that are difficult to read or decipher will

be deemed incorrect. Make sure that you have enough time towards the end of the assignment to

present your solutions carefully. Time you put in early will usually turn out to be more productive

than a last-minute effort.

• You are reminded that your submission for this assignment is to be your own individual work.

For many students, discussions with friends will form a natural part of the undertaking of the

assignment work. However, it is still an individual task. You should not share your answers (even

draft solutions) with other students. Do not post solutions (or even partial solutions) on social

media or the discussion board. It is University policy that cheating by students in any form is not

permitted, and that work submitted for assessment purposes must be the independent work of the

student concerned.

Please see https://academicintegrity.unimelb.edu.au

If you have any questions, you are welcome to post them on the LMS discussion board so long as

you do not reveal details about your own solutions. You can also email the Head Tutor, Lianglu Pan

([email protected]) or the Lecturer, Toby Murray ([email protected]). In your

message, make sure you include COMP90038 in the subject line. In the body of your message, include a

precise description of the problem.

Late Submission and Extension

Late submission will be possible, but a late submission penalty will apply: a flagfall of 2 marks, and then

1 mark per 12 hours late.

Extensions will only be awarded in extreme/emergency cases, assuming appropriate documentation is

provided simply submitting a medical certificate on the due date will not result in an extension.

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