Assignment 4 Graphs and Connect 4

Assignment 4
COMP 250 
1 General Instructions (read carefully)
• You are provided some starter code that you should fill in as requested.
Add your code only where you are instructed to do so. You
can add some helper methods. Do not modify the code in any other
way and in particular : do not change the methods or constructors that
are already given to you, do not import extra code and do not touch
the method headers. Any failure to comply with these rules will
give you an automatic 0.
• Submission instructions
 Finally, submit a .zip file containing only
the files Configuration.java and Game.java to the other folder
on myCourses.
– Submitting the wrong format of file or submitting in the
wrong folder, or wrong file name (e.g. Graph (1).java)
on myCourses will be an automatic 0 to the corresponding
exercise.
– Around one day before due date : we will run whatever you had
submitted last and give you some feedback on it, mostly to warn
1
you if you currently have a grade of zero, so you can fix it if there
is a problem. Take advantage of this by submitting early !
– Morning after due date : we will run your last submission and
warn you again if your code does not compile.
– After this, you can resubmit at any moment for a period of three
days after the initial due date, with a penalty of 20% per day.
– The last acceptable file you submit will be used to determine your
final grade for this assignment.
• The starter code includes a tester class. If your code fails those tests,
it means that there is a mistake somewhere. Even if your code
passes those tests, it may still contain some errors. We will test
your code on a more challenging set of examples. We therefore highly
encourage you to modify that tester class, expand it and share it with
other students on the myCourses discussion board. Do not include it
in your submission.
• You will automatically get 0 if your code does not compile.
• Failure to comply with any of those rules will be penalized. If
anything is unclear, it is up to you to clarify it by asking either directly
a TA during office hours, or on the discussion board on myCourses.
2
2 Introduction
There are two exercises in this assignment. Be very careful about what files
you submit on myCourses and where you upload them.
You will be less constrained for this assignment than for the previous
ones. For many of the questions, there are more than one solution possible.
You can write as many auxiliary methods as you want. However, you are
not allowed to import any additional code (and you don’t need it anyways).
The testers you are provided with are very partial. We have said it for
every assignment, but let us repeat it for this one too : you need to expand
this tester.
Each question has a weight associated to it. This is only to serve as an
indicator and may be later changed without notice.
3 Exercise 1 : Graphs (28 points)
3.1 Structure of the code provided to you
A Graph has a number of nodes nbNodes and is characterized by its adjacency
matrix adjacency. adjacency[i][j] states whether or not there is an edge
from the i-th node to the j-th node.
The Graphs you will be dealing with are undirected, i.e.
adjacency[i][j] == adjacency[j][i] is always true. They may contain
self-loops and they may be non-connected.
3.2 Questions
Question 1. (2 points each) Code addEdge and removeEdge. It takes as
input a pair of nodes and adds / removes the edge between the i-th and the
j-th nodes.
Question 2. (4 points) Code nbEdges. g.nbEdges() returns the number
of edges of g.
Question 3. (10 points) Code cycle. It takes as input an integer, g.cycle(i)
returns true if the i-th node is part of a cycle (and false otherwise).
Question 4. (10 points) Code shortestPath. It takes as input a pair
of integers, g.shortestPath(i,j) returns the length of the shortest path
joining the i-th node to the j-h node. If there is no such path, it returns
g.nbNodes +1.
3
4 Exercise 2 : Connect 4 (72 points)
I hope you have learnt things that are of interest to you during this class and
in particular throughout the assignments. We are going to finish this final
assignment by a small game that I used to play a lot with my sister when we
were younger : Connect 4 (Puissance 4 pour les francophones).
For those of you who don’t know about this game, I refer you to the
wikipedia page for more details but let me sum up how the game works :
you have a vertical grid that is 7-cell-long and 6-cell-high. Each player has
some disks of a specific color (yellow for one player, red for the other one).
At each round, the player whose turn it is picks a column and puts a disk in
that column. The disk falls to the lowest available cell. After one player has
played, it is the other player’s turn.
The game can end in two cases. First, if one player has won, i.e. he
is the first one to have four disks of his colour aligned (either horizontally,
vertically or diagonally). Second, if there is no space left in the grid, in which
case there is no winner.
4.1 Structure of the code that is provided to you
You have two classes provided to you : Configuration and Game.
The class Configuration corresponds to the grid. It has three fields.
First board is the grid per say. It is initiated in the constructor to an array
int[7][6] (containing only 0’s by default). The field available is an array
int[7] such that available[i] is the first available row in the i-th column.
If the i-th column is already full, its value is 6. Finally the field spaceLeft
is true if you still have some place somewhere on the grid to put a disk.
The class Configuration already has a method print() coded for you.
It prints the grid, the disks that are contained in it and the number of the
columns. For instance, it could print :
| 0 | 1 | 2 | 3 | 4 | 5 | 6 |
+---+---+---+---+---+---+---+
| | | | | | | |
| | | | | | | |
| | | | | 1 | | |
| | | | 1 | 2 | | |
| | | 1 | 2 | 1 | | |
| 1 | 2 | 2 | 2 | 1 | 1 | 2 |
The row at the bottom is the 0-th row. So for instance the disk of the 2nd
player in the 6th column corresponds to board[6][0] and the disks of the
4
1st player that are in column 4 correspond to board[4][0], board[4][1]
and board[4][3].
The class Game corresponds to the game per say. It does not have any
fields. You will have several methods to code in it. However, you are already
provided with the method play that you can use at the end.
At the end of the tester, there are two lines that are commented. Once
you have finished coding every methods, you can uncomment these two lines,
these will allow you to play against the computer. It is a very good way of
testing further your code.
4.2 Questions in Configuration
Question 5. (8 points) First code addDisk in Configuration. As the
name suggests, c.addDisk(5,1) adds a disk corresponding to player 1 in
the column 5 and at the first row available. addDisk takes as input the
number of the column in which to add a disk and the number of the player
whose disk is to be added. Here you can assume that the column has space
left.
For instance, the grid showed in previous subsection would be updated
to
| 0 | 1 | 2 | 3 | 4 | 5 | 6 |
+---+---+---+---+---+---+---+
| | | | | | | |
| | | | | | | |
| | | | | 1 | | |
| | | | 1 | 2 | | |
| | | 1 | 2 | 1 | 1 | |
| 1 | 2 | 2 | 2 | 1 | 1 | 2 |
Don’t forget to update the other fields.
Question 6. (16 points) You can now code isWinning. It takes as argument the last column that was played and the player who played last (so you
know that the disk that is last in that column belongs to said player) and it
returns true if, and only if, the player managed to make a line of four disk
in the last round. The line can be horizontal, vertical or diagonal.
For example, consider a configuration c with the following board :
5
| 0 | 1 | 2 | 3 | 4 | 5 | 6 |
+---+---+---+---+---+---+---+
| | | | | | | |
| | | | | | | |
| | | | | 1 | 2 | |
| | | | 1 | 2 | 1 | |
| | | 1 | 2 | 1 | 1 | |
| 1 | 2 | 2 | 2 | 1 | 1 | 2 |
c.isWinning(5,2) should return true, but c.isWinning(3,1) should return false.
Question 7. (8 points) Code canWinNextRound. It takes as argument the
player p whose turn it is and returns i such that if player p places its disk
in column i, he wins the game. If there are no such column, i is equal to -1.
If there are two such columns, it returns the smallest one.
For example, consider a configuration c with the following board :
| 0 | 1 | 2 | 3 | 4 | 5 | 6 |
+---+---+---+---+---+---+---+
| | | | | | | |
| | | | | | | |
| | | | | 1 | | |
| | | | 1 | 2 | 2 | |
| | | 1 | 2 | 1 | 1 | |
| 1 | 2 | 2 | 2 | 1 | 1 | 2 |
c.canWinNextRound(2) should return 5 but c.canWinNextRound(1) should
return -1.
Question 8. (16 points) Code canWinTwoTurns. It takes as argument the
player p whose turn it is and returns i such that if player p places its disk
in column i, whatever the other player does on next round, player p can win
when it’s his turn again. Note that this requires that the other player does
not win in between.
You can assume here that player p has no winning move at this turn.
For instance, consider a configuration c with the following board :
6
| 0 | 1 | 2 | 3 | 4 | 5 | 6 |
+---+---+---+---+---+---+---+
| | | | | | | |
| | | | | | | |
| 2 | | 2 | 2 | | | |
| 1 | | 1 | 2 | | | |
| 1 | | 2 | 1 | 1 | | |
| 1 | 1 | 2 | 2 | 1 | | |
c.canWinTwoTurns(1) should return -1 because player 1 has no way of making sure he is going to win in two rounds. c.canWinTwoTurns(2) should
return 1 because if player 2 plays column 1, then there are two cases : either
player 1 plays also in column 1 in which case by playing (again) in column 1,
player 2 will complete a horizontal line, or player 1 does not play in column
1, in which case player 2 can play again in column 1 and complete a diagonal
line.
4.3 Questions in Game
Question 9. (10 points) Code getNextMove. This method asks in which
column the player wants to add his disk. It takes as argument a BufferedReader
(where the player is writing his answer), a Configuration (the grid on which
the player is playing) and an integer (the number of the player whose turn
it is). You are allowed to print whatever you want on the screen (the grid,
a warning if the other player has a winning move that should be prevented
etc). But what is important is that if the player asks for a column with no
space left, you should keep asking him until he gives a valid one.
Consider a configuration c with the following board :
| 0 | 1 | 2 | 3 | 4 | 5 | 6 |
+---+---+---+---+---+---+---+
| | | | 1 | 2 | | |
| | | 1 | 2 | 1 | 1 | |
| | | 2 | 2 | 1 | 1 | 2 |
| | | 1 | 1 | 2 | 2 | 1 |
| | | 2 | 2 | 1 | 1 | 2 |
| | | 1 | 2 | 2 | 1 | 2 |
Consider a BufferedReader keyboard, then getNextMove(keyboard, c, 2)
should ask player 2 for his move. If player 2 requests column 3, you should
ask again. If player 2 then requests column 4, ask again. If player 2 then
7
requests column 3 (again), ask (again). If then player 2 requests column 2,
return 2.
To test it, you should uncomment the corresponding lines in the tester.
Question 10. (14 points) Code movePlayer1. Player 1 is played by the
computer. It takes as input an integer which is the column that the other
player played last and a Configuration which is the current state of the
grid. Here is the strategy that the player 1 is following :
1. if player 1 has a winning move, he plays that winning move.
2. if player 1 has a move that can make him win in two rounds, he plays
that move.
3. if none of these are possible, player 1 tries to put his disk right above
the one player 2 played last (call it last). If there is no more space
left in column last, he tries the column on the left (last -1), then on
the right (last +1). If that is still not possible, he plays the columns
last -2 then last +2 etc. Player 1 skips every step that would correspond to a non-existing column. For instance, if last = 5, he would
try the columns in the following order : 5, 4, 6, 3, 2, 1, 0.
You have just coded an AI ! It is not a very clever AI, but it is one
nonetheless. You can now play against this AI \o/
8