辅导CO2226留学生、辅导Creative Computing、Java程序调试、讲解Java 讲解SPSS|解析R语言编程

University of London
Computing and Information Systems/Creative Computing
CO2226 Software engineering, algorithm design and analysis
Coursework assignment 2 2018–19
Submission details
What to hand in
Marks will be awarded for correct code (i.e. for code that produces results; if your code does
not produce the correct result, no marks will be awarded for that part of the question).
You must submit one Java file called Ass226.java. For example, if your student
number ID is 101031722, your file will be named: Ass226101031722.java
When this file is compiled it must produce a single file Ass226.class e.g.
Ass226101031722.class. When run, this must produce the answers to all the coursework
assignment questions by implementing the appropriate algorithms.
Your java file may contain other classes, but they must all be included in the single java file;
please do not submit multiple Java files as you will get a mark of zero – we only need one
file. You must write your code assuming all data files are in the same directory as the program.
Failure to do this will lead to a mark of zero. If you cannot complete a particular question the
answer should say ‘NOT DONE’. Your program should take the text files described below as
command-line arguments.
To run your program, the examiners will type (pay attention that the filenames are referenced
without the file extension):
java Ass226101031722 cities cities_lon_lat randomGraph
Your output should look like this:
Name: Joe Doe
Student ID: 101031722
Question 1: 2
Question 2: 200 205
Question 3: 4
Question 4: 15
Question 5: [350, 352]
Question 6: 5.12
Question 7: 29.47
Execution Time: 32094 milliseconds
These are just sample answers to give you an idea of what output format is expected from your
program. The examiners will change the data files to test your programs so make sure your
program works with files containing fewer/more cities. Try deleting some lines from the files and
see if your program gives different answers.2
Efficiency
You will be penalised if your program runs too slowly (5 marks for every minute over 5 minutes
on a machine with Intel Core i7 processor with12 gigabytes of RAM).
Try to speed up your program, by not recomputing values that you have already computed.
Instead, store them (rather than recomputing).
Use System.nanoTime(); to time your program. (Read the value at the beginning and end
of your program and subtract and divide by a million.)
IF YOU DO NOT INCLUDE THE EXECUTION TIME OF YOUR PROGRAM YOU WILL
SCORE ZERO.
IF YOU DO NOT USE THE DATA PROVIDED YOU WILL SCORE ZERO.
ALL SOLUTIONS SHOULD INVOLVE CALLS TO YOUR GRAPH INSTANCE METHODS;
DO NOT TRY TO CHEAT BY FINDING ANSWERS ELSEWHERE.
CO2226 Coursework assignment 2 – preparation and pre-assignment tasks
Finding the shortest paths in unweighted graphs (breadth-first search)
Find out about Adjacency matrices for representing Graphs. Here is a program to help you
become familiar with them:
import java.util.HashSet;
import java.util.ArrayList;
public class graph
{
double [] [] adj;
graph (double [] [] a)
{
adj= new double [a.length][a.length];
for (int i=0;ifor (int j=0;jadj[i][j]=a[i][j];
}
public HashSet neighbours(int v)
{
HashSet h = new HashSet ();
for (int i=0;ireturn h;
}
public HashSet vertices()
{
HashSet h = new HashSet ();
for (int i=0;ireturn h;
}3
ArrayList addToEnd (int i, ArrayList path)
// returns a new path with i at the end of path
{
ArrayList k;
k=(ArrayList)path.clone();
k.add(i);
return k;
}
public HashSet > shortestPaths1(HashSet
> sofar, HashSet visited, int end)
{
HashSet > more = new HashSet >();
HashSet > result = new HashSet >();
HashSet newVisited = (HashSet )
visited.clone();
boolean done = false;
boolean carryon = false;
for (ArrayList p: sofar)
{
for (Integer z: neighbours(p.get(p.size()-1)))
{
if (!visited.contains(z))
{
carryon=true;
newVisited.add(z);
if (z==end) {
done=true;
result.add(addToEnd(z,p));
}
else
more.add(addToEnd(z,p));
}
}
}
if (done) return result; else
if (carryon)
return shortestPaths1(more,newVisited,end);
else
return new HashSet >();
}
public HashSet > shortestPaths( int first,
int end)
{
HashSet > sofar = new HashSet
>();
HashSet visited = new 4
HashSet();
ArrayList starting = new ArrayList();
starting.add(first);
sofar.add(starting);
if (first==end)
return sofar;
visited.add(first);
return shortestPaths1(sofar,visited,end);
}
public static void main(String [] args)
{
double [ ] [] a = {
{0.0, 1.0, 1.0, 0.0},
{0.0, 0.0, 1.0, 1.0},
{0.0, 1.0, 0.0, 1.0},
{0.0, 1.0, 1.0, 0.0}
};
graph g = new graph(a);
for (int i=0;i{for (int j=0;jif (i!=j) System.out.println(i + " to " + j +": "+
g.shortestPaths(i,j));
}
}
}
Draw a picture of the graph and see if you agree with the output. Play with the program and
alter the graph in order to check that you understand how the program works.
The cities distance problem
Study the following files of data about main cities:
cities.csv. This file has two fields in the following order: id code for the country and
the name of the city in English, and it is used for describing cities.
randomGraph.csv. This file contains three fields, the id code of the source city, the id
code of the destination city and the cost for taking this route.
cities_lon_lat.csv. This file has three fields in the following order: the id code for the
city, the city’s longitude and the city’s latitude.
Examine the following program:
import java.io.*;
import java.util.Scanner;
import java.util.*;
class Assignment2
{
static int N= 500;
static double [][] edges = new double[N][N];
static TreeMap cityNames = new TreeMap
();5
static ArrayList convert(ArrayList m)
{
ArrayList z= new ArrayList();
for (Integer i:m)
z.add(cityNames.get(i));
return z;
}
static HashSet> convert
(HashSet> paths)
{
HashSet > k= new HashSet
>();
for (ArrayList p:paths)
k.add(convert(p));
return k;
}
public static void main(String[] args) throws Exception
{
for(int i=0;ifor(int j=0;jedges[i][j]=0.0;
Scanner s = new Scanner(new FileReader("randomGraph"));
String z =s.nextLine();
while (s.hasNext())
{
z =s.nextLine();
String[] results = z.split(",");
edges[Integer.parseInt(results[0])]
[Integer.parseInt(results[1])]=1.0;
edges[Integer.parseInt(results[1])]
[Integer.parseInt(results[0])]=1.0;
}
s = new Scanner(new FileReader("cities"));
z =s.nextLine();
while (s.hasNext())
{
z =s.nextLine();
String[] results = z.split(",");
cityNames.put(Integer.parseInt(results[0]),
results[1]);
}
graph G= new graph(edges);
int st =Integer.parseInt(args[0]);
int fin = Integer.parseInt(args[1]);
System.out.println("Shortest path from " + =
cityNames.get(st) + " to " +
cityNames.get(fin) + " is" +
convert(G.shortestPaths(st,fin)));
}
}6
Dijkstra’s algorithm (Finding the shortest path in a weighted graph)
Watch Dijkstra’s Algorithm (YouTube video) and Dijkstra's Algorithm again.
Study Dijkstra's algorithm MIT Lecture 17 Video.
Study the pseudo code below for Dijkstra's Algorithm to find a shortest path from
start to end:
Set S = {start};
//S is the set of vertices for which the shortest paths from
start have already been found
HashMap Q = Map each Vertex to Infinity
(Double.POSITIVE_INFINITY), except map start -> 0;
// Q.get(i) represents the shortest distance found from start
to i found so far
ArrayList [] paths;
For each i
set path[i] to be the path just containing start.
while (Q is not empty)
{
let v be the key of Q with the smallest value;
//I've given you a method int findSmallest(HashMap
t) for this
if (v is end and Q does not map v to infinity)
return paths[end]; let w be the value of v in Q;
add v to S;
for (each neighbour u of v) do
{
if (u not in S)
{
let w1 be the the weight of the (v,u) edge + w;
if w1 < the value of u in Q, then do the following:
{
update Q so now the value of u is w1
update paths(u) to be paths(v) with u stuck on the
end
}
}
remove v from Q;
}
}7
Task 1
Implement Dijkstra’s Algorithm using the pseudo-code above; namely, put a function
dijkstra into the graph class.

Here are some hints:
int findSmallest(HashMap t)
{
Object [] things= t.keySet().toArray();
double val=t.get(things[0]);
int least=(int) things[0];
Set k = t.keySet();
for (Integer i : k)
{
if (t.get(i) < val)
{
least=i;
val=t.get(i);
}
}
return least;
}
Now, fill in the following bits:
public ArrayList dijkstra (int start, int end)
{
int N= ...;
HashMap Q = new HashMap
();
ArrayList [] paths = new ArrayList [N];
for (int i=0;i{
Q.put(i,...);
paths[i]=new ArrayList ();
paths[i].....;
}
HashSet S= new HashSet();
S.add(...);
Q.put(start,....);
while (!Q.isEmpty())
{
int v = findSmallest(...);
if (v==end && ...) return ....;
double w = Q.get(...);
S.add(...);
for(int u: neighbours(v))
if (...)
{
double w1= ....;
if (w1 < Q.get(u))
{8
Q.put(u,...);
paths[u]= addToEnd(...);
}
}
Q.remove(...);
}
return new ArrayList ();
}
Task 2
Test your implementation using the following test program:
class testDijk
{
public static void main(String [] args) throws Exception
{
int N=1000;
double edges[][]=new double[N][N];
for(int i=0;ifor(int j=0;jedges[i][j]=0.0;
Scanner s = new Scanner(new FileReader("randomGraph"));
String z;
while (s.hasNext())
{
z =s.nextLine();
String[] results = z.split(",");
edges[Integer.parseInt(results[0])]
[Integer.parseInt(results[1])]=Double.parseDouble(
results[2]);
}
graph G= new graph (edges);
System.out.println(G.dijkstra(Integer.parseInt(args[0]),
Integer.parseInt(args[1])));
}
}
Use this randomGraph file (please note that this example is from a different scenario which
refers to tube stations so you might need to make some adjustments when reading in the
data).
Each line of the file has three values: the first two are vertices and the thirds is the weight
of the edge between them.
When you run:
java testDijk 0 999
You should get:
[0, 492, 665, 114, 452, 999]9
CO2226 Coursework assignment 2 – main questions
Please note that cities in the questions are referred to by their name. As part of the
assignment, you will need to resolve them into their ISO codes (e.g. 300 for Athens). In
the case of a direct link, please ignore this option and look for a route that includes
at least one extra link.
1. How many shortest paths exist between the cities of Athens and Tehran? A
shortest path here means a path with a minimal number of vertices. (Note: Use
the shortestPaths method above.)
[15 marks]
2. Which pair of cities have the highest number of shortest paths between them?
Just give the city IDs.
[10 marks]
3. How many shortest paths do they have?
[10 marks]
4. How long are each of these shortest paths?
Hint: You may wish to use the following method.
static ArrayList firstElement (HashSet
> s)
{
return ( ArrayList)s.toArray()[0];
}
[10 marks]
5. Which set of cities are furthest away from the city of Toronto in terms of number
of stops? (Just print out the set of numbers corresponding to the cities).
[15 marks]
6. What is the length in terms of sum of the weights of the edges of the shortest path
(in terms of the sum of the weights of the edges) between the cities of Rome and
New York? (use Dijkstra's Algorithm).
[20 marks]
7. What is the length (in km) of the shortest path (in terms of distance) between the
cities of Lisbon and Manila? (Note: Use Dijkstra's Algorithm).
[20 marks]10
You will need to use the following method (and the relevant data from the cities_lon_lat
file).
static double realDistance(double lat1, double lon1,
double lat2, double lon2)
{
int R = 6371;
// km (change this constant to get miles)
double dLat = (lat2-lat1) * Math.PI / 180;
double dLon = (lon2-lon1) * Math.PI / 180;
double a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(lat1 * Math.PI / 180 ) * Math.cos(lat2 *
Math.PI / 180 )* Math.sin(dLon/2) * Math.sin(dLon/2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-
a)); double d = R * c;
return d;
}
For finding the distance in km between any two points on the Earth's surface with given
latitude and longitude: the latitude and longitude of each city is given in the cities_lon_lat
file. You will have to use this to compute the adjacency matrix for the weighted graph
representation of the cities problem. We need the ad[i][j] to be the distance from city i to
city j now.
You will also need to write a method for finding the length of path by adding up all the
weights of the edges in the path.
[Total: 100 marks]
[END OF COURSEWORK ASSIGNMENT 2]