I attempted this problem and for some reason its not coming out correct. Given an array of strings, find how many possible solutions exist to the maze where the strings consist of one "R" (the rat), one "C" (the cheese), multiple "X's (blocks that cannot be passed through), and "."'s (possible pathways). The task is to find the number of possible routes the rat can take to get to the cheese without ever increasing the (Euclidean) distance between itself and the cheese at any point on its path. What looks wrong with my code?
public class RatRoute {
private static String[] enc;
private static int count;
private static int[] r;
private static int[] c;
// Test the program
public static void main(String[] args) {
String[] test = {
".R...",
"..X..",
"....X",
"X.X.X",
"...C."};
int num1 = numRoutes(test);
System.out.println(num1);
}
// Set variables, and call recursive function
public static int numRoutes(String[] enc) {
RatRoute.enc = enc;
r = findR(enc);
c = findC(enc);
recursiveHelper(r[0], r[1]);
return count;
}
// Recursive
public static void recursiveHelper(int x, int y) {
/*System.out.println();
System.out.println();
for (int k = 0; k < enc.length; k++) {
System.out.println(enc[k]);
}*/
if(isBlock(x,y)) {
return;
} else if (isBigger(x,y)) {
return;
} else if (isCheese(x, y)) {
count++;
//System.out.println("Found the Cheese! Path number: " + count);
//System.out.println();
return;
}
enc[x] = currentPath(x,y);
recursiveHelper(x + 1, y);
recursiveHelper(x, y + 1);
recursiveHelper(x, y - 1);
recursiveHelper(x - 1, y);
enc[x] = returnPath(x,y);
}
// Change the most recently traveled coordinates into a block
public static String currentPath(int x, int y) {
char[] Chars = enc[x].toCharArray();
Chars[y] = 'X';
String newString = String.valueOf(Chars);
return newString;
}
// Turn path already traveled from blocks back into a usable path to travel (undo the currentPath method)
public static String returnPath(int x, int y) {
char[] Chars = enc[x].toCharArray();
Chars[y] = '.';
String newString = String.valueOf(Chars);
return newString;
}
// Check if the next movement is into the cheese
public static boolean isCheese(int x, int y) {
if (enc[x].charAt(y) == 'C') {
return true;
} else {
return false;
}
}
// Check if the next movement is into a block, or outside the given array
public static boolean isBlock(int x, int y) {
if (x == -1 || y == -1
|| x >= enc.length || y >= enc[x].length()) {
return true;
} else if (enc[x].charAt(y) == 'X') {
//System.out.println(x + "," + y);
return true;
} else {
return false;
}
}
// See if the distance between the rat and the cheese has gotten larger or smaller
public static boolean isBigger(int x, int y) {
double rx = r[0]; double ry = r[1];
double cx = c[0]; double cy = c[1];
double originalDist = Math.sqrt(Math.pow(rx-cx, 2) + Math.pow(ry-cy, 2));
double newDist = Math.sqrt(Math.pow(x-cx, 2) + Math.pow(y-cy, 2));
//System.out.println("Orginal Distance: " + originalDist);
//System.out.println("New Distance: " + newDist);
if (newDist > originalDist) {
return true;
} else {
return false;
}
}
// Find the variables for the original position of the rat
public static int[] findR(String[] enc) {
for (int i = 0; i < enc.length; i++) {
for (int j = 0; j < enc[i].length(); j++) {
if (enc[i].charAt(j) == 'R') {
int[] coordinates = {i, j};
//System.out.println(coordinates[0] + "," + coordinates[1]);
return coordinates;
} else {
}
}
}
int[] other = {-1, -1};
return other;
}
// Find the variables for the original position of the rat
public static int[] findC(String[] enc) {
for (int i = 0; i < enc.length; i++) {
for (int j = 0; j < enc[i].length(); j++) {
if (enc[i].charAt(j) == 'C') {
int[] coordinates = {i, j};
//System.out.println(coordinates[0] + "," + coordinates[1]);
return coordinates;
} else {
}
}
}
int[] other = {-1, -1};
return other;
}
}
Let's start from an useful observation:
[...] without ever increasing the (Euclidean) distance between itself and
the cheese at any point on its path.
Basically what it means is that whenever the rat gets closer to the cheese it never goes back to a farther position.
So let's say the rat x coordinate is 3 and the cheese x coordinate is 5 the rat cannot "go left" (i.e. x = 2) because this would cause it to be farther than before form the cheese.
Because of this, a nice way to simple the problem is to find the directions the rat can go. In your example the rat is up-left from the cheese, so it will only be able to move down or right because otherwise it would get farther from the cheese. When the rat x matches the cheese x it won't be able to move right or left anymore, same goes for the y.
With your code, if:
r[0] - c[0] = 0 // the rat cannot move on the x any more.
r[1] - c[1] = 0 // the rat cannot move on the y any more.
When
r[0] - c[0] == 0 && r[1] - c[1] == 0
The rat reached the cheese! In this case, we can increment the counter because a successful route was found.
Now let's put this in practice with recursion.
From the code you posted, we start with a given c (found with findC(enc)) and a given r (found with findR(enc))
So the recursive method would look like this:
private void findRoutesFrom(int[] r) {
// what directions can the rat go?
// if the cheese is on the right of the mouse, xDirection
// would be +1.
int xDirection = (int) Math.signum(c[0] - r[0]);
// if the cheese is below of the mouse, yDirection
// would be +1.
int yDirection = (int) Math.signum(c[1] - r[1]);
// Now, if xDirection is != 0 the rat can attempt to move
// in that direction, checking if there's not a block
if(xDirection != 0 && !isBlock(r[0] + xDirection, r[1])) {
// if it can move in that direction, then use recursion to
// find all the possible paths form the new position
findRoutesFrom(new int[]{r[0] + xDirection, r[1]});
}
// same goes for yDirection
if(yDirection != 0 && !isBlock(r[0], r[1] + yDirection)) {
findRoutesFrom(new int[]{r[0], r[1] + yDirection});
}
// if the rat reaches the cheese, increase the successful
// paths counter
if(xDirection == 0 && yDirection == 0) {
count++;
}
}
That's it! Because of the Eculedean distance constraint it is enough to check if the directions are != 0 because, when that is false, then the rat cannot move on that direction any more.
Related
I am coding a simple tic-tac-toe for a high-school mini project, but I need it to be within a strict data volume (not more than 112 lines). I thought checking for each row, column and cross would be long, so is there any alternative to do so (You should see a [[[HERE]]] comment)? (Btw, I already know it looks awful) Thanks in advance!
public class TTTGame {
//OPTIONS v
public static final String draw = "DRAW"; // <- Definitions for different states
public static final String circles = "CIRCLES"; // BOT
public static final String crosses = "CROSSES"; // PLAYER
public static final String getCrosses = "X"; //<- Symbols to display
public static final String getCircles = "O";
//OPTIONS ^
//DO NOT MODIFY UNDER THIS LINE (Just kidding, do whatever u want) v
public static int[][] board = {
{0,0,0},
{0,0,0},
{0,0,0},
};
public static final int empty = 0; // Definition of the values
public static final int cross = 1;
public static final int circle = 2;
public static int turns = 0; //Just here to count turns, nothing special
public static void main(String[]args) { //Main process
board[1][1] = circle;
display();
while (true) {
PlayerTurn();
if (checkStop()||checkWinner()!=null) {display();GStop();break;}
BotTurn();
if (checkStop()||checkWinner()!=null) {display();GStop();break;}
display();
turns += 1;
}
}
private static void GStop() { //Force stop the match function
System.out.println("Winner : " + checkWinner());
System.exit(1);
}
private static boolean checkStop() { //Check if match is already full / completed (Draw)
for (int x = 0; x < 3; x++)
for (int y = 0; y < 3; y++)
if (board[x][y]==empty) return false;
return true;
}
#Nullable
private static String checkWinner() { //Check Winner
// [[[ HERE ]]] ---------------
return null;
}
private static void PlayerTurn() { //Player turn
int x; Scanner c = new Scanner(System.in);
while (true) {
x = c.nextInt();
x = x-1;
if ((x>=0)&&(x < 9)) {
if (board[x / 3][x % 3] == empty) {
board[x / 3][x % 3] = cross;
break;
} else System.out.println("Already chosen");
} else System.out.println("Invalid");
}
}
private static void BotTurn() { //Bot turn -> (Modify these to change the AI behaviour, here's a very simple one);
boolean choose = true;
for (int y = 0; y < 3 ; y++)
for (int x = 0; x < 3; x++)
if (board[y][x] == empty&&choose) {
board[y][x] = circle;
choose = false;
}
}
private static void display() { //Display the board
int nn = 1;
String a = "z";
for (int y = 0; y < 3 ; y++) {
for (int x = 0; x < 3; x++) {
if (board[y][x] == 0) a = "*";
if (board[y][x] == 1) a = getCrosses;
if (board[y][x] == 2) a = getCircles;
System.out.print(a + " ");
}
System.out.print(" "); //Indications
for (int xn = 0; xn < 3; xn++) {
System.out.print(nn);
nn+=1;
System.out.print(" ");
}
System.out.println(" ");
}
}
}
How about this idea: (neither the only nor the best nor the most performant solution... just an idea)
You can use the sum of each row, diagonal and column to determine if the either player one (all 1s) or player two (all 2s) wins. Therefore you only need to set the empty field to be higher than 6.
For example let's say your board looks like this:
7 1 1 -> 7+1+1 = 9 // no one wins
2 2 2 -> 2+2+2 = 6 // player two wins, he has 3 * 2 in a row
1 7 2 -> 1+7+2 =10 // no win here
if all three numbers where 1s (sum == 3) your player one wins.
It is "cumbersome" to implement, but as I said it is just an idea:
// first we check every column
for( int x=0; x<board[y].length; x++){
int sum = 0;
for( int y=0; y<board.length; y++){
sum += board[y][x];
}
if(sum == 3 || sum == 6){
return true;
}
}
// then every row
for( int y=0; y<board.length; y++){
int sum = 0;
for( int x=0; x<board[y].length; x++){
sum += board[y][x];
}
if(sum == 3 || sum == 6){
return true;
}
}
// and finally the diagonals (if we ever reach that part)
int sum= board[0][0] + board[1][1] + board[2][2];
if(sum == 3 || sum == 6){
return true;
}
sum= board[0][2] + board[1][1] + board[2][0];
if(sum == 3 || sum == 6){
return true;
}
you could also return 1 when the sum == 3 and the first player wins or 2 when player two wins.
so im having problems polishing up my program. this program is supposed to create a 1D array with a user input. then it creates a box of 'O's like this..
N = 4
OOOO
OOOO
OOOO
OOOO
the user inputs coordinates based on the box and the 'O' is changed to an 'X'.
the program is supposed to repeat itself after the coordinates are selected while remembering the position of X and including it in the next loop.
i tried implementing a while loop but it seems that code just loops over the Array without remembering the last position of X.
how could i change the code so it does what i need it to do?
public static void makeArray(int M) {
String input = "";
boolean repeat = false;
int N = InputNumber(input);
String[] Board = new String[N];
M = (int) Math.sqrt(N);
String A = "O";
String B = "X";
System.out.println("Printing " + (M) + " x " + (M) + " board...");
System.out.println("Done.");
System.out.println();
while (!repeat) {
int X = Xvalue(M);
int Y = Yvalue(M);
int C = convertIndex(X, Y, M);
System.out.println("Marking location " + X + "," + Y + ")");
for (int i = 0; i < (Board.length); i++) {
{
Board[i] = A;
if ((i % M == 0)) {
System.out.println();
}
if (i == C) {
Board[i] = Board[i].replace(A, B);
}
if (i == C && C == -1) {
repeat = true;
}
}
System.out.print(Board[i]);
}
System.out.println();
}
}
public static int convertIndex(int x, int y, int N) {
int valX = (x - 1) * N;
int valY = y;
int targetIndex = valX + valY;
return (targetIndex - 1);
}
public static int Xvalue(int M) {
boolean repeat = false;
int X = 0;
while (!repeat) {
System.out.print("Please enter the X-coordinate: ");
String InputX = new Scanner(System.in).nextLine();
X = Integer.parseInt(InputX);
if (X > M) {
System.out.println();
System.out.println("Error, please enter a valid X Coordinate...");
repeat = false;
} else {
repeat = true;
}
}
return X;
}
public static int Yvalue(int M) {
boolean repeat = false;
int Y = 0;
while (!repeat) {
System.out.println();
System.out.print("Please enter the Y-coordinate: ");
String InputY = new Scanner(System.in).nextLine();
Y = Integer.parseInt(InputY);
if (Y > M) {
System.out.println("Error, please enter a valid Y Coordinate...");
repeat = false;
} else {
repeat = true;
}
}
return Y;
}
The trouble with you loop is that it defines every element in you your array before it prints them:
while (!repeat) {
//...
for (int i = 0; i < (Board.length); i++) {
{
Board[i] = A; //Makes each element "O"
//...
if (i == C) { //Makes only the current cooridinate "X"
Board[i] = Board[i].replace(A, B);
}
//...
}
System.out.print(Board[i]);
}
}
To fix it so that old X's are retained, you need to remove assignment Board[i] = A;. But you'll still need to initialize your board, or else you'll have null strings. So you need to add something before the loop like:
String[] Board = new String[N];
M = (int) Math.sqrt(N);
String A = "O";
String B = "X";
//initialize board
for (int i = 0; i < Board.length; i++)
Board[i] = A;
Try using a char[][] instead of a String[]. Then you can just plugin the coordinates the user inputs (e.g. board[x][y] = B). This better represents what you're showing the user as well.
This saves you from having to loop through your String[] and then finding the right character to change. Remember, Strings are immutable, so you'd have to reassign the entire string after replacing the right character. With the char[][] you simply assign 'X' to the right coordinates.
EDIT:
Since a single array is required, you should be able to do the following (instead of looping):
board[x] = board[x].substring(0, y) + A + board[x].substring(y + 1);
This question already has answers here:
How do I do a deep copy of a 2d array in Java?
(7 answers)
Closed 8 years ago.
So I am trying to make an algorithm for finding full paths in NxN grid. For example in 1x1 grid there is 1 possible path, in 2x2 grid there is 1, in 3x3 there is 2 and in 4x4 there is 8. The idea is to find scenarios where we can fill every spot of the grid by moving.
I have made a recursive function for the job and here is the code:
public static int getRoutesHelp(int[][] table, int x, int y)
{
if(x > table.length-1 || x < 0 || y < 0 || y > table.length-1)
return 0;
if(table[x][y] == 1)
return 0;
table[x][y] = 1;
if(isDeadEnd(table, x, y))
{
if(isTableFull(table))
return 1;
}
else
{
int a = getRoutesHelp(table, x-1, y);
int d = getRoutesHelp(table, x, y+1);
int b = getRoutesHelp(table, x+1, y);
int c = getRoutesHelp(table, x, y-1);
return a+b+c+d;
}
return 0;
}
public static int getRoutes(int size)
{
int[][] table = new int[size][size];
// init table
for(int i = 0; i < size; i++)
{
for(int a = 0; a < size; a++)
{
table[i][a] = 0;
}
}
return getRoutesHelp(table, 0 ,0);
}
So basically I start from 0.0 and start moving to all possible directions and by repeating this I get the amount of successful routes. The problem is that after the assignment of int d the original table is somehow filled with 1 but it should be empty as far as I understand because java passes a copy of the table right? I've been fighting with this for like 4 hours and can't really find the problem so any help is appreciated. Empty slots in table are marked with 0 and filled slots with 1.
EDIT: I managed to fix the issue I had with the copying and now my other problem is that with 5x5 grid my algorithm returns 52 routes and it should be 86. So it works with 4x4 grid okay, but once I move further it breaks.
Added the isDeadEnd function here
public static boolean isDeadEnd(int[][] table, int x, int y)
{
int toCheck[] = new int[4];
toCheck[0] = x-1; // left
toCheck[1] = y-1; // top
toCheck[2] = x+1; // right
toCheck[3] = y+1; // bottom
int valuesOfDirections[] = new int[4]; // left, top, right, bottom
for(int i = 0; i < 4; i++)
{
int tarkastettava = toCheck[i];
if(tarkastettava > table.length-1 || tarkastettava < 0)
{
valuesOfDirections[i] = 1;
}
else
{
if(i == 0 || i == 2)
{
valuesOfDirections[i] = table[tarkastettava][y];
}
else
{
valuesOfDirections[i] = table[x][tarkastettava];
}
}
}
for(int i = 0; i < 4; i++)
{
if(valuesOfDirections[i] == 0)
{
return false;
}
}
return true;
}
Come to think of it, you probably can do a simple backtrack here:
table[x][y] = 1;
if(isDeadEnd(table, x, y)) {
if(isTableFull(table))
return 1;
}
table[x][y] = 0;
}
And later:
int res = a + b + c + d;
if (res == 0) {
// backtrack here too
table[x][y] = 0;
}
return res;
im creating a code for Minesweeper and trying to implement a GUI. But the problem is that when i run the code and play the game, the position i click on the board reveals the y,x coordinate of that location on the answer board rather than the x, y coordinate. Ive been trying to fix this but i cant see to figure it out. i think it maybe is the way i create the board, but i tried everything i could think of.
class Board{
public MsGUI gui;
private static char[][] userBoard;
private static char[][] solutionBoard;
private static int boundSize = 5;
public Board(){
userBoard = new char[][] {{'-','-','-','-','-'},
{'-','-','-','-','-'},
{'-','-','-','-','-'},
{'-','-','-','-','-'},
{'-','-','-','-','-'}};
solutionBoard = new char[][] {{'0','2','B','2','0'},
{'0','3','B','3','0'},
{'1','3','B','3','1'},
{'B','1','3','B','2'},
{'1','1','2','B','2'}};
return;
}
private static void printBoard(char[][] board){
for (int x = 0; x < boundSize; x++){
for(int y = 0; y < boundSize; y++){
System.out.print(" " + Character.toString(board[x][y]));
}
System.out.println("");
}
System.out.println("");
}
public void flagCell(int xCoordinate, int yCoordinate){
userBoard[xCoordinate][yCoordinate] = 'F';
}
public boolean isFlagged(int xCoordinate,int yCoordinate){
if(userBoard[xCoordinate][yCoordinate] == 'F'){
return true;
}
else{
return false;
}
}
public int getHeight() {
return userBoard.length;
}
public int getWidth(){
return userBoard[0].length;
}
public char getValue(int xCoordinate, int yCoordinate) {
return userBoard[xCoordinate][yCoordinate];
}
private static boolean checkIfAlreadyMarked(int xCoordinate, int yCoordinate)
{
boolean marked = false;
if (Character.toString(userBoard[xCoordinate][yCoordinate]).equals("-") == false)
{
marked = true;
}
return marked;
}
public void revealCell(int xCoordinate, int yCoordinate){
int count = 0;
for(int i = 0;i < userBoard.length;i++){
for(int J = 0;J < userBoard[i].length;J++){
if(userBoard[i][J] != '-'){
count = count + 1;
}
}
if(count == 19){
gui.win("you won");
return;
}
}
if(solutionBoard[xCoordinate][yCoordinate] == 'B'){
userBoard[xCoordinate][yCoordinate] = solutionBoard[xCoordinate][yCoordinate];
gui.lose("You lost. Better luck next time!");
return;
}
if(solutionBoard[xCoordinate][yCoordinate] != '0'){
userBoard[xCoordinate][yCoordinate] = solutionBoard[xCoordinate][yCoordinate];
}else{
userBoard[xCoordinate][yCoordinate] = solutionBoard[xCoordinate][yCoordinate];
for(int i = 1; i > -2; i--){
if(xCoordinate-i >= solutionBoard.length || xCoordinate-i < 0)
continue;
for(int z = 1; z > -2; z--){
if(yCoordinate-z >= solutionBoard[xCoordinate].length || yCoordinate-z < 0)
continue;
else if(userBoard[xCoordinate-i][yCoordinate-z] == 'F' || userBoard[xCoordinate-i][yCoordinate-z] != '-')
continue;
else{
revealCell(xCoordinate-i, yCoordinate-z);
}
}
}
}
}
public void unflagCell(int xCoordinate, int yCoordinate){
userBoard[xCoordinate][yCoordinate]='-';
}
public static void main(String[] args){
Board b = new Board();
b.gui = new MsGUI(b);
b.gui.setVisible(true);
}
}
The way you are initializing the solutionBoard is not what you expect it to be.
If you get solutionBoard[0], you're not accessing the first column(which would be consistent with what I think is your understanding), but the first row(first item of the two-dimensional array): {'0','2','B','2','0'}
So if you want to have x for a row index and y for a column index and still keep this "readable" initialization, you'll have to swap the indices whenever you access the array.
But this will only help you with one problem - human readable array assignment in the beginning, but I think you'll regret this decision in the future.
EDIT:
You can have the array initialized as you want and still use readable format like this:
String readableBoard =
"0 2 B 2 0;" +
"0 3 B 3 0;" +
"1 3 B 3 1;" +
"B 1 B B 2;" +
"1 1 2 B 2";
char[][] board = initBoard(readableBoard);
....
private char[][] initBoard(String readableBoard){
char[][] board = new char[5][5];
String[] rows = readableBoard.split(";");
String[] fields = null;
for (int y = 0; y<rows.length;y++){
fields = rows[y].split(" ");
for (int x = 0; x<fields.length; x++){
board[x][y]=fields[x].charAt(0);
}
}
return board;
}
Now when you call
board[2][0]
You'll get 'B'
If you look at your nested for loops, you're printing off the columns instead of what I assume to be the desired rows. Try switching your for loops to iterate over y, then x.
for (int y = 0; y < boundSize; y++){
for(int x = 0; x < boundSize; x++){
System.out.print(" " + Character.toString(board[x][y]));
}
System.out.println("");
}
System.out.println("");
I have been racking my brain around this problem for days now... I can't see what my algorithm could possibly be missing. This is the problem here.
From what I gather I am getting the points in a somewhat circular ccw order. So I've implemented a version of the graham scan that goes through looking to find the convex hull by ensuring it is using points that always give right-hand turns.
My algorithm works for all given test input and all input I can come up for it but it just won't get accepted by the online judge which is what is required for the assignment to be 'complete'.
Anyways, here is my code I would be forever in your debt if someone can find what I'm missing.
import java.util.Scanner;
import java.util.Vector;
import java.util.Arrays;
import java.util.Comparator;
public class Main {
public Main() {}
public void handlePoints(Point[] points) throws Exception {
int m = 1;
Vector<Point> convexHull = new Vector<Point>();
// This is THE ONLY gaurunteed point to be in the hull - and it is the lowest left point so that's ok.
convexHull.add(points[0]);
// Can be removed if ill-suited.
convexHull.add(points[1]);
for (int i = 2; i < points.length; i++) {
// Find the next valid point on the hull.
while (counterClockWise(convexHull.elementAt(m-1), convexHull.elementAt(m), points[i]) <= 0) {
convexHull.removeElementAt(m);
if (m > 1) {
m -= 1;
}
// All points are colinear
else if (i == points.length - 1) {
break;
}
else {
convexHull.add(points[i]);
i++;
}
}
convexHull.add(points[i]);
m++;
}
if (convexHull.size() <= 3) {
throw new Exception();
}
String test = "" + convexHull.size() + '\n';
for (Point p : convexHull) {
test += p.x + " " + p.y + '\n';
}
System.out.print(test);
}
// Simply calculated whether or not the 3 points form a countedClockWise turn.
public int counterClockWise(Point p1, Point p2, Point p3) {
return ((p2.x - p1.x) * (p3.y - p1.y)) - ((p2.y - p1.y) * (p3.x - p1.x));
}
// Rearranges the array to maintain its order but be started and ended by the point with the lowest y value
private static Point[] moveLowestToFront(Point[] array) {
// Rearrange for y:
int lowestY = 99999;
int lowestIndex = 0;
for (int i = 0; i < array.length; i++) {
if (array[i].y < lowestY) {
lowestY = array[i].y;
lowestIndex = i;
}
}
// Scan through again to see if there are any competing low-y values.
int lowestX = 99999;
for (int i = 0; i < array.length; i++) {
if (array[i].y == lowestY) {
if (array[i].x < lowestX) {
lowestX = array[i].x;
lowestIndex = i;
}
}
}
Point[] rearrangedPoints = new Point[array.length];
int j = 0;
// Take from low to end cutting off repeated start point.
for (int i = lowestIndex; i < array.length - 1; i++) {
rearrangedPoints[j] = array[i];
j++;
}
// Throw the remaining and put them at the end.
for (int i = 0; i < lowestIndex; i++) {
rearrangedPoints[j] = array[i];
j++;
}
// End the array with the repeated first point.
rearrangedPoints[array.length - 1] = array[lowestIndex];
return rearrangedPoints;
}
public static void main(String[] args) throws Exception {
Scanner sc = new Scanner(System.in);
Main convexHullFinder = new Main();
int numDataSets = sc.nextInt();
System.out.println(numDataSets);
for (int z = 0; z < numDataSets; z++) {
int numPoints = sc.nextInt();
Vector<Point> points = new Vector<Point>();
// Read in all the points for this set.
points.add(new Point(sc.nextInt(), sc.nextInt()));
int j = 1;
for (int i = 1; i < numPoints; i++) {
Point p = new Point(sc.nextInt(), sc.nextInt());
// Remove repeated points.
if (p.x < 0 || p.y < 0) {
throw new Exception();
}
if ( (p.x == points.elementAt(j-1).x) && (p.y == points.elementAt(j-1).y) ) {}
else {
points.add(p);
j++;
}
}
Point[] reducedPoints = points.toArray(new Point[points.size()]);
// Rearrange the set to start and end on the lowest Y point.
reducedPoints = moveLowestToFront(reducedPoints);
if (numPoints >= 3) {
convexHullFinder.handlePoints(reducedPoints);
}
else {
throw new Exception();
}
try {
System.out.println(sc.nextInt());
}
catch (Exception e) {
}
}
}
}
class Point {
public int x;
public int y;
public Point(int x, int y) {
this.x = x;
this.y = y;
}
}
From the sounds of it, the points are sorted such that Graham Scan applies. Therefore, I think your stack operation (handlePoints) is probably not right.
I'm more used to Andrew's algorithm (a modification of Graham Scan), but I'm fairly sure you shouldn't be adding points to the convex hull both inside and outside of the while loop. The reason being I'm fairly sure the purpose of the while loop remains the same regardless of which algorithm is used. It's to remove invalid points from the convex hull. However, there's a chance that you're adding points during the while loop.
I'm not sure that's all that needs fixed though, but I don't have anything set up to run Java at the moment.