There are tones of solutions for Knights tour or shortest path for Knights movement from source cell to destination cell. most of the solutions are using BFS which seems the best algorithm.
Here is my implementation using HashMap:
public class Knight_HashMap {
static HashMap<String, Position> chessboard = new HashMap<String, Position>();
static Queue<Position> q = new LinkedList<Position>();
static int Nx, Ny, Kx, Ky, Cx, Cy;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("insert Board dimentions: Nx, Ny");
Nx = sc.nextInt();
Ny = sc.nextInt();
System.out.println("inset Knight's location: Kx, Ky");
Kx = sc.nextInt();
Ky = sc.nextInt();
System.out.println("insert destination location: Cx, Cy");
Cx = sc.nextInt();
Cy = sc.nextInt();
sc.close();
// Assume the position for simplicity. In real world, accept the values using
// Scanner.
Position start = new Position(Kx, Ky, 0); // Positionition 0, 1 on the chessboard
Position end = new Position(Cx, Cy, Integer.MAX_VALUE);
chessboard.put(Arrays.toString(new int[] { Kx, Ky }), new Position(Kx, Ky, 0));
q.add(start);
while (q.size() != 0) // While queue is not empty
{
Position pos = q.poll();
if (end.equals(pos)) {
System.out.println("Minimum jumps required: " + pos.depth);
return;
} else {
// perform BFS on this Position if it is not already visited
bfs(pos, ++pos.depth);
}
}
}
private static void bfs(Position current, int depth) {
// Start from -2 to +2 range and start marking each location on the board
for (int i = -2; i <= 2; i++) {
for (int j = -2; j <= 2; j++) {
Position next = new Position(current.x + i, current.y + j, depth);
if (isValid(current, next)) {
if (inRange(next.x, next.y)) {
// chessboard.put(Arrays.toString(new int[] { next.x, next.y }), next);
// Skip if next location is same as the location you came from in previous run
if (current.equals(next))
continue;
Position position = chessboard.get(Arrays.toString(new int[] { next.x, next.y }));
if (position == null) {
position = new Position(Integer.MAX_VALUE, Integer.MAX_VALUE, Integer.MAX_VALUE);
}
/*
* Get the current position object at this location on chessboard. If this
* location was reachable with a costlier depth, this iteration has given a
* shorter way to reach
*/
if (position.depth > depth) {
chessboard.put(Arrays.toString(new int[] { current.x + i, current.y + j }),
new Position(current.x, current.y, depth));
// chessboard.get(current.x + i).set(current.y + j, new Position(current.x,
// current.y, depth));
q.add(next);
}
}
}
}
}
}
private static boolean isValid(Position current, Position next) {
// Use Pythagoras theorem to ensure that a move makes a right-angled triangle
// with sides of 1 and 2. 1-squared + 2-squared is 5.
int deltaR = next.x - current.x;
int deltaC = next.y - current.y;
return 5 == deltaR * deltaR + deltaC * deltaC;
}
private static boolean inRange(int x, int y) {
return 0 <= x && x < Nx && 0 <= y && y < Ny;
}
}
class Position {
public int x;
public int y;
public int depth;
Position(int x, int y, int depth) {
this.x = x;
this.y = y;
this.depth = depth;
}
public boolean equals(Position that) {
return this.x == that.x && this.y == that.y;
}
public String toString() {
return "(" + this.x + " " + this.y + " " + this.depth + ")";
}
}
This works well with small dimension but with 10^9 x 10^9 I face outOfMemory exception. I also tried with java 2d array, ArrayList, HashMap alongside with a LinkedList as Queue. but for that dimension 10^9 x 10^9 with any data structure, I face outOfMemory exception.
Is there possibility of optimization to avoid outOfMemory or anyother way/data structure to handle such huge dimension?
Note: I should mention this question is from BAPC17 contest named Knight's Marathon
