package graphs;

// Problem Title => Snake and Ladders Problem

/*PROBLEM STATEMENT

*

* Given a snake and ladder board,

* find the minimum number of dice throws required to reach the destination or last cell from source or 1st cell.

* Basically, the player has total control over outcome of dice throw and wants to find out minimum number of throws required to reach last cell.

* If the player reaches a cell which is base of a ladder,

* the player has to climb up that ladder and if reaches a cell is mouth of the snake,

* has to go down to the tail of snake without a dice throw.

snakes and ladders

*

For example, consider the board shown, the minimum number of dice throws required to reach cell 30 from cell 1 is 3.

Following are the steps:

a) First throw two on dice to reach cell number 3 and then ladder to reach 22

b) Then throw 6 to reach 28.

c) Finally through 2 to reach 30.

There can be other solutions as well like (2, 2, 6), (2, 4, 4), (2, 3, 5).. etc.

* */

/* Approach =>

* The idea is to consider the given snake and ladder board as a directed graph with number of vertices equal to the number of cells in the board.

* The problem reduces to finding the shortest path in a graph.

* Every vertex of the graph has an edge to next six vertices if next 6 vertices do not have a snake or ladder.

* If any of the next six vertices has a snake or ladder,

* then the edge from current vertex goes to the top of the ladder or tail of the snake. Since all edges are of equal weight, we can efficiently find shortest path using Breadth First Search of the graph.

* Following is the implementation of the above idea.

* The input is represented by two things, first is ‘N’ which is number of cells in the given board, second is an arrays.array ‘move[0…N-1]’ of size N. An entry move[i] is -1 if there is no snake and no ladder from i, otherwise move[i] contains index of destination cell for the snake or the ladder at i.

*/

import java.util.LinkedList;

import java.util.Queue;

public class Graph_Problem_25 {

// An entry in queue used in BFS

static class qentry {

int v;// Vertex number

int dist;// Distance of this vertex from source

}

// This function returns minimum number of dice

// throws required to Reach last cell from 0'th cell

// in a snake and ladder game. move[] is an arrays.array of

// size N where N is no. of cells on board If there

// is no snake or ladder from cell i, then move[i]

// is -1 Otherwise move[i] contains cell to which

// snake or ladder at i takes to.

static int getMinDiceThrows(int[] move, int n) {

int[] visited = new int[n];

Queue<qentry> q = new LinkedList<>();

qentry qe = new qentry();

qe.v = 0;

qe.dist = 0;

// Mark the node 0 as visited and enqueue it.

visited[0] = 1;

q.add(qe);

// Do a BFS starting from vertex at index 0

while (!q.isEmpty()) {

qe = q.remove();

int v = qe.v;

// If front vertex is the destination

// vertex, we are done

if (v == n - 1)

break;

// Otherwise, dequeue the front vertex and

// enqueue its adjacent vertices (or cell

// numbers reachable through a dice throw)

for (int j = v + 1; j <= (v + 6) && j < n; ++j) {

// If this cell is already visited, then ignore

if (visited[j] == 0) {

// Otherwise, calculate its distance and

// mark it as visited

qentry a = new qentry();

a.dist = (qe.dist + 1);

visited[j] = 1;

// Check if there is a snake or ladder at 'j'

// then tail of snake or top of ladder

// become the adjacent of 'i'

if (move[j] != -1)

a.v = move[j];

else

a.v = j;

q.add(a);

}

}

}

// We reach here when 'qe' has last vertex

// return the distance of vertex in 'qe'

return qe.dist;

}

public static void main(String[] args) {

// Let us construct the board given in above diagram

int N = 30;

int[] moves = new int[N];

for (int i = 0; i < N; i++)

moves[i] = -1;

// Ladders

moves[2] = 21;

moves[4] = 7;

moves[10] = 25;

moves[19] = 28;

// Snakes

moves[26] = 0;

moves[20] = 8;

moves[16] = 3;

moves[18] = 6;

System.out.println("Min Dice throws required is " + getMinDiceThrows(moves, N));

}

}