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.

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.

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 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.

## C++

`// C++ program to find minimum number of dice throws required to ` `// reach last cell from first cell of a given snake and ladder ` `// board ` `#include<iostream> ` `#include <queue> ` `using` `namespace` `std; ` ` ` `// An entry in queue used in BFS ` `struct` `queueEntry ` `{ ` ` ` `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 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. ` `int` `getMinDiceThrows(` `int` `move[], ` `int` `N) ` `{ ` ` ` `// The graph has N vertices. Mark all the vertices as ` ` ` `// not visited ` ` ` `bool` `*visited = ` `new` `bool` `[N]; ` ` ` `for` `(` `int` `i = 0; i < N; i++) ` ` ` `visited[i] = ` `false` `; ` ` ` ` ` `// Create a queue for BFS ` ` ` `queue<queueEntry> q; ` ` ` ` ` `// Mark the node 0 as visited and enqueue it. ` ` ` `visited[0] = ` `true` `; ` ` ` `queueEntry s = {0, 0}; ` `// distance of 0't vertex is also 0 ` ` ` `q.push(s); ` `// Enqueue 0'th vertex ` ` ` ` ` `// Do a BFS starting from vertex at index 0 ` ` ` `queueEntry qe; ` `// A queue entry (qe) ` ` ` `while` `(!q.empty()) ` ` ` `{ ` ` ` `qe = q.front(); ` ` ` `int` `v = qe.v; ` `// vertex no. of queue entry ` ` ` ` ` `// 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) ` ` ` `q.pop(); ` ` ` `for` `(` `int` `j=v+1; j<=(v+6) && j<N; ++j) ` ` ` `{ ` ` ` `// If this cell is already visited, then ignore ` ` ` `if` `(!visited[j]) ` ` ` `{ ` ` ` `// Otherwise calculate its distance and mark it ` ` ` `// as visited ` ` ` `queueEntry a; ` ` ` `a.dist = (qe.dist + 1); ` ` ` `visited[j] = ` `true` `; ` ` ` ` ` `// Check if there 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.push(a); ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` ` ` `// We reach here when 'qe' has last vertex ` ` ` `// return the distance of vertex in 'qe' ` ` ` `return` `qe.dist; ` `} ` ` ` `// Driver program to test methods of graph class ` `int` `main() ` `{ ` ` ` `// Let us construct the board given in above diagram ` ` ` `int` `N = 30; ` ` ` `int` `moves[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; ` ` ` ` ` `cout << ` `"Min Dice throws required is "` `<< getMinDiceThrows(moves, N); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to find minimum number of dice ` `// throws required to reach last cell from first ` `// cell of a given snake and ladder board ` ` ` `import` `java.util.LinkedList; ` `import` `java.util.Queue; ` ` ` `public` `class` `SnakesLadder ` `{ ` ` ` `// 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 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 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)); ` ` ` `} ` `} ` |

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## Python3

`# Python3 program to find minimum number ` `# of dice throws required to reach last ` `# cell from first cell of a given ` `# snake and ladder board ` ` ` `# An entry in queue used in BFS ` `class` `QueueEntry(` `object` `): ` ` ` `def` `__init__(` `self` `, v ` `=` `0` `, dist ` `=` `0` `): ` ` ` `self` `.v ` `=` `v ` ` ` `self` `.dist ` `=` `dist ` ` ` `'''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 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.'''` `def` `getMinDiceThrows(move, N): ` ` ` ` ` `# The graph has N vertices. Mark all ` ` ` `# the vertices as not visited ` ` ` `visited ` `=` `[` `False` `] ` `*` `N ` ` ` ` ` `# Create a queue for BFS ` ` ` `queue ` `=` `[] ` ` ` ` ` `# Mark the node 0 as visited and enqueue it ` ` ` `visited[` `0` `] ` `=` `True` ` ` ` ` `# Distance of 0't vertex is also 0 ` ` ` `# Enqueue 0'th vertex ` ` ` `queue.append(QueueEntry(` `0` `, ` `0` `)) ` ` ` ` ` `# Do a BFS starting from vertex at index 0 ` ` ` `qe ` `=` `QueueEntry() ` `# A queue entry (qe) ` ` ` `while` `queue: ` ` ` `qe ` `=` `queue.pop(` `0` `) ` ` ` `v ` `=` `qe.v ` `# Vertex no. of queue entry ` ` ` ` ` `# 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) ` ` ` `j ` `=` `v ` `+` `1` ` ` `while` `j <` `=` `v ` `+` `6` `and` `j < N: ` ` ` ` ` `# If this cell is already visited, ` ` ` `# then ignore ` ` ` `if` `visited[j] ` `is` `False` `: ` ` ` ` ` `# Otherwise calculate its ` ` ` `# distance and mark it ` ` ` `# as visited ` ` ` `a ` `=` `QueueEntry() ` ` ` `a.dist ` `=` `qe.dist ` `+` `1` ` ` `visited[j] ` `=` `True` ` ` ` ` `# Check if there a snake or ladder ` ` ` `# at 'j' then tail of snake or top ` ` ` `# of ladder become the adjacent of 'i' ` ` ` `a.v ` `=` `move[j] ` `if` `move[j] !` `=` `-` `1` `else` `j ` ` ` ` ` `queue.append(a) ` ` ` ` ` `j ` `+` `=` `1` ` ` ` ` `# We reach here when 'qe' has last vertex ` ` ` `# return the distance of vertex in 'qe ` ` ` `return` `qe.dist ` ` ` `# driver code ` `N ` `=` `30` `moves ` `=` `[` `-` `1` `] ` `*` `N ` ` ` `# 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` ` ` `print` `(` `"Min Dice throws required is {0}"` `. ` ` ` `format` `(getMinDiceThrows(moves, N))) ` ` ` `# This code is contributed by Ajitesh Pathak ` |

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Output:

Min Dice throws required is 3

Time complexity of the above solution is O(N) as every cell is added and removed only once from queue. And a typical enqueue or dequeue operation takes O(1) time.

This article is contributed by **Siddharth**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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