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# Expected number of moves to reach the end of a board | Dynamic programming

• Last Updated : 13 May, 2021

Given a linear board of length N numbered from 1 to N, the task is to find the expected number of moves required to reach the Nth cell of the board, if we start at cell numbered 1 and at each step we roll a cubical dice to decide the next move. Also, we cannot go outside the bounds of the board. Note that the expected number of moves can be fractional.
Examples:

Input: N = 8
Output:
p1 = (1 / 6) | 1-step -> 6 moves expected to reach the end
p2 = (1 / 6) | 2-steps -> 6 moves expected to reach the end
p3 = (1 / 6) | 3-steps -> 6 moves expected to reach the end
p4 = (1 / 6) | 4-steps -> 6 moves expected to reach the end
p5 = (1 / 6) | 5-steps -> 6 moves expected to reach the end
p6 = (1 / 6) | 6-steps -> 6 moves expected to reach the end
If we are 7 steps away, then we can end up at 1, 2, 3, 4, 5, 6 steps
away with equal probability i.e. (1 / 6).
Look at the above simulation to understand better.
dp[N – 1] = dp
= 1 + (dp + dp + dp + dp + dp + dp) / 6
= 1 + 6 = 7
Input: N = 10
Output: 7.36111

Approach: This problem can be solved using dynamic programming. To solve the problem, decide the states of the DP first. One way will be to use the distance between the current cell and the Nth cell to define the states of DP. Let’s call this distance X. Thus dp[X] can be defined as the expected number of steps required to reach the end of the board of length X + 1 starting from the 1st cell.
Thus, the recurrence relation becomes:

dp[X] = 1 + (dp[X – 1] + dp[X – 2] + dp[X – 3] + dp[X – 4] + dp[X – 5] + dp[X – 6]) / 6

Now, for the base-cases:

dp = 0
Let’s try to calculate dp.
dp = 1 + 5 * (dp) / 6 + dp (Why? its because (5 / 6) is the probability it stays stuck at 1.)
dp / 6 = 1 (since dp = 0)
dp = 6
Similarly, dp = dp = dp = dp = dp = 6

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``#define maxSize 50``using` `namespace` `std;` `// To store the states of dp``double` `dp[maxSize];` `// To determine whether a state``// has been solved before``int` `v[maxSize];` `// Function to return the count``double` `expectedSteps(``int` `x)``{` `    ``// Base cases``    ``if` `(x == 0)``        ``return` `0;``    ``if` `(x <= 5)``        ``return` `6;` `    ``// If a state has been solved before``    ``// it won't be evaluated again``    ``if` `(v[x])``        ``return` `dp[x];` `    ``v[x] = 1;` `    ``// Recurrence relation``    ``dp[x] = 1 + (expectedSteps(x - 1) +``                 ``expectedSteps(x - 2) +``                 ``expectedSteps(x - 3) +``                 ``expectedSteps(x - 4) +``                 ``expectedSteps(x - 5) +``                 ``expectedSteps(x - 6)) / 6;``    ``return` `dp[x];``}` `// Driver code``int` `main()``{``    ``int` `n = 10;` `    ``cout << expectedSteps(n - 1);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``    ``static` `int` `maxSize = ``50``;` `    ``// To store the states of dp``    ``static` `double` `dp[] = ``new` `double``[maxSize];``    ` `    ``// To determine whether a state``    ``// has been solved before``    ``static` `int` `v[] = ``new` `int``[maxSize];``    ` `    ``// Function to return the count``    ``static` `double` `expectedSteps(``int` `x)``    ``{``    ` `        ``// Base cases``        ``if` `(x == ``0``)``            ``return` `0``;``            ` `        ``if` `(x <= ``5``)``            ``return` `6``;``    ` `        ``// If a state has been solved before``        ``// it won't be evaluated again``        ``if` `(v[x] == ``1``)``            ``return` `dp[x];``    ` `        ``v[x] = ``1``;``    ` `        ``// Recurrence relation``        ``dp[x] = ``1` `+ (expectedSteps(x - ``1``) +``                     ``expectedSteps(x - ``2``) +``                     ``expectedSteps(x - ``3``) +``                     ``expectedSteps(x - ``4``) +``                     ``expectedSteps(x - ``5``) +``                     ``expectedSteps(x - ``6``)) / ``6``;``        ` `        ``return` `dp[x];``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `n = ``10``;``    ` `        ``System.out.println(expectedSteps(n - ``1``));``    ``}``}` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the approach``maxSize ``=` `50` `# To store the states of dp``dp ``=` `[``0``] ``*` `maxSize` `# To determine whether a state``# has been solved before``v ``=` `[``0``] ``*` `maxSize` `# Function to return the count``def` `expectedSteps(x):` `    ``# Base cases``    ``if` `(x ``=``=` `0``):``        ``return` `0``    ``if` `(x <``=` `5``):``        ``return` `6` `    ``# If a state has been solved before``    ``# it won't be evaluated again``    ``if` `(v[x]):``        ``return` `dp[x]` `    ``v[x] ``=` `1` `    ``# Recurrence relation``    ``dp[x] ``=` `1` `+` `(expectedSteps(x ``-` `1``) ``+``                 ``expectedSteps(x ``-` `2``) ``+``                 ``expectedSteps(x ``-` `3``) ``+``                 ``expectedSteps(x ``-` `4``) ``+``                 ``expectedSteps(x ``-` `5``) ``+``                 ``expectedSteps(x ``-` `6``)) ``/` `6``    ``return` `dp[x]` `# Driver code``n ``=` `10` `print``(``round``(expectedSteps(n ``-` `1``), ``5``))` `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ``static` `int` `maxSize = 50;` `    ``// To store the states of dp``    ``static` `double` `[]dp = ``new` `double``[maxSize];``    ` `    ``// To determine whether a state``    ``// has been solved before``    ``static` `int` `[]v = ``new` `int``[maxSize];``    ` `    ``// Function to return the count``    ``static` `double` `expectedSteps(``int` `x)``    ``{``    ` `        ``// Base cases``        ``if` `(x == 0)``            ``return` `0;``            ` `        ``if` `(x <= 5)``            ``return` `6;``    ` `        ``// If a state has been solved before``        ``// it won't be evaluated again``        ``if` `(v[x] == 1)``            ``return` `dp[x];``    ` `        ``v[x] = 1;``    ` `        ``// Recurrence relation``        ``dp[x] = 1 + (expectedSteps(x - 1) +``                     ``expectedSteps(x - 2) +``                     ``expectedSteps(x - 3) +``                     ``expectedSteps(x - 4) +``                     ``expectedSteps(x - 5) +``                     ``expectedSteps(x - 6)) / 6;``        ` `        ``return` `dp[x];``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `n = 10;``    ` `        ``Console.WriteLine(expectedSteps(n - 1));``    ``}``}` `// This code is contributed by AnkitRai01`

## Javascript

 ``
Output:
`7.36111`

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