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

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 N^th 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: 7 
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[7] 
= 1 + (dp[1] + dp[2] + dp[3] + dp[4] + dp[5] + dp[6]) / 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 N^th 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 1^st 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] = 0 
Let’s try to calculate dp[1]. 
dp[1] = 1 + 5 * (dp[1]) / 6 + dp[0] (Why? its because (5 / 6) is the probability it stays stuck at 1.) 
dp[1] / 6 = 1 (since dp[0] = 0) 
dp[1] = 6 
Similarly, dp[1] = dp[2] = dp[3] = dp[4] = dp[5] = 6 

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
#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

<script>

// Javascript implementation of the approach
var maxSize = 50;

// To store the states of dp
var dp = Array(maxSize);

// To determine whether a state
// has been solved before
var v = Array(maxSize);

// Function to return the count
function 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
var n = 10;
document.write( expectedSteps(n - 1).toFixed(5));

// This code is contributed by noob2000.
</script>

Output:

7.36111