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 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: 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 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] = 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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 

chevron_right


Output:

7.36111

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : mohit kumar 29, AnkitRai01