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++
// 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; } |
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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 |
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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 |
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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 |
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Output:
7.36111
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