Count of possible hexagonal walks

We are given an infinite two dimensional plane made of hexagons connected together. We can visualize this plane as a honeycomb. Element X is present on one of the cells / hexagon.
We are given N steps, the task is to calculate number of such hexagonal paths possible in which element X has to perform a walk of N steps and return back to the original hexagon, where Examples:

Input : 1
Output : Number of walks possible is/are 0
Explanation :
0 because using just one step we can move to
any of the adjacent cells but we cannot trace
back to the original hexagon.

Input : 2
Output : Number of walks possible is/are 6

Input : 4
Output : Number of walks possible is/are 90

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach :

• A hexagonal walk can be defined as walking through adjacent hexagons and returning to the original cell. We know the fact that a hexagon contains six sides i.e. a hexagon is surrounded by six hexagons. Now, we have to count number of ways we take N steps and come back to the original hexagon.
• Now, let us suppose the original hexagon (where element X was initially present) to be the origin. We need all possible ways we can take (N-k) steps such that we have some steps which would trace back to our original hexagon. We can visualize this hexagon and it’s related coordinate system from the picture below. •

• Now, let’s assume, our element X was present at 0:0 of the given picture. Thus, we can travel in six possible directions from a hexagon. Now, using the directions above we memorize all possible movements such that we trace back to the original 0:0 index. For memorizing we use a 3D array and we preprocess our answer for a given number of steps and then query accordingly.

Below is the implementation of above approach :

C++

 // C++ implementation of counting // number of possible hexagonal walks #include using namespace std;    int depth = 16; int ways; int stepNum;    void preprocess(int list[]) {     // We initialize our origin with 1     ways = 1;        // For each N = 1 to 14, we traverse in all possible     // direction. Using this 3D array we calculate the      // number of ways at each step and the total ways      // for a given step shall be found at      // ways[step number] because all the steps      // after that will be used to trace back to the      // original point index 0:0 according to the image.     for (int N = 1; N <= 14; N++)      {         for (int i = 1; i <= depth; i++)          {             for (int j = 1; j <= depth; j++)              {                 ways[N][i][j] = ways[N - 1][i][j + 1]                                  + ways[N - 1][i][j - 1]                                 + ways[N - 1][i + 1][j]                                  + ways[N - 1][i - 1][j]                                 + ways[N - 1][i + 1][j - 1]                                  + ways[N - 1][i - 1][j + 1];             }         }            // This array stores the number of ways         // possible for a given step         list[N] = ways[N];     } }    // Driver function int main() {     int list;        // Preprocessing all possible ways     preprocess(list);     int steps = 4;     cout << "Number of walks possible is/are "           << list[steps] << endl;     return 0; }

Java

 // Java implementation of counting // number of possible hexagonal walks import java.util.*;    class GFG {            static int depth = 14;     static int ways[][][] = new int;     static int stepNum;             static void preprocess(int list[])     {                    // We initialize our origin with 1         ways = 1;                 // For each N = 1 to 14, we traverse in          // all possible direction. Using this 3D         // array we calculate the number of ways         // at each step and the total ways for a         // given step shall be found at ways[step         // number] because all the steps         // after that will be used to trace back         // to the original point index 0:0          // according to the image.         for (int N = 1; N <= 14; N++)          {             for (int i = 1; i < depth; i++)              {                 for (int j = 1; j < depth; j++)                  {                     ways[N][i][j] =                              ways[N - 1][i][j + 1]                            + ways[N - 1][i][j - 1]                           + ways[N - 1][i + 1][j]                            + ways[N - 1][i - 1][j]                       + ways[N - 1][i + 1][j - 1]                       + ways[N - 1][i - 1][j + 1];                 }             }                     // This array stores the number of             // ways possible for a given step             list[N] = ways[N];         }     }                    /* Driver program to test above function */     public static void main(String[] args)      {          int list[] = new int;                         // Preprocessing all possible ways             preprocess(list);             int steps = 4;             System.out.println( "Number of walks"                            + " possible is/are "+                                    list[steps] );     } }

Python3

 # Python 3 implementation of counting # number of possible hexagonal walks    depth = 16 ways = [[[0 for i in range(17)]              for i in range(17)]             for i in range(17)]    def preprocess(list, steps):            # We initialize our origin with 1     ways = 1        # For each N = 1 to 14, we traverse in      # all possible direction. Using this 3D      # array we calculate the number of ways      # at each step and the total ways for a      # given step shall be found at ways[step      # number] because all the steps after      # that will be used to trace back to the      # original point index 0:0 according to the image.     for N in range(1, 16, 1):         for i in range(1, depth, 1):             for j in range(1, depth, 1):                 ways[N][i][j] = (ways[N - 1][i][j + 1] +                                   ways[N - 1][i][j - 1] +                                   ways[N - 1][i + 1][j] +                                   ways[N - 1][i - 1][j] +                                   ways[N - 1][i + 1][j - 1] +                                  ways[N - 1][i - 1][j + 1])            # This array stores the number of ways         # possible for a given step         list[N] = ways[N]        print("Number of walks possible is/are",                                 list[steps])     # Driver Code if __name__ == '__main__':     list = [0 for i in range(16)]     steps = 4            # Preprocessing all possible ways     preprocess(list, steps)        # This code is contributed by # Surendra_Gangwar

C#

 // C# implementation of counting // number of possible hexagonal walks using System;    class GFG {            static int depth = 14;     static int [, ,]ways = new int[16,16,16];     // static int stepNum;            static void preprocess(int []list)     {                    // We initialize our origin with 1         ways[0,8,8] = 1;                // For each N = 1 to 14, we traverse in          // all possible direction. Using this 3D         // array we calculate the number of ways         // at each step and the total ways for a         // given step shall be found at ways[step         // number] because all the steps         // after that will be used to trace back         // to the original point index 0:0          // according to the image.         for (int N = 1; N <= 14; N++)          {             for (int i = 1; i < depth; i++)              {                 for (int j = 1; j < depth; j++)                  {                     ways[N,i,j] =                              ways[N - 1,i,j + 1]                          + ways[N - 1,i,j - 1]                         + ways[N - 1,i + 1,j]                          + ways[N - 1,i - 1,j]                     + ways[N - 1,i + 1,j - 1]                      + ways[N - 1,i - 1,j + 1];                 }             }                    // This array stores the number of             // ways possible for a given step             list[N] = ways[N,8,8];         }     }                   /* Driver program to test above function */     public static void Main()      {         int []list = new int;                        // Preprocessing all possible ways             preprocess(list);             int steps = 4;             Console.WriteLine( "Number of walks"                         + " possible is/are "+                                 list[steps] );     } }    // This code is contributed by anuj_67.

Output :

Number of walks possible is/are 90

The time complexity of the above code is and the space complexity is also similar due to the 3D array used.

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