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# Count of possible hexagonal walks

• Difficulty Level : Hard
• Last Updated : 24 May, 2021

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 

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 functionint 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 walksimport 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 = 16ways = [[[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 Codeif __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 walksusing 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.

## Javascript

 

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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