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 ![N\in[1, 14]](https://geeksforgeeks.org/wp-content/uploads/2023/10/rubhabha_quicklatex.com-71fc3cf89db806c8bcd25cba7707903e_l3-1.png "Rendered by QuickLaTeX.com")
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 <iostream>using namespace std;int depth = 16;int ways[16][16][16];int stepNum;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][8][8] 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 functionint main(){ int list[15]; // 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[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][8][8] 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(String[] args) { int list[] = new int[15]; // 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 walksdepth = 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[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][8][8] 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][8][8] 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][8][8] 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[15]; // 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
<script> // Javascript implementation of counting // number of possible hexagonal walks let depth = 14; let ways = new Array(16); for (let i = 0; i < 16; i++) { ways[i] = new Array(16); for (let j = 0; j < 16; j++) { ways[i][j] = new Array(16); for (let k = 0; k < 16; k++) { ways[i][j][k] = 0; } } } let stepNum; function preprocess(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][8][8] because all the steps // after that will be used to trace back // to the original point index 0:0 // according to the image. for (let N = 1; N <= 14; N++) { for (let i = 1; i < depth; i++) { for (let 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]; } } let list = new Array(15); list.fill(0); // Preprocessing all possible ways preprocess(list); let steps = 4; document.write( "Number of walks" + " possible is/are "+ list[steps] ); </script> |
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.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
