Number of different cyclic paths of length N in a tetrahedron

• Difficulty Level : Expert
• Last Updated : 05 May, 2021

Given a tetrahedron(vertex are A, B, C, D), the task is to find the number of different cyclic paths with length n from a vertex.
Note: Considering only a single vertex B i.e. to find the number of different cyclic paths of length N from B to itself.

Examples:

Input: 2
Output: 3
The paths of length 2 which starts and ends at D are:
B-A-B
B-D-B
B-C-B

Input: 3
Output: 6

Approach: Dynamic Programming can be used to keep track of the number of paths for previous values of N. Check for the number of moves which are left and where are we when we are moving in a path. That is 4n states, each with 3 options. Observe that all the vertices A, B, C are equivalent. Let zB be 1 initially and as at 0 steps, we can reach B itself only. Let zACD be 1 as paths for reaching other vertexes A, C and D is 0. Hence the recurrence relation formed will be:

Paths for N steps to reach b is = zADC*3

At every step, zADC gets multiplied by 2 (2 states) and it is added by zB since zB is the number of paths at step n-1 which comprises of the remaining 2 states.

Below is the implementation of the above approach:

C++

 // C++ program count total number of // paths to reach B from B #include #include using namespace std;   // Function to count the number of // steps in a tetrahedron int countPaths(int n) {     // initially coming to B is B->B     int zB = 1;       // cannot reach A, D or C     int zADC = 0;       // iterate for all steps     for (int i = 1; i <= n; i++) {           // recurrence relation         int nzB = zADC * 3;           int nzADC = (zADC * 2 + zB);           // memoize previous values         zB = nzB;         zADC = nzADC;     }       // returns steps     return zB; }   // Driver Code int main() {     int n = 3;     cout << countPaths(n);       return 0; }

Java

 // Java program count total // number of paths to reach // B from B import java.io.*;   class GFG {       // Function to count the // number of steps in a // tetrahedron static int countPaths(int n) {     // initially coming     // to B is B->B     int zB = 1;       // cannot reach A, D or C     int zADC = 0;       // iterate for all steps     for (int i = 1; i <= n; i++)     {           // recurrence relation         int nzB = zADC * 3;           int nzADC = (zADC * 2 + zB);           // memoize previous values         zB = nzB;         zADC = nzADC;     }       // returns steps     return zB; }   // Driver Code public static void main (String[] args) {     int n = 3;     System.out.println(countPaths(n)); } }   // This code is contributed by ajit

Python3

 # Python3 program count total number of # paths to reach B from B   # Function to count the number of # steps in a tetrahedron def countPaths(n):           # initially coming to B is B->B     zB = 1       # cannot reach A, D or C     zADC = 0       # iterate for all steps     for i in range(1, n + 1):           # recurrence relation         nzB = zADC * 3           nzADC = (zADC * 2 + zB)           # memoize previous values         zB = nzB         zADC = nzADC           # returns steps     return zB   # Driver code n = 3 print(countPaths(n))   # This code is contributed by ashutosh450

C#

 // C# program count total // number of paths to reach // B from B using System;   class GFG{           // Function to count the // number of steps in a // tetrahedron static int countPaths(int n) {           // initially coming     // to B is B->B     int zB = 1;       // cannot reach A, D or C     int zADC = 0;       // iterate for all steps     for (int i = 1; i <= n; i++)     {           // recurrence relation         int nzB = zADC * 3;           int nzADC = (zADC * 2 + zB);           // memoize previous values         zB = nzB;         zADC = nzADC;     }       // returns steps     return zB; }       // Driver Code     static public void Main ()     {         int n = 3;         Console.WriteLine(countPaths(n));     } }   // This code is contributed by Sach

PHP

 B     \$zB = 1;       // cannot reach A, D or C     \$zADC = 0;       // iterate for all steps     for (\$i = 1; \$i <= \$n; \$i++)     {           // recurrence relation         \$nzB = \$zADC * 3;           \$nzADC = (\$zADC * 2 + \$zB);           // memoize previous values         \$zB = \$nzB;         \$zADC = \$nzADC;     }       // returns steps     return \$zB; }   // Driver Code \$n = 3; echo countPaths(\$n);     // This code is contributed // by Sachin ?>

Javascript



Output:

6

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