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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