You are given input as order of graph n (highest number of edges connected to a node), you have to find the number of vertices in a Fibonacci cube graph of order n.
Input : n = 3 Output : 5 Explanation : Fib(n + 2) = Fib(5) = 5 Input : n = 2 Output : 3
A Fibonacci Cube Graph is similar to hypercube graph, but with a fibonacci number of vertices. In fibonacci cube graph only 1 vertex has degree n rest all has degree less than n.
Fibonacci cube graph of order n has F(n + 2) vertices, where F(n) is a n-th fibonacci number,
Fibonacii series : 1, 1, 2, 3, 5, 8, 13, 21, 34……………….
For input n as order of graph, find the corresponding fibonacci number at the position n + 2.
where F(n) = F(n – 1) + F(n – 2)
Approach : Find the (n + 2)-th fibonacci number.
Below is the implementation of above approach :
Note that the above code can be optimized to work in O(Log n) using efficient implementations discussed in Program for Fibonacci numbers
- Convert the undirected graph into directed graph such that there is no path of length greater than 1
- Detect cycle in the graph using degrees of nodes of graph
- Graph implementation using STL for competitive programming | Set 2 (Weighted graph)
- Check if a M-th fibonacci number divides N-th fibonacci number
- Fibonacci Search
- Fibonacci Coding
- Nth XOR Fibonacci number
- K- Fibonacci series
- Fibonacci Word
- Fibonacci modulo p
- Non Fibonacci Numbers
- Fibonacci Power
- Sum of Fibonacci Numbers
- Fibonacci problem (Value of Fib(N)*Fib(N) - Fib(N-1) * Fib(N+1))
- Even Fibonacci Numbers Sum
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