Given two positive integer n, k. Consider an undirected complete connected graph of n nodes in a complete connected graph. The task is to calculate the number of ways in which one can start from any node and return to it by visiting K nodes.
Input : n = 3, k = 3 Output : 2 Input : n = 4, k = 2 Output : 3
Lets f(n, k) be a function which return number of ways in which one can start from any node and return to it by visiting K nodes.
If we start and end from one node, then we have K – 1 choices to make for the intermediate nodes since we have already chosen one node in the beginning. For each intermediate choice, you have n – 1 options. So, this will yield (n – 1)k – 1 but then we have to remove all the choices cause smaller loops, so we subtract f(n, k – 1).
So, recurrence relation becomes,
f(n, k) = (n – 1)k – 1 – f(n, k – 1) with base case f(n, 2) = n – 1.
f(n, k) = (n – 1)k – 1 – (n – 1)k – 2 + (n – 1)k – 3 ….. (n – 1) (say eqn 1)
Dividing f(n, k) by (n – 1),
f(n, k)/(n – 1) = (n – 1)k – 2 – (n – 1)k – 3 + (n – 1)k – 4 ….. 1 (say eqn 2)
On adding eqn 1 and eqn 2,
f(n, k) + f(n, k)/(n – 1) = (n – 1)k – 1 + (-1)k
f(n, k) * ( (n -1) + 1 )/(n – 1) = (n – 1)k – 1 + (-1)k
Below is the implementation of this approach:
- Minimum cost path from source node to destination node via an intermediate node
- Number of unique permutations starting with 1 of a Binary String
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- Number of arrays of size N whose elements are positive integers and sum is K
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- Count number of triplets with product equal to given number with duplicates allowed | Set-2
- Count number of triplets with product equal to given number with duplicates allowed
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