Prerequisite: Hamiltonian Cycle
Given an integer n(>=2), find a permutation of numbers from 1 to n such that the sum of two consecutive numbers of that permutation is a perfect square. If that kind of permutation is not possible to print “No Solution”.
Input : 17 Output : [16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17] Explanation : 16+9 = 25 = 5*5, 9+7 = 16 = 4*4, 7+2 = 9 = 3*3 and so on. Input: 20 Output: No Solution Input : 25 Output : [2, 23, 13, 12, 24, 25, 11, 14, 22, 3, 1, 8, 17, 19, 6, 10, 15, 21, 4, 5, 20, 16, 9, 7, 18]
We can represent a graph, where numbers from 1 to n are the nodes of the graph and there is an edge between ith and jth node if (i+j) is a perfect square. Then we can search if there is any Hamiltonian Path in the graph. If there is at least one path then we print a path otherwise we print “No Solution”.
1. First list up all the perfect square numbers which we can get by adding two numbers. We can get at max (2*n-1). so we will take only the squares up to (2*n-1). 2. Take an adjacency matrix to represent the graph. 3. For each number from 1 to n find out numbers with which it can add upto a perfect square number. Fill respective cells of the adjacency matrix by 1. 4. Now find if there is any Hamiltonian path in the graph using backtracking as discussed earlier.
17 -> [16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17] 20 -> No Solution 25 -> [2, 23, 13, 12, 24, 25, 11, 14, 22, 3, 1, 8, 17, 19, 6, 10, 15, 21, 4, 5, 20, 16, 9, 7, 18]
This backtracking algorithm takes exponential time to find Hamiltonian Path. Hence the time complexity of this algorithm is exponential.
In the last part of the hampath(n) function if we just print the path rather returning it then it will print all possible Hamiltonian Path i.e. all possible representations.
Actually we will first get a representation like this for n = 15. For n<15 there is no representation. For n = 18, 19, 20, 21, 22, 24 there is also no Hamiltonian Path. For rest of the numbers it works well.
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