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# Permutation of numbers such that sum of two consecutive numbers is a perfect square

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”.

Examples:

```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]```

Method: 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”.

Approach:

```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.  ```

Implementation:

## C++

 `#include ``using` `namespace` `std;` `// Function to check if a node is safe to visit in the``// current Hamiltonian path``bool` `issafe(``int` `v, ``int` `graph[][20], ``int` `path[], ``int` `pos)``{``    ``// Check if the current node is connected to the``    ``// previous node in the path``    ``if` `(graph[path[pos - 1]][v] == 0)``        ``return` `false``;` `    ``// Check if the current node has already been visited in``    ``// the current path``    ``for` `(``int` `i = 0; i < pos; i++)``        ``if` `(path[i] == v)``            ``return` `false``;` `    ``// If the current node is connected to the previous node``    ``// and has not been visited yet, return true``    ``return` `true``;``}` `// Function to form the Hamiltonian path by visiting nodes``// recursively``bool` `formpath(``int` `graph[][20], ``int` `path[], ``int` `pos, ``int` `n)``{``    ``// If all the nodes have been visited, return true``    ``if` `(pos == n + 1)``        ``return` `true``;` `    ``// Try visiting each node and see if it forms a``    ``// Hamiltonian path``    ``for` `(``int` `v = 1; v < n + 1; v++) {``        ``if` `(issafe(v, graph, path, pos)) {``            ``path[pos] = v;``            ``if` `(formpath(graph, path, pos + 1, n) == ``true``)``                ``return` `true``;``            ``path[pos] = -1;``        ``}``    ``}``    ``// If none of the nodes form a Hamiltonian path, return``    ``// false``    ``return` `false``;``}` `// Function to find a Hamiltonian path in a given graph``void` `hampath(``int` `n)``{``    ``// Adjacency matrix to store the graph``    ``int` `graph[20][20];``    ``// Array to store the Hamiltonian path``    ``int` `path[20];``    ``// Temporary variable used to store square root of 2n-1``    ``int` `k = 0;` `    ``// If there is only 1 node in the graph, there is no``    ``// Hamiltonian path``    ``if` `(n == 1) {``        ``cout << ``"No Solution"``;``        ``return``;``    ``}` `    ``// Vector to store the squares of the numbers from 1 to``    ``// sqrt(2n-1)``    ``vector<``int``> l;``    ``int` `nsqrt = ``sqrt``(2 * n - 1);``    ``for` `(``int` `i = 1; i <= nsqrt + 1; i++) {``        ``l.push_back(i * i);``    ``}` `    ``// Initialize the graph to 0``    ``memset``(graph, 0, ``sizeof``(graph));` `    ``// Form the graph using the squares stored in the vector``    ``// 'l'``    ``for` `(``int` `i = 1; i < n + 1; i++) {``        ``for` `(``auto` `ele : l) {``            ``// Check if the difference between the square``            ``// and the current node is greater than 0 and``            ``// less than or equal to n Also, make sure that``            ``// the difference is not equal to 2 times the``            ``// current node (to avoid repeated edges)``            ``if` `((ele - i) > 0 && (ele - i) <= n``                ``&& (2 * i != ele)) {``                ``graph[i][ele - i] = 1;``                ``graph[ele - i][i] = 1;``            ``}``        ``}``    ``}` `    ``// Loop through all vertices starting from vertex 1``    ``for` `(``int` `j = 1; j < n + 1; j++) {``        ``// Reset the path array``        ``memset``(path, -1, ``sizeof``(path));``        ``// Start the path from vertex j``        ``path[1] = j;``        ``// Check if a Hamiltonian path can be formed``        ``// starting from vertex j``        ``if` `(formpath(graph, path, 2, n) == ``true``) {``            ``cout << ``"Hamiltonian Path: "``;``            ``// Print the path``            ``for` `(``int` `i = 1; i < n + 1; i++) {``                ``cout << path[i] << ``" "``;``            ``}``            ``// Return the path``            ``return``;``        ``}``    ``}` `    ``// If no solution is found, print "No Solution"``    ``cout << ``"No Solution"``;``    ``return``;``}``// Driver Function``    ``int` `main()``    ``{``        ``cout << ``"17 -> "``;``        ``hampath(17);``        ``cout << endl << ``"20 -> "``;``        ``hampath(20);``        ``cout << endl << ``"25 -> "``;``        ``hampath(25);``        ``return` `0;``    ``}`

## Java

 `import` `java.util.*;` `public` `class` `HamiltonianPath {` `    ``// Function to check if a node is safe to visit in the``    ``// current Hamiltonian path``    ``static` `boolean` `issafe(``int` `v, ``int``[][] graph, ``int``[] path,``                          ``int` `pos)``    ``{``        ``// Check if the current node is connected to the``        ``// previous node in the path``        ``if` `(graph[path[pos - ``1``]][v] == ``0``) {``            ``return` `false``;``        ``}` `        ``// Check if the current node has already been``        ``// visited in the current path``        ``for` `(``int` `i = ``0``; i < pos; i++) {``            ``if` `(path[i] == v) {``                ``return` `false``;``            ``}``        ``}` `        ``// If the current node is connected to the previous``        ``// node and has not been visited yet, return true``        ``return` `true``;``    ``}` `    ``// Function to form the Hamiltonian path by visiting``    ``// nodes recursively``    ``static` `boolean` `formpath(``int``[][] graph, ``int``[] path,``                            ``int` `pos, ``int` `n)``    ``{``        ``// If all the nodes have been visited, return true``        ``if` `(pos == n + ``1``) {``            ``return` `true``;``        ``}` `        ``// Try visiting each node and see if it forms a``        ``// Hamiltonian path``        ``for` `(``int` `v = ``1``; v < n + ``1``; v++) {``            ``if` `(issafe(v, graph, path, pos)) {``                ``path[pos] = v;``                ``if` `(formpath(graph, path, pos + ``1``, n)``                    ``== ``true``) {``                    ``return` `true``;``                ``}``                ``path[pos] = -``1``;``            ``}``        ``}``        ``// If none of the nodes form a Hamiltonian path,``        ``// return false``        ``return` `false``;``    ``}` `    ``// Function to find a Hamiltonian path in a given graph``    ``static` `void` `hampath(``int` `n)``    ``{``        ``// Adjacency matrix to store the graph``        ``int``[][] graph = ``new` `int``[n+``1``][n+``1``];``        ``// Array to store the Hamiltonian path``        ``int``[] path = ``new` `int``[n+``1``];` `        ``// If there is only 1 node in the graph, there is no``        ``// Hamiltonian path``        ``if` `(n == ``1``) {``            ``System.out.println(``"No Solution"``);``            ``return``;``        ``}` `        ``// Vector to store the squares of the numbers from 1``        ``// to sqrt(2n-1)``        ``ArrayList l = ``new` `ArrayList();``        ``int` `nsqrt = (``int``)Math.sqrt(``2` `* n - ``1``);``        ``for` `(``int` `i = ``1``; i <= nsqrt + ``1``; i++) {``            ``l.add(i * i);``        ``}` `        ``// Initialize the graph to 0``        ``for` `(``int` `i = ``1``; i < n + ``1``; i++) {``            ``for` `(``int` `j = ``1``; j < n + ``1``; j++) {``                ``graph[i][j] = ``0``;``            ``}``        ``}` `        ``// Form the graph using the squares stored in the``        ``// vector 'l'``        ``for` `(``int` `i = ``1``; i < n + ``1``; i++) {``            ``for` `(``int` `ele : l) {``                ``// Check if the difference between the``                ``// square and the current node is greater``                ``// than 0 and less than or equal to n Also,``                ``// make sure that the difference is not``                ``// equal to 2 times the current node (to``                ``// avoid repeated edges)``                ``if` `((ele - i) > ``0` `&& (ele - i) <= n``                    ``&& (``2` `* i != ele)) {``                    ``graph[i][ele - i] = ``1``;``                    ``graph[ele - i][i] = ``1``;``                ``}``            ``}``        ``}` `        ``for` `(``int` `j = ``1``; j < n + ``1``; j++) {``            ``// Reset the path array``            ``Arrays.fill(path, -``1``);``            ``// Start the path from vertex j``            ``path[``1``] = j;``            ``// Check if a Hamiltonian path can be formed``            ``// starting from vertex j``            ``if` `(formpath(graph, path, ``2``, n) == ``true``) {``                ``System.out.print(``"Hamiltonian Path: "``);``                ``// Print the path``                ``for` `(``int` `i = ``1``; i < n + ``1``; i++) {``                    ``System.out.print(path[i] + ``" "``);``                ``}``                ``// Return the path``                ``return``;``            ``}``        ``}``        ``// If no solution is found, print "No Solution"``        ``System.out.print(``"No Solution"``);``    ``}``    ``public` `static` `void` `main(String[] args)``    ``{``        ``System.out.print(``"17 -> "``);``        ``hampath(``17``);``        ``System.out.println();``        ``System.out.print(``"20 -> "``);``        ``hampath(``20``);``        ``System.out.println();``        ``System.out.print(``"25 -> "``);``        ``hampath(``25``);``    ``}``}`

## Python3

 `# Python3 program for Sum-square series using``# hamiltonian path concept and backtracking` `# Function to check whether we can add number``# v with the path in the position pos.`  `def` `issafe(v, graph, path, pos):` `    ``# if there is no edge between v and the``    ``# last element of the path formed so far``    ``# return false.``    ``if` `(graph[path[pos ``-` `1``]][v] ``=``=` `0``):``        ``return` `False` `    ``# Otherwise if there is an edge between``    ``# v and last element of the path formed so``    ``# far, then check all the elements of the``    ``# path. If v is already in the path return``    ``# false.``    ``for` `i ``in` `range``(pos):` `        ``if` `(path[i] ``=``=` `v):``            ``return` `False` `    ``# If none of the previous cases satisfies``    ``# then we can add v to the path in the``    ``# position pos. Hence return true.``    ``return` `True` `# Function to form a path based on the graph.`  `def` `formpath(graph, path, pos):` `    ``# If all the elements are included in the``    ``# path i.e. length of the path is n then``    ``# return true i.e. path formed.``    ``n ``=` `len``(graph) ``-` `1``    ``if` `(pos ``=``=` `n ``+` `1``):``        ``return` `True` `    ``# This loop checks for each element if it``    ``# can be fitted as the next element of the``    ``# path and recursively finds the next``    ``# element of the path.``    ``for` `v ``in` `range``(``1``, n ``+` `1``):` `        ``if` `issafe(v, graph, path, pos):``            ``path[pos] ``=` `v` `            ``# Recurs for next element of the path.``            ``if` `(formpath(graph, path, pos ``+` `1``) ``=``=` `True``):``                ``return` `True` `            ``# If adding v does not give a solution``            ``# then remove it from path``            ``path[pos] ``=` `-``1` `    ``# if any vertex cannot be added with the``    ``# formed path then return false and``    ``# backtracks.``    ``return` `False` `# Function to find out sum-square series.`  `def` `hampath(n):` `    ``# base case: if n = 1 there is no solution``    ``if` `n ``=``=` `1``:``        ``return` `'No Solution'` `    ``# Make an array of perfect squares from 1``    ``# to (2 * n-1)``    ``l ``=` `list``()` `    ``for` `i ``in` `range``(``1``, ``int``((``2` `*` `n``-``1``) ``*``*` `0.5``) ``+` `1``):``        ``l.append(i``*``*``2``)` `    ``# Form the graph where sum of two adjacent``    ``# vertices is a perfect square``    ``graph ``=` `[[``0` `for` `i ``in` `range``(n ``+` `1``)] ``for` `j ``in` `range``(n ``+` `1``)]` `    ``for` `i ``in` `range``(``1``, n ``+` `1``):``        ``for` `ele ``in` `l:` `            ``if` `((ele``-``i) > ``0` `and` `(ele``-``i) <``=` `n``                    ``and` `(``2` `*` `i !``=` `ele)):``                ``graph[i][ele ``-` `i] ``=` `1``                ``graph[ele ``-` `i][i] ``=` `1` `    ``# starting from 1 upto n check for each``    ``# element i if any path can be formed``    ``# after taking i as the first element.``    ``for` `j ``in` `range``(``1``, n ``+` `1``):``        ``path ``=` `[``-``1` `for` `k ``in` `range``(n ``+` `1``)]``        ``path[``1``] ``=` `j` `        ``# If starting from j we can form any path``        ``# then we will return the path``        ``if` `formpath(graph, path, ``2``) ``=``=` `True``:``            ``return` `path[``1``:]` `    ``# If no path can be formed at all return``    ``# no solution.``    ``return` `'No Solution'`  `# Driver Function``print``(``17``, ``'->'``, hampath(``17``))``print``(``20``, ``'->'``, hampath(``20``))``print``(``25``, ``'->'``, hampath(``25``))`

## C#

 `//C# Equivalent of above code``using` `System;``using` `System.Collections.Generic;` `public` `class` `HamiltonianPath``{``  ``// Function to check if a node is safe to visit in the``  ``// current Hamiltonian path``  ``static` `bool` `issafe(``int` `v, ``int``[,] graph, ``int``[] path,``                     ``int` `pos)``  ``{``    ` `    ``// Check if the current node is connected to the``    ``// previous node in the path``    ``if` `(graph[path[pos - 1], v] == 0)``    ``{``      ``return` `false``;``    ``}` `    ``// Check if the current node has already been``    ``// visited in the current path``    ``for` `(``int` `i = 0; i < pos; i++)``    ``{``      ``if` `(path[i] == v)``      ``{``        ``return` `false``;``      ``}``    ``}` `    ``// If the current node is connected to the previous``    ``// node and has not been visited yet, return true``    ``return` `true``;``  ``}` `  ``// Function to form the Hamiltonian path by visiting``  ``// nodes recursively``  ``static` `bool` `formpath(``int``[,] graph, ``int``[] path,``                       ``int` `pos, ``int` `n)``  ``{``    ``// If all the nodes have been visited, return true``    ``if` `(pos == n + 1)``    ``{``      ``return` `true``;``    ``}` `    ``// Try visiting each node and see if it forms a``    ``// Hamiltonian path``    ``for` `(``int` `v = 1; v < n + 1; v++)``    ``{``      ``if` `(issafe(v, graph, path, pos))``      ``{``        ``path[pos] = v;``        ``if` `(formpath(graph, path, pos + 1, n)``            ``== ``true``)``        ``{``          ``return` `true``;``        ``}``        ``path[pos] = -1;``      ``}``    ``}``    ` `    ``// If none of the nodes form a Hamiltonian path,``    ``// return false``    ``return` `false``;``  ``}` `  ``// Function to find a Hamiltonian path in a given graph``  ``static` `void` `hampath(``int` `n)``  ``{``    ` `    ``// Adjacency matrix to store the graph``    ``int``[,] graph = ``new` `int``[n + 1, n + 1];``    ` `    ``// Array to store the Hamiltonian path``    ``int``[] path = ``new` `int``[n + 1];` `    ``// If there is only 1 node in the graph, there is no``    ``// Hamiltonian path``    ``if` `(n == 1)``    ``{``      ``Console.WriteLine(``"No Solution"``);``      ``return``;``    ``}` `    ``// List to store the squares of the numbers from 1``    ``// to sqrt(2n-1)``    ``List<``int``> l = ``new` `List<``int``>();``    ``int` `nsqrt = (``int``)Math.Sqrt(2 * n - 1);``    ``for` `(``int` `i = 1; i <= nsqrt + 1; i++)``    ``{``      ``l.Add(i * i);``    ``}` `    ``// Initialize the graph to 0``    ``for` `(``int` `i = 1; i < n + 1; i++)``    ``{``      ``for` `(``int` `j = 1; j < n + 1; j++)``      ``{``        ``graph[i, j] = 0;``      ``}``    ``}` `    ``// Form the graph using the squares stored in the``    ``// List 'l'``    ``for` `(``int` `i = 1; i < n + 1; i++)``    ``{``      ``foreach` `(``int` `ele ``in` `l)``      ``{``        ``// Check if the difference between the``        ``// square and the current node is greater``        ``// than 0 and less than or equal to n Also,``        ``// make sure that the difference is not``        ``// equal to 2 times the current node (to``        ``// avoid repeated edges)``        ``if` `((ele - i) > 0 && (ele - i) <= n``            ``&& (2 * i != ele))``        ``{``          ``graph[i, ele - i] = 1;``          ``graph[ele - i, i] = 1;``        ``}``      ``}``    ``}` `    ``for` `(``int` `j = 1; j < n + 1; j++)``    ``{``      ``// Reset the path array``      ``Array.Fill(path, -1);``      ``// Start the path from vertex j``      ``path[1] = j;``      ` `      ``// Check if a Hamiltonian path can be formed``      ``// starting from vertex j``      ``if` `(formpath(graph, path, 2, n) == ``true``)``      ``{``        ``Console.Write(``"Hamiltonian Path: "``);``        ``// Print the path``        ``for` `(``int` `i = 1; i < n + 1; i++)``        ``{``          ``Console.Write(path[i] + ``" "``);``        ``}``        ``// Return the path``        ``return``;``      ``}``    ``}``    ` `    ``// If no solution is found, print "No Solution"``    ``Console.Write(``"No Solution"``);``  ``}``  ``public` `static` `void` `Main(``string``[] args)``  ``{``    ``Console.Write(``"17 -> "``);``    ``hampath(17);``    ``Console.WriteLine();``    ``Console.Write(``"20 -> "``);``    ``hampath(20);``    ``Console.WriteLine();``    ``Console.Write(``"25 -> "``);``    ``hampath(25);``  ``}``}`

## Javascript

 `function` `issafe(v, graph, path, pos) {``    ``if` `(graph[path[pos - 1]][v] === 0) {``        ``return` `false``;``    ``}``    ``for` `(let i = 0; i < pos; i++) {``        ``if` `(path[i] === v) {``            ``return` `false``;``        ``}``    ``}``    ``return` `true``;``}` `function` `formpath(graph, path, pos, n) {``    ``if` `(pos === n + 1) {``        ``return` `true``;``    ``}``    ``for` `(let v = 1; v < n + 1; v++) {``        ``if` `(issafe(v, graph, path, pos)) {``            ``path[pos] = v;``            ``if` `(formpath(graph, path, pos + 1, n)) {``                ``return` `true``;``            ``}``            ``path[pos] = -1;``        ``}``    ``}``    ``return` `false``;``}` `function` `hampath(n) {``    ``const graph = ``new` `Array(n + 1).fill(0).map(() => ``new` `Array(n + 1).fill(0));``    ``const path = ``new` `Array(n + 1).fill(-1);``    ``if` `(n === 1) {``        ``console.log(``"No Solution"``);``        ``return``;``    ``}``    ``const l = [];``    ``const nsqrt = Math.floor(Math.sqrt(2 * n - 1));``    ``for` `(let i = 1; i <= nsqrt + 1; i++) {``        ``l.push(i * i);``    ``}``    ``for` `(let i = 1; i < n + 1; i++) {``        ``for` `(let j = 1; j < n + 1; j++) {``            ``graph[i][j] = 0;``        ``}``    ``}``    ``for` `(let i = 1; i < n + 1; i++) {``        ``for` `(let ele of l) {``            ``if` `((ele - i) > 0 && (ele - i) <= n && (2 * i !== ele)) {``                ``graph[i][ele - i] = 1;``                ``graph[ele - i][i] = 1;``            ``}``        ``}``    ``}` `    ``for` `(let j = 1; j < n + 1; j++) {``        ``path[1] = j;``        ``if` `(formpath(graph, path, 2, n)) {``            ``console.log(``"Hamiltonian Path: "` `+ path.slice(1).join(``" "``));``            ``return``;``        ``}``    ``}``    ``console.log(``"No Solution"``);``}` `console.log(``"17 -> "``);``hampath(17);``console.log();``console.log(``"20 -> "``);``hampath(20);``console.log();``console.log(``"25 -> "``);``hampath(25);`

Output

```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]```

Discussion:

• 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.