Prufer Code to Tree Creation
What is Prufer Code?
Given a tree (represented as graph, not as a rooted tree) with n labeled nodes with labels from 1 to n, a Prufer code uniquely identifies the tree. The sequence has n-2 values.
How to get Prufer Code of a tree?
- Initialize Prufer code as empty.
- Start with a leaf of lowest label say x. Find the vertex connecting it to the rest of tree say y. Remove x from the tree and add y to the Prufer Code
- Repeat above step 2 until we are left with two nodes.
A tree with labels from 1 to n. 5 / \ 1 4 / \ 2 3 PruferCode = {} The lowest label leaf is 2, we remove it from tree and add the other vertex (connecting it to the tree) to Prufer code Tree now becomes 5 / \ 1 4 \ 3 Prufer Code becomes = {1} The lowest label leaf is 3, we remove it from tree and add the other vertex (connecting it to the tree) to Prufer code Tree now becomes 5 / \ 1 4 Prufer Code becomes = {1, 1} The lowest label leaf is 1, we remove it from tree and add the other vertex (connecting it to the tree) to Prufer code Tree now becomes 5 \ 4 Prufer Code becomes = {1, 1, 5} We have only two nodes left now, so we stop.
How to construct a tree from given Prufer Code?
Input : (4, 1, 3, 4) Output : Edges of following tree 2----4----3----1----5 | 6 Input : (1, 3, 5) Output : Edges of following tree 2----1----3----5----4
Let the length of given Prufer code be m. The idea is to create an empty graph of m+2 vertices. We remove first element from sequence. Let first element of current sequence be x. Then we find the least value which is not present in the given sequence and not yet added to the tree. Let this value be y. We add an edge from x to y and repeat this step.
Let us understand algorithm to construct tree with above first example:
Input : (4, 1, 3, 4) Step 1: First we create an empty graph of 6 vertices and get 4 from the sequence. Step 2: Out of 1 to 6, the least vertex not in Prufer sequence is 2. Step 3: We form an edge between 2 and 4. 2----4 1 3 5 6 Step 4: Next in the sequence is 1 and corresponding vertex with least degree is 5 (as 2 has been considered). 2----4 1----5 3 6 Step 5: Next in the sequence is 3 and corresponding vertex with least degree is 1 (as 1 is now not part of remaining Prufer sequence) 2----4 3----1----5 6 Step 6: Next in the sequence is 4 and corresponding vertex with least degree is 3 (as 3 has not been considered as is not present further in sequence) 2----4----3----1----5 6 Step 7: Finally two vertices are left out from 1 to 6 (4 and 6) so we join them. 2----4----3----1----5 | 6 This is the required tree on 6 vertices.
Following is the implementation.
C++
// C++ program to construct tree from given Prufer Code #include <bits/stdc++.h> using namespace std; // Prints edges of tree represented by given Prufer code void printTreeEdges( int prufer[], int m) { int vertices = m + 2; int vertex_set[vertices]; // Initialize the array of vertices for ( int i = 0; i < vertices; i++) vertex_set[i] = 0; // Number of occurrences of vertex in code for ( int i = 0; i < vertices - 2; i++) vertex_set[prufer[i] - 1] += 1; cout << "\nThe edge set E(G) is :\n" ; // Find the smallest label not present in // prufer[]. int j = 0; for ( int i = 0; i < vertices - 2; i++) { for (j = 0; j < vertices; j++) { // If j+1 is not present in prufer set if (vertex_set[j] == 0) { // Remove from Prufer set and print // pair. vertex_set[j] = -1; cout << "(" << (j + 1) << ", " << prufer[i] << ") " ; vertex_set[prufer[i] - 1]--; break ; } } } j = 0; // For the last element for ( int i = 0; i < vertices; i++) { if (vertex_set[i] == 0 && j == 0) { cout << "(" << (i + 1) << ", " ; j++; } else if (vertex_set[i] == 0 && j == 1) cout << (i + 1) << ")\n" ; } } // Driver code int main() { int prufer[] = { 4, 1, 3, 4 }; int n = sizeof (prufer) / sizeof (prufer[0]); printTreeEdges(prufer, n); return 0; } |
Java
// Java program to construct tree from given Prufer Code class GFG { // Prints edges of tree represented by given Prufer code static void printTreeEdges( int prufer[], int m) { int vertices = m + 2 ; int vertex_set[] = new int [vertices]; // Initialize the array of vertices for ( int i = 0 ; i < vertices; i++) vertex_set[i] = 0 ; // Number of occurrences of vertex in code for ( int i = 0 ; i < vertices - 2 ; i++) vertex_set[prufer[i] - 1 ] += 1 ; System.out.print( "\nThe edge set E(G) is :\n" ); // Find the smallest label not present in // prufer[]. int j = 0 ; for ( int i = 0 ; i < vertices - 2 ; i++) { for (j = 0 ; j < vertices; j++) { // If j+1 is not present in prufer set if (vertex_set[j] == 0 ) { // Remove from Prufer set and print // pair. vertex_set[j] = - 1 ; System.out.print( "(" + (j + 1 ) + ", " + prufer[i] + ") " ); vertex_set[prufer[i] - 1 ]--; break ; } } } j = 0 ; // For the last element for ( int i = 0 ; i < vertices; i++) { if (vertex_set[i] == 0 && j == 0 ) { System.out.print( "(" + (i + 1 ) + ", " ); j++; } else if (vertex_set[i] == 0 && j == 1 ) System.out.print((i + 1 ) + ")\n" ); } } // Driver code public static void main(String args[]) { int prufer[] = { 4 , 1 , 3 , 4 }; int n = prufer.length; printTreeEdges(prufer, n); } } // This code is contributed by Arnab Kundu |
Python3
# Python3 program to construct # tree from given Prufer Code # Prints edges of tree represented # by given Prufer code def printTreeEdges(prufer, m): vertices = m + 2 # Initialize the array of vertices vertex_set = [ 0 ] * vertices # Number of occurrences of vertex in code for i in range (vertices - 2 ): vertex_set[prufer[i] - 1 ] + = 1 print ( "The edge set E(G) is :" ) # Find the smallest label not present in # prufer. j = 0 for i in range (vertices - 2 ): for j in range (vertices): # If j+1 is not present in prufer set if (vertex_set[j] = = 0 ): # Remove from Prufer set and print # pair. vertex_set[j] = - 1 print ( "(" , (j + 1 ), ", " ,prufer[i], ") " ,sep = " ",end = " ") vertex_set[prufer[i] - 1 ] - = 1 break j = 0 # For the last element for i in range (vertices): if (vertex_set[i] = = 0 and j = = 0 ): print ( "(" , (i + 1 ), ", " , sep = " ", end=" ") j + = 1 else if (vertex_set[i] = = 0 and j = = 1 ): print ((i + 1 ), ")" ) # Driver code prufer = [ 4 , 1 , 3 , 4 ] n = len (prufer) printTreeEdges(prufer, n) # This code is contributed by SHUBHAMSINGH10 |
C#
// C# program to construct tree from given Prufer Code using System; class GFG { // Prints edges of tree represented by given Prufer code static void printTreeEdges( int [] prufer, int m) { int vertices = m + 2; int [] vertex_set = new int [vertices]; // Initialize the array of vertices for ( int i = 0; i < vertices; i++) vertex_set[i] = 0; // Number of occurrences of vertex in code for ( int i = 0; i < vertices - 2; i++) vertex_set[prufer[i] - 1] += 1; Console.Write( "\nThe edge set E(G) is :\n" ); // Find the smallest label not present in // prufer[]. int j = 0; for ( int i = 0; i < vertices - 2; i++) { for (j = 0; j < vertices; j++) { // If j+1 is not present in prufer set if (vertex_set[j] == 0) { // Remove from Prufer set and print // pair. vertex_set[j] = -1; Console.Write( "(" + (j + 1) + ", " + prufer[i] + ") " ); vertex_set[prufer[i] - 1]--; break ; } } } j = 0; // For the last element for ( int i = 0; i < vertices; i++) { if (vertex_set[i] == 0 && j == 0) { Console.Write( "(" + (i + 1) + ", " ); j++; } else if (vertex_set[i] == 0 && j == 1) Console.Write((i + 1) + ")\n" ); } } // Driver code public static void Main(String[] args) { int [] prufer = { 4, 1, 3, 4 }; int n = prufer.Length; printTreeEdges(prufer, n); } } // This code has been contributed by 29AjayKumar |
Javascript
<script> // JavaScript program to construct tree from given Prufer Code // Prints edges of tree represented by given Prufer code function printTreeEdges(prufer,m) { let vertices = m + 2; let vertex_set = new Array(vertices); // Initialize the array of vertices for (let i = 0; i < vertices; i++) vertex_set[i] = 0; // Number of occurrences of vertex in code for (let i = 0; i < vertices - 2; i++) vertex_set[prufer[i] - 1] += 1; document.write( "<br>The edge set E(G) is :<br>" ); // Find the smallest label not present in // prufer[]. let j = 0; for (let i = 0; i < vertices - 2; i++) { for (j = 0; j < vertices; j++) { // If j+1 is not present in prufer set if (vertex_set[j] == 0) { // Remove from Prufer set and print // pair. vertex_set[j] = -1; document.write( "(" + (j + 1) + ", " + prufer[i] + ") " ); vertex_set[prufer[i] - 1]--; break ; } } } j = 0; // For the last element for (let i = 0; i < vertices; i++) { if (vertex_set[i] == 0 && j == 0) { document.write( "(" + (i + 1) + ", " ); j++; } else if (vertex_set[i] == 0 && j == 1) document.write((i + 1) + ")\n" ); } } // Driver code let prufer=[4, 1, 3, 4]; let n = prufer.length; printTreeEdges(prufer, n); // This code is contributed by rag2127 </script> |
The edge set E(G) is : (2, 4) (5, 1) (1, 3) (3, 4) (4, 6)
Time Complexity: O(n2) where n is the number of vertices in the tree.
Auxiliary Space: O(n)
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