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Possible edges of a tree for given diameter, height and vertices

  • Difficulty Level : Hard
  • Last Updated : 14 May, 2021
Geek Week

Find a tree with the given values and print the edges of the tree. Print “-1”, if the tree is not possible.

Given three integers n, d and h. 

n -> Number of vertices. [1, n]
d -> Diameter of the tree (largest 
   distance between two vertices).
h -> Height of the tree (longest distance
   between vertex 1 and another vertex)

Examples : 

Input : n = 5, d = 3, h = 2 
Output : 1 2
         2 3
         1 4
         1 5
Explanation :  

We can see that the height of the tree is 2 (1 -> 
2 --> 5) and diameter is 3 ( 3 -> 2 -> 1 -> 5).
So our conditions are satisfied.

Input :  n = 8, d = 4, h = 2
Output : 1 2
         2 3
         1 4
         4 5
         1 6
         1 7
         1 8
Explanation :



  1. Observe that when d = 1, we cannot construct a tree (if tree has more than 2 vertices). Also when d > 2*h, we cannot construct a tree.
  2. As we know that height is the longest path from vertex 1 to another vertex. So build that path from vertex 1 by adding edges up to h. Now, if d > h, we should add another path to satisfy diameter from vertex 1, with a length of d – h.
  3. Our conditions for height and diameter are satisfied. But still some vertices may be left. Add the remaining vertices at any vertex other than the end points. This step will not alter our diameter and height. Chose vertex 1 to add the remaining vertices (you can chose any).
  4. But when d == h, choose vertex 2 to add the remaining vertices.

C++




// C++ program to construct tree for given count
// width and height.
#include <bits/stdc++.h>
using namespace std;
 
// Function to construct the tree
void constructTree(int n, int d, int h)
{
    if (d == 1) {
 
        // Special case when d == 2, only one edge
        if (n == 2 && h == 1) {
            cout << "1 2" << endl;
            return;
        }
        cout << "-1" << endl; // Tree is not possible
        return;
    }
 
    if (d > 2 * h) {
        cout << "-1" << endl;
        return;
    }
 
    // Satisfy the height condition by add
    // edges up to h
    for (int i = 1; i <= h; i++)    
        cout << i << " " << i + 1 << endl;
     
    if (d > h) {
 
        // Add d - h edges from 1 to
        // satisfy diameter condition
        cout << "1"
            << " " << h + 2 << endl;
        for (int i = h + 2; i <= d; i++) {
            cout << i << " " << i + 1 << endl;
        }
    }
 
    // Remaining edges at vertex 1 or 2(d == h)
    for (int i = d + 1; i < n; i++)
    {
        int k = 1;
        if (d == h)
            k = 2;
        cout << k << " " << i + 1 << endl;
    }
}
 
// Driver Code
int main()
{
    int n = 5, d = 3, h = 2;
    constructTree(n, d, h);
    return 0;
}

Java




// Java program to construct tree for given count
// width and height.
class GfG {
 
// Function to construct the tree
static void constructTree(int n, int d, int h)
{
    if (d == 1) {
 
        // Special case when d == 2, only one edge
        if (n == 2 && h == 1) {
            System.out.println("1 2");
            return;
        }
        System.out.println("-1"); // Tree is not possible
        return;
    }
 
    if (d > 2 * h) {
        System.out.println("-1");
        return;
    }
 
    // Satisfy the height condition by add
    // edges up to h
    for (int i = 1; i <= h; i++)    
        System.out.println(i + " " + (i + 1));
     
    if (d > h) {
 
        // Add d - h edges from 1 to
        // satisfy diameter condition
        System.out.println("1" + " " + (h + 2));
        for (int i = h + 2; i <= d; i++) {
            System.out.println(i + " " + (i + 1));
        }
    }
 
    // Remaining edges at vertex 1 or 2(d == h)
    for (int i = d + 1; i < n; i++)
    {
        int k = 1;
        if (d == h)
            k = 2;
        System.out.println(k + " " + (i + 1));
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 5, d = 3, h = 2;
    constructTree(n, d, h);
}
}

Python3




# Python3 code to construct tree for given count
# width and height.
 
# Function to construct the tree
def constructTree(n, d, h):
    if d == 1:
 
        # Special case when d == 2, only one edge
        if n == 2 and h == 1:
            print("1 2")
            return 0
         
        print("-1")    # Tree is not possible
        return 0
     
    if d > 2 * h:
        print("-1")
        return 0
         
    # Satisfy the height condition by add
    # edges up to h
    for i in range(1, h+1):
        print(i," " , i + 1)
     
    if d > h:
 
        # Add d - h edges from 1 to
        # satisfy diameter condition
        print(1,"  ", h + 2)
        for i in range(h+2, d+1):
            print(i, " " , i + 1)
             
    # Remaining edges at vertex 1 or 2(d == h)
    for i in range(d+1, n):
        k = 1
        if d == h:
            k = 2
        print(k ," " , i + 1)
 
# Driver Code
n = 5
d = 3
h = 2
constructTree(n, d, h)
 
# This code is contributed by "Sharad_Bhardwaj".

C#




// C# program to construct tree for 
// given count width and height.
using System;
 
class GfG
{
 
    // Function to construct the tree
    static void constructTree(int n, int d, int h)
    {
        if (d == 1)
        {
 
            // Special case when d == 2,
            // only one edge
            if (n == 2 && h == 1)
            {
                Console.WriteLine("1 2");
                return;
            }
             
            // Tree is not possible
            Console.WriteLine("-1");
            return;
        }
 
        if (d > 2 * h)
        {
            Console.WriteLine("-1");
            return;
        }
 
        // Satisfy the height condition
        // by add edges up to h
        for (int i = 1; i <= h; i++)
            Console.WriteLine(i + " " + (i + 1));
 
        if (d > h)
        {
 
            // Add d - h edges from 1 to
            // satisfy diameter condition
            Console.WriteLine("1" + " " + (h + 2));
            for (int i = h + 2; i <= d; i++)
            {
                Console.WriteLine(i + " " + (i + 1));
            }
        }
 
        // Remaining edges at vertex 1 or 2(d == h)
        for (int i = d + 1; i < n; i++)
        {
            int k = 1;
            if (d == h)
                k = 2;
            Console.WriteLine(k + " " + (i + 1));
        }
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int n = 5, d = 3, h = 2;
        constructTree(n, d, h);
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// Javascript program to construct tree for
// given count width and height.
 
// Function to construct the tree
function constructTree(n, d, h)
{
    if (d == 1)
    {
         
        // Special case when d == 2,
        // only one edge
        if (n == 2 && h == 1)
        {
            document.write("1 2", "<br>");
            return;
        }
           
        // Tree is not possible
        document.write("-1", "<br>");
        return;
    }
 
    if (d > 2 * h)
    {
        document.write("-1", "<br>");
        return;
    }
 
    // Satisfy the height condition
    // by add edges up to h
    for(var i = 1; i <= h; i++)
        document.write(i + " " + (i + 1), "<br>");
 
    if (d > h)
    {
         
        // Add d - h edges from 1 to
        // satisfy diameter condition
        document.write("1" + " " + (h + 2), "<br>");
        for(var i = h + 2; i <= d; i++)
        {
            document.write(i + " " + (i + 1), "<br>");
        }
    }
 
    // Remaining edges at vertex 1 or 2(d == h)
    for(var i = d + 1; i < n; i++)
    {
        var k = 1;
        if (d == h)
            k = 2;
             
        document.write(k + " " + (i + 1), "<br>");
    }
}
 
// Driver Code
var n = 5, d = 3, h = 2;
 
constructTree(n, d, h);
 
// This code is contributed by bunnyram19
 
</script>

Output : 

1 2
2 3
1 4
1 5

 

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