Find if a degree sequence can form a simple graph | Havel-Hakimi Algorithm
Given a sequence of non-negative integers arr[], the task is to check if there exists a simple graph corresponding to this degree sequence. Note that a simple graph is a graph with no self-loops and parallel edges.
Examples:
Input: arr[] = {3, 3, 3, 3}
Output: Yes
This is actually a complete graph()
Input: arr[] = {3, 2, 1, 0}
Output: No
A vertex has degree n-1 so it’s connected to all the other n-1 vertices.
But another vertex has degree 0 i.e. isolated. It’s a contradiction.
Approach: One way to check the existence of a simple graph is by Havel-Hakimi algorithm given below:
- Sort the sequence of non-negative integers in non-increasing order.
- Delete the first element(say V). Subtract 1 from the next V elements.
- Repeat 1 and 2 until one of the stopping conditions is met.
Stopping conditions:
- All the elements remaining are equal to 0 (Simple graph exists).
- Negative number encounter after subtraction (No simple graph exists).
- Not enough elements remaining for the subtraction step (No simple graph exists).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function that returns true if // a simple graph exists bool graphExists(vector< int > &a, int n) { // Keep performing the operations until one // of the stopping condition is met while (1) { // Sort the list in non-decreasing order sort(a.begin(), a.end(), greater<>()); // Check if all the elements are equal to 0 if (a[0] == 0 and a[a.size() - 1] == 0) return true ; // Store the first element in a variable // and delete it from the list int v = a[0]; a.erase(a.begin() + 0); // Check if enough elements // are present in the list if (v > a.size()) return false ; // Subtract first element from next v elements for ( int i = 0; i < v; i++) { a[i]--; // Check if negative element is // encountered after subtraction if (a[i] < 0) return false ; } } } // Driver Code int main() { vector< int > a = {3, 3, 3, 3}; int n = a.size(); graphExists(a, n) ? cout << "Yes" : cout << "NO" << endl; return 0; } // This code is contributed by // sanjeev2552 |
Python3
# Python implementation of the approach # Function that returns true if # a simple graph exists def graphExists(a): # Keep performing the operations until one # of the stopping condition is met while True : # Sort the list in non-decreasing order a = sorted (a, reverse = True ) # Check if all the elements are equal to 0 if a[ 0 ] = = 0 and a[ len (a) - 1 ] = = 0 : return True # Store the first element in a variable # and delete it from the list v = a[ 0 ] a = a[ 1 :] # Check if enough elements # are present in the list if v> len (a): return False # Subtract first element from next v elements for i in range (v): a[i] - = 1 # Check if negative element is # encountered after subtraction if a[i]< 0 : return False # Driver code a = [ 3 , 3 , 3 , 3 ] if (graphExists(a)): print ( "Yes" ) else : print ( "No" ) |
Yes
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Improved By : sanjeev2552