Find if an undirected graph contains an independent set of a given size

Given an undirected graph, check if it contains an independent set of size k. Print ‘Yes’ if there exists an independent set of size k. Print ‘No’ otherwise.

Independent Set: An independent set in a graph is a set of vertices which are not directly connected to each other.

Example 1:



Input : K = 4,
        graph = [[1, 1, 0, 0, 0],
                 [1, 1, 1, 1, 1],
                 [0, 1, 1, 0, 0],
                 [0, 1, 0, 1, 0],
                 [0, 1, 0, 0, 1]]
Output : Yes

The above graph contains an independent set of size 4 (vertices 1, 2, 3, 4 are not directly connected to each other). Hence the output is ‘Yes’.

Example 2:

Input : K = 4,
        graph = [[1, 1, 0, 0, 0],
                 [1, 1, 1, 1, 1],
                 [0, 1, 1, 0, 0],
                 [0, 1, 0, 1, 1],
                 [0, 1, 0, 1, 1]]
Output : No

The above graph doesn’t contain an independent set of size 4. Hence output is ‘No’.

Approach:

  • Initialize a variable sol with boolean False value.
  • Find all the possible sets of vertices of size K from the given graph.
  • If an independent set of size k is found, change the value of sol to True and return.
  • Else continue checking for other possible sets.
  • In the end, if sol is True, print ‘Yes’ else print ‘No’.

Below is the implementation of the above approach:

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# Python3 code to check if a given graph
# contains an independent set of size k
  
# Function to construct a set of given size k
def func(graph, arr, k, index, sol):
      
    # Check if the selected set is independent or not. 
    # Change the value of sol to True and return 
    # if it is independent
    if k == 0:
        if check(graph, arr) == True:
            sol[0] = True
            return
          
    else:
        # Set of size k can be formed even if we don't 
        # include the vertex at current index.
        if index >= k:
            return (func(graph, arr[:] + [index], k-1, index-1, sol) or
                    func(graph, arr[:], k, index-1, sol))
          
        # Set of size k cannot be formed if we don't
        # include the vertex at current index.
        else:
            return func(graph, arr[:] + [index], k-1, index-1, sol)
  
# Function to check if the given set is 
# independent or not
# arr --> set of size k (contains the
# index of included vertex)
def check(graph, arr):
      
    # Check if each vertex is connected to any other
    # vertex in the set or not
    for i in range(len(arr)):
        for j in range(i + 1, len(arr)):
              
            if graph[arr[i]][arr[j]] == 1:
                return False
                  
    return True
  
# Driver Code    
graph = [[1, 1, 0, 0, 0],
         [1, 1, 1, 1, 1],
         [0, 1, 1, 0, 0],
         [0, 1, 0, 1, 0],
         [0, 1, 0, 0, 1]]
          
k = 4
arr = []     # Empty set 
sol = [False]
  
func(graph, arr[:], k, len(graph)-1, sol)
  
if sol[0] == True:
    print("Yes")
else:
    print("No")

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Output:

Yes

Time Complexity: O({V\choose k} * k^2) where V is the number of vertices in the graph and k is the given size of set.



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