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

Suppose we have a given undirected graph; we have to check whether it contains an independent set of size l. If there is any independent set of size l then return Yes otherwise No.

We have to keep in mind that an independent set in a graph is defined as a set of vertices which are not directly connected to each other.

So, if the input is like L = 4,

then the output will be yes

To solve this, we will follow these steps −

• Define a function is_valid() . This will take graph, arr

• for i in range 0 to size of arr, do

  • for j in range i + 1 to size of arr, do

    • if graph[arr[i], arr[j]] is same as 1, then

      • return False

• return True

• Define a function solve() . This will take graph, arr, k, index, sol

• if k is same as 0, then

  • if is_valid(graph, arr) is same as True, then

    • sol[0] := True

    • return

  • otherwise,

    • if index >= k, then

      • return(solve(graph, arr[from index 0 to end] concatenate a list with [index], k-1, index-1, sol) or solve(graph, arr[from index 0 to end], k, index-1, sol))

    • otherwise,

      • return solve(graph, arr[from index 0 to end] concatenate a list with [index], k-1, index-1, sol)

Example

Let us see the following implementation to get better understanding −

 Live Demo

def is_valid(graph, arr):
   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
def solve(graph, arr, k, index, sol):
   if k == 0:
      if is_valid(graph, arr) == True:
         sol[0] = True
         return
      else:
         if index >= k:
            return (solve(graph, arr[:] + [index], k-1, index-1, sol) or solve(graph, arr[:], k, index-1, sol))
         else:
            return solve(graph, arr[:] + [index], k-1, index-1, sol)

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 = []
sol = [False]
solve(graph, arr[:], k, len(graph)-1, sol)
if sol[0]:
   print("Yes")
else:
   print("No")

Input

[[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]],
4

Output

Yes