However, it can be solved more efficiently than the O(n2 2n) time that would be given by a naive brute force algorithm that examines every vertex subset and checks whether it is an independent set.
Međutim, on može biti rešen efikasnije od O(n2 2n)što bi bila složenost naivnog algoritma grube sile koji ispituje svaki podskup čvorova i proverava da li je on nezavisan skup.