is(V, E, k)
{
count = 0;
for each v in V {
choose x[v] in { 0 , 1 };
count = count + x[v];
}
if (count < k) {
failure;
}

for each u in V such that x[u] == 1 {
for each v in V such that x[v] == 1 {
if muchia {u,v} apartine E {
failure;
}
}
}
success;
}