Problem with graph We have the following graph: 
 
I want to find a minimal vertex cover. 
I thought that the set $C^\star=\{1,2,3,7\}$ is a minimal vertx cover. Is this correct? How could we prove it? 
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Then I want to find a vertex cover using the following approximation algorithm 
C <- 0 
while E ≠ 0 do 
    choose e = (u,v) ∈ E arbitrary 
    C <- C U {u,v} 
    G <- G - {u,v} 
return C

I have done the following: 
Suppose we start with e=(1,4) then C={1,4} and the graph is the following: 
 
Then when we choose e=(5,7) we have C={1,4,5,7} and the graph will look as follows: 
 
Then we can choose e=(2,3) and then we have C={1,2,3,4,5,7} and the graph will be: 
 
Have we finished now? 
Is a minimal vertex cover the set $C=\{1,2,3,4,5,7\}$ ? 
Does this mean that no matter which edge we choose at each step the minimal vertex cover will contain $6$ vertices? 
$$$$  
After that I want to find a maximal independent set of vertices and a maximal clique. Could you give me some hints how we could find them? 
 A: 
I thought that the set $C^⋆=\{1,2,3,7\}$ is a minimal vertex cover. Is this correct? How could we prove it?

No, because it does not cover all edges (in particular, none of the edges $\{4, 5\}, \{5, 6\}, \{4, 6\}$ are covered by any vertex in $C^*$).

Have we finished now? Is a minimal vertex cover the set $C=\{1,2,3,4,5,7\}$? Does this mean that no matter which edge we choose at each step the minimal vertex cover will contain $6$ vertices?

Yes, you've correctly implemented the greedy $2$-approximation algorithm. It is not minimal though: observe that $C$ has a subset $\{1, 3, 4, 5, 7\}$, which is still a vertex cover. So this greedy algorithm is not guaranteed to be optimal.

After that I want to find a maximal independent set of vertices and a maximal clique. Could you give me some hints how we could find them?

As a general heuristic, try a modified greedy algorithm. When picking vertices, don't just pick any random one; try picking the one with the lowest (or highest) degree. Then remove it (and anything else that we know is safe to remove) from the graph and repeat.
