It was always my impression that a mixed graph was graph where some edges were directional and others non-directional. There are directional and non-directional graphs, so logically, a mixed would contain both?
Also, I always thought bidirectional edges are the SAME as undirected edges, as undirected edges mean you can go either way, there are not restrictions. I think it's just semantics. If you have all your other nodes connected by directional edges, then a bidirectional edge for one exclusive set of two nodes is just logical.
I have the answer for c = 1/2.
Just consider the white edges. As you might know, for every (undirected) graph you can can delete at most half of edges to get a bipartite graph. We'll call the partitions A and B. So there are at least half of the white edges between the group A and B. So if we color all the vertices of which ever group to white, there are at least half of the white edges which is connected to at least one white vertex. So we can choose to paint all the vertices of A to white, or all the vertices of B to white. (We'll choose it later)
Now that we partitioned the vertices into two groups, we want to see which partition should we paint to black, the other partition would be colored to white, and as mentioned, for whites the conditions is held.
The black edges between two partitions, will be connected to at least one black vertex whichever way. Now see the black edges inside the partitions. If black edges in partition A is more, we'll color all the vertices of the A to black. And if black edges in partition B is more, we'll do otherwise.
You can see for white and black the conditions is held.
Proof for the theorem used in this solution: Given a graph, partition it into 2 groups whichever way. If there was a vertex in one group, which more than half of its neighbors are in its group, move the vertex to the other partition. You can see this operation will end, and when ended, for each vertex at least half of its neighbors is in the other group. So at least half of the edges is in-between the partitions.
Update: In this proof, by neighbors we mean the number of edges connected to one vertex.
Best Answer
In the first graph, no more edges can be added, because the pairs $(v_1,v_2)$ and $(v_1,v_4)$ have degree sum $3$, and every other pair is already adjacent. So this graph is its own closure.
For the second graph, we can add edges $v_1v_3$ and $v_2v_4$ immediately, since for both pairs the degree sum is $4$. Now the degree sum of $v_1,v_4$ is $5$, so we can add this edge too (note that we could not add this edge initially). There are no more non-adjacent pairs, so we've reached the closure.
The point of this definition is that a simple graph is Hamiltonian if and only if its closure is Hamiltonian - this is the Bondy-Chvátal theorem.