title: Is O(V+E) equivalent to O(V^2)?
body: <p>My question is about whether <code>O(V+E) = O(V^2)</code>.</p>
  
  <p>Basically, if <code>O(V+E)</code> is linear time such that <code>V+E = n</code>, wouldn't <code>O(V^2)</code> also be linear time? </p>
  
  <p>I assume the worst-case/upper bound for <code>O(V+E)</code> is an edge between each vertex, which would result in <code>(V-1)^2</code> edges. I also assumed that that can be considered <code>V^2</code>, so I would think that that would be equivalent to <code>O(V^2)</code>.</p>