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    <blockquote> <p>Is this correct, or am I missing something obvious?</p> </blockquote> <p>Two vertices having the same edge; e.g.,...</p> <pre><code>A -- B -- C Weights: B = 2; A = 1; C = 1 </code></pre> <p><code>{ A, C }</code> and <code>{ B }</code> are both weighted minimal vertex covers by your definition.</p> <p>Only <code>{ B }</code> is a standard minimal vertex cover.</p> <p><strong>EDIT:</strong>... a better example showing a different reason:</p> <pre><code>A -- B -- C -- D Weights: B = 2; C = 2; A = 1; D = 1 </code></pre> <p><code>{ A, C }</code>, <code>{ B, D }</code>, <code>{ B, C }</code> are all standard minimal vertex covers.</p> <p>Only <code>{ A, C }</code> and <code>{ B, D }</code> are weighted minimal vertex covers by your definition. Intuitively, this is because the <code>{ B, C }</code> vertex cover counts the <code>B -- C</code> edge twice.</p> <hr> <p>The first counter example disproves that all weighted MVCs (as per your definition) are standard MVCs. The second counter example disproves that all standard MVCs are weighted MVCs.</p> <p>After some thought... you are correct in that the weighted MVC for a tree is any VC with a cost equal to the number of edges. </p> <p>Finding a weighted MVC is actually quite simple. If you draw the tree out, and pick all the vertexes from every second level (doesn't matter if you start from the first or second level), you end up with a valid weighted MVC by your definition (since all edges are covered, no edge is counted twice).</p> <p>...more generally, the set of all weighted MVCs is the set of all VCs not containing neighbours. For example, in a tree where no child is a parent, the set of leaf nodes is also a valid weighted MVC.</p>
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    1. COI don't see how these are counter examples. I understand that in the first case, only `{B}` is a minimal vertex cover, but we're not going for the minimal VC, we are going for the minimal weight VC and by that definition, `{A, C}` is a valid answer. However, if we ran the algorithm for just the minimal VC, we would get `{B}` and that would also be a minimal weight VC.
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    2. COFor the second example, I don't see why the fact that the `B -- C` edge is counted twice makes it an incorrect answer. The definition of a VC is that each edge should have `atleast` one vertex in the cover.
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    3. CO@efficiencyIsBliss Erm... *"I understand that in the first case, only `{B}` is a minimal vertex cover, but we're not going for the minimal VC, we are going for the minimal weight VC and by that definition, `{A, C}` is a valid answer."* That's *exactly* what my answer says. That's exactly why it's a counter-example to the claim in your question *"Therefore, we don't need any special algorithm to find the answer, we only need to find the minimum size vertex cover (cover with minimum vertices)."*
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