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I have recently come across a result that showed that a given problem is in FPT when parameterized by the treewidth of a graph. However, they did this by showing that the problem is in FPT when parameterized by the treewidth of an augmented version of the graph that had treewidth of at most 2k+1 when the treewidth of the original graph is k.

Why is it possible do this?

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Let $k$ be the parameter. Is an algorithm has a running time of the form $O(f(2k+1) \cdot \text{poly}(n))$ for a suitable function $f$, then it also has a running time of $O(g(k) \cdot \text{poly}(n) )$ for $g(k) = f(2k+1)$.

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