complexity
All checks were successful
build pdf / build (push) Successful in 4s

This commit is contained in:
2025-09-29 15:32:40 +08:00
parent a276cd658f
commit 9aedda643f
2 changed files with 37 additions and 4 deletions

View File

@@ -356,11 +356,16 @@ for each 2-approx mincut $C$ in $(G,w_\lambda)$:\\
return the optimal $(C,F)$
\end{algo}
\paragraph{time for $\lambda^*$} $L(\lambda)-b\lambda$ is pwl concave. The number of segments is at most $3^m$. We need almost linear time to find the solution to a fixed $\lambda$. So parametric seach gives complexity $m^{1+o(1)} O(\log 3^m)$.
\note{need to check this}
To compute $\lambda^*$ we need to use parametric search.
\begin{lemma}[\cite{salowe_parametric}]
Let $S(n)$ be the complexity of solving the Lagrangian dual problem for fixed $\lambda$ (where $n$ is the size of the input), then one can compute $\lambda^*$ using parametric search in $O(S(n)^2)$ time.
\end{lemma}
It follows directly from the preceding lemma that $\lambda^*$ can be computed in $\tilde O(m^2)$ time.
\paragraph{time for the rest parts} Reweighting takes linear time.
Finding $<2$-approx mincut takes $\tilde O(n^3)$. FPTAS for knapsack takes $O(\frac{1}{\e}m^2)$. The total complexity is $O(\frac{1}{\e}m^2n^3)$.
Reweighting the graph takes linear time.
Finding $<2$-approx mincut takes $\tilde O(n^3)$.
An $1+\e$ approximate solution to knapsack can be found in time $\tilde O(m+\frac{1}{\e^2})$ \cite{10.1145/3618260.3649730}.
The total complexity is $\tilde O(mn^3+\frac{n^3}{\e^2})$.
\bibliographystyle{plain}
\bibliography{ref}