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@@ -121,6 +121,13 @@ s.t.&   &   \sum_i D_i d(s_i,t_i)&=1    &   &\\
 \item \ip{} $\geq$ \lp{}. Given any feasible solution to \ip{}, we can scale all $x_e$ and $y_i$ simultaneously with factor $1/\sum_i D_i y_i$. The scaled solution is feasible for \lp{} and gets the same objective value.
 \item \lp{} $=$ \dual{}. by duality.
 \item \metric{} $=$ \lp{}. It is easy to see \metric{} $\geq$ \lp{} since any feasible metric to \metric{} induces a feasible solution to \lp{}. In fact, the optimal solution to \lp{} also induces a feasible metric. Consider a solution $x_e,y_i$ to \lp{}. Let $d_x$ be the shortest path metric on $V$ using edge length $x_e$. It suffices to show that $y_i=d_x(s_i,t_i)$. This can be seen from a reformulation of \lp{}. The constraint $\sum_i D_i y_i=1$ can be removed and the objective becomes $\sum_e c_e x_e / \sum_i D_i y_i$. This reformulation does not change the optimal solution. Now suppose in the optimal solution to \lp{} there is some $y_i$ which is strictly smaller than $d_x(s_i,t_i)$. Then the denominator $\sum_i D_i y_i$ in the objective of our reformulation can be larger, contradicting to the optimality of solution $x_e,y_i$.
+
+\begin{theorem}[Japanese Theorem]
+$D$ is a demand matrix. $D$ is routable in $G$ iff $\forall l:E\to \R^+$, $\sum_e c_e l(e)\geq \sum_{uv} D(u,v) d_l(u,v)$, where $d_l(s,t)$ is the short path distance based on $l$.
+\end{theorem}
+\begin{proof}
+$D$ is routable iff $\opt(\dual{})\geq 1$
+\end{proof}
 \end{enumerate}
 
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