From 26536bcef3965d03c18e1715f88ca8a7c79bf759 Mon Sep 17 00:00:00 2001
From: Yuga SUDA <suda-yuga@g.ecc.u-tokyo.ac.jp>
Date: Mon, 19 Jan 2026 10:53:33 +0900
Subject: [PATCH] [match_transport] Fix notation typos in OT constraints and
 dual update

---
 lectures/match_transport.md | 24 ++++++++++++------------
 1 file changed, 12 insertions(+), 12 deletions(-)

diff --git a/lectures/match_transport.md b/lectures/match_transport.md
index 6743d2fc..e614467c 100644
--- a/lectures/match_transport.md
+++ b/lectures/match_transport.md
@@ -68,8 +68,8 @@ Given a **cost function** $c \colon X \times Y \rightarrow \mathbb{R}$, the (dis
 $$
     \begin{aligned}
         \min_{\mu \geq 0}& \sum_{(x,y) \in X \times Y} \mu_{xy}c_{xy} \\
-        \text{s.t. }& \sum_{x \in X} \mu_{xy} = n_x \\
-        & \sum_{y \in Y} \mu_{xy} = m_y 
+        \text{s.t. }& \sum_{y \in Y} \mu_{xy} = n_x \\
+        & \sum_{x \in X} \mu_{xy} = m_y 
     \end{aligned}
 $$
 
@@ -110,8 +110,8 @@ Hence, our problem is
 $$
 \begin{aligned}
 \min_{\mu \in \mathbb{Z}_+^{X \times Y}}& \sum_{(x,y) \in X \times Y} \mu_{xy}|x-y|^{1/\zeta} \\
-\text{s.t. }& \sum_{x \in X} \mu_{xy} = n_x \\
-& \sum_{y \in Y} \mu_{xy} = m_y 
+\text{s.t. }& \sum_{y \in Y} \mu_{xy} = n_x \\
+& \sum_{x \in X} \mu_{xy} = m_y 
 \end{aligned}
 $$
 
@@ -495,12 +495,12 @@ $$
 |x - y'| + |x' - y| = |x - y| + |x' - y'|
 $$
 
-Letting $\alpha := \frac{|x - y|+|x' - y|}{|x - y'| - |x' - y|} \in (0,1),$ we have $|x - y| = \alpha|x - y'| +(1-\alpha) |x' - y| $ and $|x' - y'| = (1-\alpha)|x - y'| +\alpha |x' - y|. $ 
+Letting $\alpha := \frac{|x - y| - |x - y'|}{|x - y| - |x' - y'|} \in (0,1),$ we have $|x - y'| = \alpha|x' - y'| +(1-\alpha) |x - y|$ and $|x' - y| = (1-\alpha)|x' - y'| +\alpha |x - y|. $ 
 
 Hence, by strict concavity of $h,$
 
 $$
-h(|x-y|)+ h(|x'-y'|) <\alpha h(|x - y'|) +(1-\alpha) h(|x' - y|) +  (1-\alpha) h(|x - y'|) +\alpha h(|x' - y|) = h(|x-y'|) + h(|x'-y|).
+h(|x-y'|)+ h(|x'-y|) > \alpha h(|x' - y'|) +(1-\alpha) h(|x - y|) +  (1-\alpha) h(|x' - y'|) +\alpha h(|x - y|) = h(|x'-y'|) + h(|x-y|).
 $$
 
 Therefore, as in the first case, we can strictly improve the cost among $x,y,x',y'$ by uncrossing the pairs.
@@ -1666,8 +1666,8 @@ Let's recall the formulation
 $$
 \begin{aligned}
 V_P = \min_{\mu \geq 0}& \sum_{(x,y) \in X \times Y} \mu_{xy}c_{xy} \\
-\text{s.t. }& \sum_{x \in X} \mu_{xy} = n_x \\
-& \sum_{y \in Y} \mu_{xy} = m_y 
+\text{s.t. }& \sum_{y \in Y} \mu_{xy} = n_x \\
+& \sum_{x \in X} \mu_{xy} = m_y 
 \end{aligned}
 $$
 
@@ -1695,8 +1695,8 @@ Then we  can formulate the following problem and its dual
 $$
  \begin{aligned}
 W_P = \max_{\mu \geq 0}& \sum_{(x,y) \in X \times Y} \mu_{xy}y_{xy} \\
-\text{s.t. }& \sum_{x \in X} \mu_{xy} = n_x \\
-& \sum_{y \in Y} \mu_{xy} = m_y 
+\text{s.t. }& \sum_{y \in Y} \mu_{xy} = n_x \\
+& \sum_{x \in X} \mu_{xy} = m_y 
 \end{aligned}
 $$
 
@@ -1941,7 +1941,7 @@ exam_assign_OD.plot_hierarchies(subpairs)
 
 We proceed to describe and implement the algorithm to compute the dual solution. 
 
-As already mentioned, the algorithm starts from the matched pairs $(x_0,y_0)$ with no subpairs and assigns the (temporary) values $\psi_{x_0} = c_{x_0 y_0}$ and $\psi_{y_0} = 0,$ i.e. the $x$ type sustains the whole cost of matching. 
+As already mentioned, the algorithm starts from the matched pairs $(x_0,y_0)$ with no subpairs and assigns the (temporary) values $\phi_{x_0} = c_{x_0 y_0}$ and $\psi_{y_0} = 0,$ i.e. the $x$ type sustains the whole cost of matching. 
 
 
 
@@ -1961,7 +1961,7 @@ $$
 \leq \min (c_{x_0 y_j} + c_{x_i y_0} - c_{x_0 y_0} , c_{x_i y_j}) -  c_{x_j y_j} , \quad \text{for all } 1 \leq i < j \leq p.
 $$
 
-Then for all $i \in [p]$ compute the adjustment $ \Delta_i = \sum_{k = i+1}^p \beta_k + \phi_{x_p} - \phi_{x_1}$ and modify the dual variables 
+Then for all $i \in [p]$ compute the adjustment $ \Delta_i = \sum_{k = i+1}^p \beta_k + \phi_{x_p} - \phi_{x_i}$ and modify the dual variables 
 
 $$
 \begin{aligned}