From b3a497f71d39add0ffd708e57eee66dd7f3aa19e Mon Sep 17 00:00:00 2001
From: "Jip J. Dekker" <jip@dekker.one>
Date: Tue, 18 May 2021 19:33:16 +1000
Subject: [PATCH] Fix the diagonals for the MIP N-Queens

---
 chapters/2_background.tex | 31 ++++++++++++++++---------------
 1 file changed, 16 insertions(+), 15 deletions(-)

diff --git a/chapters/2_background.tex b/chapters/2_background.tex
index 327904e..ae0497c 100644
--- a/chapters/2_background.tex
+++ b/chapters/2_background.tex
@@ -842,11 +842,14 @@ can then be rewritten as linear \glspl{constraint} using the \glspl{variable}
   following model shows a integer program of this problem.
 
   \begin{align}
+    \text{given} \hspace{2em} & N = {1,\ldots,n} & \\
     \text{maximise} \hspace{2em} & 0 & \\
-    \text{subject to} \hspace{2em} & q_{i} \in \{1,\ldots{},n\} & \forall_{i=1}^{n} \\
-    & y_{ij} \in \{0,1\} & \forall_{i=1}^{n} \forall_{j=1}^{n} \\
-    \label{line:back-mip-channel} & x_{i} = \sum_{j=1}^{n} j * y_{ij} & \forall_{i=1}^{n} \\
-    \label{line:back-mip-row} & \sum_{i=1}^{n} y_{ij} \leq 1 & \forall_{j=1}^{n}
+    \text{subject to} \hspace{2em} & q_{i} \in N & \forall_{i \in N} \\
+    & y_{ij} \in \{0,1\} & \forall_{i,j \in N} \\
+    \label{line:back-mip-channel} & x_{i} = \sum_{j \in N} j * y_{ij} & \forall_{i \in N} \\
+    \label{line:back-mip-row} & \sum_{i \in N} y_{ij} \leq 1 & \forall_{j \in N} \\
+    \label{line:back-mip-diag1} & \sum_{i,j \in N. i + j =k} y_{ij} \leq 1 & \forall_{k=3}^{2n-1} \\
+    \label{line:back-mip-diag2} & \sum_{i,j \in N. i - j =k} y_{ij} \leq 1 & \forall_{k=2-n}^{n-2}
   \end{align}
 
 
@@ -857,10 +860,8 @@ can then be rewritten as linear \glspl{constraint} using the \glspl{variable}
   \(q\). The \cref{line:back-mip-channel} is used to connect the \(q\) and \(y\)
   \glspl{variable} and make sure that their values correspond.
   \Cref{line:back-mip-row} ensures that only one queen is placed in the same
-  row.
-
-  \jip{TODO: Fix diagonals}
-
+  row. Finally, \cref{line:back-mip-diag1,line:back-mip-diag2} constrain all
+  diagonals to contain only one queen.
 \end{example}
 
 \subsection{Boolean Satisfiability}%
@@ -900,13 +901,13 @@ most efficient way to solve the problem.
   \gls{sat} encoding for this problem is the following.
 
   \begin{align}
-    \text{given} \hspace{2em} & n & \\
-    \text{find} \hspace{2em} & q_{ij} \in \{\text{true}, \text{false}\} & \forall_{i=1}^{n}\forall_{j=1}^{n} \\
-    \label{line:back-sat-at-least}\text{subject to} \hspace{2em} & \exists_{j=1}^{n} q_{ij} &  \forall_{i=1}^{n} \\
-    \label{line:back-sat-row}& \neg q_{ij} \lor \neg q_{ik} & \forall_{i=1}^{n} \forall_{j=1}^{n} \forall_{k=j}^{n}\\
-    \label{line:back-sat-col}& \neg q_{ij} \lor \neg q_{kj} & \forall_{i=1}^{n} \forall_{j=1}^{n} \forall_{k=i}^{n}\\
-    \label{line:back-sat-diag1}& \neg q_{ij} \lor \neg q_{(i+k)(j+k)} & \forall_{i=1}^{n} \forall_{j=1}^{n} \forall_{k=1}^{min(n-i, n-j)}\\
-    \label{line:back-sat-diag2}& \neg q_{ij} \lor \neg q_{(i+k)(j-k)} & \forall_{i=1}^{n} \forall_{j=1}^{n} \forall_{k=1}^{min(n-i, j)}
+    \text{given} \hspace{2em} & N = {1,\ldots,n} & \\
+    \text{find} \hspace{2em} & q_{ij} \in \{\text{true}, \text{false}\} & \forall_{i,j \in N} \\
+    \label{line:back-sat-at-least}\text{subject to} \hspace{2em} & \exists_{j \in N} q_{ij} &  \forall_{i \in N} \\
+    \label{line:back-sat-row}& \neg q_{ij} \lor \neg q_{ik} & \forall_{i,j \in N} \forall_{k=j}^{n}\\
+    \label{line:back-sat-col}& \neg q_{ij} \lor \neg q_{kj} & \forall_{i,j \in N}  \forall_{k=i}^{n}\\
+    \label{line:back-sat-diag1}& \neg q_{ij} \lor \neg q_{(i+k)(j+k)} & \forall_{i,j \in N} \forall_{k=1}^{min(n-i, n-j)}\\
+    \label{line:back-sat-diag2}& \neg q_{ij} \lor \neg q_{(i+k)(j-k)} & \forall_{i,j \in N} \forall_{k=1}^{min(n-i, j)}
   \end{align}
 
   The encoding of the problem uses a Boolean \gls{variable} for every position