From 0fece14db65e1107e5619515bcd03372f7c1078f Mon Sep 17 00:00:00 2001
From: NotXia <35894453+NotXia@users.noreply.github.com>
Date: Mon, 11 Dec 2023 17:35:23 +0100
Subject: [PATCH] Add FAIRK1 constraint programming

---
 .../module1/img/_constraint_graph.drawio      |  61 ++++++
 .../module1/img/_constraint_graph.pdf         | Bin 0 -> 16694 bytes
 .../module1/main.tex                          |   1 +
 .../module1/sections/_constraints.tex         | 193 +++++++++++++++++-
 4 files changed, 254 insertions(+), 1 deletion(-)
 create mode 100644 src/fundamentals-of-ai-and-kr/module1/img/_constraint_graph.drawio
 create mode 100644 src/fundamentals-of-ai-and-kr/module1/img/_constraint_graph.pdf

diff --git a/src/fundamentals-of-ai-and-kr/module1/img/_constraint_graph.drawio b/src/fundamentals-of-ai-and-kr/module1/img/_constraint_graph.drawio
new file mode 100644
index 0000000..8c90f64
--- /dev/null
+++ b/src/fundamentals-of-ai-and-kr/module1/img/_constraint_graph.drawio
@@ -0,0 +1,61 @@
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diff --git a/src/fundamentals-of-ai-and-kr/module1/main.tex b/src/fundamentals-of-ai-and-kr/module1/main.tex
index bcf94a7..ab06501 100644
--- a/src/fundamentals-of-ai-and-kr/module1/main.tex
+++ b/src/fundamentals-of-ai-and-kr/module1/main.tex
@@ -14,5 +14,6 @@
     \input{sections/_games.tex}
     \input{sections/_planning.tex}
     \input{sections/_constraints.tex}
+    \eoc
 
 \end{document}
\ No newline at end of file
diff --git a/src/fundamentals-of-ai-and-kr/module1/sections/_constraints.tex b/src/fundamentals-of-ai-and-kr/module1/sections/_constraints.tex
index 3db952d..ac69171 100644
--- a/src/fundamentals-of-ai-and-kr/module1/sections/_constraints.tex
+++ b/src/fundamentals-of-ai-and-kr/module1/sections/_constraints.tex
@@ -62,4 +62,195 @@ The search tree is constructed depth-first.
 At level $i$, a value is assigned to the variable $X_i$ and 
 constraints involving $X_i$ are propagated
 to restrict the domain of the remaining unassigned variables.
-If the domain of a variable becomes empty, the path is considered a failure and the algorithm backtracks.
\ No newline at end of file
+If the domain of a variable becomes empty, the path is considered a failure and the algorithm backtracks.
+
+\begin{example}
+    Consider the variables: 
+    \[ X_1 :: [1, 2, 3] \hspace{1cm} X_2 :: [1, 2, 3] \hspace{1cm} X_3 :: [1, 2, 3] \]
+    and the constraints $X_1 < X_2 < X_3$.
+
+    We assign the variables in lexicographic order, at each step we have that:
+    \begin{enumerate}
+        \item $X_1 = 1 \hspace{1cm} X_2 :: [\cancel{1}, 2, 3] \hspace{1cm} X_3 :: [\cancel{1}, 2, 3]$
+        \item $X_1 = 1 \hspace{1cm} X_2 = 2 \hspace{1cm} X_3 :: [\cancel{1}, \cancel{2}, 3]$
+        \item $X_1 = 1 \hspace{1cm} X_2 = 2 \hspace{1cm} X_3 = 3$
+    \end{enumerate}
+
+    In another scenario, assume that we choose $X_1 = 2$ as first step:
+    \begin{enumerate}
+        \item $X_1 = 2 \hspace{1cm} X_2 :: [\cancel{1}, \cancel{2}, 3] \hspace{1cm} X_3 :: [\cancel{1}, \cancel{2}, 3]$
+        \item $X_1 = 2 \hspace{1cm} X_2 = 3 \hspace{1cm} X_3 :: [\cancel{1}, \cancel{2}, \cancel{3}]$
+    \end{enumerate}
+    As the domain of $X_3$ is empty, search on this branch fails and backtracking is required.
+\end{example}
+
+
+
+\subsection{Look ahead}
+
+\begin{description}
+    \item[Partial look ahead (PLA)] \marginnote{Partial look ahead (PLA)}
+        When a variable $X_i$ has been assigned, it is propagated to unassigned variables (as in forward checking).
+        Then, for each unassigned variable $X_h$, we check its values against the ones of the variables $X_{h+1}, \dots, X_{n}$ in front of it 
+        (note that the check is not done in both directions).
+        
+        When checking the values of an unassigned variable $X_p :: D_p$ against $X_q :: D_q$, 
+        we maintain a value $x \in D_p$ only if there is at least a corresponding value $y \in D_q$ such that 
+        an assignment $X_p=x$ and $X_q=y$ does not violate the constraints. 
+        $y$ is called support \marginnote{Support} of $x$. 
+
+        \begin{example}
+            Consider the variables and constraints: 
+            \[ X_1 :: [1, 2, 3] \hspace{0.5cm} X_2 :: [1, 2, 3] \hspace{0.5cm} X_3 :: [1, 2, 3]  \hspace{1cm}  X_1 < X_2 < X_3 \]
+
+            We assign the variables in lexicographic order. At each step we have that:
+            \begin{enumerate}
+                \item $X_1 = 1 \hspace{1cm} X_2 :: [\cancel{1}, 2, \cancel{3}] \hspace{1cm} X_3 :: [\cancel{1}, 2, 3]$ \\
+                    Here, we assign $X_1=1$ and propagate to unassigned constraints.
+                    Then, we check the constraints of $(X_2, X_3)$: $X_2 = 3$ is not possible as it does not have a support in $X_3$.
+                \item $X_1 = 1 \hspace{1cm} X_2 = 2 \hspace{1cm} X_3 :: [\cancel{1}, \cancel{2}, 3]$
+                \item $X_1 = 1 \hspace{1cm} X_2 = 2 \hspace{1cm} X_3 = 3$
+            \end{enumerate}
+        \end{example}
+
+    \item[Full look ahead (FLA)] \marginnote{Full look ahead (FLA)}
+        Same as partial look ahead, but the checks on unassigned variables are bidirectional.
+
+        \begin{example}
+            Consider the variables and constraints: 
+            \[ X_1 :: [1, 2, 3] \hspace{0.5cm} X_2 :: [1, 2, 3] \hspace{0.5cm} X_3 :: [1, 2, 3]  \hspace{1cm}  X_1 < X_2 < X_3 \]
+
+            We assign the variables in lexicographic order. At each step we have that:
+            \begin{enumerate}
+                \item $X_1 = 1 \hspace{1cm} X_2 :: [\cancel{1}, 2, \cancel{3}] \hspace{1cm} X_3 :: [\cancel{1}, \cancel{2}, 3]$ \\
+                    Here, we assign $X_1=1$ and propagate to unassigned constraints.
+                    Then, we check the constraints of:
+                    \begin{itemize}
+                        \item $(X_2, X_3)$: $X_2 = 3$ is not possible as it does not have a support in $X_3$.
+                        \item $(X_3, X_2)$: $X_3 = 2$ is not possible as it does not have a support in $X_2$.
+                    \end{itemize} 
+                    Note that checking in a different order might result in different (but correct) domains.
+                \item $X_1 = 1 \hspace{1cm} X_2 = 2 \hspace{1cm} X_3 = 3$
+            \end{enumerate}
+        \end{example}
+\end{description}
+
+
+
+\section{Search heuristics}
+When searching, we have two degrees of freedom:
+\begin{enumerate}
+    \item Selection of the variable to assign.
+    \item Selection of the value to assign.
+    \item Strategy for propagation.
+\end{enumerate}
+
+Heuristics can be used to make better selections (for the points 1. and 2.):
+\begin{descriptionlist}
+    \item[Variable selection] \marginnote{Variable selection heuristics}
+        \phantom{}
+        \begin{description}
+            \item[First-fail (minimum remaining values)] \marginnote{First-fail}
+                Choose the variable with the smallest domain.
+
+            \item[Most-constrained principle] \marginnote{Most-constrained principle}
+                Choose the variable that appears the most in the constraints. 
+                In this way, harder variables are ideally assigned first.
+        \end{description}
+
+    \item[Value selection] \marginnote{Value selection heuristics}
+        The choice of the value to assign to a variable does not have general rules.
+\end{descriptionlist}
+
+Moreover, heuristics can be:
+\begin{descriptionlist}
+    \item[Static] \marginnote{Static heuristics} 
+        Computed before the search. 
+    
+    \item[Dynamic] \marginnote{Dynamic heuristics} 
+        Can be computed at each assignment.
+\end{descriptionlist}
+
+
+
+\section{Consistency techniques}
+
+Consistency techniques reduce the original problem by removing domain values that are surely impossible.
+This class of methods can be applied statically before the search or after each assignment.
+
+\begin{description}
+    \item[Constraint graph] \marginnote{Constraint graph}
+        Graph where nodes are variables and arcs are constraints.
+        \begin{itemize}
+            \item There is a directed arc from $X_i$ to $X_j$ if there is a binary constraint $X_i \circ X_j$.
+            \item There is a self-loop to $X_i$ if there is a unary constraint on $X_i$.
+        \end{itemize}
+
+    \item[Node consistency (level 1 consistency)] \marginnote{Node consistency}
+        A node $X_i$ is consistent if all of its values in the domain satisfy the constraints.
+
+        A graph is node consistent if all its nodes are consistent.
+
+    \item[Arc consistency (level 2 consistency)] \marginnote{Arc consistency}
+        An arc from $X_i :: D_i$ to $X_j :: D_j$ is consistent if each value of $D_i$ has a support in $D_j$ (as in look ahead).
+
+        A graph is arc consistent if all its arcs are consistent.
+
+        \begin{description}
+            \item[AC-3 algorithm] \marginnote{AC-3 algorithm}
+                A possible algorithm to achieve arc consistency.
+                Uses a queue to store arcs to explore:
+                \begin{enumerate}
+                    \item At the beginning, all arcs are in the queue. 
+                    \item When evaluating an arc $(X_i, X_j)$, 
+                        values in the domain of $X_i$ are removed if they don't have a support in the domain of $X_j$.
+                    \item If the domain of $X_i$ has been modified, 
+                        for all the neighbors $X_k$ of $X_i$, the arc $(X_k, X_i)$ is added to the queue.
+                    \item Repeat until empty queue (quiescence).
+                \end{enumerate} 
+            \end{description}
+
+        \begin{example}
+            Note: when doing the exercises, it is sufficient to iterate until none of the nodes change, without using a queue.
+            
+            Consider the variables and constraints: 
+            \[ X_1 :: [1, 2, 3] \hspace{0.5cm} X_2 :: [1, 2, 3] \hspace{0.5cm} X_3 :: [1, 2, 3]  \hspace{1cm}  X_1 < X_2 < X_3 \]
+            The constraint graph and the iterations to achieve arc consistency are the following:
+
+            \begin{minipage}{0.7\textwidth}
+                \begin{enumerate}
+                    \item We check the following pairs:
+                        \begin{itemize}
+                            \item $(X_1, X_2)$. Here, $X_1 :: [1, 2, \cancel{3}]$.
+                            \item $(X_2, X_1)$. Here, $X_2 :: [\cancel{1}, 2, 3]$.
+                            \item $(X_2, X_3)$. Here, $X_2 :: [\cancel{1}, 2, \cancel{3}]$.
+                            \item $(X_3, X_2)$. Here, $X_3 :: [\cancel{1}, \cancel{2}, 3]$.
+                            \item $(X_1, X_3)$. Here, $X_1 :: [1, 2, \cancel{3}]$ (no changes).
+                            \item $(X_3, X_1)$. Here, $X_3 :: [\cancel{1}, \cancel{2}, 3]$ (no changes).
+                        \end{itemize}
+                        The resulting domains are:
+                        \[ X_1 :: [1, 2, \cancel{3}] \hspace{0.5cm} X_2 :: [\cancel{1}, 2, \cancel{3}] \hspace{0.5cm} X_3 :: [\cancel{1}, \cancel{2}, 3] \]
+                    \item We check the following pairs:
+                        \begin{itemize}
+                            \item $(X_1, X_2)$. Here, $X_1 :: [1, \cancel{2}, \cancel{3}]$.
+                            \item Other pairs do nothing.
+                        \end{itemize}
+                        The resulting domains are:
+                        \[ X_1 :: [1, \cancel{2}, \cancel{3}] \hspace{0.5cm} X_2 :: [\cancel{1}, 2, \cancel{3}] \hspace{0.5cm} X_3 :: [\cancel{1}, \cancel{2}, 3] \]
+                \end{enumerate}
+            \end{minipage}
+            \begin{minipage}{0.20\textwidth}
+                \includegraphics[width=\linewidth]{img/_constraint_graph.pdf}
+            \end{minipage}
+        \end{example}
+
+    \item[Path consistency (level 3 consistency)] \marginnote{Path consistency}
+        A path $(X_i, X_j, X_k)$ is path consistent if $X_i$ and $X_j$ are node consistent, $(X_i, X_j)$ is arc consistent and
+        for every pair of values in the domains of $X_i$ and $X_j$, there is a support in the domain of $X_k$.
+
+    \item[K-consistency] \marginnote{K-consistency}
+        Generalization of arc/path consistency.
+        If a problem with $n$ variables is $n$-consistent, the solution can be found without search.
+
+        Usually it is not applicable as it has exponential complexity.
+\end{description}
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