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Created FAIKR modules
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src/fundamentals-of-ai-and-kr/module1/sections/_search.tex
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src/fundamentals-of-ai-and-kr/module1/sections/_search.tex
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\chapter{Search problems}
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\section{Search strategies}
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\begin{description}
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\item[Solution space] \marginnote{Solution space}
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Set of all the possible sequences of actions an agent may apply.
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Some of these lead to a solution.
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\item[Search algorithm] \marginnote{Search algorithm}
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Takes a problem as input and returns a sequence of actions that solves the problem (if exists).
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\end{description}
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\subsection{Search tree}
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\begin{description}
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\item[Expansion] \marginnote{Expansion}
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Starting from a state, apply a successor function and generate a new state.
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\item[Search strategy] \marginnote{Search strategy}
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Choose which state to expand.
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Usually is implemented using a fringe that decides which is the next node to expand.
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\item[Search tree] \marginnote{Search tree}
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Tree structure to represent the expansion of all states starting from a root
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(i.e. the representation of the solution space).
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Nodes are states and branches are actions.
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A leaf can be a state to expand, a solution or a dead-end.
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\Cref{alg:search_tree_search} describes a generic tree search algorithm.
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.25\textwidth]{img/_search_tree.pdf}
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\caption{Search tree}
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\end{figure}
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Each node contains:
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\begin{itemize}
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\item The state
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\item The parent node
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\item The action that led to this node
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\item The depth of the node
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\item The cost of the path from the root to this node
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\end{itemize}
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\end{description}
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\begin{algorithm}
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\caption{Tree search} \label{alg:search_tree_search}
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\begin{lstlisting}
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def treeSearch(problem, fringe):
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fringe.push(problem.initial_state)
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# Get a node in the fringe and expand it if it is not a solution
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while fringe.notEmpty():
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node = fringe.pop()
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if problem.isGoal(node.state):
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return node.solution
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fringe.pushAll(expand(node, problem))
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return FAILURE
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def expand(node, problem):
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successors = set()
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# List all neighboring nodes
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for action, result in problem.successor(node.state):
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s = new Node(
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parent=node, action=action, state=result, depth=node.dept+1,
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cost=node.cost + problem.pathCost(node, s, action)
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)
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successors.add(s)
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return successors
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\end{lstlisting}
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\end{algorithm}
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\subsection{Strategies}
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\begin{description}
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\item[Non-informed strategy] \marginnote{Non-informed strategy}
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Domain knowledge not available. Usually does an exhaustive search.
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\item[Informed strategy] \marginnote{Informed strategy}
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Use domain knowledge by using heuristics.
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\end{description}
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\subsection{Evaluation}
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\begin{description}
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\item[Completeness] \marginnote{Completeness}
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if the strategy is guaranteed to find a solution (when exists).
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\item[Time complexity] \marginnote{Time complexity}
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time needed to complete the search.
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\item[Space complexity] \marginnote{Space complexity}
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memory needed to complete the search.
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\item[Optimality] \marginnote{Optimality}
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if the strategy finds the best solution (when more solutions are possible).
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\end{description}
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\section{Non-informed search}
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\subsection{Breadth-first search (BFS)}
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\marginnote{Breadth-first search}
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Always expands the less deep node. The fringe is implemented as a queue (FIFO).
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\begin{center}
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\def\arraystretch{1.2}
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\begin{tabular}{c | m{10cm}}
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\hline
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\textbf{Completeness} & Yes \\
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\hline
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\textbf{Optimality} & Only with uniform cost (i.e. all edges have same cost) \\
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\hline
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\textbf{\makecell{Time and space\\complexity}}
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& $O(b^d)$, where the solution depth is $d$ and the branching factor is $b$ (i.e. each non-leaf node has $b$ children) \\
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\hline
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\end{tabular}
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\end{center}
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The exponential space complexity makes BFS impractical for large problems.
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.30\textwidth]{img/_bfs.pdf}
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\caption{BFS visit order}
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\end{figure}
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\subsection{Uniform-cost search}
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\marginnote{Uniform-cost search}
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Same as BFS, but always expands the node with the lowest cumulative cost.
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\begin{center}
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\def\arraystretch{1.2}
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\begin{tabular}{c | m{10cm}}
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\hline
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\textbf{Completeness} & Yes \\
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\hline
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\textbf{Optimality} & Yes \\
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\hline
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\textbf{\makecell{Time and space\\complexity}}
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& $O(b^d)$, with solution depth $d$ and branching factor $b$ \\
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\hline
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\end{tabular}
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\end{center}
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.50\textwidth]{img/_ucs.pdf}
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\caption{Uniform-cost search visit order. $(n)$ is the cumulative cost}
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\end{figure}
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\subsection{Depth-first search (DFS)}
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\marginnote{Depth-first search}
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Always expands the deepest node. The fringe is implemented as a stack (LIFO).
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\begin{center}
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\def\arraystretch{1.2}
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\begin{tabular}{c | m{10cm}}
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\hline
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\textbf{Completeness} & No (loops) \\
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\hline
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\textbf{Optimality} & No \\
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\hline
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\textbf{Time complexity}
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& $O(b^m)$, with maximum depth $m$ and branching factor $b$ \\
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\hline
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\textbf{Space complexity}
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& $O(b \cdot m)$, with maximum depth $m$ and branching factor $b$ \\
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\hline
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\end{tabular}
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\end{center}
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.30\textwidth]{img/_dfs.pdf}
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\caption{DFS visit order}
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\end{figure}
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\subsection{Depth-limited search}
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\marginnote{Depth-limited search}
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Same as DFS, but introduces a maximum depth.
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A node at the maximum depth will not be explored further.
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This allows to avoid infinite branches (i.e. loops).
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\subsection{Iterative deepening}
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\marginnote{Iterative deepening}
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Rus a depth-limited search by trying all possible depth limits.
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It is important to note that each iteration is executed from scratch (i.e. a new execution of depth-limited search).
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\begin{algorithm}
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\caption{Iterative deepening}
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\begin{lstlisting}
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def iterativeDeepening(G):
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for c in range(G.max_depth):
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sol = depthLimitedSearch(G, c)
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if sol is not FAILURE:
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return sol
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return FAILURE
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\end{lstlisting}
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\end{algorithm}
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Both advantages of DFS and BFS are combined.
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\begin{center}
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\def\arraystretch{1.2}
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\begin{tabular}{c | m{10cm}}
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\hline
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\textbf{Completeness} & Yes \\
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\hline
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\textbf{Optimality} & Only with uniform cost \\
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\hline
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\textbf{Time complexity}
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& $O(b^d)$, with solution depth $d$ and branching factor $b$ \\
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\hline
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\textbf{Space complexity}
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& $O(b \cdot d)$, with solution depth $d$ and branching factor $b$ \\
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\hline
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\end{tabular}
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\end{center}
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\section{Informed search}
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\marginnote{Informed search}
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Informed search uses evaluation functions (heuristics) to reduce the search space and
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estimate the effort needed to reach the final goal.
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\subsection{Best-first search}
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\marginnote{Best-first seacrh}
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Uses heuristics to compute the desirability of the nodes (i.e. how close they are to the goal).
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The fringe is ordered according the estimated scores.
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\begin{description}
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\item[Greedy search / Hill climbing]
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\marginnote{Greedy search / Hill climbing}
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The heuristic only evaluates nodes individually and does not consider the path to the root
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(i.e. expands the node that currently seems closer to the goal).
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\begin{center}
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\def\arraystretch{1.2}
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\begin{tabular}{c | m{9cm}}
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\hline
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\textbf{Completeness} & No (loops) \\
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\hline
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\textbf{Optimality} & No \\
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\hline
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\textbf{\makecell{Time and space\\complexity}}
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& $O(b^d)$, with solution depth $d$ and branching factor $b$ \\
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\hline
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\end{tabular}
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\end{center}
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% The complexity can be reduced depending on the heuristic.
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\begin{figure}[ht]
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\centering
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\includegraphics[width=0.65\textwidth]{img/_greedy_best_first_example.pdf}
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\caption{Hill climbing visit order}
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\end{figure}
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\item[A$^\textbf{*}$]
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\marginnote{A$^*$}
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The heuristic also considers the cumulative cost needed to reach a node from the root.
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The score associated to a node $n$ is:
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\[ f(n) = g(n) + h'(n) \]
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where $g$ is the depth of the node and $h'$ is the heuristic that computes the distance to the goal.
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\begin{description}
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\item[Optimistic/Feasible heuristic]
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\marginnote{Optimistic/Feasible heuristic}
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Given $t(n)$ that computes the true distance of a node $n$ to the goal.
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An heuristic $h'(n)$ is optimistic (i.e. feasible) if:
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\[ h'(n) \leq t(n) \]
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In other words, $h'$ is optimistic if it always underestimates the distance to the goal.
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\end{description}
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\begin{theorem}
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If the heuristic used by A${^*}$ is optimistic $\Rightarrow$ A${^*}$ is optimal
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\end{theorem}
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\begin{proof}
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Consider a scenario where the queue contains:
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\begin{itemize}
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\item A node $n$ whose child is the optimal solution
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\item A sub-optimal solution $G_2$
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\end{itemize}
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\begin{center}
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\includegraphics[width=0.5\textwidth]{img/_a_start_optimality.pdf}
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\end{center}
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We want to prove that A$^*$ will always expand $n$.
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Given an optimistic heuristic $f(n) = g(n) + h'(n)$ and
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the true distance of a node $n$ to the goal $t(n)$,
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we have that:
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\[
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\begin{split}
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f(G_2) &= g(G_2) + h'(G_2) = g(G_2) \text{, as } G_2 \text{ is a solution: } h'(G_2)=0 \\
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f(G) &= g(G) + h'(G) = g(G) \text{, as } G \text{ is a solution: } h'(G)=0
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\end{split}
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\]
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Moreover, $g(G_2) > g(G)$ as $G_2$ is suboptimal.
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Therefore, $\bm{f(G_2) > f(G)}$.
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Furthermore, as $h'$ is feasible, we have that:
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\[
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\begin{split}
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h'(n) \leq t(n) &\iff g(n) + h'(n) \leq g(n) + t(n) = g(G)=f(G) \\
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&\iff \bm{f(n) \leq f(G)}
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\end{split}
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\]
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In the end, we have that $f(G_2) > f(G) \geq f(n)$.
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So we can conclude that A$^*$ will never expand $G_2$ as:
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\[ f(G_2) > f(n) \]
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\end{proof}
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\begin{center}
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\def\arraystretch{1.2}
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\begin{tabular}{c | m{9cm}}
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\hline
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\textbf{Completeness} & Yes \\
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\hline
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\textbf{Optimality} & Only if the heuristic is optimistic \\
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\hline
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\textbf{\makecell{Time and space\\complexity}}
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& $O(b^d)$, with solution depth $d$ and branching factor $b$ \\
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\hline
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\end{tabular}
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\end{center}
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In generally, it is better to use heuristics with large values (i.e. heuristics that don't underestimate too much).
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\begin{figure}[ht]
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\centering
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\includegraphics[width=0.65\textwidth]{img/_a_start_example.pdf}
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\caption{A$^*$ visit order}
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\end{figure}
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\end{description}
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\section{Graph search}
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\marginnote{Graph search}
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Differently from a tree search, searching in a graph requires to keep track of the explored nodes.
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\begin{algorithm}
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\caption{Graph search} \label{alg:search_graph_search}
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\begin{lstlisting}
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def graphSearch(problem, fringe):
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closed = set()
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fringe.push(problem.initial_state)
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# Get a node in the fringe and
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# expand it if it is not a solution and is not closed
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while fringe.notEmpty():
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node = fringe.pop()
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if problem.isGoal(node.state):
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return node.solution
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if node.state not in closed:
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closed.add(node.state)
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fringe.pushAll(expand(node, problem))
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return FAILURE
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\end{lstlisting}
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\end{algorithm}
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\subsection{A$^\textbf{*}$ with graphs}
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\marginnote{A$^*$ with graphs}
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The algorithm keeps track of closed and open nodes.
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The heuristic $g(n)$ evaluates the minimum distance from the root to the node $n$.
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\begin{description}
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\item[Consistent heuristic (monotone)] \marginnote{Consistent heuristic (monotone)}
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An heuristic is consistent if for each $n$, for any successor $n'$ of $n$ (i.e. nodes reachable from $n$ by making an action)
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holds that:
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\[
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\begin{cases}
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h(n) = 0 & \text{if the corresponding status is the goal} \\
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h(n) \leq c(n, a, n') + h(n') & \text{otherwise}
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\end{cases}
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\]
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where $c(n, a, n')$ is the cost to reach $n'$ from $n$ by taking the action $a$.
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In other words, $f$ never decreases along a path.
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In fact:\\
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\begin{minipage}{.48\linewidth}
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\[
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\begin{split}
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f(n') &= g(n') + h(n') \\
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&= g(n) + c(n, a, n') + h(n') \\
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&\geq g(n) + h(n) \\
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&= f(n)
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\end{split}
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\]
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\end{minipage}
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\begin{minipage}{.48\linewidth}
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\centering
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\includegraphics[width=0.3\textwidth]{img/monotone_heuristic.png}
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\end{minipage}
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\begin{theorem}
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If $h$ is a consistent heuristic, A$^*$ on graphs is optimal.
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\end{theorem}
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\end{description}
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