Finish TUT2
This commit is contained in:
Lance Li
+189
-46
@@ -220,22 +220,22 @@ Consider the search problem of Figure 1. Execute UCS (without cycle-checking) st
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\tikzexternalenable
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\tikzexternalenable
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\begin{tikzpicture}
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\begin{tikzpicture}
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\node[draw,circle] (S) at (0, 0) {$S$};
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\node[draw,circle] (S) at (0, 0) {$S$};
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\node[draw,circle] (P) at (1, -1) {$P$};
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\node[draw,circle] (P) at (1, -1) {$P$};
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\node[draw,circle] (D) at (2, 2) {$D$};
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\node[draw,circle] (D) at (2, 2) {$D$};
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\node[draw,circle] (E) at (4, 2) {$E$};
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\node[draw,circle] (E) at (4, 2) {$E$};
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\node[draw,circle] (H) at (4, 0) {$H$};
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\node[draw,circle] (H) at (3, 0.5) {$H$};
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\node[draw,circle] (Q) at (3, -1) {$Q$};
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\node[draw,circle] (Q) at (3, -1) {$Q$};
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\node[draw,circle] (G) at (6, 0) {$G$};
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\node[draw,circle] (G) at (6, 0) {$G$};
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\draw[thick,-latex] (S) -- (D) node[midway, above left] {$3$};
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\draw[thick,-latex] (S) -- (D) node[midway, above left] {$3$};
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\draw[thick,-latex] (S) -- (P) node[midway, below left] {$1$};
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\draw[thick,-latex] (S) -- (P) node[midway, below left] {$1$};
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\draw[thick,-latex] (S) -- (E) node[midway, below ] {$9$};
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\draw[thick,-latex] (S) -- (E) node[midway, below ] {$9$};
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\draw[thick,-latex] (P) -- (Q) node[midway, below ] {$15$};
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\draw[thick,-latex] (P) -- (Q) node[midway, below ] {$15$};
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\draw[thick,-latex] (D) -- (E) node[midway, above ] {$2$};
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\draw[thick,-latex] (D) -- (E) node[midway, above ] {$2$};
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\draw[thick,-latex] (E) -- (H) node[midway, left] {$1$};
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\draw[thick,-latex] (E) -- (H) node[midway, right] {$1$};
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\draw[thick,-latex] (H) -- (Q) node[midway, above left] {$4$};
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\draw[thick,-latex] (H) -- (Q) node[midway, right] {$4$};
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\draw[thick,-latex] (Q) -- (G) node[midway, below ] {$1$};
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\draw[thick,-latex] (Q) -- (G) node[midway, below ] {$1$};
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\end{tikzpicture}
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\end{tikzpicture}
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\tikzexternaldisable
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\tikzexternaldisable
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@@ -246,7 +246,7 @@ Consider the search problem of Figure 1. Execute UCS (without cycle-checking) st
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\centering
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\centering
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\renewcommand{\arraystretch}{1.15}
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\renewcommand{\arraystretch}{1.15}
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\begin{tabular}{r | l}
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\begin{tabular}{c | c}
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(Node, Path) & Frontier \\ \hline
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(Node, Path) & Frontier \\ \hline
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& $\{ (S, 0) \}$ \\
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& $\{ (S, 0) \}$ \\
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$(S, S)$ & $\{ (SP, 1), (SD, 3), (SE, 9) \}$ \\
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$(S, S)$ & $\{ (SP, 1), (SD, 3), (SE, 9) \}$ \\
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@@ -277,8 +277,8 @@ A tracing example of executing A* (without cycle-checking) on the problem of fig
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\draw[thick,-latex] (A) -- (B) node[midway, left] {$4$};
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\draw[thick,-latex] (A) -- (B) node[midway, left] {$4$};
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\draw[thick,-latex] (A) -- (C) node[midway, above ] {$1$};
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\draw[thick,-latex] (A) -- (C) node[midway, above ] {$1$};
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\draw[thick,-latex] (B.east) -- (C.south west) node[midway, below ] {$2$};
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\draw[thick,-latex] (B.east) -- (C.south west) node[midway, below right] {$2$};
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\draw[thick,-latex] (C.west) -- (B.north east) node[midway, above ] {$2$};
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\draw[thick,-latex] (C.west) -- (B.north east) node[midway, above left] {$2$};
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\draw[thick,-latex] (B) -- (D) node[midway, below ] {$6$};
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\draw[thick,-latex] (B) -- (D) node[midway, below ] {$6$};
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\draw[thick,-latex] (C) -- (D) node[midway, right] {$9$};
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\draw[thick,-latex] (C) -- (D) node[midway, right] {$9$};
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@@ -326,54 +326,197 @@ Each cost is depicted as the cost to get that node plus the heuristic cost at th
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$A$ & $\{ (C,AC, 1+7=8), {\color{blue}(B,AB, 4+3=7)} \}$ \\ \hline
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$A$ & $\{ (C,AC, 1+7=8), {\color{blue}(B,AB, 4+3=7)} \}$ \\ \hline
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$B$ & $\{ {\color{blue}(C,AC, 1+7=8)}, (D,ABD, 10+0=10) \}$ \\ \hline
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$B$ & $\{ {\color{blue}(C,AC, 1+7=8)}, (D,ABD, 10+0=10) \}$ \\ \hline
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$C$ & $\{ {\color{blue}(B,ACB, 3+3=6)}, (D,ABD, 10+0=10) \}$ \\ \hline
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$C$ & $\{ {\color{blue}(B,ACB, 3+3=6)}, (D,ABD, 10+0=10) \}$ \\ \hline
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$B$ & $\{ (C,ACBC, 5+7=12), {\color{blue}(D,ACBD, 9+0=9)} \}$ \\ \hline
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$B$ & $\{ {\color{blue}(D,ACBD, 9+0=9)}, (D,ABD, 10+0=10) \}$ \\ \hline
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\end{tabular}
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\end{tabular}
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\end{table} \end{solution}
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\end{table} \end{solution}
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\item Prove that the A* Search algorithm with cycle-checking is optimal when a consistent heuristic is utilized.
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\item Prove that the A* Search algorithm with cycle-checking is optimal when a consistent heuristic is utilized.
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% TODO: Add solution
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\begin{solution}
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The proof uses the following notation:
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\begin{listu}
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\item The function $\hat{g}(n)$ denotes the current best known path cost from the initial state to node $n$. Note that the value of $\hat{g}(n)$ is updated during the execution of the algorithm.
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\item The function $g^(n)$ denotes the minimum path cost from an initial state to state $n$.
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\item The function $h(n)$, known as a heuristic, denotes the approximated path cost from state $n$ to the (nearest) goal state.
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\item The functions $f(n)$ and $\hat{f}(n)$ are evaluation functions that are adopted by the informed search algorithm in question. For example, in the case of the greedy best-first search algorithm, $f(n) = \hat{f}(n) = h(n)$; in the case of the A* search algorithm, $f(n) = g(n) + h(n)$ and $\hat{f}(n) = \hat{g}(n) + h(n)$. $\hat{f}_{\mathrm{pop}}(n)$ denotes the value of $\hat{f}(n)$ when it is popped from the frontier.
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\end{listu}
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% \begin{proof}
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\textit{Proof.}
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Mathematically, we would want to prove that $\hat{f}_{\mathrm{pop}}(n) = f(s_g)$, i.e., when the goal node $s_f$ is popped from the frontier, we would have found the optimal path to it. Let \[ s_0, s_1, \dots, s_{g-1}, s_g \] be the path from the start node $s_0$ leading to the goal node $s_g$.
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\begin{listu}
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\item \textbf{Base case:} $\hat{f}_{\mathrm{pop}}(s_0) = f(s_0) = h(s_0)$.
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\item \textbf{Inductive step:} Assume that for all $s_0, s_1, \dots, s_k$, $\hat{f}_{\mathrm{pop}}(s_i) = f(s_i)$. We know that
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\begin{align*}
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\hat{f}_{\mathrm{pop}}(s_{k+1}) & = \hat{g}(s_{k+1}) + h(s_{k+1}) \\
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& \ge g(s_{k+1}) + h(s_{k+1}) \\
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& = f(s_{k+1}) \tag{1}\label{eq:tut2-2-1}
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\end{align*}
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In order to make sure that each $s_{k+1}$ is only explored after when we pop $s_k$, the condition of $f(s_i) \le f(s_{i+1})$ is required, leading to the need for the consistency of $h$. By popping $s_k$, we have
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\begin{align*}
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\hat{f}_{\mathrm{pop}}(s_{k+1})
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& = \min \left\{ \hat{f}(s_{k+1}), \hat{g}_{\mathrm{pop}}(s_k) + c(s_k, s_{k+1}) + h(s_{k+1}) \right\} \\
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& \le \hat{g}_{\mathrm{pop}}(s_k) + c(s_k, s_{k+1}) + h(s_{k+1}) \\
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& = g(s_k) + c(s_k, s_{k+1}) + h(s_{k+1})
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& \text{from our IH} \\
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& = g(s_{k+1}) + h(s_{k+1}) \\
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& = f(s_{k+1}) \tag{2}\label{eq:tut2-2-2}
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\end{align*}
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From equations \eqref{eq:tut2-2-1} and \eqref{eq:tut2-2-2}, we obtain $\hat{f}_{\mathrm{pop}}(s_{k+1}) = f(s_{k+1})$. Hence, by induction, whenever we pop a node from the frontier, the optimal path to the node would have been found. \qed
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\end{listu}
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% \end{proof}
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\end{solution}
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\item In the search problem below, we have listed 5 heuristics. Indicate whether each heuristic is admissible and/or consistent in the table below.
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\item In the search problem below, we have listed 5 heuristics. Indicate whether each heuristic is admissible and/or consistent in the table below.
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\begin{remark}
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\begin{figure}[ht!]
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\begin{listu}
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\item Consistent: $h(v_i) \le h(v_j) + c(v_i, v_j)$
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\item Admissible: $h(v) \le c^*(v)$
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\end{listu}
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\end{remark}
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\begin{table}[ht!]
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\centering
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\centering
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\begin{tabular}{||c|c|c|c|c|c|c||}
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\tikzexternalenable
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\hline
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\begin{tikzpicture}
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& S & A & B & G & Admissible & Consistent \\ \hline \hline
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\node[draw,circle] (S) at (0, 0) {$S$};
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$h_1$ & 0 & 0 & 0 & 0 & \color{primary} True & \color{primary} True \\ \hline
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\node[draw,circle] (A) at (2, 2) {$A$};
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$h_2$ & 8 & 1 & 1 & 0 & \color{primary} True & \color{primary} False \\ \hline
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\node[draw,circle] (B) at (2, 0) {$B$};
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$h_3$ & 9 & 3 & 2 & 0 & \color{primary} True & \color{primary} False \\ \hline
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\node[draw,circle] (G) at (4, 0) {$G$};
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$h_4$ & 6 & 3 & 1 & 0 & \color{primary} True & \color{primary} True \\ \hline
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$h_5$ & 8 & 4 & 2 & 0 & \color{primary} False & \color{primary} False \\ \hline
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\draw[thick,-latex] (S) -- (A) node[midway, above] {$6$};
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\end{tabular}
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\draw[thick,-latex] (A) -- (B) node[midway, left] {$1$};
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\end{table}
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\draw[thick,-latex] (A) -- (G) node[midway, above] {$4$};
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\draw[thick,-latex] (B) -- (G) node[midway, below] {$2$};
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\end{tikzpicture}
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\tikzexternaldisable
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\caption*{Figure 3: Graphy for question 3.}
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\end{figure}
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\begin{solution}
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\begin{remark} \color{primary}
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\begin{listu}
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\item Consistent: $h(v_i) \le h(v_j) + c(v_i, v_j)$
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\item Admissible: $h(v) \le c^*(v)$
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\end{listu}
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\end{remark}
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\begin{table}[ht!]
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\centering
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\begin{tabular}{||c|c|c|c|c|c|c||}
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\hline
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& S & A & B & G & Admissible & Consistent \\ \hline \hline
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$h_1$ & 0 & 0 & 0 & 0 & \color{primary} True & \color{primary} True \\ \hline
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$h_2$ & 8 & 1 & 1 & 0 & \color{primary} True & \color{primary} False \\ \hline
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$h_3$ & 9 & 3 & 2 & 0 & \color{primary} True & \color{primary} False \\ \hline
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$h_4$ & 6 & 3 & 1 & 0 & \color{primary} True & \color{primary} True \\ \hline
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$h_5$ & 8 & 4 & 2 & 0 & \color{primary} False & \color{primary} False \\ \hline
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\end{tabular}
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\end{table}
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\end{solution}
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\item Given that a heuristic $h$ is such that $h(s_g) = 0$ where $s_g$ is any goal state, prove that if $h$ is consistent, then it must be admissible
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\item Given that a heuristic $h$ is such that $h(s_g) = 0$ where $s_g$ is any goal state, prove that if $h$ is consistent, then it must be admissible
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% TODO: Add solution
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\begin{solution}
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\textit{Proof.} The proof can be done by induction on $k(s_i)$, which denotes the number of actions required to reach the goal from a node si to the goal node $s_g$.
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\item \begin{enumerate}[label=\alph*)]
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\begin{listu}
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\item \textbf{Base case:} $k = 1$, i.e. the node $s_i$ is one step away from $s_g$
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Since the heuristic function $h$ is consistent, \[
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h(s_i) \le c(s_i, s_g) + h(s_g)
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\]
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Since $h(s_g) = 0$, we have \[
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h(s_i) \le c(s_i, s_g) = h^*(s_i)
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\]
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Therefore, $h$ is admissible.
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\item \textbf{Inductive step:}
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\begin{center}
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\tikzexternalenable
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\begin{tikzpicture}
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\node[draw,circle,minimum size=1cm] (si1) at (0, 0) {$s_{i+1}$};
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\node[draw,circle,minimum size=1cm] (si) at (1.5, 1.5) {$s_i$};
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\node[draw,circle,minimum size=1cm] (sg) at (3, 0) {$s_g$};
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\draw[thick,-latex] (si) -- (si1);
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\draw[thick,-latex] (si) -- (sg);
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\draw[thick,-latex] (si1) -- (sg);
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\end{tikzpicture}
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\tikzexternaldisable
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\end{center}
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Suppose that our assumption holds for every node that is $k-1$ actions away from $s_g$, and let us observe a node $s_i $that is $k$ actions away from $s_g$; that is, the optimal path from $s_i$ to $s_g$ has $k > 1$ steps. We can write the optimal path from $s_i$ to $s_g$ as \[
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s_i \to s_{i+1} \to \cdots \to s_{g-1} \to s_g
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\]
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Since $h$ is consistent, we have \[
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h(s_i) \le c(s_i, s_{i+1}) + h(s_{i+1})
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\]
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Now, note that since $s_{i+1}$ is on a least-cost path from $s_i$ to $s_g$, we must have that the path $s_{i+1} \to s_{i+2} \to \cdots \to s_{g-1} \to s_g$ is a least-cost path from $s_{i+1}$ to $s_g$ as well. By our induction hypothesis, we have \[
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h(s_{i+1}) \le h^*(s_{i+1})
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\]
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Combining the two inequalities, we have \[
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h(s_i) \le c(s_i, s_{i+1}) + h^*(s_{i+1})
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\]
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Note that $h^*(s_{i+1})$ is the cost of the optimal path from $s_{i+1}$ to $s_g$; by our previous observation (that $s_{i+1} \to s_{i+2} \to \cdots \to s_{g-1} \to s_g$ is a least-cost path from $s_{i+1}$ to $s_g$), we have that the cost of the optimal path from $s_i$ to $s_g$, i.e. $h^*(s_i)$, equals $c(s_i, s_{i+1}) + h^*(s_i)$, which concludes the proof. \qed
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\end{listu}
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\end{solution}
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\item \begin{enumerate}
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\item Provide a counter-example to show that the \textbf{Greedy Best-First Search} algorithm without cycle-checking is incomplete.
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\item Provide a counter-example to show that the \textbf{Greedy Best-First Search} algorithm without cycle-checking is incomplete.
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\begin{solution}
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\begin{solution}
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% \begin{center} \begin{tikzpicture}
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Consider the following search space, where
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% \node[draw,circle] (S0) at (0, 0) {$S_0$};
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% \end{tikzpicture} \end{center}
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\begin{minipage}[t]{0.45\linewidth} \begin{listu}
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% TODO: Add solution
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\item $h(s_0) = 3$
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\item $h(s_1) = 4$
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\item $h(s_2) = 5$
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\item $h(s_g) = 0$
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\end{listu} \end{minipage}
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\begin{minipage}[t]{0.45\linewidth} \begin{center} \begin{tikzpicture}[baseline=(current bounding box.north)]
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\node[draw,circle] (s0) at (0, 0) {$S_0$};
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\node[draw,circle] (s1) at (1.5, 1.5) {$S_1$};
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\node[draw,circle] (s2) at (1.5, 0) {$S_2$};
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\node[draw,circle] (sg) at (3, 0) {$S_g$};
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\draw[thick] (s0) -- (s1);
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\draw[thick] (s1) -- (s2);
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\draw[thick] (s2) -- (sg);
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\end{tikzpicture} \end{center} \end{minipage}
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\end{solution}
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\end{solution}
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\item Provide a counter-example to show that Greedy Best-First Search (with or without cycle-checking) is not optimal.
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\item Provide a counter-example to show that Greedy Best-First Search (with or without cycle-checking) is not optimal.
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% TODO: Add solution
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\begin{solution}
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Consider the following search space, where
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\begin{minipage}[t]{0.45\linewidth} \begin{listu}
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\item $h(s_0) = 9$
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\item $h(s_1) = 1$
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\item $h(s_g) = 0$
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\item $h(s_g') = 0$
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\end{listu} \end{minipage}
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\begin{minipage}[t]{0.45\linewidth} \begin{center} \begin{tikzpicture}[baseline=(current bounding box.north)]
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\node[draw,circle] (sp) at (0, 0) {$S_g'$};
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\node[draw,circle] (s0) at (1.5, 1.5) {$S_0$};
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|
\node[draw,circle] (s1) at (1.5, 0) {$S_1$};
|
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|
\node[draw,circle] (sg) at (3, 0) {$S_g$};
|
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|
|
||||||
|
\draw[thick,-latex] (sp) -- (s0) node[midway, left] {$10$};
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|
\draw[thick,-latex] (s0) -- (s1) node[midway, right] {$8$};
|
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|
\draw[thick,-latex] (s1) -- (sg) node[midway, below] {$1$};
|
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|
\end{tikzpicture} \end{center} \end{minipage}
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|
|
||||||
|
With either variant of the greedy best-first search algorithm, when $s_0$ is explored, $s_g'$ would be added to the front of the frontier and then explored next, resulting in the algorithm returning the non-optimal $s_0 \to s_g'$ path.
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|
\end{solution}
|
||||||
\end{enumerate}
|
\end{enumerate}
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||||||
\end{enumerate}
|
\end{enumerate}
|
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Reference in New Issue
Block a user