Q3 proof-read, clarified, and re-organized.

A4/main.tex

@@ -84,41 +84,39 @@ Let ${\sf OPT}$ be the value of the optimal solution of the metric traveling sal
 
 \solution
 
-\textbf{Variables and Assumptions}: To begin, we will define our variables and state our assumptions, let \(G = (E, V)\) be the complete graph where \(V\) represents the spatial nodes to be visited and \(E\) be a series of edges that connect all vertices with each other. Each edge is assigned a weight \(c_{ij}\) for the corresponding vertices \(i\) and \(j\). Furthermore, we will assume that the triangle inequality holds for all triangles formed by all edges. In other words, \(\forall i, j, k \in V, c_{ik} \leq c_{ij} + c_{jk}\) essentially stating that for any given vertices \(i\), \(j\), \(k\), the direct edge from \(i\) to \(k\) is never worse than the sequence of edges from \(i\) to \(j\), to \(k\). Let \verb|GRD| be the described greedy algorithm given in the question. We let function \(c(S)\) be the sum of all edge weights for traversing all vertices of a graph \(S\). Lastly, we will assume \verb|OPT| the cost of the a optimal solution to traveling salesman problem (TSP). Our objective is to show that, given \(S_g\) solution generated by \verb|GRD| \(S_g\) is at worst, \(c(S_g) \leq 2 \times \verb|OPT|\).
+\textbf{Variables and Assumptions}: To begin, we will define our variables and state our assumptions, let \(G = (E, V)\) be the complete graph where \(V\) represents the spatial nodes to be visited and \(E\) be a series of edges that connect all vertices with each other. Each edge is assigned a weight \(c_{ij}\) for the corresponding vertices \(i\) and \(j\). Furthermore, we will assume that the triangle inequality holds for all triangles formed by all edges. In other words, \(\forall i, j, k \in V, c_{ik} \leq c_{ij} + c_{jk}\) essentially stating that for any given vertices \(i\), \(j\), \(k\), the direct edge from \(i\) to \(k\) is never worse than the sequence of edges from \(i\) to \(j\), to \(k\). Let \verb|GRD| be the given greedy algorithm and \(S_g = E_g, V_g\) be the graph representation of the sequence of vertices and edges to take as the solution (traversal graph) produced by such an algorithm. We let function \(c(S)\) be the sum of all edge weights for traversing all vertices of a graph \(S\). Lastly, we will assume \verb|OPT| is the cost of a optimal solution to traveling salesman problem (TSP) where such a traversal graph is \(S_o = (E_o, V_o)\). Our objective is to show that, \(S_g\) is at worst, \(c(S_g) \leq 2 \times \verb|OPT|\).
 
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-\textbf{Claim 1}: To begin, notice that the \(c(S_o)\) such that \(c(S_o) = \verb|OPT|\), is a cycle that traverses all vertices minimally and cyclically where each node is traversed exactly once with the exception of the starting node. Then, see that \(c(S_o)\) may be trivially converted into a tree graph by simply removing any edge in \(S_o\) and arbitrarily selecting a vertex to become the root of the tree. Furthermore, see that the traversal of such a tree costs will not cost more than the traversal of the original cycle \(S_o\). In other words, \(S_o = (E_o, V_o), \forall e \in E_o, c(S_o - \{e\}) \leq c(S_o)\).
+\textbf{Prim's Minimum Spanning Tree Algorithm Review}: Very briefly, the Prim's minimum spanning tree (MST) algorithm begins by arbitrarily selecting a vertex from a graph, and iteratively selecting the next vertex with the lowest edge weight connecting to the current set of selected vertices.
-
-% TODO Double check if the definition of the w function makes sense.
 
 \medskip
 
-\textbf{Prim's Minimum Spanning Tree Algorithm Review}: Very briefly,the Prim's minimum spanning tree (MST) algorithm begins by arbitrarily selecting a vertex from a graph, and iteratively selecting the next vertex with the lowest edge weight connecting to the current set of selected vertices.
+\textbf{Claim 1}: To begin, notice that \(c(S_o)\) is a cycle that traverses all vertices minimally and cyclically where each node is traversed exactly once with the exception of the starting node. Then, see that \(c(S_o)\) may be trivially converted into a tree graph by simply removing any edge in \(S_o\) and arbitrarily selecting a vertex to become the root of the tree. Such a tree is spanning (all vertices connected). Furthermore, see that the traversal of such a tree costs will not cost more than the traversal of the original cycle \(S_o\). In other words, \(\forall e \in E_o, c(S_o - \{e\}) \leq c(S_o)\).
 
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-\textbf{Claim 2}: See that the cycle graph \(S_e = (E_e, V_e)\) generated by \verb|GRD| will always result in requiring half of the edges to traverse all nodes via connected edges when compared to traversing a MST. To see this, we assert that \verb|GRD| produces a traversal graph (vertices representing nodes and edges representing the edge taken to reach each vertex) \(S_g = (E_g, V_g)\) that is no different from a graph produced by Prim's MST algorithm \(S_p = (E_p, V_p)\) from \(G\), after running a depth first search (DFS) on \(S_p\), and removing the duplicates, connecting edges that traversed to the duplicate vertices directly to the subsequent vertex after the removed vertex.
+\textbf{Claim 2}: See that the greedy traversal graph \(S_g\) will always result in requiring half of the number of edges to traverse all nodes via connected edges when compared to traversing a MST. To see this, we assert that \verb|GRD| produces a traversal graph \(S_g\) that is no different from a graph produced by Prim's MST algorithm \(S_p = (E_p, V_p)\) from \(G\), after running a depth first search (DFS) on \(S_p\), and removing the duplicates, connecting edges that traversed to the duplicate vertices directly to the subsequent vertex after the removed vertex.
 
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-This is because \verb|GRD| is substantially different only in the step of adding the selected vertex to the current graph. Where in Prim's, the algorithm selects the vertex \(s \in G\) associated with the lowest weighted edge that connects to a vertex \(i\) in the partial solution \(S_{pp} = (E_{pp}, V_{pp})\) and proceeding to the next iteration, \verb|GRD| selects the next vertex and edge identically, however, instead of moving to the next iteration, \verb|GRD| connects \(s \in G\) to the next node \(i\) is linked to \(k \in S_{pp}\). In other words, where Prim's may resolve to connect \(i \rightleftarrows s\) such that \( E_{pp} = \{\ldots, \{i, k\}, \{i, s\}, \ldots\} \), and a traversal by DFS results in a sequence \(\ldots \rightarrow i \rightarrow k \rightarrow i \rightarrow s \rightarrow \ldots \). \verb|GRD| resolves the newly selected vertex such that \( E_{pp} = \{\ldots, \{i, s\}, \{s, k\}, \ldots\} \), effectively changing \(i \rightleftarrows k\) to \(i \rightleftarrows s \rightleftarrows k \) where the traversal is trivially \( \ldots \rightarrow i \rightarrow s \rightarrow k \rightarrow \ldots \) thus maintaining the chain form of the graph. From this breakdown, we can see that the Prim's approach requires double the edges for the full traversal in contrast against the \verb|GRD| algorithm. 
+This is because \verb|GRD| is substantially different only in the step of adding the selected vertex to the current graph. Where in Prim's, the algorithm selects the vertex \(s \in G\) associated with the lowest weighted edge that connects to a vertex \(i\) in the partial solution \(S_{pp} = (E_{pp}, V_{pp})\) and proceeding to the next iteration, \verb|GRD| selects the next vertex and edge in the same fashion, however, instead of moving to the next iteration, \verb|GRD| connects the selected node, \(s \in G\), to the next node \(i\) was linked to \(k \in S_{pp}\). In other words, where Prim's may resolve to connect \(i \rightarrow s\) such that \( E_{pp} = \{\ldots, \{i, k\}, \{i, s\}, \ldots\} \), and a traversal by DFS results in a sequence \(\ldots \rightarrow i \rightarrow k \rightarrow i \rightarrow s \rightarrow \ldots \). \verb|GRD| resolves the newly selected vertex such that the partial traversal graph for \verb|GRD| is \(S_{gp} = (E_{gp}, V_{gp}) \) where \( E_{gp} = \{\ldots, \{i, s\}, \{s, k\}, \ldots\} \), effectively changing \(i \rightarrow k\) to \(i \rightarrow s \rightarrow k \) thus maintaining the chain form of the graph. 
 
 \smallskip
 
-However, to see that the methods are analogous and thus, comparable, the DFS traversal \(\ldots \rightarrow i \rightarrow k \rightarrow i \rightarrow s\ \rightarrow \ldots \) can be simplified into \(i \rightarrow k \rightarrow s\) (removing the second appearance of \(i\) in the sequence) without worsening the total weight required for the traversal by the triangle inequality (TI) assumption. To prove this, we may focus on \(k \rightarrow i \rightarrow s\), and see that \(c_{ks} \leq c_{ki} + c_{is}\) (TI assumption).
+We can then see that Prim + DFS does create double the edges than \verb|GRD|, since the DFS traversal \(\ldots \rightarrow i \rightarrow k \rightarrow i \rightarrow s\ \rightarrow \ldots \) can be simplified into \( \ldots \rightarrow i \rightarrow k \rightarrow s \rightarrow \ldots \) (removing the second appearance of \(i\) in the sequence) without worsening the total weight required for the traversal by the triangle inequality (TI) assumption. To prove this, we may focus on \(k \rightarrow i \rightarrow s\), and see that \(c_{ks} \leq c_{ki} + c_{is}\) (TI assumption).
 
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-From this, we can see that Prim's algorithm is known to generate a MST, and to traverse such a tree in it's entirety is double the cost of the cyclical traversal path provided by \verb|GRD|. In other words, we have proven \(2 c(S_p) = (c(S_g))\) or \(c(S_p) = \frac{1}{2}(c(S_g))\).
+From this, we can see that Prim's algorithm is known to generate a MST, and to traverse such a tree one vertex after another is double the cost of the cyclical traversal path provided by \verb|GRD|. In other words, we have proven \(2 c(S_p) = (c(S_g))\) or \(c(S_p) = \frac{1}{2}(c(S_g))\).
 
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-\textbf{Proof of 2-Approximation}:
+\textbf{Proof Algorithm is 2-Approximation}:
 
 \begin{align}
-	c(S_p) &\leq c(S_o - \{e\}) &(\text{Prim's Algorithm generates MST}) \\
+	c(S_p) &\leq c(S_o - \{e\}) &(\text{Prim's is MST therefore, costs \(\leq\) other spanning trees.}) \\
-	\frac{1}{2} c(S_g) &\leq c(S_o - \{e\}) &(\text{Claim 2}) \\
+	\frac{1}{2} c(S_g) &\leq c(S_o - \{e\}) &(\text{Claim 2 substitution}) \\
 	\frac{1}{2} c(S_g) &\leq c(S_o - \{e\}) \leq c(S_o) &(\text{Claim 1}) \\
 	c(S_g) &\leq 2c(S_o)
 \end{align}