More notes and etc

Author: AustinTSchaffer

OMSCS/Courses/GA/08.2 - NP - Knapsack.md

@@ -52,5 +52,4 @@ Recall that Knapsack's DP solution requires $O(nB)$ time, which is not polynomia
 
 
 ## Ungraded Homework
-Prove that Knapsack is NP-Complete: [[8.99.1 - Knapsack is NP-Complete (TODO)]]
-	
+Prove that Knapsack is NP-Complete: [[8.99.1 - Knapsack is NP-Complete]]

OMSCS/Courses/GA/Practice Problems/8.10f - NP-C Generalization - Set Cover.md

@@ -4,12 +4,11 @@ tags:
   - Algorithms
   - Practice
 ---
-# 8.20 - Dominating Set (TODO)
+# 8.20 - Dominating Set
 > In an undirected graph $G=(V,E)$, we say $D \subseteq V$ is a dominating set if every $v \in V$ is either in $D$ or adjacent to at least one member of $D$. In the DOMINATING SET problem, the input is a graph and a budget $b$, and the aim is to find a dominating set in the graph of size at most $b$, if one exists. Prove that this problem is NP-complete.
 
-This feels similar to Homework 10 and also Set Cover from 8.10f ([[8.10 - NP-Completeness by Generalization]]). In this problem, we can reproject the edges of $G$ in a Vertex Cover problem as vertices in $G'$ in a Set Cover / "Allocations" problem.
+This feels similar to Homework 10 and also Set Cover from 8.10f ([[8.10 - NP-Completeness by Generalization]]). In 8.10f and HW10, we can reproject the edges of $G$ in a Vertex Cover problem as vertices in $G'$ in a Set Cover / "Allocations" problem. We don't get to restrict the set of vertices from $G$ that Dominating Set is allowed to consider. Also in a Bipartite graph, the only way to ensure a dominating set would be to select at least all vertices in one of the partitions of the graph.
 
-We don't get to restrict the set of vertices from $G$ that Dominating Set is allowed to consider. Also in a Bipartite graph, the only way to ensure a dominating set would be to select at least all vertices in one of the partitions of the graph.
-
-What other problems are similar and available in the course? [[06.1 - NP - Theory Guidance]]
+One note is that in min-VC, isolated vertices are never selected. In min-DS, isolated vertices are always selected. Therefore, we definitely need to remove degree-0 vertices from G so they don't show up in the DS problem's solution.
 
+Another note is that only one vertex in a given clique needs to be selected in min-DS. However, in min-VC, $n-1$ vertices from a clique of size $n$ must be selected in order to cover all edges within that clique. A helpful note from the TA mentioned projecting each edge as a vertex within $G$, keeping the edge. This "locks" the vertex behind that edge, requiring that the edge be covered in order to also "dominate" the vertex. This requires some additional output transformation, when one of those artificial vertices is selected for the dominating set.

…oblems/8.15 - Maximum Common Subgraph.md …8.15 - Maximum Common Subgraph (TODO).mdOMSCS/Courses/GA/Practice Problems/8.15 - Maximum Common Subgraph.md renamed to OMSCS/Courses/GA/Practice Problems/8.15 - Maximum Common Subgraph (TODO).md OMSCS/Courses/GA/Practice Problems/8.15 - Maximum Common Subgraph.md renamed to OMSCS/Courses/GA/Practice Problems/8.15 - Maximum Common Subgraph (TODO).md

@@ -0,0 +1,30 @@
+---
+tags:
+  - OMSCS
+  - Algorithms
+  - Practice
+---
+# 8.99.1 - Knapsack is NP-Complete
+Prove that the Knapsack problem is NP-Complete. Relevant lectures:
+- [[02.1 - Dynamic Programming 2 - Knapsack]]
+- [[08.2 - NP - Knapsack]]
+
+Can't we just reduce from SSS? The book refers to these as being the same problem.
+
+- Input Transformation
+	- Convert each value from the SSS problem to a knapsack item with value/weight equal to the SSS value.
+	- Choose the variant of knapsack which disallows repeating items.
+	- Set $B=g$
+	- Set $g=g$
+- Output Transformation
+	- If no return no.
+	- Otherwise, return $S$.
+	- ~~This one is interesting, because Knapsack does not ever return "No". Instead it attempts to maximize the value gained from the available items, within capacity $B$.~~ Nevermind there's a search variant of Knapsack.
+
+That said, knapsack is a maximizing problem, not a search problem. How do we validate that the result is the max?
+
+From the textbook, a goal $g$ is added to make this problem a search problem.
+
+> We are also given a weight capacity $W$ and a goal $g$ (the former is present in the original optimization problem, the latter is added to make it a search problem).
+
+So in the search variant of the knapsack problem, we simply add $g=g$ to ensure that the goal is the same as the capacity.

OMSCS/Courses/GA/Practice Problems/8.20 - Dominating Set (TODO).md

@@ -14,7 +14,7 @@ tags:
 - **8.14** – Clique & Independent Set
 	- [[8.14 - Clique + IS]]
 - **8.15** – Maximum Common Subgraph
-	- [[8.15 - Maximum Common Subgraph]]
+	- [[8.15 - Maximum Common Subgraph (TODO)]]
 - **8.17** – problem Π
 	- [[8.17 - Problem Pi]]
 - **8.20** – Dominating Set