More PPs

Author: AustinTSchaffer

OMSCS/Courses/GA/Practice Problems/8.13 - NP-Complete Spanning Tree Problems (TODO).md

@@ -0,0 +1,50 @@
+---
+tags:
+  - OMSCS
+  - Algorithms
+  - Practice
+---
+# 8.13 - NP-Complete Spanning Tree Problems
+Determine which of the following problems are NP-complete and which are solvable in polynomial time. In each problem you are given an undirected graph $G=(V,E)$, along with...
+
+## Subproblem (a)
+> ...a set of nodes $L \subseteq V$, and you must find a spanning tree such that its set of leaves includes the set $L$.
+
+- $G'$ is a copy of $G$
+- For all vertices $v \in L$, remove $v$ from $G'$
+- Find a spanning tree $T=(V-L,E')$ of $G'$
+- For all vertices $u \in L$, find an edge from $u$ to any vertex $v \in (V-L)$, where $(u,v) \in E$. Add $u$ to and $(u,v)$ to $T$.
+
+If any of these steps fail, the problem is not possible. 
+
+## Subproblem (b)
+> ...a set of nodes $L \subseteq V$, and you must find a spanning tree such that its set of leaves is precisely the set $L$.
+
+This one is NP-Complete, because you can use a graph G and $L=[s,t]$ in order to find an s-t Rudrata Path in G.
+
+## Subproblem (c)
+> ...a set of nodes $L \subseteq V$, and you must find a spanning tree such that its set of leaves is included in the set $L$.
+
+This variant is more permissive than Subproblem B, but you can still use $L=[s,t]$ in order to find an s-t Rudrata Path in G. You can't have a spanning tree with less than 2 leaves, so the permissiveness of this problem variant is immaterial.
+
+## Subproblem (d)
+> ...an integer $k$, and you must find a spanning tree with $k$ or fewer leaves.
+
+This one is NP-Complete, because you can use a graph G and $k=2$ in order to find a Rudrata Path in G. You can't have a tree with less than 2 leaves, so the "or fewer" permissiveness of this problem variant is immaterial.
+
+## Subproblem (e)
+> ...an integer $k$, and you must find a spanning tree with $k$ or more leaves.
+
+The issue with this problem is you need to select a subset of vertices $L \subseteq V$ where $|L| \ge k$ such that $\forall_{u \in L} \exists_{v \in (V-L)} (u,v) \in E$. Essentially, you can't select a vertex to join $L$ if it only connects to vertices which are already in L. If you have a "star" graph $G^*$ with $n$ vertices, and $k=n-1$, the obvious thing to do is let $L$ contain all vertices which aren't the central point. However, if a naive algorithm selects the central vertex first, it won't be able to find a 2nd vertex to add to $L$, requiring backtracking. In the worst case, it keeps coming back to that central vertex, only to find that it can't grow $L$.
+
+We can validate a solution to this variant of the problem by checking the degree of all vertices, and making sure that there are at least $k$ vertices with degree 1. We can also run DFS on the graph to make sure every vertex is reachable from a single starting vertex, and then ensuring that no back-edges exist in $G$. Therefore, this variant of the problem is definitely in NP.
+
+Unclear how to proceed. Thankfully this problem variant isn't even on the list of recommended practice problems.
+
+## Subproblem (f)
+> ...an integer $k$, and you must find a spanning tree with exactly $k$ leaves.
+
+This one is NP-Complete, because you can use a graph G and $k=2$ in order to find a Rudrata Path in G.
+
+## Hint
+All the NP-completeness proofs are by generalization, except for one.

OMSCS/Courses/GA/Practice Problems/8.13 - NP-Complete Spanning Tree Problems.md

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+---
+tags:
+  - OMSCS
+  - Algorithms
+  - Practice
+---
+# 8.15 - Maximum Common Subgraph
+
+Show that the following problem is NP-complete.
+- **Maximum Common Subgraph (MCS)**
+- **Input:** Two graphs $G_1=(V_1,E_1)$ and $G_2 = (V_2,E_2)$; a budget $b$.
+- **Output:** Two set of nodes $V'_1 \subseteq V_1$ and $V'_2 \subseteq V_2$ whose deletion leaves at least $b$ nodes in each graph, and makes the two graphs identical.
+
+## MCS in NP Notes
+How do we validate the solution in P-time?
+
+- Count vertices, make sure they're the same.
+- Count edges, make sure they're the same.
+- Get vertex degrees, make sure they have the same distribution of vertex degrees.
+- If undirected
+	- Check number/sizes of connected components.
+- If directed
+	- Check SCCs
+- Wait, this one might not be verifiable in P-time.
+
+The Graph Isomorphism problem is unknown as whether it's in P or NP-Complete. In the general case, this question if MCS in NP is indeterminate. In the context of this course/problem, we assume that the graphs $G_1$ and $G_2$ were derived from the same set of vertices, and therefore $G_1'$ and $G_2'$ have the same vertex labels.
+
+## MCS in NP
+A solution to the MCS problem can be verified through a multi-step polynomial-time process.
+
+- Construct $G'_1$ and $G'_2$ by removing $V_1'$ from $V_1$ and removing $V_2'$ from $V_2$. This step is at worst $O(n^2)$.
+- Count the number of vertices in both $G_1'$ and $G_2'$. Make sure the are the same number and equal to $b$. This step is $O(n)$.
+ - For all vertices in $G_1'$, check that the vertex exists in $G_2'$. This step is at worst $O(n^2)$.
+ - Count the number of edges in both $G'_1$ and $G'_2$. Make sure this is the same. This step is $O(n+m)$.
+ - For each edge in $G_1'$, check to make sure that the edge exists in $G_2'$. This step is $O((n+m)^2)$
+
+The overall verification runtime is $O((n+m)^2)$, which is polynomial in the size of $I$. Therefore, the MSC problem is in NP.
+
+## Reduction Idea
+- We can easily reduce from Clique problem or the IS problem. We will use the IS problem because I always use the Clique problem.
+- Set $G_1=(V_1,E_1)=G=(V,E)$
+- Create an IS of size $|V|$ by making a $G_2=(V_2,\emptyset)$, where $|V_2|=|V|$.
+- Set $b=g$
+- Since $G_2$ contains all possible ISets from 1 to $|V|$, the solution to the MCS problem must be 2 ISs of size $b$, where one of the ISs came from $G$.
+- Return $S=V-V'_1$ as a solution to the IS problem.
+
+**A big note**, I originally wrote to create an IS of only size $g$. This presents a problem with the note from earlier, in that the Graph Isomorphism problem is unknown as being part of P or NP-Complete. Therefore, we must create an IS with the same size as $G$, so that an IS in $G$ of size $g$ can be found which has the same vertex labels as the artificially constructed IS.
+
+## Reduction from IS to MCS
+We will demonstrate that the IS problem can be reduced to the MCS problem, demonstrating that the MCS problem is in NP-Hard.
+
+### Input Transformation
+Take an instance $I=(G=(V,E), g)$ of the IS problem.
+
+- Create $G_1$, a perfect copy of $G$. $O(n+m)$
+- Create $G_2$, a copy of $G$ containing all the vertices of $G$, but no edges. $O(n)$
+- Set $b=g$. This is $O(1)$
+
+Use $I'=(G_1,G_2,b)$ as the instance of the MCS problem. This transformation requires $O(n+m)$ time, which is polynomial in $|I|$.
+
+### Output Transformation
+If there is no solution to the MCS problem, return "no" as the solution to the IS problem.
+
+If there is a solution to the MCS problem, we are given $S=(V'_1, V'_2)$. Create the solution to the IS problem $S'$ by removing all vertices in $V_1'$ from $V$. This requires $O(n^2)$ time at worst.
+
+### Justification
+The IS problem seeks to find a subset of vertices $S \subseteq V$ which have no edges between them in $G$, where $|S|=g$. We can find that subset using the MCS problem, by artificially constructing $G_2$ which only contains vertices, and is therefore an IS of size $|V|$. All independent sets in $G$ of any size also exist in $G_2$. Therefore, if we find a common subgraph between $G$ and $G_2$, that common subgraph is an IS. If that subgraph is of size $b=g$, then we have a solution to the IS problem.
+
+If the IS problem has a solution, then there exists an IS in G of size $g$. Therefore, there exists a solution to the MCS problem, because that IS can pair up with the corresponding vertices of $G_2$ to form 2 identical subgraphs.
+
+If the IS problem doesn't have a solution, there there does not exist an IS in G of size $g$. Therefore, there does not exist a solution to the MCS problem, because there is no way to extract a subgraph of G of size $b$ which matches any subgraph of $G_2$ of size $b$.
+
+Therefore, the MCS problem has a solution if-and-only-if the IS problem has a solution.
+
+## Conclusion
+We have demonstrated that the MCS problem is in NP, by demonstrating a polynomial time solution verification algorithm. We have also shown that the MCS problem is in NP-Hard, by demonstrating that there exists a reduction from the IS problem to the MCS problem. Therefore, the MCS problem is in NP-Complete.

OMSCS/Courses/GA/Practice Problems/8.15 - Maximum Common Subgraph (TODO).md

@@ -10,11 +10,11 @@ tags:
 - **8.10(b-f)** – _Proving **NP**-Completeness by generalization._
 	- [[8.10 - NP-Completeness by Generalization]]
 - **8.13(a-d, f)** – Determine which of the following problems are **NP**-Complete...
-	- [[8.13 - NP-Complete Spanning Tree Problems (TODO)]]
+	- [[8.13 - NP-Complete Spanning Tree Problems]]
 - **8.14** – Clique & Independent Set
 	- [[8.14 - Clique + IS]]
 - **8.15** – Maximum Common Subgraph
-	- [[8.15 - Maximum Common Subgraph (TODO)]]
+	- [[8.15 - Maximum Common Subgraph]]
 - **8.17** – problem Π
 	- [[8.17 - Problem Pi]]
 - **8.20** – Dominating Set