More notes and PPs

Author: AustinTSchaffer

OMSCS/Courses/GA/04.2 - Graphs - 2-SAT.md

@@ -83,7 +83,7 @@ Simplification Example
 - **Satisfiable:** $X=[F,T,T,F]$
 
 ## Graph of Implications
-- Take $f$ with all clauses of size 2.\
+- Take $f$ with all clauses of size 2.
 	- $n$ variables
 	- $m$ clauses
 - Create a directed graph

OMSCS/Courses/GA/06.6 - NP - Known NP-Complete Problems.md

@@ -0,0 +1,128 @@
+---
+tags:
+  - OMSCS
+  - Algorithms
+---
+# 06.6 - NP - Known NP-Complete Problems
+
+## SAT
+- **Input:** $f$
+	- a boolean expression in CNF
+	- $n$ variables and $m$ clauses
+- **Output:**
+	- A satisfying assignment $S$ of $f$ if one exists, such that $f$ evaluates as true.
+	- No otherwise.
+
+## 3SAT
+- **Input:** $f$
+	- a boolean expression in CNF
+	- $n$ variables and $m$ clauses
+	- Each clause has up to 3 literals
+- **Output**
+	- A satisfying assignment $S$ of $f$ if one exists, such that $f$ evaluates as true.
+	- No otherwise.
+
+## Clique
+$S$ is an Clique of $G=(V,E)$ if $\forall_{a,b \in S}((a,b) \in E)$. Put simply, the induced subgraph $S$ of $G$ contains edges between all pairs of vertices.
+
+- **Input:**
+	- $G=(V,E)$
+	- goal $g$
+- **Output:**
+	- $S \subset V$ where $S$ is a clique of size $|S| \ge g$, if one exists, and "NO" otherwise.
+
+## Independent Set (IS)
+$S$ is an IS of $G=(V,E)$ if $\forall_{a,b \in S}((a,b) \notin E)$. Put simply, the induced subgraph $S$ of $G$ contains no edges.
+
+- **Input:**
+	- Undirected $G=(V,E)$
+	- Goal $g$
+- **Output:**
+	- Independent set $S$ where $|S| \ge g$ (if one exists), "No" otherwise
+
+## Vertex Cover (VC)
+$S$ "covers" $G$ if $\forall_{a,b \in E} \space (a \in S) \vee (b \in S)$. Put simply, all edges in $G$ connect to at least one vertex in $S$.
+
+- **Input**
+	- $G=(V,E)$
+	- budget $b$
+- **Output**
+	- vertex cover $S$ of size $|S| \le b$ if one exists
+	- No otherwise
+
+## Subset Sum (SSS)
+- **Input**
+	- A set of values $V$
+	- A goal $g$
+- **Output**
+	- A subset of $S \subseteq V$ whose sum equals $g$
+	- No otherwise
+
+## Rudrata Path
+This is the hamiltonian cycle problem, but renamed because he was centuries too late to claim the name of the problem.
+
+- **Input:**
+	- A simple graph $G=(V,E)$
+- **Output:**
+	- A path through $G$ which touches all vertices exactly once
+	- No otherwise
+
+## Rudrata st-Path
+This is an extension of the Rudrata Path problem.
+
+- **Input:**
+	- A simple graph $G=(V,E)$
+	- A pair of vertices $s$ and $t$
+- **Output:**
+	- A path through $G$ which touches all vertices exactly once
+	- The path starts at $s$ and ends at $t$
+	- No otherwise
+
+## Rudrata Cycle
+This is an extension of the Rudrata Path problem.
+
+- **Input:**
+	- A simple graph $G=(V,E)$
+- **Output:**
+	- A cycle in $G$ which touches all vertices exactly once
+	- No otherwise
+
+## Integer Linear Programming (ILP)
+- **Input:** A linear program (LP) in standardized form
+	- $\max \space c^Tx$
+	- s.t. $Ax \le b$
+	- s.t. $x \ge 0$
+- **Output:**
+	- An assignment to the variables of the LP, which maximizes the objective function within the given constraints, where the variables are all integers.
+	- No otherwise.
+
+## Zero-One Equations
+This is a special case of ILP.
+
+- **Input:**
+	- Matrix $A$ of size $m \times n$, containing only 0s and 1s
+- **Output:**
+	- An $n \times 1$ vector $x$
+		- $x$ contains only 0s and 1s
+		- such that $Ax=1$
+		- where $1$ is an $m \times 1$ vector containing only 1s
+	- No otherwise.
+
+## 3D Matching
+> You've heard of bipartite, now get ready for TRIPARTITE.
+
+- **Input**
+	- A set of triples defining the potential 3-groupings between 3 categories, each of size $n$ 
+- **Output**
+	- A set of $n$ disjoint triples from the given set of triples, such that all items are covered.
+	- No otherwise
+
+## Traveling Salesman Problem
+- **Input**
+	- A graph $G=(V,E)$
+	- A set of edge weights/costs/lengths $w$
+	- A budget $b$
+	- A starting vertex $s$ (optional?)
+- **Output**
+	- A cycle in $G$ (which starts and ends at $s$, if provided), which has total cost $\le b$.
+	- No otherwise.

OMSCS/Courses/GA/Office Hours/2026-04-12 - TA Office Hours - Exam 3.md

@@ -0,0 +1,79 @@
+---
+tags:
+  - OMSCS
+  - Algorithms
+---
+# 2026-04-12 - TA Office Hours - Exam 3
+- The Dual LP is a "certificate of optimality" for the Primal LP
+- We use the Dual to prove Feasibility and Boundedness
+- When checking for infeasibility, "try to find at least one contradiction"
+	- In the example $(\max \space x,x \le -3, x \ge 0)$, we can see that $x$ has no valid values.
+	- In the example $(\max \space x+y, x+y \le 1, x+y \ge 3, x,y \ge 0)$, then if we change it to standard form, we have the following conflicting expressions:
+		- $x+y \le 1$
+		- $-(x+y) \le 3$
+	- We can't always check $(0,0)$
+- Primal is Unbounded -> Dual is Infeasible
+- Primal is Infeasible -> Dual is Unbounded OR Infeasible
+- Primal is Bounded -> Dual is Bounded AND Feasible
+- Dual of Dual is the Primal
+- LP's cannot be Infeasible AND Bounded.
+
+## Optimal Solution Example
+- $\max 5x+3y$
+- s.t.
+	- $5x - 2y \ge 0$
+	- $x+y \le 7$
+	- $x \le 5$
+	- $x, y \ge 0$
+
+## Max Flow as LP
+![[Pasted image 20260412153326.png]]
+
+![[Pasted image 20260412153443.png]]
+
+## Graph Transformation Ideas (NP Proofs)
+- Transforming each vertex/edge
+- Duplicate the original graph
+- Add a structure to every existing vertex.
+	- Bowtie -> clique to every vertex
+	- The kite problem -> add a tail to every vertex
+- Create a "complement" graph
+	- $\overline{G}$
+	- Clique and IS are complements of each other
+- Be mindful to remember the budget/target.
+
+## Graph Transformation Pitfalls
+- Attempting to find the structure in the graph, where finding that structure is NP-Complete.
+- Adding edges to every vertex, which may break the induced subgraph
+- Forgetting the budget and/or target
+- Pseudo-polynomial runtimes that reference the budget or target. If you use $O(g^2)$, make sure to assert $g \le n$.
+- Removing added vertices. Example, in the Bowtie problem, the original G may have contained a bowtie. Removing all added vertices doesn't necessarily mean you've converted the bowtie solution to a clique. Make sure the output matches the output for the NP-Complete problem.
+
+## SAT Transformation Ideas
+- $(A)$ - Add a unit clause
+- $(A \vee \overline{A})$ - Add a tautology.
+- "Exactly one of A or B" - $(A \vee B)(\overline{A} \vee \overline{B})$
+- "At most one of A or B" - $(\overline{A} \vee \overline{B})$
+- "At least one of A or B" - $(A \vee B)$
+
+## SAT Transformation Pitfalls
+- Cannot add true/false constants
+- Need to maintain CNF
+- Variables do not exist in the boolean formula (they're called literals)
+- Literals do not exist in the satisfying assignment (they're variables)
+
+## 3SAT Verification Runtime
+- $\le 3$ literals per clause
+- We can lookup variable assignment in $O(1)$
+- There are $m$ clauses.
+- We need at most 3 lookups per clause
+- $O(3m) = O(m)$
+
+The SAT runtime require $O(nm)$ runtime, since we don't have a restriction on the number of literals per clause, apart from the number of variables ($n$) in the problem.
+
+## Set Transformations
+- Create the equivalent sets (e.g. Homework 10)
+- Watch out for set membership checks ($O(n)$)
+- "Don't worry about optimizing everything with boolean arrays"
+- Watch for terms $n$ and $m$. Use $|S|$, $|V|$, $|D|$, etc.
+

OMSCS/Courses/GA/Office Hours/images/Pasted image 20260412151229.png

@@ -0,0 +1,13 @@
+---
+tags:
+  - OMSCS
+  - Algorithms
+  - Practice
+---
+# 6.17 - Making Change (TODO)
+> Given an unlimited supply of coins of denominations $x_1,x_2,...,x_n$, we wish to make change for a value $v$; that is, we wish to find a set of coins whose total value is $v$. This might not be possible: for instance, if the denominations are 5 and 10 then we can make change for 15 but not for 12.
+> 
+> Give an $O(nv)$ dynamic-programming algorithm for the following problem.
+
+**Input:** $[x_1,x_2,...,x_n];\space v$
+**Question:** Is it possible to make change for $v$ using coins of denominations $x_1,...,x_n$?

OMSCS/Courses/GA/Office Hours/images/Pasted image 20260412153326.png

@@ -0,0 +1,92 @@
+---
+tags:
+  - OMSCS
+  - Algorithms
+  - Practice
+---
+# 7.25 - The Dual of Maximum Flow
+Consider the following network with edge capacities.
+
+```mermaid
+graph LR
+
+S --[1]--> A
+A --[1]--> B
+S --[3]--> B
+A --[2]--> T
+B --[1]--> T
+```
+
+1. Write the problem of finding the maximum flow from S to T as a linear program.
+2. Write down the dual of this linear program. There should be a dual variable for each edge of the network and for each vertex other than $S,T$.
+
+## Max-Flow as LP
+This is a laborious process in which we remodel the flow network as a series of equalities, which must then be converted to inequalities to satisfy the format of an LP. We can introduce variables for the in/out flow for each vertex ($S_o,A_i,A_o,B_i,B_o,T_i \ge 0$), but someone on the course forum showed that we can shortcut a little bit by encoding those flows directly from edge to edge.
+
+- Objective: $\max \{AT + BT\}$
+- s.t.
+	- $SA,SB,AB,AT,BT \ge 0$
+	- $SA \le 1$
+	- $SB \le 3$
+	- $AB \le 1$
+	- $AT \le 2$
+	- $BT \le 1$
+	- $SA = AB + AT$
+		- $SA - AB - AT = 0$
+		- $SA - AB - AT \le 0$
+		- $-SA + AB + AT \le 0$
+	- $SB + AB = BT$
+		- $SB+AB-BT=0$
+		- $SB+AB-BT \le 0$
+		- $-SB-AB+BT \le 0$
+
+$$
+	\begin{matrix}
+		\max\{c^Tx\} \\
+		Ax \le b \\
+		x \ge 0 \\
+		
+	\end{matrix} \space\space
+	x=\begin{bmatrix}SA \\ SB \\ AB \\ AT \\ BT\end{bmatrix} \space\space
+	c=\begin{bmatrix}0 \\ 0 \\ 0 \\ 1 \\ 1\end{bmatrix}
+	\space\space
+	b=\begin{bmatrix}1 \\ 3 \\ 1 \\ 2 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0\end{bmatrix}
+	\space\space
+	A=\begin{bmatrix}
+		1 & 0 & 0 & 0 & 0 \\
+		0 & 1 & 0 & 0 & 0 \\
+		0 & 0 & 1 & 0 & 0 \\
+		0 & 0 & 0 & 1 & 0 \\
+		0 & 0 & 0 & 0 & 1 \\
+		1 & 0 & -1 & -1 & 0 \\
+		-1 & 0 & 1 & 1 & 0 \\
+		0 & 1 & 1 & 0 & -1 \\
+		0 & -1 & -1 & 0 & 1 \\
+	\end{bmatrix}
+$$
+
+## Dual LP
+$$
+	\begin{matrix}
+		\min\{b^Ty\} \\
+		A^Ty \ge c \\
+		y \ge 0 \\
+		
+	\end{matrix} \space : \space
+	y=\begin{bmatrix}y_1 \\ y_2 \\ y_3 \\ y_4 \\ y_5 \\ y_6 \\ y_7 \\ y_8 \\ y_9 \end{bmatrix} \space\space
+	c=\begin{bmatrix}0 \\ 0 \\ 0 \\ 1 \\ 1\end{bmatrix}
+	\space\space
+	b=\begin{bmatrix}1 \\ 3 \\ 1 \\ 2 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0\end{bmatrix}
+	\space\space
+	A=\begin{bmatrix}
+		1 & 0 & 0 & 0 & 0 \\
+		0 & 1 & 0 & 0 & 0 \\
+		0 & 0 & 1 & 0 & 0 \\
+		0 & 0 & 0 & 1 & 0 \\
+		0 & 0 & 0 & 0 & 1 \\
+		1 & 0 & -1 & -1 & 0 \\
+		-1 & 0 & 1 & 1 & 0 \\
+		0 & 1 & 1 & 0 & -1 \\
+		0 & -1 & -1 & 0 & 1 \\
+	\end{bmatrix}
+$$

OMSCS/Courses/GA/Office Hours/images/Pasted image 20260412153443.png

@@ -5,3 +5,29 @@ tags:
   - Practice
 ---
 # 7.26 - Satisfiable System of Linear Inequalities (TODO)
+
+In a satisfiable system of linear inequalities
+
+$$\begin{matrix}
+a_{11}x_1 + ... + a_{1n}x_n & \le & b_1 \\
+& \vdots & \\
+a_{m1}x_1 + ... + a_{mn}x_n & \le & b_m \\
+\end{matrix}
+$$
+
+we describe the $j$th inequality as *forced-equal* if it is satisfied with equality by *every* solution $x=(x_1,...,x_n)$ of the system. Equivalently, $\sum_i a_{ji}x_i \le b_j$ is *not* forced-equal if there exists an $x$ that satisfies the whole system and such that $\sum_i a_{ji}x_i \lt b_j$.
+
+For example, in
+$$\begin{matrix}
+x_1 + x_2 & \le & 2 \\
+-x_1 - x_2 & \le & -2 \\
+x_1 & \le & 1 \\
+-x_2 & \le & 0 \\
+\end{matrix}
+$$
+
+the first two inequalities are forced-equal, while the third and fourth are not. A solution $x$ to the system is called *characteristic* if, for every inequality $I$ that is not forced-equal, $x$ satisfies $I$ without equality. In the instance above, such a solution is $(x_1,x_2)=(-1,3)$ for which $x_1 \lt 1$ and $-x_2 \lt 0$ while $x_1+x_2=2$ and $-x_1-x_2=-2$.
+
+1. Show that any satisfiable system has a characteristic solution.
+2. Given a satisfiable system of linear inequalities, show how to use linear programming to determine which inequalities are forced-equal, and to find a characteristic solution.
+