Expanded on 2 existing practice problems

Author: AustinTSchaffer

OMSCS/Courses/GA/Practice Problems/3.16 - CS Curriculum.md

@@ -264,7 +264,7 @@ This requires that we iterate over the $n$ vertices in $TO$. For each vertex $v$
 
 This works. In fact, this process effectively finds the highest number of edges that lead backward from a given node (+1).
 
-## Algorithm
+## Algorithm 1
 We use the TopoSort algorithm to generate an array of the vertices of $G$ in topological order, calling it $TO$. We also generate the reverse of $G$, $G^R=(V, E^R)$.
 
 We initialize an array of length $n$, naming this array $ST$. This array holds the minimum semester during which a given course can be taken.
@@ -284,3 +284,25 @@ This is an adaptation of the "longest path" algorithm. See notes above.
 - Finding the max value in ST: $O(n)$
 
 Overall: $O(n+m)$
+
+## Algorithm 2
+**This is a minor simplification of Algorithm 1 which does not require $G^R$.**
+
+We use the TopoSort algorithm to generate an array of the vertices of $G$ in topological order, calling it $TO$.
+
+We initialize an array of length $n$, naming this array $ST$. This array holds the minimum **semester** during which a given course can be **taken**. Each course $v$ starts with $ST[v]=1$.
+
+For each course $v$ in V in $TO$, we check the outgoing edges to each course $u$ which explicitly requires $v$ as a prerequisite. We update $TO[u]$ to the max of $TO[u]$ and $TO[v]+1$.
+
+Now, we have the semesters that each course was taken packaged within $ST$. To find the minimum number of semesters required to complete this CS program, we take the maximum semester number found in $ST$.
+
+## Justification of Correctness
+This is an adaptation of the "longest path" algorithm. See notes above. We're simply propagating forward the information about the longest path through a given DAG $G$.
+
+## Runtime
+- Running TopoSort: $O(n+m)$
+- Initializing $ST$: $O(n)$
+- Iterating over $v \in V$ in $TO$, and $(v,u) \in E$, $ST[u] = \max\{ST[u], ST[v]+1\}$ for each of a course's dependents: $O(n+m)$
+- Finding the max value in ST: $O(n)$
+
+Overall: $O(n+m)$

OMSCS/Courses/GA/Practice Problems/3.7 - Bipartite Graphs.md

@@ -87,7 +87,7 @@ E -.B-.-> A
 D -.B-.-> B
 ```
 
-### Algorithm 2 (From TA)
+### Algorithm 3 (From TA)
 We can convert this problem to a 2-SAT problem. See [[04.2 - Graphs - 2-SAT]] for details on that algorithm. To encode this problem as a 2-SAT problem, we need to define the problem in terms of the conjunctive AND of multiple OR clauses, each with 2 literals. How do we do this?
 
 In a bipartite graph, all edges must connect one vertex $v \in S_1$ with a vertex in $u \in S_2$. $V = S_1 \cup S_2$ and $S_1 \cap S_2=\emptyset$. Therefore, we need to model vertex $v$ as being in one set or the other set, and vertex $u$ as being in the opposite set. This can be accomplished as an exclusive-or (XOR).
@@ -104,6 +104,19 @@ $$\big(v_a \vee v_b\big) \wedge \big(\overline{v_a} \vee \overline{v_b}\big) $$
 
 Taking the AND across all of these CNF-XOR conditions produces a 2-SAT problem. The conversion from $G$ to $f$ takes $O(n+m)$ time. The 2-SAT algorithm takes $O(n+m)$ time.
 
+### Algorithm 4 (From Classmate)
+This is an adaptation of my original idea of using BFS. We can circumvent the issue with running BFS n-times by making n disjoint subgraphs based on the output of DFS. All of these steps should execute in linear time.
+
+1. Run $DFS(G)$ to produce $ccnum[]$, identifying all connected components. O(n+m) 
+2. For each unique ccnum value $k$ build subgraph $G_k=(V_k,E_k)$.
+	1. add vertex $v$ to $V_k$ if $ccnum[v] = k$
+	2. add edge $e=(u,v)$ to $E_k$ if $ccnum[u] = ccnum[v] = k$. O(n+m)
+3. Initialize $color[v] = -1$ for all $v \in V$. O(n)
+4. For each subgraph $G_k$ pick any vertex $s \in V_k$ as a source. Assign $color[s] = 0$. Run $BFS(G_k, s)$, returning $dist_k[]$. The time complexity for this step is O(n+m) total across all subgraphs.
+5. For each vertex v in $V_k$ assign $color[v] = dist_k[v] \mod 2$. O(n)
+6. For each edge $e=(u,v) \in E_k$ check if $color[u] = color[v]$. If yes return “not bipartite.” O(m)
+7. If all subgraphs pass return “bipartite” with $V1 = \{v : color[v] = 0\}$ and $V2 = \{v : color[v] = 1\}$. O(n)
+
 ## 3.7b - Odd Cycles
 We can't have an odd-length cycle in a bipartite graph, because each edge must connect a vertex $v \in S_1$ with a vertex in $u \in S_2$, where $V = S_1 \cup S_2$ and $S_1 \cap S_2=\emptyset$.