More practice problems

Author: AustinTSchaffer

OMSCS/Courses/GA/00.7 - Common Course Runtimes.md

@@ -67,3 +67,4 @@ In this course, we will always treat graphs as adjacency lists, with an indexabl
 	- If the graph is a clique
 		- $m=n(n-1)/2=O(n^2)$
 		- $O(n+m)=O(n+n^2)=O(n^2)$
+o

OMSCS/Courses/GA/Office Hours/2026-03-15 - TA Office Hours - Exam 2.md

@@ -7,8 +7,8 @@ tags:
 This OH covers the following strategies, outlined below. It also covers the following practice problems:
 - [[4.3 - Squares]]
 - [[5.9 - Properties of Weighted Undirected Graphs]]
-- [[7.22 -]]
-- [[7.24 -]]
+- [[7.22 - Fast Max-Flow Recomputation (TODO)]]
+- [[7.24 - Direct Bipartite Matching (TODO)]]
 
 ## Graph Strategies
 - Check for connectivity with BFS or DFS
@@ -34,7 +34,7 @@ This OH covers the following strategies, outlined below. It also covers the foll
 
 ## Max-Flow Strategies
 - Convert to flow network
-	- [[7.3 -]]
+	- [[7.3 - Cargo Plane (TODO)]]
 	- Adding new source and sink
 - Updating capacities
 	- infinite capacity

OMSCS/Courses/GA/Practice Problems/4.3 - Squares.md

@@ -82,7 +82,5 @@ However, this doesn't work for fully-connected graphs. In a fully connected grap
 1. Run FW
 2. Find $iter[1][\cdot][\cdot]=2$
 	1. Index 1 is the number of intermediate nodes
-	2. 
-3. 
 
 ![[Pasted image 20260316155849.png]]

OMSCS/Courses/GA/Practice Problems/5.20 - Perfect Matching Tree.md

@@ -41,7 +41,7 @@ I believe this works, but it doesn't use a black box from [[04.0.1 - Graphs - Bl
 
 ## Justifications
 - Iterating over vertices by descending order of preorder number prioritizes vertices which are "fringe" vertices. The vertex with the highest preorder number has only one edge.
-- We walk "up" the tree from those fringe vertices, towards the arbitrary starting vertex, removing connected pairs from the tree as we go.
+- We walk "up" the *tree* from those fringe vertices, towards the arbitrary starting vertex, removing connected pairs from the tree as we go.
 - We run into an issue when 2 vertices $v_1$ and $v_2$ share the same parent ($u=prev[v_1]=prev[v_2]$), and are independently selected as $v \in V_{pre}$ in our algorithm. This indicates that the DFS from $s$ encountered a branch at $u$. $u$ is the root of of a subtree $T_u$ of $G$, and removing $u$ from $T_u$ produces at least 2 additional subtrees. At least 2 of these subtrees, symbolized $T_{u,a}$ and $T_{u,b}$, are both independently incapable of producing a $PM$ set of edges. Therefore, both must include an edge to $u$ in order to produce a $PM$ set. However, this would require that the $PM$ set of $G$ contains $u$ at least twice, which is a contradiction in the definition of what a $PM$ set of edges is.
 
 ## Runtime
@@ -50,4 +50,12 @@ I believe this works, but it doesn't use a black box from [[04.0.1 - Graphs - Bl
 - Creating $V_{pre}$: $O(n)$
 - Iterating over $V_{pre}$: $O(n)$
 
-Overall: $O(n+m)$
+Overall: $O(n+m)$
+
+## Notes from TA Solution
+> It's a little bit more creative \[than running MST\].
+
+- We know by the problem statement that there needs to be $\ge 2$ leaves. If the graph only has 2 connected vertices, then the graph only has 1 edge.
+- Tim mentions the "degree" of each vertex in the resulting $G=(V,E')$. In the resulting graph, each $v \in V$ will have degree 1.
+- The approach is based on taking the leaf edges and then re-evaluating the graph. Take a leaf and the vertex it connects to. Remove both from the tree. Keep going until you have an empty graph or an isolated vertex.
+- The TA solution follows the procedure outlined in the **Exploration** section above, which unfortunately does not use a black-box algorithm.

OMSCS/Courses/GA/Practice Problems/5.22 - Fast MST Recomputation.md

@@ -8,17 +8,28 @@ tags:
 ![[Pasted image 20260316211257.png]]
 
 ## 5.22a
-Do nothing. $e$ is already not in the MST of G, and now it's a worse option.
+> $e$ is not currently part of the MST, and the new weight of $e$ is now higher.
+
+Do nothing. $e$ is already not in the MST of G, and now it's a worse option. $T$ is still the MST.
 
 ## 5.22b
+> $e$ is not currently part of the MST, and the new weight of $e$ is now lower.
+
 - Add $e$ to $E'$
 - Find the cycle in $T$ containing $e$
 - Remove max edge in $C$.
 
 ## 5.22c
-Do nothing. $e$ is already in the MST of G, and now $e$ is/was an even better choice across the cut of $S$ and $\overline{S}$, where $e=(u,v)$ and $u \in S$ and $v \in \overline{S}$.
+> $e$ is currently part of the MST, and the new weight of $e$ is now lower.
+
+Do nothing. $e$ is already in the MST of G, and now $e$ is/was an even better choice across the cut of $S$ and $\overline{S}$, where $e=(u,v)$ and $u \in S$ and $v \in \overline{S}$. $T$ is still the MST.
+
+## 5.22d
+> $e$ is currently part of the MST, and the new weight of $e$ is now higher.
 
-## 5.23d
 $e=(u,v)$ defines a $cut(S,\overline{S})$ where $u \in S$ and $v \in \overline{S}$. We need to find $\hat{e}=\min\{ cut(S, \overline{S}) \}$. Add $\hat{e}$ to $E'$. Remove $e$ from $E$.
 
-How do we find $\hat{e}=\min\{ cut(S, \overline{S}) \}$ in linear time? If we remove $e$ from $E'$, we can run DFS on $T$, to produce a `ccnum[]`, which identifies $S$ and $\overline{S}$. We can then iterate over $\hat{e}=(u,v) \in E$, filtering out any $\hat{e}$ where $ccnum[u]=ccnum[v]$. Those options for $\hat{e}$ do not cross $cut(S, \overline{S})$, and therefore create a cycle within $S$ or $\overline{S}$, and do not complete the MST. For the options for $\hat{e}$ that remain, we take the one with the lowest edge weight.
+How do we find $\hat{e}=\min\{ cut(S, \overline{S}) \}$ in linear time? If we remove $e$ from $E'$, we can run DFS on $T$, to produce a `ccnum[]`, which identifies $S$ and $\overline{S}$. We can then iterate over $\hat{e}=(u,v) \in E$, filtering out any $\hat{e}$ where $ccnum[u]=ccnum[v]$. Those options for $\hat{e}$ do not cross $cut(S, \overline{S})$, and therefore create a cycle within $S$ or $\overline{S}$, and do not complete the MST. For the options for $\hat{e}$ that remain, we take the one with the lowest edge weight.
+
+## Notes from TA Solution
+**For 5.22b**, instead of adding $e$ first, we can find the cycle that $e$ **would be** apart of if $e$ is added to the MST. For the given edge $e=(u,v)$, we run a DFS from $u$. This will find $v$ through some other path $P$. We can then reconstruct that path to find the heaviest edge $e' \in P$. If $e'$ is heaver than $e$, we remove $e'$ from $T$ and add $e$ to $T$.

OMSCS/Courses/GA/Practice Problems/5.23 - Light Spanning Trees.md

@@ -13,7 +13,9 @@ tags:
 1. We make a copy of $G$, $G'=(V',E')$, removing all $v \in U$ from $V'$.
 2. We check for connectivity by running DFS on $G'$. If $G'$ is not connected, then we can't make a tree where all of the vertices in $U$ are leaves of a light spanning tree of $G$.
 3. We run Kruskal's on $G'$ to produce an MST of $G'$: $T=(V',E'')$
-4. We then iterate over all $u \in U$, and find the minimum edge $e_u$ between $u$ and a vertex in $V'$. This edge represents the $\min\{cut(V', \{u\}\}$. We add $e_u$ to a new set $E_u$.
+4. We then iterate over all $u \in U$, and find the minimum edge $e_u$ between $u$ and a vertex in $V'$. This edge represents the $\min\{cut(V', \{u\}\}$.
+	1. If $e_u$ does not exist, then $u$ must only connect to other vertices in $U$. Therefore, we cannot satisfy the prompt.
+	2. Otherwise, we add $e_u$ to a new set $E_u$.
 5. We add all edges in $E_u$ to $E''$
 
 ## Justification

OMSCS/Courses/GA/Practice Problems/7.18 - Max Flow Variations.md

@@ -0,0 +1,20 @@
+---
+tags:
+  - OMSCS
+  - Algorithms
+  - Practice
+---
+# 7.18 - Max Flow Variations (TODO)
+![[Pasted image 20260317230911.png]]
+
+## 7.18a
+> There are many sources and many sinks, and we wish to maximize the total flow from all sources to all sinks.
+
+If there are multiple sources, we can add a super source $SS$ that connects to all sources $S$. The capacity of the edges from SS to each $s \in S$ should be at least the sum of the flow leaving $s$. The capacity of these edges can be made infinite. Same for multiple sinks. Just make one big sink.
+
+Before sending the flow to the asker, remove the edges which connect from the super source, and the edges which connect to the super sink.
+
+## 7.18b
+> Each vertex also has a capacity on the maximum flow that can enter it.
+
+For each vertex, duplicate it, and add an edge between the 2 copies of the vertex. That edge should have capacity equal to the max flow that can enter the vertex.

OMSCS/Courses/GA/Practice Problems/7.18 -.md

@@ -0,0 +1,29 @@
+---
+tags:
+  - OMSCS
+  - Algorithms
+  - Practice
+---
+# 7.19 - Max Flow Validation
+> Suppose someone presents you with a solution to a max-flow problem on some network. Give a linear time algorithm to determine whether the solution does indeed give a maximum flow.
+
+- What is the solution to a max flow problem?
+- [[04.0.1 - Graphs - Black Box Algorithms]]
+- The outputs are
+	- `flow[]` - A list of edges representing the amount capacity used per each indexed edge such that the flow is maximized from the starting vertex to the terminating vertex
+	- `C` - the total flow.
+
+We need to reconstruct the residual network and check if there exists a path from $s$ to $t$ in the residual network (an "st-path").
+
+How to reconstruct the residual:
+
+- From: [[05.2 - Max Flow - MF MC]]
+- Original flow network is directed $G=(V,E)$ with capacities $c_e \in C : e \in E$
+- Construct a residual graph containing the vertices of $G$ but no edges ($G^f=(V, \emptyset)$)
+- For each edge $e=(v,w) \in E$
+	- Add $vw$ to $G^f$
+	- Add $wv$ to $G^f$
+	- Set $c^f_{vw}=c_{vw}-f_e$
+	- Set $c^f_{wv}=f_e$
+
+We can then run DFS to check for an st-path in $G^f$. If none exists, then the flow through $G$ is already maximized.