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+# CSE5313 Computer Vision (Lecture 16: Exam Review)
+
+## Exam Review
+
+### Information flow graph
+
+Parameters:
+
+- $n$ is the number of nodes in the initial system (before any node leaves/crashes).
+- $k$ is the number of nodes required to reconstruct the file $k$.
+- $d$ is the number of nodes required to repair a failed node.
+- $\alpha$ is the storage at each node.
+- $\beta$ is the edge capacity **for repair**.
+- $B$ is the file size.
+
+#### Graph construction
+
+Source: System admin.
+
+Sink: Data collector.
+
+Nodes: Storage servers.
+
+Edges: Represents transmission of information. (Number of $\mathbb{F}_q$ elements is weight.)
+
+Main observation:
+
+- $k$ elements (number of servers required to reconstruct the file)  The message size is $B$. from $\mathbb{F}_q$ must "flow" from the source (system admin) to the sink (data collector).
+- Any cut $(U,\overline{U})$ which separates source from sink must have capacity at least $k$.
+
+### Bounds for local recoverable codes
+
+#### Turan's Lemma
+
+Let $G$ be a graph with $n$ vertices. Then there exists an induced directed acyclic subgraph (DAG) of $G$ on at least $\frac{n}{1+\operatorname{avg}_i(d^{out}_i)}$ nodes, where $d^{out}_i$ is the out-degree of vertex $i$.
+
+#### Bound 2
+
+Consider the induced acyclic graph $G_U$ on $U$ nodes.
+
+By the definition of $r$-locally recoverable code, each leaf node in $G_U$ must be determined by other nodes in $G\setminus G_U$, so we can safely remove all leaf nodes in $G_U$ and the remaining graph is still a DAG.
+
+Let $N\subseteq [n]\setminus U$ be the set of neighbors of $U$ in $G$.
+
+$|N|\leq r|U|\leq k-1$.
+
+Complete $n$ to be of the size $k-1$ by adding elements not in $U$.
+
+$|C_N|\leq q^{k-1}$
+
+Also $|N\cup U'|=k-1+\lfloor\frac{k-1}{r}\rfloor$
+
+All nodes in $G_U$ can be recovered from nodes in $N$.
+
+So $|C_{N\cup U'}|=|C_N|\leq q^{k-1}$.
+
+Therefore, $\max\{|I|:C_I<q^k,I\subseteq [n]\}\geq |N\cup U'|=k-1+\lfloor\frac{k-1}{r}\rfloor$.
+
+Using reduction lemma, we have $d= n-\max\{|I|:C_I<q^k,I\subseteq [n]\}\leq n-k-1-\lfloor\frac{k-1}{r}\rfloor+2=n-k-\lceil\frac{k}{r}\rceil +2$.
+
+### Reed-Solomon code
+