# 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