Expander code - AbsoluteAstronomy.com

Coding theory

Coding theory is the study of the properties of codes and their fitness for a specific application. Codes are used for data compression, cryptography, error-correction and more recently also for network coding...

, expander codes are a type of linear block code that arises by using bipartite

Bipartite

Bipartite means having two parts, or an agreement between two parties. More specifically, it may refer to any of the following:-Mathematics:* 2 * bipartite graph* bipartite cubic, a type of cubic function-Medicine:...

expander graph

Expander graph

In combinatorics, an expander graph is a sparse graph that has strong connectivity properties, quantified using vertex, edge or spectral expansion as described below...

s. Along with concatenated codes, expander codes are interesting since they can construct binary

Binary

- Mathematics :* Binary numeral system, a representation for numbers using only two digits * Binary function, a function in mathematics that takes two arguments- Computing :* Binary file, composed of something other than human-readable text...

codes (codes using just 0 and 1) with constant positive rate and relative distance. Furthermore, expander codes can be both encoded and decoded in time proportional to the block length of the code. In fact, expander codes are the only known asymptotically good codes which can be both encoded and decoded from a constant fraction of errors in polynomial time.

This article is based on Dr. Venkatesan Guruswami's course notes .

Expander codes

In coding theory

Coding theory

, an expander code is a

linear block code whose parity check matrix is the adjacency matrix of a bipartite expander graph

Expander graph

In combinatorics, an expander graph is a sparse graph that has strong connectivity properties, quantified using vertex, edge or spectral expansion as described below...

. These codes have good relative distance (

, where

and

are properties of the expander graph as defined later), rate (

), and decodability (algorithms of running time

exist).

Definition

Consider a bipartite graph

Bipartite graph

In the mathematical field of graph theory, a bipartite graph is a graph whose vertices can be divided into two disjoint sets U and V such that every edge connects a vertex in U to one in V; that is, U and V are independent sets...

, where

and

are the vertex sets and

is the set of edges connecting vertices in

to vertices of

. Suppose every vertex in

has degree

Degree (mathematics)

In mathematics, there are several meanings of degree depending on the subject.- Unit of angle :A degree , usually denoted by ° , is a measurement of a plane angle, representing 1⁄360 of a turn...

(the graph is

-regular),

, and

. Then

is a

expander graph if every small enough subset

has the property that

has at least

distinct neighbors in

. Note that this holds trivially for

. When

and

for a constant

, we say that

is a lossless expander.

Since

is a bipartite graph, we may consider its

adjacency matrix. Then the linear code

generated by viewing the transpose of this matrix as a parity check matrix is an expander code.

It has been shown that nontrivial lossless expander graphs exist. Moreover, we can explicitly construct them.

Rate

The rate of

is its dimension divided by its block length. In this case, the parity check matrix has size

, and hence

has dimension at least

Distance

Suppose

. Then the distance of a

expander code

is at least

Proof

Note that we can consider every codeword

as a subset of vertices

, by saying that vertex

if and only if the

th index of the codeword is a 1. Then

is a codeword iff every vertex

is adjacent to an even number of vertices in

. (In order to be a codeword,

, where

is the parity check matrix. Then, each vertex in

corresponds to each column of

. Matrix multiplication over

then gives the desired result.) So, if a vertex

is adjacent to a single vertex in

, we know immediately that

is not a codeword. Let

denote the neighbors in

, and

denote those neighbors of

which are unique, i.e., adjacent to a single vertex of

Lemma 1

For every

of size

Proof

Trivially,

, since

implies

follows since the degree of every vertex in

. By the expansion property of the graph, there must be a set of

edges which go to distinct vertices. The remaining

edges make at most

neighbors not unique, so

Corollary

Every sufficiently small

has a unique neighbor. This follows since

Lemma 2

Every subset

with

has a unique neighbor.

Proof

Lemma 1 proves the case

, so suppose

. Let

such that

. By Lemma 1, we know that

. Then a vertex

is in

iff

, and we know that

, so by the first part of Lemma 1, we know

. Since

, and hence

is not empty.

Corollary

Note that if a

has at least 1 unique neighbor, i.e.

, then the corresponding word

corresponding to

cannot be a codeword, as it will not multiply to the all zeros vector by the parity check matrix. By the previous argument,

. Since

is linear, we conclude that

has distance at least

Encoding

The encoding time for an expander code is upper bounded by that of a general linear code -

by matrix multiplication. A result due to Spielman shows that encoding is possible in

time.

Decoding

Decoding of expander codes is possible in

time when

using the following algorithm.

Let

be the vertex of

that corresponds to the

th index in the codewords of

. Let

be a received word, and

. Let

is even

, and

is odd

. Then consider the greedy algorithm:
----
Input: received codeword



 intialize y' to y

 while there is a v in R adjacent to an odd number of vertices in V(y')

     if there is an i such that o(i) > e(i)

         flip entry i in y'

     else

         fail

Output: fail, or modified codeword

.
----

Correctness

We must show that the algorithm terminates with the correct codeword when the received codeword is within half the code's distance of the original codeword. Let the set of corrupt variables be

, and the set of unsatisfied (adjacent to an odd number of vertices) vertices in

. The following lemma will prove useful.

Lemma 3

, then there is a

with

Proof

By Lemma 1, we know that

. So an average vertex has at least

unique neighbors (recall unique neighbors are unsatisfied and hence contribute to

), since

, and thus there is a vertex

with

.

So, if we have not yet reached a codeword, then there will always be some vertex to flip. Next, we show that the number of errors can never increase beyond

Lemma 4

If we start with

, then we never reach

at any point in the algorithm.

Proof

When we flip a vertex

and

are interchanged, and since we had

, this means the number of unsatisfied vertices on the right decreases by at least one after each flip. Since

, the initial number of unsatisfied vertices is at most

, by the graph's

-regularity. If we reached a string with

errors, then by Lemma 1, there would be at least

unique neighbors, which means there would be at least

unsatisfied vertices, a contradiction.

Lemmas 3 and 4 show us that if we start with

(half the distance of

), then we will always find a vertex

to flip. Each flip reduces the number of unsatisfied vertices in

by at least 1, and hence the algorithm terminates in at most

steps, and it terminates at some codeword, by Lemma 3. (Were it not at a codeword, there would be some vertex to flip). Lemma 4 shows us that we can never be farther than

away from the correct codeword. Since the code has distance

(since

), the codeword it terminates on must be the correct codeword, since the number of bit flips is less than half the distance (so we couldn't have traveled far enough to reach any other codeword).

Complexity

We now show that the algorithm can achieve linear time decoding. Let

be constant, and

be the maximum degree of any vertex in

. Note that

is also constant for known constructions.

Pre-processing: It takes time to compute whether each vertex in has an odd or even number of neighbors.
Pre-processing 2: We take time to compute a list of vertices in which have .
Each Iteration: We simply remove the first list element. To update the list of odd / even vertices in , we need only update entries, inserting / removing as necessary. We then update entries in the list of vertices in with more odd than even neighbors, inserting / removing as necessary. Thus each iteration takes time.
As argued above, the total number of iterations is at most .

This gives a total runtime of

time, where

and

are constants.

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