Linear Encoding

In Section 2.6.1.2 we showed the achievability of the capacity region as defined in Theorem 2.2 by using general random coding functions at different nodes of the network. In this section we restrict our attention to linear encoding schemes. The advantage of using linear encoding scheme is that the decoding process becomes much

easier. In this case, the equivalent transfer function of the network from any source to any destination, having the erasure locations at that destination, is linear. Hence, decoding at the destination is simply forming and solving a linear system of equations. In this section we show that linear encoders achieve capacity. Let us first define the linear block coding scheme with block length of n:

Recall that W(s) _{= {1,2, . . . ,d2}nRs_{e} is the message set for information source}

s∈ S. We assume that different messages are equiprobable and independent of each other. For anyw(s) _{∈ W}(s)_{, letb(w}(s)_{) be the length-nR}

sbinary expansion ofw(s)−1. The encoding operation is as follows:

Each node i ∈ V transmits n linear combinations of the non-erased symbols re- ceived from its incoming edges and the binary representation of the message it wants to transmit across the network. More precisely, node i generates a random binary matrix Bi of size n ×n(dI(i) +Ri), where dI(i) is the in-degree of node i and Ri is the rate of the codebook used at node i (in the case where i is not a source of information Ri = 0). Each element of Bi is drawn i.i.d. Bernoulli(1/2). For a given sequence y, let ey be a sequence derived by replacing every e with 0. Note that ey

and y have the same length.14 _{If node i receives Y}n

i =yni on its incoming edges and wants to transmit message w(i)_{, then it sends out x}

i = Bi·[b(w(i)),yein]†. (Since the input-output relation at each node is linear, setting the erased symbols equal to zero is the same as finding linear combinations of only the non-erased bits.)

Each destination d knows all the matrices Bi and also the erasure locations Zn on all the links across the network, since each received and transmitted symbol at any node is a linear combination of the elements of vector b(w) = (4 b(w(s)_{), s ∈ S).}

Therefore, each destination receives a collection of linear combinations of elements of b(w). Using {Bi}i∈V and Zn, destination node d can construct the matrix that

corresponds to the linear input-output relation of the network. We denote this matrix

14_{The corresponding mapping from alphabet GF(q)∪ {e}to GF(q) again replacesewith 0. This} variation is useful for packet erasure networks.

byF({Bi}, Zn), givingYfdn(w) = F({Bi}, Zn)·b(w)†. Note that matrixF is a function of different nodes’ encoding matrices{Bi} and Zn.

Now, upon receiving Yn

d =y∈ {0,1, e}ndI(d), the destination nodedlooks (solves) for the message vector w ∈ W =4 Q_s∈SW(s) _{such that F({M}

i}, Zn)·b(w)† = ye. If there is a unique wwith this property, node d declares it as the transmitted message vector; otherwise it declares an error. Note that the actual transmitted message vector, say w₀ ∈ W, always satisfies the above property. Therefore, an error occurs only if there is another message vector w6=w₀ such that Yn

d (w) = Ydn(w0) = y.

2.7.1

Achievability Result for Linear Encoding

Looking at the achievability proof and probability of error analysis for general random coding in Sections 2.6.1.2 and 2.6.1.3, it can be easily verified that the linear case requires the same error events (2.12). Since the erasure vector Zn _{is available at the} destination, there is no difference betweenYei and Yi, and we can determine one from the other. By expanding the conditional error event E(w) given A(_δn) for different cuts in the network, all of the relations up to step (d) of equation (2.16) go through for the linear case. In fact, the relations up to step (d) only require the independence of encoding functions for different nodes of the network, which holds for the linear case. Now, we look at the following probability in (2.16):

Pi = Pr4 µ _\ j:(i,j)∈[Vx,Vc x] {Yn ij(w) =Yijn(w0)} ¯ ¯ ¯ ¯A(δn), {(w(i), Yin(w))6= (w (i) 0 , Yin(w0))} ¶ . (2.20) As in the general random coding argument, for a fixediwe haveYn

ij(w) = Yijn(w0) for

all j such that (i, j) ∈ [Vx,Vxc], only if Xin(w) and Xin(w0) differ in locations where

an erasure occurs on all the edges of the interest. Because of strong typicality, the number of these location is at most n(Q_{j: (i,j)∈[Vx,V}c

x]²ij +δ). Therefore X n

i (w) and

Xn

i(w0) should be the same in at least n(1−

Q

j: (i,j)∈[Vs,Vc

by our encoding scheme this means that Bi·([w(i), Yin(w)]†−[w (i) 0 , Yin(w0)]†) | {z } z

should be zero in at least n(1−Q_{j: (i,j)∈[Vs,V}c

s]²ij +δ) specific locations. Also note that since (w(i)_{, Y}n

i (w)) 6= (w

(i)

0 , Yin(w0)), z is a non-zero vector. From the above

argument we have

Pi ≤ Pr

µ

Bi·zbe 0 in at leastnαi specific locations

¯ ¯ ¯ ¯z 6= 0 ¶ (a)

≤ 2−nαi _{= 2}−n(1−Qj: (i,j)∈[Vx,V_xc]²ij−δ), (2.21)

where αi = 1−

Q

j: (i,j)∈[Vx,Vc

x]²ij −δ, and (a) follows from the following Lemma and its Corollary. Proof of this lemma is provided in [57].

Lemma 2.4. Let X be a non-zero vector of size n×1 from some finite field GF(q).

Suppose that A is a random matrix of size m×n with i.i.d. components distributed

uniformly overGF(q). Then, the coordinates ofY =A·X are i.i.d. uniform random

variables over GF(q).

Corollary 2.5. The probability that Y =A·X is zero in k specific coordinates equals

q−k_.

Now note that by replacing Pi in (2.20) and (2.16) with its bound from (2.21), we get the same bound as (2.18) for random linear codes. Therefore, linear operations are sufficient for achieving the capacity.