The Char-q-y network

3.4 Network Coding Solution but No Routing Solution

4.1.1 The Char-q-y network

In [12], Joseph Connelly and Kenneth Zeger presented a network named as the Char-q network which has a vector linear solution for any message dimension if and only if the characteristics of the finite field divides q. Inspired by the Char-q network, we construct a network that we name as the Char-q-y network, where y is a label of a source node. This Char-q-y network will be used for each and every proof presented in this chapter. The source labelled by y is distinguished from the rest because no terminal demandsy. We show that if the middle edges of the network do not transmit any information generated by the source y, then the Char-q-y has a scalar linear linear solution over any finite field. But if the middle edges transmit any symbol generated by y, then the Char-q-y network has a vector linear solution if and only if the characteristic of the finite field dividesq. The purpose of constructing such a network is that we will attach this network to another network in such a way that if the other network does not receive any information about y from one of Char-q-y network’s middle edges, then the other network would render linearly unsolvable for a particular message dimension (thereby forcing the characteristic of the finite field to be a divisor of q for the network to have a vector linear solution for that particular message dimension).

We first give a description of the Char-q-y network. Forq= 2, the Char-q-y network is presented in Fig. 4.1. It has q+ 3 sources and q+ 3 terminals. Inq-y of the Char-q-y network, q is a positive

4.1 Networks Char-q-y, G1, G2, andG3

x₁ x₂ x₃

t₅ t₁ t₂ t₃ t₄

x₃

x₃

x₂ x₁

x₂ x₁

m₂ m₃ m₄ m₅

n₂ n₃ n₄ n₅

e₁ e₂ e₃ e₄ e₅

y a

a a y

n₁ m₁

Figure 4.1: The Char-q-y network for q = 2. The network has 5 sources and 5 terminals. Out of the 5 sources, 2 sources are labelled asaandy, and the rest 3 sources are labelled as x1, x2,andx3. The Char-2-y network has 5 middle edges: e₁, e₂, e₃, e₄ and e₅. The demands of each terminal is shown below the terminal’s label. Note that the sourcey is not demanded by any of the terminals.

integer (a product of primes), and y is a source label. The source nodes in the Char-q-y network are labelled as: a, y, x1, x2, . . . , xq+1 (the reason of why the first two labels a and y are different from the rest will be clear when we will use this network to prove theorems), and the terminal nodes are labelled as: t₁, t₂, . . . , t_q+3. The intermediate nodes of the Char-q-y network are union of these two sets: {m₁, m₂, . . . , m_q+3}and {n₁, n₂, . . . , n_q+3}. The list of the edges are given below.

• for 1≤i≤q+ 3: ei = (mi, ni) (these are the middle edges).

• for 1≤i≤q+ 1 and i=q+ 3: (a, mi).

• fori= 1 and 4≤i≤q+ 3: (y, m_i).

• for 1≤i≤q+ 1, 2≤j ≤q+ 3, andj6=i+ 1: (x_i, m_j).

• for 1≤i≤q+ 2: (ni, ti) and (nq+3, ti).

TH-2118_136102023

• for 1≤i≤q+ 2: (ni, tq+3).

• for 1≤i≤q: (x_i, t₁).

• fori= 2,3: (y, t_i).

• (a, tq+2).

The demands of the terminals are given below.

• t₁ demandsx_q+1.

• for 1≤i≤q+ 1: t_i+1 demandsx_i.

• tq+3 demandsa.

Among the local coding matrices of the Char-q-y network, let the encoding matrices (local coding matrices associated with each source-edge pair) be denoted by C_{s,e} where sis a source and eis an edge, let the decoding matrices (local coding matrices associated with each edge-terminal pair) be denoted by C{e,t} where e is an edge and t is a terminal, and let all other local coding matrices be denoted by C{e_i,ej} where both ei and ej are adjacent edges. We now define the following matrices.

(i) M_i =C_{(a,mi),ei}C_{a,(a,mi)} for 1≤i≤q+ 1 and i=q+ 3.

(ii) Ai =C{(y,m_i),ei}C{y,(y,m_i)} fori= 1 and 4≤i≤q+ 3.

(iii) W_(j,i) =C_{(xi,mj),ej}C_{xi,(xi,mj)} for 1≤i≤q+ 1, 2≤j≤q+ 3, andj6=i+ 1.

(iv) T_i1 =C_{(ni,ti),ti}C_{ei,(ni,ti)} for 1≤i≤q+ 2.

(v) T_i2 =C_{{(nq+3,ti),ti}}C_{{eq+3,(nq+3,ti)}} for 1≤i≤q+ 2.

(vi) Zi =C_{{(ni,tq+3),tq+3}}C_{{ei,(ni,tq+3)}} for 1≤i≤q+ 2.

Let the message carried by the edge ei be denoted by Yei for 1 ≤ i ≤ q + 3. Below we list the expressions of these messages. Let the message vector generated by the source a be denoted by a, the message vector generated by the source y be denoted by y, and the messgae generated by x_i for 1≤i≤q+ 1 be denoted by x_i. Then,

4.1 Networks Char-q-y, G1, G2, andG3

Ye2 =M2a+

q+1

X

i=2

W_(2,i)xi (4.2)

Y_e3 =M₃a+W_(3,1)x₁+

q+1

X

i=3

W_(3,i)x_i (4.3)

for 4≤j≤q+ 1 : Yej =Mja+Ajy+

q+1

X

i=1,i6=(j−1)

W_(j,i)xi (4.4)

Yeq+2=Aq+2y+

q

X

i=1

W_(q+2,i)xi (4.5)

Y_eq+3=M_q+3a+A_q+3y+

q+1

X

i=1

W_(q+3,i)x_i. (4.6)

We now prove the following lemma.

Lemma 25. Over a finite field whose characteristic does not divide q, for any positive integer d, the Char-q-y network has a d-dimensional vector linear solution if and only if A1 is zero.

Proof: First consider the ‘only if’ part. Due to the demands of the terminalt₁, from equations (4.1) and (4.6), we have the following equations.

T11M1+T12Mq+3 = 0 (4.7)

T₁₁A₁+T₁₂A_q+3 = 0 (4.8)

T₁₂W_(q+3,q+1)=I. (4.9)

Due to the demands of terminalt2, from equations (4.2) and (4.6), we have the following equations.

T₂₁M₂+T₂₂M_q+3 = 0 (4.10)

T₂₂W_(q+3,1)=I (4.11)

for 2≤i≤q+ 1 : T21W_(2,i)+T22W_(q+3,i)= 0. (4.12) Due to the demands of terminalt₃, from equations (4.3) and (4.6), we have the following equations.

T₃₁M₃+T₃₂M_q+3= 0 (4.13)

T31W_(3,1)+T32W_(q+3,1) = 0 (4.14)

T₃₂W_(q+3,2)=I (4.15)

for 3≤i≤q+ 1 : T31W_(3,i)+T32W_(q+3,i)= 0. (4.16)

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Due to the demands of the terminaltj for 4≤j≤q+ 1, from equations (4.4) and (4.6), we have:

Tj1Mj+Tj2Mq+3 = 0 (4.17)

T_j1A_j +T_j2A_q+3= 0 (4.18)

Tj2W(q+3,j−1) =I (4.19)

for 1≤i≤q+ 1, i6=j−1 : T_j1W_(j,i)+T_j2W_(q+3,i) = 0. (4.20) Due to the demands of the terminalt_q+2, from equations (4.5) and (4.6), we have:

T_(q+2)1Aq+2+T_(q+2)2Aq+3= 0 (4.21)

for 1≤i≤q: T_(q+2)1W_(q+2,i)+T_(q+2)2W_(q+3,i)= 0 (4.22)

T_(q+2)2W_(q+3,q+1)=I. (4.23)

Due to the demands of the terminaltq+3, from equations (4.1)-(4.5), we have:

Z₁M₁+Z₂M₂+...+Z_q+1M_q+1=I (4.24) Z₁A₁+Z₄A₄+...+Z_q+2A_q+2= 0 (4.25) for 1≤i≤q+ 1 :

q+2

X

j=2,j6=i+1

Z_jW_(j,i)= 0. (4.26)

From equations (4.9), (4.11), (4.15), (4.19) and (4.23), we get: Ti2 is invertible for 1≤i≤q+ 2, and W_(q+3,i) is invertible for 1≤ i ≤ q+ 1. Then, from equations (4.12), (4.14), (4.16), (4.20) and (4.22): T_i1 is invertible for 2≤ i ≤q+ 2, and W_(j,i) is invertible for 2 ≤ j ≤ q+ 2, 1 ≤ i ≤q+ 1, i6=j−1.

From equations (4.7), (4.10), (4.13) and (4.17), we have:

for 2≤i≤q+ 1 : M_i =−T_i1⁻¹T_i2M_q+3. (4.27) Substituting equation (4.27) in equation (4.24), we get:

Z₁M₁−(Z₂T₂₁⁻¹T₂₂+· · ·+Z_q+1T_(q+1)1⁻¹ T_(q+1)2)M_q+3 =I. (4.28) From equations (4.18), and (4.21), we have:

for 4≤i≤q+ 2 : Ai =−T⁻¹Ti2Aq+3. (4.29)

4.1 Networks Char-q-y, G1, G2, andG3

Substituting equation (4.29) in equation (4.25), we get:

Z₁A₁−(Z₄T₄₁⁻¹T₄₂+· · ·+Z_q+2T_(q+2)1⁻¹ T_(q+2)2)A_q+3= 0. (4.30) From equations (4.12), (4.14), (4.16), (4.20) and (4.22), we have:

for 2≤j≤q+ 2,1≤i≤q+ 1, i6=j−1 :W_(j,i)=−T_j1⁻¹Tj2W_(q+3,i). (4.31) Substituting equation (4.31) in equation (4.26), for 1≤i≤q+ 1 we have:

q+2

X

j=2,j6=i+1

ZjT_j1⁻¹Tj2W_(q+3,i)= 0. (4.32)

Since W_(q+3,i) for 1≤i≤q+ 1 has been already shown to be invertible, for 1≤i≤q+ 1, we must have:

q+2

X

j=2,j6=i+1

ZjT_j1⁻¹Tj2 = 0. (4.33)

Expanding equation (4.33) for each value of 1≤i≤q+ 1, we have:

Z₃T₃₁⁻¹T₃₂+Z₄T₄₁⁻¹T₄₂+· · ·+Z_q+2T_(q+2)1⁻¹ T_(q+2)2= 0 (4.34) Z₂T₂₁⁻¹T₂₂+Z₄T₄₁⁻¹T₄₂+· · ·+Z_q+2T_(q+2)1⁻¹ T_(q+2)2= 0 (4.35)

... ... (4.36)

Z₂T₂₁⁻¹T₂₂+Z₃T₃₁⁻¹T₃₂+Z₄T₄₁⁻¹T₄₂+· · ·+Z_q+1T_(q+1)1⁻¹ T_(q+1)2= 0. (4.37) Substituting equation (4.37) in equation (4.28) we get:

Z1M1=I (4.38)

Adding the q + 1 equations shown in equations (4.34)-(4.37), i.e. by performing the operation Pq+1

i=1

Pq+2

j=2,j6=i+1Z_jT_j1⁻¹T_j2, we have:

q(Z₂T₂₁⁻¹T₂₂+Z₃T₃₁⁻¹T₃₂+Z₄T₄₁⁻¹T₄₂+· · ·+Z_q+2T_(q+2)1⁻¹ T_(q+2)2) = 0. (4.39) Since the characteristic of the finite field does not divide q, we must have q 6= 0 in the finite field.

Then, from equation (4.39), we must have:

Z2T₂₁⁻¹T22+Z3T₃₁⁻¹T32+Z4T₄₁⁻¹T42+· · ·+Zq+2T_(q+2)1⁻¹ T_(q+2)2= 0. (4.40)

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For each value of 1≤i≤q+ 1, subtracting equation (4.33) from (4.40), we get:

for 2≤j ≤q+ 2 : Z_jT_j1⁻¹T_j2 = 0. (4.41) Substituting the values set by equation (4.41) in equation (4.30), we get:

Z1A1 = 0 (4.42)

Since Z₁ is invertible due to equation (4.38), we must haveA₁= 0. This proves the only if part.

To show the ‘if’ part, we present a scalar linear solution of the Char-q-y network. In this case all the local coding matrices are elements of the underlying finite field. Chose suitable coding coefficients such that the middle edges carry the following information.

Y_e1 =a

for 2≤i≤q+ 1 : Y_ei =a+

q+1

X

j=1,j6=i−1

x_i

Yeq+2 =

q

X

j=1

xi

Yeq+3 =a+

q+1

X

j=1

xj.

It can be easily seen that if the middle edges carry information as shown above, all the terminals can compute its desired information.

We now proof the following lemma.

Lemma 26. Over a finite field whose characteristic dividesq, the Char-q-y network has scalar linear solution even when A_i6= 0 for i= 1 and 4≤i≤q+ 3.

Proof: Let the characteristic of the finite field be p. Chose suitable coding coefficients such

4.1 Networks Char-q-y, G1, G2, andG3

that the middle edges carry the following information.

Y_e1 =a+y (4.43)

Y_e2 =a+x₂+x₃+· · ·+x_q+1 (4.44) Ye3 =a+x1+x3+· · ·+xq+1 (4.45) for 4≤i≤q+ 1 : Yei =a+y+

q+1

X

j=1,j6=i−1

xi (4.46)

Y_eq+2 =y+

q

X

j=1

x_i (4.47)

Y_eq+3 =a+y+

q+1

X

j=1

x_j. (4.48)

Terminalstifor 1≤i≤q+2 receives its desired symbols by subtracting the sum ofYei and the symbols received from the direct edges, fromYeq+3. Terminal tq+3 retrieves aby the operation: Pq+2

i=1Yi, as

q+2

X

i=1

Y_i = (p+ 1)a+py+

q+1

X

j=1

px_j =a. (4.49)