Network G 2 - Network Coding Solution but No Routing Solution

3.4 Network Coding Solution but No Routing Solution

4.1.3 Network G 2

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

and (4.61), we have:

g(Y₂₂,x)¯ ≤ 3(2n−1)

2 = 3n−3

2 = 3n−2 +1

2 (4.62)

g(Y₂₂,y)¯ ≤ 3(2n−1)

2 = 3n−3

2 = 3n−2 +1

2. (4.63)

Since the rank functiong() is integer valued by definition, from equations (4.62) and (4.63), we have:

g(Y22,x)¯ ≤3n−2 (4.64)

g(Y₂₂,y)¯ ≤3n−2. (4.65)

Substituting values from equation (4.64) and (4.65) in equation (4.58), we get:

6n−4≥3d= 3(2n−1) = 6n−3. (4.66)

Equation (4.66) results in 3≥4, which is a contradiction.

We now show the ‘if’ part of the proof. We show that G₁ has a scalar linear solution (thereby having a vector linear solution for any message dimension) if the characteristic of the finite field divides q. Let the edges Ye¯i for 1 ≤ i ≤ q+ 3 carry the messages as indicated by equations (4.43)-(4.48).

Then, the terminals ¯t5 to ¯tq+7 can retrieve its desired information (the terminals in the Char-q-¯y part retrieves its desired information as shown in Lemma 26). Now, in the M-network part, let Y₁₁ = ¯a, Y₁₃= ¯b,Y₂₂= ¯x, and Y₂₃= ¯y. Then, it can be easily seen that terminals ¯t₁,t¯₂ and ¯t₃ can retrieve its desired information. The terminal ¯t₄ receives ¯afromY₁₁, ¯bfromZ₄, ¯a+ ¯yfrom Y_e_¯₁, and as a result it can deduce ¯y as well (by subtracting ¯a from ¯a+ ¯y).

u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5 u1,u2,u3,u4,u5

v₃ v₂

v₁

u₁ u₂ u₃ u₄ u₅

a b c s w x y z

a,r,x a,r,y a,r,z a,s,x a,s,y a,s,z a,w,x a,w,y a,w,z b,r,x b,r,y b,r,z b,s,x b,s,y b,s,z b,w,x b,w,y b,w,z c,r,x c,r,y c,r,z c,s,x c,s,y c,s,z c,w,x c,w,y c,w,z

Figure 4.3: The generalized M-networkN_m form= 3.

v₃ v₂

v₁

u₁ u₂ u₃ u₄ u₅

x₂ x₁

x₃ x₁ x₂ x₃

x₁ x₂ x₃

t₂₇ t₂₆ t₂₅ t₂₄ t₂₃ t₂₂ t₂₁ t₂₀ t₁₉ t₁₈ t₁₇ t₁₆ t₁₅ t₁₄ t₁₃ t₁₂ t₁₁ t₁₀ t₉ t₈ t₇ t₆ t₅ t₄ t₃ t₂ t₁

a,r,x a,r,y a,r,z a,s,x a,s,y a,s,z a,w,x a,w,y a,w,z b,r,x b,r,y b,r,z b,s,x b,s,y b,s,z b,w,x b,w,y b,w,z c,r,x c,r,y c,r,z c,s,x c,s,y c,s,z c,w,x c,w,y c,w,z e₁

w w

a b c s w x y z

x a

a x

w w c c c a y

Figure 4.4: The network G₂ forq⁰ = 2.

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

Theq⁰+ 3 source nodes of the Char-q⁰-xnetwork are: a, x, x1, . . . , x_q⁰₊₁ (note that theN₃ network and Char-q⁰-x network has sourcesaand x in common). Edgee1 is the middle edge of the Char-q⁰-x network which has paths from aandx, but not from any other sources. The message carried bye₁ is denoted by Y_e₁.

The additional edges that are not part of N₃ and Char-q⁰-x, are listed below:

• (w, t₇), (w, t₈), (w, t₉), (w, t₁₆), (w, t₁₇), (w, t₁₈).

• (c, t₁₉), (c, t₂₀), (c, t₂₁), (c, t₂₂), (c, t₂₃), (c, t₂₄).

• (a, t25).

• (y, t₂₆).

• (head(e₁), t₂₅).

We first develop some general equations that hold for the network G₂. Letf be the function that maps the network G₂ to a discrete polymatrid D2 such that G₂ is a discrete polymatroidal network with respect to D2. Let ρ be the rank function ofD2, and let ρ_max≤d. Now letg=ρ◦f. Then we have the following equations:

g(Y₁₁, a) +g(Y₂₂, r) +g(Y₃₃, x)

=g(Y11, a, Y22, r, Y33, x) [using Lemma 5 repetitively] (4.67)

≤g(Y11, a, Y22, r, Y33, x, Z4,1, Z5,1)

≤g(Y₁₁, Y₂₂, Y₃₃, Z_4,1, Z_5,1) [due the demands of t₁]

≤5d [since rank of any element is less than or equal to d]. (4.68) Similar to equation (4.68), we have the following equations:

g(Y₁₁, a) +g(Y₂₂, r) +g(Y₃₃, y)≤5d (4.69) g(Y11, a) +g(Y22, r) +g(Y33, z)≤5d (4.70) g(Y₁₁, a) +g(Y₂₂, s) +g(Y₃₃, x)≤5d (4.71) g(Y₁₁, a) +g(Y₂₂, s) +g(Y₃₃, y)≤5d (4.72) g(Y11, a) +g(Y22, s) +g(Y33, z)≤5d (4.73)

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g(Y11, b) +g(Y22, r) +g(Y33, x)≤5d (4.74) g(Y₁₁, b) +g(Y₂₂, r) +g(Y₃₃, y)≤5d (4.75) g(Y₁₁, b) +g(Y₂₂, r) +g(Y₃₃, z)≤5d (4.76) g(Y11, b) +g(Y22, s) +g(Y33, x)≤5d (4.77) g(Y₁₁, b) +g(Y₂₂, s) +g(Y₃₃, y)≤5d (4.78) g(Y11, b) +g(Y22, s) +g(Y33, z)≤5d (4.79) g(Y11, c) +g(Y22, w) +g(Y33, z)≤5d. (4.80) We also have the following inequalities:

g(Y11, a) +g(Y11, b) +g(Y11, c)≥5d (4.81) g(Y22, r) +g(Y22, s) +g(Y22, w)≥5d (4.82) g(Y₃₃, x) +g(Y₃₃, y) +g(Y₃₃, z)≥5d. (4.83) We prove one of equations (4.81)-(4.83) and the rest can be prove similarly.

g(Y11, a) +g(Y11, b) +g(Y11, c)

≥g(Y₁₁, a, b) +g(Y₁₁) +g(Y₁₁, c) [applying rule [P3] of Definition 4]

≥g(Y11, a, b, c) + 2g(Y11)

=g(a, b, c) + 2g(Y₁₁)

= 5d.

Adding equations (4.81)-(4.83), we have:

g(Y₁₁, a) +g(Y₁₁, b) +g(Y₁₁, c) +g(Y₂₂, r) +g(Y₂₂, s) +g(Y₂₂, w) +g(Y₃₃, x) +g(Y₃₃, y) +g(Y₃₃, z)≥15d or, (g(Y11, a) +g(Y22, r) +g(Y33, x)) + (g(Y11, b) +g(Y22, s) +g(Y33, y)) +g(Y11, c)

+g(Y₂₂, w) +g(Y₃₃, z)≥15d. (4.84) But as equations (4.68), (4.78) and (4.80) hold, from equation (4.84), we have:

g(Y11, a) +g(Y22, r) +g(Y33, x) = 5d (4.85)

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

g(Y11, b) +g(Y22, s) +g(Y33, y) = 5d (4.86) g(Y₁₁, c) +g(Y₂₂, w) +g(Y₃₃, z) = 5d. (4.87) Similarly, rearranging equation (4.84), we get the following equalities:

g(Y11, a) +g(Y22, r) +g(Y33, y) = 5d (4.88) g(Y₁₁, a) +g(Y₂₂, s) +g(Y₃₃, x) = 5d (4.89) g(Y₁₁, a) +g(Y₂₂, s) +g(Y₃₃, y) = 5d (4.90) g(Y11, b) +g(Y22, r) +g(Y33, x) = 5d (4.91) g(Y₁₁, b) +g(Y₂₂, r) +g(Y₃₃, y) = 5d (4.92) g(Y11, b) +g(Y22, s) +g(Y33, x) = 5d. (4.93) Subtracting equations (4.85) from (4.91), we get:

g(Y₁₁, a) =g(Y₁₁, b). (4.94)

Subtracting equations (4.85) from (4.89), we get:

g(Y22, r) =g(Y22, s). (4.95)

Subtracting equations (4.85) from (4.88), we get:

g(Y33, x) =g(Y33, y). (4.96)

As equations (4.68), (4.78) and (4.80) holds, from equation (4.84) we also have:

(g(Y₁₁, a) +g(Y₂₂, r) +g(Y₃₃, x)) + (g(Y₁₁, b) +g(Y₂₂, s) +g(Y₃₃, y)) +g(Y₁₁, c)

+g(Y22, w) +g(Y33, z) = 15d. (4.97) Rearranging terms in equation (4.97), we have:

(g(Y₁₁, a) +g(Y₁₁, b) +g(Y₁₁, c)) + (g(Y₂₂, r) +g(Y₂₂, s) +g(Y₂₂, w)) + (g(Y₃₃, x)

+g(Y33, y) +g(Y33, z)) = 15d. (4.98)

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As equations (4.81)-(4.83) holds, we must have:

g(Y₁₁, a) +g(Y₁₁, b) +g(Y₁₁, c) = 5d (4.99) g(Y₂₂, r) +g(Y₂₂, s) +g(Y₂₂, w) = 5d (4.100) g(Y33, x) +g(Y33, y) +g(Y33, z) = 5d. (4.101) Applying equations (4.94)-(4.96) to equations (4.99)-(4.101), we have:

2g(Y₁₁, a) +g(Y₁₁, c) = 5d (4.102) 2g(Y22, r) +g(Y22, w) = 5d (4.103) 2g(Y₃₃, x) +g(Y₃₃, z) = 5d. (4.104) Multiplying equation (4.70) by 2 and then adding to equation (4.87), we have:

2(g(Y11, a) +g(Y22, r) +g(Y33, z)) +g(Y11, c) +g(Y22, w) +g(Y33, z)≤15d or, 5d+ 5d+ 3g(Y₃₃, z)≤15d [substituting equations (4.102) and (4.103)]

or, 3g(Y33, z)≤5d or,g(Y33, z)≤ 5d

3 . (4.105)

We now derive one more equation that must hold if the characteristic of the finite field does not divideq⁰. Note that in such a caseY_e₁ in the Char-q⁰-xnetwork is independent ofx(from Lemma 25), and is a function of only a. So due to the demands of terminalt₂₅, we have:

g(Y₁₁, a, c) +g(Y₂₂, w) +g(Y₃₃, x)

=g(Y11, Y22, Y33, a, c, w, x) [using Lemma 5 repetitively]

≤g(Y₁₁, Y₂₂, Y₃₃, a, c, w, x, Z_4,27, Z_5,27)

=g(Y₁₁, Y₂₂, Y₃₃, a, c, w, x, Z_4,27, Z_5,27, Y_e₁)

=g(Y11, Y22, Y33, a, Z4,27, Z5,27, Ye1) [due to demands oft25]

=g(Y₁₁, Y₂₂, Y₃₃, a, Z_4,27, Z_5,27)

=g(Y22, Y33, Z4,27, Z5,27) +g(Y11, a|Y₂₂, Y33, Z4,27, Z5,27)

≤g(Y22, Y33, Z4,27, Z5,27) +g(Y11, a)

≤4d+g(Y₁₁, a). (4.106)

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

It can be seen that due to terminals t1, t14, and t27 all of the source messages are to be retrieved from {Y_ii, Yij|i= 1,2,3 and j= 4,5}. Then like equation (3.7) it can be shown thatY11=d. Then,

g(Y11, a, c) +g(Y11, a)

=g(Y₁₁, a, c) +g(Y₁₁, b) [from equation (4.94)]

≥g(Y11, a, c, b) +g(Y11) [using rule [P3] of Definition 4]

≥4d.

Then we have

g(Y₁₁, a, c)≥4d−g(Y₁₁, a). (4.107) Substituting equation (4.107) in equation (4.106), we have:

g(Y₁₁, w) +g(Y₃₃, x)≤2g(Y₁₁, a). (4.108) We first show thatG₂ has no scalar linear solution.

Lemma 28. The network G₂ has no scalar linear solution over any finite field.

Proof: Note that equations (4.106)-(4.108) cannot be used as they hold only if the characteristic of the finite field divides q⁰; and current lemma is to be shown to be true over all finite fields. Let us assume that the network has a scalar linear solution. Then,ρmax≤1 whereρ is the rank function of D2, with respect to whichG₂ is a discrete polymatroidal network.

Since d= 1, and the rank function of a discrete polymatroid is always an integer, from equation (4.105) we have: g(Y33, z)≤1. Since 1 =g(z)≤g(Y33, z), we must have:

g(Y33, z) = 1. (4.109)

Substituting equation (4.109) in equation (4.87), we have:

g(Y11, c) +g(Y22, w) = 4. (4.110)

Since rank of any element is less than or equal to 1, we have g(Y11, c) ≤ 2 andg(Y22, w)≤ 2. Then equation (4.110) implies:

g(Y11, c) = 2. (4.111)

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Substituting equation (4.111) in equation (4.102), we have:

2g(Y₁₁, a) = 3 or, g(Y₁₁, a) = 3

2. (4.112)

Equation (4.112) is a contradiction as the rank function always outputs an integer.

We now prove the following lemma.

Lemma 29. The networkG₂ has a2-dimensional vector linear solution if and only if the characteristic of the finite field divides q⁰.

Proof:

Consider the ‘only if’ part. We show that if the characteristic of the finite field does not divideq⁰ then network G₂ has no 2-dimensional vector linear solution. We prove this result by contradiction.

Assume that G₂ has a 2-dimensional vector linear solution even when the characteristic of the finite field does not divideq⁰. So we have ρmax =d= 2 for the discrete polymatroidD2.

Since the rank function of a discrete polymatroid is integer valued, from equation (4.105), we have:

g(Y33, z)≤3. (4.113)

Substituting equation (4.113) in equation (4.104), we have:

g(Y33, x)≥ 7

2. (4.114)

Then it mus that

g(Y33, x)≥4. (4.115)

Since rank of an element is less than or equal to 2, we must have:

g(Y33, x) = 4. (4.116)

Substituting equation (4.116) in equation (4.104), we have:

g(Y₃₃, z) = 2. (4.117)

Substituting equation (4.117) in equation (4.87), we have:

g(Y , c) +g(Y , w) = 8. (4.118)

4.1 Networks Char-q-y,G₁,G₂, and G₃ 81

a1a2 a= x1x2 x=x3 x32x31

a+x2+x3

a+x1+x3 x1112x x1 x21x22 x2

x1 x2

x3x1x2x3 a+x+x1+x2+x3x+x1+x2

aa x ===

a+x v3

u3u4u5 z1z2 x1 x2

y1 y2 y1y2z2 z1

u2 b2

r1s1 r2 r1w2 w1

a2 c1c2w2 w1 r2s2 s2 s1s=

u1 a1b1 b1b2 b= c1c2 a1a2 a= x1x2 x=

Ye1 y=z=r=w=c=

u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅ u₁,u₂,u₃,u₄,u₅

rcayrrr cc a,r,xa,r,ya,r,za,s,xa,s,ya,s,za,w,xa,w,ya,w,zb,r,xb,r,yb,r,zb,s,xb,s,yb,s,zb,w,xb,w,yb,w,zc,r,xc,r,yc,r,zc,s,xc,s,yc,s,zc,w,xc,w,zc,w,y

Figure 4.5: A 2-dimensional vector linear solution ofG₂ forq⁰ = 2 when the characteristic dividesq⁰.

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Since rank of an element is less than or equal to 2, we must have:

g(Y₁₁, c) = 4 (4.119)

g(Y₂₂, w) = 4. (4.120)

Substituting equation (4.119) in equation (4.102), we have (note d= 2):

g(Y₁₁, a) = 3. (4.121)

Substituting equations (4.116), (4.120), and (4.121) in equation (4.108), we have: 8 ≤ 6, which is a contradiction.

To prove the ‘if’ part we present a 2-dimensional coding scheme over a finite field whose characteristic divides q⁰. In fig. 4.5 we show a 2-dimensional vector linear solution of G₂ when q⁰ = 2. This coding scheme can easily be extended for any value of q⁰. (For a different value ofq⁰, only a decoding matrix in the Char-q⁰-x network changes (see equation (4.49).)

Now, consider the following lemma.

Lemma 30. The networkG₂ has a5-dimensional vector linear solution if and only if the characteristic of the finite field divides q⁰.

Proof: Consider the ‘only if’ part. We show that if the characteristic of the finite field does not divide q⁰ then network G₂ has no 5-dimensional vector linear solution. We prove this result by contradiction. Assume that G₂ has a 5-dimensional vector linear solution when the characteristic of the finite field does not divideq⁰. So we have d= 5 for the discrete polymatroidD₂.

Since the rank function of a discrete polymatroid is integer valued, from equation (4.105) we have:

g(Y₃₃, z)≤8. (4.122)

From equation (4.104) we get that 25−g(Y₃₃, z) must be divisible by 2 (otherwise g(Y₃₃, x) would not be an integer). Henceg(Y₃₃, z) must be an odd number. For similar reasoning, from equations (4.102) and (4.103) we get thatg(Y₁₁, c) andg(Y₂₂, w) must be odd numbers.

Then, since 5 =g(z)≤g(Y₃₃, z), either g(Y₃₃, z) = 5 or g(Y₃₃, z) = 7.

Case I:g(Y33, z) = 5.

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

Substituting g(Y33, z) = 5 in equation (4.87), we get:

g(Y₁₁, c) +g(Y₂₂, w) = 20. (4.123) Since rank of any union of two elements is less than or equal to 10, we must have

g(Y11, c) =g(Y22, w) = 10. (4.124) But equation (4.124) is a contradiction because as we have argued, g(Y₁₁, c) and g(Y₂₂, w) must be odd numbers.

Case II:g(Y33, z) = 7.

Substituting g(Y33, z) = 7 in equation (4.104) we have:

g(Y₃₃, x) = 9. (4.125)

Substituting g(Y₃₃, z) = 7 in equation (4.87), we get:

g(Y11, c) +g(Y22, w) = 18. (4.126) Since neither of g(Y11, c) and g(Y22, w) can be equal to 10 (as 10 is an even number), we must have:

g(Y₁₁, c) = 9 (4.127)

g(Y22, w) = 9. (4.128)

Substituting equation (4.127) in equation (4.102) we have:

g(Y₁₁, a) = 8. (4.129)

Substituting equations (4.125), (4.128), and (4.129) in equation (4.108), we have: 18≤16, which is a contradiction.

To prove the ‘if’ part we now design a 5-dimensional vector linear solution when the characteristic of the finite field divides q⁰. We first note that G₂ has a 3-dimensional vector linear solution over all finite fields. This is because, from Theorem 11 of Chapter 3 we know that N₃ has a 3-dimensional vector linear solution over all finite fields, and from Lemma 25 we know that the Char-q⁰-x network has a vector linear solution for any message dimension over all finite fields whose characteristic divides q⁰. Now, from Lemma 29 we get that G₂ has a 2-dimensional vector linear solution over a finite field

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whose characteristic divides q⁰. So a 5-dimensional vector linear solution can easily be constructed.

Dalam dokumen On the Role of the Message Dimension and the Characteristic of the Finite Field in Linear Network Coding (Halaman 95-106)