interference alignment - (for interference channels)

Interference Alignment

(for Interference Channels)

Bobak Nazer, Boston University

Wireless Information Theory Summer SchoolUniversity of Oulu

July 28, 2011

Outline

I. K-User Interference Channels

II. Alignment via Linear Precoding

III. Ergodic Alignment

IV. Lattice Alignment for Fixed Channels

K-User Interference Channel

1

2

1

2


1

2

K

1

2

K

• K transmitter-receiver pairs share a common wireless channel.


1

2

K

H

1

2

K


• Receivers observe noisy linear combinations of transmitted signals:

Yk =

K∑

ℓ=1

hkℓXℓ + Zk


1

2

K

h11 h12 · · · h1K

h21 h22 · · · h2K

......

. . ....

hK1 hK2 · · · hKK

1

2

K


• Receivers observe noisy linear combinations of transmitted signals:

Yk =

K∑

ℓ=1

hkℓXℓ + Zk

Strong Interference

1

2

K

1

2

K

Strong Interference

1

2

K

1

2

K

• Strategy: First, decode and subtract interfering signals. Then,recover desired codeword.

Strong Interference

1

2

K

1

2

K

• Strategy: First, decode and subtract interfering signals. Then,recover desired codeword.

• Optimal if interference is very strong.(Carleial ’75, Sato ’81, Han-Kobayashi ’81, Sankar-Erkip-Poor ’08)

Weak Interference

1

2

K

1

2

K

• Strategy: Treat interfering signals as additional noise.

Weak Interference

1

2

K

1

2

K

• Strategy: Treat interfering signals as additional noise.

• Optimal if interference is very weak. (Motahari-Khandani ’07,

Shang-Kramer-Chen ’07, Annapureddy-Veeravalli ’08)

K-User Interference Channel – Problem Statement

w1 E1X1

w2 E2X2

...

wK EKXK

H

Z1

Y1

Z2

Y2

ZK

YK

D1 w1

D2 w2

...

DK wK

• Transmitter ℓ maps message wℓ ∈ {1, 2, . . . , 2nRℓ} into

complex-valued codeword Xnℓ = (Xℓ[1], . . . ,Xℓ[n]) obeying power

constraint∑n

i=1 |Xℓ[i]|2 ≤ nP .

• Receiver k observes Yk[i] =∑K

ℓ=1 hkℓXℓ[i] + Zk[i]. Noise Zk[i] isi.i.d. CN (0, N).

• What rates R1, . . . , RK are sustainable with vanishing probability oferror P

(

{w1 6= w1} ∪ · · · ∪ {wK 6= wK})

?

K-User Interference Channel – Problem Statement

w1 E1X1

w2 E2X2

...

wK EKXK

h11 h12 · · · h1Kh21 h22 · · · h2K...

.... . .

...hK1 hK2 · · · hKK

Z1

Y1

Z2

Y2

ZK

YK

D1 w1

D2 w2

...

DK wK

• Transmitter ℓ maps message wℓ ∈ {1, 2, . . . , 2nRℓ} into

complex-valued codeword Xnℓ = (Xℓ[1], . . . ,Xℓ[n]) obeying power

constraint∑n

i=1 |Xℓ[i]|2 ≤ nP .

• Receiver k observes Yk[i] =∑K

ℓ=1 hkℓXℓ[i] + Zk[i]. Noise Zk[i] isi.i.d. CN (0, N).

• What rates R1, . . . , RK are sustainable with vanishing probability oferror P

(

{w1 6= w1} ∪ · · · ∪ {wK 6= wK})

?

K-User Interference Channel – Symmetric Case

w1 E1X1

w2 E2X2

...

wK EKXK

h11 h12 · · · h1Kh21 h22 · · · h2K...

.... . .

...hK1 hK2 · · · hKK

Z1

Y1

Z2

Y2

ZK

YK

D1 w1

D2 w2

...

DK wK

• Equal rates R1 = · · · = RK = R.

• Direct gains hkk = 1 and cross-gains hkℓ = β.

• Very Strong Case: β is large enough so that R = log(1 + PN ).

• Weak Case: β is small enough so that receivers can treatinterference as noise.

• How do these thresholds on β scale with K?

K-User Interference Channel – Symmetric Case

w1 E1X1

w2 E2X2

...

wK EKXK

1 β · · · β

β 1 · · · β...

.... . .

...

β β · · · 1

Z1

Y1

Z2

Y2

ZK

YK

D1 w1

D2 w2

...

DK wK

• Equal rates R1 = · · · = RK = R.

• Direct gains hkk = 1 and cross-gains hkℓ = β.

• Very Strong Case: β is large enough so that R = log(1 + PN ).

• Weak Case: β is small enough so that receivers can treatinterference as noise.

• How do these thresholds on β scale with K?

Symmetric Very Strong Case

• Receiver k must cancel out interference before decoding wk.

• Simple approach: Each receiver decodes all K − 1 undesiredmessages and removes them.

• Multiple-access capacity region requires that:

R ≤1

K − 1log

(

1 +β2(K − 1)P

N + P

)

• Set equal to R = log(1 + PN ) and solve for β2:

β2 ≥

(

(1 + PN )K−1 − 1

)

(N + P )

(K − 1)P

• β threshold increases exponentially with K.

Symmetric Weak Case

• Receiver treats all of the interference as noise. Resulting rate is

R = log

(

1 +P

N + (K − 1)β2P

)

.

• Genie-aided bounds show this is the capacity if

P ≤

√

K−1β2 − 2(K − 1)

2(K − 1)2β2

• Implies that β threshold must fall with K:

β2 ≤1

4(K − 1)

• See Shang-Kramer-Chen ’07, Motahari-Khandani ’07,

Annapureddy-Veeravalli ’08 for more general results.

Interference-Free Capacity

1 X1h11

Z1

Y1 1

• Interference-free capacity:

RFREEk = log

(

1 +|hkk|

2P

N

)

Interference-Free Capacity

1 1

• Interference-free capacity:

RFREEk = log

(

1 +|hkk|

2P

N

)

Time Division

1

2

K

1

2

K

• Eliminate interference through time division:

RTDMAk =

1

Klog

(

1 +K|hkk|

2P

N

)

Can each user get half the cake?

1

2

K

1

2

K

• Is it possible for each user to communicate as if there is only oneother user?

RHALFk =

1

2log

(

1 +2|hkk|

2P

N

)


1

2

K

1

2

K

• Maddah-Ali - Motahari - Khandani ’08: Proposed interference alignmentfor the MIMO X channel.

• Cadambe-Jafar ’08: Alignment can get “half the cake” for theinterference channel as the SNR→∞:

limP→∞

RIAk

log (1 + P )=

1

2


1 V1

2V2

KVK

1

2

K

• Maddah-Ali - Motahari - Khandani ’08: Proposed interference alignmentfor the MIMO X channel.

• Cadambe-Jafar ’08: Alignment can get “half the cake” for theinterference channel as the SNR→∞:

limP→∞

RIAk

log (1 + P )=

1

2

Interference Alignment – Fixed Channels

w1 E1X1

w2 E2X2

...

wK EKXK

h11 h12 · · · h1Kh21 h22 · · · h2K...

.... . .

...hK1 hK2 · · · hKK

Z1

Y1

Z2

Y2

ZK

YK

D1 w1

D2 w2

...

DK wK

• Receiver see statistically equivalent channels: Yk =

K∑

ℓ=1

Xℓ + Zk

• Interference channel capacity depends only on marginal channels.=⇒ If one user can decode wℓ, they all can.

• Multiple-access capacity: R =1

Klog

(

1 +KP

N

)


w1 E1X1

w2 E2X2

...

wK EKXK

1 1 · · · 1

1 1 · · · 1...

.... . .

...

1 1 · · · 1

Z1

Y1

Z2

Y2

ZK

YK

D1 w1

D2 w2

...

DK wK

• Receiver see statistically equivalent channels: Yk =

K∑

ℓ=1

Xℓ + Zk

• Interference channel capacity depends only on marginal channels.=⇒ If one user can decode wℓ, they all can.

• Multiple-access capacity: R =1

Klog

(

1 +KP

N

)


w1 E1X1

w2 E2X2

...

wK EKXK

h11 h12 · · · h1Kh21 h22 · · · h2K...

.... . .

...hK1 hK2 · · · hKK

Z1

Y1

Z2

Y2

ZK

YK

D1 w1

D2 w2

...

DK wK

• Receivers see Yk = Xk + j

K∑

ℓ 6=k

Xℓ + Zk

• Transmitters send only real-valued signals.

• Receivers ignore imaginary part of observed signal to get:

R =1

2log

(

1 +2P

N

)


w1 E1X1

w2 E2X2

...

wK EKXK

1 j · · · j

j 1 · · · j...

.... . .

...

j j · · · 1

Z1

Y1

Z2

Y2

ZK

YK

D1 w1

D2 w2

...

DK wK

• Receivers see Yk = Xk + j

K∑

ℓ 6=k

Xℓ + Zk

• Transmitters send only real-valued signals.

• Receivers ignore imaginary part of observed signal to get:

R =1

2log

(

1 +2P

N

)

Interference Alignment – Time-Varying Channels

w1 E1X1[t]

w2 E2X2[t]

...

wK EKXK [t]

H

Z1[t]

Y1[t]

Z2[t]

Y2[t]

ZK [t]

YK [t]

D1 w1

D2 w2

...

DK wK

• What about general fixed H? Only partially understood (e.g. highSNR, special cases).

• Much more is known for time-varying channels.

• Assume every transmitter and receiver knows H[t] causally (i.e.knows H[t] before time t).


w1 E1X1[t]

w2 E2X2[t]

...

wK EKXK [t]

H[t]

Z1[t]

Y1[t]

Z2[t]

Y2[t]

ZK [t]

YK [t]

D1 w1

D2 w2

...

DK wK

• What about general fixed H? Only partially understood (e.g. highSNR, special cases).

• Much more is known for time-varying channels.

• Assume every transmitter and receiver knows H[t] causally (i.e.knows H[t] before time t).


w1 E1X1[t]

w2 E2X2[t]

w3 E3X3[t]

Hodd

1 1 −1−1 1 1

1 −1 1

Z1[t]

Y1[t]

Z2[t]

Y2[t]

Z3[t]

Y3[t]

D1 w1

D2 w2

D3 w3

• Example from Cadambe-Jafar ’09.

• Separate coding over Hodd and Heven: R =1

3log

(

1 +3P

N

)

• Joint coding over Hodd and Heven: R =1

2log

(

1 +2P

N

)


w1 E1X1[t]

w2 E2X2[t]

w3 E3X3[t]

Heven

1 −1 11 1 −1−1 1 1

Z1[t]

Y1[t]

Z2[t]

Y2[t]

Z3[t]

Y3[t]

D1 w1

D2 w2

D3 w3

• Example from Cadambe-Jafar ’09.

• Separate coding over Hodd and Heven: R =1

3log

(

1 +3P

N

)

• Joint coding over Hodd and Heven: R =1

2log

(

1 +2P

N

)

Joint Coding

• t odd: Y1[t] = X1[t] +X2[t]−X3[t] + Z1[t]

• t even: Y1[t] = X1[t]−X2[t] +X3[t] + Z1[t]

• Joint Coding: Send new symbol every odd time.Repeat symbols on even times.

• Decoding: Y1[t− 1] + Y1[t] = 2X1[t− 1] + Z1[t− 1] + Z1[t]

• Effective SNR: 4P/2N = 2P/N .

• Two channel uses per symbol: R =1

2log

(

1 +2P

N

)

• Same strategy works for users 2 and 3.

Outline





Main References

This section is almost entirely drawn from:

• V. Cadambe and S. A. Jafar, Interference Alignment and Degrees ofFreedom of the K-User Interference Channel. IEEE Transactions onInformation Theory, vol. 54, no. 8, pp. 3425-3441, August 2008.

For a comprehensive overview of interference alignment, see:

• Syed A. Jafar, Interference Alignment: A New Look at SignalDimensions in a Communication Network, Foundations and Trendsin Communications and Information Theory, Vol. 7, No. 1, pages:1-136.


w1 E1X1[t]

w2 E2X2[t]

...

wK EKXK [t]

H[t]

Z1[t]

Y1[t]

Z2[t]

Y2[t]

ZK [t]

YK [t]

D1 w1

D2 w2

...

DK wK

• At each time t, each channel gain hkℓ[t] is drawn according to anindependent distribution with uniform phase.

• Milder assumptions possible.

• Every transmitter and receiver knows H[t] causally (i.e. knows H[t]before time t).

Symbol Extension

• Joint coding over m time slots:

H[1] = {hkℓ[1]}H[2] = {hkℓ[2]}

...H[m] = {hkℓ[m]}

xℓ ,

Xℓ[1]Xℓ[2]...

Xℓ[m]

yk ,

Yk[1]Yk[2]...

Yk[m]

• Convenient to represent this problem with diagonal matrices:

Dkℓ ,

hkℓ[1] 0 · · · 00 hkℓ[2] · · · 0...

.... . .

...0 0 · · · hkℓ[m]

yk =

K∑

ℓ=1

Dkℓxℓ + zk

• Can visualize m = 3 in 3D:

t = 1

t = 2

t = 3

Non-Aligned Signaling over 3 Time Slots

Tx 1

Tx 2

Tx 3

Rx 1

Rx 2

Rx 3

Non-Aligned Signaling over 3 Time Slots

Tx 1

Tx 2

Tx 3

Rx 1

Rx 2

Rx 3

Total Degrees of Freedom

DoF =3 vectors

3 channel uses

= 1

Each user gets 1/3 the cake.

Aligned Signaling over 3 Time Slots

Tx 1

Tx 2

Tx 3

Rx 1

Rx 2

Rx 3


Tx 1

Tx 2

Tx 3

Rx 1

Rx 2

Rx 3

Total Degrees of Freedom

DoF =4 vectors

3 channel uses

=4

3

Each user gets 4/9 the cake.


Tx 1

v1A

Tx 2

Tx 3

Rx 1

Rx 2

Rx 3

• Set v1A = 1√3[ 1 1 1 ]T


Tx 1

v1A

Tx 2v2

Tx 3

Rx 1

Rx 2

Rx 3

• Set v1A = 1√3[ 1 1 1 ]T

• v2 aligns with v1A at Rx 3 :

D32v2 = D31v1A

v2 = D−132 D31v1A


Tx 1

v1A

Tx 2v2

Tx 3

v3

Rx 1

Rx 2

Rx 3

• Set v1A = 1√3[ 1 1 1 ]T


D32v2 = D31v1A

v2 = D−132 D31v1A

• v3 aligns with v2 at Rx 1 :

D13v3 = D13v2

v3 = D−113 D12v2


Tx 1

v1A

v1B

Tx 2v2

Tx 3

v3

Rx 1

Rx 2

Rx 3

• Set v1A = 1√3[ 1 1 1 ]T


D32v2 = D31v1A

v2 = D−132 D31v1A

• v3 aligns with v2 at Rx 1 :

D13v3 = D13v2

v3 = D−113 D12v2

• v1B aligns with v3 at Rx 2 :

D21v1B = D23v3

v1B = D−121 D23v3

Getting Half the Cake

• Collect signaling vectors into matrices Vℓ = [vℓ1 vℓ2 · · · vℓm].

• Receiver k allocates subspace Ik as interference space.

• Alignment conditions:

Receiver 1D11V1 ∩ I1 = ∅D12V2 ⊆ I1

...D1KVK ⊆ I1

Receiver 2D21V1 ⊆ I2

D22V2 ∩ I2 = ∅...

D2KVK ⊆ I2

· · ·

Receiver KDK1V1 ⊆ IKDK2V2 ⊆ IK

...DKKVK ∩ IK = ∅

• Want m/2 dimensions for Vk and Ik.

• Not feasible in general.

Getting Half the Cake – Asymptotic Alignment

• Enumerate all S , K(K − 1) cross-channels with a single index:

T ={

Ti

}

={

Dkℓ : k 6= ℓ}

• Use the same signaling vectors V(m) at every transmitter.

• Define signaling vectors recursively:

V(0) = {1}

V(m) ={

vi, T1vi, . . . , TSvi : vi ∈ V(m−1)

}

={

Tα11 Tα2

2 · · ·TαSS 1 : α1 + α2 + · · ·+ αS ≤ m

}

• Size of signaling space:

∣

∣

∣V(m)

∣

∣

∣=

(

m+ S

m

)

Getting Half the Cake – Asymptotic Alignment

• Interference space at receiver k is Ik =⋃

ℓ 6=k

DkℓV(m) ⊂ V(m+1)

• Desired signal space at receiver k is DkkV(m).

• V(m) only contains products of cross-channels so there is no overlapbetween desired signal space and interference space.

• Number of vectors is nearly the same for large m:

∣

∣

∣V(m)

∣

∣

∣

∣

∣

∣V(m+1)

∣

∣

∣

=

(m+Sm

)

(m+1+Sm+1

) =m+ 1

m+ 1 + S

m→∞−−−−→ 1

• Desired signal space asymptotically gets half the dimensions:

∣

∣

∣DkkV

(m)∣

∣

∣

∣

∣

∣DkkV(m)

∣

∣

∣+∣

∣

∣Ik

∣

∣

∣

m→∞−−−−→

1

2

K-User Interference Channel – Degrees-of-Freedom Region

1

2

K

1

2

K

• Everyone can get half the cake dk = limP→∞

Rk

log(1 + P )=

1

2

• If all but user k quiet: Rk = log(1 + |hkk|2P )

• Time share to get degrees-of-freedom region:

dk + dℓ ≤ 1, ∀k 6= ℓ.

Outline





Main Reference

This section is almost entirely drawn from:

• B. Nazer, M. Gastpar, S. A. Jafar, and S. Vishwanath, ErgodicInterference Alignment,

Ergodic Interference Alignment

1

2

K

1

2

K

• We can get (slightly more than) half the interference-free rateat any SNR!

REIAk =

1

2EH

[

log

(

1 +2|hkk|

2P

N

)]

Key Idea

1. At time t with channel H, user k transmits signal Xk.

H =

h11 h12 · · · h1Kh21 h22 · · · h2K...

.... . .

...hK1 hK2 · · · hKK

Key Idea


H =

h11 h12 · · · h1Kh21 h22 · · · h2K...

.... . .

...hK1 hK2 · · · hKK

2. When complementary matrix HC occurs, retransmit signal Xk.

HC =

h11 −h12 · · · −h1K−h21 h22 · · · −h2K...

.... . .

...−hK1 −hK2 · · · hKK

Key Idea


H =

h11 h12 · · · h1Kh21 h22 · · · h2K...

.... . .

...hK1 hK2 · · · hKK


HC =

h11 −h12 · · · −h1K−h21 h22 · · · −h2K...

.... . .

...−hK1 −hK2 · · · hKK

± δ

Key Idea


H =

h11 h12 · · · h1Kh21 h22 · · · h2K...

.... . .

...hK1 hK2 · · · hKK


HC =

h11 −h12 · · · −h1K−h21 h22 · · · −h2K...

.... . .

...−hK1 −hK2 · · · hKK

± δ

3. Otherwise, transmit new signals and wait for their HC .

Ergodic Alignment at the Receivers

Y1(t)Y2(t)...

YK(t)

Y1(tC)Y2(tC)

...

YK(tC)


Y1(t)Y2(t)...

YK(t)

Y1(tC)Y2(tC)

...

YK(tC)

Y1(t) + Y1(tC)Y2(t) + Y2(tC)

...

YK(t) + YK(tC


h11 h12 · · · h1K

h21 h22 · · · h2K

......

. . ....


X + Z(t)

h11 −h12 · · · −h1K

−h21 h22 · · · −h2K

......

. . ....

−hK1 −hK2 · · · hKK

± δ

X + Z(tC)

Y1(t) + Y1(tC)Y2(t) + Y2(tC)

...

YK(t) + YK(tC


h11 h12 · · · h1K

h21 h22 · · · h2K

......

. . ....


X + Z(t)

h11 −h12 · · · −h1K

−h21 h22 · · · −h2K

......

. . ....

−hK1 −hK2 · · · hKK

± δ

X + Z(tC)

2h11 0 · · · 00 2h22 · · · 0...

.... . .

...

0 0 · · · 2hKK

± δ

X + Z(t) + Z(tC)


Sum of channel observations is (nearly) interference-free:

H+HC =

2h11 0. . .

0 2hKK

± δ

Worst case SINR:

2P(

|hkk|2 − 2δ(Re(hkk) + Im(hkk)) + δ2

)

1 + 4δ2(K − 1)P


Sum of channel observations is (nearly) interference-free:

H+HC =

2h11 0. . .

0 2hKK

± δ

Worst case SINR:

limδ↓0

2P(

|hkk|2 − 2δ(Re(hkk) + Im(hkk)) + δ2

)

1 + 4δ2(K − 1)P=

2|hkk|2P

N

Pairing Up Channels

• We need to match up almost every matrix with its complement.

Pairing Up Channels


• Want a finite set of possible matrices H for analysis:

Pairing Up Channels



1. Quantize each channel coefficient to precision δ(closest point in δ(Z + jZ)).

Pairing Up Channels




2. Set threshold hMAX. Throw out any matrix with |hkℓ| > hMAX.

Pairing Up Channels





• Choose δ, hMAX to get desired rate gap.

Pairing Up Channels





• Choose δ, hMAX to get desired rate gap.

• Since phase is i.i.d. uniform, P(H) = P(HC).

Convergence in Type

Sequence of quantized channel matrices Hn is ǫ-typical if:

∣

∣

∣

∣

1

n#(H|Hn)− P (H)

∣

∣

∣

∣

≤ ǫ ∀H ∈ H

Lemma (Csiszar-Korner 2.12)

For any i.i.d. sequence of quantized channel matrices, Hn, theprobability of the set of all ǫ-typical sequences, An

ǫ , is lower boundedby:

P(Anǫ ) ≥ 1−

|H|

4nǫ2

Convergence in Type

Rate

Convergence in Type

Rate

Channel Thresholding

Convergence in Type

Rate


Channel Quantization

Convergence in Type

H1 H1CH2 H2CH3 H3CH4 H4C

Rate


Channel Quantization

Convergence in Type

H1 H1CH2 H2CH3 H3CH4 H4C

Rate


Channel QuantizationUnpaired Matrices

Achievable Rate

Theorem

Each user can achieve at least half its interference-free capacity at anysignal-to-noise ratio:

Rk =1

2E[

log(

1 + 2|hkk|2Pk

)]

>1

2RFREE

k

Network Transformation

1

2

K

1

2

K

Network Transformation

1

2

K

12

12

12

1

2

K

Rayleigh Fading

−20 −10 0 10 20 300

1

2

3

4

5

6

SNR in dB

Rat

e pe

r U

ser

Upper Bound

Ergodic Alignment

5 User TDMA

10 User TDMA

• Channel coefficients i.i.d. Rayleigh. Equal transmit power per user.

When does ergodic alignment reach capacity?

• If all channel gains have fixed, equal magnitudes (and time-varyingi.i.d. uniform phase), ergodic alignment reaches capacity:

C =1

2log

(

1 +2P

N

)

Symmetric Case

• In general, we should waterfill power allocation over channel states.

• Is this enough?


• For Rayleigh fading, we get a very weak interference channel withsome constant probability ρ > 0.

• Ignore all interference in weak interference case. Get RWEAKk .

• Otherwise, use ergodic alignment to get REAk .

• Each user gets Rk = ρRWEAKk + (1− ρ)REA

k > REAk

• We need to mix between decoding, ignoring, and aligninginterference.

• Open Question: Does this come to within a constant gap of thecapacity region?


• Jafar ’09: Whenever the channel is in a bottleneck state, ergodicalignment achieves the capacity.

• Example: K transmitter-receiver pairs randomly placed in a square.Signal strength governed by distance. As K →∞, ergodicalignment achieves capacity.

Outline





Main References

Nested lattice framework in this section is almost entirely drawn from:

• U. Erez and R. Zamir, Achieving 12 log(1 + SNR) on the AWGN

channel with lattice encoding and decoding, IEEE Transactions onInformation Theory, vol. 50, pp. 2293-2314, October 2004.

• U. Erez, S. Litsyn, and R. Zamir, Lattices which are good for (al-most) everything, IEEE Transactions on Information Theory, vol. 51,pp. 3401-3416, October 2005.

• R. Zamir, Lattices are everywhere, in Proceedings of the 4th AnnualWorkshop on Information Theory and its Applications, La Jolla, CA,February 2009.

See Ram Zamir’s lattice tutorials (http://www.eng.tau.ac.il/~zamir/ ) or my ISIT 2011 tutorial (http://iss.bu.edu/bobak/tutorial_isit11.pdf ) for moreinformation.

http://www.eng.tau.ac.il/~zamir/

http://iss.bu.edu/bobak/tutorial_isit11.pdf

Point-to-Point Channels

w Ex pY |X

yD w

The Usual Suspects:

• Message w ∈ {0, 1}k

• Encoder E : {0, 1}k → X n

• Input x ∈ X n

• Estimate w ∈ {0, 1}k

• Decoder D : Yn → {0, 1}k

• Output y ∈ Yn

• Memoryless Channel p(y|x) =n∏

i=1

p(yi|xi)

• Rate R =k

n.

• (Average) Probability of Error: P{w 6= w} → 0 as n→∞. Assumew is uniform over {0, 1}k .

i.i.d. Random Codes

• Generate 2nR codewordsx = [X1 X2 · · · Xn] independentlyand elementwise i.i.d. according tosome distribution pX

p(x) =

n∏

i=1

pX(xi)

• Bound the average error probabilityfor a random codebook.

• If the average performance overcodebooks is good, there must existat least one good fixed codebook.

0 1 2 3 4 · · · q − 1

0

1

2

3

4

...

q − 1

Joint Typicality Decoding

Decoder looks for a codeword that is jointly typical with the receivedsequence y

Error Events

1. Transmitted codeword x is not jointly typicalwith y.=⇒ Low probability by the

Weak Law of Large Numbers.

2. Another codeword x is jointly typical with y.

Cuckoo’s Egg Lemma

Let x be an i.i.d. sequence that is independent from the receivedsequence y.

P

{

(x,y) is jointly typical}

≤ 2−n(I(X;Y )−3ǫ)

See Cover and Thomas.

Point-to-Point Capacity

• We can upper bound the probability of error via the union bound:

P{w 6= w} ≤∑

w 6=w

P

{

(x(w),y) is jointly typical.}

≤ 2−n(I(X;Y )−R−3ǫ) ← Cuckoo’s Egg Lemma

• If R < I(X;Y ), then the probability of error can be driven to zeroas the blocklength increases.

Theorem (Shannon ’48)

The capacity of a point-to-point channel is C = maxpX

I(X;Y ).

Linear Codes

• Linear Codebook: A linear map between messages and codewords(instead of a lookup table).

q-ary Linear Codes

• Represent message w as a length-k vector over Fq.

• Codewords x are length-n vectors over Fq.

• Encoding process is just a matrix multiplication, x = Gw.

x1x2...xn

=

g11 g12 · · · g1kg21 g22 · · · g2k...

.... . .

...gn1 gn2 · · · gnk

w1

w2...wk

• Recall that, for prime q, operations over Fq are just mod qoperations over the reals.

• Rate R =k

nlog q

Random Linear Codes

• Linear code looks like a regularsubsampling of the elements of Fn

q .

• Random linear code: Generateeach element gij of the generatormatrix G elementwise i.i.d.according to a uniform distributionover {0, 1, 2, . . . , q − 1}.

• How are the codewords distributed?0 1 2 3 4 · · · q − 1

0

1

2

3

4

...

q − 1

Fq

Fq

Codeword Distribution

It is convenient to instead analyze the shifted ensemble x = Gw ⊕ v

where v is an i.i.d. uniform sequence. (See Gallager.)

Shifted Codeword Properties

1. Marginally uniform over Fnq . For a given message w, the codeword x

looks like an i.i.d. uniform sequence.

P{x = x} =1

qnfor all x ∈ F

nq

2. Pairwise independent. For w1 6= w2, codewords x1, x2 areindependent.

P{x1 = x1, x2 = x2} =1

q2n= P{x1 = x1}P{x2 = x2}

Achievable Rates

• Cuckoo’s Egg Lemma only requires independence between the truecodeword x(w) and the other codeword x(w). From the unionbound:

P{w 6= w} ≤∑

w 6=w

P

{

(x(w),y) is jointly typical.}

≤ 2−n(I(X;Y )−R−3ǫ)

• This is exactly what we get from pairwise independence.

• Thus, there exists a good fixed generator matrix G and shift v forany rate R < I(X;Y ) where X is uniform.

Removing the Shift

w Ex

z

yD w

• For a binary symmetric channel (BSC), the output can be written asthe modulo sum of the input plus i.i.d. Bernoulli(p) noise,

y = x⊕ z

y = Gw ⊕ v ⊕ z

• Due to this symmetry, the probability of error depends only on therealization of the noise vector z. For a BSC, x = Gw is a good codeas well.

• We can now assume the existence of good generator matrices forchannel coding.

Point-to-Point AWGN Channels

• Codewords must satisfy powerconstraint:

‖x‖2 ≤ nP .

• i.i.d. Gaussian noise with varianceN :

z ∼ N (0, NI) .

• Shannon ’48: Channel capacity:

C =1

2log

(

1 +P

N

)

w Ex

zy

D w

(Cover and Thomas,Elements of Information Theory)

• In high dimensions, noise starts to look spherical.

Lattices

• A lattice Λ is a discrete subgroup ofRn.

• Can write a lattice as a lineartransformation of the integervectors,

Λ = {Bs : s ∈ Zn} ,

for some B ∈ Rn×n.

Lattice Properties

• Closed under addition:λ1, λ2 ∈ Λ =⇒ λ1 + λ2 ∈ Λ.

• Symmetric: λ ∈ Λ =⇒ −λ ∈ Λ Zn is a simple lattice.

Lattices

• A lattice Λ is a discrete subgroup ofRn.

• Can write a lattice as a lineartransformation of the integervectors,

Λ = {Bs : s ∈ Zn} ,

for some B ∈ Rn×n.

Lattice Properties

• Closed under addition:λ1, λ2 ∈ Λ =⇒ λ1 + λ2 ∈ Λ.

• Symmetric: λ ∈ Λ =⇒ −λ ∈ Λ BZn

Voronoi Regions

• Nearest neighbor quantizer:

QΛ(x) = argminλ∈Λ

‖x− λ‖2

• The Voronoi region of a lattice pointis the set of all points that quantizeto that lattice point.

• Fundamental Voronoi region V:points that quantize to the origin,

V = {x : QΛ(x) = 0}

• All Voronoi regions are just shifts ofV

Nested Lattices

• Two lattices Λ and ΛFINE are nestedif Λ ⊂ ΛFINE

• Nested Lattice Code: All latticepoints from ΛFINE that fall in thefundamental Voronoi region V of Λ.

• V acts like a power constraint

Rate =1

nlog

(

Vol(V)

Vol(VFINE)

)

Nested Lattice Codes from q-ary Linear Codes

• Choose an n× k generatormatrix G ∈ F

n×kq for q-ary code.

• Integers serve as coarse lattice,Λ = Z

n.

• Map elements {0, 1, 2, . . . , q − 1}to equally spaced points between−1/2 and 1/2.

• Place codewords x = Gw intothe fundamental Voronoi regionV = [−1/2, 1/2)n

0 1 2 3 4 · · · q − 1

0

1

2

3

4

...

q − 1

Fq

Fq

(− 12,− 1

2) ( 1

2,− 1

2)

(− 12, 12) ( 1

2, 12)

Modulo Operation

• Modulo operation with respect tolattice Λ is just the residualquantization error,

[x] mod Λ = x−QΛ(x) .

• Mimics the role of mod q in q-aryalphabet.

• Distributive Law:[

x1 + [x2] mod Λ]

mod Λ

= [x1 + x2] mod Λ

Modulo Operation

• Modulo operation with respect tolattice Λ is just the residualquantization error,

[x] mod Λ = x−QΛ(x) .

• Mimics the role of mod q in q-aryalphabet.

• Distributive Law:[

x1 + [x2] mod Λ]

mod Λ

= [x1 + x2] mod Λ

mod Λ

mod Λ AWGN Channel

w Ex

zy

mod Λy

D w

• Codebook lives on Voronoi region V of coarse lattice Λ.

• Take mod Λ of received signal prior to decoding.

• What is the capacity of the mod Λ channel?

Using random codes: C =1

nmaxp(x)

I(x; y)

mod Λ AWGN Channel Capacity

w Ex

zy

mod Λy

D w

nC = maxp(x)

I(x; y)

= maxp(x)

(

h(y)− h(y|x))


w Ex

zy

mod Λy

D w

nC = maxp(x)

I(x; y)

= maxp(x)

(

h(y)− h(y|x))

= maxp(x)

(

h(y)− h(

[z] mod Λ))

Distributive Law


w Ex

zy

mod Λy

D w

nC = maxp(x)

I(x; y)

= maxp(x)

(

h(y)− h(y|x))

= maxp(x)

(

h(y)− h(

[z] mod Λ))

Distributive Law

≥ maxp(x)

(

h(y)− h(z))

Point Symmetry of Voronoi Region


w Ex

zy

mod Λy

D w

nC = maxp(x)

I(x; y)

= maxp(x)

(

h(y)− h(y|x))

= maxp(x)

(

h(y)− h(

[z] mod Λ))

Distributive Law

≥ maxp(x)

(

h(y)− h(z))

Point Symmetry of Voronoi Region

= maxp(x)

(

h(y)−n

2log(2πeN)

)

Entropy of Gaussian Noise


w Ex

zy

mod Λy

D w

• Channel output entropy upper bounded by the logarithm of theVoronoi region volume:

h(y) ≤ log(Vol(V)) with equality if y ∼ Unif(V)

• y = [x+ z] mod Λ is uniform over V if x is uniform over V.

• Random coding over the Voronoi region V can achieve:

C =1

nlog(Vol(V)) −

1

2log(2πeN)

Power Constraints and Second Moments

w Ex

zy

mod Λy

D w

• Must scale lattice Λ so that the uniform distribution over theVoronoi region V meets the power constraint P .

• Set second moment σ2Λ =

1

nVol(V)

∫

V‖x‖2dx equal to P .

Normalized Second Moment: G(Λ) =σ2Λ

(Vol(V))2/n

=⇒1

nlog(Vol(V)) =

1

2log

(

σ2Λ

G(Λ)

)

=1

2log

(

P

G(Λ)

)


w Ex

zy

mod Λy

D w

• Usual i.i.d. random coding over V combined with the union bound:

C ≥1

nlog(Vol(V))−

1

2log(2πeN)

=1

2log

(

P

G(Λ)

)

−1

2log(2πeN)

=1

2log

(

P

N

)

−1

2log(2πeG(Λ))

What is G(Λ)?

w Ex

zy

mod Λy

D w

• The normalized second moment G(Λ) is a dimensionless quantitythat captures the shaping gain.

• Integer lattice is not so bad, G(Zn) = 1/12.

• Capacity under mod Zn is at least

C ≥1

2log

(

P

N

)

−1

2log

(

2πe

12

)

≈1

2log

(

P

N

)

− 0.255

Asymptotically Good G(Λ)

Theorem (Zamir-Feder-Poltyrev ’94)

There exists a sequence of lattices Λ(n) such that limn→∞

G(Λ(n)) =1

2πe.

n = 1 n = 2

· · ·

n→∞

• Best possible normalized second moment is that of a sphere.

• Using a sequence Λ(n) with an asymptotically good G(Λ(N)) allowsto approach

R =1

2log

(

P

N

)

−1

2log

(

2πe

2πe

)

=1

2log

(

P

N

)

Linear Codes for mod Λ Channels

• Instead of an “inner” randomcodes, we can use a q-ary linearcode.

• This is exactly a nested lattice.

• Each codeword has a uniformmarginal distribution over thegrid.

• Rate loss due to finiteconstellation which goes to 0 asq →∞.

• Codewords are pairwiseindependent so we can apply theunion bound.

0 1 2 3 4 · · · q − 1

0

1

2

3

4

...

q − 1

Fq

Fq

(− 12,− 1

2) ( 1

2,− 1

2)

(− 12, 12) ( 1

2, 12)

MMSE Scaling

• Erez-Zamir ’04: Prior to taking mod Λ, scale by α.

y = [αy] mod Λ

= [αx+ αz] mod Λ

= [x+ αz− (1− α)x] mod Λ

Effective Noise

• For now, ignore that the effective noise is not independent of thecodeword. Effective noise variance NEFFEC = α2N + (1− α)2P .

• Optimal choice of α is the MMSE coefficient αMMSE =P

N + P.

NEFFEC = α2MMSEN + (1− αMMSE)

2P =PN

N + P

C =1

2log

(

P

NEFFEC

)

=1

2log

(

1 +P

N

)

Dithering

• Now the noise is dependent on the codeword.

• Dithering can solve this problem (just as in the discrete case).

• Map message to a codeword t.

• Generate a random dither vector d uniformly over V.

• Transmitter sends a dithered codeword:

x = [t+ d] mod Λ

• x is now independent of the codeword t.

Decoding – Remove Dither First

• Transmitter sends dithered codeword x = [t+ d] mod Λ.

• After scaling the channel output y by α, the decoder subtracts thedither d.

y = [αy − d] mod Λ

= [αx+ αz− d] mod Λ

= [x− d+ αz− (1− α)x] mod Λ

=[

[t+ d] mod Λ− d+ αz− (1− α)x]

mod Λ

= [t+ αz− (1− α)x] mod Λ Distributive Law

• Effective noise is now independent from the codeword t.

• By the probabilistic method, (at least) one good fixed dither exists.No common randomness necessary.

Summary

• Linear code embedded in the integer lattice:

R =1

2log

(

P

N

)

−1

2log

(

2πe

12

)

• Linear code embedded in the integer lattice, MMSE scaling:

R =1

2log

(

1+P

N

)

−1

2log

(

2πe

12

)

• Linear code embedded in a good shaping lattice, MMSE scaling:

R =1

2log

(

1+P

N

)

Theorem (Erez-Zamir ’04)

Nested lattice codes can achieve the AWGN capacity.

Two-Way Relay Channel

w1Has

Wants w2 w1

Has

Wants

w2Relay

Two-Way Relay Channel – Time-Division

w1 w2

w1 w2w1

w1 w2w1 w2

w1 w1 w2w1 w2

(a)

(b)

(c)

(d)

Two-Way Relay Channel – Network Coding

w1 w2

w1 w2w1

w1 w2w1 w2

(a)

(b)

(c)

w1 ⊕w2

Two-Way Relay Channel – Physical-Layer Network Coding

w1 w2

w1 w2

(a)

(b)

w1 ⊕w2

AWGN Two-Way Relay Channel – Symmetric Rates

w1Has

Wants w2 w1

Has

Wants

w2Relay

• Upper Bound:

R ≤1

2log

(

1 +P

N

)

• Decode-and-Forward: Relay decodes w1,w2 and transmits w1 ⊕w2.

R =1

4log

(

1 +2P

N

)

• Compress-and-Forward: Relay transmits quantized y.

R =1

2log

(

1 +P

N

P

3P +N

)


zMAC

yMAC

Relay

xBC

z2z1

User 1

x1w1

w2

User 2

x2 w2

w1

• Equal power constraints P .

• Equal noise variances N .

• Equal rates R.

• Upper Bound:

R ≤1

2log

(

1 +P

N

)

• Decode-and-Forward: Relay decodes w1,w2 and transmits w1 ⊕w2.

R =1

4log

(

1 +2P

N

)

• Compress-and-Forward: Relay transmits quantized y.

R =1

2log

(

1 +P

N

P

3P +N

)


0 5 10 15 200

0.5

1

1.5

2

2.5

3

3.5

SNR in dB

Rat

e pe

r U

ser

Upper Bound

Compress

Decode

Decoding the Sum of Lattice Codewords

Encoders use the same nestedlattice codebook.

Transmit dithered codewords:

x1 = t1

x2 = t2

t1 E1x1

t2 E2x2

z

yD v

v = [t1 + t2] mod Λ

Decoder recovers modulo sum.

[y] mod Λ

= [x1 + x2 + z] mod Λ

= [t1 + t2 + z] mod Λ

=[

[t1 + t2] mod Λ + z]

mod Λ Distributive Law

= [v + z] mod Λ

R =1

2log

(

P

N

)

Decoding the Sum of Lattice Codewords – MMSE Scaling



x1 = [t1 + d1] mod Λ

x2 = [t2 + d2] mod Λ

t1 E1x1

t2 E2x2

z

yD v

v = [t1 + t2] mod Λ

Decoder scales by α, removes dithers, recovers modulo sum.

[αy − d1 − d2] mod Λ

= [α(x1 + x2 + z)− d1 − d2] mod Λ

= [x1 + x2 − (1− α)(x1 + x2) + z− d1 − d2] mod Λ

=[

[t1 + t2] mod Λ− (1− α)(x1 + x2) + z]

mod Λ

= [v − (1− α)(x1 + x2) + z] mod Λ

Effective Noise NEFFEC = (1− α)22P + α2N

Decoding the Sum of Lattice Codewords – MMSE Scaling

• Effective noise after scaling is NEFFEC = (1− α)22P + α2N .

• Minimized by setting α to be the MMSE coefficient:

αMMSE =2P

N + 2P

• Plugging in, get

NEFFEC =2NP

N + 2P

• Resulting rate is

R =1

2log

(

P

NEFFEC

)

=1

2log

(

1

2+

P

N

)

• Getting the full “one plus” term is an open challenge. Does notseem possible with nested lattices.

Finite Field Computation over a Gaussian MAC

Map messages to lattice points:

t1 = φ(w1)

t2 = φ(w2)


x1 = [t1 + d1] mod Λ

x2 = [t2 + d2] mod Λ

w1 E1x1

w2 E2x2

z

yD u

u = w1 ⊕w2

• Integer coarse lattice Λ = Zn, φ(w) = [γGw] mod Z

n where γ isa scalar and G is the generator matrix for the q-ary code.

• General coarse lattice Λ = BZn, φ(w) = [BγGw] mod Λ

• Mapping between finite field messages and lattice codewordspreserves linearity:

φ−1(

[t1 + t2] mod Λ)

= w1 ⊕w2


w1Has

Wants w2 w1

Has

Wants

w2Relay



• Equal rates R.

• Upper Bound:

R ≤1

2log

(

1 +P

N

)

• Compute-and-Forward: Relay decodes w1 ⊕w2 and retransmits.

R =1

2log

(

1

2+

P

N

)

• See Wilson-Narayanan-Pfister-Sprintson ’10.


zMAC

yMAC

Relay

xBC

z2z1

User 1

x1w1

w2

User 2

x2 w2

w1



• Equal rates R.

• Upper Bound:

R ≤1

2log

(

1 +P

N

)

• Compute-and-Forward: Relay decodes w1 ⊕w2 and retransmits.

R =1

2log

(

1

2+

P

N

)

• See Wilson-Narayanan-Pfister-Sprintson ’10.


0 5 10 15 200

0.5

1

1.5

2

2.5

3

3.5

SNR in dB

Rat

e pe

r U

ser

Upper Bound

Compute

Compress

Decode

Compute-and-Forward Illustration

w2

w1x1

x2

z

yw1 ⊕w2

Random i.i.d. codes are not good for computation

2nR codewords each. 2n2R possible sums of codewords.

Random i.i.d. codes are not good for computation

2nR codewords each. 2n2R possible sums of codewords.

x1

x2

z

y

Unequal Power Constraints – Double Nesting

• What if the power constraintsare not equal?

• Idea fromNam-Chung-Lee ’10:

• Draw the codewords from thesame fine lattice ΛFINE.

• Use two nested coarse latticesΛ1 and Λ2 to enforce thepower constraints P1 and P2.

Unequal Power Constraints – Double Nesting

t1 E1x1

t2 E2x2

z

yD v

v = [t1 + t2] mod Λ2

• Encoder 1 sends x1 = [t1 + d1] mod Λ1. Coarse lattice Λ1 hassecond moment P1.

• Encoder 2 sends x2 = [t2 + d2] mod Λ2. Coarse lattice Λ2 hassecond moment P2 > P1.

• Decoder performs MMSE scaling, remove dithers, recovers mod Λ2

sum.

R1 =1

2log

(

P1

P1 + P2+

P1

N

)

R2 =1

2log

(

P2

P1 + P2+

P2

N

)

AWGN Two-Way Relay Channel

zMAC

yMAC

Relay

xBC

z2z1

User 1

x1w1

w2

User 2

x2 w2

w1

• User powers P1, P2.

• MAC noise variance NMAC.

• Relay power PBC .

• Broadcast noise variances

N1, N2.

Theorem (Nam-Chung-Lee ’10)

Rate region is within 1/2 bit of:

R1 ≤ min

(

1

2log

(

P1

P1 + P2

+P1

NMAC

)

,1

2log

(

1 +PBC

N2

))

R2 ≤ min

(

1

2log

(

P2

P1 + P2

+P2

NMAC

)

,1

2log

(

1 +PBC

N1

))

Moreover, “constant gap” goes to zero as powers increase.

Many-to-One Interference Channel – Symmetric Very Strong Case

• Equal rates R.

• Only receiver 1 seesinterference:

y1 = x1 + βK∑

ℓ=2

xℓ + z1

• How big does β have to be toachieve R = 1

2 log(

1 + PN

)

?(i.e. “very strong” case)

w1 E1x1

w2 E2x2

β

wK EKxK

β.

.

.

.

.

.

z1y1

z2y2

zKyK

D1 w1

D2 w2

DK wK

• Scheme A: Decode w2, . . . ,wK at receiver 1 and remove prior todecoding w1.

R ≤1

2(K − 1)log

(

1 +β2(K − 1)P

N + P

)

• Scheme B: Decode w2 ⊕ · · · ⊕wK at receiver 1 and remove prior todecoding w1.




xℓ = [tℓ + dℓ] mod Λ

w1 E1x1

w2 E2x2

β

wK EKxK

β.

.

.

.

.

.

z1y1

z2y2

zKyK

D1 w1

D2 w2

DK wK

Decoder scales by β−1, removes dithers, recovers modulo sum.

[

β−1y1 −K∑

ℓ=2

dℓ

]

mod Λ =

[ K∑

ℓ=2

(xℓ − dℓ) + β−1(x1 + z1)

]

mod Λ

=

[

[

K∑

ℓ=2

tℓ

]

mod Λ + β−1(x1 + z1)

]

mod Λ


[

β−1y1 −K∑

ℓ=2

dℓ

]

mod Λ =

[

[

K∑

ℓ=2

tℓ

]

mod Λ + β−1(x1 + z1)

]

mod Λ

• Effective noise variance NEFFEC = β−2(P +N).

• Can decode mod Λ sum of lattice points at rate R = 12 log

( β2PP+N

)

.

• Setting equal to “very strong” condition R = 12 log

(

1 + PN

)

we get

β2 =(P +N)2

PN

• How can we recover w1?

• We need to first subtract the real sum of the codewords. So far, weonly have the modulo-sum.

Successive Cancellation of Sums

• First, add back in dithers to get modulo sum of codewords:[

[

K∑

ℓ=2

tℓ

]

mod Λ +[

K∑

ℓ=2

dℓ

]

mod Λ

]

mod Λ =[

K∑

ℓ=2

xℓ

]

mod Λ

• Subtract from y1 to expose the coarse lattice point nearest to thereal sum

∑Kℓ=2 xℓ:

β−1y1 −[

K∑

ℓ=2

xℓ

]

mod Λ = QΛ

(

K∑

ℓ=2

xℓ

)

+ β−1(x1 + z1)

• Coarse lattice point easier to decode than fine lattice point:

QΛ

(

QΛ

(

K∑

ℓ=2

xℓ

)

+ β−1(x1 + z1)

)

= QΛ

(

K∑

ℓ=2

xℓ

)

w.h.p.

• Finally, get back the real sum

[

K∑

ℓ=2

xℓ

]

mod Λ +QΛ

(

K∑

ℓ=2

xℓ

)

=

K∑

ℓ=2

xℓ

Successive Cancellation of Sums

• We now have the sum of interfering codewords and can cancel itseffects:

y1 − β

K∑

ℓ=2

xℓ = x1 + z1

• Can apply standard MMSE lattice decoding to recover lattice pointt1 and then map back to w1.

• Overall, structured coding permits

β2 ≥(P +N)2

PN

• Compare to decoding interfering codewords in their entirety:

β2 ≥

(

(1 + PN )K−1 − 1

)

(N + P )

(K − 1)P

• Originally shown in Sridharan-Jafarian-Vishwanath-Jafar ’08 usingspherical shaping region. Nested lattice scheme is new.

Many-to-One Interference Channel – Approximate Capacity

w1 E1x1 h11

w2 E2x2

h12

......

wK EKxK hKK

h1K

h22

z1y1

z2y2

zKyK

D1 w1

D2 w2

DK wK

Lattice Codes

......

• Deterministic model by Avestimehr-Diggavi-Tse ’11 shows how todecompose by signal scale.

Theorem (Bresler-Parekh-Tse ’10)

Lattices codes combined with the deterministic model can approachthe capacity region to within (3K + 3)(1 + log(K + 1)) bits per user.

Interference Channel – Symmetric Very Strong Case

w1 E1x1

w2 E2x2

...

wK EKxK

H

z1y1

z2y2

zKyK

D1 w1

D2 w2

...

DK wK

• Equal rates R. How big does β have to be to achieveR = 1

2 log(

1 + PN

)

? (i.e. “very strong” case)

• Can use the many-to-one decoder at every receiver to get

β2 ≥(P +N)2

PNDoes not depend on K.

• What about asymmetric interference channels?

Interference Channel – Symmetric Very Strong Case

w1 E1x1

w2 E2x2

...

wK EKxK

1 β · · · β

β 1 · · · β...

.... . .

...

β β · · · 1

z1y1

z2y2

zKyK

D1 w1

D2 w2

...

DK wK

• Equal rates R. How big does β have to be to achieveR = 1

2 log(

1 + PN

)

? (i.e. “very strong” case)

• Can use the many-to-one decoder at every receiver to get

β2 ≥(P +N)2

PNDoes not depend on K.

• What about asymmetric interference channels?

Interference Channel

w1 E1x1

w2 E2x2

...

wK EKxK

H

z1y1

z2y2

zKyK

D1 w1

D2 w2

...

DK wK

• Not clear how to map to a deterministic model using lattices.

• “Real” interference alignment scheme of Motahari et al. ’08 uses alattice structure to get K/2 DoF (up to a set of measure one)

• Some special cases at finite SNR: Jafarian-Viswanath ’09,’10,

Ordentlich-Erez ’11

Conclusions

• Interference alignment can lead to dramatically higher rates forinterference channels.

• Many unanswered questions: delay, channel state information, etc.

• Many other applications: secrecy (see work of Ulukus and Yener),distributed storage, etc.

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interference alignment - (for interference channels)

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