Multilevel Topological Interference Management: A TIM-TIN Perspective
MMultilevel Topological Interference Management: A TIM-TINPerspective
Chunhua Geng, Hua Sun, Syed A. Jafar ∗ Abstract
The robust principles of treating interference as noise (TIN) when it is sufficiently weak, andavoiding it when it is not, form the background of this work. Combining TIN with the topolog-ical interference management (TIM) framework that identifies optimal interference avoidanceschemes, we formulate a TIM-TIN problem for multilevel topological interference management,wherein only a coarse knowledge of channel strengths and no knowledge of channel phases isavailable to transmitters. To address the TIM-TIN problem, we first propose an analytical base-line approach, which decomposes a network into TIN and TIM components, allocates the signalpower levels to each user in the TIN component, allocates signal vector space dimensions to eachuser in the TIM component, and guarantees that the product of the two is an achievable numberof signal dimensions available to each user in the original network. Next, a distributed numericalalgorithm called ZEST is developed. The convergence of the algorithm is demonstrated, leadingto the duality of the TIM-TIN problem (in terms of GDoF). Numerical results are also providedto demonstrate the superior sum-rate performance and fast convergence of ZEST.
The capacity of wireless interference networks is a rapidly evolving research front, spurred in partby exciting breakthroughs such as the idea of interference alignment [2] which provides fascinating ∗ Chunhua Geng (email: [email protected]) is with MediaTek USA Inc., Irvine, CA. Hua Sun (email:[email protected]) is with the Department of Electrical Engineering, University of North Texas, Denton, TX. SyedA. Jafar (email: [email protected]) is with the Center of Pervasive Communications and Computing (CPCC) in theDepartment of Electrical Engineering and Computer Science (EECS) at the University of California Irvine, Irvine,CA. A part of this work was presented in IEEE Information Theory Workshop (ITW) 2013 [1]. a r X i v : . [ c s . I T ] F e b heoretical insights and shows much promise under idealized conditions. The connection to practicalsettings however remains tenuous. This is in part due to the following two factors. First, becauseof the assumption of precise channel knowledge, idealized studies often get caught in the minutiaeof channel realizations, e.g., rational versus irrational values, that have little bearing in practice.Second, by focusing on the degrees-of-freedom (DoF) of fully connected networks, these studiesignore the most critical aspect of interference management in practice – the differences of signalstrengths due to path loss and fading (in short, network topology). Indeed, the DoF metric treatsevery channel as essentially equally strong (capable of carrying exactly 1 DoF). So the desiredsignal has to actively avoid every interferer, whereas in practice each user needs to avoid only a fewsignificant interferers and the rest are weak enough to be safely ignored. Therefore, by trivializingthe topology of the network, the DoF studies of fully connected networks make the problem muchharder than it needs to be. Non-trivial solutions to this harder problem invariably rely on muchmore channel knowledge than is available in practice. Thus, the two limiting factors re-enforce eachother.Evidently, in order to avoid these pitfalls, one should shift focus away from optimal ways ofexploiting precise channel knowledge (which is rarely available), and toward powerful even optimalways of exploiting a coarse knowledge of interference network topology . This line of thought mo-tivates robust models of interference networks where only a coarse knowledge of channel strengthlevels is available to the transmitters and no channel phase knowledge is assumed. This is the multilevel topological interference management framework. It is a generalization of the elementarytopological interference management (TIM) framework introduced in [3], wherein the transmitterscan only distinguish between channels that are connected (strong) and not connected (weak). Existing wireless interference networks are mainly based on two robust interference managementprinciples — 1) ignore interference that is sufficiently weak, and 2) avoid interference that is not.In slightly more technical terms, ignoring interference translates into treating interference as noise (TIN) [4, 5], and avoiding interference translates into access schemes such as TDMA/FDMA/CDMA.Recent work has explored the optimality of both of these principles.2.
TIN:
The optimality of the first principle, treating interference as noise when it is sufficientlyweak, is discussed extensively. In [6, 7, 8], it is shown that in a so-called “noisy interference”regime, TIN achieves the exact sum capacity of interference channels. In [9], for general K -user interference channels, it provides a broadly applicable TIN-optimality condition underwhich TIN is optimal from a generalized degrees-of-freedom (GDoF) perspective and achievesa constant gap (of no more than log(3 K ) bits) to the entire capacity region. More specifically,under a fully asymmetric setting, the TIN-optimality condition identified in [9] stipulatesthat if for each user the desired signal strength is no less than the sum of the strongestinterference from this user and the strongest interference to this user (all values in dB scale),then power control and TIN achieves the whole GDoF region of this network. Remarkably, thisresult holds even if perfect channel knowledge is assumed everywhere. The TIN-optimalityresult is also generalized to other channel models (e.g., X channels [10, 11], parallel channels[12], compound networks [13], MIMO channels [14], and cellular networks [15, 16]) and isreformulated from a combinatorial perspective [17].2. TIM:
The optimality of the second principle, avoidance , has been investigated most recentlyby [3], as the TIM problem. With channel knowledge at the transmitters limited to a coarseknowledge of network topology (which links are stronger/weaker than the effective noisefloor), TIM is shown in [3] to be essentially an index coding problem [18]. TIM subsumeswithin itself the TDMA/FDMA/CDMA schemes as trivial special cases, but is in generalmuch more capable than these conventional approaches. Remarkably, for the class of linearschemes, which are found to be optimal in most cases studied so far, and within whichTIM is equivalent to the index coding problem, TIM is essentially an optimal allocation ofsignal vector spaces based on an interference alignment perspective [19]. Variants of theTIM problem have also been investigated, such as those under short coherent time [20],with alternating connectivity [21, 22], with multiple antennas [23], with transmitter/receivercooperation [24, 25], with reconfigurable antennas [26], with network topology uncertainty[27], and with confidential messages [28]. 3 .2 TIM-TIN: Joint view of signal vector spaces and signal power levels
The two principles – avoiding versus ignoring interference – which are mapped to TIM and TIN,respectively, naturally correspond to interference management in terms of signal vector spaces andsignal power levels . TIM uses the interference alignment perspective [3, 19] to optimally allocatesignal vector subspaces among the interferers. Note that in order to resolve the desired signalfrom interference based on the signal vector spaces, the strength of each signal is irrelevant. Whatmatters is only that desired signal and the interference occupy linearly independent spaces. TIN, onthe other hand, optimally allocates signal power levels among users by setting the transmit powerlevels at transmitters and the noise floor levels at receivers. Thus TIN depends very much on thestrengths of signals relative to each other. Associating TIM with signal vector space allocationsand TIN with signal power level allocations within the multilevel TIM framework, we refer to thejoint allocation of signal vector spaces and signal power levels as the TIM-TIN problem.
TIM-TIN Problem:
With only a coarse knowledge of channel strengths available to the trans-mitters, we wish to carefully allocate not only the beamforming vector directions (signal vectorspaces) but also the transmit powers (signal power levels) to each of those beamforming vectors.The necessity of a joint TIM-TIN perspective is evident as follows. In vector space allocationschemes used for DoF studies, the signal space containing the interference is entirely rejected (zero-forced). This is typically fine for linear DoF studies because all signals are essentially equallystrong, every substream carries one DoF, so any desired signal projected into the interference spacecannot achieve a non-zero DoF. However, once we account for the difference in signal strengths inthe GDoF framework, the signal vector space dimensions occupied by interference may not be fully occupied in terms of power levels if the interference is weak. So, non-zero GDoF may be achievedby desired signals projected into the same dimensions as occupied by the interference, where inter-ference is weaker than desired signal. It is this aspect that we wish to exploit in this work. It isworthwhile noticing that within the multilevel TIM framework, in general the solution based on acombination of TIM and TIN is not optimal. In [29], it has been shown that for K -user symmetric interference channels, the GDoF optimal solution relies on rate splitting and superposition encod-4ng at transmitters and (partial) interference decoding at receivers. The appeal of joint TIM-TINmainly lies in its implementation simplicity and wide applicability in existing wireless networks.
First, to address the TIM-TIN problem, an analytical baseline approach is presented. Because ofthe minimal channel knowledge requirements in the TIM and TIN settings, a robust combination ofthe two, denoted as
TIM-TIN decomposition presents itself. Any given network is decomposed intoa TIM component and a TIN component, containing only strong and weak interferers, respectively,and a direct multiplication of the signal dimensions available in each is shown to be achievable inthe original network. In other words, the TIM solution identifies the fraction of the signal spacethat is available to each user, and within each of these available signal space dimensions, the TINapproach identifies the fraction of signal levels that are available to the same user. A product of thetwo fractions therefore identifies the net fraction of signal dimensions available to each user in thisdecomposition based approach. The optimality of this decomposition approach is also discussedfor some non-trivial network settings.Next, a distributed numerical approach is developed for the TIM-TIN problem, which onlyneeds local channel measurements to update transmit powers and beamforming vectors. The pro-posed algorithm, called ZEST, utilizes the reciprocity of wireless networks, and is guaranteed to beconvergent in terms of GDoF. As a byproduct, the duality of the TIM-TIN problem is established.We also numerically validate the superior GDoF performance and fast convergence of ZEST.
Notations : For a positive integer Z , [ Z ] (cid:44) { , , ..., Z } . Z + and Z − denote the sets of non-negative integers and non-positive integers, respectively. For vectors u and v , we say that u dominates v if u ≥ v , where ≥ denotes componentwise inequality. For a matrix A , det( A ) denotesits determinant, span( A ) represents the space spanned by the column vectors of A , and A ( i, j )denotes the entry of A in the i -th row and j -th column. All logarithms are to the base 2. Within the multilevel TIM framework, for a K -user interference with arbitrary channel strengths, the optimalGDoF region is still open. System Model
In this work, we consider a K -user complex Gaussian interference channel, where Transmitter k ( k ∈ [ K ]) intends to communicate with Receiver k and all the transmitters and receivers areequipped with one antenna. Following [9, 30], the channel model is given by Y k ( t ) = K (cid:88) i =1 √ P α ki e jθ ki X i ( t ) + Z k ( t ) , ∀ k ∈ [ K ] , (1)where at each time index t , X i ( t ) is the transmitted symbol of Transmitter i (subject to a unit powerconstraint, i.e., E [ | X i ( t ) | ] ≤ Y k ( t ) is the received signal of Receiver k , and Z k ( t ) ∼ CN (0 ,
1) isthe additive white Gaussian noise (AWGN) at Receiver k . In (1), P > α ki ≥ i and Receiver k , and θ ki is the corresponding channel phase. The definitions of messages, achievable rate of user k ( R k )and channel capacity region ( C ) are all standard. The GDoF region is defined as D (cid:44) (cid:110) ( d , d , ..., d K ) : d i = lim P →∞ R i log P , ∀ i ∈ [ K ] , ( R , R , ..., R K ) ∈ C (cid:111) . (2)In the multilevel TIM framework, only a coarse knowledge of channel strength levels is avail-able to the transmitters and no channel phase knowledge is assumed. The channel strength levelknowledge at transmitters can either be perfect or quantized. We also assume that receivers haveperfect channel state information. Apparently, multilevel TIM is a generalization of the elementaryone. It also should be noted that unlike most previous works in pursuit of the coarse DoF metricwhere all non-zero channels are essentially treated as approximately equally strong (i.e., each non-zero channel carries one DoF), in the multilevel TIM framework, the main challenge lies in how toleverage the disparate channel strengths, and the more general GDoF metric is of interest. Thisprogressive refinement (from DoF to GDoF) has been shown instrumental for capacity approxi-mation of Gaussian interference networks in recent works [9, 10, 30, 31], where the GDoF resultusually further serves as a stepping stone for the capacity characterization within a constant gap.Below we define the problem of multilevel TIM with quantized channel strength levels, or6uantized multilevel TIM (QM-TIM) in more details. Note that in practice, the desired signalstrength and interfering signal strength usually fall into different ranges, so it is reasonable toassume that desired links and interfering links use different quantization levels. For direct channels,the channel strength levels are assumed to be large enough to guarantee a satisfying interference-free achievable rate. As a result, for direct links the quantized channel strength levels are alwaysnormalized to one, i.e., α ii = 1, ∀ i ∈ [ K ]. While for interfering links , for better interferencemanagement, they are quantized by l + 1 levels with quantization thresholds t , t , ..., t l , where0 ≤ t < t < ... < t l < ∞ . Hereafter, we denote the QM-TIM problem with the above quantizationconfiguration by QM-TIM( t , t , ..., t l ). With those notations, the original TIM problem is a specialcase of QM-TIM, which can be denoted by QM-TIM(0). As another example, the simplest setting ofQM-TIM beyond the elementary one is QM-TIM( t , t ). One natural choice for the two quantizedthresholds could be t = 0 and t = 0 .
5. In this case, we have the following three kinds of interferinglinks: 1) Weak interfering links: the interfering links that are no stronger than the noise floor; 2)Medium interfering links: the interfering links whose channel strength level value falls into therange from 0 to 0.5; 3) Strong interfering links: the interfering links whose channel strength levelis no less than 0.5.
In this section, we formulate the TIM-TIN problem within the multilevel TIM framework formally.As mentioned before, with only a coarse knowledge of channel strengths available to the transmit-ters, in the TIM-TIN problem, we allocate not only the beamforming vectors but also the transmitpowers to each of those beamforming vectors, in order to jointly optimize both signal vector spaceand signal power level allocations.For a K -user interference channel in (1), over n channel uses, Transmitter i sends out b i ( b i ≤ n ) With a little abuse of notations, in QM-TIM, we also use α ij to denote the quantized channel strength level forthe link between Transmitter j and Receiver i , ∀ i, j ∈ [ K ]. Here the unit quantized channel strength level is a result of normalization and imposes no loss of generality. Morespecifically, assume that we have multiple quantization levels for the direct links, and 1 is the highest quantizationthreshold, which represents a satisfying SNR value for the direct links (i.e., without interference, a satisfying achievablerate can be guaranteed). Through appropriate system design, it is natural to expect that all the direct links achievethis satisfying SNR value, thus the actual channel strength levels for all the direct links should be no less than 1.After quantization we have α ii = 1, ∀ i ∈ [ K ]. s i,l and is transmitted along an n × v i,l , l ∈ [ b i ]. Assume that all symbols s i,l are drawn from independentGaussian codebooks, each with zero mean and unit power, and the beamforming vectors v i,l arescaled to have unit norm. Receiver k obtains an n × n channel uses y k = K (cid:88) i =1 b i (cid:88) l =1 √ P α ki e jθ ki √ P r i,l v i,l s i,l + z k , (3)where z k is an n × k , P r i,l is the transmit power for l -th data stream of User i . Due to the unit power constraint, we require r i,l ≤
0. For User k , the covariance matrix of the desired signal is Q Dk = (cid:80) b k l =1 ( v k,l v † k,l ) P r k,l + α kk .The covariance matrix of the net interference-plus-noise is Q N + Ik = (cid:80) i (cid:54) = k Q ki + I , where I is an n × n identity matrix, and Q ki = (cid:80) b i l =1 ( v i,l v † i,l ) P r i,l + α ki is the covariance matrix of the interferencefrom Transmitter i (cid:54) = k . Given the beamforming vectors of each transmitter and power allocationsof all data streams, as in the TIM-TIN problem the receivers do not attempt to decode interferencefrom unintended transmitters , for User k ∈ [ K ] the achievable rate per channel use is given by R k = 1 n I ( s k, , s k, , ..., s k,b k ; y k )= 1 n (cid:104) h ( y k ) − h ( y k | s k, , s k, , ..., s k,b k ) (cid:105) = 1 n (cid:26) log (cid:104) det( Q Dk + Q N + Ik ) (cid:105) − log (cid:104) det( Q N + Ik ) (cid:105)(cid:27) , and the achievable GDoF value d k is d k = lim P →∞ R k log P = lim P →∞ log (cid:104) det( Q Dk + Q N + Ik ) (cid:105) − log (cid:104) det( Q N + Ik ) (cid:105) n log P . (4)Next, we simplify the achievable GDoF expression into a more intuitive form. Consider aterm of the type log (cid:104) det( I + (cid:80) mi =1 P κ i v i v † i ) (cid:105) , where v i ( i ∈ [ m ]) is an n × κ ≥ κ ≥ ... ≥ κ m ≥
0. Consider the vectors v i ’s one by one. For v , we relabel it as v Π(1) and correspondingly its power exponent κ as κ Π(1) . For v , if it fallsinto span( v Π(1) ), we remove it and then proceed to v ; otherwise, we relabel it as v Π(2) and8orrespondingly κ as κ Π(2) . We repeat this operation for each vector. In other words, for v i , ifit falls into span( v Π(1) , v Π(2) , ..., v Π( l ) ) (i.e., the space spanned by all previous linearly independentvectors obtained from { v , v , ..., v i − } ), we remove it and then proceed to v i +1 ; otherwise, werelabel it as v Π( l +1) and correspondingly its power exponent κ i as κ Π( l +1) . Finally, we have γ ≤ n linearly independent beamforming vectors V Π = { v Π(1) , v Π(2) , ..., v Π( γ ) } and their associated powerexponents P Π = { κ Π(1) , κ
Π(2) , ..., κ Π( γ ) } . With those definitions, we have the following lemma. Lemma 1
Suppose that v i , i ∈ [ m ] are n × vectors, and κ ≥ κ ≥ ... ≥ κ m ≥ . We have log (cid:104) det( I + m (cid:88) i =1 P κ i v i v † i ) (cid:105) = γ (cid:88) i =1 κ Π( i ) log P + o (log( P )) . (5)The proof of Lemma 1 is given in Appendix A. Now we can proceed to the following lemma. Lemma 2
In the TIM-TIN problem, given the beamforming vectors and the power allocations foreach user, zero-forcing with successive cancellation (ZF-SC) achieves the maximal GDoF value ofeach user given by (4). The proof of Lemma 2 is deferred to Appendix B. With Lemma 2, to maximize the achievableGDoF in the TIM-TIM problem, the remaining challenge is choosing beamforming vectors andtheir powers for each user judiciously. To address this problem, in the following we develop twoapproaches, i.e., an analytical decomposition approach and a numerical distributed approach. (cid:2204) (cid:2779),(cid:2778) (cid:2204) (cid:2780),(cid:2778) (cid:2204) (cid:2778),(cid:2778) (cid:2204) (cid:2778),(cid:2779) (cid:2204) (cid:2779),(cid:2779)
Figure 1: The received signal at Receiver 1, where the length of the vector represents the received power ofthe carried symbol. Here the number of channel uses n is 2. Note that with the ZF-SC receiver, each user only successively decodes and cancels the (possible multiple) desireddata streams from its own transmitter, but does not decode interfering signals from others. xample 1 To help understand Lemma 1 and 2, consider a -user interference channel, in whichover channel uses User , and deliver , and data streams, respectively. Given thebeamforming vectors, the transmitted power allocated to each symbol and channel strength levels foreach link, the received signal at Receiver is depicted in Fig. 1, where v , and v , are aligned alongone direction. The length of the vector represents the received power of the carried symbol. We have r , + α > r , + α > r , + α > r , + α > r , + α > . Define d (cid:48) k = lim P →∞ log[det( Q Dk + Q N + Ik )]log P and d (cid:48)(cid:48) k = lim P →∞ log[det( Q N + Ik )]log P . Following Lemma 1, we have d (cid:48) = r , + α + r , + α and d (cid:48)(cid:48) = r , + α + r , + α . So the achievable GDoF value of User is d = d (cid:48) − d (cid:48)(cid:48) r , + α + r , + α ) − ( r , + α + r , + α )]2 (6) Next, we illustrate how to achieve this GDoF value via a ZF-SC receiver. To decode s , , wefirst zero force the strongest interference s , and then treat all the other interference as noise. Theachievable GDoF value of data stream s , is d , = ( r , + α − max { r , + α , r , + α , r , + α } )2 = 12 ( r , + α − r , − α ) After recovering s , , we subtract it off from the received signal and then decode s , . Similarly,we first zero force the strongest interference s , (and its aligned counterpart s , ) and then treatthe remaining interference s , as noise. The achievable GDoF value of data stream s , is d , = ( r , + α − r , − α ) . The achievable GDoF value for User is the sum of d , and d , , whichequals (6). Also note that the achievable GDoF value does not depend on the decoding order, i.e.,if we reverse the decoding order of s , and s , , we still achieve the same GDoF value for User 1. In this section, for the TIM-TIN problem we present an analytical baseline approach, denotedby
TIM-TIN decomposition . The basic idea of this approach is as follows. Any given networkcan be decomposed into a TIM component and a TIN component, each containing all the desiredlinks and non-overlapping interfering links, such that in total, these two components cover all the10nterfering links. In other words, denote the sets of all interfering links in the original network, TIMcomponent, and TIN component by I , I TIM , and I TIN , respectively. We have I TIM ∩ I
TIN = φ and I TIM ∪ I
TIN = I . First, consider the TIM component only. Assume that all the links areequally strong. Applying the TIM solution yields an achievable GDoF tuple ( d , TIM , ..., d K, TIM ),which identifies the fraction of the signal space available to each user. Next, consider the TINcomponent only. Applying appropriate power control at each transmitter and treating interferenceas noise at each receiver, we obtain an achievable GDoF tuple ( d , TIN , ..., d K, TIN ), which identifieswithin the available signal space dimensions assigned to each user, the fraction of signal levels thatare available to each of them. Finally, the product of the two above fractions, i.e., the GDoF tuple( d , TIN × d , TIM , ..., d K, TIN × d K, TIM ), is achievable, identifying the net fraction of signal dimensionsavailable to each user by this decomposition approach. Note that the decomposition is quiteflexible, i.e., any interfering link can be considered in either TIM or TIN component (but not bothsimultaneously). Therefore, for one interference channel, we have multiple possible decompositions.For the TIM-TIN decomposition approach, we have the following theorem.
Theorem 1
For one specific TIM-TIN decomposition in a general K -user interference channel,let D TIM be the achievable GDoF region of the TIM component via signal space approach (i.e.,interference alignment and ZF), and D TIN be the achievable GDoF region of the TIN component viasignal level approach (i.e., power control and TIN). Then, the following GDoF region is achievablein the original K -user interference channel, ¯ D = (cid:110) ( d ,d , ..., d K ) : d i = d i, TIM × d i, TIN , ∀ i ∈ [ K ] , ∀ d TIM = ( d , TIM , ..., d K, TIM ) ∈ D TIM , ∀ d TIN = ( d , TIN , ..., d K, TIN ) ∈ D TIN (cid:111) (7)
The whole achievable GDoF region based on the TIM-TIN decomposition approach is given by D TIM − TIN = Convex Hull (cid:16) (cid:83)
T IM−T IN ¯ D (cid:17) , where T IM − T IN denotes the set of all the possibleTIM-TIN decompositions and the convex hull operation comes from time-sharing.Proof : Here, the key is to prove (7). In a specific TIM-TIN decomposition, for User k ∈ [ K ], denote the set of its interferers in the TIM component by I k . To achieve the GDoF tuple11 d , TIM × d , TIN , ..., d K, TIM × d K, TIN ) in the original channel, the beamforming vectors of each userare the same as those yield the GDoF tuple d TIM in the TIM component, and the power allocationfor (all the data streams of) each user follows from the solution that yields the GDoF tuple d TIN inthe TIN component. At the receiver, User k zero-forces the interference from the users in I k andtreats the remaining interference as noise, which achieves the GDoF value d k, TIM × d k, TIN . (cid:4) Example 2
Consider a 5-user interference channel within the QM-TIM(0,0.5) framework in Fig. 2(a).The network is decomposed into a TIN component and a TIM component as shown in Fig. 2(b) andFig. 2(c), respectively. For the TIN component, which contains all the medium interfering linksand satisfies the TIN-optimality condition of [9], according to Theorem 1 in [9] we obtain that itsoptimal symmetric GDoF value is . . In the TIM component, which contains all the strong in-terfering links, the symmetric GDoF value is . [3]. Therefore, through this decomposition, in theoriginal network the symmetric GDoF value . × . . is achievable. The achievable scheme isgiven explicitly in Fig. 2(d). In this scheme, n = 2 and b i = 1 , ∀ i ∈ { , ..., } . More specifically, theachievable scheme uses a 2 dimensional space and 4 beamforming vectors, where any two of themare linearly independent and W and W are aligned along the same vector. The transmit powerallocations are r = 0 , r = − . , r = − . , r = − . and r = − . . It is easy to verify thatevery user achieves a GDoF value . . • Receiver 1 first zero forces the interference from Transmitter 4 (to simplify notations, inthe following for each Receiver k we denote the interference from Transmitter i (cid:54) = k by I i ).Then, in the remaining signal dimension, it treats the interference I as noise. Therefore, theachievable GDoF value for Receiver is (1 − . / . . • Receiver 2 zero forces I and treats I and I as noise to get (0 . − . / . GDoF. • Receiver 3 zero forces I and I and treats I as noise to get (0 . − . / . GDoF. • Receiver 4 zero forces I and treats I as noise to get (0 . − . / . GDoF. • Receiver 5 zero forces I to get . / . GDoF.
As mentioned before, within the multilevel TIM framework, in general the solution based on acombination of TIM and TIN (including the decomposition approach presented in this section) is12 T T R R R T R T R (a) T T T R R R T R T R (b) T T T R R R T R T R (c) (cid:2179) (cid:2780) (cid:2179) (cid:2782) (cid:2179) (cid:2779) (cid:2179) (cid:2778) (cid:2179) (cid:2781) (d) Figure 2: (a) A 5-user interference channel. The red solid lines and dashed blue lines represent strong andmedium interfering links, respectively. The weak interfering links are omitted to avoid cluttering the graph.(b) The TIN component with all medium interfering links. (c) The TIM component with all strong interferinglinks. (d) The achievable scheme to achieve the symmetric GDoF value 0.3 in the original network. not optimal from an information theoretic perspective. However, as shown in the following, thisrobust decomposition approach works rather well when the quantized channel strength levels forcross links are concentrated around the bottom half of the signal levels, i.e., QM-TIM( t , t , ..., t l )where t l ≤ .
5. For this setting, it characterizes the symmetric GDoF value to a constant factorthat is no larger than 2.
Theorem 2
For QM-TIM ( t , t , ..., t l ) where t l ≤ . , the TIM-TIN decomposition approach char-acterizes the symmetric GDoF value d sym within a factor of − t l ≤ . The proof of Theorem 2 is relegated to Appendix C.
Remark 1
The setting of QM-TIM ( t , t , ..., t l ) where t l ≤ . is justified by the conjecture thatthe optimal allocation of limited quantization bins for interfering links would be more concentratednear the noise floor. Intuitively, this is because the opportunities to communicate exist only wherethe desired signal significantly dominates noise/interference strengths, especially for settings withchannel uncertainties where one might be forced to treat interference as noise. Although in generalthe optimal channel quantization is still an interesting open problem (which is beyond the scope ofthis paper), the above conjecture is partially settled for the -user Z interference channel in [32]. Example 3
Consider the 5-user interference channel in Fig. 2(a) again. In Example 2, we haveshown that the symmetric GDoF value . is achievable. Since . is an outer bound, the sym-metric GDoF value of the original network can be characterized to a factor of . In fact, we can mprove this factor further. One can verify that if the medium interfering link between Transmitter and Receiver is moved from the TIN component to the TIM component, the symmetric GDoFvalue for the new TIN component increases to and the new TIM component still achieves thesymmetric GDoF value . Therefore, the achievable symmetric GDoF value via this new TIM-TINdecomposition is improved upon to , and the symmetric GDoF value of the network in Fig. 2(a)is characterized to a factor of . T K T K ` T K +1 T K + S T K + S +1 T K + S + M T K + S + M +1 T K ` S ` M ` T K ` S ` M T K ` S ` T K ` S R K ` S R K R K ` R K ` S ` R K ` S ` M R K ` S ` M ` R K +1 R K + S R K + S +1 R K + S + M R K + S + M +1 M mediuminterferersM mediuminterferersinterferersS stronginterferersS strong … … … … … … … … Figure 3: The symmetric multilevel neighboring interference channel with an infinite number of users. Toavoid cluttering the figure, only the direct links for users with indexes { K − S − M − , ..., K + S + M + 1 } and the interring links for Receiver k are shown. The red solid lines and blue dashed lines represent strongand medium interfering links, respectively. Finally, we demonstrate that TIM-TIN decomposition achieves the optimal symmetric GDoFvalue for one broad class of multilevel neighboring interference channel, which is a natural gener-alization of the cellular blind interference alignment problem (or wireless index coding problem)in [19]. In order to limit the number of parameters while still covering broad classes of networksettings, here we mainly study symmetric cases, i.e., where relative to its own position, each receiverhas the same set of strong and medium interfering links. More specifically, consider the channeldepicted in Fig. 3, which is a locally connected interference channel with an infinite number of userswithin the QM-TIM(0,0.5) framework. For each receiver k , there are 2( S + M ) + 1 transmittersconnected to it with channel strength level larger than the effective noise floor. One of them is the14esired Transmitter k . The 2 S transmitters with indices { k − S, ..., k − } and { k + 1 , ..., k + S } are connected to Receiver k with strong interfering links, and the 2 M transmitters with indices { k − S − M, ..., k − S − } and { k + S + 1 , ..., k + S + M } are connected to Receiver k with mediuminterfering links. For such networks, we have the following result. Theorem 3
For the above symmetric multilevel neighboring interference channel, the symmetricGDoF value is d sym = S + M +1 , M ≤ S S +1) , M > S (8) which is achievable by TIM-TIN decomposition. The proof details are provided in Appendix D. It is notable that for the symmetric neighboringinterference channel, the signal space approach (with one-to-one alignments, see Appendix D)always achieves 1 / ( S + M +1) GDoF. When M > S +1, according to Theorem 3, the decompositionapproach outperforms the pure signal space approach in terms of GDoF, and with M increasingthe gap between these two strictly increases. Remark 2
The result in Theorem 3 can be extended to some asymmetric cases directly. Forinstance, suppose that the number of strong interferers for each user k is still S , whose indicesare still { k − S, ..., k − } and { k + 1 , .., k + S } . However, different users have different numbers ofmedium interferers. For User k , the indices of the medium interferers are { k − S − M U k , ..., k − S − } and { k + S + 1 , ..., k + S + M D k } . If ∀ k , M U k > S and M D k > S , the symmetric GDoF value forsuch asymmetric multilevel neighboring interference channels remains as S +1) . The converse andachievability arguments both follow from the proof of Theorem 3. The TIM-TIN decomposition approach presented in Section 4 is a centralized analytical method,which requires the coarse channel strength information of all links in the network together forjoint signal vector space and signal power level allocation. In this section, we devise a distributednumerical algorithm to address the TIM-TIN problem, which only requires local measurements on15he signal strengths at each user. The proposed algorithm is built upon a distributed power controlalgorithm based on the duality of TIN [33], whose key ingredient is restated below.
Lemma 3 (Lemma 1 in [33]) In a general K -user interference channel, assume that a valid powerallocation ( r , ..., r K ) , r i ≤ , ∀ i ∈ [ K ] , achieves a GDoF tuple ( d , ..., d K ) . In its reciprocalchannel using the power allocation (¯ r , ..., ¯ r K ) , where ¯ r k = − max j : j (cid:54) = k { , α kj + r j } , ∀ k ∈ [ K ] , (9) the achieved GDoF tuple ( ¯ d , ..., ¯ d K ) dominates ( d , ..., d K ) , i.e., ¯ d k ≥ d k , ∀ k ∈ [ K ] . The proposed TIM-TIN distributed numerical algorithm, which is called ZEST, is specified atthe top of next page. The convergence of the ZEST algorithm is given by the following theorem.
Theorem 4
In the ZEST algorithm, −→ d ( m )Σ converges. The proof of Theorem 4 is presented in Appendix E, where we show that −→ d ( m ) ≤ ←− d ( m ) switch ≤ ←− d ( m ) ≤ −→ d ( m ) switch ≤ −→ d ( m +1) . (10)Remarkably, the proof of Theorem 4 leads to the duality of the TIM-TIN problem naturally. Theorem 5 (Duality of TIM-TIN)
In the TIM-TIN problem, any K -user interference channel andits reciprocal channel have the same achievable GDoF region.Proof : Through (10), one can find that for any channel with arbitrary beamforming vectors andpower allocations, in its reciprocal channel, we can always construct some beamforming vectors andtheir associated power allocations, such that the obtained GDoF tuple in the reciprocal channeldominates that achieved in the original channel. Since in the TIM-TIN problem, the achievable In the TIN scheme, assume that the allocated power to User i ∈ [ K ] is P r i , r i ≤
0. From the GDoF perspective,we refer to the power exponent vector ( r , ..., r K ) as the power allocation. Note that in steps 3) and 5) of the proposed ZEST algorithm, when the beam l ∈ [ b k ] of User k ∈ [ K ] updatesits power allocation following Lemma 3, it treats all the remaining received beams after ZF and SC (including theother desired beams of User k ) as interference. Also note that since in steps 2) and 4) a successive cancellation isadopted in the lexicographic order, the beam l of User k does not receive interference from beam s of User k , where s, l ∈ [ b k ] and s < l . lgorithm 1 ZEST: ZE ro-forcing with S uccessive cancellation and power control for T IM-TIN1) Let m = 1. Set n and b k , and randomly choose unit-norm beamforming vectors −→ v ( m ) k,l andpower allocations −→ r ( m ) k,l that satisfy the unit power constraint, k ∈ [ K ], l ∈ [ b k ].2) In the original channel, update the receiving vectors −→ u ( m ) k,l to the unit-norm ZF-SC receivingvectors that achieve the maximal GDoF value for each user (See Lemma 2. Without loss ofgenerality, the cancellation is taken in the lexicographic order). Compute the achievable GDoFtuple −→ d ( m ) and the achievable sum-GDoF value −→ d ( m )Σ .3) Reverse the direction of the communication. Calculate the power allocation ←− r ( m ) k,l for each datastream in the reciprocal channel following (9), and set the beamforming and receiving vectors ←− v ( m ) k,l and ←− u ( m ) k,l as follows ←− v ( m ) k,l = −→ u ( m ) k,l , ←− u ( m ) k,l = −→ v ( m ) k,l , ∀ k ∈ [ K ] , ∀ l ∈ [ b k ]Compute the achievable GDoF tuple ←− d ( m ) switch and the achievable sum-GDoF value ←− d ( m )Σ , switch (using receivers with the reverse lexicographic cancellation order).4) In the reciprocal channel, update the receiving vectors ←− u ( m ) k,l to the unit-norm ZF-SC receivingvectors that achieve the maximal GDoF value for each user (Again, the cancellation is taken in thelexicographic order). Compute the achievable GDoF tuple ←− d ( m ) and the achievable sum-GDoFvalue ←− d ( m )Σ .5) Reverse the direction of the communication. Calculate the power allocation −→ r ( m +1) k,l for eachdata stream in the original channel following (9), and set −→ v ( m +1) k,l = ←− u ( m ) k,l , −→ u ( m +1) k,l = ←− v ( m ) k,l , ∀ k ∈ [ K ] , ∀ l ∈ [ b k ]Compute the achievable GDoF tuple −→ d ( m ) switch and the achievable sum-GDoF value −→ d ( m )Σ , switch (using receivers with the reverse lexicographic cancellation order). Then let m = m + 1.6) Repeat steps 2) through 5) until the achievable sum GDoF value (i.e., −→ d ( m )Σ ) converges or m reaches a predefined threshold.GDoF region for any interference channel must be upper-bounded, the original channel and itsreciprocal channel have the same achievable GDoF region. (cid:4) Example 4
To help interpret how the ZEST algorithm works, consider the -user interferencechannel in Fig. 4(a). It is known that the sum-GDoF value of this channel is . [3, 18, 19]. Inprevious literatures, the optimal solution is obtained through a centralized design. Here we show howto achieve this sum-GDoF value through the distributed ZEST algorithm. Set n = 2 and b k = 1 , k ∈ { , ..., } . Following the ZEST algorithm, in step 1), we randomly generate the beamformingvector and assign the power to each data stream for every user. As shown in Fig. 4(a), the notation nterferenceDesired0 : V : V : V : V : V : V : V : V : V : V : V : V : V : V ` : V ` : V ` : V ` : V ` : V n = 1 (a) Interference Desired n = 10 : V ? : V ? V ? : V ? : V ? : V ? : V ? : V ? : V ? V ? ` : V ? ` : V ? ` : V ? ` : V ? V ? (b) InterferenceDesired n = 20 V V V V V V V V V V V V V V V (c) Figure 4: Applying ZEST to a 5-user interference channel, where the solid blue and dash black links representdirect and cross links, respectively. In this channel, all the direct and interfering links are with channelstrength level 1, and all the other links are with channel strength level 0 and thus omitted to avoid clutteringthe graph. (a) The transmission scheme in the original channel in step 1), (b) the transmission scheme inthe reciprocal channel in step 3), (c) the transmission scheme in the original channel in step 5). “ − . v ” at the left side of Transmitter 1 denotes that the beamforming vector of User 1 is v and its allocated power level is −→ r (1)1 , = − . . All the other notations follow similarly. In step2), each receiver updates its receiving vector. For instance, at Receiver 1, to obtain the maximalachievable GDoF value, the stronger interfering data stream with beamforming vector v is zero-forced and the weaker one is treated as noise. Therefore, we have −→ u (1)1 , = v ⊥ (where v ⊥ denotesthe × vector orthogonal to v ), and the achievable GDoF value is (0 . − . / . . Afterupdating all users, the achievable GDoF tuple is −→ d (1) = (0 . , . , , , . . Next, in step 3), wereverse the communication direction and update the power allocation for each data stream as shownin Fig. 4(b). For instance, for User 1, after zero-forcing the stronger interference, the remaininginterference level is 0.3. According to Lemma 3, we have ←− r (1)1 , = − . . After updating powers forall users, we obtain ←− d (1) switch = (0 . , . , , . , . . Proceed to step 4) and update the receivingvector in the reciprocal channel. As shown in Fig. 4(b), now each receiver receives one desireddata stream and one interfering stream. It is easy to obtain the zero-forcing receiving vector foreach user and have ←− d (1) = (0 . , . , . , . , . . Following step 5), reverse the communicationdirection again and update the power for each user. At each transmitter, after zero forcing the only ne interfering stream, each user sees no interference. Hence as depicted in Fig. 4(c), −→ r (2) i, = 0 , i ∈ { , ..., } and −→ d (1) switch = (0 . , . , . , . , . . One can further verify that since then the ZESTalgorithm converges, and the final solution given by this distributed algorithm is exactly the sameas that obtained via the centralized design. To further validate the GDoF performance of the proposed ZEST algorithm, we consider a random5-user interference channel. We assume that the channel strength levels of all direct links are alwaysequal to 1. Motivated by cellular networks where users suffer strong interference from neighboringcells, we assume that at Receiver i , the interference from Transmitter i − i + 1 are stronginterference, and the others are weak. For the strong interference, we assume that their channelstrength levels fall into a uniform distribution of [ x, , − x ], where x ≥ .
5. Following [13, 17, 33],we keep the channel strength levels α ij fixed and scale the parameter P in each random channelrealization, and we always assume that every transmitter is subject to a unit peak power constraintand the noise variance at each receiver is normalized to one. Since all the direct channels are withchannel strength level 1, P in fact denotes the SNR of the desired link for each user.We compare the achievable sum-GDoF of the proposed ZEST algorithm, the well-known dis-tributed interference alignment algorithm Max-SINR [34], the state-of-the-art power control al-gorithm SAPC (i.e., SINR approximation power control) [35], TDMA (i.e., the orthogonal schemewith equal time sharing among all users) and the full power transmission (i.e., every user alwaysutilizes full power to transmit its own signal). It is notable that Max-SINR and SAPC optimizes thesignal space allocation and signal level allocation, respectively. Among all the schemes consideredhere, only ZEST jointly optimizes the signal space and signal level allocation for data transmission.For both ZEST and Max-SINR, we set the number of channel uses n as 2 and the number ofscalar data streams for each user d k as 1, ∀ k ∈ { , ..., } . We also note that for both ZEST andMax-SINR, different initializations may yield different sum-rates, particularly for ZEST in low and Here we consider a cyclic setting, i.e., when i = 1, i − i = 5, i + 1 = 1. The Max-SINR algorithm is originally proposed for MIMO interference channels. Here we adopt the algorithmfor SISO interference channels with multiple channel uses. In our experiment, for both ZEST and Max-SINR, in each channel realiza-tion we start from multiple random initializations and pick the largest yielded sum-rate as the finalsolution. When the SNR value P is less than 30 dB, we set the number of random initializationsas 30, and 10 otherwise. How to smartly choose the initialization of ZEST to improve sum-rate inlow and medium SNR regimes is an interesting open question.In our experiments, we consider two specific x values, i.e., x = 0 . x = 0 .
75, where the lattermodels the settings with more diverse channel strengths between strong and weak interfering links.For the two different x values, the averaged sum-rate of all algorithms over 200 random channelrealizations are given in Fig. 5(a) and 5(b), respectively. It can be seen that in both cases ZESTachieves the largest sum-GDoF value (i.e., the steepest slope in the high SNR regime) among allthe schemes. More interestingly, ZEST outperforms SAPC, TDMA and the full power transmissionalmost over the entire SNR range. Compared with Max-SINR, ZEST is particularly favorable inthe settings with more disparate interference strengths (e.g., when x = 0 . x = 0 . As noted in Section 5.1, the GDoF-based algorithm ZEST in general converges much faster thanMax-SINR and SAPC, which are optimization algorithms both based on the classical SINR metric. We point out that the convergence of Max-SINR is still open. In our experiment, we note that for Max-SINR,with a sufficient number of iterations, different initializations usually converge to the same sum-rate. In practice,when the number of iterations is limited, different initializations may lead to different final solutions though. In[36] a convergent Max-SINR algorithm is developed, which in fact jointly optimizes the signal vector space andsignal power level allocations. However, the proposed algorithm in [36] is based on the duality of SINR in multiuserMIMO networks under an artificial sum power constraint [37]. While in ZEST, the convergence is guaranteed underthe practical individual user power constraint . But due to the non-convexity of the problem, the convergent pointdepends on the initialization. For SAPC, following [35] we always set the initial power of each user as its maximaltransmit power.
10 20 30 40 50 60
SNR(dB) s u m r a t e ( bp s ) ZESTMax-SINRTDMASAPCFull Power (a)
SNR(dB) s u m r a t e ( bp s ) ZESTMax-SINRTDMASAPCFull Power (b)
Figure 5: Sum-rate performance of ZEST, Max-SINR, TDMA, SAPC, and the full power transmission, when(a) x = 0 .
5, and (b) x = 0 .
75, where the latter models the settings with more diverse channel strengthsbetween strong and weak interfering links. iteration s u m r a t e ( bp s ) ZESTMax-SINRSAPC
Figure 6: A representative example for the convergence behavior of ZEST, max-SINR, and SAPC.
We have similar observations for the GDoF-based power control algorithm iGPC (iterative GDoF-duality-based power control) [33], which usually exhibits a much faster convergence rate thanSAPC. An interesting question one may ask is if the outputs of the GDoF-based algorithms, whichconverge faster, could be used as initializations to speed up the convergence of their conventionalcounterparts.Here, we consider using the output beamforming vectors of ZEST as the initialization of Max-SINR, and the output power allocations of iGPC as the initialization of SAPC. We note thatin our experiment, compared with conventional initialization methods (i.e., random initializationin Max-SINR, and maximum power initialization in SAPC [35]), using the GDoF-based solution21s a starting point usually means starting from a higher sum-rate, and thus helps speed up theoptimizations in Max-SINR and SAPC in many cases. However, the answer to the question askedabove is not always positive. Two counter examples are given in Fig. 7. The main observation hereis that neither Max-SINR nor SAPC is guaranteed to converge monotonically. Therefore, startingfrom a higher sum rate does not always lead to faster convergence for these two algorithms. iteration s u m r a t e ( bp s ) Max-SINR (Random init.)Max-SINR (ZEST init.) (a) iteration s u m r a t e ( bp s ) SAPC (Max power init.)SAPC (iGPC init.) (b)
Figure 7: Examples where using the GDoF-based solutions as initializations does not speed up the con-vergence of conventional algorithms: (a) Max-SINR and (b) SAPC, as these two algorithms do not alwaysconverge monotonically.
In this paper, we formulate a joint signal vector space and signal power level allocation problem (i.e.,the TIM-TIN problem) under the assumption that only a coarse knowledge of channel strengths andno knowledge of channel phases is available to the transmitters. A decomposition of the probleminto TIN and TIM components is proposed as a baseline. A distributed numerical algorithm calledZEST is developed as well. The convergence of the ZEST algorithm leads to the duality of theTIM-TIN problem. The joint TIM-TIN approach is promising to be a fundamental building blockin existing and future wireless networks, due to its robustness to channel knowledge at transmitters,implementation simplicity (e.g., no need to decode any interference, and being implemented in adistributed fashion) and potential superior performance. This line of research is still in its infancythough. It is hoped that this work could inspire more future research in this area. Future direction22nclude, e.g., translating theoretical insights obtained in this work into the design of practical large-scale wireless networks, such as device-to-device networks and heterogeneous cellular networks.
Appendix A: Proof of Lemma 1
Let x i ∼ CN (0 , P κ i ) be independent Gaussian variables. Denote by z an n × m > γ , denote the n × v i (cid:42) V Π as v Π (cid:48) ( j ) , j ∈ [ m − γ ]. We havelog (cid:104) det (cid:16) I + m (cid:88) i =1 P κ i v i v † i (cid:17)(cid:105) = h (cid:16) m (cid:88) i =1 v i x i + z (cid:17) + o (log( P )) (11)= h (cid:16) γ (cid:88) i =1 v Π( i ) x Π( i ) + m − γ (cid:88) j =1 v Π (cid:48) ( j ) x Π (cid:48) ( j ) + z (cid:17) + o (log( P )) (12)= h (cid:16) γ (cid:88) i =1 v Π( i ) x Π( i ) + z (cid:17) + o (log( P )) (13)= log (cid:104) det (cid:16) I + γ (cid:88) i =1 P κ Π( i ) v Π( i ) v † Π( i ) (cid:17)(cid:105) + o (log( P )) , (14)where (13) is due to the facts that v Π (cid:48) ( j ) , ∀ j ∈ [ m − γ ] is a linear combination of the vectorsin V Π = { v Π(1) , v Π(2) , ..., v Π( γ ) } , and the term (cid:80) m − γj =1 v Π (cid:48) ( j ) x Π (cid:48) ( j ) becomes insignificant when P approaches infinity. More specifically, as P → ∞ , for the term v i ( x i + x j + ... + x k ) ( i < j < ... < k ),only the symbol x i with the dominant power exponent κ i matters, implying that for the vector v i we can ignore all the other independent symbols with equal or smaller power exponents in the limitof P → ∞ .The following is essentially the same as the proof of Lemma 1 in [38]. Define V Π (cid:44) [ v Π(1) v Π(2) ... v Π( γ ) ]with size n × γ , and the diagonal matrix P Π (cid:44) diag[ P κ Π(1) P κ Π(2) ... P κ Π( γ ) ] with size γ × γ . We23ave log (cid:104) det( I + γ (cid:88) i =1 P κ Π( i ) v Π( i ) v † Π( i ) ) (cid:105) = log (cid:104) det( I + V Π P Π V † Π ) (cid:105) (15)= log (cid:104) det( I + V † Π V Π P Π ) (cid:105) (16)= log (cid:104) det( P Π ) (cid:105) + log (cid:104) det( P − + V † Π V Π ) (cid:105) (17)= γ (cid:88) i =1 κ Π( i ) log P + O (1) (18) Appendix B: Proof of Lemma 2
Recall that in Section 3, from vectors V = { v , ..., v m } and their associated power exponents R = { κ , ..., κ m } , we obtain γ ≤ n linearly independent beamforming vectors V Π = { v Π(1) , v Π(2) , ..., v Π( γ ) } and their associated power exponents P Π = { κ Π(1) , κ
Π(2) , ..., κ Π( γ ) } . Define these operations as N v and N κ , respectively, i.e., N v ( V , R ) = V Π and N κ ( V , R ) = P Π . Denote by κ Σ , N κ ( V , R ) the sum ofall entries in N κ ( V , R ).To prove lemma 2, without loss of generality, we only need to consider User 1 and assume thatthe successive cancellation is taken in the lexicographic order. According to the chain rule, we have R = 1 n I ( s , , s , , ..., s ,b ; y ) = b (cid:88) i =1 n I ( s ,i ; y | s , , ..., s ,i − ) (cid:124) (cid:123)(cid:122) (cid:125) (cid:44) R ,i (19)Let d ,i = lim P →∞ R ,i log P , ∀ i ∈ [ b ]. We have d = b (cid:88) i =1 d ,i (20)For Receiver 1, denote the sets of the beamforming vectors and associated power exponents forall the received data streams as V and R , respectively. Consider each term in the right hand sideof (20). Start with d , . We have the following two cases.24 r , + α ∈ N κ ( V , R ): In this case, we have v , ∈ N v ( V , R ). From Lemma 1, we have R , = 1 n (cid:104) h ( y ) − h ( y | s , ) (cid:105) = 1 n (cid:104) κ Σ , N κ ( V , R ) − κ Σ , N κ ( V \ v , , R \{ r , + α } ) (cid:105) log P + o (log( P ))Therefore, in the GDoF sense, we have d , = κ Σ , N κ ( V , R − κ Σ , N κ ( V \ v , , R \{ r , α } ) n , which isachievable by zero-forcing all the data streams falling into span( N v ( V , R ) \ v , ) and treatingthe remaining interference as noise. • r , + α / ∈ N κ ( V , R ): In this case, v , / ∈ N v ( V , R ). We have R , = o (log( P )) and d , = 0, which is trivially achievable (by ZF and TIN).After decoding s , , we subtract it out from the received signal and then consider the second term inthe right hand side of (20), i.e., d , . Similarly, we can argue that by zero-forcing certain interferingdata streams for s , and treating others as noise, d , is achievable. Repeating this subtract-and-decode argument until all the desired data streams for User 1 are decoded, we establish that d isachievable via the ZF-SC receiver and complete the proof. Appendix C: Proof of Theorem 2
In the achievability, the original network is decomposed into a TIN component containing all theinterfering links with channel strength levels no stronger than t l and a TIM component containingall the other interfering links. First, consider the TIN component. When t l ≤ .
5, the TINcomponent satisfies the TIN-optimality condition identified in [9] (recall that the channel strengthlevel of the direct link is normalized to 1). Following Theorem 1 in [9], its symmetric GDoF value d TINsym ≥ − t l . Next, for the TIM component, assume that given the optimal signal space solution,the optimal symmetric GDoF value is denoted by d TIMsym . Finally, according to Theorem 1 in thispaper, the symmetric GDoF value d TIMsym × d TINsym is achievable.For the converse, d TIMsym and d TINsym are both outer bounds for the original network, since removinginterfering links from the channel does not decrease GDoF. Therefore, min { d TIMsym , d
TINsym } can serveas an outer bound for the symmetric GDoF value of the original network. We have d TINsym × d TIMsym ≤ sym ≤ min { d TINsym , d
TIMsym } , and the symmetric GDoF value d sym can be characterized to a factor β = min { d TIMsym , d
TINsym } d TINsym × d TIMsym ≤ min { d TIMsym , } (1 − t l ) × d TIMsym = 11 − t l , (21)which is no larger than 2. Appendix D: Proof of Theorem 3
First, consider the achievability. When M is no larger than S , the achievable scheme is to treatall the medium interfering links as strong interfering links and apply the one-to-one alignment(see Theorem 6 of [19]). Note that this scheme falls into the category of TIM-TIN decomposition,where the TIN component contains no interfering links and the TIM component contains all themedium and strong interfering links. Otherwise, when M is larger than S , we use the followingdecomposition to achieve the optimal symmetric GDoF value: let the TIN and TIM componentcontain all the medium interfering links and all the strong interfering links, respectively. For theTIN component, the achievable symmetric GDoF value is , and for the TIM component, theachievable symmetric GDoF value is S +1 [19]. Therefore, in the original network, the symmetricGDoF value S +1) is achievable.Next, consider the converse. We start with a useful lemma. Lemma 4
Consider a 3-user interference channel within the QM-TIM(0,0.5) framework, where i, j, k ∈ { , , } , i (cid:54) = j , j (cid:54) = k , and i (cid:54) = k . Denote by l ij the link between Transmitter j and Receiver i , M the set of all medium interfering links, and S the set of all strong interfering links. If l ij ∈ S ,and l ki , l ik , l kj , l jk ∈ {S ∪ M} , then the sum GDoF value of this channel is .Proof: The achievability is straightforward. In the following we only consider the converse. Withoutloss of generality, we assume i = 1, j = 2, and k = 3. To obtain the desired outer bound, we firstset α = 0. This does not hurt the sum capacity because regardless of the channel strength level ofthe cross link l , we can always provide the message W to Receiver 2 through a genie and removethis interfering link.Without perfect channel knowledge at transmitters, the channel can be regarded as a compound26hannel, and its capacity is upper bounded by any possible channel realization. Recall that accord-ing to the definition of QM-TIM(0,0.5) given in Section 2, for both strong and medium interferinglinks, their channel strength levels can be set as the threshold value 0.5. Consider a specific channelrealization where α = α = α = α = 1, α = α = α = α = 0 .
5, and all the links havethe same channel phase. The capacity of the original channel is upper bounded by this case.For any reliable decoding scheme, Receiver 1 can always decode its own message W . Afterdecoding W , Receiver 1 can subtract it from its received signal and has the same signal as Receiver2. So Receiver 1 can also decode W . Now consider Transmitters 1 and 2. We find that they havethe same channel vectors to Receiver 1 and 3. It implies that the sum capacity of the originalchannel is upper bounded by that of a 2-user interference channel with transmitters { T , , T } andreceivers { R , R } , where T , is a combination of Transmitter 1 and 2. The sum-GDoF value ofthis 2-user interference channel (where both desired links have channel strength level 1 and bothcross links have channel strength level 0.5) is known to be 1 [30]. Therefore, we establish the desiredouter bound. (cid:4) Now consider the following two cases.
Case I ( M ≤ S ) : For the converse, consider any consecutive S + M + 1 users. Without loss ofgenerality, assume that the user indices range from 1 to S + M + 1. For these users, we intend toprove the outer bound d + d + ... + d S + M +1 ≤ … … … … … … T T M T M +1 T S +1 T S + M +1 R S + M +1 R S +1 R M +1 R M R G G G T S +2 R S +2 Figure 8: When M ≤ S , for the converse we consider this ( S + M + 1)-user interference channel, where thechannel strength levels of the red solid lines and blue dashed lines are 1 and 0 .
5, respectively.
Towards this end, first remove all the users other than the considered S + M + 1 users, which27oes not hurt the sum capacity of users 1 to S + M + 1. Next, in the remaining network, dividethe S + M + 1 users into three subgroups as shown in Fig. 8: • G : this subgroup includes users 1 to M ; • G : this subgroup includes users M + 1 to S + 1; • G : this subgroup includes users S + 2 to S + M + 1.To derive the desired outer bound, consider the channel realization below. Assume that all thelinks have the same channel phase. For the direct links, recall that their channel strength levelsare all equal to 1. For the medium interfering links, we set their channel strength levels to beexactly 0 .
5. For the strong interfering links, we set their channel strength levels to be either 1 or0 . G , we assume that the cross links between them andthe receivers in G and G are all with channel strength level 1, while the cross links between themand the receivers in G are all with channel strength level 0 .
5. Next, for the transmitters in G , thecross links between them and all the receivers are with channel strength level 1. Finally, for thetransmitters in G , the cross links between them and the receivers in G and G are all with channelstrength level 1, while the cross links between them and the receivers in G are all with channelstrength level 0 . G i i ∈ { , , } , areequipped with the same received signal. Thus removing all of them but one cannot hurt the sumcapacity. Also note that for all the transmitters in the same subgroup G i , i ∈ { , , } , they have thesame channel vectors to all the remaining three receivers. Thus combining all the transmitters ineach subgroup into one transmitter does not hurt the sum capacity either. Therefore, the networkis reduced to a 3-user interference channel where α = α = 0 . Case II (
M > S ) : For the converse, consider any consecutive 2( S + 1) users. Without loss ofgenerality, assume the user indices range from 1 to 2( S + 1). For these users, we intend to show d + d + ... + d S +1) ≤
1. Similar to the previous case, we first remove all the other users. In theremaining network, divide the 2( S + 1) users into two subgroups as shown in Fig. 9:28 G : this subgroup includes users 1 to S + 1; • G : this subgroup includes users S + 2 to 2( S + 1). … … … … T R G G T T S +1 T S +2 T S +1 T S +2 R S +2 R S +1 R S +1 R S +2 R Figure 9: When
M > S , for the converse we consider this 2( S + 1)-user interference channel, where thechannel strength levels of the red solid lines and blue dashed lines are 1 and 0 .
5, respectively.
Again, assume that all the links have the same channel phase. For the direct links, their channelstrength levels are all 1. For the medium interfering links, we set their channel strength levels to beexactly 0 .
5. Next, we set the channel strength levels of the strong interfering links to be either 1 or0 . G i , the cross links between them and the receiversin the same subgroup G i are all with channel strength level 1, and the cross links between them andall the receivers in the other subgroup G j are with channel strength level 0 .
5, where i, j ∈ { , } and i (cid:54) = j . Removing all the receivers but one in each subgroup G i , i ∈ { , } , cannot hurt thesum capacity. Combining all the transmitters in each subgroup G i , i ∈ { , } , into one transmittercannot hurt the sum capacity either. Finally, we end up with a 2-user interference channel with α = α = 1, α = α = 0 . Appendix E: Proof of Theorem 4
As the sum-GDoF of an interference channel must be upper bounded by a finite value, to prove thistheorem, we only need to show that the achievable sum-GDoF via the ZEST algorithm monotoni-cally increases after each update, i.e., −→ d ( m )Σ ≤ ←− d ( m )Σ , switch ≤ ←− d ( m )Σ ≤ −→ d ( m )Σ , switch ≤ −→ d ( m +1)Σ . Towards29his end, we show that the GDoF tuple obtained in each step satisfies −→ d ( m ) ( a ) ≤ ←− d ( m ) switch ( b ) ≤ ←− d ( m ) ( c ) ≤ −→ d ( m ) switch ( d ) ≤ −→ d ( m +1) , (22)where (b) and (d) follow from Lemma 2 directly, as in these two steps, the receiver is updated tothe ZF-SC receiver that achieves the maximal GDoF.Next, consider (a). Let B = (cid:80) Kk =1 b k . In the m -th iteration, define an indicator functionI {|−→ u ( m ) † k,l −→ v ( m ) j,s |(cid:54) =0 } = , |−→ u ( m ) † k,l −→ v ( m ) j,s | (cid:54) = 00 , |−→ u ( m ) † k,l −→ v ( m ) j,s | = 0 (23)Next, define G j,sk,l = α kj I {|−→ u ( m ) † k,l −→ v ( m ) j,s |(cid:54) =0 } , which is the effective channel strength level from datastream s of User j to data stream l of User k in the original channel. Also define a B × B matrix G (cid:18) (cid:80) k − n =1 b n + l, (cid:80) j − m =1 b m + s (cid:19) = G j,sk,l .Recall that a successive cancellation procedure is adopted at each receiver. According to theZEST algorithm given in Section 5, without loss of generality, we have assumed that the cancellationis taken in the lexicographic order. Therefore, for Receiver k ∈ [ K ], the effective channel strengthlevel from data stream p of User k to data stream q of User k is 0, for p, q ∈ [ b k ] and p < q . Setthe corresponding entries of G as 0, i.e., G (cid:18) k − (cid:88) n =1 b n + q, k − (cid:88) n =1 b n + p (cid:19) = 0 , ∀ k ∈ [ K ] , ∀ p, q ∈ [ b k ] , p < q, (24)and denote the obtained matrix by −→ G . Next, for the K -user original channel in the m -th iterationwith beamforming vectors −→ v ( m ) j,s and ZF-SC receiving vectors −→ u ( m ) k,l , we construct a counterpart B -user interference channel with the channel strength level matrix −→ G , which is denoted by IC o .For IC o , −→ G ( j, i ) denotes the channel strength level from Transmitter i to Receiver j . Assume thatin IC o , the allocated power to Transmitter i is −→ r i = −→ r ( m ) j,s where i = (cid:80) j − l =1 b l + s . By treatinginterference as noise at each receiver, we obtain the achievable GDoF tuple of IC o ( d ,o , ..., d B,o )and (cid:80) i (cid:48) j i = i j d i,o = n × −→ d ( m ) j , where i j = (cid:80) j − l =1 b l + 1, i (cid:48) j = (cid:80) jl =1 b l , and −→ d ( m ) j is the j -th entry of −→ d ( m ) . 30imilarly, for the reciprocal channel in the m -th iteration with beamforming vectors ←− v ( m ) j,s = −→ u ( m ) j,s and receiving vectors ←− u ( m ) k,l = −→ v ( m ) k,l , we construct a counterpart B -user interference channelwith the channel strength level matrix ←− G = −→ G T , which is the reciprocal channel of IC o anddenoted by IC r . Assume that in IC r , the allocated power to Transmitter i is ←− r i = ←− r ( m ) j,s where i = (cid:80) j − l =1 b l + s . By treating interference as noise at each receiver, we obtain the achievable GDoFtuple of IC r ( d ,r , ..., d B,r ) and (cid:80) i (cid:48) j i = i j d i,r = n × ←− d ( m ) j, switch , where ←− d ( m ) j, switch is the j -th entry of ←− d ( m ) switch . According to Lemma 3, we have i (cid:48) j (cid:88) i = i j d i,o ≤ i (cid:48) j (cid:88) i = i j d i,r ⇒ −→ d ( m ) j ≤ ←− d ( m ) j, switch , ∀ j ∈ [ K ] , (25)and hence prove (a). The proof of (c) follows similarly. Therefore, we establish (22) and completethe proof. References [1] C. Geng, H. Sun, and S. Jafar, “Multilevel topological interference management,” IEEE Infor-mation Theory Workshop (ITW), 2013.[2] S. Jafar, “Interference Alignment: A new look at signal dimensions in a communication net-work”, Foundations and Trends in Communications and Information Theory, vol. 7, no. 1, pp.1-136.[3] S. Jafar, “Topological interference management through index coding,”
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