Interference Alignment and the Degrees of Freedom for the K User Interference Channel

Interfer ence Alignment and Spatial De grees of Freedom for the K User Interfer ence Channel V i veck R. Cadamb e, Syed A. Jafar Electrical Engineering and Com puter Science University of California Irv ine, Irvine, Califor nia, 92 697, USA Email: vcad ambe@uci.ed u, s yed@uci.ed u Abstract While the best kno wn ou terbound for the K user interference channel states that there cannot be more than K/ 2 degree s of freedom, it has been conjectured that in general the constant interference channel with any number of users has only one degree of f reedom. In this paper , we explore the spatial degrees of fr eedom per o rthogonal time and frequenc y dimension for the K user wireless interference channel where the channel coefﬁcien ts take distinct v alues across frequenc y slots but are ﬁxed in time. W e answer ﬁv e closely related questions. First, we sho w t hat K/ 2 de grees of freedom can be achie ved by channel design, i.e. if t he nod es are allo wed to choose the be st constant, ﬁnite and no nzero channel coef ﬁcient values. S econd, we show that if channel coef ﬁcients can not be controlled by the nodes but are se lected by n ature, i.e., ran domly drawn from a co ntinuous distribution, the t otal number of sp atial degree s of freedom for t he K user interference channel is almost surely K/ 2 per orthogonal t ime and frequency dimension. Thus, only half the spatial degrees of fr eedom are lost due to distri buted processing of transmitted and recei ved signals on the interference channel. Third, we sho w that interference alignment and zero forcing sufﬁce to achie ve all the deg rees of freedom in all cases. Fourth, w e sho w that the degrees of freedom D directly lead to an O ( 1) capacity characterization of the form C ( S N R ) = D log(1 + S N R ) + O (1) for the multiple access channel, the broadcast channel, the 2 user interference channel, the 2 user MIMO X channel and t he 3 user interference channel with M > 1 antennas at each node. It is not known if this relationship is true for all networks in general, and the K user interference channel with a single antenna at all nodes i n particular . Fifth, we consider the degree of freedom ben eﬁts from cogniti ve sharing of messages on the 3 user interference channel. If o nly one of t he three messages is made available non-causally to all the nodes except its intended receiv er the degree s of freedom are not in creased. Ho wever , if two mess ages are shared amon g all n odes (except their intended recei vers) then t here are two de grees of freedom. W e ﬁnd that unlike t he 2 user interference channel, on the 3 user interference channe l a cogniti ve t ransmitter is not equi v alent to a cognitiv e receiv er from a degrees of freedom perspecti ve. If on e recei ver has cogniti ve kno wledge of all the othe r users’ messages the degrees of freedom are the same as without cogniti ve message sharing. Ho wev er , if one transmitter has cognitiv e kno wl edge of all t he other users’ messages then the degree s of freedom are increased from 3 / 2 to 2 . I . I N T RO D U C T I O N The capacity of ad -hoc w ireless networks is the muc h sou ght afer “holy-g rail” of ne twork infor mation th eory [1]. While capacity characteriz ations have been foun d fo r cen tralized n etworks (Gaussian multiple ac cess and broadc ast networks with multip le antennas), similar c apacity ch aracterizations for m ost distributed commun ication scenarios (e.g . interfer ence networks) remain lon g standing open pro blems. In the absence of precise capac ity characterizatio ns, r esearchers have pur sued asymptotic and/or approximate capacity characterizations. R ecent work has f ound the asymptotic scaling laws of n etwork capacity as the n umber of nodes in creases in a large network [2 ], [3]. Howe ver , very little is known about the cap acity r egion o f smaller (ﬁnite) d ecentralized networks. An importan t step in this direction is the recent app roximate characterizatio n of the c apacity region of the 2 user inter ference channel that is accurate within one bit of the true capac ity region [4]. Ap proxim ate ch aracterizations of capacity regions would also be invaluable for most op en p roblems in network informatio n theor y and m ay b e the key to improving our un derstandin g of wireless n etworks. It can be argued th at the most prelimin ary f orm o f capacity chara cterization fo r a n etwork is to characterize its d egrees o f f reedom . The degrees of freedom re present the rate of growth of network capacity with the lo g of the signal to no ise ra tio ( SNR). In most cases, the spatial degrees of freedo m turn out to be the nu mber of non-in tefering paths that can be created in a wireless n etwork throug h signal p rocessing at the transmitter s and receivers. While time, frequency and space all of fer degrees of freed om in the form of orthogon al dimensio ns over which communication can take place, spatial degrees o f f reedom are especially interesting in a d istributed network. Potentially a wireless network may have as many spatial dime nsions as the numb er of transm itting a nd receiving antennas. Howe ver, the ability to access and resolve spatial d imensions is limited by the distributed nature of the network. There fore, characterizin g th e d egrees o f free dom for d istributed wireless network s is by itself a non -trivial problem . For examp le, consid er an interf erence n etwork with n single-anten na transmitters and n single-an tenna receivers where each tran smitter has a message for its co rrespond ing receiver . For n = 2 it is kn own that this interferen ce network has only 1 d egree of freedom [ 5], [6]. Th ere a re n o known resu lts to show that more than 1 degrees of free dom are ach iev ab le on the interfere nce channel with any number of users. It is con jectured in [ 7] that the K user interference ch annel h as only 1 degree o f f reedom . Y et, the b est kn own outer bound for the n umber of degrees of freed om with K interfering nodes is K / 2 , also presented in [7] . The unr esolved gap between th e inner an d ou terboun ds highlights our lack of understandin g of the capacity of wireless n etworks becau se ev en the number o f d egrees of f reedom, wh ich is the most basic characterizatio n of the network capacity , remains an open problem . It is this op en problem th at we p ursue in this pap er . T o gain a b etter u nderstand ing of the interference ch annel, we ﬁrst consider th e p ossibility th at the transmitters and receivers can place themselves optimally , i.e., the nodes can choose their chann els. Thus, the ﬁrst objective o f this pa per is to answer the question : Question 1: What is the maximum nu mber of degr ees of fr eedom for the K user interference channel if we are allowed to choose the best (ﬁnite a nd n o n-zer o) chan nel coefﬁcient values ? While the scenario above of f ers new insights, in pr actice it is more c ommon that the channel coef ﬁcients ar e ch osen by natu re. The nod es control their co ding schemes, i.e. the transmitted symb ols, but not the ch annel coefﬁcients, which may be assumed to be r andomly drawn f rom a continu ous distrib u tion and causally known to all the nodes. In this context we ask the main question o f this paper: Question 2: What is the number of de grees of fr eedom for the K user interfer ence chan nel per o rthogonal time and frequency dimension ? Note th at the normaliza tion by the num ber of o rthogo nal time and freque ncy dim ensions is necessary be cause we wish to char acterize the spatial degrees of freedom. Spatial degrees of freedom ha ve been characterized f or s ev eral multiuser com munication scen arios with mu ltiple antenna n odes. The ( M , N ) poin t to point MIMO channel has min( M , N ) degrees o f fr eedom [ 8], [9], the ( M 1 , M 2 , N ) multiple access chan nel has min( M 1 + M 2 , N ) degrees of freedom [10], the ( M , N 1 , N 2 ) broad cast channel has min( M , N 1 + N 2 ) degrees of freedom [1 1]–[13 ], and the ( M 1 , M 2 , N 1 , N 2 ) interference chan nel has min( M 1 + M 2 , N 1 + N 2 , max( M 1 , N 2 ) , max( M 2 , N 1 )) d egrees of f reedom [6 ], wher e M i (or M when only one transmitter is presen t) and N i (or N when only one rece i ver is present) indicate the numb er of antennas at the i th transmitter an d receiver , respectively . If one tries to extrapolate these re sults in to an under standing of the degrees of freedo m for fully con nected (all channel c oefﬁcients are non-zero ) wireless n etworks with a ﬁnite numbe r of nodes, one could arr iv e at the following (incorr ect) intuiti ve inferences: • The number of degr ees of fr eedom for a wir e less network with perfect chann e l knowledge at all nodes is an inte ger . • The d e gr ees of fr eedom of a wir eless network with a ﬁ nite n umber of no des is not higher th an the max imum number o f co-loca ted antenn as at any node. The degrees o f freedo m characterizatio ns for the p oint to point, multiple access, bro adcast and interf erence scenarios described above are all consistent with both these statements. Note that the results of [14] ind icate that e ven with single an tenna sour ces, destinations and relay nodes the network can have m ore than one degree of freedom. Howe ver, f or this distributed or thogon alization r esult it is assume d that the numb er of relay nodes a pproach es inﬁnity . Thus it do es no t contra dict the in tuition above which is for ﬁnite networks. Multih op ne tworks with half - duplex relay nod es may also lead to fractional degrees of f reedom due to the n ormalization associated with the half-du plex constra int. This is typ ically becau se of the absence o f a direct link across h ops, i.e. some chan nel coefﬁcients ar e zero . For multih op networks with orthogon al hop s [ 15] ha s shown that the full N degrees of freedom a re achievable even if each interm ediate hop consists of N (distributed) single an tenna relay nodes as lon g as the initial source no de and th e ﬁn al destination n odes are equipped with N antenn as ea ch. Note that the r esult of [15] is also con sistent with the inferences described above. Also, we note that channels with specialized structures or coo peration among nod es may b e able to achieve high er degrees of free dom than chan nels whose coefﬁcients are ran domly selected from con tinuous distributions [16]. Perhaps biased by these results, most work on degrees o f freedom for wireless networks has focused on either networks wher e som e no des a re eq uipped with mu ltiple antenn as [6 ], [15 ] or networks with single antenna nodes where so me form of coo peration open s up the possibility that the single a ntenna no des may be able to achieve MIMO behavior [5], [7], [17]. Networks of single an tenna no des w ith n o co operation b etween the tran smitters o r receivers could be consider ed uninteresting from the degrees of f reedom perspective as th e above mentioned intuitive statements would sug gest that these networks c ould only have 1 d egree of free dom. In oth er words, on e mig ht argu e that w ith a single antenna at each nod e it is imp ossible to av oid interferen ce and therefore it is impo ssible to create multiple non -interferin g p aths necessary fo r degrees of freed om. The 2 user interf erence network with a sing le antenna at each node is a goo d example of a network wh ich adher es to all the above intuitiv e infere nces, where indeed it can be rigorously shown that ther e is only one degree of freedom. Studying a K user in terference ch annel where all c hannel coefﬁcients are equal will also lead to only one degree o f freedom , as will th e K user interfer ence channel with i.i.d. chan nel coefﬁcients an d no k nowledge of channel coef ﬁcients at the transmitters [ 14]. Similarly , if all re ceiv ers observe signals that ar e d egraded versions of, say , receiver 1’ s signal then it c an ag ain be argued that th e MAC su m c apacity wh en receiver 1 decodes all messages is a n o uterbou nd to the inter ference channe l sum capacity ( Carleial’ s ou terboun d [18]). Thus the degrees of freedom can not be more than th e nu mber of antenn as at r eceiv er 1 . Finally , the co njecture that the K user in terference ch annel has only 1 degree of freedom is also consistent with this in tuition [7]. Clear evidence that the intuiti ve conclusions mentioned above do not apply to all wireless n etworks is provid ed by the recent degree of freedo m region char acterization for the 2 user X ch annel in [19 ]–[22 ]. Th e 2- user X channel is identical to the 2-u ser inte rference cha nnel with the exception that each transmitter in th e X channel has an indepen dent m essage for each receiver . Thus, unlike the in terference chann el which has o nly 2 messages, the X chann el h as 4 messages to be comm unicated between two transmitters and two rec ei vers. Sur prisingly , it was shown in [ 22] that the X chan nel, with only a sing le antenna at all no des has 4 / 3 d egrees of freedom per orthog onal time/freque ncy dimension if the c hannels are time/fr equency selectiv e. This is intere sting for se veral reasons. First, it shows that the degrees of freedom can take non-in teger values. Secon d, it shows that the degre es of fre edom of a distributed wireless network can be hig her than th e max imum num ber of co -located antennas a t any node in the network. Finally , th e achiev ab ility p roof for the non-in teger degrees of freedom for the X chann el uses the novel co ncept of intefe rence-align ment [2 0]–[2 3]. Interfere nce align ment r efers to the simple id ea that signal vectors can be aligned in such a m anner that they c ast overlapping shadows at th e r eceiv ers where th ey constitute interf erence wh ile they continue to be d istinct at the receiv ers wh ere they are desired . Th e possibility of imp licit interferen ce align ment was ﬁrst obser ved by Maddah- Ali, Motah ari and Khan dani in [20] . T he ﬁrst explicit interfer ence a lignment s cheme w a s pre sented in [23] where it was shown to b e sufﬁcient to achie ve the full degrees of freedom for the MIMO X c hannel. Interfer ence align ment was subsequently used in [21], [22] to sho w achiev ability o f all po ints within th e degrees of freedo m region of the MIMO X channel. Interferen ce a lignment was also independen tly discovered in the co ntext of the compou nd broadcast chan nel in [ 24]. Since th e distinctio n between the X channel and the interferen ce channel is quite sig niﬁcant, it is no t imm ediately obvious whether the results fo und for the X ch annel hav e any implications for the inter ference channel. For instance, the achiev ability schem es with intef erence alignment prop osed in [21] utilize the br oadcast an d mu ltiple access channels inherent in the X cha nnel. Howe ver , the interference chann el does not have broadcast and multiple acce ss compon ents as each tr ansmitter ha s a message for o nly one unique recei ver . Th erefore, in this paper we answer the following question. Question 3: What ar e the de g r ees of fr eed om b eneﬁts fr om interference alignmen t on the K u ser interfer ence network? The degree s of freedo m can be vie wed as a capacity char acterization that is accurate to within o (log ( ρ )) wher e ρ represen ts the signal to noise ratio (SNR). I n order to pursue increasingly accur ate capacity character izations, in this p aper we explore the n otion of O (1) capacity of a network. Th e O (1) capacity is an appro ximation accurate to within a bou nded con stant of the a ctual ca pacity region . The constant term can depend on ly o n the channel gains and is inde pendent of the transmit p owers of the users. T he O (1) capa city is a m ore accu rate descrip tion of the network capacity than the degrees of freedo m of a network. I nterestingly , fo r the point to point MIMO channel the O (1 ) capac ity C ( ρ ) is directly related to the degrees of freedom D a s C ( ρ ) = D log(1 + ρ ) . This lead s us to the third set of qu estions that we pu rsue in this paper . Question 4: Is the O (1) capa city C ( ρ ) of the m u ltiple access and br oadcast channels, as well as th e 2 user interfer en ce and X channels r ela ted to the d e gr ees of fr ee d om D as C ( ρ ) = D log(1 + ρ ) ? Does the sa me r ela tionship hold for the K u ser interfer en ce chann e l? Finally , we exp lore the ben eﬁts in terms of degrees of fr eedom, from the cog nitiv e sharing of messages on the interferen ce channel. Based on the cog nitive radio mo del introduced in [2 5]–[2 7] co gnitive message sharing r efers to the form of coo peration where a message is made available no n-causally to some transmitters and/o r re ceiv ers besides the in tended source a nd destination of the message. It was shown in [17] that f or the 2 user in terference channel with single an tennas at eac h n ode, cogn iti ve message sharing ( from one transmitter to another ) does not produ ce any gain in th e degrees o f freedo m. The r esult was extended in [22] to the 2 user interferen ce chann el with mu ltiple anten na nodes and equal n umber of an tennas at each node , to show that there is n o gain in degrees of freedom whether a message is s hared with the transmitter, receiver or both the transmitter and re ceiv er of the other user . [22] also e stablishes an in teresting du ality relation ship wher e it is shown that from the degrees of freedo m perspective cognitive transmitters ar e equ ivalent to cogn itive receiver s , i.e. sharing a m essage with another u ser’ s transmitter is equiv alent to shar ing a message with the other user’ s receiv er . It is not clear if sim ilar resu lts will hold f or the 3 user interference chann el, and it form s the last set of questions that we add ress in this pape r . Question 5 : F or the 3 user interfer ence chann el, what are the beneﬁ ts of cogn itive message sharing? Ar e cognitive transmitters equivalen t to cognitive receiver s in the manner shown for the 2 user interfer ence channel? A. Overview of Results The answer to the ﬁrst que stion is p rovided in Sectio n III an d may be sum marized in g eneral terms as follows: “ Re gar dless of how many speakers and listener s are located within earshot of each other , each speaker can speak half the time a nd be hear d without any interfer ence by its inten ded listener ” This result may seem impossible at ﬁrst. For example, h ow c an a to tal dur ation of 1 hour be sha red b y 10 0 speakers such that e ach speaker speaks for 30 min utes and is hea rd interferen ce free by its in tended listener when all systems are located within ear shot o f each oth er? And yet, this seem ingly im possible result is made possible b y the con cept of interfer ence alig nment. A simp le scheme is explained in Section I II whe re it is assumed the speakers and listener s can choose their locations. The answer to the second question is pr ovided in Th eorems 1 and 2. W e show that the K user interfere nce channel with single antenn as at all nodes has (almost surely) a total of K/ 2 degrees of freedom per or thgonal time and fr equency dimension when the channels are dr awn randomly from a continuo us distrib ution. The imp lications of this result f or our under standing of the capacity of wireless networks are qu ite profou nd. It shows that we ha ve grossly und erestimated the ca pacity of wireless networks. For example, at high SNR the true capacity is high er by 5 0%, 90 0%, and 4900 % than anything p reviously shown to be a chiev a ble for network s with 3 , 20 , and 1 00 interfering users, respectiv ely . Inte rference is o ne of the principal ch allenges faced by wireless networks. Howe ver, we h av e shown that with perfec t chann el k nowledge th e fr equency selecti ve interfer ence chann el is not interference limited. In fact, after the ﬁrst two u sers, ad ditional users do not co mpete for degree s of freedom and each ad ditional user is able to achie ve 1 / 2 degree of freedom without hurtin g the previously existing u sers. What makes this result ev en mor e remark able is that linear scaling of degrees o f f reedom with u sers is achieved without co operatio n in the form of message sharing that may allo w MIMO behavior . Note that it has been sho wn pr e viously fo r th e 2 user interferen ce channel th at un idirectiona l message sharin g (e.g . from transmitter 1 to tran smitter 2 ) does not allow H [13] H [23] H [32] H [33] H [31] H [21] v [1] 2 x [1] 2 H [31] v [1] 1 x [1] 1 H [22] v [2] x [2] H [33] v [3] x [3] H [32] v [2] x [2] H [31] v [1] 2 x [1] 2 H [23] v [3] x [3] H [21] v [1] 1 x [1] 1 H [11] v [1] 2 x [1] 2 H [11] v [1] 1 x [1] 1 H [13] v [3] x [3] H [12] v [2] x [2] v [1] 2 x [1] 2 v [1] 1 x [1] 1 v [3] x [3] v [2] x [2] H [12] H [21] H [11] H [22] Fig. 1. Interfe rence alignment on t he 3 use r interference c hannel to achie ve 4 / 3 degre es of freedom higher degrees of freedo m [17], [22] an d even b i-directiona l message shar ing (through f ull d uplex noisy chann els between the transmitters an d full duplex noisy chann els between the receivers) will n ot increase th e d egrees of freedom if the co st of message sharing is consider ed [7 ], [28]. Theref ore it is qu ite sur prising that the K user interferen ce chann el has K / 2 degrees of fr eedom even withou t any message sharing. T o summar ize, Th eorem 1 shows th at only half the de gr ees of fr eed om ar e lost due to distrib uted pr ocessing at the transmitters and r eceivers on the interfer ence chann el . The answer to question 3 is provided b y the achievability pr oof f or Theorem 1 where we ﬁn d th at, similar to the 2 user X chann el, interference alig nment sufﬁces to ach iev e all the degrees of f reedom on the K u ser interferenc e channel as well. Thus, interferen ce alignment is as relev ant for the K user interferenc e channel wh ere it achieves the full K / 2 degrees o f f reedom, as it is fo r th e 2 user X chan nel where it achieves the f ull 4 / 3 degrees of fre edom. Interestingly , interference alignmen t does f or wireless networks wh at MI MO techno logy has done for the po int to point wir eless ch annel. In both cases the capacity , originally limited to log(1 + S N R ) , is shown to be capable o f linearly increasing with the n umber of antennas. While MIMO tech nology requires n odes equipp ed with multiple antennas, inte rference alig nment works with the distributed antenn as naturally av ailable in a network acro ss the interfering transm itters and receivers. Figure 1 sho ws ho w interfer ence alignment ap plies to th e 3 user in terference channel. In this ﬁgur e we illustrate how 4 degrees of freed om are achieved over a 3 symbol extension of the ch annel with 3 sing le an tenna users, so that a total of 4 / 3 d egrees of fre edom ar e ach iev ed p er channel use. The achievability proof f or 3 / 2 degrees o f freedom is m ore inv olved and is provided in Section IV -A. User 1 achiev es 2 degrees of freed om by transmitting two indep endently coded streams alo ng the b eamfor ming vectors v [1] 1 , v [1] 2 while u sers 2 and 3 ach ie ve one degree of freedom b y send ing their indep endently encoded d ata streams alon g the beamfor ming vectors v [2] , v [3] , r espectively . The bea mformin g vectors are chosen as follows. • At rec ei ver 1 , the interferen ce from transmitters 2 and 3 are p erfectly aligne d. • At receiver 2 , th e interf erence from transmitter 3 aligns itself along on e of the d imensions of the two- dimensiona l interference signal from transmitter 1 . • Similarly , at receiver 3 , th e interf erence from tran smitter 2 alig ns itself along one o f the dimension s of interferen ce from transmitter 1 . For the answer to question 4 , we show that for the multiple access, b roadcast, a nd 2 u ser interferen ce and X channels, the total degrees of freedom D and the O (1) capacity C ( ρ ) are indeed rela ted as C ( ρ ) = D log (1 + ρ ) . Thus, the two descriptions ar e eq uiv alen t. Howe ver , fo r the 3 user interference chan nel with single antenna no des it appears unlikely that such a relationship exists. The e vidence in this p aper raises th e interesting p ossibility th at the sum capacity of the K user interferen ce ch annel with single antenna nodes may hav e a different for m than the multiple access, broadcast, and 2 u ser interference and X c hannels in that the d ifference b etween the true cap acity C ( ρ ) an d the degrees of fr eedom app roximation D log(1 + ρ ) may not be boun ded. Finally , the answer to qu estion 5 is p rovided in Th eorem 5. W e show that sharing one message with all other transmitters an d/or receiv ers does no t increase the degrees of freedom f or the 3 user interfer ence channel. Sharing two message s with all other transmitter s and/or receivers o n the other hand ra ises the degrees of freed om from 3 / 2 to 2 . Interesting ly , we ﬁnd that the eq uiv alan ce established between the cogn iti ve transmitters and cognitive receivers on the 2 user interference channel d oes n ot d irectly apply to the 3 user interference cha nnel. In tuitiv ely , this may b e understoo d as follows. A cognitive transmitter on the 3 user interference cha nnel can be more useful than a cognitive recei ver . This is because a cognitive transmitter with no message of its own can still increase the degrees of fr eedom by canceling interf erence from its cognitively acquired message at oth er rece i vers. In oth er words a cog nitiv e transmitter with no message of its own, still lends a transmit anten na to the transmitter whose message it shares. On the o ther han d, a cognitiv e recei ver with no message o f its own is useless. I I . S Y S T E M M O D E L Consider th e K user interference chann el, c omprised of K transmitters and K receivers. W e ass ume coding may occur over multiple o rthogo nal freq uency and ti me dim ensions and the rates as well as the degrees of freedom are normalized by th e num ber of orthog onal time and freq uency dimensions. Each nod e is equipp ed w ith only one antenna (multiple an tenna nodes are consider ed later in this p aper). The chann el o utput at the k th receiver over the f th frequen cy slot and the t th time slot is de scribed as follows: Y [ k ] ( f , t ) = H [ k 1] ( f ) X [1] ( f , t ) + H [ k 2] ( f ) X [2] ( f , t ) + · · · + H [ kK ] ( f ) X [ K ] ( f , t ) + Z [ k ] ( f , t ) where, k ∈ { 1 , 2 , · · · , K } is the user in dex, f ∈ N is the fre quency slot index, t ∈ N is th e time slot index, Y [ k ] ( f , t ) is the output sign al of the k th receiver , X [ k ] ( f , t ) is the input signal of the k th transmitter, H [ kj ] ( f ) is the c hannel fade coe fﬁcient fro m tr ansmitter j to receiver k over the f th frequen cy slot and Z [ k ] ( f , t ) is the additive white Gaussian noise (A WGN) term at the k th receiver . The chan nel co efﬁcients vary across frequen cy slots but are assume d constant in time. W e assume all noise terms are i.i.d . ( indepen dent identically distributed) zero mean complex Gaussian with unit variance. W e assume all cha nnel coefﬁcients H [ kj ] ( f ) are known a-p riori to all transmitters and receiv ers. Note that since the channel coefﬁcients do not vary in time, only causal channel knowledge is re quired. If w e allow n on-causal channel knowledge then the c hannel m odel above is eq uiv alen tly represented as coding entirely in the time domain, i.e. over o nly one f requency slot. T o av oid degener ate chann el condition s (e.g . all channel coefﬁcients are equ al or ch annel coefﬁcients ar e equa l to either ze ro or inﬁnity) we assume that the channel coefﬁcient values are drawn i.i.d. from a continuou s distribution and the absolute value of all the ch annel coefﬁcients is bounded between a non- zero minimum v alue and a ﬁnite maximum v alue. Since the channel values are assumed con stant in time , the time ind ex t is sometimes supp ressed for co mpact notation . W e assume tha t tran smitters 1 , 2 , · · · , K hav e independ ent m essages W 1 , W 2 , · · · , W K intended f or receivers 1 , 2 , · · · , K , respectively . The total power across all transm itters is assumed to be equal to ρ per ortho gonal time and freq uency dime nsion. W e indicate the size of the message set by | W i ( ρ ) | . For cod ew o rds spanning f 0 × t 0 channel uses (i. e. u sing f 0 frequen cy slots and t 0 time slots), the rates R i ( ρ ) = log | W i ( ρ ) | f 0 t 0 are ach iev ab le if the probab ility o f erro r for all messages can be simultaneo usly m ade ar bitrarily small by ch oosing an approp riately large f 0 t 0 . The capacity region C ( ρ ) of the three user inter ference c hannel is the set of all achievable rate tup les R ( ρ ) = ( R 1 ( ρ ) , R 2 ( ρ ) , · · · , R K ( ρ )) . A. De gr ees of F r eedo m Similar to the degrees of freedom region deﬁnition for the MIMO X chann el in [22] we deﬁne the degrees of freedom r egion D for the K user interf erence chan nel as fo llows: D =  ( d 1 , d 2 , · · · , d K ) ∈ R K + : ∀ ( w 1 , w 2 , · · · , w K ) ∈ R K + w 1 d 1 + w 2 d 2 + · · · + w K d K ≤ lim sup ρ →∞ " sup R ( ρ ) ∈C ( ρ ) [ w 1 R 1 ( ρ ) + w 2 R 2 ( ρ ) + · · · + w K R K ( ρ )] 1 log( ρ ) #  (1) I I I . I N T E R F E R E N C E A L I G N M E N T T H RO U G H C H A N N E L D E S I G N W ith the exceptio n o f this section, througho ut this paper we assume that the channel coef ﬁcients are determined by natu re, i.e. we d o not control the channel values, and we only contr ol the codin g scheme, i.e. the transmitted symbols. However , in this section we take a different perspective to gain a dditional in sights in to the problem . W e wish to know what is the best we can d o if w e are allowed to pick all the ch annel co efﬁcient values subject to the only co nstraint that the coefﬁcient values are ﬁnite, no n-zero constants. It is importan t tha t we can only pick non-ze ro channe l coefﬁcient values because the K / 2 outerbound app lies if and only if all channel coef ﬁcients have non-ze ro values. For example, if we are allowed to set some ch annel coefﬁcients to zero th e p roblem becom es trivial because by setting all interfering links to zero we can easily a chieve K degrees of fr eedom over th e K non-in terfering channels. A. Interference Alignment by Cho ice o f Channel Coefﬁcients As we show next, we can achieve K/ 2 degrees of freedo m for the K user inter ference channel with non-zer o channel coefﬁcients if we ar e allowed to pick the values of the channel coefﬁ cients. The proof is quite simple. W e consider a tw o symbol extension o f the channel, i.e. cod ing ov er two frequen cy slots, where the channel is deﬁned by 2 × 2 diagon al channel matrices th at we choose as follows H [ ij ] = " 1 0 0 − 1 # if i 6 = j (2) H [ ij ] = " 1 0 0 1 # if i = j (3) Each user transmits h is cod ed symbols alo ng the beamform ing v ector v [ i ] = " 1 1 # (4) This ensure s that all the interfer ence terms at each rece i ver appear along the directio n vector [1 − 1] T while the desired signal at each r eceiv er app ears along the direction [1 1] T . Thus the d esired signal an d interference are orthog onal so that each user is able to achieve one de gree of freedom for h is message. Since K d egrees o f freed om are achieved over the 2 symbol extension of th e c hannel th e d egrees of fr eedom eq ual K / 2 . T hus, it is interesting to note that the K / 2 o uterbou nd is tigh t for s ome interfe rence ch annels with n on-zero chan nel coefﬁcients. Sinc e joint processing at all transmitters and at all receivers would resu lt in K degrees of freedom on the K user interferen ce channel, we o bserve that if we are allowed to pick the chann el coefﬁcients then the maximum pena lty for distributed signal pr o cessing is the loss o f half th e d e gr ee s o f fr e e dom . It remains to be shown if this b ound is tight wh en channel coefﬁcients are chosen by nature, i.e. mod eled as random variables drawn f rom a continuou s distribution. As we show in the ne xt section, the outerbound of K/ 2 is almost surely tight for the K user interfe rence channel. B. Interference a lignment th r ough choice of pr o pagation delays - Ca n everyone spe ak h a lf the time with no interfer en ce? W e end this section with ano ther interesting e xample of interference alignm ent. Con sider the K user in terference channel where there is a propag ation delay fr om each tr ansmitter to each re ceiv er . Let T ij represent the signal propag ation d elay from transmitter i to receiver j . Su ppose th e locations of the transmitters and receivers can be conﬁgured such th at the dela y T ii from each transmitter to its intend ed receiver is an e ven multiple of a basic symbol duration T s , while the signal propag ation delays T ij , ( i 6 = j ) f rom each transmitter to all unintended receiv ers are odd multiple s o f the symbo l du ration. The commu nication strategy is the following. All transmissions occu r simultaneou sly at e ven symbo l durations. Note that with this policy , each receiver sees its own transmitter’ s signal interferen ce-free o ver e ven time periods, while it sees all interfering signals simultaneou sly over odd time period s. Thus each user is ab le to achieve 1 / 2 d e gr ees of fr eed om and the total d e gr ees of fr eed om achieved is equal to K/ 2 . I V . D E G R E E S O F F R E E D O M F O R T H E K U S E R I N T E R F E R E N C E C H A N N E L - I N T E R F E R E N C E A L I G N M E N T T H RO U G H P R E C O D I N G Hencefor th, we assume that the chan nel co efﬁcients ar e not con trolled by the nodes but rather selected by nature. Thus, th e ch annels do n ot au tomatically align the interfer ence and any inter ference alignment can only be accomplished th rough code design. The f ollowing theorem presents the main result o f this section. Theorem 1 : The n umber of d egrees of freedom for the K u ser interfer ence channel with sing le antennas at all nodes is K/ 2 . max d ∈D d 1 + d 2 + · · · + d K = K/ 2 (5) The co n verse argu ment for the theorem f ollows directly fro m the outerboun d f or the K user inte rference ch annel presented in [7]. The ach iev ability proof is pr esented next. Since the proof is r ather in volved, we present ﬁrst the constructive proof for K = 3 . The p roof for general K ≥ 3 is then provided in Ap pendix I. A. Achievability Pr oof for Theorem 1 with K = 3 W e sho w th at ( d 1 , d 2 , d 3 ) = ( n +1 2 n +1 , n 2 n +1 , n 2 n +1 ) lies in the de grees of free dom r egion ∀ n ∈ N . Sin ce the d egrees of fre edom region is closed, this automatically implies that max ( d 1 ,d 2 ,d 3 ) ∈D d 1 + d 2 + d 3 ≥ sup n 3 n + 1 2 n + 1 = 3 2 This r esult, in conjunctio n with the co n verse ar gumen t proves the theore m. T o show that ( n +1 2 n +1 , n 2 n +1 , n 2 n +1 ) lies in D , we co nstruct an inter ference alignmen t schem e using only 2 n + 1 frequen cy slots. W e collectively denote the 2 n + 1 symbo ls tr ansmitted over the ﬁrst 2 n + 1 freq uency slots at each time instant as a supersym bol. W e call this the (2 n + 1) symbo l e xtension of the channel. W ith the extended channel, the signal vector at the k th user’ s receiver can b e expressed as ¯ Y [ k ] = ¯ H [ k 1] ¯ X [1] + ¯ H [ k 2] ¯ X [2] + ¯ H [ k 3] ¯ X [3] + ¯ Z [ k ] , k = 1 , 2 , 3 . where ¯ X [ k ] is a (2 n + 1) × 1 column vector rep resenting the 2 n + 1 symbol extension of the transm itted sym bol X [ k ] , i.e ¯ X [ k ] ( t ) △ =        X [ k ] (1 , t ) X [ k ] (2 , t ) . . . X [ k ] (2 n + 1 , t )        Similarly ¯ Y [ k ] and ¯ Z [ k ] represent 2 n + 1 symbol exten sions of the Y [ k ] and Z [ k ] respectively . ¯ H [ kj ] is a diago nal (2 n + 1 ) × (2 n + 1) matrix rep resenting the 2 n + 1 symbol extension of the ch annel i.e ¯ H [ kj ] △ =        H [ kj ] (1) 0 . . . 0 0 H [ kj ] (2) . . . 0 . . . · · · . . . . . . 0 0 · · · H [ kj ] (2 n + 1 )        Recall that we a ssume that the ch annel coefﬁcient v alues for each f requency s lot a re chosen indep endently from a continuo us distrib ution. Thus, all the diag onal channe l matrices ¯ H [ kj ] are comp rised of all d istinct diagon al elements with p robability 1 . W e show that ( d 1 , d 2 , d 3 ) = ( n + 1 , n, n ) is achievable on this extended channel implying that ( n +1 2 n +1 , n 2 n +1 , n 2 n +1 ) lies in the degrees of freedo m region of the original ch annel. In the extended channel, message W 1 is enco ded at transm itter 1 into n + 1 indep endent streams x [1] m ( t ) , m = 1 , 2 , . . . , ( n + 1) sent alo ng vectors v [1] m so th at ¯ X [1] ( t ) is ¯ X [1] ( t ) = n +1 X m =1 x [1] m ( t ) v [1] m = ¯ V [1] X [1] ( t ) where X [1] ( t ) is a ( n + 1) × 1 column vector and ¯ V [1] is a (2 n + 1) × ( n + 1) dimensional matrix. Similar ly W 2 and W 3 are each encoded into n ind ependen t streams by tra nsmitters 2 an d 3 as X [2] ( t ) an d X [3] ( t ) respectively . ¯ X [2] ( t ) = n X m =1 x [2] m ( t ) v [2] m = ¯ V [2] X [2] ( t ) ¯ X [3] ( t ) = n X m =1 x [3] m ( t ) v [3] m = ¯ V [3] X [3] ( t ) The rece i ved signal at the i th receiver can then be written as ¯ Y [ i ] ( t ) = ¯ H [ i 1] ¯ V [1] X [1] ( t ) + ¯ H [ i 2] ¯ V [2] X [2] ( t ) + ¯ H [ i 3] ¯ V [3] X [3] ( t ) + ¯ Z [ i ] ( t ) In this ach iev ab le schem e, receiver i elimin ates interferen ce b y zero-fo rcing all ¯ V [ j ] , j 6 = i to deco de W i . At receiver 1, n + 1 desired strea ms ar e deco ded after zero -forcing the interfe rence to achieve n + 1 degrees of freed om. T o obtain n + 1 in terference free dim ensions from a 2 n + 1 dimensio nal receiv ed signal vector ¯ Y [1] ( t ) , th e dimension of th e in terference sh ould be no t more than n . This can be en sured by pe rfectly alignin g the interfer ence from transmitters 2 and 3 as fo llows. ¯ H [12] ¯ V [2] = ¯ H [13] ¯ V [3] (6) At the same time, receiver 2 z ero-for ces the in terference from ¯ X [1] and ¯ X [3] . T o extrac t n interfe rence-fr ee dimensions from a 2 n + 1 dime nsional vector, th e dimension of the interferen ce has to be n ot more th an n + 1 . i.e. rank h ¯ H [21] ¯ V [1] ¯ H [23] ¯ V [3] i ≤ n + 1 This ca n be a chieved by choosing ¯ V [3] and ¯ V [1] so th at ¯ H [23] ¯ V [3] ≺ ¯ H [21] ¯ V [1] (7) where P ≺ Q , means that th e set o f column v ectors of matrix P is a sub set of th e set o f column vectors of matrix Q . Sim ilarly , to decode W 3 at rec eiv er 3 , we wish to choose ¯ V [2] and ¯ V [1] so th at ¯ H [32] ¯ V [2] ≺ ¯ H [31] ¯ V [1] (8) Thus, we wish to pick vectors ¯ V [1] , ¯ V [2] and ¯ V [3] so that equatio ns (6), (7), (8) are satisﬁed. Note tha t the chann el matrices ¯ H [ ij ] have a full rank of 2 n + 1 almost surely . Since multiplyin g by a f ull rank matrix (or its inverse) does no t affect the c onditions rep resented by equations (6), (7) and (8), they can be equivalently e xpressed as B = TC (9) B ≺ A (10) C ≺ A (11) where A = ¯ V [1] (12) B = ( ¯ H [21] ) − 1 ¯ H [23] ¯ V [3] (13) C = ( ¯ H [31] ) − 1 ¯ H [32] ¯ V [2] (14) T = ¯ H [12] ( ¯ H [21] ) − 1 ¯ H [23] ( ¯ H [32] ) − 1 ¯ H [31] ( ¯ H [13] ) − 1 (15) Note that A is a (2 n + 1) × ( n + 1) matrix. B and C are (2 n + 1) × n matrices. Since all cha nnel matrices are in vertible, we can choose A , B an d C so that th ey satisfy equ ations (9)-(1 1) and then use equation s (12)- (15) to ﬁnd ¯ V [1] , ¯ V [2] and ¯ V [3] . A , B , C are p icked as f ollows. Let w be th e (2 n + 1) × 1 column vector w =        1 1 . . . 1        W e now choose A , B and C as: A = [ w Tw T 2 w . . . T n w ] B = [ Tw T 2 w . . . T n w ] C = [ w Tw . . . T n − 1 w ] It ca n be easil y v eriﬁed that A , B and C satisfy t he thr ee equa tions (9)-(11). T herefor e, ¯ V [1] , ¯ V [2] and ¯ V [3] satisfy the in terferenc e alignment equations in (6), ( 7) and ( 8). Now , con sider the r eceiv ed signal vectors at Receiv er 1 . T he desired signal arrives along the n + 1 vectors ¯ H [11] ¯ V [1] while th e interfer ence arrives a long the n vector s ¯ H [12] ¯ V [2] and the n vectors ¯ H [13] ¯ V [3] . As enf orced by equation (6) th e interferenc e vectors are per fectly aligned . Ther efore, in or der to prove that there are n + 1 interferen ce free dime nsions it suf ﬁces to show tha t the column s of the square, (2 n + 1) × (2 n + 1) d imensional matrix h ¯ H [11] ¯ V [1] ¯ H [12] ¯ V [2] i (16) are lin early indep endent almost su rely . Multip lying by the fu ll rank matrix ( ¯ H [11] ) − 1 and substituting the values of ¯ V [1] , ¯ V [2] , eq uiv ale ntly we need to show that alm ost sure ly S △ =  w Tw T 2 w . . . T n w Dw DTw DT 2 w . . . DT n − 1 w  (17) has linearly independent column vecto rs where D = ( ¯ H [11] ) − 1 ¯ H [12] is a diagonal matrix. In o ther words, we need to sh ow det( S ) 6 = 0 with pr obability 1. The proof is obtained b y contradiction. If possible, let S be sing ular with non-ze ro probab ility . i.e, Pr( | S | = 0) > 0 . Fu rther, let the diagon al entries of T b e λ 1 , λ 2 , . . . λ 2 n +1 and the diagona l entries of D be κ 1 , κ 2 . . . κ 2 n +1 . Th en the following equation is tru e with non-zero pro bability . | S | =            1 λ 1 λ 2 1 . . . λ n 1 κ 1 κ 1 λ 1 . . . κ 1 λ n − 1 1 1 λ 2 λ 2 2 . . . λ n 2 κ 2 κ 2 λ 2 . . . κ 2 λ n − 1 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 λ 2 n +1 λ 2 2 n +1 . . . λ n 2 n +1 κ 2 n +1 κ 2 n +1 λ 2 n +1 . . . κ 2 n +1 λ n − 1 2 n +1            = 0 Let C ij indicate the co factor of the i th row and j th column of | S | . Expanding the determ inant along the ﬁrst row , we g et | S | = 0 ⇒ C 11 + λ 1 C 12 + . . . λ n 1 C 1( n +1) + κ 1  C 1( n +2) + λ 1 C 1( n +3) + . . . + λ n − 1 1 C 1(2 n +1)  = 0 None of ‘co -factor’ terms C 1 j in the above expan sion depend λ 1 and κ 1 . If all values other th an κ 1 are gi ven, then the a bove i s a linear equation in κ 1 . Now , | S | = 0 imp lies on e of th e fo llowing tw o events 1) κ 1 is a root of the lin ear equ ation. 2) All the co efﬁcients form ing the linear eq uation in κ 1 are equ al to 0 , so that th e sing ularity cond ition is tr i vially satisﬁed fo r all v alues of κ 1 . Since κ 1 is a random variable drawn from a continu ous distribution, the proba bility of κ 1 taking a value which is equal to the ro ot of this linear equation is zero. There fore, the seco nd event happens with probability g reater than 0 a nd we c an wr ite, Pr ( | S | = 0) > 0 ⇒ Pr( C 1( n +2) + λ 1 C 1( n +3) + . . . + λ n − 1 1 C 1(2 n +1) = 0) > 0 Consider th e equation C 1( n +2) + λ 1 C 1( n +3) + . . . + λ n − 1 1 C 1(2 n +1) = 0 Since the terms C 1 j do not depend on λ 1 , the above equation is a polynomial o f degree n in λ 1 . Again, as b efore, there are two possibilities. The ﬁrst p ossibility is that λ 1 takes a value equal to one of the n roots of the above equation. Since λ 1 is drawn from a continu ous distribution, the pro bability o f th is event happen ing is zer o. The second possibility is that all the co efﬁcients of the above polyn omial are zero with non -zero pro bability an d w e can write Pr( C 1( n +2) + . . . + κ 1 λ n 1 C 1(2 n +1) = 0) > 0 ⇒ Pr( C 1(2 n +1) = 0) > 0 W e have now shown that if the determin ant of the (2 n + 1) × (2 n + 1) matrix S is equal to 0 with n on-zero probab ility , then the determina nt of following 2 n × 2 n matrix (obtain ed by stripp ing off the ﬁrst r ow and last column of S ) is equal to 0 with non- zero probability . det     1 λ 2 λ 2 2 . . . λ n 2 κ 2 κ 2 λ 2 . . . κ 2 λ n − 2 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 λ 2 n +1 λ 2 2 n +1 . . . λ n 2 n +1 κ 2 n +1 κ 2 n +1 λ 2 n +1 . . . κ 2 n +1 λ n − 2 2 n +1     = 0 with p robab ility gre ater than 0 . Repeating the above a rgument and eliminating th e ﬁrst ro w an d last column at each stage we get det     1 λ n +1 λ 2 n +1 . . . λ n n +1 . . . . . . . . . . . . . . . 1 λ 2 n +1 λ 2 2 n +1 . . . λ n 2 n +1     = 0 with p robability greater than 0 . But this is a V andermonde matrix and its d eterminan t Y n +1 ≤ i 1 d 1 − d 2 − d 3 +1 2 d 2 − d 1 − d 3 +1 2 d 3 − d 1 − d 2 +1 2 d 1 + d 2 + d 3 − 1 0 It is easily veriﬁed that th e values of α i are no n-negative for all ( d 1 , d 2 , d 3 ) ∈ D and that they add up to one. Thu s, all points in D are con vex combinations of achiev able po ints J, K, L , N an d (0 , 0 , 0) . Since conve x combin ations are achiev able by time sharing between the end points, this implies th at D ⊂ D ′ . T og ether with the con verse, we have D = D ′ and the proof is complete. Note that the p roof p resented above uses codin g over multiple freq uency slots where the chan nel coefﬁcients take distinct values. W e now examine the possible ramiﬁcations of this assumption both from a theoretical as well as a practical p erspective. From a theoretical pe rspective the assump tion of frequency selective c hannels is intriguing becau se it is not clear if K/ 2 degrees of f reedom will be achieved with constan t chan nels over only one frequ ency slot. Ther efore th e validity of the conjecture in [7] that the interf erence chan nel with constan t channe l coefﬁcients has only 1 d egree of fre edom for any nu mber of users still remain s undeter mined. The issue is analogou s to the 2 user X channel with a single antenn a at all no des. It is shown in [22] th at the time/f requency varying MIMO X channel has 4 / 3 degrees o f f reedom per time/frequen cy dimension. However , it is not known wh ether the X chann el with sing le antenna no des and constant chan nel coefﬁcients can achieve more than 1 degre e of freed om. From a practical p erspective, we present several observations. 1) The assumptio n that th e ch annel coefﬁcients vary over frequency is not r estrictiv e as it hold s true in practice fo r almost all wireless channe ls. Moreover , note that it is no t necessary that the channel coefﬁcients are indepen dent across f requen cy slots. It sufﬁces that they are ch osen accord ing to a continuo us joint distribution. 2) W e have shown that by codin g over 2 n + 1 f requen cy slots, we can achieve 3 n +1 2 n +1 degrees of freedo m on the 3 user interfere nce ch annel. The fact that only a ﬁnite number o f fr equency slots sufﬁce to achieve a certain number of degrees of freedom may be signiﬁcant in practice. On the other hand if non-ca usal ch annel knowledge is not an issue and the channel is time varying then o nly one frequency slot sufﬁces for th is achiev ability proof . 3) Recall that for the 2 user X chan nel, time and freq uency variations are not needed when more than 1 antenna is present at each node. Sim ilarly , we will show in Section VI tha t with M > 1 antenna s at each n ode th e 3 user interf erence chan nel with co nstant chan nel matrices ha s 3 M / 2 degrees o f freed om. Before co nsidering the MIMO case with con stant chann el matrices we visit the issue of O (1) capacity . V . T H E O (1) C A PAC I T Y O F W I R E L E S S N E T W O R K S Consider a multiuser wireless c hannel with transmit po wer ρ , noise power norma lized to unity , and sum capacity C ( ρ ) . The degrees of f reedom d provide a capacity approx imation that is accurate within o (log( ρ )) , i.e., C ( ρ ) = d log ( ρ ) + o (log( ρ )) (20) where th e little ”o” notatio n is d eﬁned as follows: f ( x ) = o ( g ( x )) ⇔ lim x →∞ f ( x ) g ( x ) = 0 . (21) Similarly , o ne can deﬁne a capacity characteriza tion C 1 ( ρ ) , that is accu rate to within an O (1) term, lim sup ρ →∞   C ( ρ ) − C 1 ( ρ )   < ∞ . (22) so th at we c an wr ite C ( ρ ) = C 1 ( ρ ) + O (1) (23) While the O (1) notatio n implies an asymptotic appro ximation as ρ → ∞ , it is easy to see that for all communi- cation n etworks, if the O (1) capacity ch aracterization is known, then one can ﬁnd a capacity ch aracterization that is within a constan t of the capacity for all ρ . This is becau se the capacity C ( ρ ) is a non -negativ e, monotonically increasing f unction of the tr ansmit p ower ρ . T his is seen as follows. Let C 1 ( ρ ) be an O (1) c apacity cha racterization. Mathematically , ∃ ρ o , C o < ∞ , such that sup ρ ≥ ρ o   C ( ρ ) − C 1 ( ρ )   < C o . (24) Then we can co nstruct a capacity char acterization C ( ρ ) th at is accurate to within a constant for all ρ as follows: C ( ρ ) △ = ( C 1 ( ρ o ) ∀ ρ ≤ ρ o C 1 ( ρ ) ∀ ρ > ρ o ) such that the absolute value of the d ifference between the ca pacity C ( ρ ) and C ( ρ ) is bound ed above b y max { C o , C 1 ( ρ ) } . Clearly , the O (1) ca pacity provides in g eneral a more accu rate capa city characterizatio n than the degrees of freedom deﬁnition. Howe ver , it turns out tha t in most cases the two are dir ectly r elated. For example, it is well known that for the fu ll rank MIMO cha nnel with M inpu t antennas and N outpu t antennas, transmit power ρ and i.i.d. zero mean unit variance add iti ve white Gaussian noise (A WGN) at each rece i ver , the capacity C ( ρ ) may be expressed as: C ( ρ ) = min( M , N ) log(1 + ρ ) + O (1 ) = d log (1 + ρ ) + O (1) . (25) As form alized by the fo llowing theor em, a similar relatio nship between the degrees of freedom and the O (1) capacity ch aracterization also holds for most multiuser communic ation channels. Theorem 3 : For the MI MO multiple access ch annel, the MI MO broadcast channel, the two user MIMO inter- ference channel and the 2 user MIMO X channel, an O (1) ch aracterization of the sum ca pacity can be obtain ed in ter ms of th e to tal nu mber of degrees of freedo m as fo llows: C ( ρ ) = d log(1 + ρ ) + O (1 ) . (26) Pr oof: Sin ce the pr oof is qu ite simple, we only present a b rief ou tline as follows. For the MIMO MAC and BC, the outerbound o n su m capacity obtained from full coop eration among th e d istributed n odes is d log(1 + ρ ) + O (1) . The innerbo und obtain ed fro m zer o f orcing is also d log(1 + ρ ) + O (1 ) so that we can write C ( ρ ) = d lo g(1 + ρ ) + O (1) . For the two user MI MO inte rference cha nnel and the 2 user MIMO X ch annel the outerboun d is obtained following an extension of C arlieal’ s outerbo und which results in a MIMO MA C channel. The innerbound is obtained from zero forcing. Since both of these boun ds are with in O (1) of d lo g(1 + ρ ) we c an similar ly write C ( ρ ) = d log (1 + ρ ) + O (1 ) . Finally , consider the K user inter ference channel with single antennas at each node. In this case we have on ly shown: ( K/ 2 − ǫ ) log (1 + ρ ) + O (1) ≤ C ( ρ ) ≤ ( K / 2) lo g (1 + ρ ) + O (1) , ∀ ǫ > 0 . (27) Consider a h ypothetical cap acity f unction C ( ρ ) = K / 2 log(1 + ρ ) − c p log(1 + ρ 2 ) . Such a c apacity f unction would also satisfy th e inner a nd outerbo unds provided above for the K u ser interfer ence channel and h as D = K / 2 d egrees of freedo m. Howe ver , this hypoth etical capacity functio n does not have a O (1) cap acity characterization equal to C ( ρ ) = K / 2 log(1 + ρ ) as th e difference between C ( ρ ) and C ( ρ ) is u nboun ded. T o claim that the O (1) capacity of the 3 user interfer ence ch annel is (3 / 2) log(1 + ρ ) we need to show an innerbound o f ( K/ 2) log(1 + ρ ) + O (1) . Since o ur ach iev ab le schemes are b ased on interferen ce alignment and zero forcin g, the natural qu estion to ask is whether an interf erence alignmen t and zero forcing based sch eme c an achieve exactly K/ 2 d egrees of freed om. The fo llowing e xplanatio n uses the K = 3 case to su ggest that the answer is negative. Consider an ach ie vable scheme that uses a M symb ol extension of the channel. Now , con sider a point ( α 1 , α 2 , α 3 ) that ca n be ac hiev ed over this extended cha nnel using in terference align ment and zero-for cing alo ne. If possible, let the total degrees of freedom ov er th is extended channel be 3 M / 2 . i.e. α 1 + α 2 + α 3 = 3 M / 2 . It can be argued along the same lines as the co n verse part of Theorem 1 th at ( α i , α j ) is ach iev ab le in the 2 user in terference chann el for ∀ ( i, j ) ∈ { (1 , 2 ) , (2 , 3) , (3 , 1) } . Ther efore α 1 + α 2 ≤ M α 2 + α 3 ≤ M α 1 + α 3 ≤ M It can be easily seen th at the only point ( α 1 , α 2 , α 3 ) that satisﬁes th e above inequalities and achieves a total of 3 M / 2 degrees of freedom is ( M 2 , M 2 , M 2 ) . Th erefore, any scheme that achieves a total of 3 M / 2 degrees of freedom over the extended chann el achieves the point ( M 2 , M 2 , M 2 ) . W e assume tha t the messages W i are encoded along M / 2 independen t stream s similar to the coding scheme in the p roof of Theorem 1 i.e. ¯ X [ i ] = M / 2 X m =1 x [ i ] m v [ i ] m = ¯ V [ i ] X [ i ] Now , at recei ver 1 , to d ecode a n M / 2 dimension al sign al using zero -forcin g, the dimension o f the interference has to b e at most M / 2 . i.e., rank [ ¯ H [13] ¯ V [3] ¯ H [12] ¯ V [2] ] = M / 2 (28) Note th at since ¯ V [2] has M / 2 linearly indep endent column vector s and ¯ H [12] is f ull ran k with probab ility 1 , rank ( ¯ H [12] ¯ V [2] ) = M / 2 . Similarly the dimen sion of the interf erence from tra nsmitter 3 is also equ al to M / 2 . Therefo re, the two vector spaces on the left hand side of equation ( 28) m ust have full intersection , i.e span ( ¯ H [13] ¯ V [3] ) = span ( ¯ H [12] ¯ V [2] ) (29) span ( ¯ H [23] ¯ V [3] ) = span ( ¯ H [21] ¯ V [1] ) (At receiver 2) (30) span ( ¯ H [32] ¯ V [2] ) = span ( ¯ H [31] ¯ V [1] ) (At receiver 3) (31) where span ( A ) represents the space spann ed by the colum n vectors of m atrix A The above equation s imply that span ( ¯ H [13] ( ¯ H [23] ) − 1 ¯ H [21] ¯ V [1] ) = span ( ¯ H [12] ( ¯ H [32] ) − 1 ¯ H [31] ¯ V [1] ) ⇒ spa n ( ¯ V [1] ) = span ( T ¯ V [1] ) where T = ( ¯ H [13] ) − 1 ¯ H [23] ( ¯ H [21] ) − 1 ¯ H [12] ( ¯ H [32] ) − 1 ¯ H [31] . T he above equation implies that there exists at least on e eigenv ector e o f T in span ( ¯ V [1] ) . Note that since all cha nnel matrices are dia gonal, the set of eigenv ectors of all channel m atrices, th eir inverses and their products ar e all identica l to the set of co lumn vector s of the identity matrix. i.e vectors of the for m [0 0 . . . 1 . . . 0] T . Therefor e e is an eigen vector for all channel matrices. Since e lies in span ( ¯ V [1] ) , equ ations (29)-(3 1) imply that e ∈ span ( ¯ H [ ij ] ¯ V [ i ] ) , ∀ i , j ∈ { 1 , 2 , 3 } ⇒ e ∈ sp an ( ¯ H [11] ¯ V [1] ) ∩ span ( ¯ H [12] ¯ V [2] ) Therefo re, at receiv er 1, the desired signa l ¯ H [11] ¯ V [1] is n ot linearly indepen dent with the inter ference ¯ H [21] ¯ V [2] . Therefo re, receiver 1 can not d ecode W 1 completely by merely ze ro-for cing the interf erence signa l. Evidently , interferen ce alignm ent in the manner d escribed ab ove c annot a chieve exactly 3 / 2 d egrees of freedom on the 3 user interferen ce channel with a single antenna at all nod es. Thus, the degrees of freedom fo r the 3 user interfe rence chann el with M = 1 do not au tomatically lead us to the O (1) ca pacity . Th e possibility that the sum capacity of th e 3 user in terference channel with single antennas at all nod es may no t b e o f the form (3 / 2 ) lo g(1 + ρ ) + O (1) is in teresting because it suggests th at th e 3 user interferen ce ch annel capac ity may not be a straighfo rward extensio n o f the 2 u ser interfe rence channel capacity characterizatio ns. W e explore this interesting aspect of the 3 u ser interfer ence channel furth er in the context of multiple an tenna nodes. Our g oal is to ﬁnd o ut if e xactly 3 M / 2 degre es of fr eedom may be achieved with M antennas at each nod e. As shown by th e following theorem, indeed we can ach iev e exactly 3 M / 2 degrees of freed om so that the O (1 ) capacity ch aracterization for M > 1 is indeed re lated to the d egrees of freed om as C ( ρ ) = (3 M / 2) log(1 + ρ ) . V I . D E G R E E S O F F R E E D O M O F T H E 3 U S E R I N T E R F E R E N C E CH A N N E L W I T H M > 1 A N T E N NA S A T E AC H N O D E A N D C O N S TA N T C H A N N E L C O E FFI C I E N T S The 3 user MIMO in terference ch annel is interesting fo r two r easons. First we wish to show that with multiple antennas we can ach iev e 3 M / 2 degrees of freedom with constan t channe l matric es, i.e., multiple freq uency slots are not req uired. Sec ond, we wish to show th at exactly 3 M / 2 degre es of f reedom are ach ie ved b y zero forcin g and interfer ence align ment which gives us a lowerboun d on sum cap acity of 3 M / 2 log(1 + ρ ) + O (1) . Since the outerbo und o n sum capac ity is also 3 M / 2 lo g(1 + ρ ) + O (1) we have an O (1) appro ximation to the capacity o f the 3 user MIMO interfer ence channel with M > 1 antennas at all no des. Theorem 4 : In a 3 user inte rference cha nnel with M > 1 anten nas at each transmitter and each r eceiv er and constant co efﬁcients, the sum cap acity C ( ρ ) may b e character ized as: C ( ρ ) = (3 M / 2) lo g(1 + ρ ) + O (1) (32) The pr oof is p resented in Appendice s II and III. V I I . C O G N I T I V E M E S S AG E S H A R I N G O N T H E 3 U S E R I N T E R F E R E N C E C H A N N E L Cognitive message s haring refers to a form of co operation between tra nsmitters and/or recei vers whe re th e message of on e user is made available non-causally to the tran smitter or rece i ver of a nother user . Degrees of fre edom with cognitive co operation are considere d in [17] and [23] . I t is shown in [22] that for the tw o user interf erence c hannel with equ al n umber M of anten nas at all n odes there is no gain in de grees of freed om wh en one user has a co gnitive transmitter, a cog nitiv e rec ei ver or both. In all these cases th e total degrees of free dom eq uals M . Howev er th e full 2 M degrees of fr eedom are ob tained if bo th users have a cognitive tra nsmitter , or both users have a cognitive receiver , o r one user has a co gnitive transmitter and the other user h as a cog nitiv e receiv er . I n this section we generalize this result to the th ree user interference chan nel. T o gener alize the resu lt we introdu ce some n otation. Let T i be deﬁned as the set of messages a vailable non- causally at tran smitter i and re ceiv er i . Let us also deﬁne R i as the set containin g the message inten ded for receiver i and also the m essages n on-causally available at receiver i . W ith no co gnitive sharing o f messages T i = R i = { W i } . Further, let u s deﬁne W i = T i ∪ R i . The f ollowing theo rem presents the to tal numb er o f degrees of freedom for some intere sting cogn itiv e message sharing scen arios. Theorem 5 : The total number of degrees of freed om η ⋆ for the 3 user interference c hannel under various cognitive message sharing scen arios are determined as follows: 1) If on ly one message (e.g. W 1 ) is shared amon g all nodes the degrees of freedo m are unchanged . W 1 = { W 1 } , W 2 = { W 1 , W 2 } , W 3 = { W 1 , W 3 } ⇒ η ⋆ = 3 / 2 . (33) Note that this includes all scenarios wh ere message W 1 is mad e av ailab le to only the tran smitter , on ly the receiver or both tra nsmitter an d rece i ver of users 2 and 3 . In all these cases, there is no beneﬁt in ter ms o f degrees of freedom. 2) If two messages (e.g . W 1 , W 2 ) are shared amon g all nodes then we have 2 degrees of freedom. W 1 = { W 1 , W 2 } , W 2 = { W 1 , W 2 } , W 3 = { W 1 , W 2 , W 3 } ⇒ η ⋆ = 2 (34) Note that the m essages W 1 , W 2 may be shared through cogn iti ve transmitters, receivers or both. 3) If on ly one receiv er (e.g. r eceiv er 3) is fully cog nitiv e then we have 3 / 2 degrees of f reedom. W 1 = { W 1 } , W 2 = { W 2 } , T 3 = { W 3 } , R 3 = { W 1 , W 2 , W 3 } ⇒ η ⋆ = 3 / 2 (35) 4) If on ly one transmitter (e.g . transmitter 3) is f ully cog nitiv e then we ha ve 2 degrees of freedo m. W 1 = { W 1 } , W 2 = { W 2 } , T 3 = { W 1 , W 2 , W 3 } , R 3 = { W 3 } ⇒ η ⋆ = 2 (36) The last two c ases are sig niﬁcant as they show the distinction between cognitive transmitters and co gnitive receivers that was no t visible in the two user interference channel stud ied in [ 22]. I n [ 22] it was shown that fro m a degre e of f reedom p erspective, cog nitiv e transmitters are eq uiv a lent to cognitive rec eiv ers for the two u ser interferenc e channel. Howev er , cases 3 and 4 ab ove show that cogn iti ve tra nsmitters may be more p owerful than cogn iti ve receivers. Intu iti vely , a cognitive receiver with no me ssage of its own is useless whe reas a cognitive transmitter with n o message of its own is still useful. Pr oof: 1) Consider the case where W 1 is shared with either the tran smitter or receiver (o r both) of user 2 and user 3 . Now , with W 1 = φ we have a tw o user inter ference channe l with no cognitive m essage sharing wh ich gives us the oute rboun d d 2 + d 3 ≤ 1 . For the next ou terboun d, set W 2 = φ and let the transmitter of user 2 coope rate with the tra nsmitter of user 1 as a two anten na tr ansmitter . W e now have a two user cognitive interfer ence channel with user 1 as the primary u ser (with two transmit antennas and o ne recei ve antenna) and user 3 with the cognitiv e transmitter, co gnitive re ceiv er or both. Using the standard MA C outerbound argumen t it is easily seen that by red ucing the noise at receiver 1 we m ust be able to decode both messages W 1 , W 3 at receiver 1 . This giv es us the outerbo und d 1 + d 3 ≤ 1 . Similarly , we obtain the outerbou nd d 1 + d 2 ≤ 1 . Adding up the three outerbou nd we have η ⋆ ≤ 3 / 2 . Since 3 / 2 degrees of fr eedom are achie vable e ven with out any c ognitive cooper ation, we ha ve η ⋆ = 3 / 2 . Note that this result is easily extended to K users, i.e. with only one message shared am ong all nodes the degrees of fr eedom are not increased . 2) For achie vability , set W 3 = φ and let transmitter 3 stay silent. Then we have a cognitive two user in terference channel with two shared me ssages which has 2 degrees of freedo m as established in [22] . For th e converse argument let tran smitter 1 and 2 cooper ate as a two antenn a transmitter T 12 and rec eiv er 1 an d 2 co operate as a two antenna rece i ver R 12 . Then we have a two user cog nitiv e in terference channel wh ere the prim ary user has two transmit and two receive antenna s while th e cognitive user has a single transmit anten na and a single receive anten na. Once again, the standard MA C ou terboun d argumen t is u sed to show that by reducing noise at the primar y receiver , we must be able to deco de a ll messages at the primar y receiver . This g iv es us the o uterbo und d 1 + d 2 + d 3 ≤ 2 . Since the inn er and outerbou nds agree, η ⋆ = 2 . 3) Achiev ability of 3 / 2 degrees of f reedom is trivial as no co gnitive message sharing is r equired. For the converse, setting W 1 , W 2 , W 3 to φ o ne at a time lead s to th e two user co gnitive interferenc e channel with on e shared message for which the degrees of freedo m are bou nded above b y 1 . Ad ding the three outer bound s we co nclude that η ⋆ = 3 / 2 . This result is also easily extended to the K u ser in terference chann el. 4) For achie vability , set W 3 = φ . Then we have a two user interferen ce channel with a cognitive helper (tra nsmitter 3 ) wh o knows both u ser’ s messages. T wo degrees of freed om are achieved easily on this ch annel as tr ansmitters 1 and 3 coop erate to zero fo rce the tran smission of W 1 at receiver 2 , while transmitter s 2 and 3 cooperate to zero force th e transmission of W 2 at receiv er 1 . By eliminating interference at receivers 1 and 2 , we have two degrees of freedom. The co n verse follows directly from th e converse f or par t 2 , so that we have η ⋆ = 2 . V I I I . C O N C L U S I O N W e have sh own that with pe rfect channel knowledge the K user interferen ce channel has K / 2 spatial d egrees of freedom . Conventional wisdo m has so far been consistent with the con jecture that distributed inter fering systems cannot ha ve m ore tha n 1 degree of freedom and therefore the best kn own ou terboun d K / 2 has not been consider ed signiﬁcant. This pessimistic outloo k has for long invited researcher s to tr y to prove that more than 1 degree of freedom is not possible while ignorin g the K / 2 o uterbou nd. Th e p resent result shifts the focus onto the outerboun d by proving that it is tight if perfect an d g lobal channel knowledge is av a ilable. Thus, the present result co uld guide future research alo ng an o ptimistic p ath in the same mann er that MIMO technolo gy has shaped our v ie w of the capacity of a wire less chann el. There are se veral prom ising directions for future work. From a p ractical persp ectiv e it is important to explore to what extent interference alignme nt c an be ac complished with limited ch annel knowledge. Simpler achiev ability schem es are another promising avenue of resear ch. For example, the interfer ence align ment scheme ba sed on different propaga tion delay s that we presented in this paper is an exciting possibility as it on ly requires a careful p lacing of interfering nod es to satisfy certain de lay constra ints. A P P E N D I X I A C H I E V A B I L I T Y F O R T H E O R E M 1 F O R A R B I T R A RY K Let N = ( K − 1)( K − 2) − 1 . W e show that ( d 1 ( n ) , d 2 ( n ) , . . . d K ( n )) lies in th e degrees of f reedom region of the K user interference chan nel for any n ∈ N where d 1 ( n ) = ( n + 1) N ( n + 1) N + n N d i ( n ) = n N ( n + 1) N + n N , i = 2 , 3 . . . K This im plies that max ( d 1 ,d 2 ,...d K ) ∈D d 1 + d 2 + · · · d K ≥ sup n ( n + 1) N + ( K − 1) n N ( n + 1) N + n N = K / 2 W e pr ovide an ach ie vable sch eme to sho w that (( n + 1 ) N , n N , n N . . . n N ) lies in the degrees of f reedom region of an M n = ( n + 1 ) N + n N symbol extension of the or iginal chann el which automatically imp lies the desired result. In the extend ed channel, the signal vector at the k th user’ s recei ver can be expressed as ¯ Y [ k ] ( t ) = K X j =1 ¯ H [ kj ] ¯ X [ j ] ( t ) + ¯ Z [ k ] ( t ) where ¯ X [ j ] is an M n × 1 column vector r epresenting the M n symbol extension of the transmitted symbol X [ k ] , i.e ¯ X [ j ] ( t ) △ =        X [ j ] (1 , t ) X [ j ] (2 , t ) . . . X [ j ] ( M n , t ))        Similarly ¯ Y [ k ] and ¯ Z [ k ] represent M n symbol extensions of the Y [ k ] and Z [ k ] respectively . ¯ H [ kj ] is a diagonal M n × M n matrix r epresenting the M n symbol extension of the ch annel i.e ¯ H [ kj ] △ =        H [ kj ] (1) 0 . . . 0 0 H [ kj ] (2) . . . 0 . . . · · · . . . . . . 0 0 · · · H [ kj ] ( M n )        Recall that the d iagonal elements of ¯ H [ kj ] are drawn in depend ently fro m a continuous d istribution and are th erefore distinct with probab ility 1 . In a man ner similar to the K = 3 c ase, m essage W 1 is en coded at transmitter 1 into ( n + 1) N indepen dent streams x [1] m ( t ) , m = 1 , 2 , . . . , ( n + 1) N along vectors v [1] m so that ¯ X [1] ( t ) is ¯ X [1] ( t ) = ( n +1) N X m =1 x [1] m ( t ) v [1] m = ¯ V [1] X [1] ( t ) where X [1] ( t ) is a ( n + 1) N × 1 column vector and ¯ V [1] is a M n × ( n + 1) N dimensiona l matrix. Similarly W i , i 6 = 1 is enco ded into n K indepen dent streams by transmitter i as ¯ X [ i ] ( t ) = n N X m =1 x [ i ] m ( t ) v [ i ] m = ¯ V [ i ] X [ i ] ( t ) The rece i ved signal at the i th receiver can then be written as ¯ Y [ i ] ( t ) = K X j =1 ¯ H [ ij ] ¯ V [ j ] X [ j ] ( t ) + ¯ Z [ i ] ( t ) All rec ei vers decod e th e d esired signal by zero-fo rcing the interferenc e vector s. At recei ver 1 , to obtain ( n + 1) N interferen ce free d imensions correspo nding to th e de sired signal from an M n = ( n + 1) N + n N dimensiona l recei ved signal v ector ¯ Y [1] , th e dimen sion of the interfer ence sh ould b e n ot mor e than n N . T his can be ensur ed by perfectly aligning the interfere nce from tran smitters 2 , 3 . . . K as follows ¯ H [12] ¯ V [2] = ¯ H [13] ¯ V [3] = ¯ H [14] ¯ V [4] = . . . = ¯ H [1 K ] ¯ V [ K ] (37) At the same time, receiver 2 zero- forces the interferen ce from ¯ X [ i ] , i 6 = 2 . T o extract n N interferen ce-free dimensions from a M n = ( n + 1 ) N + n N dimensiona l vector, th e dimension of the interferenc e has to be n ot m ore than ( n + 1 ) N . This ca n be a chieved by choosing ¯ V [ i ] , i 6 = 2 so th at ¯ H [23] ¯ V [3] ≺ ¯ H [21] ¯ V [1] ¯ H [24] ¯ V [4] ≺ ¯ H [21] ¯ V [1] . . . ¯ H [2 K ] ¯ V [ K ] ≺ ¯ H [21] ¯ V [1] (38) Notice that the above relations align th e interfere nce from K − 2 transmitters within the in terference from transm itter 1 at rece i ver 2 . Similarly , to de code W i at receiver i when i 6 = 1 we wish to cho ose ¯ V [ i ] so that the following K − 2 r elations are satisﬁed. ¯ H [ ij ] ¯ V [ j ] ≺ ¯ H [ i 1] ¯ V [1] , j / ∈ { 1 , i } (39) W e now wish to pic k vectors ¯ V [ i ] , i = 1 , 2 . . . K so tha t equ ations (37), (38) an d (39) are satisﬁed. Sin ce ch annel matrices ¯ H [ ij ] have a full rank of M n almost s urely , equations (3 7), (38) and ( 39) can be equiv alen tly expressed as ¯ V [ j ] = S [ j ] B j = 2 , 3 , 4 . . . K At receiv er 1 (40) T [2] 3 B = B ≺ ¯ V [1] T [2] 4 B ≺ ¯ V [1] . . . T [2] K B ≺ ¯ V [1]              At rec ei ver 2 (41) T [ i ] 2 B ≺ ¯ V [1] T [ i ] 3 B ≺ ¯ V [1] . . . T [ i ] i − 1 B ≺ ¯ V [1] T [ i ] i +1 B ≺ ¯ V [1] . . . T [ i ] K B ≺ ¯ V [1]                              At receiv er i wher e i = 3 . . . K (42) where B = ( ¯ H [21] ) − 1 ¯ H [23] ¯ V [3] (43) S [ j ] = ( ¯ H [1 j ] ) − 1 ¯ H [13] ( ¯ H [23] ) − 1 ¯ H [21] , j = 2 , 3 , . . . K (44) T [ i ] j = ( ¯ H [ i 1] ) − 1 ¯ H [ ij ] S [ j ] i, j = 2 , 3 . . . K , j 6 = i (45) Note that T [2] 3 = I , the M n × M n identity m atrix. W e now ch oose ¯ V [1] and B so that the y satisfy th e ( K − 2)( K − 1) = N + 1 relations in ( 41)-(42) and then u se equ ations in (40) to determine ¯ V [2] , ¯ V [3] . . . ¯ V [ K ] . Th us, our g oal is to ﬁnd matr ices ¯ V [1] and B so th at T [ i ] j B ≺ ¯ V [1] for all i, j = { 2 , 3 . . . K } , i 6 = j . Let w be the M n × 1 co lumn vector w =        1 1 . . . 1        W e need to cho ose n ( K − 1)( K − 2) − 1 = n N column vectors for B . The sets of column vectors of B an d ¯ V [1] are chosen to be e qual to the sets B and ¯ V [1] where B =     Y m,k ∈{ 2 , 3 ,.. .K } ,m 6 = k , ( m,k ) 6 =(2 , 3)  T [ m ] k  α mk  w : ∀ α mk ∈ { 0 , 1 , 2 . . . n − 1 }    ¯ V [1] =     Y m,k ∈{ 2 , 3 ,.. .K } ,m 6 = k , ( m,k ) 6 =(2 , 3)  T [ m ] k  α mk  w : ∀ α mk ∈ { 0 , 1 , 2 . . . n }    For example, if K = 3 we get N = 1 . B and ¯ V [1] are cho sen as B = h w T [3] 2 w . . . ( T [3] 2 ) n − 1 w i ¯ V [1] = h w T [3] 2 w . . . ( T [3] 2 ) n w i T o clarify the n otation furth er , c onsider the case wh ere K = 4 . Assuming n = 1 , B consists o f exactly one element i.e B = { w } . Th e set ¯ V [1] consists of all 2 N = 2 5 = 32 co lumn vectors of the form ( T [2] 4 ) α 24 ( T [3] 2 ) α 32 ( T [3] 4 ) α 24 ( T [4] 3 ) α 43 ( T [4] 2 ) α 42 w where all α 24 , α 32 , α 34 , α 42 , α 43 take v alues 0 , 1 . B and ¯ V [1] can be veriﬁed to have n N and ( n + 1 ) N elements respectively . ¯ V [ i ] , i = 2 , 3 . . . K are cho sen using equations (40). Clear ly , for ( i, j ) = (2 , 3) , T [ i ] j B = B ≺ ¯ V [1] Now , for i 6 = j, i, j = 2 . . . K , ( i, j ) 6 = (2 , 3) T [ i ] j B = (  Y m,k ∈{ 2 , 3 ,.. .N } ,m 6 = k, ( m,k ) 6 =(2 , 3)  T [ m ] k  α mk  w : ∀ ( m, k ) 6 = ( i, j ) , α mk ∈ { 0 , 1 , 2 . . . n − 1 } , α ij ∈ { 1 , 2 , . . . n } ) ⇒ T [ i ] j B ∈ ¯ V [1] ⇒ T [ i ] j B ≺ ¯ V [ 1 ] Thus, th e interfer ence alignmen t equations (40)-(4 2) are satisﬁed. Throu gh interferen ce alignment, we h a ve now en sured th at the d imension of the interf erence is small enoug h. W e n ow need to verify that the comp onents of the desired signal are linearly indepen dent of the comp onents of the interference so th at the signal strea m can be completely d ecoded by z ero-for cing the interfer ence. Con sider the received sig nal vecto rs at Recei ver 1 . T he d esired signal arrives a long the ( n + 1 ) N vectors ¯ H [11] ¯ V [1] . As enf orced by equ ations (40), the in terference vectors from transm itters 3 , 4 . . . K are perf ectly a ligned with the interf erence from transmitter 2 and therefor e, all interferen ce arrives along th e n N vectors ¯ H [12] ¯ V [2] . I n order to p rove that th ere are ( n + 1) N interferen ce f ree dimension s it suf ﬁces to sho w that the columns of the squ are, M n × M n dimensiona l matrix h ¯ H [11] ¯ V [1] ¯ H [12] ¯ V [2] i (46) are linearly indepen dent almost surely . Multiplyin g the above M n × M n matrix with ( ¯ H [11] ) − 1 and substituting for ¯ V [1] and ¯ V [2] , we get a matrix wh ose l th row has entries of the fo rms Y ( m,k ) ∈{ 2 , 3 .. .K } ,m 6 = k , ( m,k ) 6 =(2 , 3) ( λ mk l ) α mk and d l Y ( m,k ) ∈{ 2 , 3 .. .K } ,m 6 = k , ( m,k ) 6 =(2 , 3) ( λ mk l ) β mk where α mk ∈ { 0 , 1 , . . . n − 1 } and β mk ∈ { 0 , 1 , . . . n } an d λ mk l , d l are drawn indepen dently from a contin uous distribution. T he same itera ti ve argument as in section IV -A can b e used. i.e. expa nding the correspo nding dete r- minant along the ﬁrst row , the linear in depende nce condition b oils down to on e of the fo llowing occurrin g with non-ze ro probability 1) d 1 being equ al to one of the roots o f a linear eq uation 2) The coe fﬁcients of the above men tioned linear equation bein g equa l to zero Thus the iterative argument can be extended here, strip ping th e last row and last co lumn at each iteration an d th e linear indepe ndence co ndition can b e shown to be eq uiv alen t to th e lin ear indepen dence of a n N × n N matrix whose rows are of the form Q ( m,k ) ∈{ 2 , 3 .. .K } ,m 6 = k ( λ mk l ) α mk where α pq ∈ { 0 , 1 , . . . n − 1 } . Note that this matrix is a more ge neral version of the V and ermond e m atrix obtained in section IV -A. So th e argument for the K = 3 case does not extend here. Howe ver , the iterative pro cedure which elimin ated the last row an d the last colu mn at ea ch iteration, can be contin ued. For example, expanding the determ inant along the ﬁrst row , the singularity con dition simpliﬁes to one o f 1) λ mk l being e qual to one of the roo ts of a ﬁnite degree polynom ial 2) The coe fﬁcients of the above men tioned po lynomial being equal to zer o Since the probab ility of condition 1 occurrin g is 0 , condition 2 must occu r with n on-zero pro bability . Condition 2 leads to a polyno mial in an other random variable λ pq l and thus the iterative p rocedu re can be continued until th e linear indep endence con dition is s hown to b e eq uiv a lent almo st sur ely to a 1 × 1 m atrix being equal to 0 . Assuming, without loss o f generality , th at we placed the w in the ﬁrst row (this correspon ds to the term α mk = 0 , ∀ ( m, k ) ), the linear independ ence condition bo ils down to the condition tha t 1 = 0 with non-z ero probability - an obvious contradictio n. Thus the matrix h ¯ H [11] ¯ V [1] ¯ H [12] ¯ V [2] i can be shown to be non-sing ular with prob ability 1 . Similarly , the desired signal can be chosen to be linearly in depend ent o f th e interfer ence at all o ther receivers almost surely . Thus ( ( n +1) N ( n +1) N + n N , n N ( n +1) N + n N , · · · , n N ( n +1) N + n N ) lies in the d egrees of fr eedom r egion of the K user interferen ce channel and therefore, the K user interfere nce channel has K/ 2 degrees of freedo m. A P P E N D I X I I P R O O F O F T H E O R E M 4 F O R M E V E N Pr oof: The outerbo und is straightf orward as be fore. T o prove achiev ability we ﬁr st c onsider the case wh en M is even. Throu gh an achiev ab le scheme, we show that there are M / 2 non-in terfering paths between transmitter i an d receiver i for ea ch i = 1 , 2 , 3 r esulting in a total of 3 M / 2 p aths in the network. T ransmitter i tran smits message W i for receiver i using M / 2 indep endently encoded streams over vectors v [ i ] i.e X [ i ] ( t ) = M / 2 X m =1 x [ i ] m ( t ) v [ i ] m = V [ i ] X i ( t ) , i = 1 , 2 , 3 The signa l received at receiver i can be written as Y [ i ] ( t ) = H [ i 1] V [1] X 1 ( t ) + H [ i 2] V [2] X 2 ( t ) + H [ i 3] V [3] X 3 ( t ) + Z i ( t ) All r eceiv ers cance l the interfer ence by zer o-forc ing and th en dec ode the desired message. T o dec ode the M / 2 streams alon g the co lumn vectors of V [ i ] from the M componen ts of th e received vector, the d imension of the interferen ce h as to be less than or equal to M / 2 . The following three interferen ce alig nment equa tions ensure that the d imension of the interfer ence is eq ual to M / 2 at all the recei vers. span ( H [12] V [2] ) = span ( H [13] V [3] ) (47) H [21] V [1] = H [23] V [3] (48) H [31] V [1] = H [32] V [2] (49) where span ( A ) represen ts the vector space spanned by the co lumn vectors of matrix A W e now wish to choose V [ i ] , i = 1 , 2 , 3 so that the above equations are satisﬁed. Sinc e H [ ij ] , i, j ∈ { 1 , 2 , 3 } h a ve a f ull rank of M alm ost surely , the above equations can be equivalently r epresented as span ( V [1] ) = span ( EV [1] ) (50) V [2] = FV [1] (51) V [3] = GV [1] (52) where E = ( H [31] ) − 1 H [32] ( H [12] ) − 1 H [13] ( H [23] ) − 1 H [21] F = ( H [32] ) − 1 H [31] G = ( H [23] ) − 1 H [21] Let e 1 , e 2 , . . . e M be the M e igenv ectors of E . Then we set V 1 to b e V [1] = [ e 1 . . . e ( M / 2) ] Then V [2] and V [3] are f ound usin g equatio ns (50)-(5 2). Clearly , V [ i ] , i = 1 , 2 , 3 satisfy the desired interfere nce alignment e quations (4 7)-(49). Now , to decode th e message using zero-fo rcing, we need the desired sign al to be linearly in depend ent of the in terferenc e at the r eceiv ers. For examp le, at receiver 1 , we n eed the colu mns o f H [11] V [1] to be linearly indepen denden t with the columns of H [21] V [2] almost surely . i.e we need the matrix below to b e of full ran k almost su rely h H [11] V [1] H [12] V [2] i Substituting v alue s for V [1] and V [2] in th e ab ove ma trix, and multiplying by full rank matrix ( H [11] ) − 1 , th e line ar indepen dence condition is equiv alent to the condition that the column vectors of h e 1 e 2 . . . e ( M / 2) Ke 1 . . . Ke ( M / 2) i are linear ly indep endent almost surely , where K = ( H [11] ) − 1 H [12] F . This is easily seen to b e true beca use K is a ran dom ( full ran k) linear transform ation. T o g et an intuitive understan ding of th e line ar in depend ence condition , consider th e case o f M = 2 . Let L rep resent the line along which lies th e ﬁrst eigenvector of the rando m 2 × 2 matrix E . The prob ability of a random r otation ( and scaling ) K o f L b eing co llinear with L is zero. Using a similar argu ment, we can show that matrices h H [22] V [2] H [21] V [1] i and h H [33] V [3] H [31] V [1] i have a fu ll rank o f M almost surely a nd th erefore re ceiv ers 2 and 3 can decode th e M / 2 stream s o f V [2] and V [3] using zero -forcing . Thus, a to tal 3 M / 2 interf erence free transmission s per cha nnel-use are achievable with probab ility 1 and the proo f is co mplete. A P P E N D I X I I I P R O O F O F T H E O R E M 4 F O R M O D D Pr oof: Consider a two time-slot symbol extension of th e chann el, with th e same chanel coefﬁcients over the two symbo ls. It can be expr essed as ¯ Y [ k ] = ¯ H [ k 1] ¯ X [1] + ¯ H [ k 2] ¯ X [2] + ¯ H [ k 3] X [3] + ¯ Z [ k ] i = 1 , 2 , 3 where ¯ X [ i ] is a 2 M × 1 vector that rep resents the two symbol extension of th e transmitted M × 1 symbol symbo l X [ k ] , i.e ¯ X [ k ] ( t ) △ = " X [ k ] (1 , 2 t + 1) X [ k ] (1 , 2 t + 2) # where X [ k ] ( t ) is an M × 1 v ector rep resenting the vector transmitted at time s lot t by transmitter k . Similarly ¯ Y [ k ] and ¯ Z [ k ] represent the two sym bol extensions o f th e the r eceiv ed sy mbol Y [ k ] and the noise v ector Z [ k ] respectively at rec eiv er i . ¯ H [ ij ] is a 2 M × 2 M block d iagonal matr ix repr esenting the extension o f the c hannel. ¯ H [ ij ] △ = " H [ ij ] (1) 0 0 H [ ij ] (1) # W e will now show ( M , M , M ) lies in the d egrees of freedom region o f this extended channel cha nnel with an achiev able scheme, implying that that a total of 3 M / 2 degrees o f f reedom ar e achievable over the original cha nnel. T ransmitter k tra nsmits message W i for re ceiv er i u sing M ind epende ntly encoded streams over vectors v [ k ] i.e ¯ X [ k ] = M X m =1 x [ k ] m v [ k ] m = ¯ V [ k ] X [ k ] where ¯ V [ k ] is a 2 M × M matr ix and ¯ X [ k ] is a M × 1 vector representing M indep endent streams. The following three interfer ence alignment equatio ns ensure that the dim ension of the interference is equal to M at receivers 1 , 2 and 3 . rank [ ¯ H [21] ¯ V [2] ] = rank [ ¯ H [31] ¯ V [3] ] (53) ¯ H [12] ¯ V [1] = ¯ H [32] ¯ V [3] (54) ¯ H [13] ¯ V [1] = ¯ H [23] ¯ V [2] (55) The above equatio ns imply that span ( ¯ V [1] ) = span ( ¯ E ¯ V [1] ) (56) ¯ V [2] = ¯ F ¯ V [1] (57) ¯ V [3] = ¯ G ¯ V [1] (58) where E = ( H [13] ) − 1 H [23] ( H [21] ) − 1 H [31] ( H [32] ) − 1 H [12] F = ( H [13] ) − 1 H [23] G = ( H [12] ) − 1 H [32] and ¯ E , ¯ F and ¯ G are 2 M × 2 M block -diagon al matrices representin g th e 2 M sym bol extension of E , F an d G respectively . Let e 1 , e 2 , . . . , e M , be the eig en vectors of E . Then , we pick ¯ V [1] to be ¯ V [1] = " e 1 0 e 3 . . . 0 e M 0 e 2 0 . . . e M − 1 e M # (59) As in the even M case, ¯ V [2] and ¯ V [3] are then determined b y using equations (56)-(5 8). Now , we need the desired signal to be linearly independent o f the interference at all the receiv ers. At rece i ver 1 , the d esired linear in depend ence condition bo ils down to span ( ¯ V [1] ) ∩ span ( ¯ K ¯ V [1] ) = { 0 } where K = ( H 11 ) − 1 H [21] ( F ) − 1 and ¯ K is the two-sym bol diagon al extension of K . Notice that K is an M × M matrix. The linear independ ence co ndition is equ iv alen t to saying that all the columns of the f ollowing 2 M × 2 M matrix ar e indepe ndent. " e 1 0 e 3 . . . 0 e M Ke 1 0 Ke 3 . . . 0 Ke M 0 e 2 0 . . . e M − 1 e M 0 Ke 2 0 . . . Ke M − 1 Ke M # (60) W e now argue that the prob ability of the column s of the above matrix being linear ly dep endent is zer o. L et c i , i = 1 , 2 . . . 2 M den ote the c olumns of the above matrix. Sup pose the columns c i are linear ly depen dent, then ∃ α i s.t 2 M X i =1 α i c i = 0 Let P = { e 1 , e 3 . . . e M − 2 , Ke 1 , . . . Ke M − 2 } Q = { e 2 , e 4 . . . e M − 1 , Ke 2 , . . . Ke M − 1 } Now , there are two possibilities 1) α M = α 2 M = 0 . This implies that either one of the following sets of vectors is linearly depend ent. Note that both sets are ca n be exp ressed as the union of a) A set of ⌊ ( M / 2 ) ⌋ eigen vectors of E b) A random tran sformation K of this set. An ar gument along the same lines as the even M case leads to the co nclusion that the prob ability of th e u nion of the two sets listed above being linearly depend ent in a M dimensiona l space is zer o. 2) α 2 M 6 = 0 or α M 6 = 0 This implies th at α M e M + α 2 M Ke M ∈ span ( P ) ∩ spa n ( Q ) ⇒ spa n ( { Ke M , e M } ) ∩ sp an ( P ) ∩ span ( Q ) 6 = { 0 } Also rank ( span ( P ) ∪ sp an ( Q )) = rank ( P ) + rank ( Q ) − rank ( P ∩ Q ) ⇒ r ank ( P ∩ Q ) = 2 M − 2 − r ank ( span ( P ) ∪ span ( Q )) Note that P and Q ar e M − 1 dimension al spaces. (The case where their dimen sions are less than M − 1 is handled in the ﬁrst part). Also, P and Q are d rawn from comp letely different set of vector s. Therefore , the union of P , Q has a r ank of M alm ost sure ly . Equivalently span ( P ) ∩ span ( Q ) has a dim ension of M − 2 almost su rely . Since th e set { e M , Ke M } is drawn fro m an eigen vector e M that d oes n ot exist in either P or Q , the proba bility of the 2 dim ensional spa ce span ( { e , Ke M } ) inte rsecting with the M − 2 dimension al space P ∩ Q is zero. For example, if M = 3 , let L indica te the line form ed by th e intersection of the th e two plan es span ( { e 1 , Ke 1 } ) and span ( { e 2 , Ke 2 } ) . Th e p robability that line L lies in the plan e for med by span ( { e 3 , Ke 3 } ) . Thus, th e probab ility that the desired signal lies in the span of the inte rference is ze ro at receiver 1 . Similarly , it can be argu ed that th e desired signal is indep endent of the inter ference at receiv ers 2 and 3 almost sur ely . Therefo re ( M , M , M ) is a chiev a ble over the two-symbol extended chann el. Thus 3 M / 2 degrees o f freedom are achievable over the 3 user interferen ce chann el with M anten na a t each transmitting and re ceiving node. R E F E R E N C E S [1] S. 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Interference Alignment and the Degrees of Freedom for the K User Interference Channel

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