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IEEE COMMUNICATIONS LETTERS, VOL. 17, NO. 12, DECEMBER 2013

Computational Complexity of Interference Alignment for Symmetric MIMO Networks Liangping Ma, Senior Member, IEEE, Tianyi Xu, Member, IEEE, and Gregory Sternberg, Senior Member, IEEE

Abstract—It is known that the problem of maximizing the sum degrees of freedom (DoF) for an arbitrary MIMO network without symbol extension is NP-hard in the number of transmitterreceiver pairs. The DoF achievability problem is also NP-hard if each transmitter/receiver has at least three antennas, and is polynomial-time solvable if each transmitter/receiver has no more than two antennas. It was conjectured that for the special case of the symmetric network (where all the transmitters have the same number of antennas and all the receivers have the same number of antennas), polynomial-time algorithms might exist for these two problems. In this paper, we show that the conjecture holds only in a very limited sense. We also show that for two important special cases of the symmetric network, both problems are polynomial-time solvable if the number of antennas at each transmitter/receiver is no more than two, and generally are NPhard otherwise. Index Terms—Computational complexity, NP-hard, MIMO, interference alignment.

I. I NTRODUCTION

I

NTERFERENCE alignment [1][2][3][4] recently has received much attention. The work of Cadambe and Jafar [2] shows that with symbol extension, interference alignment achieves a total degrees of freedom (DoF) that grows linearly with the number of users, where a user is defined as a transmitter-receiver pair. The critical assumption is that the extended symbols are transmitted over independently faded channels. However, in practice the channel diversity is limited, making this assumption hard to satisfy [5]. A more practical setting for interference alignment is MIMO networks without symbol extension. Assume that the number of users are K, transmitter k has Mk antennas, receiver k has Nk antennas, and user k achieves dk DoF, where k = 1, 2, . . . , K. It is shown in [6] that the maximization of the sum DoF problem for a network as such is NP-hard, and that the DoF achievability problem is NP-hard if the number of antennas at each transmitter/receiver is at least three and is polynomial-time solvable if the number of antennas at each transmitter/receiver is not more than two. It was conjectured in [6] that there might exist polynomial-time algorithms that solve these problems for some special cases, for example, the symmetric network (i.e., arbitrary channel matrices, Mk = M , Nk = N , ∀k = 1, . . . , K). In this paper, we show that the computational complexity results for the symmetric network are the same as those for the arbitrary MIMO network, except for the problem Manuscript received September 9, 2013. The associate editor coordinating the review of this letter and approving it for publication was J. Choi. L. Ma and T. Xu are with InterDigital Communications, Inc., San Diego, CA 92121, USA. {} G. Sternberg is with InterDigital Communications, Inc., King of Prussia, PA 19406, USA. Digital Object Identifier 10.1109/LCOMM.2013.111013.132057

of maximizing the sum DoF with M ≤ 2 and N ≤ 2, which is polynomial-time solvable. When the channel matrices are generic (being generic has a precise meaning as defined in [7] and can be understood as that the entries of a channel matrix are independent and drawn from continuous distributions), it is shown that easily checkable (and in fact polynomial-time) tight upper bounds on the achievable sum DoF are derived for (i) a symmetric network with the constraints that dk = d, ∀k = 1, . . . , K and that M and N are divisible by d [8], and for (ii) a fully symmetric network, where M = N and dk = d, ∀k = 1, . . . , K [9]. In this letter, we show that without the assumption that the channels are generic, the two problems for the two networks become NP-hard in most cases. This paper provides two guidelines to practical interference alignment: (1) for the problems proved to be NP-hard in this paper, it is impossible to find polynomial-time algorithms unless P=NP, and (2) for those proved polynomial-time solvable, polynomial-time algorithms are suggested. The remainder of this paper is organized as follows. Section II gives the system model, Section III presents the main results, and Section IV concludes the paper. II. S YSTEM M ODEL Consider a MIMO network or interference channel with K users or equivalently K transmitter-receiver pairs. Let Mk and Nk be the number of antennas at transmitter k and receiver k, respectively, ∀k = 1, . . . , K. The received signal at receiver k is expressed as yk =

K 

Hkj xj + zk ,

(1)

j=1

where Hkj ∈ CNk ×Mj is the channel matrix (which is arbitrary, not necessarily generic as defined in [7]) from transmitter j to receiver k, xj ∈ CMj ×1 is the signal vector transmitted by transmitter j, and zk ∈ CNp ×1 is a circular Gaussian noise vector with i.i.d. CN (0, 1) entries. We set xj = Vj sj , where Vj ∈ CMj ×dj is the transmit beamforming matrix at transmitter j such that Vj∗ Vj = Idj ×dj , where Vj∗ is the conjugate transpose of Vj , and sj ∈ Cdj ×1 the symbol vector transmitted by transmitter j. Let Uk ∈ CNk ×dk be the receive beamforming matrix at receiver k such that U∗k Uk = Idk ×dk . Then, with transmit beamforming and receive beamforming, the received signal at receiver k in (1) is transformed to U∗k yk =

c 2013 IEEE 1089-7798/13$31.00 

K  j=1

U∗k Hkj Vj sj + U∗k zk .

(2)

MA et al.: COMPUTATIONAL COMPLEXITY OF INTERFERENCE ALIGNMENT FOR SYMMETRIC MIMO NETWORKS

Two interference alignment problems [6] are considered in this paper. The first problem is to maximize the sum DoF, namely, K 

max

{Vk ,Uk }K k=1

s.t.

k=1

(3)

dk U∗k Hkj Vj

rank(U∗k Hkk Vk )

=

0dk ×dj , ∀j = k

=

dk , ∀k = 1, . . . , K. (5)

(4)

The second problem is the achievability of a given DoF vector (d1 , . . . , dK ): find {Vk , Uk }K k=1 that achieves (d1 , . . . , dK )

(6)

subject to the constraints (4)–(5). Remark II.1. There is a connection between the sum DoF maximization problem and the DoF achievability problem, by noting that the former can be solved by solving a series of the latter, each for a particular DoF tuple (d1 , . . . , dK ). We will use this connection to prove Theorem III.9. III. M AIN R ESULTS We will reuse some of the results from [6]. For convenience, we quote them in the following. Lemma III.1. (Theorem 1 of [6]) The DoF maximization problem in (3) is NP-hard. Moreover, the DoF achievability problem in (6) is NP-hard if each transmitter/receiver has at least 3 antennas. Lemma III.2. (Theorem 2 of [6]) The DoF achievability problem in (6) is polynomial-time solvable if each transmitter/receiver has at most 2 antennas. A. Symmetric MIMO Network For a symmetric MIMO network, Mk = M, Nk = N , and there is no restriction on dk ’s [6]. We define the following interference alignment problems: • A Max sum DoF(M, N, K) problem is the DoF maximization problem in (3) with Mk = M and Nk = N , ∀k = 1, . . . , K; • A DoF achievability(M, N, [d1 , . . . , dK ], K) problem is the DoF achievability problem in (6) with Mk = M and Nk = N , ∀k = 1, . . . , K. Imposing additional constraints on the problems defined above will give the interference alignment problems for two important special symmetric MIMO networks, as we will see later in the paper. For the general symmetric MIMO network here, we have the following results. Theorem III.3. The Max sum DoF(M, N, K) problem is NP-hard in K. Proof: In the proof of the first part of Theorem 1 in [6] (which is Lemma III.1 in this paper), it is shown that the maximum independent set problem for an arbitrary graph is polynomial-time reducible to an interference alignment problem where each user has a single antenna. The interference alignment problem is also an instance of the interference alignment problem of symmetric MIMO networks. Since

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the maximum independent set problem is NP-complete, the Max sum DoF(M, N, K) problem is NP-hard. Remark III.1. In the above theorem, the problem of maximizing the sum DoF is NP-hard in K for any combination of M ≥ 1 and N ≥ 1 (in the sense that there exists a set of channel matrices that renders the problem NP-hard). This is because first, by the proof the problem is NP-hard for M = 1 and N = 1. Second, we can construct a (M  , N  ) network out of the (M = 1, N = 1) network, where M  ≥ 1 and N  ≥ 1, by adding auxiliary antennas with zero channel gains such that each transmitter has M  antennas and each receiver has N  antennas. Since the problem is NP-hard for M = 1 and N = 1, the problem is also NP-hard for any M  and N  . The other theorems of NP-hardness in the paper should be understood similarly as similar arguments can be made. Theorem III.4. The DoF achievability(M, N, [d1 , . . . , dK ], K) problem is NP-hard in K if M ≥ 3 and N ≥ 3. Proof: In the proof (Appendix A) of Theorem 1 in [6], an interference alignment problem (DoF achievability problem in (6) with dk = 1, ∀k = 1, . . . , K) is constructed for any given 3-colorability problem such that the 3-colorability problem is polynomial-time reducible to the DoF achievability problem. Because the 3-colorability problem is NP-complete, so the DoF achievability problem with dk = 1 is also NP-complete and the more general DoF achievability problem in (6) is NPhard. In the constructed network (called Network-1) in [6], there are two types of users: main users and dummy users. Each main transmitter/receiver has 3 antennas, and each dummy transmitter/receiver has either 2 or 3 antennas. We modify the constructed network by adding 1 more antenna to each dummy transmitter/receiver that has 2 antennas, resulting in a network (called Network-2) where each transmitter/receiver has 3 antennas. We claim that the DoF achievability problem for Network-2 is NP-complete, because firstly, Network-1 is a special case of Network-2, i.e., a Network-2 with one of the antennas in the dummy transmitter/receiver turned off (or equivalently the channel gains for all the scalar channels associated with that antenna being zero), and secondly, the DoF achievability problem for Network-1 is NP-complete. Thus, the DoF achievability(3, 3, [1, . . . , 1], K) problem is NP-hard in K. Therefore, the more general problem, namely, DoF achievability(M, N, [d1 , . . . , dK ], K) with M ≥ 3 and N ≥ 3, is NP-hard. Theorem III.5. The DoF achievability(M, N, [d1 , . . . , dK ], K) problem is polynomial-time solvable if M ≤ 2 and N ≤ 2. Proof: This is a direct consequence of Lemma III.2. Remark III.2. Comparing Theorems III.3 through III.5 with Lemmas III.1 and III.2, we see that the computational complexity results for symmetric MIMO networks are the same as those for the general MIMO network. B. Symmetric Arbitrary MIMO Network with dk = d, and M and N Being Divisible by d We now consider a symmetric arbitrary MIMO network with the restrictions that dk = d, ∀k = 1, . . . , K and that

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M and N are divisible by d. This case is important because, with the assumption of generic channel matrices, it is one of the few cases for which a tight upper bound on the achievable sum DoF and in fact a polynomial-time solution are known [8]. For this case, we define the following interference alignment problems: • A Max sum DoF d div(M, N, K) problem is the Max sum DoF(M, N, K) problem defined in Section III-A with the restrictions that dk = d, ∀k = 1, . . . , K and that M and N are divisible by d; • A DoF achievability d div(M, N, d, K) problem is the DoF achievability(M, N, [d1 , . . . , dK ], K) problem defined in Section III-A with the restrictions that dk = d, ∀k = 1, . . . , K and that M and N are divisible by d. We first consider the DoF achievability problem with M ≥ 3 and N ≥ 3. Theorem III.6. The DoF achievability d div(M, N, d, K) problem with M ≥ 3 and N ≥ 3 is NP-hard in K. Proof: This is clear by noting that any integer is divisible by 1 and that in the proof of Theorem III.4, the DoF achievability problem for the constructed network Network-2 is to check if each user can achieve DoF= 1, which is also an instance of the DoF achievability d div(M, N, d, K) problem with M ≥ 3 and N ≥ 3. The rest of the proof follows the same line of reasoning as in the proof of Theorem III.4. We next consider the maximization of the sum DoF problem with M ≥ 3 and N ≥ 3. Theorem III.7. The Max sum DoF d div(M, N, K) problem with M ≥ 3 and N ≥ 3 is NP-hard in K. Proof: For the Max sum DoF d div(M, N, K) problem, by definition dk = d. Therefore, when the sum DoF is maximized, so is the individual DoF d. Let d∗ be the maximum value for d (the result of the Max sum DoF d div(M, N, K) problem divided by K). Consider the DoF achievability d div(M, N, d∗ , K) problem with M ≥ 3 and N ≥ 3. We can immediately solve the DoF achievability d div(M, N, d∗ , K) problem by solving the Max sum DoF d div(M, N, K) problem. Hence the former is no harder than the latter. Since the former is NP-hard by Theorem III.6, so is the latter. Remark III.3. The method used in the proof is similar to what is used to show that a decision problem is no harder than an optimization problem, say, in [10, pp968-969]. Remark III.4. Here we do not modify the proof of the first part of Theorem 1 in [6] as we have done in the proof of Theorem III.3, because in the maximization of the sum DoF problem constructed in [6] different users may have different DoFs (more specifically either 0 or 1), whereas in Theorem III.7 all users are required to have the same DoF. We now consider the case where the number of antennas at each transmitter/receiver is no more than 2. Note that for a symmetric network, the cases where some transmitters (or receivers) have no more than 2 antennas while other transmitters (or receivers) have at least 3 antennas are not valid by definition.

IEEE COMMUNICATIONS LETTERS, VOL. 17, NO. 12, DECEMBER 2013

Theorem III.8. The DoF achievability d div(M, N, d, K) problem with M ≤ 2 and N ≤ 2 is polynomial-time solvable in K. Proof: Again, this is a direct consequence of Lemma III.2. Theorem III.9. The Max sum DoF d div(M, N, K) problem with M ≤ 2 and N ≤ 2 is polynomial-time solvable in K. Proof: We show that the interference alignment problem in the theorem is at most twice as hard as the DoF achievability d div(M, N, d, K) problem. Let d∗ be the maximum value of d (the result of the Max sum DoF d div(M, N, K) problem divided by K). Since M ≤ 2 and N ≤ 2, the only possible values for d∗ are 0, 1, 2. In the first step, according to Theorem III.8, there exists a polynomial-time algorithm that can decide whether a solution to the DoF achievability d div(M, N, 2, K) problem exists. If the answer is yes, then the same solution solves the Max sum DoF d div(M, N, K) problem and yields d∗ = 2. Otherwise, we go to the second step, where there exists a polynomial-time algorithm that can decide whether a solution to the DoF achievability d div(M, N, 1, K) problem exists. If the answer is yes, then the same solution solves the Max sum DoF d div(M, N, K) problem and yields d∗ = 1. Otherwise, d∗ = 0. Remark III.5. In the proof of the first part of Theorem 1 in [6], we can generalize the constructed interference network to one with Mk = M and Nk = N . Therefore, the DoF maximization problem in (3) is NP-hard even if Mk = M ≤ 2 and Nk = N ≤ 2. By Theorem III.9, if we force dk = d, then the problem becomes polynomial-time solvable. Thus, it is the variability of dk that results in the NP-hardness of the DoF maximization problem in (3) with Mk = M ≤ 2 and Nk = N ≤ 2. There is an intuitive explanation for this. To illustrate, consider the case where M = N = 2. If we use the decision problem to solve the optimization problem as described in Remark II.1, the number of decision problems will be equal to the number of DoF vectors (d1 , . . . , dK ), dk = 0, 1, 2, ∀k = 1, . . . , K, which is equal to 3K and thus not polynomial in K. Remark III.6. It is shown in [6] that the DoF achievability(M, N, [d1 , . . . , dK ], K) problem with M ≤ 2 and N ≤ 2 is equivalent to a 2-SAT problem, which is solvable in O(K 4 ) time. Therefore, both problems in Theorems III.8 and III.9 can be solved in O(K 4 ) time. Additionally, the proofs suggest polynomial-time algorithms. Remark III.7. Comparing Theorems III.6 through III.9 with Theorems III.3 through III.5, we see that the addition of the constraints that dk = d, ∀k = 1, . . . , K and that M and N are divisible by d only changes the computational complexity result for the maximization of the sum DoF problem with M ≤ 2 and N ≤ 2, which is evident by comparing Theorem III.9 with Theorem III.3. C. Fully Symmetric Arbitrary MIMO Networks In a fully symmetric arbitrary MIMO network, Mk = Nk = M , dk = d, ∀k = 1, 2, . . . , K. The interference alignment

MA et al.: COMPUTATIONAL COMPLEXITY OF INTERFERENCE ALIGNMENT FOR SYMMETRIC MIMO NETWORKS

problem for a fully symmetric MIMO network is an important case because, again with the assumption of generic channel matrices, it is another one of the few cases for which a tight upper bound on the achievable sum of DoF and a polynomial time solution are known [9]. Note that a fully symmetric MIMO network is not a special case of the network described earlier in Section III-B due to the lack of constraint on the divisibility of M by d. Nevertheless, the proofs of the results for a fully symmetric MIMO network are very similar to those given in Section III-B due to the fact that the constructed interference alignment problems can be generalized to either case. We define the following interference alignment problems: • A Max sum DoF fully(M, K) problem is the Max sum DoF(M, N, K) problem defined in Section III-A with the restrictions that dk = d, ∀k = 1, . . . , K and that M = N ; • A DoF achievability fully(M, d, K) problem is the DoF achievability(M, N, [d1 , . . . , dK ], K) problem defined in Section III-A with the restrictions that dk = d, ∀k = 1, . . . , K and that M = N . Theorem III.10. The DoF achievability fully(M, d, K) problem with M ≥ 3 is NP-hard in K. Proof: This is clear by noting that in the proof of Theorem III.4, each transmitter/receier in the constructed network Network-2 has 3 antennas and the DoF achievability problem is to check whether each user can achieve a DoF equal to 1, which is also an instance of the DoF achievability fully(M, d, K) problem with M ≥ 3. The rest of the proof follows the same line of reasoning as that in the proof for Theorem III.4. Theorem III.11. The Max sum DoF fully(M, K) problem with M ≥ 3 is NP-hard in K. Proof: The proof follows the same line of reasoning as in the proof for Theorem III.7 except that we now consider the connection between the Max sum DoF fully(M, K) problem and the DoF achievability fully(M, d, K) problem. Theorem III.12. The DoF achievability fully(M, d, K) problem with M ≤ 2 is polynomial-time solvable in K. Proof: Again, this is a direct consequence of Lemma III.2. Theorem III.13. The Max sum DoF fully(M, K) problem with M ≤ 2 is polynomial-time solvable in K. Proof: The proof follows the same line of reasoning as in the proof for Theorem III.9 except that we now solve the Max sum DoF fully(M, K) problem by solving at most two

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instances of the DoF achievability fully(M, d, K) problem. Remark III.8. Comparing Theorems III.10 through III.13 with Theorems III.6 through III.9, we see that the computational complexity results for fully symmetric networks are the same as those for the symmetric networks with the constraints that dk = d, ∀k = 1, . . . , K and that M and N are divisible by d. IV. C ONCLUSION We consider the conjecture that there might exist polynomial-time algorithms that solve the maximization of the sum DoF problem or the DoF achievability problem for symmetric MIMO networks, and show that the conjecture holds only in a very limited sense. In particular, both problems are polynomial-time solvable if the number of antennas at each transmitter/receiver is no more than two, and generally are NP-hard otherwise. By requiring the DoF of each user to be the same, the maximization of the sum DoF problem for a symmetric network changes from a NP-hard problem to a polynomial-time solvable one for the case where each transmitter/receiver has no more than two antennas. We also consider the complexity of two special symmetric MIMO networks, and show that without the assumption of generic channels, both problems become NP-hard in most cases. R EFERENCES [1] M. Maddah-Ali, A. Motahari, and A. Khandani, “Communication over MIMO X channels: interference alignment, decomposition, and performance analysis,” IEEE Trans. Inf. Theory, vol. 54, pp. 3457–3470, Aug. 2008. [2] V. R. Cadambe and S. A. Jafar, “Interference alignment and degrees of freedom of the k-user interference channel,” IEEE Trans. Inf. Theory, vol. 54, no. 8, pp. 3425–3441, Aug. 2008. [3] V. Cadambe and S. Jafar, “Interference alignment and the degrees of freedom of wireless X networks,” IEEE Trans. Inf. Theory, vol. 55, pp. 3893–3908, Sept. 2009. [4] K. Gomadam, V. Cadambe, and S. Jafar, “A distributed numerical approach to interference alignment and applications to wireless interference networks,” IEEE Trans. Inf. Theory, vol. 57, pp. 3309–3322, June 2011. [5] G. Bresler and D. Tse, “Degrees-of-freedom for the 3-user Gaussian interference channel as a function of channel diversity,” in 2009 Allerton Conference on Communication, Control, and Computing. [6] M. Razaviyayn, M. Sanjabi, and Z. Luo, “Linear transceiver design for interference alignment: complexity and computation,” IEEE Trans. Inf. Theory, vol. 58, pp. 2896–2910, May 2012. [7] C. M. Yetis, T. Gou, S. A. Jafar, and A. H. Kayran, “On feasibility of interference alignment in MIMO interference networks,” IEEE Trans. Signal Process., vol. 58, no. 9, pp. 4771–4782, Sep. 2010. [8] M. Razaviyayn, G. Lyubeznik, and Z. Luo, “On the degrees of freedom achievable through interference alignment in a MIMO interference channel,” IEEE Trans. Signal Process., vol. 60, pp. 812–821, Feb. 2012. [9] G. Bresler, D. Cartwright, and D. Tse, “Settling the feasibility of interference alignment for the MIMO interference channel: the symmetric case,” arXiv:1104.0888v1, Apr. 2011. [10] T. H. Cormen, C. E. Leiserson, R. L. Rivest, and C. Stein, Introduction to Algorithms, 2nd ed. The MIT Press, 2007.