Capacity of a Two-way Function Multicast Channel

Transcription

1 Capacity of a Two-way Function Multicast Channel 1 Seiyun Shin, Student Member, IEEE and Changho Suh, Member, IEEE Abstract We explore the role of interaction for the problem of reliable computation over two-way multicast networks. Specifically we consider a four-node network in which two nodes wish to compute a modulosum of two independent Bernoulli sources generated from the other two, and a similar task is done in the other direction. The main contribution of this work lies in the characterization of the computation capacity region for a deterministic model of the network via a novel transmission scheme. One consequence of this result is that, while we can get an interaction gain over the one-way non-feedback computation capacities, we can sometime get all the way to perfect-feedback computation capacities simultaneously in both directions. This result draws a parallel with the recent result developed in the context of two-way interference channels [1]. Index Terms Computation Capacity, Interaction, Network Decomposition, Perfect-feedback, Two-way Function Multicast Channel I. INTRODUCTION The inherent two-way nature of communication links provides an opportunity to enable interaction among nodes. It allows the nodes to efficiently exchange their messages by adapting their transmitted signals to the past received signals that can be fed back through backward communication links. This problem was first studied by Shannon in []. However, we are still lacking in our understanding of how to treat two-way information exchanges, and the underlying difficulty has impeded progress on this field over the past few decades. Since interaction is enabled through the use of feedback, feedback is a more basic research topic that needs to be understood beforehand. The history of feedback traces back to Shannon who showed that feedback has no bearing on capacity for memoryless point-to-point channels [3]. Subsequent work

2 demonstrated that feedback provides a gain for point-to-point channels with memory [4], [5] as well as for many multi-user channels [6] [8]. For many scenarios, however, capacity improvements due to feedback are rather modest. On the contrary, one notable result in [9] has changed the traditional viewpoint on the role of feedback. It is shown in [9] that feedback offers more significant capacity gains for the Gaussian interference channel. Subsequent works [1], [10], [11] show more promise on the use of feedback. In particular, [1] demonstrates a very interesting result: Not only feedback can yield a net increase in capacity, but also we can sometime get perfect-feedback capacities simultaneously in both directions. We seek to examine the role of feedback for more general scenarios in which nodes now intend to compute functions of the raw messages rather than the messages themselves. These general settings include many realistic scenarios such as sensor networks [1] and cloud computing scenarios [13], [14]. For an idealistic scenario where feedback links are perfect with infinite capacities and are given for free, Suh-Gastpar [15] have shown that feedback provides a significant gain also for computation. However, the result in [15] assumes a dedicated infinite-capacity feedback link as in [9]. As an effort to explore a net gain that reflects feedback cost, [16] investigated a two-way setting of the function multicast channel considered in [15] where two nodes wish to compute a linear function modulo-sum of the two Bernoulli sources generated from the other two nodes. The two-way setting includes a backward computation demand as well, thus well capturing feedback cost. A scheme is proposed to demonstrate that a net interaction gain can occur also in the computation setting. However, the maximal interaction gain is not fully characterized due to a gap between the lower and upper bounds. In particular, whether or not one can get all the way to perfect-feedback computation capacities in both directions as in the two-way interference channel [1] has been unanswered. In this work, we characterize the computation capacity region of the two-way function multicast channel via a new capacity-achieving scheme. In particular, we consider a deterministic model [17] which well captures key properties of the wireless Gaussian channel. As a result, we answer the above question positively. Specifically, we demonstrate that for some channel regimes to be detailed later; see Corollary 1, the new scheme simultaneously achieves the perfect-feedback computation capacities in both directions. As in the two-way interference channel [1], interestingly, this occurs even when feedback offers gains in both directions and thus feedback w.r.t. one direction must compete with the traffic in the other direction. Our achievability builds upon the scheme in [1] where feedback allows the exploitation of effectively future information as side information via retrospective decoding to be detailed later; see Remark.

3 3 Fig. 1. Four-node ADT deterministic network. A key distinction relative to [1] is that in our computation setting, the retrospective decoding occurs in a nested manner for some channel regimes; this will be detailed when describing our achievability. We also employ network decomposition in [18] for ease of achievability proof. II. MODEL Consider a four-node Avestimehr-Diggavi-Tse ADT deterministic network as illustrated in Fig. 1. This network is a full-duplex bidirectional system in which all nodes are able to transmit and receive signals simultaneously. Our model consists of the forward and backward channels which are assumed to be orthogonal. For simplicity, we focus on a setting in which both forward and backward channels are symmetric but not necessarily the same. In the forward channel, n and m indicate the number of signal bit levels or resource levels for direct and cross links respectively. The corresponding values for the backward channel are denoted by ñ, m. With N uses of the network, node k k = 1, wishes to transmit its own message Sk K, while node k k = 1, wishes to transmit its own message S K k. We assume that S1 K, SK, S K are

4 4 independent and identically distributed according to Bern 1. Here we use shorthand notation to indicate the sequence up to K or K, e.g., S K 1 k where q = maxm, n and V k F m := S 11,, S 1K. Let X k F q be an encoded signal of node be part of X k visible to node j k. Similarly let Xk F q be an encoded signal of node k where q = max m, ñ and Ṽk be part of X k visible to node j k. The received signals at node k and k are then given by Y 1 =G q n X 1 G q m X, Y = G q m X 1 G q n X, 1 Ỹ 1 = G q ñ X1 G q m X, Ỹ = G q m X1 G q ñ X, where G and G are shift matrices and operations are performed in F : [G] ij = 1 {i = j + 1} 1 i, j q, [ G] ij = 1 {i = j + 1} 1 i, j q. The encoded signal X ki of node k at time i is a function of its own message and past received signals: X ki = f ki S1 K, Ỹ i 1 k. We define Ỹ i 1 k := {Ỹkt} t=1 i 1 where Ỹkt denotes node k s received signal at time t. Similarly the encoded signal Xki of node k at time i is a function of its own message and past received sequences: Xki = f ki S K k, Y i 1 k. From the received signal Y N k, node k wishes to compute modulo- sums of S K 1 and SK i.e., {S 1i S i } K i=1. Similarly node k wishes to compute { S 1j S j } K j=1 from its received signals Ỹ k N. We say that a computation rate pair R, R is achievable if there exists a family of codebooks and encoder/decoder functions such that the decoding error probabilities go to zero as code length N tends to infinity. Here R := K N and R := K N. The capacity region C is the closure of the set of achievable rate pairs. III. MAIN RESULTS Theorem 1 Two-way Computation Capacity: The computation capacity region C is the set of R, R such that R C pf, 3 R C pf, 4 R + R m + m, 5 R + R n + ñ, 6 where C pf and C pf indicate the perfect-feedback computation capacities in the forward and backward channels respectively see 9 and 10 in Baseline for detailed formulas. Proof: See Section IV and V for the achievability and the converse proofs respectively.

5 5 For comparison to our result, we state two baselines: 1 The capacity region for the non-interactive scenario in which there is no interaction among the signals arriving from different nodes; the capacity for the perfect-feedback scenario in which feedback is given for free to aid computations in both direction. Baseline 1 Non-interaction Computation Capacity [18]: Let α := m n and α := m ñ. The computation capacity region C no for the non-interactive scenario is the set of R, R such that R C no and R C no where C no = C no = min { m, 3 n}, α < 1, min { n, 3 m}, α > 1, n, α = 1, min { m, 3ñ}, α < 1, min { ñ, 3 m}, α > 1, ñ, α = Here C no and C no denote the non-feedback computation capacities of forward and backward channels respectively. Baseline Perfect-feedback Computation Capacity [15]: The computation capacity region C pf the perfect-feedback scenario is the set of R, R such that R C pf and R C pf where 3n, α < 1, C pf = 3m, α > 1, n, α = 1, 3ñ, α < 1, C pf = 3 m, α > 1, ñ, α = 1. With Theorem 1 and Baseline 1, one can readily see that feedback offers a gain in terms of capacity region as long as α / [ 3, 3 ], α / [ 3, 3 ]. A careful inspection reveals that there are channel regimes in which one can enhance C no or C no without sacrificing the other counterpart. This implies a net interaction gain. Definition 1 Interaction Gain: We say that an interaction gain occurs if one can achieve R, R = C no + δ, C no + δ for some δ 0 and δ 0 such that maxδ, δ > 0. A tedious yet straightforward calculation with this definition leads us to identify channel regimes which exhibit an interaction gain. See the regimes marked in light blue in Fig.. for 9 10

6 6 Interaction gain No interaction gain No feedback gain Interaction gain No interaction gain Perfectfeedback Perfectfeedback Fig.. Gain-vs-nogain picture: The plot is over two key parameters: α and α, where α is the ratio of the interference-to-noise ratio in db to the signal-to-noise ratio in db of the forward channel and α is the corresponding quantity of the backward channel. The parameter γ is the ratio of the backward signal-to-noise ratio in db to the forward signal-to-noise ratio in db, and is fixed to be a value greater than or equal to 1 in the plot. Dark pink region: feedback does not increase capacity in either direction and thus interaction is not useful. Light pink: feedback does increase capacity but interaction cannot provide such increase. Light blue: feedback can be offered through interaction and there is a net interaction gain. Dark blue: interaction is so efficient that one can achieve perfect-feedback capacities simultaneously in both directions. We also find the regimes in which feedback does increase capacity but interaction cannot provide such increase, meaning that whenever δ > 0, δ mush be δ and vice versa. The regimes are α < 3, α < 3 and α > 3, α > 3. One can readily check that this follows from the cut-set bounds 5 and 6. Achieving perfect-feedback capacities: It is noteworthy to mention that there exist channel regimes in which both δ and δ can be strictly positive. This implies that for the regimes, not only feedback does not sacrifice one transmission for the other, it can actually improve both simultaneously. As in the two-way interference channel [1], this occurs for the computation setting as well: more interestingly, the gains δ and δ can reach up to the maximal feedback gains, reflected in C pf C no and C pf C no respectively. The dark blue regimes in Fig. indicate such channel regimes when γ := ñ n 1. Note that such regimes depend on γ. The amount of feedback that one can send is limited by available resources, which is affected by the channel asymmetry parameter γ.

7 7 The following corollary identifies channel regimes in which achieving perfect-feedback capacities in both directions is possible. Corollary 1: Consider a case in which feedback helps for both forward and backward channels: C pf > C no and C pf > C no. Under the case, the channel regimes in which C = C pf are: I 0 α 3, α > 3 { Cpf C no m C pf,},, C pf C no n C pf II α > 3, 0 α { Cpf C no ñ C pf,} 3,. C pf C no m C pf Proof: A tedious yet straightforward calculation with Theorem 1 completes the proof. Remark 1 Why the Perfect-feedback Regimes?: The rationale behind achieving perfect-feedback capacities in both directions bears a resemblance to the one found in the two-way interference channel [1]: Interaction enables full-utilization of available resources, whereas the dearth of interaction limits that of those. We will elaborate this for the considered regime in Corollary 1 : 0 α 3, α > 3. We first note that the total number of available resources for the forward and backward channels depend on n and m in this regime. In the non-interaction case, observe from Baseline 1 that some resources are under-utilized; specifically one can interpret n C no and m C no as the remaining resource levels resource holes that can potentially be utilized to aid function computations. It turns out feedback can maximize resource utilization by filling up such resource holes under-utilized in the non-interactive case. Note that C pf C no represents the amount of feedback that needs to be sent for achieving C pf. Hence, the condition C pf C no m C pf similarly C pf C no n C pf in Corollary 1 implies that as long as we have enough resource holes, we can get all the way to perfectfeedback capacity. We will later provide an intuition as to why feedback can do so while describing our achievability; see Remark in particular. IV. PROOF OF ACHIEVABILITY Our achievability proof consists of three parts. We will first provide two achievable schemes for two toy examples in which the key ingredients of our achievability idea are well presented. Once the description of the two schemes is done, we will then outline the proof for generalization while leaving the detailed proof in Appendix A.

8 8 A. Example: m, n = 1,, m, ñ =, 1 We first review the perfect-feedback scheme [15], which we will use as a baseline for comparison to our achievable scheme. It suffices to consider the case of m, n = 1,, as the other case of m, ñ =, 1 follows similarly by symmetry. 1 Perfect-feedback strategy: The perfect-feedback scheme for m, n = 1, consists of two stages; the first stage has two time slots; and the second stage has one time slot. See Fig. 3. Observe that the bottom level at each receiving node naturally forms a modulo- sum function F l, say F l := a l b l where a l or b l denotes a source symbol of node 1 or. For ease of presentation, we use a simple notation F l. In the first stage, we send forward symbols at node 1 and. In time 1, node 1 sends a 1, a ; similarly, node sends b, b 1. Then node 1 gets F := a b ; and node gets F 1 := a 1 b 1. As in the first time slot, node 1 and deliver a 3, a 4 and b 4, b 3 respectively at time. Then node 1 and obtain F 4 and F 3 respectively. Note that until the end of time, F 1, F 3 are not delivered yet to node 1. Similarly F, F 4 are missing at node. Feedback can however accomplish the computation of these interested functions. With feedback, each transmitting node can now obtain the desired functions which were obtained only at one receiving node. Exploiting a feedback link from node to node 1, node 1 can get F 1, F 3. Similarly, node can get F, F 4 from node 1. Now the strategy in Stage is to forward all of these fed-back functions at time 3. Node 1 then gets F 1 cleanly on the top level. On the bottom level, it gets a mixture of the two desired functions: F 3 F. Note that F in the mixture was already obtained at time 1. Hence using F, node 1 can decode F 3. Similarly, node can obtain F, F 4. In summary, node 1 and can compute four modulo- sum functions during three time slots, thus achieving R = 4 3 = C pf. In our model, however, feedback is provided in the limited fashion since feedback signals are delivered only through the backward channel. There are two different types of transmissions for using the backward channel. The channel can be used 1 for backward-message computation, or for sending feedback signals. Usually, unlike the perfect-feedback case, the channel use for one purpose limits that for the other, and this tension incurs a new challenge. We develop an achievable scheme that can completely resolve the tension, thus achieving the perfect-feedback performance. Achievability: Like the perfect-feedback case, our scheme consists of two stages. The first stage consists of L time slots; and the second stage consists of L time slots. During the first stage, the number 4L and 4L 1 of fresh symbols are transmitted through the forward and backward channels respectively. And no fresh symbols are transmitted in the second stage. In this example, we claim that the following

9 9 Fig. 3. A perfect-feedback scheme for m, n = 1, model. rate pair is achievable: R, R = 4L 3L, 4L 1 3L = 4 3, 4L 4 3L. In other words, during the total 3L time slots, our scheme ensures 4L and 4L 4 forward and backward-message computations. As L, we get the desired result: R, R 4 3, 4 3 = C pf, C pf. Stage 1 : In this stage, each node superimposes fresh symbols and feedback symbols. Details are as follows. Time 1 & : Node 1 sends a 1, a ; and node sends b, b 1. Node 1 and then get a 1, F and b, F 1 respectively. Now observe that F 1 and F are not delivered yet to node 1 and respectively. In an attempt to satisfy these demands, note that the perfect-feedback strategy is to feed back F from node 1 to node, and to feed back F 1 from node to node 1. A similar feedback strategy is employed in our backward channel. A distinction is that fresh backward symbols are superimposed. Specifically node 1 delivers ã F, ã 1 a 1, the mixtures of the fresh symbols ã, ã 1 and the feedback signals F, a 1. Similarly node delivers b 1 F 1, b b. Note that unlike the perfect-feedback strategy, not only F on top is transmitted, but also a 1 on bottom is transmitted. And these two signals F, a 1 are exactly the ones received at time 1. Node 1 and then receive b 1 F 1, F a and ã F, F 1 b 1 respectively. Notice here that exploiting its own symbols, each node can obtain the interested function. Node 1 can obtain F from F a by exploiting its own symbol a. Similarly, node can obtain F 1. In time, we repeat this w.r.t. new symbols. As a result, node 1 and get a 3, F 4 and b 4, F 3. respectively, while node 1 and get b 3 F 3, F 4 a 4 and ã 4 F 4, F 3 b 3. Similar to the first time slot, node 1 and utilize their own symbols as side information to obtain F 4 and F 3 respectively.

10 10 Time l: For time l = 3,, L, the transmission signals at node 1 and are: node 1 : a l 1 F + l a l, 11 a l bl 1 F l 1 a l 1 ã l 4 node : b l F + l 1 b l 1 b l 1 ã l F l b l b. 1 l 4 1 Similarly, for time l = 3,, L, node 1 and deliver: node 1 : ãl ã l 1 + F l ã l 1 a l 1 F l a l ã l F l, 13 node : b l 1 F + l 1 b l bl b l F l 1 b l 1 b l 1 F l Note that the transmitted signal of each node consists of two parts: Fresh symbols, e.g., a l 1, a l at node 1; and feedback signals, e.g., F l a l, b l 1 F l 1 a l 1 ã l 4. We note that ã l 4 and b l 4 1 in 11 and 1 respectively is valid from time 5. For the last two time slots, node 1 and do not send any fresh backward symbols. Instead, they mimic the perfect-feedback scheme: In time l, l = L 1, L, node 1 feeds back F l on top; and node feeds back F l 1 on top. Note that until time L, a total of 4L forward symbols are delivered a l 1, a l, b l 1, b l, for l = 1,, L. Similarly, a total of 4L 1 backward symbols are delivered. One can readily check that node 1 and can obtain {F l } L l=1 and {F l 1} L l=1 respectively. Similarly, node 1 and can obtain { F l } L 1 l=1 and { F l 1 } L 1 l=1 respectively. Recall that among the total 4L and 4L 4 forward and backward functions, {F l 1 } L l=1 and {F l} L l=1 are not delivered yet to node 1 and respectively. Similarly { F l 1 } L 1 l=1 and { F l } L 1 l=1 are missing at node 1 and respectively. For ease of understanding, Fig. 4 illustrates a simple case of L =. In time 3, node 1 gets a 5 F a, F 6 ã 1 ; and node gets b 6 F 1 b 1, F 5 b. Note that using their own symbols ã 1 and b, node 1 and can obtain F 6 and F 5 respectively. In time 4, we repeat the same w.r.t. new symbols. As a result, node 1 and obtain a 7 F 4 a 4, F 8 and b 8 F 3 b 3, F 7. In the last two time slots time 3 and 4, node 1 and get F 5, F 7 and F 6, F 8 respectively. Stage : During the next L time slots in the second stage, we accomplish the computation of desired functions that each node did not obtain yet. Recall that the transmission strategy in the perfect-feedback scenario is simply to forward all of the received signals at each node. And the received signals are in the form of modulo- sum functions of interest see Fig. 3. In our model, however, the received signals

11 11 Fig. 4. An achievable scheme for m, n = 1,, m, ñ =, 1, and L =. include symbols generated from the other-side nodes. For instance, the received signal at node 1 in time 1 is b 1 F 1, which contains the backward symbol b 1. Hence, unlike the perfect-feedback scheme, forwarding the signal directly from node 1 to node 1 is not guaranteed for node 1 to decode the desired function F 1. To address this, we take a recently developed approach [1]: Retrospective decoding. The key feature of this approach is that the successive refinement is done in a retrospective manner and this leads us to resolve the issue mentioned above. The outline of the strategy is as follows: Node 1 and start to decode F 4L 3, F 4L 1 and F 4L, F 4L respectively. Here one key point to emphasize is that these decoded functions act as side information. And it turns out that this information enables the other-side nodes to

12 1 obtain the desired functions w.r.t. the past symbols. Specifically the decoding order reads: F 4L 3, F 4L, F 4L 1, F 4L F 4L 1 3, F 4L 1, F 4L 1 1, F 4L 1 F 5, F 6, F 7, F 8 F 1, F, F 3, F 4 F 1, F, F 3, F 4. With the refinement in time L + l l = 1,, L i.e., the lth time of Stage, node 1 and can decode: node 1 : F 4L l 1 3, F 4L l 1 1, node : F 4L l 1, F 4L l 1. Subsequently, node 1 and decode: node 1 : F 4L l 3, F 4L l 1 F 4L l+1 3, node : F 4L l, F 4L l F 4L l+1. Note that after one more refinement in time L + l + 1, F4L l+1 3 and F 4L l+1 in the above can be canceled out at node 1 and, so they can finally decode the interested functions. Specifically, the enabling transmission strategy is as follows: Time L + 1: Taking the perfect-feedback strategy for F 4L 3, F 4L 1, F 4L, F 4L, one can readily see that node 1 and can decode F 4L 3, F 4L 1 and F 4L, F 4L respectively. Time L + l: With newly decoded functions in time L + l 1, l =,, L, the transmission signals at node 1 and are: F 4L l 1 3 node 1 : F 4L l 1 1 F + 4L l 3 F 4L l b 4L l 1 3 F 4L l 1 3 b 4L l b4l l 1 1 F 4L l 1 1 b 4L l + F 4L l F, 15 4L l 1 F 4L l 1 F node : F 4L l 1 F 4L l 1 ã 4L l 3 + ã4l l 1 4L l ã 4L l 1 F 4L l 1 ã 4L l 1 F 4L l 3 + F 4L l 1 F. 16 4L l 1 3 Notice that the signals in the first bracket are newly decoded functions; those in the second bracket are received those at time L l 1 1, L l 1 on top; and those in the third bracket are modulo- sum functions decoded at Stage 1 e.g., even-index functions for node 1. This transmission

13 13 leads node 1 and to decode F 4L l 1 3, F 4L l 1 1 and F 4L l 1, F 4L l 1 using their own symbols and previously decoded functions. Similarly, for time L + l, l = 1,, L, node 1 and deliver: node 1 : F 4L l a 4L l 1 3 F 4L l a 4L l F 4L l 1 1 a 4L l 1 1 F 17 4L l a 4L l F 4L l + F 4L l F 4L l 1 a 4L l 3 F, 4L l+1 a 4L l+1 F 4L l+1 node : F 4L l 1 + b 4L l 1 F 4L l 3 b 4L l 3 F 4L l 1 b 4L l 1 F 18 4L l 1 b 4L l 1 F 4L l 3 + F 4L l 1 F 4L l 1 3 b 4L l F. 4L l+1 3 b 4L l+1 3 F 4L l+1 3 Notice that the signals in the third bracket are modulo- sum functions decoded at Stage 1 and the summation of those and received signals on top. Specifically a 4L l 3 F 4L l+1 a 4L l+1 and b 4L l F 4L l+1 3 b 4L l+1 3 red colored in the third bracket are the received signals at time L l 1. This transmission leads node 1 and to decode F 4L l 3, F 4L l 1 F 4L l+1 3 and F 4L l, F 4L l F 4L l+1 using their own symbols and previously decoded functions. For ease of illustration, we elaborate how the decoding works in the case of L =. We exploit the received signals at time 3 = L 1 and 4 = L at node 1 and. As they get modulo- sums of forward symbols directly, the transmission strategy of node 1 and at time 5 = L + 1 is exactly the same as that in the perfect-feedback scheme: Forwarding F 5, F 7 and F 6, F 8 respectively. Then node 1 and obtain F 5, F 7 F 6 and F 6, F 8 F 5. Using F 6 received at time 3, node 1 can decode F 7. Similarly node can decode F 8. Now in the backward channel, with the newly decoded F 5, F received at time 1 and a 5 F received at time 3, node 1 can construct: F b 5 b = a 5 F a F 5 F. This constructed signal is sent on the top level. Furthermore, with the newly decoded F 7, a 1, F 4, F 6 received at time 1, and 3 and a 7 F 4 4

14 14 received at time 4, node 1 can construct: This is sent on the bottom level. F 4 b 7 b 4 F 6 a 1 = a 7 F 4 a 4 F 7 F 4 F 6 a 1. In a similar manner, node encodes F 1 a 6 a 1, F 3 a 8 a 3 F 5 b. Sending all of the encoded signals, node 1 and then get F 1 a 6 a 1, F 3 F a 5 and F b 5 b, F 4 F 1 b 6 respectively. Observe that from the top, node 1 can finally decode F 1 of interest by using a 6, a 1 own symbols. From the bottom, node 1 can also obtain F 3 from F 3 F a 8 a 3 a 5 by utilizing F received at time 1 and a 8, a 3, a 5 own symbols. Similarly, node can decode F, F 4. With the help of the decoded functions, node 1 and can then construct signals that can aid decoding the desired functions at the other-side nodes. Node 1 uses newly decoded F 1 and b 1 F 1 received at time 1 to generate F 1 ã 1 on top; using b 3 F 3, F, F 3, it also constructs F 3 ã 3 F on bottom. In a similar manner, node encodes F b, F 4 b 4 F 1. Forwarding all of these signals at time 6, node 1 and get F 1 ã 1, F 3 F ã 3 ã and F b, F 4 F 1 b 4 b 1 respectively. Here using their past decoded functions and own symbols, node 1 and can obtain F 1, F 3 and F, F 4. Consequently, during 6 time slots, 8 modulo- sum functions w.r.t. forward symbols are computed, while 4 backward functions are computed. This gives R, R = 4 3, 3. And one can easily check that for an arbitrary number of L, R, R = 4L 3L, 4L 1 3L we get the desired rate pair i.e., R, R 4 3, 4 3 = C pf, C pf. = 4 3, 4L 4 3L is achievable. Note that as L, Remark How to achieve the perfect-feedback bound?: As in the two-way interference channel [1], the key point in our achievability lies in exploiting the following three types of information as side information: 1 past received signals; own message symbols; and 3 future decoded functions. Recall our achievability in Fig. 4 that the encoding strategy is to combine own symbols with past received signals, e.g., in time 1 node 1 encodes ã F, ã 1 a 1, which is the mixture of its own symbols ã, ã 1 and the received signals F, a 1. And the decoding strategy is to utilize past received signals, e.g., in time 1, node 1 exploits its own symbol a to decode F. The most interesting part that is also highlighted in the two-way interference channel [1] is the utilization of the last type of information: Future decoded functions. For instance, with b 1 F 1 received at time 1 only, node 1 cannot help node 1 to decode F 1. However, note that our strategy is to forward F 1 ã 1 at node 1 in time 6. Here the signal is the summation of b 1 F 1 and F 1. And F 1 is actually the

15 15 function that node 1 wishes to decode in the end; it can be viewed as a future function since it is not available at the moment. So the approach is to defer the decoding procedure for F 1 until F1 becomes available at node 1; see in Fig. 4 that F 1 is computed at time 5 a deferred time slot in the second stage. The decoding procedure for F 3 and F, F 4 at node 1 and similarly follows: Deferring the decoding of these functions until F and F 1, F 3 becomes available at node 1 and respectively. The decoding of F 5, F 7 and F 6, F 8 at node 1 and precedes that of F 1, F 3 and F, F 4 at node 1 and respectively. The idea of deferring the refinement together with the retrospective decoding plays a key role to achieve the perfect-feedback bound in the limit of L. B. Example: m, n = 1,, m, ñ = 1, 0 Similar to the previous example, we first review the perfect-feedback scheme, which we will use as a baseline for comparison to our achievable scheme. We focus only on the case of m, ñ = 1, 0, as the one for m, n = 1, was already covered. 1 Perfect-feedback strategy: The perfect-feedback scheme for m, ñ = 1, 0 consists of three time slots. In time 1, we send backward symbols ã 1 and b at node 1 and respectively. Then node 1 and get b and ã 1 respectively. Node 1 can then deliver the received symbol b to node 1 through feedback. Similarly, node can get ã 1 from node. In time, with the feedback signals, node 1 and can construct F and F 1 respectively; and send them over the backward channel. Then node 1 and get F1 and F respectively. Note that until the end of time, F is not delivered to node 1. Similarly, F1 is missing at node. Using one more time slot, we can deliver these functions to the intended nodes. With feedback, node can get F from node. Sending this at time 3 allows node 1 to obtain F. Similarly, node can obtain F 1. As a result, node 1 and obtain F 1, F during three time slots. This gives a rate of 3 = C pf. As mentioned in the previous example, in the two-way setting of our interest, a challenge arises due to the tension between feedback transmission and the traffic w.r.t. the other direction. Here are achievable schemes that address this challenge. Achievability: The underlying idea is similar to that for m, n = 1,, m, ñ =, 1 : Exploiting effectively future information as side information via retrospective decoding. A key distinction relative to the prior example is that here the retrospective decoding occurs in a nested manner. Our achievability now introduces the concept of multiple layers, say M layers. Each layer consists of two stages as in the previous example. Hence there are M stages in total. For each layer, the first stage consists of L time slots; and the second stage consists of L + 1 time slots.

16 16 Fig. 5. An achievable scheme for m, n = 1,, m, ñ = 1, 0, and L, M = 1, 1. Until the end of [M L 1]th layer, in the first stage, 4L and L of fresh symbols are transmitted through the forward and backward channels respectively. From [M L 1 + 1]th layer to Mth layer, 4L fresh symbols are transmitted over the forward channel, while no fresh symbols are transmitted over the backward channel. The operation in the second stage for each layer is the same as that for the other layers: No fresh forward and backward symbols are transmitted. We claim that the proposed scheme our scheme ensures 4LM and LM L 1 forward and backward-message computations during total 3L + 1M time slots, and thus can achieve R, R = 4LM 3L+1M, LM L 1 3L+1M. Set M = L. As L tends to infinity, the rate pair becomes R, R = 4 3, 3 = C pf, C pf. Instead of describing schemes for arbitrary L, M, we will illustrate our scheme for L, M = 1, 1 and L, M =, 4, where key ingredients of our idea can be presented. While describing our schemes, we provide a guideline as to how to generalize into arbitrary L, M. Stage 1 for L, M = 1, 1 : We now illustrate the scheme for L, M = 1, 1, which consists of a single layer, stages, and 4 time slots. See Fig. 5. We noted that the first stage consists of = L time slots as we focus on the case of L = 1. The claimed achievable rate pair reads: 1, 1. In time 1, node 1 sends a 1, a ; node sends b, b 1. Then node 1 gets a 1, F. Similarly node gets b, F 1. On the other hand, node 1 and keep silent. The forward transmission strategy in time is pretty much the same as that in time 1. Node 1 and

17 17 transmit a 3, a 4 and b 4, b 3 so that node 1 and obtain a 3, F 4 and b 4, F 3. Instead of keeping silent in the backward channel, node 1 and send ã 1 F a 3 and b F 1 b 4 respectively. Note that these signals contain each node s own symbol and received signals. Note that node 1 and then get b F 1 b 4 and ã 1 F a 3 respectively. Note that until the end of time, F 1, F 3 and F, F 4 are not delivered yet to node 1 and respectively, while F 1, F are missing at both node 1 and. Stage for L, M = 1, 1 : The transmission strategy in the second stage is to accomplish the computation of desired functions that each node did not obtain yet. The stage consists of time slots. In time 3, from the received symbol at time, node 1 cancels out a 1 odd-index symbol and adds a 4 own symbol to construct the even-index function F 4 = F 4L in general, thus encoding b 1 F 4 b. In a similar manner, node encodes a F 3 ã 1. The transmission strategy for each node is to forward the encoded signal on top. Furthermore, through the bottom level, each node forwards its own symbol whose index is the other transmitting node s symbol index canceled out. For example, node forwards b 1 i.e., its own symbol of index 1, because node 1 cancels out a 1 at time 3. Similarly, node 1 forwards a. Node 1 and then get b 1 F 4 b, F 3 ã 1 and a F 3 ã 1, F 4 b respectively. Note that from the received signal on bottom, node 1 and can obtain F 3 and F 4 using ã 1 and b own symbols respectively. Furthermore, from the received signal on top, node 1 and can obtain b 1 b and a ã 1 using F 4 and F 3 previously decoded functions. From the received signal on top, node 1 and can also generate b 1 F and a F 1 using ã and b 1 respectively. Note that sending back them allows node 1 and to obtain F 1 and F respectively. In time 4, on the top level, node 1 and forward what they just computed: F1 and F. Furthermore, node 1 and forward a 1 and b on bottom level. Node 1 and then obtain F 1, a 1 b and F, b ã 1. Observe that node 1 can now obtain F 1 by adding b 1 b received on top in time 3 and a 1 b received on bottom at time 4. Similarly, node can obtain F. Through the backward channel, node 1 and simply feed back F 1 and F. Then node 1 and get F and F 1 respectively. We note that this strategy is repeated for the first two time slots of the second stage of each layer, when we consider arbitrary L, M. As a result, during 4 time slots, node 1 and obtain {F l } 4 l=1 while node 1 and obtain F 1, F. This gives 1, 1. We next introduce the next simplest case where L = where we employ the concept of multiple layers when describing an achievable scheme. Stage 1 for L, M =, 4 : So far, we discussed the scheme only for the single layer case where part of our idea can be presented. However, one can see that there is still a gap between the achievable rate pair shown above and the desired rate pair. In the previous example where m, n = 1,, m, ñ =, 1,

18 18 Fig. 6. An achievable scheme for m, n = 1,, m, ñ = 1, 0, and L, M =, 4 at Stage 1 and. our strategy to resolve this issue is to extend only the parameter L w.r.t. a single layer to infinity. A key distinction in this example relative to the prior example is that here we resolve the issue by further employing the multi-layer concept works. We claim that as both the number of layer M and the parameter L tends to infinity with M = L, the desired rate pair is achievable. Actually, the simplest multi-layer case is the one in which L, M =, 4. We will illustrate the scheme for the case in order to present our idea of employing multi-layer concept in a clear manner. The claimed achievable rate pair reads: 8 7, 7 = 8 7, 4M 7M. Furthermore as we set M without changing the value of L, we get 8 7, 4 7 which outperforms the achievable rate pair for the single layer case. Here we will illustrate for M = L = 4 and demonstrate how to extend the parameter M to infinity solely in our scheme. The proposed scheme consists of 8 = 3L + 1M time slots. And the first stage

19 19 within the first layer consists of 4 = L time slots. See Fig. 6. The strategy for the first two time slots is exactly the same as that for the previous single layer case. In time 1, node 1 sends a 1, a ; node sends b, b 1. Then node 1 gets a 1, F. Similarly node gets b, F 1. On the other hand, node 1 and keep silent. In time, node 1 and transmit a 3, a 4 and b 4, b 3 so that node 1 and obtain a 3, F 4 and b 4, F 3. Through the backward channel, node 1 and send ã 1 F a 3 and b F 1 b 4. Note that node 1 and then get b F 1 b 4 and ã 1 F a 3 respectively. A distinction occurs in the next two time slots. In time 3, node 1 and send a 5, a 6 and b 6, b 5, thus yielding a 5, F 6 and b 6, F 5 at node 1 and. Through the backward channel, they keep silent at time 3. In time 4, using F 1 b 4 b received at time, node 1 cancels out its odd-index symbol a 1 and adds the fresh symbol a 7, thus encoding a 7 b 1 b 4 b. Similarly, node encodes b 8 a a 3 ã 1. Then each node forwards the encoded signal on top. Furthermore, through the bottom level, each node forwards its own symbols whose indices are the other transmitting node s symbol indices added and canceled out. For example, node forwards b 7 b 1 i.e., its own symbols of indices 7 and 1, because node 1 adds a 7 and cancels out a 1 at time 4. Similarly, node 1 forwards a 8 a. Node 1 and then get a 7 b 1 b 4 b, F 8 a 3 ã 1 and b 8 a a 3 ã 1, F 7 b 4 b. Note that from F 8 a 3 ã 1, received on bottom node 1 can decode F 8 using ã 1 own symbol and a 3 received at time. Similarly, node can decode F 7. Note that for arbitrary L, M, this encoding and decoding strategy is repeated for every even time slot of the first stage of each layer and hence ensures a computation of modulo- sum function on bottom of each receiving node. Through the backward channel, node 1 delivers F 6 a 7 b 1 b 4 ã 3 F which is the mixture of F 6 received at time 3, a 7 b 1 b 4 b received at time 4, and 3 own symbol. Similarly, node delivers F 5 b 8 a a 3 b 4 F 1. Node 1 and then get F 5 b 8 a a 3 b 4 F 1 and F 6 a 7 b 1 b 4 ã 3 F respectively. In general, each transmitted signal at time l, l = 1,, L consists of a fresh backward message symbol and two signals received on bottom and top at time l 1 and l respectively. Note that until the end of time 4, F 1, F 3, F 5, F 7 and F, F 4, F 6, F 8 are not delivered yet to node 1 and respectively, while F 1, F, F 3, F 4 are missing at both node 1 and. Stage for L, M =, 4 : The transmission strategy in the second stage is to accomplish the computation of desired functions that each node did not obtain yet. We employ the retrospective decoding strategy introduced in the previous example. The stage consists of 3 time slots. For the next two time slots, the strategy is pretty much the same as the strategy during time 3 and 4 in the case of

20 0 L, M = 1, 1. In time 5, from the received signal at time 4 = L, node 1 cancels out its all oddindex symbols a 3, a 5 and adds the even-index symbol a 8 = a 4L, thus encoding b 5 F 8 a b 4 F 1. In a similar manner, node encodes a 6 F 7 b 1 ã 3 F using even-index symbols b 4, b 6 and the odd-index symbol b 7. The transmission strategy for each node is to forward the encoded signal on top. Furthermore, through the bottom level, each node forwards its own symbols whose indices are the other transmitting node s symbol indices added and canceled out. For example, node forwards b 8 b 5 b 3 i.e., its own symbols of indices 8, 5, and 3, because node 1 cancels out a 3, a 5 and adds a 8 at time 5. Similarly, node 1 forwards a 7 a 6 a 4. Node 1 and then get b 5 F 8 a b 4 F 1, b 7 a 4 b 1 ã 3 F and a 6 F 7 b 1 ã 3 F, a 8 b 3 a b 4 F 1 respectively. From the received signal on bottom, node 1 can decode F 7 by adding a 7 b 1 b 4 b received at time 4, F 4 received at time, and ã, ã 3 own symbols. Similarly, node can decode F 8. From the received signal on top, node 1 and use F 8, F and F 7, F 1 to generate b 5 b F 4 F 1 and a 6 a 1 F 3 F respectively. Note that sending back them allows node 1 and to obtain F 3 F and F 4 F 1 by cancelling a 6, a 1 and b 5, b own messages respectively. In time 6, node 1 and forward what they just decoded on top: F3 F and F 4 F 1. Furthermore on bottom, node 1 forwards its own symbols a 5 a whose indices are the other transmitting node s symbol indices canceled out when decoding F 4 F 1. Similarly, node forwards b 6 b 1. Node 1 and then obtain F 3 F, a 5 a F 4 F 1 and F 4 F 1, b 6 b 1 F 3 F. Observe that node 1 can now obtain F 5 by adding b 5 a b 4 F 1 received on top at time 5 and a 5 a b 4 F 1 received on bottom at time 6. Similarly, node can obtain F 6. With the newly decoded F 7 and a 7 b 1 b 4 b received at time 4, node 1 generates b 7 b 1 b 4 F. Sending this through the backward channel leads node to obtain F. Similarly, constructing a 8 a a 3 F 1 and sending this leads node 1 to obtain F 1. For arbitrary L, M a similar strategy is applied during the first two time slots of Stage m, m = 1,, M. Notice that until the end of time 6, F 1, F 3 and F, F 4 are not delivered yet to node 1 and, while F, F 4 and F 1, F 3 are missing at node 1 and. Using one more time slot, we will resolve these computations. The idea is to apply the retrospective strategy over multi-layers. In other words, we defer the transmission of the last time at Stage to time 19. It will be clear as to why we defer the transmission. Stage 3 and 4 for L, M =, 4 : The scheme for Layer is pretty much the same as that for Layer 1 except the transmission through the forward channel at time 9. See Fig. 7. Specifically, the scheme for Stage 3 time 7,, 10 is similar to that for Stage 1 and the scheme for Stage 4 time 11, 1, and

21 1 Fig. 7. An achievable scheme for m, n = 1,, m, ñ = 1, 0, and L, M =, 4 at Stage 3 and 4. time 6 is similar to that for Stage. Using the most recently received signal w.r.t. the previous layer i.e., a 8 a a 3 F 1 received at time 6 in Stage, node 1 constructs a 13 a 8 a F 1 and sends on top. The construction idea is to cancel out node 1 s odd-index symbol a 3 and to add new symbol a 13. As node 1 cancels the odd-index of 3, node adds its own symbol of index 3 i.e., b 3, and sends it with the new symbol b 13. This compensation leads node to get F 13 since node can exploit a 8 a a 3 F 1 transmitted at node in time 6 and F 3 received at time. A similar transmission strategy at the bottom of node 1 and the top of node leads node 1 to obtain F 14. Note that on top, node 1 and get a 13 a 8 a F 1 and b 14 b 7 b 1 F. These signals will be exploited in the next layer to accomplish the computation of F and F 1 mentioned at the end of Stage at node 1 and. This strategy is repeated in general for every

22 layer. Specifically, for the first stage of layer M, the construction of transmitted signals at 3rd, 5th,, [L 1]th slots uses the most recently received signals in Layer M L + 1, M L +,, M 1 respectively. Similar to the previous two stages, at the end of time 1 the second last time of Stage 4, F 9, F 11 and F 10, F 1 are not delivered yet to node 1 and, while F 6, F 8 and F 5, F 7 are missing at node 1 and. Using one more time slot, we will resolve these computations. And we defer the transmission of the last time at Stage 4 to time 6. Stage 5 and 6 for L, M =, 4 : The scheme for Layer 3 is pretty much the same as that for Layer except the transmission through the backward channel at time 13. See Fig. 8. Specifically, the scheme for Stage 5 time 13,, 16 is similar to that for Stage 3 and the scheme for Stage 6 time 17, 18 and time 7 is similar to that for Stage 4. Observe at Stage 4, node 1 and obtain F 13 and F 14 respectively. With these newly decoded functions in the previous layer, node 1 and can now construct: node 1 : b 13 b 8 b F 1, node : a 14 a 7 a 1 F. For instance, b 13 b 8 b F 1 is the summation of a 13 a 8 a F 1 received at time 9 and F 13, F 8, F decoded at time 1, 4, and 1. Sending all of the constructed signals lead node 1 and to obtain F and F 1. Notice that using the decoded functions and received signals at time 5, node 1 and can also obtain F 3 and F 4. It turns out that for sufficient large M, the number of newly decoded backward functions in the first stage of each layer is L 1. We apply the above strategy for the decoded functions during odd time slots total L time of the first stage of each layer. As a result, node 1 and get L 1 independent modulo- sum functions. Observe that at the end of time 18 the second last time of Stage 6, F 17, F 19 and F 18, F 0 are not delivered yet to node 1 and. Time 19, the last time slot of Stage : With the L 1 newly decoded functions, we accomplish the computation of desired functions that each node did not obtain yet during the next L 1 time slots. For the case of L =, with the newly decoded F, and F 1, the computation of F 1, F 3 and F, F 4 at node 1 and can now be accomplished. Using F, F 1 b 4 b received at time, and a 4 own symbol, node 1 can construct F 1 F 4 ã. Node 1 then delivers F 1 F 4 ã, F 1. Note that node 1 already got F 1 at time 6. Similarly, node delivers F F 3 b 1, F. Then node 1 gets F 1 F 4 ã, F F 3 ã 1.

23 3 Fig. 8. An achievable scheme for m, n = 1,, m, ñ = 1, 0, and L, M =, 4 at Stage 5, 6, 7, and 8. Using ã, ã 1 own symbols and F, F 4 received at time 1 and, node 1 can finally decode F 1, F 3. In a similar manner, node can decode F, F 4. Through the backward channel, they send F 3 F and F 4 F 1 received at time 6. Exploiting their decoded F 1 and F, this transmission leads node 1 and to obtain F 4 and F 3 respectively. Observe that the computations of modulo- sum functions w.r.t. symbols introduced in Layer 1 is completely accomplished. In other words, node 1 and obtain {F l } 8 l=1, while node 1 and obtain { F l } 4 l=1. Stage 7 and 8 for L, M =, 4 : We repeat the same procedure in Layer 4 w.r.t. new symbols. And the strategy in time 6 is exactly the same as that in time 19. This leads node 1 and obtain {F l } 16 l=9, while node 1 and obtain { F l } 8 l=5. Observe that at the end of time 5 the second last time at Stage 8, F 5, F 7 and F 6, F 8 are not

24 4 delivered yet to node 1 and. Time 7 & 8, the last two time slots: Remember our strategy that no fresh backward symbols are introduced from Layer 3 = M L 1+1 to 4 = M. Then one can readily check that the received signal in time 14 at node 1 is b 0 F 17. And the signal received in time 14 at node is a 19 F 18. So node 1 and can construct F 17 F 0 and F 18 F 19 using a 0 and b 19 own symbols. Sending them on top at time 7 leads node 1 and to obtain F 17, F 19 and F 18, F 0. Applying the same procedure at time 8, they can further obtain F 5, F 7 and F 6, F 8. As a result, during 8 time slots, node 1 and obtain {F l } 3 l=1 while node 1 and obtain { F l } 8 l=1. This gives 8 7, 7. One further notification is that if we repeat the strategy while introducing fresh backward symbols in the first M layers where M 4, and then apply the strategy with no fresh backward symbols for the remaining layers, one can readily check that 8 7, 4M 7M is achievable. As we solely extend M to infinity, we get 8 7, 4 7 which outperforms the performance for the single layer case. Following the aforementioned guideline, we find that this idea can be extended to arbitrary values of L, M by employing the retrospective decoding strategy over multi-layers, thus yielding: R, R = 4LM 3L+1M, LM L 1 3L+1M. Set M = L. As L tends to infinity, the rate pair becomes R, R = 4 3, 3 = C pf, C pf. C. Generalization: Proof outline We now prove the achievability for arbitrary values of m, n, m, ñ. Note that C = C no when 3 α < 1, 1 < α 3, 3 α < 1, 1 < α 3. Also by symmetry, it suffices to consider only three regimes. See Fig. 9 : R1 0 α 3, 0 α 3 ; R 3 < α < 1, 1 < α 3, α > 3 ; R3 0 α 3, α > 3. 1 Regimes in which interaction provides no gain: Referring to Fig., the channel regimes of this category are R1 and R1. A simple combination of the non-feedback scheme [18] developed in the non-interactive scenario in Baseline 1 and the interactive scheme in [16] can yield the desired result for the remaining regimes. Regimes in which interaction helps only either in forward or backward direction: It turns out the achievability in this case is also a simple combination of the non-feedback scheme [18] and the interactive scheme in [16]. The channel regimes of this category are: R, R, R, and R.

25 5 No feedback gain no need to consider Fig. 9. Regimes to check for achievability proof. By symmetry, it suffices to consider R1, R, and R3. 3 Regimes in which interaction helps both in forward and backward directions: As mentioned earlier, the key idea is to employ the retrospective decoding potentially employing the concept of multiple layers. For ease of generalization to arbitrary channel parameters in the regime, here we employ network decomposition in [18] where an original network is decomposed into elementary orthogonal subnetworks and achievable schemes are applied separately into the subnetworks. See Fig. 10 for an example of such network decomposition. The idea is to use graph coloring. The figure in Fig. 10 graphically proves the fact that m, n =, 4, m, ñ = 3, 1 model can be decomposed into the following two orthogonal subnetworks: m, n = 1,, m, ñ =, 1 model blue color; and m, n = 1,, m, ñ = 1, 0 model red color. Note that the original network is simply the concatenation of these two subnetworks. We denote the decomposition as, 4, 3, 1 1,,, 1 1,, 1, 0. As mentioned earlier, the idea is simply to apply the developed achievable schemes separately for the two subnetworks. Notice that we developed the schemes for m, n = 1,, m, ñ =, 1 and m, n = 1,, m, ñ = 1, 0 model. For the case of m, n = 1,, m, ñ =, 1, our proposed scheme achieves R, R = 4 3, 4L 4 3L. And for the case of m, n = 1,, m, ñ = 1, 0, our strategy achieves R, R = 4LM 3L+1M, LM L 1 3L+1M. Setting M = L and letting L, the first scheme achieves 4 3, 4 3, while the second achieves 4 3, 3. Thus, the separation approach gives: R, R 4 = 3, , 8 = 3 3,,

26 6 Fig. 10. A network decomposition example of m, n =, 4, m, ñ = 3, 1 model. We see that the decomposition is given by, 4, 3, 1 1,,, 1 1,, 1, 0. which coincides with the claimed rate region: {R, R : R C pf = 8 3, R C pf = }. We find that this idea can be extended to arbitrary values of m, n, m, ñ. The channel regimes of this category are the remaining regimes: R3 and R3. See Appendix A for the detailed proof. V. PROOF OF CONVERSE In this section, we provide the converse proof. Note that the bounds of 3 and 4 are the perfectfeedback bounds in [15]. For completeness, we will provide the proof for the perfect-feedback bounds. Here the proofs for the upper bounds of 3 and 5 are stated as below. The proofs of 4 and 6 can be derived similarly by symmetry. Proof of 3 Perfect-feedback Bound: The proof for the case of α = 1 is straightforward owing to the standard cut-set argument: NR ɛ N IS K 1 SK ; Y N 1 HY 1i N maxm, n. If R is achievable, then ɛ N 0 as N tends to infinity, and hence R maxm, n = n. Now consider the case where α 1. Starting with Fano s inequality, we get N 3R ɛ N I S K 1 S K ; Y N 1 + I S K 1 S K ; Y N H Y1 N H Y N 1 S1 K S K + I S K 1 S K ; Y N 1 + I S K 1 S K ; Y1 N + H Y N H Y N H Y1 N + H Y N H Y1 N, Y N S1 K S K, S K 1, S K S1 K S K, Y1 N + I S1 K S K ; Y1 N

27 7 a H Y N 1 b H Y N 1 c H Y N 1 + I + H Y N H Y1 N, Y N S1 K S K, S K 1, S K + I S1 K + H Y N H Y N 1, Y N S K 1 S K, S K + I + H Y N H Y N 1, Y N S K 1 S K, S K S K 1 ; Y N 1, Y N S K 1 S K, S K =H Y1 N + H Y N H Y1i + H Y i N maxm, n S K ; Y N 1 S K S K 1 ; Y N 1, Y N S K where a follows from the independence of S K 1 SK and S K ; b follows from Corollary below; and c follows from the independence of S K 1 and SK 1 SK given S K. If R is achievable, then ɛ N 0 as N tends to infinity, and hence R 3 maxm, n. We therefore get the desired bound. Proof of 5 Cut-set Bound: Starting with Fano s inequality, we get N R + R ɛ N I S1 K S K, S K 1 S K ; Y N, Ỹ N, S K, S K = I S1 K S K, S K 1 S K ; Y N, Ỹ N S K, S K = H Y N, Ỹ N S K, S K = H Y i, Ỹi S K, S K, Y i 1, Ỹ i 1 a = H Y i, Ỹi S K, S K, Y i 1, Ỹ i 1, X i, X i b H Y i X i + H Ỹi X i c H V 1i + H Ṽ1i N m + m where a follows from the fact that X i is a function of S, Ỹ i 1 and X i is a function of S, Y i 1 ; b and c follow from the fact that conditioning reduces entropy. If R + R is achievable, then ɛ N 0 as N tends to infinity, and hence R + R m + m. Thus, we get the desired bound. Corollary : I S1 K SK ; Y 1 N S K 1, S K I S1 K; Y 1 N, Y N S K. Proof: I S K 1 S K ; Y1 N S K 1, S K a =H S1 K S K 1, S K H H S1 K S K 1, S K H S1 K S K 1, S K H b =H S K 1 S K S K 1, S K, Y1 N S1 K S K, Y1 N, Y N, S K S1 K S K, Y1 N, Y N

28 8 =I S K 1 ; Y N 1, Y N S K where a follows from the fact that HS K 1 S K = HS K 1 = HSK 1 SK = HSK 1 SK S K ; and b follows from S K 1 Corollary 3: S K 1 Y N 1, Y N, S K S K Y N 1, Y N, S K S K. see Corollary 3 below. Proof: I S1 K ; S K Y1 N, Y N, S K 1, S K = I S1 K ; S K, Y1 N, Y N S K 1, S K I S1 K ; Y1 N, Y N S K 1, S K = I S1 K ; Y1 N, Y N S K 1, S K, S K I S1 K ; Y1 N, Y N S K 1, S K = H Y1 N, Y N S K 1, S K + H Y1 N, Y N S K 1, S K, S1 K + H a = H X 1 N, X N S K 1, S K + H X1 N, X N S K 1, S K, S1 K + H b = H X 1i, X i S K 1, S K, X1 i 1, X i 1 + H + H c = H + H d = [H [H X 1i, X i S K, S1 K, X1 i 1, X i 1, Y1 i 1, Y i 1, X 1i, X i S K, S K, X1 i 1, X i 1, Y1 i 1, Y i 1, X 1i, X i S K, X1 i 1, X i 1 X 1i S K, S K, X1 i 1, X i 1 X 1i S K, X1 i 1, X i 1 X i S K, X i 1, X i 1 + H H H Y N 1, Y N S K, S K X N 1, X N S K, S K X i 1 1, X i 1 1, i 1 X, Ỹ 1 i 1, Ỹ i 1 i 1 X, Ỹ 1 i 1, Ỹ i 1 X i S K, S1 K, X1 i 1, X i 1 X 1i S K 1, S ] K, S K, X1 i 1, X i 1 X i S K, S K 1, X i 1 1, X i 1, X 1i ] 0 where a follows from the fact that X 1, X is a function of Y 1, Y see Claim 1 below; b follows from the fact that Y1 i 1, Y i 1 is a function of X1 i 1, X i 1 i 1, X 1, of S K, Y i 1 1, Y i 1, and Ỹ i 1 1, Ỹ i 1 is a function of fact that X ki is a function of S K k, Ỹ i 1 k, k = 1,. X i 1 1, Claim 1: For α 1 i.e., m n, X 1, X is a function of Y 1, Y. Proof: Consider the case of m < n. From 1, we get Y 1 G n m Y = I n G n m X 1. i 1 X is a function i 1 X ; c and d follow from the Note that I n G n m is invertible when m n. Hence, X 1 is a function of Y 1, Y. By symmetry, X is a function of Y 1, Y. Similarly we can show this for the case of m > n. This completes the proof of converse.

29 9 VI. CONCLUSION We investigated the role of interaction for computation problem settings. Our main contribution lies in the complete characterization of the two-way computation capacity region for the four-node ADT deterministic network. As a consequence of this result, we showed that not only interaction offers a net increase in capacity, but also it leads us to get all the way to perfect-feedback computation capacities simultaneously in both directions. In view of [1], this result is another instance in which interaction provides a huge gain. One future work of interest would be to explore a variety of network communication/computation scenarios in which the similar phenomenon occurs. APPENDIX A PROOF OF GENERALIZATION TO ARBITRARY m, n, m, ñ We now prove the achievability for arbitrary m, n and m, ñ. The key idea of the proof is to use the network decomposition in [18] also illustrated in Fig. 10. This idea provides a conceptually simpler proof by decomposing a general m, n, m, ñ channel into multiple elementary subchannels and taking a proper matching across forward and backward subchannels. See Theorem stated below for the identified elementary subchannels. We will use this to complete proof in the sequel. Theorem Network Decomposition: For an arbitrary m, n channel, the following network decomposition holds: m, n 0, 1 n m 1, m, 0 α 1 ; 19 m, n 1, n 3m, 3 m n, m, n, 1 m 3n 3, n m, 1 α 3 ; 0 3 α ; 1 m, n 1, 0 m n, 1 n, α. Here we use the symbol for the concatenation of orthogonal channels and i, j l denotes the l-fold concatenation of the i, j channel. A. Proof of R1 0 α 3, 0 α 3 For the following regime, the claimed achievable rate region is: {R C pf, R C pf, R + R C no + C no }. And the following achievability w.r.t. the elementary subchannels identified in Theorem forms the basis of the proof for the regime of R1.

30 30 Fig. 11. Three types of shapes of an achievable rate region for the regime R1 0 α 3, 0 α 3. Lemma 1: The following rates are achievable: i For the pair of m, n = 1, i and m, ñ = 1, j : R, R = 4 3 i, j 1 3 i. Here 1 3 i j. ii For the pair of m, n = 1, i and m, ñ =, 3 j : R, R = 4 3 i, j 1 3 i. Here 1 3 i j. Proof: The proof builds upon a simple combination of the non-feedback scheme [18] and the interactive scheme in [16]. While it requires a tedious calculation, it contains no new ingredient, hence, we omit the detailed proof. We see that there is no feedback gain in sum capacity. This means that one bit of a capacity increase due to feedback costs exactly one bit. Depending on whether or not C pf or C pf exceeds C no + C no, we have four subcases, each of which forms a different shape of the region. See Fig. 11. I C pf C no C no, Cpf C no C no : The first case is the one which the amount of feedback for maximal improvement, reflected in C pf C no or C pf C no, is smaller than the available resources offered by the backward channel or forward channel respectively. In other words, in this case, we have a sufficient amount of resources such that one can achieve the perfect-feedback bound in one direction. By symmetry, it suffices to focus on one corner point that favors the rate of forward transmission: R, R = C pf, C no C pf C no. For efficient use of Theorem and Lemma 1, we divide the regime

31 31 R1 into the following four sub-regimes: R1-1 1 α 3, 1 α 3 ; R1-1 α 3, 0 α 1 ; R1-3 0 α 1, 1 α 3 ; and R1-4 0 α 1, 0 α 1. R1-1 1 α 3, 1 α 3 : Applying Theorem in this sub-regime, the network decomposition 0 gives: m, n 1, n 3m, 3 m n, m, ñ 1, ñ 3 m, 3 m ñ. Here we note that either 1 3 n 3m = C pf C no ñ 3 m or C pf C no m ñ; otherwise, we encounter a contradiction that C pf C no C no. Consider the case where 1 3 n 3m ñ 3 m. Then we apply Lemma 1 i for the pair of 1, n 3m and 1, ñ 3 m. Note that the condition of i holds. Applying the non-feedback schemes for the remaining subchannels gives: R = 4 3 n 3m + m n = C pf, R = 1 ñ 3 m 13 n 3m + m ñ = m C pf C no, R + R = m + m = C no + C no. Now consider the case where 1 3 n 3m m ñ. Then we apply Lemma 1 ii for the pair of 1, n 3m and, 3 m ñ. Notice that the condition of ii holds. Applying the non-feedback schemes for the remaining subchannels gives: R = 4 3 n 3m + m n = C pf, R = 1 ñ 3 m + m ñ 13 n 3m = m C pf C no, R + R = m + m = C no + C no. R1-1 α 3, 0 α 1 Theorem yield: : In this sub-regime, the network decompositions 0 and 19 in m, n 1, n 3m, 3 m n, m, ñ 0, 1ñ m 1, m.

32 3 We apply Lemma 1 i for the pair of 1, n 3m and 1, m. Note that the condition of i holds. Applying the non-feedback schemes for the remaining subchannels gives: R = 4 3 n 3m + m n = C pf, R = 0 ñ m + 1 m 13 n 3m = m C pf C no, R + R = m + m = C no + C no. For the proofs of the remaining regimes R1-3 and R1-4, we omit the details as the proofs follows similarly. II C pf C no > C no, Cpf C no C no : Similar to the first case, one can readily prove the same one-to-one tradeoff relationship in achieving one corner point R, R = C no C pf C no, C pf. Hence, we omit the detailed proof. On the other hand, we note that there is a limitation in achieving the other counterpart. Note that the maximal feedback gain C pf C no for forward computation does exceed the resource limit C no offered by the backward channel. This limits the maximal achievable rate for forward computation to be saturated by R C no + C no. Hence the other corner point reads C no + C no, 0 instead. For completeness, we will show this is indeed the case as below. By symmetry, we omit the case of II. Similar to the previous case, we provide the proofs for R1-1 and R1-. The proofs for the regimes R1-3 and R1-4 follow similarly. R1-1 1 α 3, 1 α 3 : Applying Theorem in this sub-regime, the network decomposition 0 gives: m, n 1, n 3m, 3 m n, m, ñ 1, ñ 3 m, 3 m ñ. We apply Lemma 1 i for the pair of 1, 3ñ 3 m and 1, ñ 3 m. Also, we apply Lemma 1 ii for the pair of 1, 6 m ñ and, 3 m ñ. Applying the non-feedback schemes for the remaining subchannels 1, n 3m 3 m and, 3 m n gives: R = 4 3 m + 1 n 3m 3 m + m n = m + m, 3 R = 0, R + R = m + m = C no + C no.

33 33 R1-1 α 3, 0 α 1 : In this sub-regime, the network decompositions 0 and 19 in Theorem yield: m, n 1, n 3m, 3 m n, m, ñ 0, 1ñ m 1, m. We apply Lemma 1 i for the pair of 1, 3 m and 1, m. Applying the non-feedback schemes for the remaining subchannels 1, n 3m 3 m and, 3 m n gives: R = 4 3 m + 1 n 3m 3 m + m n = m + m, 3 R = 0, R + R = m + m = C no + C no. III C pf C no > C no, Cpf C no > C no : This is the case in which there are limitations now in achieving both R = C pf and R = C pf. Due to the same argument as above, what we can maximally achieve for R or R in exchange of the other channel is C no + C no which implies R, R = C no + C no, 0 or 0, C no + C no is achievable. The proof follows exactly the same as above, so we omit the details. B. Proof of R 3 < α < 1, 1 < α 3, α > 3. For the regime of R, the claimed achievable rate region is: {R C no, R C pf, R + R C no + n}. Unlike the previous regime, there is an interaction gain for this regime. Note that the sum-rate bound exceeds C no + C no ; however, there is no feedback gain in forward channel. To provide an achievability idea, let us consider the case of 3 < α. As the achievability idea also holds for the remaining regimes, we omit the details for α >. The network decomposition 1 together with 3 C pf C no = m 3ñ in 3 < α gives: m, ñ, 1 3 C pf C no 3, ñ m. We find that the shape of the region depends on where C pf C no lies in between n C no and n. See Fig. 1. I C pf C no n C no : The first case is the one in which the amount of feedback for maximal improvement, reflected in C pf C no, is small enough to achieve the maximal feedback gain without sacrificing the performance of forward computation. Now let us show how to achieve R, R = C no, C pf.

34 34 Fig. 1. Three types of shapes of an achievable rate region for the regime R 3 < α < 1, 1 < α 3, α > 3. The decomposition idea is to pair up m, n and, 1 3 C pf C no, while applying the non-feedback schemes for the remaining backward subchannels 3, ñ m. To give an achievability idea for the first pair, let us consider a simple example of m, n =, 3 and m, ñ =, 1. See Fig. 13. We will then demonstrate the general case. Our scheme consists of two stages. The first stage consists of two time slots; and the second stage consists of a single time slot. Hence there are three time slots in total. See Fig. 1. Through the forward channel, we use the non-feedback scheme at time 1 and. In time 1, node 1 sends a 1, a on the top two levels; and node sends b, b 1. Similarly, in time, node 1 delivers a 3, a 4 ; and node delivers b 4, b 3. Note that node 1 and then can obtain F i s i = 1,, 4. Through the backward channel, node 1 and transmit ã 1, ã and b, b 1 at time 1. Then node 1 obtains F 1. Similarly, node obtains F. In time, we repeat this w.r.t. new symbols, thus achieving F 3 and F 4 at node 1 and respectively. Note that until the end of time, F, F 4 are not delivered yet to node 1. Similarly, F 1, F 3 are missing at node. Let à 1 := F 1 F 3 b 4 and B 1 := F F 4 ã 3. In time 3, node 1 sends a 5 Ã1, a 6 b b 4, a 5 ; similarly, node sends b 6 B 1, b 5 ã 1 ã 3, b 6. Node 1 then gets a 5 Ã1, F 6 B 1 b b 4, F 5 ã 1 ã 3. Note that B 1 b b 4 = ã ã 3 ã 4, so node 1 can cancel out this from F 6 B 1 b b 4, thus obtaining F 6. Furthermore, node 1 can obtain F 5 by cancelling out ã 1 ã 3. Similarly, node can obtain F 5, F 6.

35 35 Fig. 13. Illustration of achievability for the regime R via an example of m, n =, 3, m, ñ =, 1. This is an instance in which we have a sufficient amount of resources that enables achieving the perfect-feedback bound in the backward channel: Cpf C no = 1 1 = n Cno. Hence we achieve R, R = C 3 no, C pf =, 4. 3 Exploiting F 5 and ã 4, and a 5 Ã1, node 1 encodes the following: b 5 ã 4 Ã1 = F 5 ã 4 a 5 Ã1. Similarly, node encodes: a 6 b 3 B 1 = F 6 b 3 b 6 B 1. Now node 1 and deliver b 5 ã 4 Ã1, F 6 and a 6 b 3 B 1, F 5 respectively. Then node 1 gets a 6 b 3 B 1, a 5 ã 4 Ã1. From this, node 1 can obtain b 3 B 1, ã 4 Ã1 = F F 3 F 4, F 1 F 3 F 4 using a 6, a 5 own symbols. Here node 1 can decode F 4 from F 1 F 3 F 4 exploiting F 1, F 3 received at time 1 and ; subsequently, it can also decode F from F F 3 F 4 using F 3, F 4. Similarly, node can obtain F 1, F 3. As a result, node 1 and obtain F i s for i = 1,, 6, during three time slots, thus achieving R =. Furthermore, node 1 and obtain F i s for i = 1,, 4., thus achieving R = 4 3. Here one can make two key observations. The first is that backward feedback signals in time 3 is interfered by forward messages but it turns out that the interference does not cause any problem. For example the feedback signal à 1 on top is mixed with a 5, hence node 1 receives a 5 Ã1, instead of Ã1

36 36 which is desired to be fed back. Nonetheless, node 1 sending b 5 ã 4 Ã1 on top, it turns out that node can decode ã 4 Ã1, using its own message b 5 and we showed that this decoded signal turns out to help decoding the desired functions. This is possible due to the fact that node 1 decodes a 5 b 5 on bottom in general these kinds of functions are on n C no bottom levels and exploits the function to encode not a 5 ã 4 Ã1, but b 5 ã 4 Ã1. This implies that feedback and independent forward-message computation do not interfere with each other and thus one can maximally utilize available resource levels: The total number of resource level is n, so n C no levels can be exploited for feedback. In general case of, 1 3 C pf C no, the maximal feedback gain is C,1 pf C,1 no 3 C pf C no = C pf C no, which does not exceed the limit on the exploitable levels n C no under the considered regime. Hence, we achieve: R 1 = C,1 pf 3 C pf C no = C pf C no, = 4 C pf C no. Now the second observation is that the feedback transmission does not cause any interference to node 1 and. This ensures R 1 = C no. On the other hand, for the remaining subchannels 3, ñ m, we apply the non-feedback schemes to achieve R = the above, we get: C 3, no ñ m = ñ m. Combining all of R = C no, R = 4 C pf C no + ñ m = 3 m = C pf. II Cpf C no > n : In this case, we do not have a sufficient amount of resources for achieving R = C pf. The maximally achievable backward rate is saturated by C no + n and this occurs when R = 0. On the other hand, under the constraint of R = C no, what one can achieve for R is C no + n C no. III n C no < C pf C no n : This is the case in which we have a sufficient amount of resources for achieving R = C pf, but not enough to achieve R = C no at the same time. Hence aiming at R = C pf, R is saturated by n C pf C no. C. Proof of R3 0 α 3, α > 3 For the considered regime, the claimed achievable rate region reads: {R C pf, R C pf, R + R C no + m, R + R C no + n}. Remember in Remark 1 that C pf C no indicates the amount of feedback that needs to be sent for achieving C pf and we interpret m C pf as the remaining resource levels that can potentially be utilized

37 37 to aid forward computation. Whether or not C pf C no m C pf i.e., we have enough resource levels to achieve R = C pf, the shape of the above claimed region is changed. Note that the third inequality in the rate region becomes inactive when C pf C no m C pf. Similarly, the last inequality is inactive when C pf C no n C pf. Depending on the two conditions, we consider the following four subcases. I C pf C no m C pf, Cpf C no n C pf ; II C pf C no > m C pf, Cpf C no n C pf ; III C pf C no m C pf, Cpf C no > n C pf ; IV C pf C no > m C pf, Cpf C no > n C pf. As mentioned earlier, now the idea is to use the network decomposition. The following achievability w.r.t. the elementary subchannels identified in Theorem forms the basis of the proof for the regimes of R3. Lemma : The following rates are achievable: i For the pair of m, n = 0, 1 and m, ñ = 1, 0 : R, R = 1 3, 3 or R, R = 3, 1 3. ii For the pair of m, n = 1, and m, ñ = 1, 0 : R, R = 4 3, 3 = C pf, C pf. iii For the pair of m, n =, 3 i and m, ñ = 1, 0 j : R, R = i, 3 j = C pf i, C pf j. Here 3i j. iv For the pair of m, n = 1, i and m, ñ =, 1 j : R, R = 4 3 i, 4 3 j = C pf i, C pf j. Here i j and j i. v For the pair of m, n =, 3 i and m, ñ =, 1 j : R, R = i, 4 3 j = C pf i, C pf j. Here 3i j. Proof: See Appendix B. I C pf C no m C pf, Cpf C no n C pf : The first case is the one in which there are enough resources available for enhancing the capacity up to perfect-feedback capacities in both directions. Hence we claim that the following rate region is achievable: R, R = C pf, C pf. For efficient use of Theorem and Lemma, we divide the regime R3 into the following four sub-regimes: R3-1 1 α 3, 3 α ; R3-1 α 3, α ; R3-3 0 α 1, 3 0 α 1, α. α ; and R3-4

38 38 R3-1 1 α 3, 3 α : Applying Theorem in this sub-regime, the network decompositions 0 and 1 give: m, n 1, n 3m, 3 m n, m, ñ, 1 m 3ñ 3, ñ m. Here we use the fact that C pf C no m C pf is equivalent to n 3m m and that C pf C no n C pf is equivalent to m 3ñ n. Without loss of generality, let us assume n 3m m 3ñ. We now apply Lemma iv for the pair of 1, n 3m and, 1 min{ m 3ñ,n 3m}. Also we apply Lemma v for the pair of, 3 m n and, 1 m 3ñ min{ m 3ñ,n 3m}. Note that a tedious calculation guarantees the condition of v: 3m n m 3ñ min{ m 3ñ, n 3m}. Lastly we apply the non-feedback schemes for the remaining subchannels 3, ñ m. This yields: R = 4 3 n 3m + m n = 3 n = C pf, R = 4 3 min{ m 3ñ, n 3m} + 4 m 3ñ min{ m 3ñ, n 3m} + ñ m 3 = 3 m = C pf. R3-1 α 3, α : In this sub-regime, the network decompositions 0 and in Theorem yield: m, n 1, n 3m, 3 m n, m, ñ 1, 0 m ñ, 1ñ. Let a := min{n 3m, m ñ}. We first apply Lemma ii for the pair of 1, a and 1, 0 a. If a = n 3m, we next apply Lemma iii for the pair of, 3 m n 1 3 ñ and 1, 0 m ñ a. And we apply Lemma v for the pair of, ñ and, 1ñ. Now consider a = m ñ. Then we apply Lemma iii for the pair of 1, n 3m a and, 1ñ. And we apply the non-feedback schemes for the remaining subchannels, 3 m n. For both cases, we get: R = 4 3 n 3m + m n = 3 n = C pf, R = 3 m ñ ñ = 3 m = C pf. R3-3 0 α 1, 3 α : Similar to R3-, R, R = C pf, C pf holds for the sub-regime. We omit the proof.

39 39 Fig. 14. Three types of shapes of an achievable rate region for the regime R3 0 α 3, α > 3 C pf C no > m C pf, Cpf C no n C pf. and the case II R3-4 0 α 1, α : Making similar arguments as in R3-1, the following sub-regime can be similarly derived, thus showing R, R = C pf, C pf. We omit the proof. II C pf C no > m C pf, Cpf C no n C pf : In this case, there are two corner points to achieve. The first corner point is R, R = C no + m C pf, C pf. The second corner point depends on where C pf C no lies in between m C no, m and beyond. See Fig. 14. For the cases of II-1 and II-, the corner point reads R, R = R, R = C pf, m C pf C no, while for the case of II-3, R, R = C no + m, 0. Let us first focus on the first corner point where R, R = C no + m C pf, C pf. Similar to I, we consider four sub-regimes R3-1, R3-, R3-3, and R3-4. We provide the details for R3-1 and R3-. The proofs for the regimes R3-3 and R3-4 follow similarly. R3-1 1 α 3, 3 α : Applying Theorem in this sub-regime, the network decompositions 0 and 1 give: m, n 1, n 3m, 3 m n, m, ñ, 1 m 3ñ 3, ñ m.

40 40 Note that it suffices to consider the case where m 3ñ n 3m since the other case implies that m 3ñ > n 3m m. This condition holds when α > 3, so it contradicts the condition of G1 that 3 α. Now we apply Lemma iv for the pair of 1, m 3ñ and, 1 m 3ñ. Also we apply Lemma v for the pair of 1, m m 3ñ and 3, ñ m. Lastly we apply the non-feedback schemes for the remaining subchannels 1, n 3m m and, 3 m n. This yields: R = 4 3 m 3ñ m m 3ñ + 1 n 3m m + m n = m m = C no + m C pf, R = 4 3 m 3ñ + ñ m = 3 m = C pf, R + R = m + m. R3-1 α 3, α : m, n 1, n 3m, 3 m n, m, ñ 1, 0 m ñ, 1ñ. We first apply Lemma ii for the pair of 1, m ñ and 1, 0 m ñ. Also, we apply Lemma iv for the pair of 1, ñ and, 1ñ. Lastly we apply the non-feedback schemes for the remaining subchannels 1, n 3m m and, 3 m n. This yields: R = 4 3 m ñ ñ + 1 n 3m m + m n = m m, = C no + m C pf, R = 3 m ñ ñ = 3 m = C pf, R + R = m + m. We are now ready to prove the second corner point which favors R. Depending on the quantity of C pf C no, we have three subcases. II-1 m C pf < C pf C no m C no : R3-1 1 α 3, 3 α : m, n 1, n 3m, 3 m n, m, ñ, 1 m 3ñ 3, ñ m.

41 41 For the above network decomposition, we apply Lemma iv for the pair of 1, 6 m m n 3m and, 1 6 m m n 3m. This is possible due to the fact that C pf C no m C no is equivalent to 3 m m n 3m 0. Also we apply Lemma v for the pair of 1, 3n 3m 6 m m and 3, ñ m. Note that a tedious yet straightforward calculation gives 3ñ m 3n 3m 6 m m. Lastly we apply the non-feedback schemes for the remaining subchannels, 1 n 3m m and, 3 m n. This yields: R = m m n 3m + 4 3n 3m 6 m m + 1 n 3m m 3 + m n = C pf, R = m m n 3m + ñ m + 1 n 3m m = m 3 n m = m C pf C no, R + R = m + m. R3-1 α 3, α : For the following sub-regime, we showed that the following rate pair is achievable: R = 4 3 m ñ ñ + 1 n 3m m + m n = m m, = C no + m C pf, R = 3 m ñ ñ = 3 m = C pf, R + R = m + m. Here it turns out that proving achievability only via the network decomposition is a bit involved. Now the idea is to tune the scheme which yields the above rate to prove the achievability of the second corner point. We use part of the backward channel for aiding forward transmission instead of its own traffic. Specifically we utilize n 3m m number of top levels in the backward channel once in three time slots in an effort to relay forward-message signal feedback. This naive change incurs one-to-one tradeoff between feedback and independent backward-message computation, thus yielding: R = m m n 3m m = C pf, R = 3 m 1 3 n 3m m = m C pf C no, R + R = m + m. II- m C no < C pf C no m :

42 4 R3-1 1 α 3, 3 α : For the following sub-regime, we showed that the following rate pair is achievable: R = 4 3 m 3ñ m m 3ñ + 1 n 3m m + m n = m m = C no + m C pf, R = 4 3 m 3ñ + ñ m = 3 m = C pf, R + R = m + m. Now the idea is to perturb the scheme to prove achievability for the second corner point that we intend to achieve. We use part of the backward channel for aiding forward transmission instead of its own traffic. Specifically we utilize n 3m m number of top levels in the backward channel once in three time slots in an effort to relay forward-message signal feedback. This naive change incurs one-to-one tradeoff between feedback and independent backward-message computation, thus yielding: R = m m n 3m m = C pf, R = 3 m 1 3 n 3m m = m C pf C no, R + R = m + m. R3-1 α 3, α : Applying the same idea as in II-1, we get the desired result. II-3 C pf C no > m : If we sacrifice all of the m direct links in the backward channel only for the purpose of helping forward computation, one can readily see that R, R = C no + m, 0 is achievable. III C pf C no m C pf, Cpf C no > n C pf : Similarly, this case requires the proof of two corner points. The first corner point is R, R = C pf, C no + n C pf. The second corner point is depends on where C pf C no lies in between n C no, n and beyond. See Fig. 15. As the proof is similar to that in the previous case, we omit the proof. IV C pf C no > m C pf, Cpf C no > n C pf : For the following case, it suffices to consider only R3-4 0 α 1, α since: n 3m =3C pf C no > 3 m C pf = m m 3 ñ a > 1 n,

43 43 Fig. 15. Three types of shapes of an achievable rate region for the regime R3 0 α 3, α > 3 C pf C no m C pf, Cpf C no > n C pf. and the case III where a follows since we consider m 3ñ > n or equivalently, Cpf C no > n C pf. With the first and the last formula, one can see that this implies that α < 1. Similarly, m 3ñ =3 C pf C no > 3n C pf = n n 3 m b > 1 m, where b follows since we consider n 3m > m. This implies that α >. For the regime of R3-4, the network decomposition yields: m, n 0, 1 n m 1, m, m, ñ 1, 0 m ñ, 1ñ. Similar to II, III, let us first focus on finding the first corner point. It turns out that the this case takes two kinds of corner point. Either R, R = C no + m C pf, C pf or R, R = C pf, C no + n C pf. If the first corner point is C no + m C pf, C pf, the second corner point corresponds to that in II; otherwise the corner point corresponds to that in III. Hence the capacity region in this case can take six types of shapes of achievable rate regions in Fig. 14 and 15. As we already described the idea of