I Introduction
Network information theory is one of the major challenges in wireless communication, which produced limited excellent results [1]. It involves the fundamental limits of communication and information theory in networks with multiple senders, receivers, optimal coding techniques and protocols which achieve these limits. An important goal is to characterize the capacity region or optimal rate which is the set of rates in which there exist codes with reliable transmissions. These rates of tuples are known to be achievable. A lot of work was done to determine the capacities of Gaussian networks with multiple senders and receivers.
A deterministic model of the wireless channel which is linear, and easy to analyze was introduced in [2], where the capacity of linear deterministic networks with a single sourcedestination pair is determined. Most subsequent research on the linear deterministic model has been focused on deriving network coding schemes to determine the capacity of deterministic networks and then using the results to find an approximate capacity region for each corresponding Gaussian network, where the approximation error can be typically ignored in the high SignaltoNoise Ratio (SNR) regime. Using this model, the deterministic capacity region of different networks with only private messages has been characterized in [3, 4, 5, 6, 7, 8].
The authors of [3] characterized the deterministic capacity region of the multipair bidirectional wireless relay network and then, using the gleaned insights, proposed a transmission strategy for the gaussian twopair twoway full duplex relay network and found an approximate characterization of the capacity region [6]. The authors of [4] developed a new tighter upper bound based on the notion of onesided genie. They utilized this bound to completely characterize the multicast deterministic capacity of a two pair symmetric bidirectional half duplex wireless relay network with only private messages. The authors of [5] utilize the genieaided bound developed in [4] to characterize the capacity region of the deterministic Y channel with private messages only. They established the achievability by using three schemes: bidirectional, cyclic, and unidirectional communication. In [8], the approximate capacity of the gaussian Ychannel was obtained. The authors of [7], characterize the deterministic capacity region of a fournode relay network with no direct links between the nodes.
In this paper, we study the deterministic capacity region of the Ychannel with private and common messages. This work is the first to deal with common messages, as all the previous work stated above deals only with private messages. The Ychannel consists of three users and one intermediate relay, where each user aims to convey two private messages to the other two users in addition to a common message to both of them via the intermediate relay. First, we use cutset bound and genieaided bounds to obtain an outer bound on the deterministic capacity region of the network. Then, we define an achievable rate region of the network and show that it coincides with the upper bound at a certain number of levels at the users. Finally, we develop a greedy strategy, namely the Gain Ordering Scheme (GOS), to send messages over the network. This strategy is then used to characterize the achievable rate region of the network. In principle, the GOS starts by sending the bits which can be combined effectively at the relay and ends with the ones that must be sent individually.
The rest of the paper is organized as follows. We describe the system model and state our main results in section II. In Section III, we detail our achievability schemes. In Section IV, we discuss the genieaided bounds in the existence of common messages. In Section V, we mention some insights about our work and illustrate our achievability scheme with two numerical examples. Finally, Section VI presents our conclusions.
Ii System Model
Consider the multiway relaying network, shown in Fig. 1, where user aims to convey two private messages with rates and to users and , respectively, in addition to a common message to both of them with rate via the intermediate relay, for and . Due to the absence of direct links between the users, communication occurs through the relay in two phases: uplink phase and downlink phase. All nodes are assumed to be full duplex (i.e., the nodes can transmit and receive simultaneously). In a deterministic channel model [2], the channel can be defined as the number of levels between each user and the relay, where each level can be represented as a circle as Fig 2 depicts. We denote the channel between user and the relay by , where . We assume that the channel is reciprocal, i.e., () where and is the SNR. We can assume without loss of generality that
(1) 
otherwise we can relabel the nodes.
An outer bound on the deterministic capacity region of this network is given by the following theorem:
Theorem 1. The capacity region of
the deterministic Y channel is given by the rates which satisfy the following conditions
(2) 
(3) 
(4) 
(5) 
(6) 
(7) 
(8) 
(9) 
where is the rate from node to node and is the rate from node to both nodes and , where and . This capacity region is outer bounded by , where
(10) 
The following Lemma represents the main result of the paper,
Lemma 1. The Gain Ordering Scheme achieves
all the integral rate tuples in the intersection between the capacity
region stated in Theorem 1 and the following extra condition:
(11) 
The proof of Lemma 1 will be shown at the end of Section III. It is worthmentioning that, when , our achievable region coincides with the upper bound. This can be simply concluded by observing that the extra condition defined in Lemma 1 is equivalent to the upper bound equations (8) and (9) in Theorem 1.
Iii Achievability
In this section, we use the GOS to characterize the regions that can be achieved. The rationale of the GOS idea is to start by selecting the bits from the rate tuple R that can together attain the maximum gain, as defined below. Then we eliminate the transmitted bits from R and continue with the transmission of the remaining rate tuple using the intermediate gain strategy. Finally, after sending the bits which achieve the intermediate gain, we go to the minimum gain strategy with the residual rate tuple. Note that the relay does not need to decode the data and therefore, combining bits on the same level at the relay provides gains in achievable rates. However, as we shall see below, these gains differ according to the number of levels combined.
Iiia Maximum Gain Strategy
This case occurs when two users want to exchange private bits. That is, and are both nonzero for some , . Thus, sending two bits on the same level from the two users yields a gain of 2 bits/level. In particular, each user can decode its desired bit by xoring its own bit with the received signal. This scheme continues until we send . Consider a rate tuple R , where
(12) 
Let
(13)  
In the uplink users 1 and 2 use levels
to send their binary vectors
and . The relay obtains and sends it back on the same levels. Now, user 1 and 2 by knowing their information can perfectly decode their desired information from . Similarly users 1 and 3 use relay levels . Also, users 2 and 3 use levels .To guarantee that this strategy works perfectly, we should have and . Since then
(14) 
(15) 
and subsequently the levels of the relay are sufficient for this strategy to work. Now, we need to achieve
(16) 
Then, we remove the occupied relay levels, thus the available levels are now , and , where
(17) 
(18) 
Recall that in the communication between user 1 and 2, we use the relay levels . This does not consume levels in if . In this case, . Otherwise, levels are used in . Subsequently, . So,
(19) 
The residual non zero rates in the vector R can be represented by any of the next strategies.
IiiB Intermediate Gain Strategy
We use this strategy when the minimum of and is equal to zero, . In this strategy, we start by the new rate tuple in (IIIA). The gain achieved from this strategy is 1.5 bits/level. There are four different scenarios that can attain this gain. Since the common bit is intended to two users, so it is reasonable to give it a higher priority than the private one. As a result, we order the four scenarios according to the number of common bits relative to the number of private bits. In what follows, we detail the four scenarios for achieving 1.5 bits/level as follows,

Full Common Scenario
In this case, each user has at least one common message i.e., (). Let(20) Let the transmit binary vectors of users 1, 2, and 3 be , , and . Users 1 and 2 send and on relay levels . User 2 also repeats on relay levels together with user 3 which sends on the same levels. The relay receives and and then sends them back to all the users on the levels . All Users receive and . Each user now can decode its intended messages without error. User 2 can decode and by xoring its own bit with and , respectively. Then, by xoring with , user 1 can extract and afterwards, user 1 xor with to get . Similarly, user 3 can decode and from and , respectively. To guarantee the operation of this strategy, we need , (i.e., ). Since
(21) Therefore, there are enough levels for serving all bits in this scenario. After removing all the occupied levels and subtracting the rates that are achieved. The rate tuple that still needs to be achieved is
(22) and the new number of levels at each node will be as follows,
(23) (24) (25) 
Two Common One Private Scenario
This scenario applies when one of the users has and , for and . We aim to achieve with at least 3 zero private components and 1 zero common component. Depending on which 3 private rates out of the 6 private rates in are zero, we have 8 different cases. Similarly, depending on the zero common rate, we may have three different cases for zero common rate. However, as space limitations prohibit showing all the cases, we focus on one of them and the rest follows in the same way. Hence, we select the case where and . Therefore, from (1), the rate vector that needs to be achieved is(26) Now, we start this scenario by defining the following
(27) (28) Let the transmit vectors of users 1, 2, and 3 be , , and . Users 1 and 2 send and on relay levels . User 1 also repeats on relay levels together with user 3 which sends on the same levels. The relay receives and and sends them back on the same levels. Users 1 and 2 receive and . User 3 receives only. User 1, knowing , can easily extract and from and , respectively.Then, knowing , user 2 can decode from and afterwards decodes from . Finally, user 3 extracts from . This strategy works if we have enough levels at the relay that allow the transmission and reception of and , i.e., , .
(29) (30) (31) (32) (33) (34) (35) (36) Now considering the case where is the minimum in (27). The remaining vector that needs to be achieved is
(37) with number of levels at each node where,
(38) (39) if then , otherwise, . Therefore,
(40) 
One Common Two Private Scenario
For the operation of this scenario, there should be at least one of the users has nonzero common bits in addition to one of the following two conditions should be satisfied in the residual rate vector
Each of the other two users have a nonzero private bits to the user that has common bits.

User has two common bits to both users , user has private bits to user and user has private bits to user , for and .
(41) (42) (43) Notice that, since we always start with the maximum gain phase, either or must be zero because one of and should be zero. According to (26), we should have equal zero. Let the transmit vectors of users 1, 2, and 3 be and . Users 1 and 2 send and on relay levels . User 1 also repeats on relay levels together with user 3 which sends on the same levels. The relay receives and and sends them back on the same levels. Users 1 and 2 receive and . User 1, knowing , can easily extract and from and , respectively. Then, knowing , user 2 can decode from and afterwards decodes from . Finally, user 3 extracts from .
Now, we start transmitting bits, users 1 and 2 send and on relay levels . User 1 also repeats on relay levels together with user 3 which sends on the same levels. The relay receives and and sends them back on the same levels. Users 1 and 2 receive . User 1, knowing , can easily extract and from and , respectively. Then, knowing , user 2 can decode from . Finally, user 3 extracts from . This strategy works if we have enough levels at the relay, i.e., , , and by extracting from equations (39), (25) and (19), and from equations (38) , (24) and (18), respectively, we get the following equations after rearranging
(44) (45) In what follows, we prove equations (44) and (45)
Similarly, for equation (45)
After removing the occupied levels, the residual rate vector that still needs to be achieved is,
(46) with number of levels at each node where,
(47) (48) (49) 

Full Private Scenario
In this scenario, the transmission and reception strategy is similar to the cyclic communication scheme described in [5], where the users communicate in a rotational manner, either or . These two cycles can be represented by the following two equations(50) (51) Note that, as we always start with the maximum gain strategy, either or must be zero. Additionally, from our selected rate tuple, .
IiiC Minimum Gain Strategy
Finally, we want to achieve the remaining rate vector with the associated number of levels at each node in (47),(48) and (49). It’s obvious that the remaining rates in (3) can not be achieved via the maximum gain strategy nor the intermediate gain one. Hence, we send each bit on a single level achieving a gain of one bit per relay level.
Considering our case: and . Let the transmit vectors of users 1, 2 be , . User 1 transmits on relay levels . User 2 sends on relay levels . Then the relay forwards on levels , and on levels . To guarantee the transmission of these bits, we should have and in uplink, and in downlink. After combining we get,
(52) 
(53) 
(54) 
Starting from (52), we should have , since . Therefore, .
In (53), we should have , since . Therefore, .
In (54), we should have , since .
where (57) is one of the terms in the minimum expression of (11), and it can be simply shown that this term is less than the other one, that’s why we take the minimum. However, in (12), if we keep and instead of and , respectively, and follow the same way, we get the following inequality
(58) 
which is the other term in the minimum expression of (11), and again, it can be simply shown that this term is less than the other one. Hence, in order to serve the remaining bits, the condition in Lemma 1 should be satisfied.
Iv Single Sided Genie Upper Bounds for Common Messages
The relay channel can be represented as the combination of multiple access channel i.e., uplink, and broadcast channel i.e., downlink. In the traditional cutset bounds in [1], the nodes are partitioned into two sets and which represent the transmitting and receiving relays, respectively. As was mentioned in [4], if all nodes in fully cooperate and share all their side information, this cooperation is referred to as the twosided genie aided bound. As was shown in [4], applying this traditional cutset bound to the relay network produces loose bounds. Therefore, a tighter singlesided genie aided upper bound was developed in [4], where the notion of single sided genie comes from transferring information in only one direction by the genie. Due to the existence of common messages in our network, we found that this bound is not tight at some regions. To show that, we consider one of the cuts around the relay. As shown in [4], we assume that the genie transfers only all data of node to node and , i.e., . Also, it transfers all data of node to node only, i.e., . Therefore, the data sent from node to node i.e. is not known at node a priori. Also, the data sent from node to both nodes and i.e. is not known at both of them. As shown here, represents a bottleneck because we assumed that the genie transfers messages from node to node only. However, it is not known at node . This results in a looser inequality,
(59) 
In the above equation, we did not exploit the common messages information transferred by the genie from user to user . As a result, we believe that this upper bound is not tight in existence of common messages.
V Discussion
In Section IV, we argued that the singlesided genie bounds and the cutset bounds are not sufficient to characterize the capacity region of the networks with common messages. As the common message will be known at one node only, however, we should send it to the other one, and this will lead to a looser inequality (59). As a result, for the nonachievable rate tuples, we found that we need extra relay levels to send the common message as illustrated in example 2. Our results reveals that, when , the upper bound is achievable. Additionally, it is important to mention that we only need the extra condition of Lemma 1 if and only if in (43), otherwise, any rate tuple is achievable using our greedy scheme. We illustrate our work through two examples. The rate tuple in the first example is achievable while the other one violates (11) and hence, can not be achieved. Consider a reciprocal network with channel gains and the rate tuple