In parallel interference channel, the signals in different subchannels of one user is transmitted by the same transmitter; thus, the subchannels can be jointly encoded. Similarly, the signals received in different subchannels of the same user can be jointly decoded. The basic idea of the proposed transmission scheme is as follows. In weak interference subchannels, the bits is transmitted at the maximal possible data rate of the direct-link without considering the existence of cross-link interference. In strong interference subchannels, the bits that will generate interference in weak interference subchannels is retransmitted following a certain rule. The received signals in weak interference subchannels and strong interference subchannels are then jointly decoded.

### 4.1 A motivating example

To show the basic idea of our joint coding scheme, we first see a simple example. As shown in Fig. 5, the two-user parallel interference channel has three subchannels, we call them subchannels I, II, and III. The number of levels in the direct-link is three in both subchannels I and II. The number of levels in the cross-link is one and two in subchannel I and II, respectively. In subchannel III, two signal levels exist in the direct-link and three signal levels exist in the cross-link. According to the statements in Section 2.3, subchannels I and II are weak interference channels, while subchannel III is a strong interference channel.

In subchannel I, user 1 and user 2 transmit their bits on all signal levels of direct-link regardless of interference. Specifically, Tx_{1} transmits *a*
_{1}, *a*
_{2}, and *a*
_{3} and Tx_{2} transmits *b*
_{1}, *b*
_{2}, and *b*
_{3}. Obviously, there are interference at the two receivers as can be seen from Fig. 5
a. In particular, since the number of direct-link signal levels *M*
_{1,1}=3 and that of cross-link signal levels *M*
_{2,1}=1, a bit *a*
_{1}⊕*b*
_{3} is received at Rx_{2}, which means that *b*
_{3} is interfered by *a*
_{1}. Because the channel is symmetric for two users, the interference scenario is similar at Rx_{1}, i.e., *a*
_{3} is interfered by *b*
_{1}. Provided the received bits in subchannel I, only *a*
_{1} and *a*
_{2} are decodable at Rx_{1} and only *b*
_{1} and *b*
_{2} are decodable at Rx_{2}. The contaminated bits *a*
_{3} and *b*
_{3} cannot be decoded without external help.

Subchannel II is also a weak interference channel, we use similar transmission strategy as in subchannel I. The number of cross-link signal levels of this subchannel is larger, *M*
_{1,2}=*M*
_{2,1}=2. Therefore, although three bits are still transmitted for each user, two bits are interfered at each receiver, and only *a*
_{4} and *b*
_{4} can be decoded.

Since the bits are transmitted regardless of interference in weak interference subchannels, different bits can be transmitted in subchannel II and subchannel I, and they are independent. This property is essential in our joint coding scheme as we will see in the sequel.

Subchannel III is a strong interference channel, where three signal levels exist in the cross-link channel. The transmission scheme in this subchannel is critical. It determines whether the contaminated bits in subchannels I and II are decodable, and affects the spectrum utilization efficiency of the parallel interference channel.

The bits that will generate interference in weak interference subchannels are retransmitted in subchannel III, which is used to recover the contaminated bits in subchannel I and subchannel II. To avoid the interference between user 1 and user 2 in retransmission, a straightforward scheme at hand is orthogonal-based transmission schemes, so that there is no interference between user 1 and user 2. For example, in the first time slot, Tx_{1} retransmits *a*
_{1},*a*
_{4}, and *a*
_{5}. Through the cross-link, these bits arrive at Rx_{2} and can be used to cancel the interference appeared in subchannels I and II. In the second time slot, Tx_{2} retransmits *b*
_{1},*b*
_{4}, and *b*
_{5}, and Rx_{1} uses these bits for interference cancelation. However, this scheme is obviously inefficient.

A better choice is to let Tx_{1} and Tx_{2} retransmitting simultaneously in subchannel III. As shown in Fig. 5
c, both Rx_{1} and Rx_{2} obtain contaminated bits and cannot recover these bits individually. However, taking into account the received bits in subchannels I and II, these bits can be jointly decoded. For Rx_{1}, in three subchannels, totally nine bits are obtained, which can be expressed as follows:

$$ \begin{array}{ll} r_{11} &= a_{1},\\ r_{12} &= a_{2},\\ r_{13} &= a_{3} \oplus b_{1},\\ r_{14} &= a_{4},\\ r_{15} &= a_{5} \oplus b_{4},\\ r_{16} &= a_{6} \oplus b_{5},\\ r_{17} &= b_{1},\\ r_{18} &= b_{4} \oplus a_{1},\\ r_{19} &= b_{5} \oplus a_{4},\\ \end{array} $$

(12)

where *r*
_{1i
}, *i*=1,⋯,9 is the received bits at Rx_{1}. We can see from (12) that each received bit correspond to an equation and all the received bits provide us with a set of equations.

It can be seen that *a*
_{1},*a*
_{2},*a*
_{4}, and *b*
_{1} can be obtained immediately when *r*
_{11},*r*
_{12},*r*
_{14}, and *r*
_{17} is received. But the other five received bits are not simply transmitted bits from user 1 or user 2, none of which can be recovered by a single equation. Fortunately, the nine equations in (12) are linear uncorrelated, and there are only nine unknown variables in (12). By solving the set of equations, all the nine bits can be recovered. In these bits, six are transmitted by user 1, which are the desired bits. The other three bits, which are transmitted by user 2 to facilitate interference cancelation, will be discarded after decoding. Since the channel is symmetric, similar characteristic holds for Rx_{2}.

###
**Remark 1**

In subchannel III, the order of the retransmission cannot be arbitrary. For example, if we exchange the occupied levels of *a*
_{4} and *a*
_{5}, as labeled in the outer column in Fig. 5
c, *a*
_{4}⊕*b*
_{5} will be received twice at Rx_{2}. One is obtained in subchannel II, and the other is obtained in subchannel III, as indicated by the black box. In this case, part of the equations are linearly correlated, and the desired bits cannot be fully decoded.

### 4.2 Subchannel grouping

In the example above, the cross-link signal levels of weak interference subchannels are as many as that of strong interference subchannel. This condition is obviously not satisfied in most scenarios. Under the condition where there are more cross-link signal levels in weak interference subchannels than that of strong interference subchannels, the resource to recover the bits contaminated in weak subchannels is not enough. Under opposite condition, the resource is too much and will be wasted. Therefore, a preprocessing step called subchannel grouping is introduced.

Denote the total number of cross-link signal levels of all the weak interference subchannels as *N*
_{weak} and that of all the strong interference subchannels as *N*
_{strong}. If *N*
_{weak}>*N*
_{strong}, we can select part of the weak interference subchannels to participate joint coding. The aggregated number of cross-link signal levels of this part of subchannels is \(N^{\prime }_{\text {weak}}\), which satisfies \(N^{\prime }_{\text {weak}} \leq N_{\text {strong}}\). At the same time, other weak interference subchannels employ individual coding introduced in Section 3. If *N*
_{weak}<*N*
_{strong}, we can select part of the strong interference subchannels to participate joint coding. The aggregated number of cross-link signal levels of this part of subchannels is \(N^{\prime }_{\text {strong}}\), which still satisfies \(N^{\prime }_{\text {strong}} \geq N_{\text {weak}}\). At the same time, other strong interference subchannels employ individual coding. As will be seen in next part, subchannel grouping ensures to satisfy a necessary condition under which the bits in the subchannels participating joint coding can be jointly decoded.

### 4.3 Joint coding scheme for two-user case

After the subchannels for joint coding are selected, in weak interference subchannels, all the direct-link signal levels are used to transmit new bits regardless of interference. In strong interference subchannels, the bits that will generate interference in weak interference subchannels, i.e., interfering bits, are retransmitted. It is demanded that in the retransmission process, the relative level orders between any pair of bits are kept unchanged compared with the orders when they are transmitted in the weak interference subchannels.

The joint coding scheme ensures the feasibility that the transmitted bits can be jointly decoded. In the following, we will formally prove the necessary condition and the feasibility of the proposed transmission scheme. As we will see, the necessary condition guarantees that there are enough number of equations, while the feasibility comes from the requirement that these equations are linearly uncorrelated. When enough number of linearly uncorrelated equations are obtained, the bits can be decoded.

###
**Necessary Condition**

For subchannels participating joint coding, the aggregated cross-link signal levels of strong interference subchannels should be no less than the aggregated cross-link signal levels of weak interference subchannels.

###
*Proof*

To ensure the interference in weak interference subchannels which can be eliminated with the help of strong interference subchannels, at each receiver, the number of equations should be no less than the number of desired bits and interfering bits. Without loss of generality, we consider user 1. Assume that in all weak interference subchannels, there are totally *X* signal levels in the direct-link and *Y* signal levels in cross-links. Then, we have *X*+*Y* unknown bits but only have *X* linear equations. To decode these bits, we need at least *Y* more linear uncorrelated equations, which should be provided by the cross-link retransmission in strong interference subchannels. Thus, the number of the aggregated signal levels of cross-links in strong interference subchannels should be no less than *Y*. □

###
**Feasibility**

By the proposed joint coding scheme, the desired bits can be recovered at each receiver.

###
*Proof*

The cornerstone of this proof is the fact that the transmit levels of interfering bits are always lower than those of the desired bits in weak interference subchannels and vice versa in strong interference subchannels at receivers.

Assume that in a weak interference subchannel *S*
_{1}, Tx_{1} transmits bit *a*
_{
m
} and Tx_{2} transmits bit *b*
_{
n
}, these two bits collide on the same signal level at Rx_{2}. The bit *a*
_{
m
} will be retransmitted in strong interference subchannel *S*
_{2} and might collide with a bit *b*
_{
p
} at Rx_{2}. The bit *b*
_{
p
} is first transmitted by Tx_{2} in a weak interference subchannel *S*
_{3} and is retransmitted by Tx_{2} in *S*
_{2}. There are two possible cases when we check the linear correlation property between *a*
_{
m
}⊕*b*
_{
n
}=*r*
_{2k
} and *a*
_{
m
}⊕*b*
_{
p
}=*r*
_{2j
}, where *m*,*n*,*p*,*k*,*j* are integers.

The first case is that *S*
_{3}≠*S*
_{1}, i.e., *b*
_{
p
} and *b*
_{
n
} come from different weak interference subchannels. This suggests that they are different bits, for independent bits are transmitted in different weak interference subchannels. Hence, the two equations *a*
_{
m
}⊕*b*
_{
n
}=*r*
_{2k
} and *a*
_{
m
}⊕*b*
_{
p
}=*r*
_{2j
} are linearly uncorrelated.

The second case is that *S*
_{3}=*S*
_{1}, i.e., *b*
_{
p
} and *b*
_{
n
} come from the same weak interference subchannel. In what follows, we show that they must occupy different signal levels at Tx_{2}.

Assume that *a*
_{
m
} is transmitted on the *m*th level of Tx_{1} in *S*
_{1} and retransmitted on the *m*
^{′}th level of Tx_{1} in *S*
_{2}. *b*
_{
n
} is transmitted on the *n*th level of Tx_{2} in *S*
_{1} and retransmitted on the *n*
^{′}th level of Tx_{2} in *S*
_{2}. *b*
_{
p
} is transmitted on the *p*th level of Tx_{2} in *S*
_{1} and retransmitted on the *p*
^{′}th level of Tx_{2} in *S*
_{2}. In weak interference subchannels, SNR>INR; thus, we have *m*<*n*. In strong interference subchannels, SNR<INR; thus, we have *m*
^{′}>*p*
^{′}. Since the relative order in the retransmission process is the same with that in the first transmission, from *m*
^{′}>*p*
^{′}, we can derive *m*>*p*. Finally, we have the relationship *n*>*m*>*p*, which means *b*
_{
p
} and *b*
_{
n
} comes from different signal level at Tx_{2}, and they are two different independent bits. As a consequence, the two equations *a*
_{
m
}⊕*b*
_{
n
}=*r*
_{2k
} and *a*
_{
m
}⊕*b*
_{
p
}=*r*
_{2j
} must be linear uncorrelated, and all the bits can be recovered. □

To help understand the proof, we provide an example here. As shown in Fig. 6, *S*
_{1} and *S*
_{2} are respectively a weak interference subchannel and a strong interference subchannel, and *S*
_{3}=*S*
_{1}. In this example, suppose that *a*
_{
m
}=*a*
_{2}, then it follows that *b*
_{
n
}=*b*
_{3} and *b*
_{
p
}=*b*
_{1}. In subchannel *S*
_{1}, since SNR>INR, *a*
_{2} and *b*
_{3} collide on the same signal level at Rx_{2}. In subchannel *S*
_{2}, since SNR<INR, *a*
_{2} and *b*
_{1} collide on the same signal level at Rx_{2}. It follows that *a*
_{2}⊕*b*
_{3} and *a*
_{2}⊕*b*
_{1} are uncorrelated. The bits *a*
_{
k
} and *b*
_{
j
} in Fig. 6(b) are first transmitted in other weak interference sunchannels, which are not shown here.

### 4.4 Joint coding scheme for multi-user case

In a multi-user symmetric interference channel, all the direct-link channel gains are identical and so are the cross-link gains. As shown in Fig. 7, for a *K*-user symmetric interference channel, each user receives interference from other *K*−1 users. Due to the symmetry, at each receiver, the interference signal levels are aligned. If we view the aligned interference as coming from a virtual user, the joint coding and decoding in the multi-user case can be implemented as same as in the two-user case.

The proofs of necessary condition and feasibility for multi-user joint coding are the same as in the two-user case, except that we view the aligned interference from *K*−1 users as coming from one virtual user. Figure 7 is an example of a three-user parallel symmetric interference channel, where only two subchannels are shown. Subchannel I is a weak interference subchannel, and subchannel II is a strong interference subchannel. In subchannel I, as can be seen in Fig. 7
a, the modulo-2 addition of *a*
_{2}⊕*b*
_{1}⊕*c*
_{1}=*r*
_{12} is received on the second level of Rx_{1}, where *a*
_{2} is the desired bit and *b*
_{1}⊕*c*
_{1} is the interference bit. We can regard *b*
_{1}⊕*c*
_{1} as one bit *u*
_{1} that is transmitted by a virtual user Tx _{
u
}. Then, we obtain an equation *a*
_{2}⊕*u*
_{1}=*r*
_{12}. In subchannel II, as can be seen in Fig. 7
b, *b*
_{1} and *c*
_{1} are retransmitted by Tx_{2} and Tx_{3}, respectively, and the virtual interference bit *u*
_{1}=*b*
_{1}⊕*c*
_{1} will reappear at Rx_{1} with a desired bit *a*
_{
l
}, where *a*
_{
l
} is a retransmission bit that causes interference in another weak interference subchannel which is not shown here. Then, we obtain another equation *a*
_{
l
}⊕*u*
_{1}=*r*12′. Since the levels of interfering bits are always lower than the levels of desired bits in weak interference subchannels and vice versa for strong interference subchannels at receivers, even though *a*
_{2} is also retransmitted in this strong interference subchannel, the relative level order of *a*
_{
l
} and *a*
_{2} cannot be equal at Tx_{1}. Thus, the two equations *a*
_{2}⊕*u*
_{1}=*r*
_{12} and *a*
_{
l
}⊕*u*
_{1}=*r*12′ are linear uncorrelated. Then, the desired bits of user 1 can be jointly decoded at Rx_{1}. Similar equations can be obtained at Rx_{2} and Rx_{3}, and in this way, we generalize the joint coding scheme to multi-user parallel symmetric interference channels.