text:
\section{Introduction}
\IEEEPARstart{I}{n} a $K$-user multiple input multiple output (MIMO) Gaussian interference channel, a vector space interference alignment (IA) technique with neither time nor frequency diversity prevents interference on the desired user by causing the interference and user signals to use the different spaces provided by multiple Tx/Rx antennas. At the $k$-th Tx/Rx pair, a plurality of interference signals from all other $\left( K-1 \right)$ Tx/Rx pairs are aligned to reduce the occupied space, while the remaining interference-free space is preserved for desired signals. With a limited number of Tx/Rx antennas, the interference subspace dimension reduced by IA is constrained by two factors: the maximum number of interference signals transmitted from one of the interfering sources and the aligned level for interference from the remaining $\left( K-1 \right)$ interfering sources. First, the dimension of interference subspace after reduction would be no less than the value given by the first factor, since the interference signals from the same source could not be aligned. Then, the reduced dimension depends on the number of Tx/Rx antennas as it determines the capability of alignment operation. In order to further reduce the occupied interference subspace while expanding signal subspace under the limited number of Tx/Rx antennas, therefore, interference cancellation (IC) can be combined with vector space IA as another type of interference mitigation technique, which is referred to as interference alignment and cancellation (IAC) [1]. Under the help of IC, the number of interfering sources can be dramatically reduced in the receivers. Therefore, more dimension of signal subspace would be preserved by IAC, which would gain more degrees of freedom (DoFs) over IA for a $K$-user MIMO interference channel.

In the classical IA scheme, let ${{\mathbf{V}}_{j}}$ and ${{\mathbf{U}}_{k}}$ denote the precoding matrix at transmitter $j$ and the zero-forcing matrix at receiver $k$, respectively, while ${{\mathbf{H}}_{kj}}$ denotes the channel matrix from transmitter $j$ to receiver $k$, $j,k\in \left[ 1,K \right]$. Then, the alignment conditions can be summarized as
\begin{IEEEeqnarray}{lcr}
\mathbf{U}_{k}^{H}{{\mathbf{H}}_{kj}}{{\mathbf{V}}_{j}}=\mathbf{0},\text{ }\forall k\ne j \\
\operatorname{rank}\left( \mathbf{U}_{k}^{H}{{\mathbf{H}}_{kk}}{{\mathbf{V}}_{k}} \right)={{d}_{k}}
\end{IEEEeqnarray}
where ${{d}_{k}}$ denotes the target DoF to achieve at receiver $k$. The conditions in (1) and (2) assure that the interference subspace $\sum\nolimits_{j\ne k}{{{\mathbf{H}}_{kj}}{{\mathbf{V}}_{j}}}$ is complementary to the signal subspace ${{\mathbf{H}}_{kk}}{{\mathbf{V}}_{k}}$ at the receiver $k$. A few heuristic algorithms have been proposed to solve the transceivers, $\left\{ {{\mathbf{V}}_{j}} \right\}$ and $\left\{ {{\mathbf{U}}_{k}} \right\}$, in an iterative manner [2,3]. However, these algorithms cannot determine whether the transceivers exist for a given tuple of DoFs $\left( {{d}_{1}},{{d}_{2}},\cdots,{{d}_{K}} \right)$, nor is there any guarantee for converging to the optimal transceivers even when they exist. Later, the authors in [4] have shown that it is NP-hard to find $\left\{ {{\mathbf{V}}_{j}} \right\}$ and $\left\{ {{\mathbf{U}}_{k}} \right\}$ jointly by solving (1) and (2) with a given tuple of DoFs [4]. They have also proposed an iterative algorithm to solve $\left\{ {{\mathbf{V}}_{j}} \right\}$ and $\left\{ {{\mathbf{U}}_{k}} \right\}$, which can computes a local optimal solution without resorting to a priori specification of the DoF tuple. In [4], however, with the utility function of ${SINR}/{\left( 1+SINR \right)}$, the optimized performance is not guaranteed if the interferences fail to be aligned, since it is still unknown whether the IA transceivers exist or not.

In summary, the aforementioned works have raised various issues to find the IA transceivers. Obviously, IAC inherits the NP-hardness from IA in terms of finding the transceivers for a given tuple of DoFs. In our earlier work, therefore, instead of investigating the NP-hard problem for IAC, we consider suboptimal heuristics design. By proposing a symbol-to-symbol (STS) alignment structure and constructing an IAC graph, we have derived the necessary and sufficient conditions for the existence of IAC transceivers [5]. In this paper, we aim to design the closed-form transceivers when they exist for IAC in MIMO interference channel. We first propose a general design for any given tuple of DoFs $\left( {{d}_{1}},{{d}_{2}},\cdots,{{d}_{K}} \right)$ that satisfies the proposed necessary and sufficient conditions. Then, we present how the optimal IAC transceivers can be designed to achieve the maximum DoFs of $2M$.

The remainder of this paper is organized as follows: Section II presents the system model and reviews how IAC works in MIMO interference channel. Section III presents the proposed IAC with symbol-to-symbol alignment scheme and list the main results obtained in [5]. Section IV provides the general design principle for the closed-form IAC transceivers while providing the optimum design on them in Section V. Finally, conclusion is made and future works are suggested in the last section.

\section{System Model}
Fig. 1 illustrates a MIMO interference channel, in which $K$ transceiver pairs share the same resource while each transmitter and receiver are equipped with $M$ antennas. Assume that transmitter $j$ sends ${{d}_{j}}$ independent data streams to receiver $j$, incurring interferences to other receivers $\left( {{d}_{j}}\le M,\text{ }j=1,2,\cdots ,K \right)$. We can further define the total DoF in the system as ${{\operatorname{DoF}}^{sys}}=\sum\nolimits_{j=1}^{K}{{{d}_{j}}}$. Let ${{\mathbf{x}}_{j}}$ denote the transmitted signal vectors of dimension ${{d}_{j}}\times 1$ from the transmitter $j$, in which each element of the vector corresponds to one independent data symbol, denoted by ${{x}_{j\ell }}$, $\ell =1,2,\cdots ,{{d}_{j}}$. Furthermore, let ${{\mathbf{H}}_{kj}}$ denote an $M\times M$ channel matrix from transmitter $j$ to receiver $k$ with each entry drawn independently from a continuous distribution while allowing no channel extension, and ${{\mathbf{V}}_{j}}=[{{\mathbf{v}}_{j1}},{{\mathbf{v}}_{j2}},\cdots,{{\mathbf{v}}_{j{{d}_{j}}}}]$ of dimension $M\times {{d}_{j}}$ represent a transmit precoding matrix at transmitter $j$ where each column vector is applied to each data symbol $\left( j,k=1,2,\cdots,K \right)$. Meanwhile, let ${{\mathbf{U}}_{k}}$ of dimension $M\times {{d}_{k}}$ be a zero-forcing matrix at receiver $k$ and ${{\mathbf{y}}_{k}}$ denote the output signal vector of dimension ${{d}_{k}}\times 1$ at receiver $k$, which is given by
\begin{IEEEeqnarray}{lcr}
\begin{array}{l}
{{\mathbf{y}}_{k}}=\underbrace{\mathbf{U}_{k}^{H}{{\mathbf{H}}_{kk}}{{\mathbf{V}}_{k}}{{\mathbf{x}}_{k}}}_{\text{desired signal}}+\underbrace{\mathbf{U}_{k}^{H}\sum\limits_{j=1;j\ne k}^{K}{{{\mathbf{H}}_{kj}}{{\mathbf{V}}_{j}}{{\mathbf{x}}_{j}}}}_{\text{interference}}+\mathbf{U}_{k}^{H}{{\mathbf{n}}_{k}},$ $$ $$ $
\\ $ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $k=1,2,\cdots ,K
\end{array}
\end{IEEEeqnarray}
where ${{\mathbf{n}}_{k}} \sim N\left( \mathbf{0},{{\sigma }^{2}}\mathbf{I} \right)$ is zero-mean additive white Gaussian noise and ${{\mathbf{A}}^{H}}$ denotes a conjugate transpose of a matrix $\mathbf{A}$.

As each data symbol ${{x}_{j\ell }}$ is encoded by using a precoding vector ${{\mathbf{v}}_{j\ell }}$ and received through the channel as ${{\mathbf{H}}_{kj}}{{\mathbf{v}}_{j\ell }}{{x}_{j\ell }}$ at receiver $k$, the signal subspace ${{\tilde{S}}_{k}}$ can be expressed by a set of signal vectors, ${{\mathcal{S}}_{k}}=\left\{ {{\mathbf{H}}_{kk}}{{\mathbf{v}}_{k\ell }} \right\}$ for $\ell \in \left[ 1,{{d}_{k}} \right]$, while the interference subspace ${{\tilde{I}}_{k}}$ can be expressed by a set of interference vectors ${{\mathcal{I}}_{k}}=\left\{ {{\mathbf{H}}_{kj}}{{\mathbf{v}}_{j\ell }} \right\}$ for $j\in \left[ 1,K \right]$, $j\ne k$, and $\ell \in \left[ 1,{{d}_{j}} \right]$. Considering that IAC is a combined scheme of IA and IC, ${{\mathcal{I}}_{k}}$ can be also written into a union of two sets, ${{\mathcal{I}}_{k}}=\mathcal{I}_{k}^{\operatorname{IA}}\bigcup \mathcal{I}_{k}^{\operatorname{IC}}$, where $\mathcal{I}_{k}^{\operatorname{IA}}$ and $\mathcal{I}_{k}^{\operatorname{IC}}$ denote the sets of interference vectors to be mitigated by IA and IC operations, respectively.

At the receive side, IC operation works to subtract interference effect caused by the known signals, i.e., the decoded signals ${{\mathbf{\hat{x}}}_{k}}$ at receiver $k$ are sent over a backhaul link to other receivers that have not performed the decoding operation yet, so that the interference effect caused by ${{\mathbf{x}}_{k}}$ can be cancelled. For the purpose of successive cancellation, therefore, a decoding operation is performed in one receiver at a time. Without loss of generality, we assume that the decoding order follows from receiver $1$ to $K$ in the subsequent discussion. As receiver $k$ can obtain the decoded signals $\left\{ {{{\mathbf{\hat{x}}}}_{j}} \right\}_{j=1}^{k-1}$ from receivers $1$ to $\left( k-1 \right)$ through the backhaul link, an estimate of the corresponding interference $\sum\nolimits_{j=1}^{k-1}{{{\mathbf{H}}_{kj}}{{\mathbf{V}}_{j}}{{\mathbf{x}}_{j}}}$ can be cancelled from ${{\mathbf{y}}_{k}}$, $k=2,3,\cdots ,K$. As a result, a set of interference vectors mitigated by IC operation can be expressed by $\mathcal{I}_{k}^{\operatorname{IC}}=\left\{ {{\mathbf{H}}_{kj}}{{\mathbf{v}}_{j\ell }} \right\}$ for $j\in \left[ 1,k-1 \right]$ and $\ell \in \left[ 1,{{d}_{j}} \right]$.

\begin{figure}[!t]
\centering
\includegraphics[width=3.4in]{fig1.pdf}
\vspace*{-10pt}
\caption{$K\times K$ MIMO interference channel model for IAC}
\end{figure}

At the transmit side, IA operation works to precode the signals jointly, so that the interferences incurred at the receive side can be effectively aligned. As IC operation can directly subtract the interference signals in $\mathcal{I}_{k}^{\operatorname{IC}}$ at receiver $k$, such a part of interference is not considered for IA operation. In other words, IA operation is responsible for only the other part of interference which is not cancelled, i.e., we have $\mathcal{I}_{k}^{\operatorname{IA}}=\left\{ {{\mathbf{H}}_{kj}}{{\mathbf{v}}_{j\ell }} \right\}$ for $j\in \left[ k+1,K \right]$ and $\ell \in \left[ 1,{{d}_{j}} \right]$. Additionally, after a process of successive cancellation, the number of the remaining received symbols at each receiver $k$, given as $\left| {{\mathcal{S}}_{k}} \right|+\left| \mathcal{I}_{k}^{\operatorname{IA}} \right|=\sum\nolimits_{j=k}^{K}{{{d}_{j}}}$, decreases as the cancellation process progresses in order. As long as $\sum\nolimits_{j=k}^{K}{{{d}_{j}}}\le M$, the desired signals can be decoded without resorting to IA operation because a total of no more than $M$ packets can be separated by $M$ antennas. With $\left| {{\mathcal{S}}_{k}} \right|+\left| \mathcal{I}_{k}^{\operatorname{IA}} \right|\le M$ for receiver $k$, it is always true that $\left| {{\mathcal{S}}_{j}} \right|+\left| \mathcal{I}_{j}^{\operatorname{IA}} \right|\le M$ for $j\in \left[ k+1,K \right]$ and therefore, all remaining receivers $\left( k+1 \right),\left( k+2 \right),\cdots,K$ can be also decoded directly without resorting to IA operation. Therefore, the decoded signals $\left\{ {{{\mathbf{\hat{x}}}}_{j}} \right\}_{j=k}^{K}$ are not required to be shared for a successive cancellation purpose. If receiver $k$ is the first receiver which satisfies $\left| {{\mathcal{S}}_{k}} \right|+\left| \mathcal{I}_{k}^{\operatorname{IA}} \right|\le M$, a new label ${{k}_{\operatorname{IAC}}}$ is allocated to receiver $\left( k-1 \right)$ so as to indicate that only the interference received at receiver $1$ to ${{k}_{\operatorname{IAC}}}$ is required to be aligned and furthermore, only the decoded signals $\left\{ {{{\mathbf{\hat{x}}}}_{j}} \right\}_{j=1}^{{{k}_{\operatorname{IAC}}}}$ have to be shared through the backhaul link for a cancellation purpose.

Consequently, two steps are required for decoding in IAC scheme. In the first step, receivers $1$ to ${{k}_{\operatorname{IAC}}}$ will perform the decoding operation with one receiver at a time and then, the decoded signals $\left\{ {{{\mathbf{\hat{x}}}}_{j}} \right\}_{j=1}^{{{k}_{\operatorname{IAC}}}}$ are shared with receivers $\left( j+1 \right)$ to $K$ through a backhaul link for cancellation. In the second step, all other receivers, indexed by $({{k}_{\operatorname{IAC}}}+1),({{k}_{\operatorname{IAC}}}+2),\cdots,$ and $K$, decode their own signals simultaneously. Note that additional process of sending the decoded packets over the backhaul link in the first step is considered as overhead. For simplicity, here we roughly measure the corresponding overhead by counting the number of receivers, to which a decoded packet has to be sent over the backhaul link. For example, if ${{x}_{1\ell }}$ has been decoded at receiver 1, it would be sent to all other  $\left( K-1 \right)$ receivers. Thus, the overhead associated with ${{x}_{1\ell }}$ is counted as $\left( K-1 \right)$ packets. Similarly, ${{x}_{j\ell }}$ will have the overhead of $\left( K-j \right)$ packets. Note that its overhead is zero for $j>{{k}_{\text{IAC}}}$. Let ${{O}_{\operatorname{sys}}}$ denote the total overhead associated with the whole system, given as ${{O}_{\operatorname{sys}}}=\sum\nolimits_{j=1}^{{{k}_{\operatorname{IAC}}}}{\left( K-j \right){{d}_{j}}}$.

\section{Interference Alignment and Cancellation with Symbol-to-Symbol Alignment Scheme}
In our earlier work [5], instead of investigating the NP-hard problem for a joint transceiver design, we consider a design of the transmitter $\left\{ {{\mathbf{V}}_{j}} \right\}$ first and then the receiver $\left\{ {{\mathbf{U}}_{k}} \right\}$ for IAC. Aiming for a successful decoding at each receiver $k$, the transmitter $\left\{ {{\mathbf{V}}_{j}} \right\}$ should be designed so that the signal subspace ${{\tilde{S}}_{k}}$ can be complimentary to the interference subspace ${{\tilde{I}}_{k}}$. This indicates two constraints to meet: first, ${{\tilde{S}}_{k}}$ and ${{\tilde{I}}_{k}}$ should be linearly independent; second, the dimension of ${{\tilde{S}}_{k}}$ and ${{\tilde{I}}_{k}}$ should satisfy the following condition:
\begin{IEEEeqnarray}{lcr}
\dim({{\tilde{S}}_{k}})+\dim({{\tilde{I}}_{k}})\le M
\end{IEEEeqnarray}
where $\dim\left( \mathcal{A} \right)$ corresponds to the cardinality of a basis for a vector space $\mathcal{A}$. As channel matrices $\{{{\mathbf{H}}_{kj}}\}$ are given, ${{\tilde{S}}_{k}}$ and ${{\tilde{I}}_{k}}$ can be determined as long as the transmitter $\left\{ {{\mathbf{V}}_{j}} \right\}$ are solved, and then, the receiver $\left\{ {{\mathbf{U}}_{k}} \right\}$ can always be found in the left null space of ${{\tilde{I}}_{k}}$. In order to meet the two constraints given in the above for $\left\{ {{\mathbf{V}}_{j}} \right\}$, we have proposed a symbol-to-symbol (STS) alignment structure for IAC in [5], based on which an IAC graph has been constructed so that the necessary and sufficient conditions for the existence of closed-form transceiver solutions can be derived.

In this section, we first briefly review the STS alignment structure. Then, we give an illustrative example to show how IAC with STS alignment scheme works in MIMO interference channel. At last, a notion of IAC graph is reviewed and then, the associated results are also discussed.

\subsection{Symbol-to-symbol (STS) alignment structure: Overview}
The constraint on ${{\mathbf{V}}_{j}}$ in (4) motivates us to align ${{\tilde{I}}_{k}}$ onto a set of basis vectors with dimension ${{Z}_{k}}=M\text{-}\dim({{{\tilde{S}}}_{k}})$. As the interference vectors in $\mathcal{I}_{k}^{\operatorname{IC}}$ can be cancelled directly and do not burden on the dimension of ${{\tilde{I}}_{k}}$, only the interference vectors in $\mathcal{I}_{k}^{\operatorname{IA}}$ are considered for alignment. The STS alignment structure can be constructed by two steps: first, a set of basis vectors, denoted by $\mathcal{\bar{I}}_{k}^{\operatorname{SIA}}$, is directly {\it selected} among interference vectors in $\mathcal{I}_{k}^{\operatorname{IA}}$, i.e., $\mathcal{\bar{I}}_{k}^{\operatorname{SIA}}=\left\{ {{{\mathbf{\bar{i}}}}_{1,k}},\cdots,{{{\mathbf{\bar{i}}}}_{n,k}},\cdots,{{{\mathbf{\bar{i}}}}_{{{Z}_{k}},k}} \right\}\subset \mathcal{I}_{k}^{\operatorname{IA}}$ , where the superscription 'SIA' indicates the STS interference alignment, while the remaining interference vectors $\{{{\mathbf{i}}_{f,k}}\}_{f=1}^{_{\left| {{\mathcal{I}}_{k}} \right|-{{Z}_{k}}}}$ belong to $\left( \mathcal{I}_{k}^{\operatorname{IA}}-\mathcal{\bar{I}}_{k}^{\operatorname{SIA}} \right)$, i.e., $\mathcal{I}_{k}^{\operatorname{IA}}-\mathcal{\bar{I}}_{k}^{\operatorname{SIA}}=\left\{ {{\mathbf{i}}_{1,k}},\cdots,{{\mathbf{i}}_{f,k}},\cdots,{{\mathbf{i}}_{\left| \mathcal{I}_{k}^{\operatorname{IA}} \right|-{{Z}_{k}},k}} \right\}\subset \mathcal{I}_{k}^{\operatorname{IA}}$; second, alignment is implemented between single vectors corresponding to one symbol each. Once $\mathcal{\bar{I}}_{k}^{\operatorname{SIA}}$ is determined in the first step, each remaining interference vector, ${{\mathbf{i}}_{f,k}}\in \mathcal{I}_{k}^{\operatorname{IA}}-\mathcal{\bar{I}}_{k}^{\operatorname{SIA}}$, will be aligned onto only one of the basis vectors, ${{\mathbf{\bar{i}}}_{n,k}}\in \mathcal{\bar{I}}_{k}^{\operatorname{SIA}}$, i.e., $\operatorname{span}\left( {{\mathbf{i}}_{f,k}} \right)=\operatorname{span}\left( {{{\mathbf{\bar{i}}}}_{n,k}} \right)$. Each ${{\mathbf{i}}_{f,k}}$ can be pick out only once while ${{\mathbf{\bar{i}}}_{n,k}}$ can be employed repeatedly. Replacing ${{\mathbf{i}}_{f,k}}$ and ${{\mathbf{\bar{i}}}_{n,k}}$ with the detailed expression, alignment equations yielded from the STS alignment structure for user $k$ can be represented as
\begin{IEEEeqnarray}{lcr}
\begin{array}{l}
{{\Phi }_{k}} \triangleq \{ \left. \operatorname{span}( {{\mathbf{H}}_{kj}}{{\mathbf{v}}_{j\ell }} )=\operatorname{span}( {{\mathbf{H}}_{kj'}}{{\mathbf{v}}_{j'{\ell }'}} ) \right| $ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $
\\ $ $$ $$ $$ $$ ${{\mathbf{H}}_{kj'}}{{\mathbf{v}}_{j'{\ell }'}}\in \mathcal{I}_{k}^{\operatorname{IA}}-\mathcal{\bar{I}}_{k}^{\operatorname{SIA}},\text{ }{{\mathbf{H}}_{kj}}{{\mathbf{v}}_{j\ell }}\in \mathcal{\bar{I}}_{k}^{\operatorname{SIA}},\text{ }j\ne {j}' \},
\\ $ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $k=1,2,\cdots ,{{k}_{\operatorname{IAC}}}
\end{array}
\end{IEEEeqnarray}
where $j\ne {j}'$ indicates that the interferences coming from the same user cannot be aligned [7]. Then, a complete set of linear alignment equations yielded from the STS alignment structure for a whole system is given by $\Psi \triangleq \left\{ \left. {{\Phi }_{k}} \right|k\in \left[ 1,{{k}_{\operatorname{IAC}}} \right] \right\}$.

\begin{figure}[!t]
\centering
\includegraphics[width=3.4in]{fig2}
\vspace*{-10pt}
\caption{IAC example with $M=6$, ${{d}_{1}}={{d}_{3}}=3$, ${{d}_{2}}=1$ and ${{d}_{4}}={{d}_{5}}=2$ for $K=5$: ${{k}_{\operatorname{IAC}}}=3$}
\end{figure}

\subsection{Illustrative example}
To illustrate how IAC with the STS alignment works, let us consider an example for $K=5$ and $M=6$ in Fig. 2, where we assume ${{d}_{1}}={{d}_{3}}=3$, ${{d}_{2}}=1$, and ${{d}_{4}}={{d}_{5}}=2$, i.e., transmitting a total number of eleven data symbols at the same time $\left( \sum\nolimits_{j=1}^{5}{{{d}_{j}}}=11 \right)$. As $\sum\nolimits_{j=3}^{5}{{{d}_{j}}}=7>M$ and $\sum\nolimits_{j=4}^{5}{{{d}_{j}}}=4<M$, ${{k}_{\operatorname{IAC}}}=3$. It implies that the receivers 4 and 5 can decode ${{\mathbf{x}}_{4}}$ and ${{\mathbf{x}}_{5}}$ without cancellation. In the sequel, we present a set of alignment equations $\Psi =\left\{ {{\Phi }_{j}}|j=1,2,\cdots ,{{k}_{\operatorname{IAC}}} \right\}$ for this example. At receiver 1, three desired data symbols ${{x}_{11}}$, ${{x}_{12}}$, and ${{x}_{13}}$ will be decoded and thus, $\dim({{{\tilde{S}}}_{1}})=3$. In order to meet (4), all other eight symbols, $\left\{ {{x}_{2{{\ell }_{2}}}} \right\}_{{{\ell }_{2}}=1}^{{{d}_{2}}}$ , $\left\{ {{x}_{3{{\ell }_{3}}}} \right\}_{{{\ell }_{3}}=1}^{{{d}_{3}}}$, $\left\{ {{x}_{4{{\ell }_{4}}}} \right\}_{{{\ell }_{4}}=1}^{{{d}_{4}}}$, and $\left\{ {{x}_{5{{\ell }_{5}}}} \right\}_{{{\ell }_{5}}=1}^{{{d}_{5}}}$, considered as interferences, have to be aligned so that $\dim({{{\tilde{I}}}_{1}})\le M\text{-}\dim({{{\tilde{S}}}_{1}})=3$. To construct a set of basis vectors of ${{\tilde{I}}_{1}}$, let us select the interference vectors, ${{\mathbf{H}}_{12}}{{\mathbf{v}}_{21}}$, ${{\mathbf{H}}_{13}}{{\mathbf{v}}_{33}}$, and ${{\mathbf{H}}_{14}}{{\mathbf{v}}_{42}}$, out of  $\mathcal{I}_{1}^{\operatorname{IA}}$, which is represented by $\mathcal{\bar{I}}_{1}^{\operatorname{SIA}}=\left\{ {{\mathbf{H}}_{12}}{{\mathbf{v}}_{21}},{{\mathbf{H}}_{13}}{{\mathbf{v}}_{33}},{{\mathbf{H}}_{14}}{{\mathbf{v}}_{42}} \right\}$. Then we have $\mathcal{I}_{1}^{\operatorname{IA}}-\mathcal{\bar{I}}_{1}^{\operatorname{SIA}}=\left\{ {{\mathbf{H}}_{13}}{{\mathbf{v}}_{31}},{{\mathbf{H}}_{13}}{{\mathbf{v}}_{32}},{{\mathbf{H}}_{14}}{{\mathbf{v}}_{41}},{{\mathbf{H}}_{15}}{{\mathbf{v}}_{51}},{{\mathbf{H}}_{15}}{{\mathbf{v}}_{52}} \right\}$. Following the STS structure, each interference vector in $\left( \mathcal{I}_{1}^{\operatorname{IA}}-\mathcal{\bar{I}}_{1}^{\operatorname{SIA}} \right)$ will be uniquely aligned onto one of the basis vectors in $\mathcal{\bar{I}}_{1}^{\operatorname{SIA}}$. For example, assume that ${{\mathbf{H}}_{13}}{{\mathbf{v}}_{31}}$ and ${{\mathbf{H}}_{14}}{{\mathbf{v}}_{41}}$ are aligned onto ${{\mathbf{H}}_{12}}{{\mathbf{v}}_{21}}$; ${{\mathbf{H}}_{15}}{{\mathbf{v}}_{52}}$ is aligned onto ${{\mathbf{H}}_{13}}{{\mathbf{v}}_{33}}$; and ${{\mathbf{H}}_{13}}{{\mathbf{v}}_{32}}$ and ${{\mathbf{H}}_{15}}{{\mathbf{v}}_{51}}$ are aligned onto ${{\mathbf{H}}_{14}}{{\mathbf{v}}_{42}}$, constructing the following set of alignment equations, ${{\Phi }_{1}}$, for the receiver 1:
\begin{IEEEeqnarray}{lcr}
\begin{array}{l}
\operatorname{span}({{\mathbf{H}}_{12}}{{\mathbf{v}}_{21}})=\operatorname{span}({{\mathbf{H}}_{13}}{{\mathbf{v}}_{31}} ) \\
\operatorname{span}({{\mathbf{H}}_{12}}{{\mathbf{v}}_{21}})=\operatorname{span}({{\mathbf{H}}_{14}}{{\mathbf{v}}_{41}}) \\
\operatorname{span}({{\mathbf{H}}_{1\text{3}}}{{\mathbf{v}}_{\text{33}}})=\operatorname{span}({{\mathbf{H}}_{15}}{{\mathbf{v}}_{52}}) \\
\operatorname{span}({{\mathbf{H}}_{14}}{{\mathbf{v}}_{42}})=\operatorname{span}({{\mathbf{H}}_{13}}{{\mathbf{v}}_{32}}) \\
\operatorname{span}({{\mathbf{H}}_{14}}{{\mathbf{v}}_{42}})=\operatorname{span}({{\mathbf{H}}_{15}}{{\mathbf{v}}_{51}})
\end{array}
\end{IEEEeqnarray}
After decoding ${{\hat{x}}_{11}}$, ${{\hat{x}}_{12}}$, and ${{\hat{x}}_{13}}$, they would be sent to the receivers 2, 3, 4, and 5 for a successive cancellation purpose.

At receiver 2, since ${{\hat{x}}_{11}}$, ${{\hat{x}}_{12}}$, and ${{\hat{x}}_{13}}$ from receiver 1 can be subtracted by IC operation, we have $\mathcal{I}_{2}^{\operatorname{IC}}=\left\{ \left. {{\mathbf{H}}_{21}}{{\mathbf{v}}_{1{{\ell }_{1}}}} \right|{{\ell }_{1}}\in \left[ 1,{{d}_{1}} \right] \right\}$ and $\mathcal{I}_{2}^{\operatorname{IA}}=\left\{ \left. {{\mathbf{H}}_{2j}}{{\mathbf{v}}_{j{{\ell }_{j}}}} \right|j\in \left[ 3,5 \right],\text{ }{{\ell }_{j}}\in \left[ 1,{{d}_{j}} \right] \right\}$. With $\dim({{\tilde{I}}_{2}})\le M\text{-}\dim({{\tilde{S}}_{2}})=5$, we select five interference vectors in $\mathcal{I}_{2}^{\operatorname{IA}}$ to construct $\mathcal{\bar{I}}_{2}^{\operatorname{SIA}}$, given as $\mathcal{\bar{I}}_{2}^{\operatorname{SIA}}=\left\{ {{\mathbf{H}}_{23}}{{\mathbf{v}}_{31}},{{\mathbf{H}}_{23}}{{\mathbf{v}}_{33}},{{\mathbf{H}}_{24}}{{\mathbf{v}}_{41}},{{\mathbf{H}}_{24}}{{\mathbf{v}}_{42}},{{\mathbf{H}}_{25}}{{\mathbf{v}}_{52}} \right\}$. Following the STS structure again, for example, ${{\Phi }_{2}}$ can be constructed with the following alignment equations:
\begin{IEEEeqnarray}{lcr}
\begin{array}{l}
\operatorname{span}({{\mathbf{H}}_{24}}{{\mathbf{v}}_{41}})=\operatorname{span}({{\mathbf{H}}_{23}}{{\mathbf{v}}_{32}}) \\
\operatorname{span}({{\mathbf{H}}_{23}}{{\mathbf{v}}_{33}})=\operatorname{span}({{\mathbf{H}}_{25}}{{\mathbf{v}}_{51}})
\end{array}
\end{IEEEeqnarray}
As ${{\hat{x}}_{21}}$, ${{\hat{x}}_{22}}$, and ${{\hat{x}}_{23}}$ are decoded now in the receiver 2, they will be shared with the receivers 3, 4, and 5 for cancellation.

At receiver 3, we have $\mathcal{I}_{3}^{\operatorname{IC}}=\left\{ \left. {{\mathbf{H}}_{3j}}{{\mathbf{v}}_{j{{\ell }_{j}}}} \right|j\in \left[ 1,2 \right],\text{ }{{\ell }_{j}}\in \left[ 1,{{d}_{j}} \right] \right\}$ and $\mathcal{I}_{3}^{\operatorname{IA}}=\left\{ \left. {{\mathbf{H}}_{3j}}{{\mathbf{v}}_{j{{\ell }_{j}}}} \right|j\in \left[ 4,5 \right],\text{ }{{\ell }_{j}}\in \left[ 1,{{d}_{j}} \right] \right\}$. With $\dim({{\tilde{I}}_{3}})\le M\text{-}\dim({{\tilde{S}}_{3}})=3$, we select three interference vectors in $\mathcal{I}_{3}^{\operatorname{IA}}$ to construct $\mathcal{\bar{I}}_{3}^{\operatorname{SIA}}$, e.g., $\mathcal{\bar{I}}_{3}^{\operatorname{SIA}}=\left\{ {{\mathbf{H}}_{34}}{{\mathbf{v}}_{41}},{{\mathbf{H}}_{35}}{{\mathbf{v}}_{51}},{{\mathbf{H}}_{35}}{{\mathbf{v}}_{52}} \right\}$. As $\left( \mathcal{I}_{3}^{\operatorname{IA}}-\mathcal{\bar{I}}_{3}^{\operatorname{SIA}} \right)$ has only one interference vector, it ends up with only one alignment equation, i.e., $\left| {{\Phi }_{3}} \right|=\left| \mathcal{I}_{3}^{IA}-\mathcal{\bar{I}}_{3}^{\operatorname{SIA}} \right|=1$, where ${{\Phi }_{3}}=\left\{ \operatorname{span}({{\mathbf{H}}_{35}}{{\mathbf{v}}_{51}})=\operatorname{span}({{\mathbf{H}}_{34}}{{\mathbf{v}}_{42}}) \right\}$.

As ${{k}_{\operatorname{IAC}}}$ = 3 in our example, the receivers 4 and 5 can decode their desired symbols without resorting to alignment operation. Consequently, we have obtained a complete set of linear alignment equations for user $k\in [1,{{k}_{\operatorname{IAC}}}]$, $\Psi =\left\{ {{\Phi }_{1}},{{\Phi }_{2}},{{\Phi }_{3}} \right\}$. With some algebraic manipulations, $\Psi$ can be equivalently written as
\begin{IEEEeqnarray}{lcr}
\begin{array}{l}
{{\mathbf{v}}_{21}}={{\left( {{\mathbf{H}}_{12}} \right)}^{-1}}{{\mathbf{F}}_{1}}{{\mathbf{v}}_{42}} \\
{{\mathbf{v}}_{31}}={{\left( {{\mathbf{H}}_{13}} \right)}^{-1}}{{\mathbf{F}}_{1}}{{\mathbf{v}}_{42}} \\
{{\mathbf{v}}_{32}}={{\left( {{\mathbf{H}}_{13}} \right)}^{-1}}{{\mathbf{H}}_{14}}{{\mathbf{v}}_{42}} \\
{{\mathbf{v}}_{33}}={{\mathbf{F}}_{\text{2}}}{{\mathbf{v}}_{42}} \\
{{\mathbf{v}}_{41}}={{\left( {{\mathbf{H}}_{24}} \right)}^{-1}}{{\mathbf{H}}_{23}}{{\left( {{\mathbf{H}}_{13}} \right)}^{-1}}{{\mathbf{H}}_{14}}{{\mathbf{v}}_{42}} \\
{{\mathbf{v}}_{51}}={{\left( {{\mathbf{H}}_{35}} \right)}^{-1}}{{\mathbf{H}}_{34}}{{\mathbf{v}}_{42}} \\
{{\mathbf{v}}_{52}}={{\left( {{\mathbf{H}}_{15}} \right)}^{-1}}{{\mathbf{H}}_{1\text{3}}}{{\mathbf{F}}_{\text{2}}}{{\mathbf{v}}_{42}} \\
\operatorname{span}({{\mathbf{v}}_{42}})=\operatorname{span}({{\mathbf{F}}_{\text{3}}}{{\mathbf{v}}_{42}})
\end{array}
\end{IEEEeqnarray}
where ${{\mathbf{F}}_{1}}={{\mathbf{H}}_{14}}{{\left( {{\mathbf{H}}_{24}} \right)}^{-1}}{{\mathbf{H}}_{23}}{{\left( {{\mathbf{H}}_{13}} \right)}^{-1}}{{\mathbf{H}}_{14}}$, ${{\mathbf{F}}_{2}}={{\left( {{\mathbf{H}}_{23}} \right)}^{-1}}{{\mathbf{H}}_{25}}{{\left( {{\mathbf{H}}_{35}} \right)}^{-1}}{{\mathbf{H}}_{34}}$, and ${{\mathbf{F}}_{3}}={{\left( {{\mathbf{H}}_{34}} \right)}^{-1}}{{\mathbf{H}}_{35}}{{\left( {{\mathbf{H}}_{15}} \right)}^{-1}}{{\mathbf{H}}_{14}}$. Since $\operatorname{span}\left( {{\mathbf{v}}_{42}} \right)=\operatorname{span}\left( {{\mathbf{F}}_{3}}{{\mathbf{v}}_{42}} \right)$, we can set ${{\mathbf{v}}_{42}}$ to be one of eigenvectors of ${{\mathbf{F}}_{3}}$. Then, the other precoding vectors $\left\{ {{\mathbf{v}}_{2{{\ell }_{2}}}} \right\}_{{{\ell }_{2}}=1}^{{{d}_{2}}}$, $\left\{ {{\mathbf{v}}_{3{{\ell }_{3}}}} \right\}_{{{\ell }_{3}}=1}^{{{d}_{3}}}$, ${{\mathbf{v}}_{41}}$, and $\left\{ {{\mathbf{v}}_{5{{\ell }_{5}}}} \right\}_{{{\ell }_{5}}=1}^{{{d}_{5}}}$ can be found subsequently by (8).

\begin{figure}[!t]
\centering
\includegraphics[width=2.6in]{fig3}
\vspace*{-10pt}
\caption{The IAC graph corresponding to the example in Fig. 2}
\end{figure}

\subsection{IAC graph}
In order to derive the conditions for solving ${{\mathbf{V}}_{j}}$ out of $\Psi$ for a general tuple of DoFs $\left( {{d}_{1}},{{d}_{2}},\cdots,{{d}_{K}} \right)$, we have constructed an IAC graph $\mathcal{G}=\left( \mathcal{P},\mathcal{E} \right)$ in our earlier work [5], where $\mathcal{P}$ and $\mathcal{E}$ represent a set of vertices and a set of edges, respectively. We have proven that $\mathcal{G}$ and $\Psi$ have an one-to-one correspondence, i.e., each precoding vector ${{\mathbf{v}}_{j{{\ell }_{j}}}}$ and each alignment equation in $\Psi$ can be represented by one unique vertex and one unique edge in $\mathcal{G}$, respectively, and vice versa. Each edge in the IAC graph holds a label to declare the index of Tx/Rx pair over which the alignment operation is performed.

As each symbol corresponding to ${{\mathbf{i}}_{f,k}}\in \mathcal{I}_{k}^{\operatorname{IA}}-\mathcal{\bar{I}}_{k}^{\operatorname{SIA}}$ is aligned onto one of unique symbols corresponding to $\left( {{\mathbf{\bar{i}}}_{n,k}}\in \mathcal{I}_{k}^{\operatorname{IA}} \right)$ at receiver $k$, each variable may not appear in all alignment equations. Therefore, we can collect the equations that involves the same subset of variables into one subset. In other words, $\Psi$ can be divided into several independent subsets and thus, solving the independent subsets respectively is equivalent to solving $\Psi$. Correspondingly, each independent subset of $\Psi$ forms one independent connected subgraph of $\mathcal{G}$. Each connected subgraph has neither an isolated vertex, nor connection to other subgraphs. Assuming that there are $Q$ connected subgraphs, each of them denoted as  ${{\mathcal{G}}_{q}}=\left( {{\mathcal{P}}_{q}},{{\mathcal{E}}_{q}} \right)$, $q=1,2,\cdots,Q$, such that $\mathcal{G}=\left\{ {{\mathcal{G}}_{1}},{{\mathcal{G}}_{2}},\cdots,{{\mathcal{G}}_{Q}} \right\}$, $\left| \mathcal{P} \right|=\sum\nolimits_{q=1}^{Q}{\left| {{\mathcal{P}}_{q}} \right|}$, and $\left| \mathcal{E} \right|=\sum\nolimits_{q=1}^{Q}{\left| {{\mathcal{E}}_{q}} \right|}$, where ${{\mathcal{P}}_{q}}$ and ${{\mathcal{E}}_{q}}$ represent a set of vertices and a set of edges in subgraph $q$, respectively.

To give an intuitive illustration on the structure of IAC graph $\mathcal{G}$, let us revisit the example in Subsection III (B). With $\Psi $ given in (8), the IAC graph $\mathcal{G}$ can be depicted accordingly, as shown in Fig. 3. We that there exist only one subgraph in this example, namely, $\mathcal{G}=\left\{ {{\mathcal{G}}_{1}} \right\}$, and $\left| \mathcal{P} \right|=\left| \mathcal{E} \right|=8$. Furthermore, the label of each edge is consistent with the index of Tx/Rx pair that is subject to the alignment operation. Note that there exists only one loop in ${{\mathcal{G}}_{1}}$ in this example.

By analyzing the IAC graph $\mathcal{G}$, the necessary and sufficient conditions for solving $\Psi$ has been derived in [5], which can be summarized by the following proposition:
\begin{proposition}
{\it For a connected subgraph in IAC graph $\mathcal{G}$, if and only if the vertices form at most one loop, the precoding vectors involved can always be solved.}
\end{proposition}

Since there exists only one subgraph and only loop formed in the IAC graph of Fig. 3,  it is clear by Proposition 1 that the set of linear alignment equations $\Psi$ in (8) is solvable. Then, it has been further proven that the solutions $\{{{\mathbf{v}}_{j\ell }}\}$, $j\in \left[ 1,K \right]$ and $\ell \in \left[ 1,{{d}_{j}} \right]$, obtained from $\Psi$, can always guarantee the independence of ${{\tilde{S}}_{k}}$ and ${{\tilde{I}}_{k}}$. Finally, the necessary and sufficient conditions for the existence of closed-form transceivers for IAC with the STS alignment has been proven for $\left( K,M,J \right)$ Gaussian interference MAC system in [5], where $J$ is the total number of users. For $J=K$, $\left( K,M,J \right)$ MAC channel reduces to the $K$-user MIMO channel which is considered in this paper. The necessary and sufficient conditions accordingly can be stated by the following theorem:
\begin{theorem}
{\it In $K$-user MIMO interference channel, for a given tuple of DoF $\left( {{d}_{1}},{{d}_{2}},\cdots,{{d}_{K}} \right)$, the closed-form transceivers for IAC with STS alignment exist if and only if the following inequalities are satisfied:}
\end{theorem}
\begin{IEEEeqnarray}{lcr}
$ $$ ${{d}_{k}}+\max \{{{d}_{k+1}},{{d}_{k+2}},\cdots,{{d}_{K}}\}\le M,\text{ }k\in \left[ 1,{{k}_{\operatorname{IAC}}} \right] \\
$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $\sum\limits_{k={{k}_{\operatorname{IAC}}}+1}^{K}{{{d}_{k}}}\le M \\
{{d}_{1}}+\sum\limits_{k=1}^{{{k}_{\operatorname{IAC}}}}{(k-1){{d}_{k}}+}\sum\limits_{k={{k}_{\operatorname{IAC}}}+1}^{K}{( {{k}_{\operatorname{IAC}}}-1 ){{d}_{k}}}\le {{k}_{\operatorname{IAC}}}M $ $$ $$ $$ $$ $
\end{IEEEeqnarray}

\section{Design of a Closed-form IAC Transceiver}
In the given example in Fig. 2 and Fig. 3, a solvable set of alignment equations $\Psi$ is available for an illustrative purpose and the corresponding IAC graph automatically satisfies Proposition 1. For any given tuple of DoFs $\left( {{d}_{1}},{{d}_{2}},\cdots,{{d}_{K}} \right)$ that satisfies Theorem 1, however, the set of alignment equations, $\Psi$, is unknown. Different from the classical IA in which a set of alignment equations is unique for a given tuple of DoFs, our STS alignment scheme for IAC provides many different possible $\Psi$'s even for the same tuple of DoFs. This is due to the fact that there are many different possibilities of selecting the basis set $\mathcal{\bar{I}}_{k}^{\operatorname{SIA}}$ and forming the alignment equations in (5). Furthermore, not all of the possible $\Psi$'s are solvable. Therefore, our objective is to find solvable $\Psi$ for any feasible tuple of DoFs.

Meanwhile, ${{k}_{\operatorname{IAC}}}$ can be determined for the given DoFs of $\left( {{d}_{1}},{{d}_{2}},\cdots,{{d}_{K}} \right)$. Then, the associated system overhead ${{O}_{\operatorname{sys}}}$ is fixed, regardless of $\Psi$, by the definition of ${{O}_{\operatorname{sys}}}$. There will be no difference among all possible $\Psi$'s in terms of the total achievable DoFs and system overhead. Hence, we only resort to designing one possible $\Psi$ where the closed-form transmitters $\left\{ {{\mathbf{V}}_{j}} \right\}$ can be solved.

In this section, we establish a framework to design the closed-form transceiver that is addressed by [5]. In Subsection A, we present a design procedure to construct an IAC graph that gives our closed-form transceiver. In Subsection B, we detail how to determine the closed-from transceiver from the constructed IAC graph.

\subsection{Construction of IAC graph for a transceiver design}
Simply due to the enormous complexity involved with the different choices of selecting the basis set and forming the alignment equation, it is not efficient to check the solvability for all possible $\Psi $'s. Considering that $\Psi$ and $\mathcal{G}$ have one-to-one correspondence, we will design a solvable $\Psi$ by constructing a proper IAC graph $\mathcal{G}$. Following the Proposition 1, a constraint on $\mathcal{G}$ for the existence of solutions for $\left\{ {{\mathbf{V}}_{j}} \right\}$ is that each subgraph ${{\mathcal{G}}_{q}}$ does not form more than one loop. Besides, since the interference coming from the same transmitter cannot be aligned at any receiver as dictated in (5), the other constraint on $\mathcal{G}$ is that any two vertices corresponding to the same transmitter cannot be directly connected by one edge or indirectly connected through several edges with the same label.

Focusing on the main flow of the design process, we first elaborate the underlying assumptions and definitions. Considering that the decoding operation in IAC is performed in order, we assume that a design process on $\mathcal{G}$ also follows the same order, namely, from receiver $1$ to receiver $K$. Initially, suppose that there exist only vertices, i.e., without any edge in $\mathcal{G}$, which is denoted by a graph ${{\mathcal{G}}^{\left[ 0 \right]}}=\left( {{\mathcal{P}}^{\left[ 0 \right]}},{{\mathcal{E}}^{\left[ 0 \right]}} \right)$, where ${{\mathcal{E}}^{\left[ 0 \right]}}=\phi$. As a total number of ${{d}_{1}}$ desired symbols for the 1st Tx/Rx pair will be directly cancelled by IC operation in the remaining receiver, $\left\{ {{\mathbf{v}}_{1,{{\ell }_{1}}}} \right\}_{{{\ell }_{1}}=1}^{{{d}_{1}}}$ do not appear in $\Psi$ and therefore, ${{\mathcal{P}}^{\left[ 0 \right]}}=\left\{ \left. {{\mathbf{v}}_{j,{{\ell }_{j}}}} \right|\forall j\in \left[ 2,K \right],\text{ }\forall {{\ell }_{j}}\in \left[ 1,{{d}_{j}} \right] \right\}$.  Assume that each vertex in ${{\mathcal{P}}^{\left[ 0 \right]}}$ can be associated with an individual subgraph of ${{\mathcal{G}}^{\left[ 0 \right]}}$, leading to ${{Q}^{\left[ 0 \right]}}=\left| {{\mathcal{P}}^{\left[ 0 \right]}} \right|$ subgraphs, where ${{Q}^{\left[ k \right]}}$ denotes the number of subgraphs in ${{\mathcal{G}}^{\left[ k \right]}}$, $k=0,1,\cdots ,K$. Then, the edges corresponding to the alignment equations from receivers $1$ to $K$ will be added on ${{\mathcal{G}}^{\left[ 0 \right]}}$ sequentially. If one edge is added between one vertex in subgraph $i$ and the other in subgraph $j$ in the process of edge addition, two different subgraphs, $i$ and $j$, would be merged. Let $\mathcal{\overset{\scriptscriptstyle\smile}{G}}=\left( \mathcal{\overset{\scriptscriptstyle\smile}{P}},\mathcal{\overset{\scriptscriptstyle\smile}{E}} \right)$ and ${{\mathcal{G}}^{\left[ k \right]}}=\left( {{\mathcal{P}}^{\left[ k \right]}},{{\mathcal{E}}^{\left[ k \right]}} \right)$ denote a temporary IAC graph in the edge-addition process and one after the edge-addition operation for receiver $k$, respectively. Therefore, we have $\left| {{\mathcal{P}}^{\left[ k \right]}} \right|=\left| {{\mathcal{P}}^{\left[ 0 \right]}} \right|$ and $\left| {{\mathcal{E}}^{\left[ k \right]}} \right|=\sum\nolimits_{i=1}^{k}{\left| {{\Phi }_{i}} \right|}$ for $k=1,2,\cdots ,K$. Finally, IAC graph is given by $\mathcal{G}={{\mathcal{G}}^{\left[ K \right]}}$. Since the interferences received at receivers $\left( {{k}_{\operatorname{IAC}}}+1 \right)$ to $K$ do not require to be aligned, however, no edges are added for $\left\{ {{\mathcal{G}}^{\left[ k \right]}} \right\}_{k={{k}_{\operatorname{IAC}}}}^{K}$ and thus, $\mathcal{G}={{\mathcal{G}}^{\left[ {{k}_{\operatorname{IAC}}} \right]}}$ and $Q={{Q}^{\left[ {{k}_{\operatorname{IAC}}} \right]}}$.

Let ${{\eta }_{k}}$ and ${{R}_{k}}$ denote a set of vertices for the precoding vectors in interference vectors of $\mathcal{I}_{k}^{\operatorname{IA}}$ and $\mathcal{\bar{I}}_{k}^{\operatorname{SIA}}$ at receiver $k$, respectively. Furthermore, let ${{{\eta }'}_{k}}$ denote a set of vertices for the precoding vectors in $\left( \mathcal{I}_{k}^{\operatorname{IA}}-\mathcal{\bar{I}}_{k}^{\operatorname{SIA}} \right)$, i.e., ${{{\eta }'}_{k}}={{\eta }_{k}}-~{{R}_{k}}$. Furthermore, let $A_{j}^{\left[ k \right]}$ denote a set of reference vertices to which the vertices $\left\{ {{\mathbf{v}}_{j{{\ell }_{j}}}} \right\}_{{{\ell }_{j}}=1}^{{{d}_{j}}}$ have been connected, and initially we have $A_{j}^{\left[ k \right]}={{R}_{k}}\bigcap$ $\left\{ {{\mathbf{v}}_{j{{\ell }_{j}}}} \right\}_{{{\ell }_{j}}=1}^{{{d}_{j}}}$. The following procedure details how $\mathcal{G}$ is designed so as to satisfy two aforementioned constraints:

\textbf{Step 1 (Initialization):} List the total number of $\left| {{\mathcal{P}}^{\left[ 0 \right]}} \right|=\sum\nolimits_{j=2}^{K}{{{d}_{j}}}$ vertices which represent $\left\{ \left. {{\mathbf{v}}_{j{{\ell }_{j}}}} \right|\forall j\in \left[ 2,K \right],\text{} \forall {{\ell }_{j}}\in \left[ 1,{{d}_{j}} \right] \right\}$ and obtain ${{\mathcal{G}}^{\left[ 0 \right]}}=\left( {{\mathcal{P}}^{\left[ 0 \right]}},{{\mathcal{E}}^{\left[ 0 \right]}} \right)$ where ${{\mathcal{P}}^{\left[ 0 \right]}}=\left\{ \left. {{\mathbf{v}}_{j,{{\ell }_{j}}}} \right|\forall j\in \left[ 2,K \right],\text{ }\forall {{\ell }_{j}}\in \left[ 1,{{d}_{j}} \right] \right\}$ and ${{\mathcal{E}}^{\left[ 0 \right]}}=\phi $.

\textbf{Step 2 (Selecting a set of reference vertices):} At the $k$-th Tx/Rx pair, let ${{\eta }_{k}}=\left\{ \left. {{\mathbf{v}}_{{k}'{{\ell }_{{{k}'}}}}} \right|\forall {k}'\in \left[ k+1,K \right],\text{ }\forall {{\ell }_{{{k}'}}}\in \left[ 1,{{d}_{{{k}'}}} \right] \right\}$. Randomly pick up a total of $\left| \mathcal{I}_{k}^{\operatorname{IA}} \right|=M-{{d}_{k}}$ vertices from ${{\eta }_{k}}$ so as to form ${{R}_{k}}$ such that $\left| {{R}_{k}} \right|=M-{{d}_{k}}$. Initially, set ${{\mathcal{G}}^{\left[ k \right]}}\leftarrow {{\mathcal{G}}^{\left[ k-1 \right]}}$.

\textbf{Step 3 (Choosing the alignment pairs):} If $P_{q}^{\left[ k \right]}\bigcap {{{\eta }'}_{k}}\ne \phi $, pick up one vertex ${{\mathbf{v}}_{{j}'{{\ell }_{{{j}'}}}}}\in \mathcal{P}_{q}^{\left[ k \right]}\bigcap {{{\eta }'}_{k}}$. If ${{R}_{k}}-\left( A_{{{j}'}}^{\left[ k \right]}\bigcap {{R}_{k}} \right)=\phi $, go back to Step 2; otherwise, connect ${{\mathbf{v}}_{{j}'{{\ell }_{{{j}'}}}}}$ to one reference vertex ${{\mathbf{v}}_{j{{\ell }_{j}}}}\in {{R}_{k}}-\left( A_{{{j}'}}^{\left[ k \right]}\bigcap {{R}_{k}} \right)$ by an edge $e_{\left( j\ell  \right),\left( j'{\ell }' \right)}^{\left[ k \right]}$ which represents the alignment equation of $\operatorname{span}\left( {{\mathbf{H}}_{kj}}{{\mathbf{v}}_{j\ell }} \right)=\operatorname{span}\left( {{\mathbf{H}}_{kj'}}{{\mathbf{v}}_{j'{\ell }'}} \right)$. Then, we have
\[\mathcal{\overset{\scriptscriptstyle\smile}{E}}_{q}^{{}}=\left\{ \begin{array}{*{35}{l}}
   \mathcal{E}_{q}^{\left[ k \right]}\bigcup \left\{ e_{\left( j\ell  \right),\left( j'{\ell }' \right)}^{\left[ k \right]} \right\}, & {{\mathbf{v}}_{j{{\ell }_{j}}}}\in \mathcal{P}_{q}^{\left[ k \right]}  \\
   \mathcal{E}_{q}^{\left[ k \right]}\bigcup \mathcal{E}_{{{q}'}}^{\left[ k \right]}\bigcup \left\{ e_{\left( j\ell  \right),\left( j'{\ell }' \right)}^{\left[ k \right]} \right\}, & {{\mathbf{v}}_{j{{\ell }_{j}}}}\in \mathcal{P}_{{{q}'}}^{\left[ k \right]},\text{ }{q}'\ne q  \\
\end{array} \right.\]
and
\[\mathcal{\overset{\scriptscriptstyle\smile}{P}}_{q}^{{}}=\left\{ \begin{array}{*{35}{l}}
   \mathcal{P}_{q}^{\left[ k \right]}, & {{\mathbf{v}}_{j{{\ell }_{j}}}}\in \mathcal{P}_{q}^{\left[ k \right]}  \\
   \mathcal{P}_{q}^{\left[ k \right]}\bigcup \mathcal{P}_{{{q}'}}^{\left[ k \right]}, & {{\mathbf{v}}_{j{{\ell }_{j}}}}\in \mathcal{P}_{{{q}'}}^{\left[ k \right]},\text{ }{q}'\ne q  \\
\end{array} \right.\]

\textbf{Step 4 (Checking the number of formed loops):} If $\left| \mathcal{\overset{\scriptscriptstyle\smile}{E}}_{q}^{{}} \right|\le \left| \mathcal{\overset{\scriptscriptstyle\smile}{P}}_{q}^{{}} \right|$,  update ${{\mathcal{G}}^{\left[ k \right]}}$ by $\mathcal{E}_{q}^{\left[ k \right]}=\mathcal{\overset{\scriptscriptstyle\smile}{E}}_{q}^{{}}$, $\mathcal{P}_{q}^{\left[ k \right]}=\mathcal{\overset{\scriptscriptstyle\smile}{P}}_{q}^{{}}$, $\mathcal{E}_{{{q}'}}^{\left[ k \right]}\leftarrow \left\{ \begin{array}{*{35}{l}}
   \mathcal{E}_{{{q}'}}^{\left[ k \right]}, & {{\mathbf{v}}_{j{{\ell }_{j}}}}\in \mathcal{P}_{q}^{\left[ k \right]}  \\
   \phi , & {{\mathbf{v}}_{j{{\ell }_{j}}}}\in \mathcal{P}_{{{q}'}}^{\left[ k \right]},\text{ }{q}'\ne q  \\
\end{array} \right.$, and $\mathcal{P}_{{{q}'}}^{\left[ k \right]}\leftarrow \left\{ \begin{array}{*{35}{l}}
   \mathcal{P}_{{{q}'}}^{\left[ k \right]}, & {{\mathbf{v}}_{j{{\ell }_{j}}}}\in \mathcal{P}_{q}^{\left[ k \right]}  \\
   \phi , & {{\mathbf{v}}_{j{{\ell }_{j}}}}\in \mathcal{P}_{{{q}'}}^{\left[ k \right]},\text{ }{q}'\ne q  \\
\end{array} \right.$,
\\ while updated as $A_{{{j}'}}^{\left[ k \right]}\leftarrow A_{{{j}'}}^{\left[ k \right]}\bigcup \left\{ {{\mathbf{v}}_{j{{\ell }_{j}}}} \right\}$ and ${{{\eta }'}_{k}}\leftarrow {{{\eta }'}_{k}}-\{ {{\mathbf{v}}_{{j}'{{\ell }_{{{j}'}}}}} \}$; if $\left| \mathcal{\overset{\scriptscriptstyle\smile}{E}}_{q}^{{}} \right|>\left| \mathcal{\overset{\scriptscriptstyle\smile}{P}}_{q}^{{}} \right|$, go back to Step 3 to change another reference vertex for ${{\mathbf{v}}_{{j}'{{\ell }_{{{j}'}}}}}$.

\textbf{Step 5:} Repeat Steps 3 and 4 until $\mathcal{P}_{q}^{\left[ k \right]}\bigcap {{{\eta }'}_{k}}=\phi$.

\textbf{Step 6:} Repeat Steps 3 to 5 for $q\in \left[ 1,{{Q}^{\left[ k \right]}} \right]$ and then, ${{\mathcal{G}}^{\left[ k \right]}}$ is obtained.

\textbf{Step 7:} Repeat Steps 2 to 6 for $k\in \left[ 2,{{k}_{\operatorname{IAC}}} \right]$ and finally, we have $\mathcal{G}={{\mathcal{G}}^{\left[ {{k}_{\operatorname{IAC}}} \right]}}$.

In Step 3, by choosing a reference vertex ${{\mathbf{v}}_{j{{\ell }_{j}}}}\in {{R}_{k}}-\left( A_{{{j}'}}^{\left[ k \right]}\bigcap {{R}_{k}} \right)$, it guarantees that two vertices from the same transmitter ${j}'$ are not connected by one edge or indirectly connected through several edges with the same label.

In the sequel, a simple proof can be sketched to show that the proposed design has formed at most one loop in each subgraph ${{\mathcal{G}}_{q}}$. At the 1st Tx/Rx pair, following Steps 2 to 6, a total of $\left| {{R}_{1}} \right|=M-{{d}_{1}}$ reference vertices are selected. As each vertex ${{\mathbf{v}}_{{j}'{{\ell }_{{{j}'}}}}}\in {{{\eta }'}_{1}}$ can be connected to only one unique reference vertex and no edge is allowed between any two reference vertices, a total of ${{Q}^{\left[ 1 \right]}}=M-{{d}_{1}}$ subgraphs have been formed in ${{\mathcal{G}}^{\left[ 1 \right]}}$ with $\left| \mathcal{E}_{q}^{\left[ 1 \right]} \right|=\left| \mathcal{P}_{q}^{\left[ 1 \right]} \right|-1$, $q\in \left[ 1,{{Q}^{\left[ 1 \right]}} \right]$. Then, the alignment equations from 2nd to ${{k}_{\operatorname{IAC}}}$-th Tx/Rx pair are added onto ${{\mathcal{G}}^{\left[ 1 \right]}}$, giving a total number of $( \sum\nolimits_{k=2}^{{{k}_{\operatorname{IAC}}}}{\left| \Phi {}_{k} \right|}=\sum\nolimits_{k=2}^{{{k}_{\operatorname{IAC}}}}{\left( k-1 \right){{d}_{k}}}+\sum\nolimits_{{k}'={{k}_{\operatorname{IAC}}}+1}^{K}{\left( {{k}_{\operatorname{IAC}}}-1 \right){{d}_{{{k}'}}}}-\left( {{k}_{\operatorname{IAC}}}-1 \right)M )$ edges. Referring to Theorem 1, we can have $\sum\nolimits_{k=2}^{{{k}_{\operatorname{IAC}}}}{\left| \Phi {}_{k} \right|}\le M-{{d}_{1}}$, which indicates that the total number of edges added on ${{\mathcal{G}}^{\left[ 1 \right]}}$ by the 2nd to ${{k}_{\operatorname{IAC}}}$-th Tx/Rx pairs is no more than the total number of subgraphs in ${{\mathcal{G}}^{\left[ 1 \right]}}$. Furthermore, by going through Step 4, the proposed design guarantees that no more than one edge is added in each subgraph $\mathcal{G}_{q}^{\left[ k \right]}$. According to graph theory, if one additional edge is added on an acyclic subgraph, then one loop would be formed. On the other hand, if one edge is added between two vertices that belong to two different subgraphs, then two subgraphs are merged into one. Consequently, we have shown that the proposed design has constructed IAC graph $G$ to satisfy Proposition 1. In other words, the closed-form transmitters $\{{{\mathbf{V}}_{j}}\}_{j=1}^{K}$ exist and can be solved from $\Psi$.

\subsection{Solution to the closed-form IAC transceivers}
\begin{figure}[!t]
\vspace*{-10pt}
\centering{
\subfloat[${{\mathcal{G}}_{A}}=({{\mathcal{P}}_{A}},{{\mathcal{E}}_{A}})$: No-loop case$ $$ $$ $]{\includegraphics[width=1.7in]{fig4a}}
\subfloat[${{\mathcal{G}}_{B}}=({{\mathcal{P}}_{B}},{{\mathcal{E}}_{B}})$: One-loop case]{\includegraphics[width=1.7in]{fig4b}}}
\caption{Two types of subgraphs in IAC graph $\mathcal{G}$: No-loop case vs. one-loop case}
\vspace*{-10pt}
\end{figure}

In the previous subsection, we have presented a general design principle on $\mathcal{G}$ to specify a solvable $\Psi$. In this subsection, we present the procedures to solve the closed-form transceivers through $\mathcal{G}$. For the subgraphs that satisfy Proposition 1, there are two distinct cases: no-loop case (subgraph ${{\mathcal{G}}_{A}}$) and one-loop case, as illustrated in Fig. 4. For a general illustration, without loss of generality, we do not give a specific label to each edge, and constrain each vertex to any specific user for the examples in Fig. 4. For a simplicity of exposition, therefore, every vertex can be denoted by the indices $\left\{ {{\mathbf{v}}_{p}} \right\}_{p=1}^{|\mathcal{P}|}$. Then, a simplified definition of the channel matrices, without clarifying the transmitter and receiver index, will be adopted, i.e., $\{\mathbf{H}_{p}^{\left( 1 \right)}\}_{p=1}^{\left| \mathcal{P} \right|}$, $\{ \mathbf{H}_{p}^{\left( 2 \right)}\}_{p=1}^{\left| \mathcal{P} \right|}$, $\cdots$, and $\{ \mathbf{H}_{p}^{\left( {{L}_{p}} \right)}\}_{p=1}^{\left| \mathcal{P} \right|}$ denote the channel matrices for ${{\mathbf{v}}_{p}}$, while using superscripts $\left( 1 \right)$, $\left( 2 \right)$, $\cdots$, and $\left( {{L}_{p}} \right)$ to distinguish the channel matrices for ${{\mathbf{v}}_{p}}$ in the different alignment equations in (5). Additionally, $\{ \mathbf{H}_{p}^{\left( i \right)}\}$ is equal to $\{ \mathbf{H}_{p}^{\left( j \right)}\}$ for $\forall i,j\in \left\{ 1,2,\cdots ,{{L}_{p}} \right\}$ if the two different equations that are involved with $\{ \mathbf{H}_{p}^{\left( i \right)}\}$ and $\{ \mathbf{H}_{p}^{\left( j \right)}\}$, respectively, belong to the same receiver; otherwise, they are different. The corresponding set of linear alignment equations, ${{\Phi }_{A}}$, that are specified by the subgraph ${{\mathcal{G}}_{A}}=({{\mathcal{P}}_{A}},{{\mathcal{E}}_{A}})$ in Fig. 4(a) is given as
\begin{IEEEeqnarray}{lcr}
\left\{ \begin{matrix}
   \operatorname{span}\left( \mathbf{H}_{1}^{\left( 1 \right)}{{\mathbf{v}}_{1}} \right)=\operatorname{span}\left( \mathbf{H}_{2}^{\left( 1 \right)}{{\mathbf{v}}_{2}} \right)  \\
   \operatorname{span}\left( \mathbf{H}_{2}^{\left( 2 \right)}{{\mathbf{v}}_{2}} \right)=\operatorname{span}\left( \mathbf{H}_{3}^{\left( 1 \right)}{{\mathbf{v}}_{3}} \right)  \\
   \vdots   \\
   \operatorname{span}\left( \mathbf{H}_{\left| {{P}_{A}} \right|-1}^{\left( 2 \right)}{{\mathbf{v}}_{\left| {{P}_{A}} \right|-1}} \right)=\operatorname{span}\left( \mathbf{H}_{\left| {{P}_{A}} \right|}^{\left( 1 \right)}{{\mathbf{v}}_{\left| {{P}_{A}} \right|}} \right)  \\
\end{matrix} \right.
\end{IEEEeqnarray}
For any equation of $\operatorname{span}(\mathbf{H}_{{{p}_{1}}}^{\left( i \right)}{{\mathbf{v}}_{{{p}_{1}}}} )=\operatorname{span}( \mathbf{H}_{{{p}_{2}}}^{\left( j \right)}{{\mathbf{v}}_{{{p}_{2}}}})$ in (12), it can be expressed equivalently by ${{\mathbf{v}}_{{{p}_{2}}}}={{\gamma }_{{{p}_{2}},{{p}_{1}}}}{( \mathbf{H}_{{{p}_{2}}}^{\left( j \right)})^{-1}}\mathbf{H}_{{{p}_{1}}}^{\left( i \right)}{{v}_{{{p}_{1}}}}$, where ${{\gamma }_{{{p}_{2}},{{p}_{1}}}}$ is the coefficient that coordinates the length of two vectors, $\mathbf{H}_{{{p}_{1}}}^{\left( i \right)}{{\mathbf{v}}_{{{p}_{1}}}}$ and $\mathbf{H}_{{{p}_{2}}}^{\left( j \right)}{{\mathbf{v}}_{{{p}_{2}}}}$, and would be determined while solving the equations. As $\{{{\mathbf{v}}_{j}}\}_{j=2}^{_{\left| {\mathcal{{P}}_{A}} \right|}}$ can be expressed in terms of ${{\mathbf{v}}_{1}}$ in (12), ${{\mathbf{v}}_{1}}$ can be selected as any arbitrary vector, i.e., assume that ${{\mathbf{v}}_{1}}$ is an $M\times 1$ unit vector and then, $\{{{\mathbf{v}}_{j}}\}_{j=2}^{_{\left| {\mathcal{{P}}_{A}} \right|}}$ can be solved sequentially.

The following set of linear alignment equations ${{\Phi }_{B}}$ is given by the subgraph ${{\mathcal{G}}_{B}}=({{\mathcal{P}}_{B}},{{\mathcal{E}}_{B}})$ in Fig. 4(b):
\begin{IEEEeqnarray}{lcr}
\left\{ \begin{matrix}
   \operatorname{span}( \mathbf{H}_{1}^{\left( 1 \right)}{{\mathbf{v}}_{1}} )=\operatorname{span}( \mathbf{H}_{2}^{\left( 1 \right)}{{\mathbf{v}}_{2}} )  \\
   \vdots   \\
   \operatorname{span}( \mathbf{H}_{\left| {\mathcal{{P}}_{B}} \right|-2}^{\left( 2 \right)}{{\mathbf{v}}_{\left| {\mathcal{{P}}_{B}} \right|-2}} )=\operatorname{span}( \mathbf{H}_{\left| {\mathcal{{P}}_{B}} \right|-1}^{\left( 1 \right)}{{\mathbf{v}}_{\left| {\mathcal{{P}}_{B}} \right|-1}} )  \\
   \operatorname{span}( \mathbf{H}_{\left| {\mathcal{{P}}_{B}} \right|-1}^{\left( 2 \right)}{{\mathbf{v}}_{\left| {\mathcal{{P}}_{B}} \right|-1}} )=\operatorname{span}( \mathbf{H}_{1}^{\left( 2 \right)}{{\mathbf{v}}_{1}} )  \\
   \operatorname{span}( \mathbf{H}_{\left| {\mathcal{{P}}_{B}} \right|-1}^{\left( 3 \right)}{{\mathbf{v}}_{\left| {\mathcal{{P}}_{B}} \right|-1}} )=\operatorname{span}( \mathbf{H}_{\left| {\mathcal{{P}}_{B}} \right|}^{\left( 1 \right)}{{\mathbf{v}}_{\left| {\mathcal{{P}}_{B}} \right|}} )
\end{matrix} \right.
\end{IEEEeqnarray}
Considering the first $(\left| {{\mathcal{P}}_{B}} \right|-1)$ equations in (13), which are obtained around the loop in ${{\mathcal{G}}_{B}}$, it can be shown that ${{\mathbf{v}}_{1}}={{\mathbf{F}}_{1}}{{\mathbf{v}}_{1}}$, i.e., ${{\mathbf{v}}_{1}}$ is the eigenvector of ${{\mathbf{F}}_{1}}$, where ${{\mathbf{F}}_{1}}={{\gamma }_{1,\left| {{\mathcal{P}}_{B}} \right|-1}}{( \mathbf{H}_{1}^{\left( 2 \right)})^{-1}}\left( \prod\nolimits_{j=\left| {{\mathcal{P}}_{B}} \right|-1}^{2}{{{\gamma }_{j,j-1}}\mathbf{H}_{j}^{\left( 2 \right)}( \mathbf{H}_{j}^{\left( 1 \right)})^{-1}} \right)\mathbf{H}_{1}^{\left( 1 \right)}$.\footnotemark \\ For the other precoding vectors $\{{{\mathbf{v}}_{j}}\}_{j=2}^{|{\mathcal{{P}}_{B}}|}$, they could be expressed in terms of ${{\mathbf{v}}_{1}}$ by (13) as we did in the no-loop case. Therefore, once ${{\mathbf{v}}_{1}}$ is known, $\{{{\mathbf{v}}_{j}}\}_{j=2}^{|{\mathcal{{P}}_{B}}|}$ can be solved sequentially. In summary, the closed-form solutions of $\left\{ \left. {{\mathbf{v}}_{j{{\ell }_{j}}}} \right|\forall {{\ell }_{j}}\in \left[ 1,{{d}_{j}} \right],\text{ }\forall j\in \left[ 2,K \right] \right\}$ can be found through two ways as discussed in the above, depending on the type of the involved subgraph. Furthermore, as $\{{{\mathbf{v}}_{1{{\ell }_{1}}}}\}_{{{\ell }_{1}}=1}^{{{d}_{1}}}$ do not appear in $\mathcal{G}$, they can be immediately found by ${{\mathbf{V}}_{1}}\subset {{\left( {{\mathbf{H}}_{1j}} \right)}^{-1}}{{\left( \mathcal{I}_{1}^{\operatorname{IA}} \right)}^{\bot }}$ where ${{\mathcal{I}}_{1}}$ is known from $\left\{ \left. {{\mathbf{v}}_{j{{\ell }_{j}}}} \right|\forall {{\ell }_{j}}\in \left[ 1,{{d}_{j}} \right],\text{ }\forall j\in \left[ 2,K \right] \right\}$. Once the transmitters $\{{{\mathbf{V}}_{j}}\}_{j=1}^{K}$ are solved, $\{\mathcal{I}_{k}^{\operatorname{IA}}\}_{k=1}^{K}$ is determined and then, the receiver ${{\mathbf{U}}_{k}}$ can be obtained by ${{\mathbf{U}}_{k}}\subset {{\left( \mathcal{I}_{k}^{\operatorname{IA}} \right)}^{\bot }}$, $\forall k\in \left[ 1,K \right]$. In the following subsection, we present an illustrative example for the current design principle to solve a closed-form IAC transceiver that can achieve the maximum degree of freedom.
\footnotetext{The indices of product operation "$\prod\nolimits_{j=\left| {{\mathcal{P}}_{B}} \right|-1}^{2}$" incrementally decreasing from the lower limit to the upper limit, indicate the order for the matrix multiplication, i.e., the matrix with the bigger index is post-multiplied by one with the smaller one.}

\subsection{Illustrative example}
In the last subsection, we have presented the design principle for a closed-form transceiver for any feasible tuple of DoFs. Naturally, among all feasible tuples of DoFs with the same $K$ and $M$, we will be interested in finding the optimum tuple which maximizes the total achievable DoFs. In our previous work [6], we have shown that the total achievable DoF by the closed-form IAC solutions is given by $2M$. Moreover, we have also shown that it can be achieved by ${{k}_{\operatorname{IAC}}}=2$. In this subsection, we consider an illustrative example for the proposed design procedure to solve a closed-form IAC transceiver that can achieve the maximum degree of freedom. In other words, our IAC transceiver must be designed to meet the constraint on the tuple of DoFs $\left( {{d}_{1}},{{d}_{2}},\cdots ,{{d}_{K}} \right)$ which can achieve a total of $2M$ for ${{k}_{\operatorname{IAC}}}=2$. In other words, our DoF constraint must follow (9) for ${{k}_{\operatorname{IAC}}}=2$, i.e.,
\begin{IEEEeqnarray}{lcr}
\left\{ \begin{matrix}
   {{d}_{1}}+\max \left\{ {{d}_{j}} \right\}_{j=2}^{K}\le M  \\
   {{d}_{2}}+\max \left\{ {{d}_{j}} \right\}_{j=3}^{K}\le M
\end{matrix} \right.
\end{IEEEeqnarray}
Then, (10) and (11) reduces to be  $\sum\nolimits_{k=3}^{K}{{{d}_{k}}}\le M$ and ${{d}_{1}}+{{d}_{2}}+\sum\nolimits_{k=3}^{K}{{{d}_{k}}}\le 2M$, respectively. Let $\sum\nolimits_{k=3}^{K}{{{d}_{k}}}=M$ and ${{d}_{1}}+{{d}_{2}}=M$. Then, a total of $2M$ can be collected. Substituting ${{d}_{1}}+{{d}_{2}}=M$ into (14) and combining the two inequalities, we can get
\begin{IEEEeqnarray}{lcr}
\max \left\{ {{d}_{j}} \right\}_{j=3}^{K}\le \frac{M}{2}
\end{IEEEeqnarray}
Therefore, the constraints on $\left( {{d}_{1}},{{d}_{2}},\cdots ,{{d}_{K}} \right)$ can be summarized as
\begin{IEEEeqnarray}{lcr}
\left\{ \begin{array}{*{35}{l}}
   {{d}_{1}}\le M-\max \left\{ {{d}_{j}} \right\}_{j=3}^{K}  \\
   {{d}_{2}}\le M-\max \left\{ {{d}_{j}} \right\}_{j=3}^{K}  \\
   \max \left\{ {{d}_{j}} \right\}_{j=3}^{K}\le \frac{M}{2}  \\
   {{d}_{1}}+{{d}_{2}}=M  \\
   \sum\nolimits_{j=3}^{K}{{{d}_{j}}}=M  \\
\end{array} \right.
\end{IEEEeqnarray}
From the third and fifth inequalities (16), we can get $K\ge 4$.

\begin{figure}[!t]
\centering
\includegraphics[width=3.4in]{fig5}
\vspace*{-10pt}
\caption{Optimal transceiver design with ${{k}_{\operatorname{IAC}}}=2$: ${{\mathcal{G}}^{\left[ 1 \right]}}=\left( {{\mathcal{P}}^{\left[ 1 \right]}},{{\mathcal{E}}^{\left[ 1 \right]}} \right)$}
\end{figure}

In the sequel, we consider a design of the closed-form transceiver with $\left( {{d}_{1}},{{d}_{2}},\cdots ,{{d}_{K}} \right)$ which satisfies (16). Following the general design principle in the previous subsection, we design a solvable $\Psi $ by constructing an IAC graph $\mathcal{G}$. With ${{k}_{\operatorname{IAC}}}=2$, we have $\mathcal{G}={{\mathcal{G}}^{\left[ 2 \right]}}$ and therefore, we only resort to ${{\mathcal{G}}^{\left[ 1 \right]}}$ and ${{\mathcal{G}}^{\left[ 2 \right]}}$ to obtain $\mathcal{G}$. As opposed to the example in Fig. 2, which achieves a total DoFs of $11$, our target transceiver is supposed to achieve the maximum total DoFs, $2M=12$, with the same system structure of $K=5$ and $M=6$. Referring to the constraints in (16), we consider a system of ${{d}_{1}}={{d}_{2}}=3$ and ${{d}_{3}}={{d}_{4}}={{d}_{5}}=2$, which achieves a total DoF of $12$, as an illustrative design example.

We first provide the design of ${{\mathcal{G}}^{\left[ 1 \right]}}$. At receiver 1, we have ${{\eta }_{1}}=\left\{ \left. {{\mathbf{v}}_{{j}'{{\ell }_{{{j}'}}}}} \right|\forall {j}'\in \left[ 2,K \right],\text{ }\forall {{\ell }_{{{j}'}}}\in \left[ 1,{{d}_{{{j}'}}} \right] \right\}$. Considering that $\left| \mathcal{\bar{I}}_{1}^{\operatorname{SIA}} \right|=M-{{d}_{1}}={{d}_{2}}$, it inspires us to construct a set of the reference vertices as ${{R}_{1}}=\left\{ \left. {{\mathbf{v}}_{2{{\ell }_{2}}}} \right|\forall {{\ell }_{2}}\in \left[ 1,{{d}_{2}} \right] \right\}$ and then, the remaining vertices are given as ${{{\eta }'}_{1}}={{\eta }_{1}}-{{R}_{1}}=\left\{ \left. {{\mathbf{v}}_{{j}'{{\ell }_{{{j}'}}}}} \right|\forall {j}'\in \left[ 3,K \right],\text{ }\forall {{\ell }_{{{j}'}}}\in \left[ 1,{{d}_{{{j}'}}} \right] \right\}$. Initially, set ${{\mathcal{G}}^{\left[ 1 \right]}}\leftarrow {{\mathcal{G}}^{\left[ 0 \right]}}$. Suppose a total of ${{d}_{2}}$ reference vertices are artificially arranged by the index ${{\ell }_{2}}$ as shown in Fig. 5. Firstly, each vertex ${{\mathbf{v}}_{3{{\ell }_{3}}}}$, $\forall {{\ell }_{3}}\in \left[ 1,{{d}_{3}} \right]$, will be connected to one unique reference vertex ${{\mathbf{v}}_{2{{\ell }_{2}}}}$, $\forall {{\ell }_{2}}\in \left[ 1,{{d}_{2}} \right]$, with ${{\ell }_{2}}={{\ell }_{3}}\bmod \left( {{d}_{2}} \right)$, creating an edge $e_{\left( 2{{\ell }_{2}} \right),\left( 3{{\ell }_{3}} \right)}^{\left[ 1 \right]}$ which represents an alignment equation, given by  $\operatorname{span}\left( {{\mathbf{H}}_{13}}{{\mathbf{v}}_{3{{\ell }_{3}}}} \right)=\operatorname{span}\left( {{\mathbf{H}}_{12}}{{\mathbf{v}}_{2{{\ell }_{2}}}} \right)$, and then, we have $\mathcal{E}_{{{\ell }_{2}}}^{\left[ 1 \right]}=\mathcal{E}_{{{\ell }_{2}}}^{\left[ 1 \right]}\bigcup \left\{ e_{\left( 2{{\ell }_{2}} \right),\left( 3{{\ell }_{3}} \right)}^{\left[ 1 \right]} \right\}$. After finishing the connection of all vertices $\left\{ {{\mathbf{v}}_{3{{\ell }_{3}}}} \right\}_{{{\ell }_{3}}=1}^{{{d}_{3}}}$ from the 3rd Tx, we turn to connect the ones from the 4th Tx and so as the other Tx's. Finally, we obtain ${{\mathcal{G}}^{\left[ 1 \right]}}=\left( {{\mathcal{P}}^{\left[ 1 \right]}},{{\mathcal{E}}^{\left[ 1 \right]}} \right)$ with $\left| {{\mathcal{P}}^{\left[ 1 \right]}} \right|=\sum\nolimits_{j=2}^{K}{{{d}_{j}}}={{d}_{2}}+M$ and $\left| {{\mathcal{E}}^{\left[ 1 \right]}} \right|=\left| \mathcal{I}_{1}^{\operatorname{IA}}\mathcal-{\bar{I}}_{1}^{\operatorname{SIA}} \right|=M$. Furthermore, a corresponding set of linear alignment equations ${{\Phi }_{1}}$, which is specified by ${{\mathcal{G}}^{\left[ 1 \right]}}$, can be expressed as
\begin{IEEEeqnarray}{lcr}
\operatorname{span}\left( {{\mathbf{H}}_{1{j}'}}{{\mathbf{v}}_{{j}'{{\ell }_{{{j}'}}}}} \right)=\operatorname{span}\left( {{\mathbf{H}}_{12}}{{\mathbf{v}}_{2{{\ell }_{2}}}} \right),\text{ }\forall {j}'\in \left[ 3,K \right] $ $$ $
\end{IEEEeqnarray}
where ${{\ell }_{2}}=\left\{ \begin{array}{*{35}{l}}
   {{\ell }_{{{j}'}}}\bmod \left( {{d}_{2}} \right),\text{ }{j}'=3
\\ \left( \sum\nolimits_{i=3}^{{j}'-1}{{{d}_{i}}}+{{\ell }_{{{j}'}}} \right)\bmod \left( {{d}_{2}} \right),\text{ }{j}'\ge 4
\end{array} \right.$.

Since no edge is allowed between any two reference vertices, a total of ${{Q}^{\left[ 1 \right]}}=\left| {{R}_{1}} \right|={{d}_{2}}$ subgraphs have been formed in ${{\mathcal{G}}^{\left[ 1 \right]}}$ after the connection operation of vertices between ${{R}_{1}}$ and ${{{\eta }'}_{1}}$, as shown in Fig. 5. With $\left| {{R}_{1}} \right|={{d}_{2}}$ and $\left| {{{{\eta }'}}_{1}} \right|=M$, the design ensures that at least one vertex has been connected to each reference vertex and thus, there exist at least two vertices in each subgraph ${{\ell }_{2}}$, i.e., $\left| \mathcal{P}_{{{\ell }_{2}}}^{\left[ 1 \right]} \right|\ge 2$. Furthermore, referring to the first and fifth equations in (16), we have $\max \left\{ {{d}_{{{j}'}}} \right\}_{{j}'=3}^{K}\le {{d}_{2}}$, so as to ensure that the vertices from the same Tx are avoided being connected to the same reference vertex. For the illustrative example, ${{\mathcal{G}}^{\left[ 1 \right]}}=\left( {{\mathcal{P}}^{\left[ 1 \right]}},{{\mathcal{E}}^{\left[ 1 \right]}} \right)$ can be obtained as given in Fig. 6.

\begin{figure}[!t]
\centering
\includegraphics[width=1.76in]{fig6}
\vspace*{-10pt}
\caption{Optimal transceiver design of the illustrative example: ${{\mathcal{G}}^{\left[ 1 \right]}}=\left( {{\mathcal{P}}^{\left[ 1 \right]}},{{\mathcal{E}}^{\left[ 1 \right]}} \right)$}
\end{figure}

In the sequel, we provide the design of ${{\mathcal{G}}^{\left[ 2 \right]}}$. Initially, set ${{\mathcal{G}}^{\left[ 2 \right]}}\leftarrow {{\mathcal{G}}^{\left[ 1 \right]}}$. At receiver 2, we have ${{\eta }_{2}}=\{ \left. {{\mathbf{v}}_{{j}'{{\ell }_{{{j}'}}}}} \right|\forall {j}'\in \left[ 3,K \right],\text{ }\forall {{\ell }_{{{j}'}}}\in \left[ 1,{{d}_{{{j}'}}} \right] \}$. With $\left| {{{{\eta }'}}_{2}} \right|=\left| \mathcal{I}_{2}^{\operatorname{IA}}-\mathcal{\bar{I}}_{2}^{\operatorname{SIA}} \right|=\left( \sum\nolimits_{j=3}^{K}{{{d}_{j}}} \right)-\left( M-{{d}_{2}} \right)={{d}_{2}}$, we can form ${{{\eta }'}_{2}}$ by selecting one vertex ${{\mathbf{v}}_{{j}'{{\ell }_{{{j}'}}}}}\in \left( P_{{{{{\ell }'}}_{2}}}^{\left[ 2 \right]}\bigcap {{\eta }_{2}} \right)$ from each $\mathcal{P}_{{{{{\ell }'}}_{2}}}^{\left[ 2 \right]}$, ${{{\ell }'}_{2}}\in \left[ 1,{{d}_{2}} \right]$. This indicates that only one vertex in each $\mathcal{P}_{{{{{\ell }'}}_{2}}}^{\left[ 2 \right]}$ is required to be connected to ${{R}_{2}}$. ${{R}_{2}}$ can be obtained by ${{R}_{2}}={{\eta }_{2}}-{{{\eta }'}_{2}}$. For the illustrative example, we have ${{{\eta }'}_{2}}=\left\{ {{\mathbf{v}}_{31}},{{\mathbf{v}}_{32}},{{\mathbf{v}}_{41}} \right\}$ and ${{R}_{2}}=\left\{ {{\mathbf{v}}_{42}},{{\mathbf{v}}_{51}},{{\mathbf{v}}_{52}} \right\}$, as shown in Fig. 7.

The connection between ${{{\eta }'}_{2}}$ to ${{R}_{2}}$ is completed in two steps. In the first step, for the only vertex ${{\mathbf{v}}_{{j}'{{l}_{{{j}'}}}}}\in {{{\eta }'}_{2}}\bigcap P_{{{\ell }_{2}}}^{\left[ 2 \right]}$ in subgraph $\mathcal{G}_{{{\ell }_{2}}}^{\left[ 2 \right]}$, $\forall {{\ell }_{2}}\in \left[ 1,{{d}_{2}} \right]$, if there is $\left( {{R}_{2}}-A_{{{j}'}}^{\left[ 2 \right]} \right)\bigcap P_{{{\ell }_{2}}}^{\left[ 2 \right]}\ne \phi $, then the vertex ${{\mathbf{v}}_{{j}'{{l}_{{{j}'}}}}}$ is connected to one reference vertex ${{\mathbf{v}}_{j{{l}_{j}}}}\in \left( {{R}_{2}}-A_{{{j}'}}^{\left[ 2 \right]} \right)\bigcap P_{{{\ell }_{2}}}^{\left[ 2 \right]}$, creating an edge $e_{\left( j{{l}_{j}} \right),\left( {j}'{{l}_{{{j}'}}} \right)}^{\left[ 2 \right]}$ which represents alignment equation of $\operatorname{span}\left( {{\mathbf{H}}_{2j}}{{\mathbf{v}}_{j{{\ell }_{j}}}} \right)=\operatorname{span}\left( {{\mathbf{H}}_{2{j}'}}{{\mathbf{v}}_{{j}'{{\ell }_{{{j}'}}}}} \right)$. Therefore, $\mathcal{E}_{{{\ell }_{2}}}^{\left[ 2 \right]}\leftarrow \mathcal{E}_{{{\ell }_{2}}}^{\left[ 2 \right]}\bigcup \left\{ e_{\left( j{{l}_{j}} \right),\left( {j}'{{l}_{{{j}'}}} \right)}^{\left[ 2 \right]} \right\}$ and one loop has been formed in $\mathcal{G}_{{{\ell }_{2}}}^{\left[ 2 \right]}$. Now, update $A_{{{j}'}}^{\left[ 2 \right]}$ by $A_{{{j}'}}^{\left[ 2 \right]}\leftarrow A_{{{j}'}}^{\left[ 2 \right]}\bigcup \left\{ {{\mathbf{v}}_{j{{l}_{j}}}} \right\}$. In the second step, for the remaining subgraphs $\mathcal{G}_{{{{{\ell }'}}_{2}}}^{\left[ 2 \right]}$, $\forall {{{\ell }'}_{2}}\in \left[ 1,{{d}_{2}} \right]$, where $\left( {{R}_{2}}-A_{{{j}'}}^{\left[ 2 \right]} \right)\bigcap P_{{{{{\ell }'}}_{2}}}^{\left[ 2 \right]}=\phi $ for the vertex ${{\mathbf{v}}_{{j}'{{l}_{{{j}'}}}}}\in {{{\eta }'}_{2}}\bigcap P_{{{{{\ell }'}}_{2}}}^{\left[ 2 \right]}$, ${{\mathbf{v}}_{{j}'{{l}_{{{j}'}}}}}$ would be connected to one of the other subgraphs which has completed the vertex connection in the first step. Find a subgraph $\mathcal{G}_{{{\ell }_{2}}}^{\left[ 2 \right]}$ with $\left( {{R}_{2}}-A_{{{j}'}}^{\left[ 2 \right]} \right)\bigcap \mathcal{P}_{{{\ell }_{2}}}^{\left[ 2 \right]}\ne \phi $, and connect the vertex ${{\mathbf{v}}_{{j}'{{l}_{{{j}'}}}}}\in {{{\eta }'}_{2}}\bigcap \mathcal{P}_{{{{{\ell }'}}_{2}}}^{\left[ 2 \right]}$ to one reference vertex ${{\mathbf{v}}_{j{{l}_{j}}}}\in \left( {{R}_{2}}-A_{{{j}'}}^{\left[ 2 \right]} \right)\bigcap \mathcal{P}_{{{\ell }_{2}}}^{\left[ 2 \right]}$, ${{\ell }_{2}}\ne {{{\ell }'}_{2}}$, creating an edge $e_{\left( j{{l}_{j}} \right),\left( {j}'{{l}_{{{j}'}}} \right)}^{\left[ 2 \right]}$ which represents alignment equation of $\operatorname{span}\left( {{\mathbf{H}}_{2j}}{{\mathbf{v}}_{j{{\ell }_{j}}}} \right)=\operatorname{span}( {{\mathbf{H}}_{2{j}'}}{{\mathbf{v}}_{{j}'{{\ell }_{{{j}'}}}}} )$. In this case, the connection operation does not form loops, but merges two subgraphs, $\mathcal{G}_{{{{{\ell }'}}_{2}}}^{\left[ 2 \right]}$ and $\mathcal{G}_{{{\ell }_{2}}}^{\left[ 2 \right]}$, into one, denoted by $\mathcal{G}_{{{\ell }_{2}}}^{\left[ 2 \right]}$. Then, we have $\mathcal{E}_{{{\ell }_{2}}}^{\left[ 2 \right]}=\mathcal{E}_{{{\ell }_{2}}}^{\left[ 2 \right]}\bigcup \mathcal{E}_{{{{{\ell }'}}_{2}}}^{\left[ 2 \right]}\bigcup \left\{ e_{\left( j{{l}_{j}} \right),\left( {j}'{{l}_{{{j}'}}} \right)}^{\left[ 2 \right]} \right\}$. Update $A_{{{j}'}}^{\left[ 2 \right]}$ and ${{Q}^{\left[ 2 \right]}}$ by $A_{{{j}'}}^{\left[ 2 \right]}\leftarrow A_{{{j}'}}^{\left[ 2 \right]}\bigcup \left\{ {{\mathbf{v}}_{j{{l}_{j}}}} \right\}$ and ${{Q}^{\left[ 2 \right]}}\leftarrow {{Q}^{\left[ 2 \right]}}-1$, respectively. In the illustrative example, each subgraph holds a reference vertex. Therefore, the connection between ${{{\eta }'}_{2}}$ and ${{R}_{2}}$ is performed in each single subgraph and thus, only the first step is sufficient. Finally, ${{\mathcal{G}}^{\left[ 2 \right]}}$ has been obtained as in Fig. 7.

\begin{figure}[!t]
\centering
\includegraphics[width=2.0in]{fig7}
\vspace*{-10pt}
\caption{Optimal transceiver design of the illustrative example: $\mathcal{G}={{\mathcal{G}}^{\left[ 2 \right]}}$}
\end{figure}

In summary, a total of ${{d}_{2}}$ edges have been added by connecting ${{{\eta }'}_{2}}$ to ${{R}_{2}}$. One edge-addition operation in the first step forms one loop in the corresponding subgraph while the one in the second step merges the corresponding subgraph into another subgraph without forming loops. Assuming that the number of subgraphs subject to the second step is ${{q}^{*}}$, then we have ${{Q}^{\left[ 2 \right]}}={{Q}^{\left[ 1 \right]}}-{{q}^{*}}={{d}_{2}}-{{q}^{*}}$ and each of ${{Q}^{\left[ 2 \right]}}$ subgraphs has one loop. Consequently, we have designed ${{\mathcal{G}}^{\left[ 2 \right]}}$ with only one loop in each subgraph. With $\mathcal{G}={{\mathcal{G}}^{\left[ 2 \right]}}$, the IAC graph $\mathcal{G}$ satisfies Proposition 1 and a set of alignment equations $\Psi $ specified by $\mathcal{G}$ can be correspondingly obtained. Then, the closed-form transceivers, $\left\{ {{\mathbf{V}}_{j}} \right\}_{j=1}^{K}$ and $\left\{ {{\mathbf{U}}_{k}} \right\}_{k=1}^{K}$, can be found as in Section IV-B, while achieving the maximum DoF of $2M=12$ as expected by the theory.

\section{Conclusion}
For $K$-user MIMO interference channel with $M$ Tx/Rx antennas, we have presented a general design principle for a closed-form IAC transceiver with any feasible tuple of DoFs $\left( {{d}_{1}},{{d}_{2}},\cdots ,{{d}_{K}} \right)$. We also have specified the design criterion for ${{k}_{\operatorname{IAC}}}=2$, where the decoded signals of only two receivers are shared through a backhaul link for cancellation, yet achieve the theoretically possible maximum DoFs, which is $2M$.

To our best knowledge, this work is the very first attempt to propose the design that gives the closed-form IAC transceivers for any feasible tuple of DoFs $\left( {{d}_{1}},{{d}_{2}},\cdots ,{{d}_{K}} \right)$. The current design principle can be also applied to IA, as a special case of IAC. As opposed to the existing heuristic algorithms [2-4], which cannot determine the existence of transceiver, i.e., failing to guarantee the performance, our design can achieve a definite DoF for a general system configuration. Moreover, as no iterative steps are involved in the proposed design, it has reduced the computational complexity over heuristic algorithms, turning out to be more practical.

Considering the optimal design with ${{k}_{\operatorname{IAC}}}=2$, it only requires to send the decoded signals to the first two receivers through a backhaul link for cancellation and thus, the overhead is significantly reduced, while achieving a maximum total DoFs of $2M$. Recall that the maximum total achievable DoFs is limited ${2MK}/{\left( K+1 \right)}\;$ for IA with an equal tuple of DoFs, i.e., ${{d}_{1}}={{d}_{2}}=\cdots ={{d}_{K}}=d$. Furthermore, it is  With a general tuple of DoFs $({{d}_{1}},{{d}_{2}},\cdots ,{{d}_{K}})$, meanwhile, it is available only by a loose upper bound so far, which is given as $\left( 2M-1 \right)$. In other words, we have shown how much performance gain can be achieved by IAC over IA in terms of the total achievable DoF for both equal and general tuples of DoFs, without incurring too much overhead.

Lastly, we will be evaluating the performance of our proposed IAC design with the imperfect CSI and unreliable backhaul link for a practical system in our future work.

\ifCLASSOPTIONcaptionsoff
  \newpage
\fi

redpajama_set_name: RedPajamaArXiv


text:
Q: Asset fingerprinting in Play Framework Rails provides a really simple solution for fingerprinting static assets. Is there anything similar available for the Play Framework? Can't seem to find anything on Google.
Thanks.
http://guides.rubyonrails.org/asset_pipeline.html
Update: My company ended up implementing one internally.

A: you may want to take a look at the new require.js support[1], available on play 2.1 master
[1] https://github.com/playframework/Play20/commit/e3c2ffb0d4779863374c44c5ec1a588a1eca4474

A: Play 2.3 (currently on RC1) introduces a Rails-like asset pipeline with support for asset fingerprinting.

A: Play framework 2 has built-in support for CofeeScript, LESS and Closure, however none of them are considered as 'static' resources as they are compiled at the runtime.
There is no equivalent of fingerprint yet, so you'll need to write (probably quite short) action by yourself.

A: I'm not sure how you could do this through a quite short action as one answer mentions.
But you could create a wrapped Assets controller like James Ward does here:
https://github.com/jamesward/play2-cloudfront
That will just handle decorating the links with CDN information.  You could then get an MD5 of the link and append that to the file name.  So if you had foo.js it will be come foo-fp-XXXXXXXXXXXXXX.js".
I think you would then have to change the routes file to route that asset to the right file when the CDN comes looking for it.  So the route would have to strip everything in "-fp-XXXXXXXXXXXXXX" and get the filename and extension.
EDIT:
  Here is git repo where I have an example of how we do asset fingerprinting at my company.  https://github.com/Egraphs/play-assets-improvements

redpajama_set_name: RedPajamaStackExchange


Learn how to make Mizoram Koat Pitha the healthy way in a Kuzhi Paniyaram Kal instead of frying it.
Though most of the people are Christians by faith, the Mizos are very proud of their rich heritage, culture and lively traditions.
Poonam who blogs at Annapurna gave me Rice flour and oil as my secret ingredients to create a delicious dish from the Mizo Land.
I decided to use the kuzhi paniyaram kal to make this desert instead of frying it. Using this method greatly reduces the usage of oil and one can have Mizoram Koat Pitha without any guilt.
Here is the link to making a traditional Kuzhi Paniyaram – a savoury snack that can be eaten as breakfast too !!!
Once both sides are cooked and have a golden colour, remove them from the mold and serve them hot.


(MCALLEN, TX) November 14, 2018 – McAllen's Blanca E. Sanchez Elementary School was recently awarded a $2,500 grant from the Laureus Sport for Good Foundation (Laureus USA) as part of Reimagine Recess, an educational campaign that challenges the conventional perception of recess. Reimagine Recess is a collaboration between national social change organizations – the Alliance for a Healthier Generation (Healthier Generation), Laureus USA and Red Nose Day in School.
Recess has long been thought of as a period for elementary school students to play and spend time with friends. Reimagine Recess encouraged educators to redefine the time as more than physical activity and to incorporate social-emotional health topics – such as empathy, conflict resolution and relationship skills. Additionally, through Reimagine Recess' interactive resource center, educators could utilize innovative tools and build a customized recess guide to meet the unique needs of their school environment.
Schools were encouraged to share photos from their reimagined recess on social media using the hashtag #ReimagineRecess for a chance to win one of two $2,500 recess grants from Laureus USA. Sanchez was one of two schools to receive a grant. Caruthersville Elementary School in Caruthersville, Missouri won the second $2,500 grant. The grants will be used to purchase recess equipment for their schools.
McAllen Independent School District schools have already been making progress toward creating healthier environments for students. Earlier this fall, the Alliance for a Healthier Generation recognized 18 McAllen ISD schools – including Blanca Sanchez – on the annual list of America's Healthiest Schools. A total of 461 schools earned the distinction, among 26 states and the District of Columbia. This honor is granted to schools that meet a rigorous set of criteria, from availability of nutritious foods to access to quality physical education opportunities, as part of Healthier Generation's Healthy Schools Program.
For more information about Reimagine Recess and to build a custom recess guide for your school, visit ReimagineRecess.org.
The Alliance for a Healthier Generation believes every child deserves a healthy future. For more than a decade, Healthier Generation has been empowering kids to develop lifelong healthy habits by ensuring the environments that surround them provide and promote good health. Driven by our passion that all young people deserve a chance to live healthier lives, our work has helped more than 28 million kids across the country. Make a difference at HealthierGeneration.org and join us on Facebook and Twitter.
Red Nose Day USA is a fundraising campaign run by the non-profit organization Comic Relief USA, also known as Comic Relief Inc., is a registered U.S. 501(c)(3) public charity. Red Nose Day started in the U.K., built on the foundation that the power of entertainment can drive positive change, and has raised over $1 billion since the campaign's founding in 1988. Since launching in the U.S. in 2015, Red Nose Day has raised over $145 million. Money raised by Red Nose Day USA supports programs that keep children in need safe, healthy and educated both in America and abroad. Red Nose Day USA has received generous support from millions of Americans, hundreds of celebrities and many outstanding partners, including Walgreens, NBC, Mars, and the Bill & Melinda Gates Foundation. For more information about Red Nose Day and its impact, visit www.rednoseday.org. Follow @RedNoseDayUSA on Twitter, Instagram and Facebook.
Red Nose Day in School, a Red Nose Day USA program, provides educators with free resources to help students understand the issue of child poverty, develop empathy, and inspire them to create positive change — one nose at a time. To access the 2018/19 school year activities visit www.RedNoseDayinSchool.org.


text:
14 LISTS It's the End of the WorldHow do you feel about a little lite apocalypse?
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Photo: Viktor Vasnetsov
The Best Nuclear Holocaust Movies
All Genre Movies Lists
600 votes 151 voters 10.4k views 26 items
A list of the best nuclear holocaust movies ever made, ranked by movie fans with film trailers when available. This nuclear holocaust movie list is ordered by popularity, so only the greatest movies about nuclear holocaust are at the top of the list. If there are any great movies about nuclear holocaust you believe we missed, please feel free to add them to the list yourself. The films on this list are the most famous nuclear holocaust movies of all time, so use it to find some movies that you haven't seen already. Click on these nuclear holocaust movie titles for more information about the films.
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Various bits of information are included with these popular nuclear holocaust movies, such as who directed them, when they were released and who starred in the films. Whether you're watching these nuclear holocaust movies on DVD, Blu-ray or streaming, these good nuclear holocaust movies can be enjoyed no matter what.
hpliebekraft added Threads Reece Dinsdale, Jane Hazlegrove, David Brierly
Threads is a 1984 BAFTA award-winning British television drama, produced jointly by the BBC, Nine Network and Western-World Television Inc. Written by Barry Hines and directed by Mick Jackson, ...more
JohnBroussard added The Day After John Lithgow, Steve Guttenberg, JoBeth Williams
The Day After is a 1983 American television film that aired on November 20, 1983, on the ABC television network. It was seen by more than 100 million people during its initial broadcast. It is ...more
Dr. Strangelove or: How I Learned to Stop Worrying and Love the Bomb James Earl Jones, Peter Sellers, George C. Scott
Dr. Strangelove or: How I Learned to Stop Worrying and Love the Bomb, more commonly known as Dr. Strangelove, is a 1964 political satire black comedy film that satirizes fears of a nuclear ...more
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#50 of 382 The Best War Movies Ever #517 of 1,699 The Most Rewatchable Movies #78 of 710 The Best Movies Roger Ebert Gave Four Stars
hpliebekraft added Fail Safe Henry Fonda, Walter Matthau, Larry Hagman
Fail-Safe is a 1964 Cold thriller film directed by Sidney Lumet, based on the 1962 novel of the same name by Eugene Burdick and Harvey Wheeler. The film features performances by veteran actors ...more
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#769 of 1,699 The Most Rewatchable Movies #49 of 194 The Best Movies of the '60s #198 of 569 The Greatest Disaster Movies of All Time
Testament Kevin Costner, Rebecca De Mornay, Jane Alexander
Testament is a drama film based on The Last Testament by Carol Amen, directed by Lynne Littman and written by John Sacret Young. The film tells the story of how one small suburban town near the ...more
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#194 of 569 The Greatest Disaster Movies of All Time #220 of 318 The Best Political Films Ever Made #39 of 42 The Best Sports Movies About Coaches
hpliebekraft added When the Wind Blows John Mills, Peggy Ashcroft, David Dundas
When the Wind Blows is a 1986 British animated drama film directed by Jimmy Murakami based on Raymond Briggs' graphic novel of the same name. The film stars the voices of John Mills and Peggy ...more
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#335 of 600 The Best Animated Films Ever #205 of 304 The Greatest Soundtracks of All Time #541 of 575 The Greatest Movies Of The 1980s, Ranked
Terminator 2: Judgment Day Arnold Schwarzenegger, Edward Furlong, Linda Hamilton
Terminator 2: Judgment Day is a 1991 American science fiction action thriller film written, produced and directed by James Cameron. The film stars Arnold Schwarzenegger, Linda Hamilton, Robert ...more
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#143 of 336 'Old' Movies Every Young Person Needs To Watch In Their Lifetime #9 of 1,699 The Most Rewatchable Movieslisted on Every Oscar Winning Film Ever
JohnBroussard added On the Beach
On the Beach is a 1957 post-apocalyptic novel written by British-Australian author Nevil Shute after he emigrated to Australia. The novel details the experiences of a mixed group of people in ...more
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#950 of 1,302 The Best Novels Ever Written #253 of 837 Books That Changed Your Life #239 of 569 The Greatest Disaster Movies of All Time
The Terminator Arnold Schwarzenegger, Bill Paxton, Linda Hamilton
The Terminator is a 1984 American science fiction/horror action film directed by James Cameron, written by Cameron and the film's producer Gale Anne Hurd. Schwarzenegger plays the Terminator, a ...more
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#15 of 336 'Old' Movies Every Young Person Needs To Watch In Their Lifetime #45 of 1,699 The Most Rewatchable Moviesfeaturing 56 Things You Didn't Know About the Terminator Movies
Miracle Mile Mare Winningham, Anthony Edwards, Denise Crosby
Miracle Mile is a 1988 American apocalyptic thriller cult film written and directed by Steve De Jarnatt, and starring Anthony Edwards and Mare Winningham that takes place mostly in real time. It ...more
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#218 of 569 The Greatest Disaster Movies of All Time #6 of 16 Movies That Took A Hard Turn And Left Audiences Completely Baffled #94 of 108 The Best Apocalyptic And Post-Apocalyptic Movies
WarGames Matthew Broderick, William H. Macy, Ally Sheedy
WarGames is a 1983 American science-fiction film written by Lawrence Lasker and Walter F. Parkes and directed by John Badham. The film stars Matthew Broderick, Dabney Coleman, John Wood, and ...more
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#254 of 1,699 The Most Rewatchable Movies #122 of 710 The Best Movies Roger Ebert Gave Four Starslisted on The Very Best Movies About Hackers
The Sacrifice Erland Josephson, Allan Edwall, Sven Wollter
The Sacrifice is a 1986 Swedish film directed by Andrei Tarkovsky. Starring Erland Josephson, it centers on a middle-aged intellectual who attempts to bargain with God to stop an impending ...more
Akira Takeshi Kusao, Mami Koyama, Tesshō Genda
Akira is a 1988 Japanese animated cyberpunk action film directed by Katsuhiro Otomo, written by Otomo and Izo Hashimoto, and featuring the voices of Mitsuo Iwata, Nozomu Sasaki, Mami Koyama, and ...more
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#119 of 600 The Best Animated Films Ever #504 of 630 The Greatest Guilty Pleasure Moviesfeaturing Some Artists Drew All Of 'Akira' With 'The Simpsons' Characters, And It's Amazing
The Time Machine Yvette Mimieux, Rod Taylor, Alan Young
The Time Machine – also known promotionally as H.G. Wells' The Time Machine – is a 1960 Metrocolor time travel science fiction film based on the 1895 novel of the same name by H. G. Wells in ...more
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#408 of 1,699 The Most Rewatchable Movies #38 of 48 The Coolest Fictional Objects You Most Want to Own #63 of 160 The Best Dystopian And Near Future Movies
Equilibrium Christian Bale, Sean Bean, Taye Diggs
Equilibrium is a 2002 American dystopian science fiction film written and directed by Kurt Wimmer and starring Christian Bale, Emily Watson, and Taye Diggs. The film follows John Preston, an ...more
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#645 of 1,699 The Most Rewatchable Movies #64 of 156 The Best Science Fiction Action Movies #26 of 86 The 35+ Greatest Dystopian Action Movies
Regina Ruehling added Chernobyl: The Final Warning Jon Voight, Jason Robards Jr., Sammi Davis
Chernobyl: The Final Warning is a 1991 made for television movie. The film chronicles the Chernobyl disaster.
Deterrence Sean Astin, Kevin Pollak, Timothy Hutton
Deterrence is a 1999 French/American dramatic film written and directed by Rod Lurie, depicting fictional events about nuclear brinkmanship. It marks the feature directorial debut of Lurie, who ...more
A Boy and His Dog Don Johnson, Jason Robards Jr., Michael Rupert
A young man and his telepathic dog roam a post-nuclear holocaust land searching for food and sex. He encounters an underground community that has survived apparently unscathed from the ...more
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#307 of 569 The Greatest Disaster Movies of All Time #77 of 160 The Best Dystopian And Near Future Movies #542 of 567 The Greatest Sci-Fi Movies of All Time
Regina Ruehling added Chernobyl Diaries Jesse McCartney, Ingrid Bolsø Berdal, Dimitri Diatchenko
Chernobyl Diaries is a 2012 American disaster horror film directed by Brad Parker and produced by Oren Peli, who also wrote the story. It stars Jesse McCartney, Jonathan Sadowski, Devin Kelley, ...more
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#575 of 769 The Best Horror Movies of All Time #221 of 569 The Greatest Disaster Movies of All Time #14 of 30 The Best Horror Movies on Netflix
Terminator 3: Rise of the Machines Arnold Schwarzenegger, Claire Danes, Kristanna Loken
Terminator 3: Rise of the Machines is a 2003 American science fiction action film, directed by Jonathan Mostow and starring Arnold Schwarzenegger, Nick Stahl, Claire Danes, and Kristanna Loken. ...more
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#37 of 93 The Best Third Films In A Movie Series #76 of 156 The Best Science Fiction Action Movies #293 of 601 The Greatest Epic Movies Ever Made
Mollevangsvagen added O-Bi, O-Ba: The End of Civilization Krystyna Janda, Jerzy Stuhr, Leon Niemczyk
O-Bi, O-Ba: The End of Civilization is a 1985 Polish drama science fiction film written and directed by Piotr Szulkin. Starring Jerzy Stuhr and Krystyna Janda, the film takes place in a ...more
Invasion U.S.A. Peggie Castle, Noel Neill, Dan O'Herlihy
Invasion, U.S.A. is a 1952 motion picture set during the Cold War and portraying the invasion of the United States by an unnamed Communist enemy meant to be taken as the Soviet Union.
Terminator Salvation Christian Bale, Helena Bonham Carter, Bryce Dallas Howard
Terminator Salvation is a 2009 American science fiction action film directed by McG and starring Christian Bale and Sam Worthington. It is the fourth and final installment in the original ...more
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#93 of 217 The Worst Sequels Of All Time #95 of 164 Live Action Films with the Best CGI Effects #16 of 30 The Worst Movies With Great Trailers
Def-Con 4 Maury Chaykin, Lenore Zann, Tim Choate
Def-Con 4 is a 1985 post-apocalyptic film, portraying three astronauts who survive World War III aboard a space station and return to earth to find greatly changed circumstances.
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#367 of 569 The Greatest Disaster Movies of All Time #183 of 186 The Best Sci-Fi Movies of the 1980s #183 of 199 The Best Space Movies
The Bed-Sitting Room Dudley Moore, Marty Feldman, Peter Cook
The Bed-Sitting Room is a 1969 British comedy film directed by Richard Lester and based on the play of the same name. It was entered into the 19th Berlin International Film Festival. The film is ...more
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Dacia is putting the Duster Black Edition into production at a cost of £17,575. The car originally appeared as a one off at the Goodwood Moving Motor Show, but demand was sufficient to put the car into production. It's available to order now, complete with black, brushed metal vinyl finish.
It's based on the two-wheel drive Duster Ambience dCi 110 and comes with ash leather interior, 18-inch alloy wheels, Kenwood media system, sat nav, DAB, Bluetooth, iPod connectivity, DVD playback and USB connectivity. The brushed metal 'paint' is actually a vinyl wrap.
They cant meet demand for the current Duster models, how do they expect to manufacturer this one?
They can meet demand at £17.500 try get one £8999 one you can't !


The EPA currently does not list any per- and polyfluoroalkyl substances (PFAS) as hazardous substances under the Superfund program which allows the EPA and federal agencies to clean up contaminated sites. Dingell, Upton, and Kildee wrote the PFAS Action Act to designate all PFAS chemicals under the Superfund program because they pose serious risks to human health and the environment and there is a growing urgency and need to act. By making this listing, EPA will be able to direct federal resources to clean up contaminated sites and limit the spread of these dangerous substances.
"Toxic PFAS chemicals are a serious threat to public health and we must pass policies that protect our communities from contamination. This bipartisan legislation is an important first step to making sure federal agencies can clean up sites like Wurtsmith in Oscoda, Buick City in Flint and others across the country," said Kildee.
You can read a copy of the PFAS Action Act here.
Last year, the Environment Subcommittee of the Energy and Commerce Committee held a hearing on per- and polyfluoroalkyl substance (PFAS) in which an EPA official committed to sending experts and senior officials to Michigan to discuss PFAS contamination sites. The EPA had originally planned a public meeting in Michigan, but later pulled it from the schedule. In October, EPA visited sites with confirmed PFAS levels in Michigan.


If you like animals and biology, you may want to consider a profession as a veterinarian. It's a well-paying career that allows you to work with both animals and humans in an interesting and important position. Vets can work in a variety of settings from a pet clinic to a farm [source: BLS]. Read the tips listed below and learn about how to become a vet.
Educational requirements Before going to veterinary school, you must complete a university degree with a high cumulative grade point average. You will also have to achieve a high score on the Graduate Record Examination (GRE). Preferred areas of background include animal science, biology and chemistry [source: University of Illinois]. Upon admission to a veterinary program, you will have to complete four years of study to obtain a degree as a doctor of veterinary medicine [source: UWYO].
Residency program Although it's not required, it's recommended that you complete an internship or residency program upon graduation from veterinary school. This will involve working with experienced vets, and will help you familiarize yourself with the hands-on aspect of the career. These years of training can also provide specializations in certain fields of veterinary sciences [source: UWYO].
Getting licensed Most states require vets to be licensed, however licensing requirements vary from state to state. Upon completing your education, find out the licensing requirements in the state where you plan on practicing. All states require applicants to pass the North American Veterinary Licensing Exam [source: BLS].
Obtaining employment There are a number of avenues to obtain employment as a vet. Most recent graduates open their own practices, treating companion animals, such as dogs, cats and birds. However, private industries, government services and the military are all in need of veterinary services. Consider contacting your state or national government body or local farms or zoos for employment [source: UWYO].


With sandal period is coming, concerns regarding unattractive ft jump towards the forefront. Following a winter invested in footwear, skin upon feet is commonly dry as well as scaly or even have thickened places, and the look of toenails might be shocking.
Nobody wants in order to pair unsightly feet along with great-looking brand new sandals. The main problem for most of us is merely that ft often obtain put at the end of the private care 'to do' checklist.
A practical approach involves creating a commitment to take some time necessary to improve the appearance of the feet, while you would that person or every other part of the body.
1. Begin by getting right into a regular feet care program. A every day (or even weekly) "aromatherapy" feet bath offers both tension relief and also the opportunity to focus on foot look. Fill the container (large enough with regard to both you) along with very tepid to warm water and give a few drops of the favorite gas or the tablespoon associated with any fragranced fluid hand cleaning soap.
2. Examine toe nails and do something if they're unsightly. With regard to unattractive yellow-colored or heavy toenails, a organic approach is by using NonyX® Toe nail Gel, the topical, easy-to-use as well as diabetic-friendly gel utilized by podiatrists, but additionally sold within pharmacies. NonyX Gel functions breaking lower and getting rid of keratin particles – the actual discolored, yellow or even thick granular accumulation under nails which in turn causes them to look yellow, darkish or stained. NonyX Carbamide peroxide gel softens keratin debris in order that it can end up being scraped from under the actual nail, beginning following about a month (better begin! )#). Along with regular make use of, toenails appear clear as well as healthy.
3. Trim toe nails carefully and often cuticles. Toenails must always be cut straight throughout – simpler to do whenever they're damp in the shower. Toenails additionally deserve the interest of the scrub brush to wash underneath. Purchase high high quality instruments and not share pedicure or even manicure tools with other people. If cuticles really are a problem, make use of a cuticle softener after which push back again the cuticle.
four. Exfoliate as well as moisturize dried out heels as well as hardened places. There tend to be many feet moisturizers available on the market. One of these, CalleX® Dried out Heel Cream, moisturizes, softens as well as exfoliates all in a single application. CalleX Ointment can also be diabetic-friendly and leads to baby-soft ft within in just a couple weeks.
5. Make use of nail shine sparingly; Make use of sunscreen liberally. While toe nail polish supplies a temporary repair for unappealing toenails, continuous use isn't advised. Toenails have to breathe and also have occasional breaks or cracks from shine. Fresh atmosphere and sunlight are ideal for feet as well as toenails, but sunscreen is actually advised to safeguard sensitive pores and skin from dangerous rays.
Should you start now using the right products along with a real dedication, attractive feet could be yours through summer!
Item Sourcing Info: NonyX Toe nail Gel comes in the actual FOOT CARE portion of Walgreens, Ceremony Aid, CVS, Walmart, Meijer, FOCUS ON, Snyders, Longs, Duane Reade, Huge Eagle, HEB, Kerr Medication, Bruno's, ShopKo, Hannaford, Duane Reade, Sweetbay, Giant-Eagle and several grocery stores within the Southeast and many Medicine Shoppe pharmacies. Rebate coupons can be found on Xenna's web site.