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\documentclass{article}

\usepackage[a4paper, total={6in, 8in}]{geometry}

\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{booktabs}
\usepackage{url}
\usepackage{tcolorbox}

\usepackage{tikz}
\usetikzlibrary{arrows,decorations,shapes,backgrounds,patterns}
\usepackage{pgfplots}
\pgfplotsset{compat=newest}
\definecolor{green}{RGB}{32,127,43}
\usetikzlibrary{calc}

\usepackage{listings}
\lstdefinestyle{customc}{
  belowcaptionskip=1\baselineskip,
  breaklines=true,
  frame=L,
  xleftmargin=\parindent,
  language=C,
  showstringspaces=false,
  basicstyle=\footnotesize\ttfamily,
  keywordstyle=\bfseries\color{green!40!black},
  commentstyle=\itshape\color{purple!40!black},
  identifierstyle=\color{blue},
  stringstyle=\color{orange},
}
\lstset{escapechar=@,style=customc}

\title{NR LDPC Decoder}
\author{Sebastian Wagner (TCL)}
\date{\today}

\def\0{\mathbf{0}}
\def\b{\mathbf{b}}
\def\Bbb{\mathbb{B}}
\def\Bcal{\mathcal{B}}
\def\c{\mathbf{c}}
\def\C{\mathbf{C}}
\def\Cbb{\mathbb{C}}
\def\Ccal{\mathcal{C}}
\def\eqdef{\triangleq}
\def\g{\mathbf{g}}
\def\G{\mathbf{G}}
\def\Gcal{\mathcal{G}}
\def\h{\mathbf{h}}
\def\H{\mathbf{H}}
\def\Hbg{\mathbf{H}_\mathrm{BG}}
\def\Hbgo{\mathbf{H}_\mathrm{BG1}}
\def\Hbgt{\mathbf{H}_\mathrm{BG2}}
\def\I{\mathbf{I}}
\def\Kb{{K_b}}
\def\m{\mathbf{m}}
\def\Mb{{M_b}}
\def\Nb{{N_b}}
\def\Nbb{\mathbb{N}}
\def\n{\mathbf{n}}
\def\nr{{n_{\rm r}}}
\def\nt{{n_{\rm t}}}
\def\s{\mathbf{s}}
\def\SNR{\mathsf{SNR}}
\def\y{\mathbf{y}}
\def\z{\mathbf{z}}
\def\Z{\mathbf{Z}}
\def\Zc{{Z_c}}

\def\herm{\mathsf{H}}
\def\trans{\mathsf{T}}
\def\EE{\mathsf{E}}
\newcommand{\sgn}{\operatorname{sgn}}

\begin{document}

\maketitle

\begin{tikzpicture}[remember picture,overlay]
   \node[anchor=north west,inner sep=0pt] at (current page.north west)
              {\includegraphics[scale=0.5]{logo.png}};
\end{tikzpicture}

\begin{center}Currently Supported:\end{center}
\tcbox[center]{
    \begin{tabular}{lll}
      \toprule
      \textbf{BG} & \textbf{Lifting Size Z} & \textbf{Code Rate R} \\
      \midrule
        1 & all & 1/3, 2/3, 8/9 \\
        2 & all & 1/5, 1/3, 2/3 \\
      \bottomrule
    \end{tabular}
}

\paragraph{Version 1.0:}
\begin{itemize}
\item Initial version
\end{itemize}

\paragraph{Version 2.0:}
\begin{itemize}
\item Enhancements in message passing:
  \begin{itemize}
  \item LUTs replaced by smaller BG-specific parameters
  \item Inefficient load/store replaced by circular memcpy
  \end{itemize}
\item Bug fixes:
  \begin{itemize}
  \item Fixed bug in function \texttt{llr2CnProcBuf}
  \item Corrected input LLR dynamic range in BLER simulations
  \end{itemize}
\item Results:
  \begin{itemize}
  \item Size of LUTs reduced significantly (60MB to 200KB)
  \item Siginifcantly enhances execution time (factor 3.5)
  \item Improved BLER performance (all simulation results have been updated)
  \end{itemize}
\end{itemize}

\newpage
\tableofcontents

\newpage
\section{Introduction}
\label{sec:introduction}

Low Density Parity Check (LDPC) codes have been developed by Gallager in 1963 \cite{gallager1962low}. They are linear error correcting codes that are capacity-achieving for large block length and are completely described by their Parity Check Matrix (PCM) $\H^{M\times N}$. The PCM $\H$ defines $M$ constraints on the codeword $\c$ of length $N$ such that
\begin{equation}
  \label{eq:29}
  \H\c = \0.
\end{equation}
The number of information bits $B$ that can be encoded with $\H$ is given by $B=N-M$. Hence the code rate $R$ of $\H$ reads
\begin{equation}
  \label{eq:37}
  R = \frac{B}{N} = 1-\frac{M}{N}.
\end{equation}

\subsection{LDPC in NR}
\label{sec:ldpc-nr}

NR uses quasi-cyclic (QC) Protograph LDPC codes, i.e. a smaller graph, called Base Graph (BG), is defined and utilized to construct the larger PCM. This has the advantage that the large PCM does not have to be stored in memory and allows for a more efficient implementation while maintaining good decoding properties.
Two BGs $\Hbg\in\Nbb^{\Mb\times \Nb}$ are defined in NR:
\begin{enumerate}
\item $\Hbgo\in\Nbb^{46\times 68}$
\item $\Hbgt\in\Nbb^{42\times 52}$
\end{enumerate}
where $\Nbb$ is the set of integers. For instance the first 3 rows and 13 columns of BG2 are given by

\setcounter{MaxMatrixCols}{30}
\begin{equation*}
  \label{eq:33}
  \Hbgt =
  \begin{bmatrix}
    9   & 117       & 204       & 26  & \emptyset & \emptyset & 189       & \emptyset & \emptyset & 205       & 0         & 0         & \emptyset & \emptyset \\
    127 & \emptyset & \emptyset & 166 & 253       & 125       & 226       & 156       & 224       & 252       & \emptyset & 0         & 0         & \emptyset \\
    81  & 114       & \emptyset & 44  & 52        & \emptyset & \emptyset & \emptyset & 240       & \emptyset & 1         & \emptyset & 0         & 0
  \end{bmatrix}.
\end{equation*}

To obtain the PCM $\H$ from the BG $\Hbg$, each element $\Hbg(i,j)$ in the BG is replaced by a lifting matrix of size $\Zc\times \Zc$ according to
\begin{equation}
  \label{eq:35}
  \Hbg(i,j) =
  \begin{cases}
    \0 & \textrm{if}~ \Hbg(i,j)=\emptyset \\
    \I_{P_{ij}} & \textrm{otherwise}
  \end{cases}
\end{equation}
where $\I_{P_{ij}}$ is the identity matrix circularly shifted to the right by $P_{ij} = \Hbg(i,j)\mod \Zc$. Hence, the resulting PCM $\H$ will be of size $\Mb\Zc\times\Nb\Zc$.

The lifting size $\Zc$ depends on the number of bits to encode. To limit the complexity, a discrete set $\mathcal{Z}$ of possible values of $\Zc$ has been defined in \cite{3gpp2017_38212} and the optimal value $\Zc$ is calculated according to
\begin{equation}
  \label{eq:36}
  \Zc = \min_{\Z\in\mathcal{Z}}\left[Z\geq\frac{B}{\Nb}\right].
\end{equation}

The base rate of the two BGs is $1/3$ and $1/5$ for BG1 and BG2, respectively. That is, BG1 encodes $K=22\Zc$ bits and BG2 encodes $K=10\Zc$ bits. Note that the first 2 columns of BG 1 and 2 are always punctured, that is after encoding, the first $2\Zc$ bits are discarded and not transmitted.
For instance, consider $B=500$ information bits to encode using BG2, \eqref{eq:36} yields $\Zc=64$ hence $K=640$. Since $K>B$, $K-B=140$ filler bits are appended to the information bits. The PCM $\Hbgt$ is of size $2688\times 3328$ and the $640$ bits $\b$ are encoded according to \eqref{eq:29} at a rate $R \approx 0.192$. To achieve the higher base rate of $0.2$, the first $128$ are punctured, i.e. instead of transmitting all $3328$ bits, only $3200$ are transmitted resulting in the desired rate $R=640/3200=0.2$.

\subsection{LDPC Decoding}
\label{sec:ldpc-decoding}

The decoding of codeword $\c$ can be achieved via the classical message passing algorithm. This algorithm can be illustrated best using the Tanner graph of the PCM. The rows of the PCM are called check nodes (CN) since they represent the parity check equations. The parity check equation of each of these check nodes involves various bits in the codeword. Similarly, every column of the PCM corresponds to a bit and each bit is involved in several parity check equations. In the Tanner graph representation, the bits are called bit nodes (BN). Let's go back to the previous example of BG2 and assume $\Zc=2$, hence the first 3 rows and 13 columns of BG2 $\Hbgt$ read
\begin{equation*}
  \label{eq:36}
  \Hbgt =
  \begin{bmatrix}
    1 & 1         & 0         & 0 & \emptyset & \emptyset & 1         & \emptyset & \emptyset & 1         & 0         & 0         & \emptyset & \emptyset \\
    1 & \emptyset & \emptyset & 0 & 1         & 1         & 0         & 0         & 0         & 0         & \emptyset & 0         & 0         & \emptyset \\
    1 & 0         & \emptyset & 0 & 0         & \emptyset & \emptyset & \emptyset & 0         & \emptyset & 1         & \emptyset & 0         & 0
  \end{bmatrix}.
\end{equation*}
Replacing the elements according to \eqref{eq:35}, we obtain the first 6 rows and 26 columns of the PCM as
\begin{equation*}
  \label{eq:39}
  \H =
  \begin{bmatrix}
    0 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\
    1 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0\\
    0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0\\
    1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0\\
    0 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0\\
    1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1
  \end{bmatrix}.
\end{equation*}

The Tanner graph of the first 8 BNs is shown in Figure \ref{fig:tannergraph}.

\begin{figure}[ht]
  \label{fig:tannergraph}
  \centering
  \def\ww{0.3cm}
  \def\hh{0.3cm}
  \tikzstyle{cnode}=[fill=white,rectangle,draw=black,thick,inner sep=2pt, minimum height=\hh,minimum width=\ww, rounded corners=1pt,text width=\ww]
  \tikzstyle{vnode}=[fill=white,circle,draw=black,thick,inner sep=2pt, minimum height=\hh,minimum width=\ww, rounded corners=1pt,text width=\ww]
  \tikzstyle{connector}=[<->,>=latex',semithick]

  \begin{tikzpicture}
    \tikzstyle{every node}=[node distance=1.5cm,text centered]
    % Check nodes
    \node[cnode, label=above:$v_0$] (v0) {};
    \node[cnode, label=above:$v_1$, right of=v0] (v1) {};
    \node[cnode, label=above:$v_2$, right of=v1] (v2) {};
    % Variable nodes
    \node[vnode, label=below:$c_3$, below of=v1, node distance=1.5cm] (c3) {};
    \node[vnode, label=below:$c_2$, left of=c3, node distance=1.5cm] (c2) {};
    \node[vnode, label=below:$c_1$, left of=c2, node distance=1.5cm] (c1) {};
    \node[vnode, label=below:$c_0$, left of=c1, node distance=1.5cm] (c0) {};
    \node[vnode, label=below:$c_4$, right of=c3, node distance=1.5cm] (c4) {};
    \node[vnode, label=below:$c_5$, right of=c4, node distance=1.5cm] (c5) {};
    \node[vnode, label=below:$c_6$, right of=c5, node distance=1.5cm] (c6) {};

    % Draw edges
    \draw (c0) edge[connector] (v1);
    \draw (c1) edge[connector] (v0);
    \draw (c1) edge[connector] (v2);
    \draw (c2) edge[connector] (v1);
    \draw (c3) edge[connector] (v0);
    \draw (c4) edge[connector] (v0);
    \draw (c4) edge[connector] (v2);
    \draw (c5) edge[connector] (v1);
    \draw (c6) edge[connector] (v0);
    \draw (c6) edge[connector] (v2);

  \end{tikzpicture}

  \caption{Tanner graph for first 7 bits nodes and 3 check nodes from \eqref{eq:39}.}
\end{figure}

The message passing algorithm is an iterative algorithm where probabilities of the bits (being either 0 or 1) are exchanged between the BNs and CNs. After sufficient iterations, the probabilities will have either converged to either 0 or 1 and the parity check equations will be satisfied, at this point, the codeword has been decoded correctly.

\newpage
\section{LDPC Decoder Implementation}
\label{sec:ldpc-implementation}

The implementation on a general purpose processor (GPP) has to take advantage of potential instruction extension of the processor architecture. We focus on the Intel x86 instruction set architecture (ISA) and its advanced vector extension (AVX). In particular, we utilize AVX2 with its 256-bit single instruction multiple data (SIMD) format. In order to utilize AVX2 to speed up the processing at the CNs and BNs, the corresponding data has to be ordered/aligned in a specific way. The processing flow of the LDPC decoder is depicted in \ref{fig:ldpc_decoder_flow}.

\begin{figure}[ht]
  \label{fig:ldpc_decoder_flow}
  \centering
  \def\ww{0.3cm}
  \def\hh{0.3cm}
  \tikzstyle{func}=[,draw=none]
  \tikzstyle{connector}=[->,>=latex',semithick]

  \begin{tikzpicture}
    \tikzstyle{every node}=[node distance=2.5cm,text centered]
    % Check nodes
    % First iteration
    \node[func]                               (llr2llrProcBuf) {\texttt{llr2llrProcBuf}};
    \node[func, above of=llr2llrProcBuf]      (llr2CnProcBuf)  {\texttt{llr2CnProcBuf}};
    \node[func, above right of=llr2CnProcBuf] (cnProc1)        {\texttt{cnProc}};
    \node[func, below right of=cnProc1]       (cn2bnProcBuf1)  {\texttt{cn2bnProcBuf}};
    \node[func, below  of=cn2bnProcBuf1]      (bnProcPc1)      {\texttt{bnProcPc}};

    % Iterations
    \node[func, right of=cnProc1, node distance=7cm] (cnProc)       {\texttt{cnProc}};
    \node[func, below right of=cnProc]               (cn2bnProcBuf) {\texttt{cn2bnProcBuf}};
    \node[func, below of=cn2bnProcBuf]               (bnProcPc)     {\texttt{bnProcPc}};
    \node[func, below left of=cnProc]                (bn2cnProcBuf) {\texttt{bn2cnProcBuf}};
    \node[func, below  of=bn2cnProcBuf]              (bnProc)       {\texttt{bnProc}};

    % Post processing
    \node[func, below of=bnProcPc]      (llrRes2llrOut) {\texttt{llrRes2llrOut}};
    \node[func, below of=llrRes2llrOut, node distance=1cm] (llr2bit) {\texttt{llr2bit}};

    % Draw edges
    \draw (llr2llrProcBuf)  edge[connector] (llr2CnProcBuf);
    \draw (llr2CnProcBuf)   edge[connector] (cnProc1);
    \draw (cnProc1)         edge[connector] (cn2bnProcBuf1);
    \draw (cn2bnProcBuf1)   edge[connector] (bnProcPc1);

    \draw (bnProcPc)       edge[connector] (bnProc);
    \draw (bnProc)         edge[connector] (bn2cnProcBuf);
    \draw (bn2cnProcBuf)   edge[connector] node[above left] {\texttt{cnProcPc}} (cnProc);
    \draw (cnProc)         edge[connector] (cn2bnProcBuf);
    \draw (cn2bnProcBuf)   edge[connector] (bnProcPc);

    \draw (bnProcPc1)      edge[connector] (bnProc);

    \draw (bnProcPc) edge[connector] node[left] {iterations done} (llrRes2llrOut);
    \draw (llrRes2llrOut) edge[connector] (llr2bit);

    % Boxes
    \node[inner sep=0pt,above right of=cn2bnProcBuf1, node distance = 2.5cm] (ref) {};

    \draw[fill=black,opacity=.2, rounded corners] (llr2llrProcBuf.south west) rectangle ($(ref) + (-.5cm,.5cm)$);
    \draw[fill=black,opacity=.2, rounded corners] ($(ref) + (.5cm,.5cm)$) rectangle ($(bnProcPc.south east) + (.4cm,0)$);

    \node[func, above of=cnProc1, node distance=.8cm] (iter1) {\textbf{First Iteration}};
    \node[func, above of=cnProc , node distance=.8cm] (iterX) {\textbf{Subsequent Iterations}};

  \end{tikzpicture}

  \caption{LDPC Decoder processing flow.}
\end{figure}

The functions involved are described in more detail in Table \ref{tab:sum_func}.

\begin{table}[ht]
  \centering
  \begin{tabular}{ll}
    \toprule
    \textbf{Function} & \textbf{Description} \\
    \midrule
    \texttt{llr2llrProcBuf} & Copies input LLRs to LLR processing buffer \\
    \texttt{llr2CnProcBuf}  & Copies input LLRs to CN  processing buffer \\
    \texttt{cnProc}         & Performs CN signal processing \\
    \texttt{cnProcPc}       & Performs parity check \\
    \texttt{cn2bnProcBuf}   & Copies the CN results to the BN processing buffer \\
    \texttt{bnProcPc}       & Performs BN processing for parity check and/or hard-decision \\
    \texttt{bnProc}         & Utilizes the results of \texttt{bnProcPc} to compute LLRs for CN processing \\
    \texttt{bn2cnProcBuf}   & Copies the BN results to the CN processing buffer \\
    \texttt{llrRes2llrOut}  & Copies the results of \texttt{bnProcPc} to output LLRs \\
    \texttt{llr2bit}        & Performs hard-decision on the output LLRs \\
    \bottomrule
  \end{tabular}
  \caption{Summary of the LDPC decoder functions.}
  \label{tab:sum_func}
\end{table}

The input LLRs are assumed to be 8-bit and aligned on 32 bytes. CN processing is carried out in 8-bit whereas BN processing is done in 16 bit. Subsequently, the processing tasks at the CNs and BNs are explained in more detail.

\newpage
\subsection{Check Node Processing}
\label{sec:check-node-proc}

Denote $q_{ij}$ the value from BN $j$ to CN $i$ and let $\Bcal_i$ be the set of connected BNs to the $i$th CN. Then, using the min-sum approximation, CN $i$ has to carry out the following operation for each connected BN.
\begin{equation}
  \label{eq:40}
  r_{ji} = \prod_{j'\in\Bcal_i\setminus j}\sgn q_{ij'}\min_{j'\in\Bcal_i\setminus j} |q_{ij'}|
\end{equation}
where $r_{ji}$ is the value returned to BN $j$ from CN $i$. There are $\Mb = \{46,42\}$ CNs in BG 1 and BG 2, respectively. Each of these CNs is connected to only a small number of BNs. The number of connected BNs to CN $i$ is $|\Bcal_i|$. In BG1 and BG2, $|\Bcal_i|=\{3,4,5,6,7,8,9,10,19\}$ and $|\Bcal_i|=\{3,4,5,6,8,10\}$, respectively. The following tables show the number of CNs $M_{|\Bcal_i|}$ that are connected to the same number of BNs.

\begin{table}[ht]
  \centering
  \begin{tabular}{llllllllll}
    \toprule
    $|\Bcal_i|$   & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 19 \\
    \midrule
    $M_{|\Bcal_i|}^\mathrm{BG1}$ & 1 & 5  &18 & 8 & 5 & 2 & 2 & 1 & 4 \\
    $M_{|\Bcal_i|}^\mathrm{BG2}$ & 6 & 20 & 9 & 3 & 0 & 2 & 0 & 2 & 0 \\
    \bottomrule
  \end{tabular}
  \caption{Ceck node groups for BG1 and BG2.}
  \label{tab:checkNodeGroups}
\end{table}

It can be observed that each CN is at least connected to 3 BNs and there are 9 groups and 5 groups in BG1 and BG2, respectively. Denote the set of CN groups as $\Gcal$ and $M_k$ the number of CNs in group $k\in\Gcal$, e.g. for BG2 $M_4=20\Zc$. Each CN group will be processed separately. The CN processing buffer $p_C^k$ of group $k$ is defined as
\begin{equation}
  \label{eq:44}
  p_C^k = \{\underbrace{q_{11}q_{21}\dots q_{M_k 1}}_{\text 1. BN},\underbrace{q_{12}q_{22}\dots q_{M_k 2}}_{\text 2. BN},\dots,\underbrace{q_{12}q_{22}\dots q_{M_k k}}_{\text last BN}\}
\end{equation}
Hence, $|p_C^k| = kM_k$, e.g, $\Zc=128$, $|p_C^4| = 4\cdot 20\cdot 128 = 10240$.

\begin{lstlisting}[frame=single,caption={Example of CN processing for group 3 from \texttt{cnProc}.},label=code_cnproc]  % Start your code-block

  const uint8_t lut_idxCnProcG3[3][2] = {{72,144}, {0,144}, {0,72}};

  // =====================================================================
  // Process group with 3 BNs

  // Number of groups of 32 CNs for parallel processing
  M = (lut_numCnInCnGroups[0]*Z)>>5;
  // Set the offset to each bit within a group in terms of 32 Byte
  bitOffsetInGroup = (lut_numCnInCnGroups_BG2_R15[0]*NR_LDPC_ZMAX)>>5;

  // Set pointers to start of group 3
  p_cnProcBuf    = (__m256i*) &cnProcBuf   [lut_startAddrCnGroups[0]];
  p_cnProcBufRes = (__m256i*) &cnProcBufRes[lut_startAddrCnGroups[0]];

  // Loop over every BN
  for (j=0; j<3; j++)
  {
    // Set of results pointer to correct BN address
    p_cnProcBufResBit = p_cnProcBufRes + (j*bitOffsetInGroup);

    // Loop over CNs
    for (i=0; i<M; i++)
    {
      // Abs and sign of 32 CNs (first BN)
      ymm0 = p_cnProcBuf[lut_idxCnProcG3[j][0] + i];
      sgn  = _mm256_sign_epi8(*p_ones, ymm0);
      min  = _mm256_abs_epi8(ymm0);

      // 32 CNs of second BN
      ymm0 = p_cnProcBuf[lut_idxCnProcG3[j][1] + i];
      min  = _mm256_min_epu8(min, _mm256_abs_epi8(ymm0));
      sgn  = _mm256_sign_epi8(sgn, ymm0);

      // Store result
      min = _mm256_min_epu8(min, *p_maxLLR); // 128 in epi8 is -127
      *p_cnProcBufResBit = _mm256_sign_epi8(min, sgn);
      p_cnProcBufResBit++;
    }
  }

}
\end{lstlisting}

Once all results of the check node processing $r_{ji}$ have been calculated, they are copied to the bit node processing buffer.

\subsection{Bit Node Processing}
\label{sec:bit-node-processing}

Denote $r_{ji}$ the value from CN $i$ to BN $j$ and let $\Ccal_j$ be the set of connected CNs to the $j$th BN. Each BN $j$ has to carry out the following operation for every connected CN $i\in\Ccal_j$.
\begin{equation}
  \label{eq:46}
  q_{ij} = \Lambda_j + \sum_{i'\in\Ccal_j\setminus i}r_{ji'}
\end{equation}

There are $\Nb = \{68,52\}$ BNs in BG 1 and BG 2, respectively. Each of these BNs is connected to only a small number of CNs. The number of connected CNs to BN $j$ is $|\Ccal_j|$. In BG1 and BG2, $|\Ccal_j|=\{1,4,7,8,9,10,11,12,28,30\}$ and $|\Ccal_j|=\{1,5,6,7,8,9,10,12,13,14,16,22,23\}$, respectively. The following tables show the number of BNs $K_{|\Ccal_j|}$ that are connected to the same number of CNs.

\begin{table}[ht]
  \centering
  \begin{tabular}{lllllllllllllllllll}
    \toprule
    $|\Ccal_j|$ & 1&4&5&6&7&8&9&10&11&12&13 & 14 & 15 & 16 & 22 & 23 &28&30 \\
    \midrule
    $K_{|\Ccal_j|}^\mathrm{BG1}$ & 42 & 1 & 1 & 2 & 4 & 3 & 1 & 4 & 3 & 4 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\
    $K_{|\Ccal_j|}^\mathrm{BG2}$ & 38 & 0 & 2 & 1 & 1 & 1 & 2 & 1 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 & 0 & 0\\
    \bottomrule
  \end{tabular}
  \caption{Bit node groups for BG1 and BG2 for base rates 1/3 and 1/5, respectively.}
  \label{tab:bitNodeGroups}
\end{table}

The BNs that are connected to a single CN do not need to be considered in the BN processing since \eqref{eq:46} yields $q_{ij} = \Lambda_j$. It can be observed that the grouping is less compact, i.e. there are many groups with only a small number of elements.

Denote the set of BN groups as $\Bcal$ and $K_k$ the number of BNs in group $k\in\Bcal$, e.g. for BG2 $K_5=2\Zc$. Each BN group will be processed separately. The BN processing buffer $p_B^k$ of group $k$ is defined as
\begin{equation}
  \label{eq:47}
  p_B^k = \{\underbrace{r_{11}r_{21}\dots r_{K_k 1}}_{\text 1. CN},\underbrace{r_{12}r_{22}\dots r_{K_k 2}}_{\text 2. CN},\dots,\underbrace{r_{12}r_{22}\dots r_{K_k k}}_{\text last CN}\}
\end{equation}
Hence, $|p_B^k| = kK_k$, e.g, $\Zc=128$, $|p_B^5| = 5\cdot 2\cdot 128 = 1024$.

Depending on the code rate, some parity bits are not being transmitted. For instance, for BG2 with code rate $R = 1/3$ the last $20\Zc$ bits are discarded. Therefore, the last 20 columns or the last $20\Zc$ parity check equation are not required for decoding. This means that the BN groups shown in table \ref{tab:bitNodeGroups} are depending on the rate.

\begin{lstlisting}[frame=single,caption={Example of BN processing for group 3 from \texttt{bnProcPc}.},label=code_bnproc]  % Start your code-block

  // If elements in group move to next address
  idxBnGroup++;

  // Number of groups of 32 BNs for parallel processing
  M = (lut_numBnInBnGroups[2]*Z)>>5;

  // Set the offset to each CN within a group in terms of 16 Byte
  cnOffsetInGroup = (lut_numBnInBnGroups[2]*NR_LDPC_ZMAX)>>4;

  // Set pointers to start of group 3
  p_bnProcBuf  = (__m128i*) &bnProcBuf  [lut_startAddrBnGroups   [idxBnGroup]];
  p_llrProcBuf = (__m128i*) &llrProcBuf [lut_startAddrBnGroupsLlr[idxBnGroup]];
  p_llrRes     = (__m256i*) &llrRes     [lut_startAddrBnGroupsLlr[idxBnGroup]];

  // Loop over BNs
  for (i=0,j=0; i<M; i++,j+=2)
  {
    // First 16 LLRs of first CN
    ymmRes0 = _mm256_cvtepi8_epi16(p_bnProcBuf[j]);
    ymmRes1 = _mm256_cvtepi8_epi16(p_bnProcBuf[j+1]);

    // Loop over CNs
    for (k=1; k<3; k++)
    {
      ymm0 = _mm256_cvtepi8_epi16(p_bnProcBuf[k*cnOffsetInGroup + j]);
      ymmRes0 = _mm256_adds_epi16(ymmRes0, ymm0);

      ymm1 = _mm256_cvtepi8_epi16(p_bnProcBuf[k*cnOffsetInGroup + j+1]);
      ymmRes1 = _mm256_adds_epi16(ymmRes1, ymm1);
    }

    // Add LLR from receiver input
    ymm0    = _mm256_cvtepi8_epi16(p_llrProcBuf[j]);
    ymmRes0 = _mm256_adds_epi16(ymmRes0, ymm0);

    ymm1    = _mm256_cvtepi8_epi16(p_llrProcBuf[j+1]);
    ymmRes1 = _mm256_adds_epi16(ymmRes1, ymm1);

    // Pack results back to epi8
    ymm0 = _mm256_packs_epi16(ymmRes0, ymmRes1);
    // ymm0     = [ymmRes1[255:128] ymmRes0[255:128] ymmRes1[127:0] ymmRes0[127:0]]
    // p_llrRes = [ymmRes1[255:128] ymmRes1[127:0] ymmRes0[255:128] ymmRes0[127:0]]
    *p_llrRes = _mm256_permute4x64_epi64(ymm0, 0xD8);

    // Next result
    p_llrRes++;
  }
}

\end{lstlisting}

The sum of the LLRs is carried out in 16 bit for accuracy and is then saturated to 8 bit for CN processing. Saturation after each addition results in significant loss of sensitivity for low code rates.

\subsection{Mapping to the Processing Buffers}
\label{sec:mapp-cn-proc}

For efficient processing with the AVX instructions, the data is required to be aligned in a certain manner. That is the reason why processing buffers have been introduced. The drawback is that the results of the processing need to copied every time to the processing buffer of the next task. However, the speed up in computation with AVX more than makes up for the time wasted in copying data. The copying is implemented as a circular memcpy because every edge in the BG is a circular shift of a $Z\times Z$ identity matrix. Hence, a circular mempcy consists of two regular memcpys each copying a part of the $Z$ values depending on the circular shift in the BG definition. The circular shifts are stored in \texttt{nrLDPC\_lut.h} in arrays \texttt{circShift\_BGX\_ZX\_CNGX}. In the specification there are only 8 sets of cirular shifts defined. However, the applied circular shift depends on $Z$, i.e. modulo $Z$. To avoid inefficient modulo operations in loops, we store the the circular shift values for every $Z$. Moreover, for convinience the arrays are already arranged depending on the CN group (CNG).

\newpage
\section{Performance Results}
\label{sec:performance-results}

In this section, the performance in terms of BLER and decoding latency of the current LDPC decoder implementation is verified.

\subsection{BLER Performance}
\label{sec:bler-performance}

In all simulations, we assume AWGN, QPSK modulation and 8-bit input LLRs, i.e. $-127$ until $+127$. The DLSCH coding procedure in 38.212 is used to encode/decode the TB and an error is declared if the TB CRC check failed. Results are averaged over at least $10\,000$ channel realizations.

The first set of simulations in Figure \ref{fig:bler-bg2-15} compares the current LDPC decoder implementation to the reference implementation developed by Kien. This reference implementation is called \textit{LDPC Ref} and uses the min-sum algorithm with 2 layers and 16 bit for processing. Our current optimized decoder implementation is referred to as \textit{LDPC OAI}. Moreover, reference results provided by Huawei are also shown.

\begin{figure}[ht]
  \centering
  \begin{tikzpicture}
  \tikzstyle{every pin}=[fill=white,draw=black]
    \pgfplotsset{every axis legend/.append style={
        cells={anchor=west}, at={(1.05,1)}, anchor=north west}}
 %   \pgfplotsset{every axis plot/.append style={smooth}}
    \pgfplotsset{every axis/.append style={line width=0.5pt}}
    \pgfplotsset{every axis/.append style={mark options=solid, mark size=2.5pt}}

    \begin{semilogyaxis}[title={}, xlabel={$\SNR$ [dB]}, ylabel={BLER},
      grid={both}, xmin=-4, xmax=2, xtick={-4,-3.5,...,2}, ymin=0,
      ymax=1,ytickten={-5,-4,-3,-2,-1,0},legend columns=1]

      % HUAWEI merged BG2 2017-06-15
      \addplot[black, solid] plot coordinates { (-3.91839,0.01) (-3.5567,0.0001) };

      % 5 iterations
      % LDPC Ref
      \addplot[red, solid, mark=o] plot coordinates {(-1.250000,0.781300) (-1.000000,0.421000) (-0.750000,0.140400) (-0.500000,0.028900) (-0.250000,0.003300) (0.000000,0.000300) (0.250000,0.000000) (0.500000,0.000000)};
      % LDPC OAI
      \addplot[blue, solid, mark=square] plot coordinates {(-1.000000,0.693730) (-0.750000,0.370190) (-0.500000,0.137260) (-0.250000,0.038850) (0.000000,0.009740) (0.250000,0.002510) (0.500000,0.000730) (0.750000,0.000180) };
      % Matlab layered min-sum with scaling factor 1
      \addplot[green, solid, mark=triangle] plot coordinates {(-1.750000,0.709000) (-1.500000,0.360600) (-1.250000,0.105500) (-1.000000,0.015700) (-0.750000,0.001300) (-0.500000,0.000100) (-0.250000,0.000000) (0.000000,0.000000) };
      % Matlab layered min-sum with scaling factor 0.8
      %\addplot[green, solid, mark=triangle] plot coordinates {(-2.750000,0.982300) (-2.500000,0.882200) (-2.250000,0.573100) (-2.000000,0.214100) (-1.750000,0.041300) (-1.500000,0.003800) (-1.250000,0.000000) (-1.000000,0.000000) };

      % 10 iterations
      % Kien's 2-layer 16bit code
      \addplot[red, solid, mark=o] plot coordinates { (-2.750000,0.915500) (-2.500000,0.576000) (-2.250000,0.165000) (-2.000000,0.017100) (-1.750000,0.000600) (-1.500000,0.000000) (-1.250000,0.000000) (-1.000000,0.000000)};
      % LDPC OAI
      \addplot[blue, solid, mark=square] plot coordinates { (-2.750000,0.997200) (-2.500000,0.955000) (-2.250000,0.710900) (-2.000000,0.270400) (-1.750000,0.042400) (-1.500000,0.002200) (-1.250000,0.000000) (-1.000000,0.000000)};
      % Matlab layered min-sum with scaling factor 1
      \addplot[green, solid, mark=triangle] plot coordinates {(-2.750000,0.942900) (-2.500000,0.723200) (-2.250000,0.362300) (-2.000000,0.098400) (-1.750000,0.014500) (-1.500000,0.001100) (-1.250000,0.000000) (-1.000000,0.000000) };
      % Matlab layered min-sum with scaling factor 0.8
      %\addplot[green, solid, mark=triangle] plot coordinates {(-3.750000,0.994300) (-3.500000,0.927200) (-3.250000,0.651100) (-3.000000,0.252000) (-2.750000,0.042500) (-2.500000,0.002700) (-2.250000,0.000000) (-2.000000,0.000000) (-1.750000,0.000000) (-1.500000,0.000000) };

      % 20 iterations
      % Kien's 2-layer 16bit code
      \addplot[red, solid, mark=o] plot coordinates { (-2.750000,0.330300) (-2.500000,0.067800) (-2.250000,0.006000) (-2.000000,0.000100) (-1.750000,0.000000) (-1.500000,0.000000) (-1.250000,0.000000) (-1.000000,0.000000)};
      % LDPC OAI
      \addplot[blue, solid, mark=square] plot coordinates  {(-2.750000,0.337900) (-2.500000,0.058300) (-2.250000,0.004000) (-2.000000,0.000200) (-1.750000,0.000000) (-1.500000,0.000000) };
      % Matlab layered min-sum with scaling factor 1
      %\addplot[green, solid, mark=triangle] plot coordinates {(-2.750000,0.843200) (-2.500000,0.524600) (-2.250000,0.198100) (-2.000000,0.037300) (-1.750000,0.003200) (-1.500000,0.000000) };
      % Matlab layered min-sum with scaling factor 0.8
      %\addplot[green, solid, mark=triangle] plot coordinates {(-3.750000,0.872300) (-3.500000,0.544600) (-3.250000,0.186400) (-3.000000,0.027500) (-2.750000,0.001900) (-2.500000,0.000000) };

      % Parity check 50 iterations
      %\addplot[blue, solid, mark=square] plot coordinates {(-2.750000,0.214600) (-2.500000,0.029200) (-2.250000,0.001500) (-2.000000,0.000100) (-1.750000,0.000000) (-1.500000,0.000000) };

      \draw (axis cs:-3.3,0.1)  node[fill=white,draw=black] (pint0) {20 iter};
      \draw (axis cs:-2.3,0.01) node[draw,black,thick,ellipse,minimum height=0.3cm] (ell0) {}; \draw[black,thick] (pint0) -- (ell0);

      \draw (axis cs:-1.2,0.0001)   node[fill=white,draw=black] (pint1) {10 iter};
      \draw (axis cs:-1.6,0.002) node[draw,black,thick,ellipse,minimum width=0.8cm] (ell1) {}; \draw[black,thick] (pint1) -- (ell1);

      \draw (axis cs:1.3,0.2)  node[fill=white,draw=black] (pint2) {5 iter};
      \draw (axis cs:-0.4,0.01) node[draw,black,thick,ellipse,minimum width=2cm] (ell2) {}; \draw[black,thick] (pint2) -- (ell2);

      \legend{ {Huawei 2017-06-15}\\
               {LDPC Ref}\\
               {LDPC OAI}\\
               {MATLAB NMS SF=1}\\};

    \end{semilogyaxis}
  \end{tikzpicture}
  \caption{BLER vs. SNR, BG2, Rate=1/5, \{5,10,20\} Iterations, B=1280.}
  \label{fig:bler-bg2-15}
\end{figure}

From Figure \ref{fig:bler-bg2-15} it can be observed that the reference decoder outperforms the current implementation significantly for low to medium number of iterations. The reason is the implementation of 2 layers in the reference decoder, which results in faster convergence for punctured codes and hence requires less iterations to achieve a given BLER target. Note that there is a large performance loss of about 4 dB at BLER $10^{-2}$ between the Huawei reference and the current optimized decoder implementation with 5 iterations.

Moreover, there is a gap of about 1.5 dB between the results provided by Huawei and the current decoder with 20 iterations. The reason is the min-sum approximation algorithm used in both the reference decoder and the current implementation. The gap can be closed by using a tighter approximation like the min-sum with normalization or the lambda-min approach. Moreover, the gap closes for higher code rates which can be observed from Figure \ref{fig:bler-bg2-r23}. The gap is only about 0.6 dB for 50 iterations.

The Matlab results denoted \texttt{MATLAB NMS} are obtained with the function \texttt{nrLDPCDecode} provided by the MATLAB 5G Toolbox R2019b. The following options are provided to the function: \texttt{'Termination','max','Algorithm','Normalized min-sum','ScalingFactor',1}. Furthermore, the 8-bit input LLRs are adapted to fit the dynamic range of \texttt{nrLDPCDecode} which is shown in Listing \ref{ldpc_matlab}.

\begin{lstlisting}[frame=single,caption={Input adaptation for MATLAB LDPC Decoder},label=ldpc_matlab]
maxLLR = max(abs(softbits));
rxLLRs = round((softbits/maxLLR)*127);
// adjust range to fit tanh use in decoder code
softbits = rxLLRs/3.4;
\end{lstlisting}

A scaling factor (SF) of 1 has been chosen to compare the results more easily with the \textit{LDPC OAI} since the resulting check node processing is the same. However, the Matlab normelized min-sum algorithm uses layered processing and floating point operations. Thus, for the same number of iterations, the performance is significantly better than \textit{LDPC OAI}, especially for small a number of iterations.

\begin{figure}[ht]
  \centering
  \begin{tikzpicture}
  \tikzstyle{every pin}=[fill=white,draw=black]
    \pgfplotsset{every axis legend/.append style={
        cells={anchor=west}, at={(1.05,1)}, anchor=north west}}
 %   \pgfplotsset{every axis plot/.append style={smooth}}
    \pgfplotsset{every axis/.append style={line width=0.5pt}}
    \pgfplotsset{every axis/.append style={mark options=solid, mark size=2.5pt}}

    \begin{semilogyaxis}[title={}, xlabel={$\SNR$ [dB]}, ylabel={BLER},
      grid={both}, xmin=3, xmax=6.5, xtick={3,3.5,...,6.5}, ymin=0,
      ymax=1,ytickten={-5,-4,-3,-2,-1,0},legend columns=1]

      % Kien's 2-layer 16bit code
      %\addplot[red, solid] plot coordinates { (-2.750000,0.915500) (-2.500000,0.576000) (-2.250000,0.165000) (-2.000000,0.017100) (-1.750000,0.000600) (-1.500000,0.000000) (-1.250000,0.000000) (-1.000000,0.000000)};

      % Huawei
      \addplot[black, solid] plot coordinates { (3.28392,0.01) (3.73319,0.0001) };

      % LDPC opt with 16bit BN processing
      %\addplot[blue, solid, mark=square] plot coordinates {(4.000000,0.487500) (4.250000,0.163400) (4.500000,0.029800) (4.750000,0.002700) (5.000000,0.000100)};
      \addplot[blue, solid, mark=square] plot coordinates {(5.000000,0.439600) (5.250000,0.185800) (5.500000,0.062100) (5.750000,0.015000) (6.000000,0.003900)};

      %\addplot[blue, dashed, mark=triangle] plot coordinates {(4.000000,0.487500) (4.250000,0.163700) (4.500000,0.030000) (4.750000,0.002900) (5.000000,0.000100)};

      %\addplot[blue, dashed, mark=square] plot coordinates {(3.000000,0.911600) (3.250000,0.614100) (3.500000,0.230100) (3.750000,0.036900) (4.000000,0.001100) (4.250000,0.000000) (4.500000,0.000000)};
      \addplot[blue, dashed, mark=square] plot coordinates {(3.000000,0.900400) (3.250000,0.600000) (3.500000,0.216400) (3.750000,0.036000) (4.000000,0.002600) (4.250000,0.000000) };

      \legend{ {Huawei 2017-06-15}\\
               {LDPC OAI 5 iter}\\
               {LDPC OAI 50 iter}\\};

    \end{semilogyaxis}
  \end{tikzpicture}
  \caption{BLER vs. SNR, BG2, Rate=2/3, \{5,50\} Iterations, B=1280.}
  \label{fig:bler-bg2-r23}
\end{figure}

In Figure \ref{fig:bler-bg2-15-2} we compare the performance of different algorithms using at most 50 iterations with early stopping if the parity check passes. The Matlab layered believe propagation (LBP) is used with unquantized input LLRs and performs the best since no approximation is done in the processing. Both NMS and offset min-sum (OMS) use a scaling factor and offset, respectively, that has been empirically found to perform best in this simulation setting. Theirs performance is very close to the BLP and OMS is slightly better than NMS. The performance of \textit{LDPC OAI} is more than 1 dB worse mainly because of the looser approximation. Moreover, the NMS algorithm with SF=1 performs worst probably because the SF is not optimized for the input LLRs. From the results in Figure \ref{fig:bler-bg2-15-2} we can conclude that the performance of the \textit{LDPC OAI} can be significantly improved by adopting an offset min-sum approximation improving the performance to within 0.3dB of the Huawei reference curve.

\begin{figure}[ht]
  \centering
  \begin{tikzpicture}
  \tikzstyle{every pin}=[fill=white,draw=black]
    \pgfplotsset{every axis legend/.append style={
        cells={anchor=west}, at={(1.05,1)}, anchor=north west}}
 %   \pgfplotsset{every axis plot/.append style={smooth}}
    \pgfplotsset{every axis/.append style={line width=0.5pt}}
    \pgfplotsset{every axis/.append style={mark options=solid, mark size=2.5pt}}

    \begin{semilogyaxis}[title={}, xlabel={$\SNR$ [dB]}, ylabel={BLER},
      grid={both}, xmin=-4, xmax=-1, xtick={-4,-3.5,...,-1}, ymin=0,
      ymax=1,ytickten={-5,-4,-3,-2,-1,0},legend columns=1]

      % HUAWEI merged BG2 2017-06-15
      \addplot[black, solid] plot coordinates { (-3.91839,0.01) (-3.5567,0.0001) };

      % Parity check 50 iterations
      \addplot[blue, solid, mark=square] plot coordinates {(-2.750000,0.214600) (-2.500000,0.029200) (-2.250000,0.001500) (-2.000000,0.000100) (-1.750000,0.000000) (-1.500000,0.000000) };

      % Matlab layered believe propagation
      \addplot[red, solid, mark=diamond] plot coordinates {(-4.500000,0.854200) (-4.250000,0.495800) (-4.000000,0.147700) (-3.750000,0.016100) (-3.500000,0.000800) (-3.250000,0.000200) (-3.000000,0.000000) };

      % Matlab layered min-sum with scaling factor 1
      \addplot[green, dashed, mark=triangle] plot coordinates {(-2.750000,0.830100) (-2.500000,0.497700) (-2.250000,0.165800) (-2.000000,0.024000) (-1.750000,0.001900) (-1.500000,0.000000) };
      % Matlab layered min-sum with scaling factor 0.8
      %\addplot[green, solid, mark=triangle] plot coordinates {(-3.750000,0.734800) (-3.500000,0.353800) (-3.250000,0.084300) (-3.000000,0.008000) (-2.750000,0.000400) };
      \addplot[green, solid, mark=triangle] plot coordinates {(-4.500000,0.964400) (-4.250000,0.748200) (-4.000000,0.333600) (-3.750000,0.057700) (-3.500000,0.004400) (-3.250000,0.000400) };

      % Matlab layered offset min-sum with offset 0.025
      \addplot[brown, solid, mark=asterisk] plot coordinates {(-4.250000,0.688800) (-4.000000,0.253800) (-3.750000,0.035600) (-3.500000,0.002000) (-3.250000,0.000000) (-3.000000,0.000000) };

      \legend{ {Huawei 2017-06-15}\\
               {LDPC OAI}\\
               {MATLAB LBP}\\
               {MATLAB NMS SF=1}\\
               {MATLAB NMS SF=0.65}\\
               {MATLAB OMS OS=0.025}\\};

    \end{semilogyaxis}
  \end{tikzpicture}
  \caption{BLER vs. SNR, BG2, Rate=1/5, max iterations = 50, B=1280.}
  \label{fig:bler-bg2-15-2}
\end{figure}

Figure \ref{fig:bler-bg1-r89} shows the performance of BG1 with largest block size of $B=8448$ and highest code rate $R=8/9$. From Figure \ref{fig:bler-bg1-r89} it can be observed that the performance gap is only about 0.3 dB if 50 iterations are used. However, for 5 iterations there is still a significant performance loss of about 2.3 dB at BLER $10^{-2}$.

\begin{figure}[ht]
  \centering
  \begin{tikzpicture}
  \tikzstyle{every pin}=[fill=white,draw=black]
    \pgfplotsset{every axis legend/.append style={
        cells={anchor=west}, at={(1.05,1)}, anchor=north west}}
 %   \pgfplotsset{every axis plot/.append style={smooth}}
    \pgfplotsset{every axis/.append style={line width=0.5pt}}
    \pgfplotsset{every axis/.append style={mark options=solid, mark size=2.5pt}}

    \begin{semilogyaxis}[title={}, xlabel={$\SNR$ [dB]}, ylabel={BLER},
      grid={both}, xmin=6, xmax=9, xtick={6,6.5,...,9}, ymin=0,
      ymax=1,ytickten={-5,-4,-3,-2,-1,0},legend columns=1]

      % Huawei
      \addplot[black, solid] plot coordinates { (6.118717,0.01) (6.291449,0.0001) };

      % LDPC opt 5 iter
      %\addplot[blue, solid, mark=square] plot coordinates {(8.500000,0.350000) (8.750000,0.155100) (9.000000,0.062400) (9.250000,0.023000) (9.500000,0.008700) (9.750000,0.003500) (10.000000,0.000900) (10.250000,0.000300) };
      \addplot[blue, solid, mark=square] plot coordinates {(7.500000,0.858900) (7.750000,0.449500) (8.000000,0.129700) (8.250000,0.025500) (8.500000,0.002300) (8.750000,0.000300) (9.000000,0.000000) };

      % LDPC opt 50 iter
      %\addplot[blue, dashed, mark=square] plot coordinates {(6.000000,0.705333) (6.100000,0.353367) (6.200000,0.102100) (6.300000,0.015133) (6.400000,0.000967) (6.500000,0.000000)};
      \addplot[blue, dashed, mark=square] plot coordinates {(6.000000,0.970000) (6.100000,0.830800) (6.200000,0.527300) (6.300000,0.216900) (6.400000,0.045500) (6.500000,0.005600) (6.600000,0.000300) (6.700000,0.000000) (6.800000,0.000000) };

      \legend{ {Huawei}\\
               {LDPC OAI 5 iter}\\
               {LDPC OAI 50 iter}\\};

    \end{semilogyaxis}
  \end{tikzpicture}
  \caption{BLER vs. SNR, BG1, Rate=8/9 \{5,50\} Iterations, B=8448.}
  \label{fig:bler-bg1-r89}
\end{figure}

\newpage
\subsection{Decoding Latency}
\label{sec:decoding-time}

This section provides results in terms of decoding latency. That is, the time it takes the decoder to to finish decoding for a given number of iterations. To measure the run time of the decoder we use the OAI tool \texttt{time\_meas.h}. The clock frequency is about 2.9 GHZ, decoder is run on a single core and the results are averaged over $10\,000$ blocks.

The results in Table \ref{tab:lat-bg2-r15} show the impact of the number of iterations on the decoding latency. It can be observed that the latency roughly doubles if the number of iterations are doubled.

\begin{table}[ht]
  \centering
  \begin{tabular}{lrrr}
    \toprule
    \textbf{Function} & \textbf{Time [$\mu s$] (5 it)} & \textbf{Time [$\mu s$] (10 it)} & \textbf{Time [$\mu s$] (20 it)}\\
    \midrule
    % \texttt{llr2llrProcBuf} & 1.1   & 1.1   & 1.1   \\
    % \texttt{llr2CnProcBuf}  & 12.4  & 12.0  & 12.0  \\
    % \texttt{cnProc}         & 11.7  & 22.1  & 43.5  \\
    % \texttt{bnProcPc}       & 6.6   & 12.1  & 23.8  \\
    % \texttt{bnProc}         & 4.2   & 8.1   & 16.2  \\
    % \texttt{cn2bnProcBuf}   & 61.3  & 118.3 & 234.9 \\
    % \texttt{bn2cnProcBuf}   & 38.1  & 82.5  & 172.3 \\
    % \texttt{llrRes2llrOut}  & 3.5   & 3.4   & 3.4   \\
    % \texttt{llr2bit}        & 0.2   & 0.1   & 0.1   \\
    \texttt{llr2llrProcBuf} & 0.5  & 0.5  & 0.5  \\
    \texttt{llr2CnProcBuf}  & 5.0  & 4.8  & 4.9  \\
    \texttt{cnProc}         & 12.4 & 23.0 & 42.7 \\
    \texttt{bnProcPc}       & 8.4  & 14.8 & 27.0 \\
    \texttt{bnProc}         & 5.5  & 10.1 & 19.0 \\
    \texttt{cn2bnProcBuf}   & 14.9 & 24.4 & 44.0 \\
    \texttt{bn2cnProcBuf}   & 10.5 & 17.8 & 31.8 \\
    \texttt{llrRes2llrOut}  & 0.3  & 0.3  & 0.3  \\
    \texttt{llr2bit}        & 0.2  & 0.2  & 0.2  \\
    \midrule
    % \textbf{Total}          & \textbf{139.4} & \textbf{260.3} & \textbf{508.4} \\
    \textbf{Total}          & \textbf{58.5} & \textbf{97.1} & \textbf{172.6} \\
    \bottomrule
  \end{tabular}
  \caption{BG2, Z=128, R=1/5, B=1280, LDPC OAI}
  \label{tab:lat-bg2-r15}
\end{table}

Table \ref{tab:lat-bg2-i5} shows the impact of the code rate on the latency for a given block size and 5 iterations. It can be observed that the performance gain from code rate 1/3 to 2/3 is about a factor 2.

\begin{table}[ht]
  \centering
  \begin{tabular}{lrrr}
    \toprule
    \textbf{Function} & \textbf{Time [$\mu s$] (R=1/5)} & \textbf{Time [$\mu s$] (R=1/3)} & \textbf{Time [$\mu s$] (R=2/3)}\\
    \midrule
    % \texttt{llr2llrProcBuf} & 3.2   & 2.9   & 2.6   \\
    % \texttt{llr2CnProcBuf}  & 36.5  & 25.4  & 14.8  \\
    % \texttt{cnProc}         & 33.6  & 25.2  & 13.3  \\
    % \texttt{bnProcPc}       & 17.6  & 10.2  & 4.5   \\
    % \texttt{bnProc}         & 8.5   & 5.4   & 2.5   \\
    % \texttt{cn2bnProcBuf}   & 175.3 & 110.6 & 50.7  \\
    % \texttt{bn2cnProcBuf}   & 106.6 & 71.2  & 36.1  \\
    % \texttt{llrRes2llrOut}  & 10.2  & 6.3   & 3.3   \\
    % \texttt{llr2bit}        & 0.4   & 0.2   & 0.1   \\
    \texttt{llr2llrProcBuf} & 1.5  & 0.9  & 0.5  \\
    \texttt{llr2CnProcBuf}  & 6.0  & 4.1  & 2.2  \\
    \texttt{cnProc}         & 32.2 & 23.7 & 14.4 \\
    \texttt{bnProcPc}       & 21.2 & 12.1 & 5.5  \\
    \texttt{bnProc}         & 9.8  & 5.9  & 2.9  \\
    \texttt{cn2bnProcBuf}   & 23.3 & 13.9 & 6.8  \\
    \texttt{bn2cnProcBuf}   & 14.8 & 9.7  & 5.0  \\
    \texttt{llrRes2llrOut}  & 0.6  & 0.4  & 0.3  \\
    \texttt{llr2bit}        & 0.7  & 0.4  & 0.2  \\
    \midrule
    % \textbf{Total}          & \textbf{392.4} & \textbf{258.0} & \textbf{128.2} \\
    \textbf{Total}          & \textbf{111.0} & \textbf{71.8} & \textbf{38.5} \\
    \bottomrule
  \end{tabular}
  \caption{BG2, Z=384, B=3840, LDPC OAI, 5 iterations}
  \label{tab:lat-bg2-i5}
\end{table}

Table \ref{tab:lat-bg1-i5} shows the results for BG1, larges block size and different code rates. The latency difference betwee code rate 1/3 and code rate 2/3 is less than half because upper left corner of the PCM is more dense than the rest of the PCM.

\begin{table}[ht]
  \centering
  \begin{tabular}{lrrr}
    \toprule
    \textbf{Function} &  \textbf{Time [$\mu s$] (R=1/3)} & \textbf{Time [$\mu s$] (R=2/3)} & \textbf{Time [$\mu s$] (R=8/9)}\\
    \midrule
    % \texttt{llr2llrProcBuf}  & 5.5   & 4.9   & 4.6  \\
    % \texttt{llr2CnProcBuf}   & 60.6  & 34.1  & 24.4 \\
    % \texttt{cnProc}          & 102.0 & 74.1  & 56.0 \\
    % \texttt{bnProcPc}        & 26.0  & 11.0  & 6.4  \\
    % \texttt{bnProc}          & 15.7  & 7.4   & 4.5  \\
    % \texttt{cn2bnProcBuf}    & 291.0 & 140.8 & 83.1 \\
    % \texttt{bn2cnProcBuf}    & 193.6 & 100.5 & 63.0 \\
    % \texttt{llrRes2llrOut}   & 13.3  & 6.9   & 5.2  \\
    % \texttt{llr2bit}         & 0.4   & 0.2   & 0.2  \\
    \texttt{llr2llrProcBuf}  & 2.1  & 1.2  & 0.9  \\
    \texttt{llr2CnProcBuf}   & 10.6 & 5.4  & 2.9  \\
    \texttt{cnProc}          & 89.8 & 66.3 & 50.0 \\
    \texttt{bnProcPc}        & 28.1 & 12.4 & 7.1 \\
    \texttt{bnProc}          & 17.1 & 8.1  & 4.8 \\
    \texttt{cn2bnProcBuf}    & 38.7 & 17.1 & 9.3 \\
    \texttt{bn2cnProcBuf}    & 25.6 & 12.7 & 7.2 \\
    \texttt{llrRes2llrOut}   & 0.8  & 0.4  & 0.3 \\
    \texttt{llr2bit}         & 0.9  & 0.4  & 0.3 \\
    \midrule
    % \textbf{Total}           & \textbf{708.9} & \textbf{380.6} & \textbf{248.1}\\
    \textbf{Total}           & \textbf{214.6} & \textbf{124.6} & \textbf{83.6}\\
    \bottomrule
  \end{tabular}
  \caption{BG1, Z=384, B=8448, LDPC OAI, 5 iterations}
  \label{tab:lat-bg1-i5}
\end{table}

From the above results it can be observed that the data transfer between CNs and BNs takes up a significant amount of the run time. However, the performance gain due to AVX instructions in both CN and BN processing is significantly larger than the penalty incurred by the data transfers.

\section{Parity Check and Early Stopping Criteria}
It is often unnecessary to carry out the maximum number of iterations. After each iteration a parity check (PC) \eqref{eq:29} can be computed and if a valid code word is found the decoder can stop. This functionality has been implemented and the additional overhead is reasonable. The PC is carried out in the CN processing buffer and the calculation complexity itself is negligible. However, for the processing it is necessary to move the BN results to the CN buffer which takes time, the overall overhead is at most $10\%$ compared to an algorithm without early stopping criteria with the same number of iterations. The PC has to be activated via the define \texttt{NR\_LDPC\_ENABLE\_PARITY\_CHECK}.

\section{Conclusion}
\label{sec:conclusion}

The results in the previous sections show that the current optimized LDPC implementation full-fills the requirements in terms of decoding latency for low to medium number of iterations at the expense of a loss in BLER performance. To improve BLER performance, it is recommended to implement a layered algorithm and a min-sum algorithm with normalization. Further improvements upon the current implementation are detailed in the next section.

\newpage
\section{Future Work}
\label{sec:future-work}

The improvements upon the current LDPC decoder implementation can be divided into two categories:
\begin{enumerate}
\item Improved BLER performance
\item Reduced decoding latency
\end{enumerate}

\subsection{Improved BLER Performance}
\label{sec:impr-bler-perf}

The BLER performance can be improved by using a tighter approximation than the min-sum approximation. For instance, the min-sum algorithm can be improved by adding a correction factor in the CN processing . The min-sum approximation in \eqref{eq:40} is modified as
\begin{equation}
  \label{eq:50}
  r_{ji} = \prod_{j'\in\Bcal_i\setminus j}\sgn q_{ij'}\min_{j'\in\Bcal_i\setminus j} |q_{ij'}| + w(q_{ij'})
\end{equation}
The correction term $w(q_{ij'})$ is defined as
\begin{equation}
  \label{eq:51}
  w(q_{ij'}) =
  \begin{cases}
     c & \textrm{if}~  \\
    -c & \textrm{if}~ \\
     0 & \textrm{otherwise}
  \end{cases}
\end{equation}
where the constant $c$ is of order $0.5$ typically.

\subsection{Reduced Decoding Latency}
\label{sec:reduc-decod-latency}

The following improvements will reduce the decoding latency:

\begin{itemize}
\item Adapt to AVX512
\item Optimization of CN processing
\item Implement 2/3-layers for faster convergence
\end{itemize}

\paragraph{AVX512:}
The computations in the CN and BN processing can be further accelerated by using AVX512 instructions. This improvement will speed-up the CN and BN processing by a approximately a factor of 2.

\paragraph{Optimization of CN Processing:}
It can be investigated if CN processing can be improved by computing two minima regardless of the number of BNs. Susequently, the (absolute) value fed back to the BN is one of those minima.

\paragraph{Layered processing:}
The LDPC code in NR always punctures the first 2 columns of the base graph. Hence, the decoder inserts LLRs with value 0 at their place and needs to retrieve those bits during the decoding process. Instead of computing all the parity equations and then passing the results to the BN processing, it is beneficial to first compute parity equations where at most one punctured BN is connected to that CN. If two punctured BNs are connected than according to \eqref{eq:40}, the result will be again 0. Thus in a first sub-iteration those parity equation are computed and the results are send to BN processing which calculates the results using only those rows in the PCM. In the second sub-iteration the remaining check equation are used.
The convergence of this layered approach is much fast since the bit can be retrieved more quickly while the decoding complexity remains the same. Therefore, for a fixed number of iterations the layered algorithm will have a significantly better performance.

\newpage
\bibliographystyle{IEEEtran}
\bibliography{./references}

\end{document}

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\documentclass{beamer}

\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{amsmath}
\usepackage{bm}

\usepackage{tabularx}
\usepackage{graphicx}
\usepackage{epstopdf}
\usepackage{multirow}

\graphicspath{{../../images/}}

\usetheme{Madrid}
\usebeamercolor{sidebartab}
\usefonttheme{professionalfonts}

\title[M.Sc. Thesis 2015]{Spatial Summarization of Image Collections}
\author{Diego A. Ballesteros Villamizar}
\institute[ETHZ]{ETH Zürich}
\date{February 8th, 2016}

\DeclareMathOperator*{\argmin}{argmin}
\DeclareMathOperator*{\argmax}{argmax}

\AtBeginSection[]
{
  \begin{frame}<beamer>
    \frametitle{Outline}
    \tableofcontents[currentsection]
  \end{frame}
}

\begin{document}

\begin{frame}
  \titlepage
\end{frame}

\section{Augmented features}

\begin{frame}{Leftover question}
  \begin{itemize}
    \item Does using a feature matrix $\mathbf{X}' = \mathbf{X} \mid \mathbb{I}$ improve the results?
    \begin{table}
      \begin{tabular}{l|lllll}
        \hline
        & \multicolumn{5}{c}{K} \\
        \hline
        \multirow{5}{*}{L} & & 0 & 2 & 5 & 10 \\
        & 0 & $17.38 \pm 1.81$ & $18.75 \pm 2.95$ & $18.82 \pm 2.58$ & $18.91 \pm 2.40$ \\
        & 2 & $22.66 \pm 4.58$ & $28.53 \pm 4.36$ &&\\
        & 5 & $25.40 \pm 4.77$ && $31.59 \pm 2.38$ &\\
        & 10 & $31.13 \pm 2.92$ &&& $30.49 \pm 3.51$ \\
      \end{tabular}
    \end{table}
    \item Not really, the best score so far is $34.35 \pm 2.15$ with $\mathbf{X} = \mathbb{I}$.
    \item Running time is significantly slower, because of the increased number of features $M = N + 4$.
  \end{itemize}
\end{frame}

\section{Sampling the distribution}

\begin{frame}{Sampling from the model}
  \begin{itemize}
    \item Using the best model, i.e. without features and with $L = 5, K=5$.
    \item How does the resulting distribution look?
    \item How to use the distribution to recommend sets?
  \end{itemize}
\end{frame}

\begin{frame}{Exact sampling}
  \begin{itemize}
    \item With $N = 10$, it is possible to calculate the probabilities from the model for all $2^{10} = 1024$ possible sets.
    \item Evaluating the model on all sets $S \subseteq V$and then normalizing the probability distribution.
    \item Takes only seconds to evaluate.
  \end{itemize}
\end{frame}

\begin{frame}{Distribution of set size ($100k$ samples)}
  \begin{figure}
    \centering
    \includegraphics[height=0.8\textheight]{length_histogram_exact}
  \end{figure}
\end{frame}

\begin{frame}{Distribution of sets with $|S| = 2$ ($100k$ samples)}
  \begin{figure}
    \centering
    \includegraphics[height=0.8\textheight]{pairs_histogram_exact}
  \end{figure}
\end{frame}

\begin{frame}{Gibbs sampling}
  \begin{itemize}
    \item What about a method that scales? For example if $N = 30$, then there are $2^{30} = 1073741824$ sets.
    \item Gibbs sampling as presented in \cite{gotovos15sampling}.
    \item Run for $1M$ iterations, remove the first half of iterations are burn-in.
    \item Running time is a couple of minutes.
  \end{itemize}
\end{frame}

\begin{frame}{Distribution of set size ($100k$ samples)}
  \begin{figure}
    \centering
    \includegraphics[height=0.8\textheight]{length_histogram_gibbs}
  \end{figure}
\end{frame}

\begin{frame}{Distribution of sets with $|S| = 2$ ($100k$ samples)}
  \begin{figure}
    \centering
    \includegraphics[height=0.8\textheight]{pairs_histogram_gibbs}
  \end{figure}
\end{frame}

\begin{frame}{Gibbs Sampling Performance}
  \begin{figure}
    \centering
    \includegraphics[height=0.8\textheight]{gibbs_performance}
  \end{figure}
\end{frame}

\section{Extending the location set}

\begin{frame}{More mean-shift clusters}
  \begin{itemize}
    \item Original dataset has over 160k photos.
    \item When clustered using mean-shift with a bandwidth of approximately 100m, there are over 2k clusters.
    \item Previous tests were done using the 10 top clusters according number of photos per cluster. This covered only over 30k photos.
    \item Does the approach scale if the number of clusters is increased?
    \item 50 clusters cover over 100k photos.
    \item 12k paths are present, in comparison there were 8k paths with 10 clusters.
  \end{itemize}
\end{frame}

\begin{frame}{Baselines}
  \begin{table}
    \centering
    \begin{tabular}{@{}lll@{}}
      \hline
      \textbf{Model} & \textbf{Accuracy} & \textbf{MRR} \\
      \hline
      Modular & $9.21 \pm 1.02$ & $27.00 \pm 1.01$ \\
      Markov & $19.72 \pm 1.23$ & $34.50 \pm 1.00$ \\
      \textbf{Markov with rejection} & $\mathbf{22.36 \pm 1.41}$ & $\mathbf{38.65 \pm 1.09}$ \\
      Proximity & $12.76 \pm 0.70$ & $27.71 \pm 0.88$ \\
      Proximity with rejection & $14.74 \pm 0.64$ & $31.34 \pm 1.14$ \\
      \hline
    \end{tabular}
  \end{table}
  \begin{itemize}
    \item Similar trend as with $N = 10$, Markov with rejection is the best model and it's significantly better than the modular model.
  \end{itemize}
\end{frame}

\begin{frame}{FLDC - Facility Location Diversity and Coherence}
  \begin{itemize}
    \item Use the best model from the case for $N = 10$.
    \item The model with a diversity term, i.e. submodular, and a coherence term, i.e. supermodular. Without features, i.e. $\mathbf{X} = \mathbb{I}$.
    \item Running on $k = 10$ folds, with a noise factor of 50.
    \item Latent dimensions are: $0 \leq L \leq 20, 0 \leq M \leq 20$.
  \end{itemize}
\end{frame}

\begin{frame}{Evaluation}
  \begin{table}
    \begin{tabular}{l|lllll}
      \hline
      & \multicolumn{5}{c}{K} \\
      \hline
      \multirow{5}{*}{L} & & 0 & 2 & 10 & 20 \\
      & 0 & $9.21 \pm 1.02$ & $16.24 \pm 1.43$ & $16.12 \pm 1.29$ & $16.44 \pm 1.77$\\
      & 2 & $12.09 \pm 1.36$ & $17.05 \pm 0.91$ &&\\
      & 10 & $12.94 \pm 1.39$ && $19.28 \pm 0.73$ &\\
      & 20 & $9.17 \pm 2.27$ &&& $\mathbf{21.53 \pm 1.24}$\\
    \end{tabular}
  \end{table}
  \begin{itemize}
    \item Model with diversity and coherence term has performance close to the markov model with rejection.
  \end{itemize}
\end{frame}

\begin{frame}{Distribution of set size ($100k$ samples)}
  \begin{figure}
    \centering
    \includegraphics[height=0.8\textheight]{length_histogram_gibbs_50}
  \end{figure}
\end{frame}

\begin{frame}{Frequency of data}
  \begin{columns}
    \begin{column}{0.5\columnwidth}
      \begin{table}
        \centering
        \caption{Frequency of Item Sets}
        \begin{tabular}{ll}
          \hline
          Set & Frequency \\
          \hline
          $[0, 7]$ & 61\\
          $[27, 28]$ & 39\\
          $[13, 7]$ & 36\\
          $[11, 14]$ & 27\\
          $[0, 13]$ & 25\\
          $[2, 7]$ & 23\\
          $[10, 2]$ & 22\\
          $[25, 34]$ & 22\\
          $[2, 15]$ & 21\\
          $[10, 15]$ & 21\\
          \hline
        \end{tabular}
      \end{table}
    \end{column}

    \begin{column}{0.5\columnwidth}
      \begin{table}
        \centering
        \caption{Locations}
        \begin{tabular}{ll}
          \hline
          Index & Location \\
          \hline
          0 & Grossmünster \\
          2 & Bürkliterrasse \\
          7 & Fraumünster \\
          10 & Quaibrücke \\
          11 & Hauptbahnhof \\
          13 & Rathaus \\
          14 & Urania-Sternwarte \\
          15 & Bellevue \\
          25 & Frau Gerolds Garten \\
          27 & Zoo \\
          28 & Restaurant Masoala (Zoo) \\
          34 & Stadion Letzigrund \\
          \hline
        \end{tabular}
      \end{table}
    \end{column}
  \end{columns}
\end{frame}

\begin{frame}{References}
  \bibliographystyle{acm}
  \bibliography{../references}
\end{frame}

\end{document}


\documentclass[11pt,twocolumn]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{bibentry}
\usepackage{cite}
\usepackage{float}
\usepackage[margin=1in]{geometry}
\usepackage{tikz}
\usepackage[section]{placeins}

\floatstyle{boxed}
\restylefloat{figure}

\usetikzlibrary{arrows,backgrounds,decorations.pathmorphing,positioning}

\newcommand{\comment}[1]{}

\title{Canary}
\author{Andrew Carter}

\begin{document}
\maketitle

\section{Problem Statement}
Determine whether a graph $G$ has a minor isomorphic to $H$.
In particular we consider $G$ to have a minor $H$ if we can find a function $f : H \to 2^G$
 such that for all $h \in H$, $f(h)$ is non-empty and connected,
  and if there is an edge between $h_i$ and $h_j$
 then there exists a $g_k \in f(h_k)$  and $g_j \in f(h_j)$ that are connected.

\section{General Algorithm}
Let $G_k$ be the set of vertices in $G$ that are assigned to $h_k$ (i.e. $G_k = f(h_k)$).

Vertices $g \in G$ may be in one of three states, a vertex may have no constraints, and is therefore considered unassigned.
A vertex maybe be constrained to a specific $G_k$, and are therefore considered assigned to $h_k$.
Or vertices may be constrainted to a binary union of $G_k \cup G_j$,
 and will be referenced as ``may'' be in $G_k$ or ``may'' be in $G_j$.

Start by considering all vertices in $G$ as unassigned.
Go through each vertex $h_k \in H$ and assign it to a vertex in $G$.

Then for each previously assigned vertex $h_j \in H$ that has an edge between $h_j$ and $h_k$ find a path in $G$ that:
\begin{enumerate}
\item Starts at a vertex in $G_j$
\item Through vertices that may be in $G_j$
\item Through vertices that are unassigned
\item Through vertices that may be in $G_k$
\item Ends at a vertex in $G_k$
\end{enumerate}
All vertices on the path that may be in $G_j$ are then put into $G_j$,
  likewise all vertices on the path that may in $G_i$ are then put into $G_i$.
All unassigned vertices on the path are in $G_j \cup G_k$.

Repeat until either the vertices and edges in $H$ are exhausted, or until no path is possible.
If no path is possible backtrack, trying a diffrent path or vertex assignment until all paths and assignments have been tried.

The middle steps may contain $0$ vertices.
Additionally the starting vertex and ending vertex are irrelevant when consider unique paths as they don't affect the change in
 constraints.
\section{Optimizations}
We can apply diffreent optimizations to prevent duplicate or strictly suboptimal paths from being consider.
This prunes the search space, hopefully speeding up termination.

\subsection{Minimal Initial Assignment}
If $h_k$ was assigned to vertex $g \in G$, and then reassigned, then no future assignment need assign $g$ to $h_k$.
\subsection{Minimal Path Length}
If a path has connected vertices $g_j, g_k$ then the path should also contain the edge $g_j, g_k$.

\begin{figure}[H]
\begin{tikzpicture}
  [
   node distance=5mm and 12mm,
   rn/.style={circle,thick,draw=red,inner sep=0pt,minimum size=6mm},
   gn/.style={circle,thick,draw=green,inner sep=0pt,minimum size=6mm},
   bn/.style={circle,thick,draw=blue,inner sep=0pt,minimum size=6mm},
   yn/.style={circle,thick,draw=yellow,inner sep=0pt,minimum size=6mm},
   cn/.style={circle,thick,draw=cyan,inner sep=0pt,minimum size=6mm},
   mn/.style={circle,thick,draw=magenta,inner sep=0pt,minimum size=6mm}
  ]
   \node[bn] (n1)               {1};
   \node[mn] (n3) [above right=of n1] {3};
   \node[mn] (n4) [right=of n1] {4};
   \node[rn] (n2) [right=of n4] {2};

   \draw[-,color=magenta] (n1) -- (n3);
   \draw[-,color=magenta] (n3) -- (n4);
   \draw[-,color=magenta] (n4) -- (n2);
   \draw[-] (n1) -- (n4);

\end{tikzpicture}
\caption{\label{fig:min path 1.1}Non-Minimal Path}
\end{figure}

In Figure \ref{fig:min path 1.1} we see that the path from 1 to 2 through both 3 and 4 is needlessly long.
We could take the edge directly between 1 and 3, resulting in strictly less constraints.

\subsection{Minimal Path Length on Path Starts and Ends}
This can be extended to starting vertices, if vertex $g_j \in G_j$,
  and connected to $g_k$ which is on a path from $G_j$ to $G_k$.
Then the path should contain an edge from a $g \in G_j$ to $g_k$.

Similarly for ending vertices.

\begin{figure}[H]
\begin{tikzpicture}
  [
   node distance=5mm and 12mm,
   rn/.style={circle,thick,draw=red,inner sep=0pt,minimum size=6mm},
   gn/.style={circle,thick,draw=green,inner sep=0pt,minimum size=6mm},
   bn/.style={circle,thick,draw=blue,inner sep=0pt,minimum size=6mm},
   yn/.style={circle,thick,draw=yellow,inner sep=0pt,minimum size=6mm},
   cn/.style={circle,thick,draw=cyan,inner sep=0pt,minimum size=6mm},
   mn/.style={circle,thick,draw=magenta,inner sep=0pt,minimum size=6mm}
  ]
   \node[bn] (n1)               {1};
   \node[mn] (n3) [below right=of n1] {3};
   \node[bn] (n4) [above right=of n1] {4};
   \node[mn] (n5) [right=of n3] {5};
   \node[mn] (n6) [right=of n3] {6};
   \node[rn] (n2) [below right=of n6] {2};

   \draw[-,color=magenta] (n1) -- (n3);
   \draw[-,color=blue] (n1) -- (n4);
   \draw[-,color=magenta] (n3) -- (n5);
   \draw[-,color=magenta] (n5) -- (n6);
   \draw[-,color=magenta] (n6) -- (n2);
   \draw[-] (n4) -- (n6);

\end{tikzpicture}
\caption{\label{fig:min path 2.1}Non-Minimal Path}
\end{figure}

In Figure \ref{fig:min path 2.1}, if vertex 4 is already assigned to vertex 1,
 then the path from 1 to 2 through vertices 3, 5, and 6 is needlessly long.
Instead the path could go through 4 (which is already assigned), then directly to 6 and onto 2.
Again this would result in strictly less constraints.

\subsection{Retroactive Minimal Path Length}
Furthermore if $g_j$ is not assigned to $h_j$, but known to be contained in $G_j \cup G_k$,
  then it follows that if $g_k$ is not connected to a vertex in $G_j$ then $g_j \in G_k$.
Otherwise the previous optimization would be violated.

\begin{figure}[H]
\begin{tikzpicture}
  [
   node distance=5mm and 12mm,
   rn/.style={circle,thick,draw=red,inner sep=0pt,minimum size=6mm},
   gn/.style={circle,thick,draw=green,inner sep=0pt,minimum size=6mm},
   bn/.style={circle,thick,draw=blue,inner sep=0pt,minimum size=6mm},
   yn/.style={circle,thick,draw=yellow,inner sep=0pt,minimum size=6mm},
   cn/.style={circle,thick,draw=cyan,inner sep=0pt,minimum size=6mm},
   mn/.style={circle,thick,draw=magenta,inner sep=0pt,minimum size=6mm}
  ]
   \node[bn] (n1)               {1};
   \node[mn] (n6) [above right=of n1] {6};
   \node[cn] (n7) [below right=of n1] {7};
   \node[mn] (n4) [right=of n6] {4};
   \node[cn] (n5) [right=of n7] {5};
   \node[rn] (n2) [right=of n4] {2};
   \node[gn] (n3) [right=of n5] {3};

   \draw[-,color=magenta] (n1) -- (n6);
   \draw[-,color=magenta] (n6) -- (n4);
   \draw[-,color=magenta] (n4) -- (n2);
   \draw[-,color=cyan] (n1) -- (n7);
   \draw[-,color=cyan] (n7) -- (n5);
   \draw[-,color=cyan] (n5) -- (n3);
   \draw[-] (n4) -- (n5);

\end{tikzpicture}
\caption{\label{fig:min path 3.1}Retroactive Minimal Path Length}
\end{figure}
In Figure \ref{fig:min path 3.1} nodes 1, 2, and 3 are all inital assignments for 3 different vertices in the minor.
 There is a path created between 1 and 2 as shown in magenta through vertices 6 and 4.
 Both vertices may be assigned to either 4 or 6.
 Similarly there is a path between 1 and 3 as shown in cyan through vertices 7 and 5.
\begin{figure}[H]
\begin{tikzpicture}
  [
   node distance=5mm and 12mm,
   n/.style={circle,thick,draw=black,inner sep=0pt,minimum size=6mm},
   rn/.style={circle,thick,draw=red,inner sep=0pt,minimum size=6mm},
   gn/.style={circle,thick,draw=green,inner sep=0pt,minimum size=6mm},
   bn/.style={circle,thick,draw=blue,inner sep=0pt,minimum size=6mm},
   yn/.style={circle,thick,draw=yellow,inner sep=0pt,minimum size=6mm},
   cn/.style={circle,thick,draw=cyan,inner sep=0pt,minimum size=6mm},
   mn/.style={circle,thick,draw=magenta,inner sep=0pt,minimum size=6mm}
  ]
   \node[bn] (n1)               {1};
   \node[bn] (n6) [above right=of n1] {6};
   \node[n] (n7) [below right=of n1] {7};
   \node[bn] (n4) [right=of n6] {4};
   \node[cn] (n5) [right=of n7] {5};
   \node[rn] (n2) [right=of n4] {2};
   \node[gn] (n3) [right=of n5] {3};

   \draw[-,color=blue] (n1) -- (n6);
   \draw[-,color=blue] (n6) -- (n4);
   \draw[-,color=magenta] (n4) -- (n2);
   \draw[-] (n1) -- (n7);
   \draw[-] (n7) -- (n5);
   \draw[-,color=cyan] (n5) -- (n3);
   \draw[-,color=cyan] (n4) -- (n5);

\end{tikzpicture}
\caption{\label{fig:min path 3.2}Assigning 4 to vertex 1}
\end{figure}

In Figure \ref{fig:min path 3.2} we see that if we were to assign 4 to vertex 1,
  that there would be a more minimal path from 1 to 3 which does not need to use vertex 7.

\begin{figure}[H]
\begin{tikzpicture}
  [
   node distance=5mm and 12mm,
   rn/.style={circle,thick,draw=red,inner sep=0pt,minimum size=6mm},
   gn/.style={circle,thick,draw=green,inner sep=0pt,minimum size=6mm},
   bn/.style={circle,thick,draw=blue,inner sep=0pt,minimum size=6mm},
   yn/.style={circle,thick,draw=yellow,inner sep=0pt,minimum size=6mm},
   cn/.style={circle,thick,draw=cyan,inner sep=0pt,minimum size=6mm},
   mn/.style={circle,thick,draw=magenta,inner sep=0pt,minimum size=6mm}
  ]
   \node[bn] (n1)               {1};
   \node[mn] (n6) [above right=of n1] {6};
   \node[cn] (n7) [below right=of n1] {7};
   \node[rn] (n4) [right=of n6] {4};
   \node[gn] (n5) [right=of n7] {5};
   \node[rn] (n2) [right=of n4] {2};
   \node[gn] (n3) [right=of n5] {3};

   \draw[-,color=magenta] (n1) -- (n6);
   \draw[-,color=magenta] (n6) -- (n4);
   \draw[-,color=red] (n4) -- (n2);
   \draw[-,color=cyan] (n1) -- (n7);
   \draw[-,color=cyan] (n7) -- (n5);
   \draw[-,color=green] (n5) -- (n3);
   \draw[-] (n4) -- (n5);

\end{tikzpicture}
\caption{\label{fig:min path 3.3}Implied Assignments}
\end{figure}
Therefore in the original position we must assign 4 to vertex 2, and likewise 5 is assigned vertex 3 as shown in Figure \ref{fig:min path 3.3}.

% No bibliography, change if/when citations are added
\bibliographystyle{alpha}
\nocite*{}
\nobibliography{bib/main}
\end{document}


%!TEX root = ../../template.tex
\section{Tomographic algorithms and reconstruction techniques}%
\label{sec:tomographic_algorithms_and_reconstruction_techniques}

Tomography is the cross-sectional imaging of an object through the use
of transmitted or reflected waves, captured by the object exposure to
the waves from a set of known angles. It has many different applications
in science, industry, and most prominently, medicine. Since the
invention of the Computed Tomography (\gls{CT}) machine in 1972, by
Hounsfield~\cite{Gunderman2006}, tomographic imaging techniques have had
a revolutionary impact, allowing doctors to see inside their patients,
without having to subject them to more invasive
procedures~\cite{Kak2001}.

The central thought around tomographic image reconstruction is to
recreate the information contained inside a target physical body,
without having to cut it or open it in any way. The theory is based on
Radon's idea that it is possible to "\emph{represent a function written
in $\mathbb{R}$ in the space of straight lines ($\mathbb{L}$) through
its line integrals}"~\cite{Radon1986}.

Say we have a body that we want to fully characterise without cutting
open or destroying in any way. Now imagine we can traverse it with some
kind of radiation, ray by ray, and that we are able to measure the rays
after they traverse the target. What we would capture would be relative
to the emitted radiation, of course, but it would also contain
information on how that ray had interacted with the target body's
matter. In the case of the ubiquitously used X-Ray radiation, the
measurement would be one of the total attenuation "imprinted" onto the
ray by the target body's molecules, in the ray's particular direction.
If said body is heterogeneous, the total attenuation can be derived by
the infinitesimal sum of all different attenuation phenomena caused by
the object's several different constituents (the same can be said of a
homogeneous object, but in that case there is only one type of
attenuation present). This means that each one of the rays contains
information regarding the constitution of said body.

The question that arises is thus "\emph{how we can use this information
to create a spatially accurate representation of this target's interior
composition?}". The answer to this question lies on many factors, but
the most prominent of which are surely choosing the quantity that we are
trying to find (that characterises the object) and assembling the
projections (that is what we call the line integrals in tomographic
imaging) in a way that allows solving an equation system for the
aforementioned quantity. This assembly, a matrix of projections
organised by their angles and position within the detector, is called
sinogram. All tomography methods revolve around finding the relationship
between it and the system's geometrical description~\cite{Bruyant2002,
Kak2001, Herman1973, Herman1995, Herman2009, Defrise2003}.

Let's consider the case in which we deal with a single ray of solar
light entering the atmosphere at a given point. Since the atmosphere
contains numerous absorbers and comparable atmospheric effects, the ray
changes from the point where it enters the atmosphere to the point at
which it is measured by a detector. Total absorption will depend on the
pollutant species, their cross-section and their concentration, since it
obeys Lambert-Beer's law. Looking from another angle, this absorption
is also the line integral that we will use to reconstruct our image.
With \gls{DOAS}, it is possible to measure several pollutants at the
same time, but for simplicity (and since it is one of the most studied
compounds in the field), let's consider that the single pollutant in our
atmospheric mixture is NO$_2$.

The problem of tomographic reconstruction can be approached in a number
of ways, depending mostly on the authors. In my literary search, I have
found that Kak and Slaney~\cite{Kak2001} have certainly explained this
problem in one of the clearer ways available. Therefore, I shall base
the rest of my presentation in their writings, and complement with other
authors' notes wherever necessary.

Considering the coordinate system displayed in
Figure~\ref{fig:coordinates}. In this schematic, the object is
represented by the function $f(x, y)$. The  $(\theta, t)$ parameters can
be used to define any line in this schematic. Line AB in particular can
be written:

\begin{equation}
    \label{eq:lineAB}
    x \cdot \cos(\theta) + y \cdot \sin(\theta) = t
\end{equation}

\begin{figure}[htpb]
    \centering
    \includegraphics[width=0.7\textwidth]{img/png/coordinates.png}
    \caption{Schematic representation for coordinate setting. The image
    depicts a parallel projection setting~\cite{Kak2001a}.}
    \label{fig:coordinates}
\end{figure}

And if we were to write a line integral along this line, it would look
like Equation~\ref{eq:lineABIntegral}, the Radon transform of function
$f(x, y)$:

\begin{equation}
    \label{eq:lineABIntegral}
    P_{\theta}(t) = \int_{-\infty}^{\infty} f(x, y) \cdot \delta(x \cdot
    \cos(\theta) + y \cdot \sin(\theta) - t) dxdy
\end{equation}

Where $\delta$, the delta function, is defined in
Equation~\ref{eq:delta}.

\begin{equation}
    \label{eq:delta}
    \delta (\phi) =
    \begin{cases}
            1, & \phi = 0\\
            0, & otherwise
    \end{cases}
\end{equation}

As I have mentioned previously, a projection is a set of line integrals
such as $P_{\theta}(t)$. Geometry plays a very important role in how the
integrals are written and solved for reconstruction. The simplest case
is the one where the set is acquired in a row, describing what is called
a parallel geometry. Another more complex case is when a single point
source is used as origin for all rays, forming a fan. This is called a
fan-beam array. There are other possible geometries, but they fall out of
the scope of this work and will therefore not be addressed any further.

The Fourier Slice Theorem (\gls{FST}) is the most important component of
the most important algorithm in tomographic inversion, the Filtered
BackProjection algorithm (\gls{FBP}). \gls{FST} is based on the equality
relation between the
two-dimensional Fourier Transform (\gls{FT}) of the object function and
the one-dimensional \gls{FT} of the object's projection at an angle
$\theta$. Let's start by writing the 2D \gls{FT} for the object
function, Equation~\ref{eq:objectFT}, and the 1D \gls{FT} of projection
P$_\theta$, in Equation~\ref{eq:1dFTproj}.

\begin{equation}
    \label{eq:objectFT}
    F(u, v) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y)
    \cdot \exp \left [ -j2\pi (ux + vy) \right ] dx dy
\end{equation}

\begin{equation}
    \label{eq:1dFTproj}
    S_{\theta}(\omega) = \int_{-\infty}^{\infty} P_{\theta} \cdot \exp\left[
    -j2 \pi \omega t \right]
\end{equation}

For simplicity, let's consider the 2D \gls{FT} at the line defined by
$v=0$ in the frequency domain. We rewrite the 2D \gls{FT} integral as:

\begin{equation}
    \label{eq:v0}
    F(u, 0) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y)
    \cdot \exp \left[  -j 2\pi  \omega ux \right] dx dy
\end{equation}

Notice that $y$ is not present in the phase factor of the \gls{FT}
expression anymore, and this means we can rearrange the integral as:

\begin{equation}
    \label{eq:v02}
    F(u, 0) = \int_{-\infty}^{\infty} \left[ \mathbf{\int_{-\infty}^{\infty}
    f(x, y) dy }\right] \cdot \exp \left[  -j 2\pi  \omega ux \right] dx
\end{equation}

Now, the \textbf{bold} part of Equation~\ref{eq:v02} is similar to
Equation~\ref{eq:lineABIntegral}. It is precisely that equation,
considering $\theta=0$ and a constant value of $x$, as in
Equation~\ref{eq:p0}.

\begin{equation}
    \label{eq:p0}
    P_{\theta=0} (x) = \int_{-\infty}^{\infty} f(x, y) dy
\end{equation}

This in turn can be substituted in Equation~\ref{eq:v02}, finally
arriving at:

\begin{equation}
    \label{eq:FTP}
    F(u, 0) = \int_{-\infty}^{\infty} P_{\theta=0} (x) \cdot \exp \left[
    -j 2\pi ux \right] dx
\end{equation}

And this is the one-dimensional \gls{FT} for the projection at angle
$\theta=0$. Finally, the enunciation of the Fourier Slice Theorem:
\begin{center}
    \begin{minipage}{0.8\textwidth}

        \noindent\textbf{\emph{The Fourier Transform of a parallel
                projection  of an image $f(x, y)$ taken at angle
                $\theta$ gives a slice of the two-dimensional Fourier
                Transform, $F(u, v)$, subtending an angle $\theta$ with
                the $u$-axis (see Figure~\ref{fig:fst})}}

    \end{minipage}
\end{center}

\begin{figure}[htpb]
    \centering
    \includegraphics[width=.8\textwidth]{img/png/fst.png}
    \caption{The \gls{FST}, a schematic
    representation~\cite{Asl2013a}.}
    \label{fig:fst}
\end{figure}

If one takes the \gls{FST} into account, the idea behind the \gls{FBP}
seems to appear almost naturally. Say one has a single projection and
its Fourier transform. From the \gls{FST}, this projection is the same
as the object's two-dimensional \gls{FT} in a single line. A crude
reconstruction of the original object would result if someone were to
place this projection in its right place in the Fourier domain and then
perform a two-dimensional \gls{IFT}, while assuming every other
projection to be 0. The result, in the image space, would be as if
someone had smeared the object in the projections direction.

What is really needed for a correct reconstruction is to do this many
times, with many projections. This brings a problem with the method:
smearing the object in all directions will clearly produce a wrong
\emph{accumulation} in the center of the image, since every projection
passes through the middle (remember we are still talking about parallel
geometry projections) and are summed on top of each other, but on the
outer edges, this does not occur. If one does not address this, the
image intensity levels in the reconstructed image will be severely
overestimated in the center and underestimated in the edges (due to
normalization). The solution is conceptually easy: we multiply the
Fourier transform by a weighting filter proportional to its frequency
($\omega$) and that encompasses its relevance in the global scheme of
projections. If there are $K$ projections, then it is adequate for this
value to be $\frac{2\pi\lvert\omega\rvert}{K}$. As an algorithm,
\gls{FBP} can be written as in Algorithm~\ref{alg:fbp}.

\begin{algorithm}
    \caption{The Filtered BackProjection Algorithm}
    \label{alg:fbp}
    \SetAlgoLined
    \KwResult{A reconstructed image of the projected object.}
    \For{$\theta \gets 0$ \KwTo $180$ \KwBy $\frac{180}{K}$}{
        measure projection $P_{theta}(t)$\;
        FT($P_{\theta}(t)$), rendering $S_{\theta}(\omega)$\;
        Multiply by $\frac{2\pi\lvert{\omega}\rvert}{K}$\;
        Sum the \gls{IFT} of the result in the image space\;
    }

\end{algorithm}

Parallel projections, in which the object is scanned linearly from
multiple directions, have the advantage of having a relatively simple
reconstruction scheme. However, they usually result in acquisition times
which are in the order of minutes. A faster way of collecting the data
is one where all radiation emanates from a single point-source, which
rotates around the target object (as well as the detectors). There are
two types of fan-beam projections: equiangular and equally spaced. In
this project, I have only worked with equiangular processes, so I will
not include an explanation for equally spaced fan-beam projections. The
reader may find this well described (much better than I would be able
to) in ~\cite{Kak2001} and ~\cite{Herman1973}.

Consider Figure~\ref{fig:equiangular}. If our projection data were
acquired through a parallel ray geometry, we would be able to say that
ray SA belonged to a projection $P_{\theta}(t)$, in which $\theta$ and
$t$ would be written:

\begin{equation}
    \label{eq:theta_and_t}
    \theta = \beta + \gamma \quad \text{ and } \quad t = D \cdot \sin \gamma
\end{equation}

In Equation~\ref{eq:theta_and_t}, $D$ is the distance between the source
$S$ and the origin $O$; $\gamma$ is the angle of a ray within a fan and
$\beta$ is the angle that the source $S$ makes with a reference axis.
Through these relationships one can \emph{translate} the parallel
projection's FBP algorithm to the fan-beam case, which involves several
complex geometric transformations, although the overall rationale is
exactly the same.

\begin{figure}[htpb]
    \centering
    \includegraphics[width=.8\textwidth]{img/png/fig319.png}
    \caption{Schematic representation of an equiangular fan-beam
    projection, taken from~\cite{Kak2001}.}
    \label{fig:equiangular}
\end{figure}

Another particularity of fan-beam projection data is the fact that they
can be sorted into a parallel projection. For that, one starts with the
premise that if one were to substitute the fan geometry for parallel
beams, most of the fan-beam rays would also appear in some projection of
the parallel setup. This re-sorting algorithm starts with
Equation~\ref{eq:theta_and_t}. Now, if we call a fan-beam projection
taken at angle $\beta$ $R_{\beta}(\gamma)$, and a parallel projection
taken at angle $\theta$ $P_{\theta}(t)$, one could thus write
Equation~\ref{eq:parallel_vs_fanbeam}, which can already be used to
re-sort any fan-beam projection into parallel beam geometry.

\begin{equation}
    \label{eq:parallel_vs_fanbeam}
    R_{\beta}(\gamma) = P_{\beta + \gamma}(D \cdot \sin \gamma)
\end{equation}

Let's call the angular interval between fan-beam projections can be
written $\delta\beta$, and the angular interval of rays within each fan
is written $\delta\gamma$. In the case that they are the same
($\beta=\gamma=\alpha$), then it it is the case that they can both be
replaced by multiples of that interval in
Equation~\ref{eq:parallel_vs_fanbeam}, which becomes
Equation~\ref{eq:parallel_vs_fanbeam2}.

\begin{equation}
    \label{eq:parallel_vs_fanbeam2}
    R_{m \cdot \alpha}(n \cdot\alpha) = P_{m \cdot\alpha + n
    \cdot\alpha}(D \cdot \sin n \cdot\alpha)
\end{equation}

Or, in non-mathematical notation, the n\textsuperscript{th} ray of the
m\textsuperscript{th} radial projection (R) is the same as the
n\textsuperscript{th} ray in the (m+n)\textsuperscript{th} parallel
projection. Although being much simpler than directly applying the
\gls{FBP} algorithm to the fan-beam projection data, this method has a
limitation, which is the non-uniformity of the generated parallel
projections. This can usually be corrected through
interpolation~\cite{Kak2001a}.


\include{preamble}

\begin{document}
	\title{Standard \LaTeX Preamble and Usage Examples}
	\section{Algorithm Example}
		\begin{algorithmic}	[1]
			\LState $counter\gets Sketch.total$
			\For{$h = 1$ \textbf{to} $H$}
			\LState $i\gets Sketch[h].hash(k_1)$
			\If{$Sketch[h][i].total \leq counter$}
			\LState{$index\gets h$}
			\LState{$hash\gets i$}
			\LState{$counter\gets Sketch[h][i].total$}
			\EndIf
			\EndFor
			\LState $result\gets Sketch[index][hash][k_2]$
		\end{algorithmic}
	\section{Theory Example}
		\begin{lemma}
			\label{thr:collision}
			Let $h_1...h_n : \{1...k\} \to \{1...v\}$, $H$ be a set of sets $H_i$ where $h_i(x) \in H_i$, $S = \{h_i^{-1}(x) | x \in \bigcup H\}$.\\
			\begin{displaymath}
			\forall H_i \in H, |H_i| = \sum_{j}^{j \in S} \begin{cases}0 \textnormal{ if } h_i(j) \notin H_i\\
			1 \textnormal{ otherwise}\end{cases}.
			\end{displaymath}
			\\
		\end{lemma}

		\begin{corollary}
			\label{cor:collision}
			As a consequence of Lemma~\ref{thr:collision},
			\begin{displaymath}
			\forall H_i \in H, \sum_{j}^{j \in H_i}j = \sum_{j}^{j \in S}{\begin{cases}0 \textnormal{ if } h(j) \notin H_i\\
				h(j) \textnormal{ otherwise}\end{cases}}.
			\end{displaymath}
			Therefore, if we were to remove an arbitrary element $x$ from $S$, then the sum would become:
			\begin{displaymath}
			\forall H_i \in H, \sum_{j}^{j \in H_i}j = \sum_{j}^{j \in S-\{x\}}{\begin{cases}0 \textnormal{ if } h(j) \notin H_i\\
				h(j) \textnormal{ otherwise}\end{cases}} + h_i(x).
			\end{displaymath}
			Finally, this allows us to conclude that in the case of the sum of all elements in $S-\{x\}$ which have images in $H_i$ is equal to $\sum_{j}^{j \in H_i}{j - h_i(x)}$.
		\end{corollary}
	\section{Tikz Example}

		\begin{tikzpicture}

		\node[server](server 1){};
		\node[server, right of= server 1](server 2){};
		\node[server, right of= server 2](server 3){};
		\node[server, right of= server 3](server 4){};
		\node[server, right of= server 4](server 5){};
		\node[server, right of= server 5](server 6){};
		\node[server, right of= server 6](server 7){};
		\node[server, right of= server 7](server 8){};

		\node[l3 switch, above of =server 5, xshift=0.1cm,yshift=0.3cm]
		(l3 switch){};

		\draw[thick,darkgray!10!gray] (server 1.north)--(l3 switch);
		\draw[thick,darkgray!10!gray] (server 2.north)--(l3 switch);
		\draw[thick,darkgray!10!gray] (server 3.north)--(l3 switch);
		\draw[thick,darkgray!10!gray] (server 4.north)--(l3 switch);
		\draw[thick,darkgray!10!gray] (server 5.north)--(l3 switch);
		\draw[thick,darkgray!10!gray] (server 6.north)--(l3 switch);
		\draw[thick,darkgray!10!gray] (server 7.north)--(l3 switch);
		\draw[thick,darkgray!10!gray] (server 8.north)--(l3 switch);

		\node[xshift=-1.05cm,yshift=0.2cm,left of = server 8,align=left](lev1){Computing Servers};
		\node[xshift=-0.2cm,yshift=0.3cm,above of = lev1,align=left](lev2){L4 Open vSwitch};

		%\begin{scope}
		%	\node[my cloud, above of=l3 switch,font=\large] (it) {Internet}
		%	\draw[thick,darkgray!10!gray] (l3 switch)--(it);
		%\end{scope}

		\begin{scope}
		\node[xshift=0.1cm, yshift=1cm,scale=0.2, above of=l3 switch] (brouter) {\router{}} edge[very thick,darkgray!10!gray] ([xshift=0.1cm,yshift=0.5cm]l3 switch);

		\node[yshift=0.65cm,my cloud, minimum width=1.25cm, minimum height=1.55cm, above of=brouter, font=\large] (it) {Internet}	edge[very thick,darkgray!30!gray] (brouter);

		%	\draw[very thick,darkgray!30!gray](brouter)--([xshift=0.1cm,yshift=0.125cm]border 1-a.north);
		\end{scope}

		\end{tikzpicture}

\end{document}