Support for IEEEeqnarray and all its variants

[krishnakumarg1984]

IEEEeqnarray is a lifesaver during those cases where standard AMSmath/mathtools environments do not fit in.

This document explains all the commands, environments use-cases and variants thereof.

Describe the solution you'd like
Full-range support for all the commands/environments as in standard equation, align etc.

Describe alternatives you've considered
Stay away from IEEEeqnarray, since not many know about it. But this is an official IEEE tool, and sometimes is a cleaner workaround for long 2 column equations, spacing, alignment etc.

[lervag]

Thanks, this seems like something we could add. But it would make my life much easier if you could provide a LaTeX sample with examples for each case that you think should be covered.

[krishnakumarg1984]

Yes. I shall do that.

[krishnakumarg1984]

@lervag Here is a document that illustrates all the IEEEeqnarray features

Click here to see example.

\documentclass{article}
\usepackage{IEEEtrantools}
\usepackage{mathtools}

\begin{document}

As a better alternative to multline:

\begin{IEEEeqnarray}{rCl}
    a & = & b + c + d + e + f
    + g + h + i + j \nonumber\\
      && +\> k + l + m + n + o + p
      \label{eq:dont_use_multline}
\end{IEEEeqnarray}

Same as above, but with multiple equations:

\begin{IEEEeqnarray}{rCl}
  a & = & b + c
  \\
  & = & d + e + f + g + h
  + i + j + k \nonumber\\
  && +\> l + m + n + o
  \\
  & = & p + q + r + s
\end{IEEEeqnarray}

\begin{IEEEeqnarray*}{rCl"s}
  \log \frac{1}{a}
  & = & -\log a
  & (binary, wrong) \\
  & = & -{\log a}
  & (unary, correct)
\end{IEEEeqnarray*}

\begin{IEEEeqnarray*}{rCl's}
  a & = & - b - b - c
  & (default unary) \\
  & = & {-} {b} - b - c
  & (default unary, no effect) \\
  & = & -\> b - b - c
  & (changed to binary) \\
  & = & - \log b - b - d
  & (default binary) \\
  & = & {-} {\log b} - b - d
  & (changed to unary) \\
  & = & - \log b - b {-} d
  & (changed $-d$ to unary)
\end{IEEEeqnarray*}

Further configuration
\begin{IEEEeqnarray}{rCl}
  \IEEEeqnarraymulticol{3}{l}{%
    a + b + c + d + e + f + g
    + h + i + j
  }\nonumber\\*%
  & = & k + l + m + n + o + p + q
  + r + s + t + u \nonumber\\*
\end{IEEEeqnarray}

\begin{IEEEeqnarray}{rCl}
  \IEEEeqnarraymulticol{3}{l}{
    h^{-}(X|Y)
  }\nonumber\\\quad
  & \le & \frac{n+1}{e}
  - h(X|Y) \nonumber\\
  && + \int p(y) \log \left(
    \frac{\mathsf{E}\bigl[|X|^2
      \big| Y=y\bigr]}{n}
  \right)
  \nonumber\\*
\end{IEEEeqnarray}

Spacing adjustment example:

\begin{IEEEeqnarray}{c}
  a = \sum_{k=1}^n\sum_{\ell=1}^n
  \sin \bigl(2\pi \, b_k \,
  c_{\ell} \, d_k \, e_{\ell} \,
  f_k \, g_{\ell} \, h  \bigr)
  \IEEEeqnarraynumspace
  \label{eq:labelc1}
\end{IEEEeqnarray}

Same result as above:
\begin{IEEEeqnarray}{c}
  a = \sum_{k=1}^n\sum_{\ell=1}^n
  \sin \bigl(2\pi \, b_k \,
  c_{\ell} \, d_k \, e_{\ell} \,
  f_k \, g_{\ell} \, h  \bigr)
  \nonumber\\*
  \label{eq:labelc2}
\end{IEEEeqnarray}

\begin{IEEEeqnarray}{rCl}
    a & = & b + c \\
      & = & d + e \IEEEyesnumber
      \IEEEyessubnumber
\end{IEEEeqnarray}

\begin{IEEEeqnarray*}{l}
  a + b + c + d + e + f
  \\ \qquad
  +\> g + h + i + j + k + l
  \qquad \\
  \IEEEeqnarraymulticol{1}{r}{
    +\> m + n + o + p + q  }
  \IEEEyesnumber
\end{IEEEeqnarray*}

\begin{IEEEeqnarray}{rCl}
  a & = & d + e + f + g + h
  + i + j + k + l + m + n
  \IEEEeqnarraynumspace
\end{IEEEeqnarray}

\begin{IEEEeqnarray}{rCl}
  \IEEEeqnarraymulticol{3}{l}{
    a + b + c + d + e + f
    + g + h
  }\nonumber\\* \quad
  & = & i + j
  \\
  & = & k + l + m
\end{IEEEeqnarray}

\begin{IEEEeqnarray}{rCl}
  \IEEEeqnarraymulticol{3}{l}{
    a + b + c + d + e + f
    + g + h
  }\nonumber\\* \qquad\qquad
  & = & i + j
  \label{eq:label45}
  \\
  & = & k + l + m
\end{IEEEeqnarray}


\begin{IEEEeqnarray}{rCl}
  a & = & b + c
  \\
  & = & d + e + f + g + h
  + i + j + k \nonumber\\
  && +\> l + m + n + o
  \label{eq:add_space}
  \\
  & = & p + q + r + s
\end{IEEEeqnarray}

\begin{IEEEeqnarray*}{rCl}
  a
  & = & b_{1}  \\
  & = & b_{2}  \IEEEyesnumber\\
  & = & b_{3}    \\
  & = & b_{4}  \IEEEyesnumber*\\
  & = & b_{5}    \\
  & = & b_{6}    \\
  & = & b_{7}  \IEEEnonumber\\
  & = & b_{8}    \\
  & = & b_{9}  \IEEEnonumber*\\
  & = & b_{10}   \\
  & = & b_{11} \IEEEyessubnumber*\\
  & = & b_{12}   \\
  & = & b_{13} \IEEEyesnumber\\
  & = & b_{14}   \\
  & = & b_{15}
\end{IEEEeqnarray*}

\begin{IEEEeqnarray}{rCl}
  \IEEEyesnumber\label{eq:block}
  \IEEEyessubnumber*
  a
  & = & b_{24} \\
  & = & b_{25}
  \label{eq:subeq_b}\\
  & = & b_{26}
\end{IEEEeqnarray}

\begin{IEEEeqnarray}{rCl}
  a & = & b + c
  \label{eq:block2_eq1}\\
  & = & d + e
  \label{eq:block2_eq2}
\end{IEEEeqnarray}

\begin{IEEEeqnarray}{c}
  \IEEEyessubnumber*
  f = g - h + i
  \label{eq:block2_eq3}
\end{IEEEeqnarray}

\begin{IEEEeqnarray}{rCl}
  a & = & b + c
  \label{eq:block3_eq1}\\
  & = & d + e
  \label{eq:block3_eq2}\\
  \noalign{\noindent and
    \vspace{2\jot}}
  f & = & g - h + i
  \label{eq:block3_eq3}
\end{IEEEeqnarray}

\begin{IEEEeqnarraybox}{t.t.t}
    \textbf{Item} &
    \textbf{Color} &
    \textbf{Count} \\
    cars & green & 17 \\
    trucks & red & 4 \\
    bikes & blue & 25
  \end{IEEEeqnarraybox}

\begin{equation}
  P_U(u) = \left\{ \,
    \begin{IEEEeqnarraybox}[][c]{l?s}
      \IEEEstrut
      0.1 & if $u=0$, \\
      0.3 & if $u=1$, \\
      0.6 & if $u=2$.
      \IEEEstrut
    \end{IEEEeqnarraybox}
  \right.
  \label{eq:example_left_right1}
\end{equation}

\begin{equation*}
  P_U(u) = \left\{ \,
    \begin{IEEEeqnarraybox}{l?s}
      0.1 & if $u=0$, \\
      0.3 & if $u=1$, \\
      0.6 & if $u=2$.
    \end{IEEEeqnarraybox}
  \right.
\end{equation*}

\begin{equation*}
  \begin{IEEEeqnarraybox}{c'c;v;c'c'c}
    D_1 & D_2 && X_1 & X_2
    & X_3
    \\\hline
    0 & 0 && +1 & +1 & +1\\
    0 & 1 && +1 & -1 & -1\\
    1 & 0 && -1 & +1 & -1\\
    1 & 1 && -1 & -1 & +1
  \end{IEEEeqnarraybox}
\end{equation*}

\begin{equation*}
  \begin{IEEEeqnarraybox}[
     \IEEEeqnarraystrutmode
    ]{c'c;v;c'c'c}
    D_1 & D_2 && X_1 & X_2 & X_3
    \\\hline
    0 & 0 && +1 & +1 & +1\\
    0 & 1 && +1 & -1 & -1\\
    1 & 0 && -1 & +1 & -1\\
    1 & 1 && -1 & -1 & +1
  \end{IEEEeqnarraybox}
\end{equation*}

\begin{equation*}
  \begin{IEEEeqnarraybox}[
     \IEEEeqnarraystrutmode
     \IEEEeqnarraystrutsizeadd{3pt}
     {1pt}
    ]{c'c/v/c'c'c}
    D_1 & D_2 & & X_1 & X_2 & X_3
    \\\hline
    0 & 0 && +1 & +1 & +1\\
    0 & 1 && +1 & -1 & -1\\
    1 & 0 && -1 & +1 & -1\\
    1 & 1 && -1 & -1 & +1
  \end{IEEEeqnarraybox}
\end{equation*}

\begin{equation}
  \left.
  \begin{IEEEeqnarraybox}[\IEEEeqnarraystrutmode
    \IEEEeqnarraystrutsizeadd{2pt}{2pt}][c]{rCl}
    x & = & a + b\\
    y & = & a - b
  \end{IEEEeqnarraybox}
  \, \right\}  \iff  \left\{ \,
  \begin{IEEEeqnarraybox}[
    \IEEEeqnarraystrutmode
    \IEEEeqnarraystrutsizeadd{7pt}
    {7pt}][c]{rCl}
    a & = & \frac{x}{2} + \frac{y}{2}
    \\
    b & = & \frac{x}{2} - \frac{y}{2}
  \end{IEEEeqnarraybox}
  \right.
  \label{eq:example_left_right2}
\end{equation}

\begin{equation}
  \left\{
    \begin{IEEEeqnarraybox}[
      \IEEEeqnarraystrutmode
      \IEEEeqnarraystrutsizeadd{2pt}
      {2pt}
      ][c]{rCl}
      \dot{x} & = & f(x,u)
      \\
      x+\dot{x} & = & h(x)
    \end{IEEEeqnarraybox}
  \right.
\end{equation}

\begin{IEEEeqnarray}{rrCl}
  & \dot{x} & = & f(x,u)
  \\*
  \smash{\left\{
      \IEEEstrut[8\jot]
    \right.}
  & x+\dot{x} & = & h(x)
  \\*
  & x+\ddot{x} & = & g(x)
\end{IEEEeqnarray}

\begin{IEEEeqnarray}{rCl+x*}
    a & = & b + c \\
    & = & d+ e \label{eq:qed}
    \\* &&& \nonumber\IEEEQEDhere
  \end{IEEEeqnarray}


\end{document}

[lervag]

Thanks! However, it is still unclear exactly which features are necessary here. Does it suffice to detect the various IEEEeqnarray environments as math environments? That is, these:

\begin{IEEEeqnarray}{argument}
\end{IEEEeqnarray}

\begin{IEEEeqnarray*}{argument}
\end{IEEEeqnarray*}

\begin{IEEEeqnarraybox}[opt1][opt2]{argument}
\end{IEEEeqnarraybox}

The last environment seems it is not really a math environment in itself?

[lervag]

I'm closing this due to lack of activity. Feel free to reopen and respond.

[krishnakumarg1984]

Sigh. Thanks to Jetfrog for requesting support on this topic.