Defining a new command, every time a command is called

Question

Short summary:

I would like to create a command that creates a new command each time it is called, based on a counter. Then I could use PGF's for-loops to go through the sequence of commands. The non-working example is:

\newcounter{group}
\newcommand{\defgroup}[1]{
   \stepcounter{group}
   \newcommand{\problem\Alph{group}}[0]{#1}
}
\defgroup{$1*1=$}
\defgroup{$1+1=$}

However, it turns out you cannot do \newcommand{\problem\Alph{group}}. Is there a way to do this?

Longer explanation:

I am writing a jigsaw activity, which requires a sequence of related versions of a document (A,B,C,D,E), each with a different problem. This is currently done as follows (MWE)

\documentclass[12pt]{exam}
\newcounter{group}
\newcommand{\groupname}{\stepcounter{group}\textbf{\Huge \Alph{group}\\}}
\newcommand{\newgroup}[1]{
    \newpage
    \groupname
    \setcounter{section}{0}
    \section{Here is your problem!}
    \begin{center}
        #1
    \end{center}
    \section{Now copy down other groups problems!}
}

\begin{document}
    \newgroup{$1+1=$}
    \newgroup{$1*1=$}
    \newgroup{$1^1=$}
\end{document}

What I would like to do is be able to to reference the other groups in the second section of each version. Thus, I want group A's document to look like:

A

1. Here is your problem: 1+1 =

2. Now copy down other groups problems:

   B: 1*1=

   C: 1^1 =

Answer 1

You can use \csdef from the etoolbox:

\documentclass{article}

\usepackage{etoolbox}

\newcounter{group}
\newcommand{\defgroup}[1]{
   \stepcounter{group}
   \csdef{problem\Alph{group}}{#1}
}

\begin{document}

\defgroup{$1*1=$}
\defgroup{$1+1=$}

\problemA

\problemB

\end{document}

Answer 2

\expandafter\newcommand\csname problem\Alph{group}\endcsname{#1}%

will define a macro named \problemX where X is the Alph - variant of the group counter value. The \expandafter will generate the macro name first, then then \newcommand grabs this macro name in order to define the relevant macro.

Please note that this approach will fail as soon as the group counter is reset to zero.

\documentclass[12pt]{exam}
\newcounter{group}



\newcommand{\defgroup}[1]{%
  \stepcounter{group}%
  \expandafter\newcommand\csname problem\Alph{group}\endcsname{#1}%
}


\begin{document}

    \defgroup{$1+1=$}
    \defgroup{$1*1=$}
    \defgroup{$1^1=$}

    \problemA

    \problemB

    \problemC

    %\setcounter{group}{0} This will fail x
   \defgroup{$1+1=$}
   \defgroup{$1*1=$}
   \defgroup{$1^1=$}

\end{document}

Answer 3

As I commented with regard to your longer explanation, a complication is that you want to have knowledge of group B and Group C problems before ever invoking the \newgroup that would define them. In fact, as of group A, you don't even know how many other groups will follow.

Thus, here, I force all problems to be defined at once in a comma separated list, and then the particulars can be invoked as needed. The answer will automatic adjust to the number of problems defined. (so go ahead, try \MakeProblemSheets{$1+1=$,$1*1=$,$1^1=$,$1/1=$} as an alternative).

\documentclass[12pt]{exam}
\usepackage{listofitems}
\newcounter{group}
\newcommand{\Groupname}[1]{\setcounter{group}{#1}\textbf{\Huge \Alph{group}\\}}
\newcommand{\Newgroup}[1]{
    \newpage
    \Groupname{#1}
    \setcounter{section}{0}
    \section{Here is your problem!}
    \begin{center}
        \problems[#1]
    \end{center}
    \section{Now copy down other groups problems!}
    \begin{center}
    \foreachitem\i\in\problems{%
      \ifnum\icnt=#1\relax\else
        \setcounter{group}{\icnt}
        \makebox[.3in][l]{\Alph{group}:}%
        \makebox[.7in][l]{\i}\\\fi
    }
    \end{center}
}
\newcommand\MakeProblemSheets[1]{%
  \readlist\problems{#1}
  \foreachitem\Prob\in\problems{\Newgroup{\Probcnt}}
}
\begin{document}
\MakeProblemSheets{$1+1=$,$1*1=$,$1^1=$}
\end{document}