How to reset the counter back to one - TeX

topnush replying to user 3.14159

the new version is perfect. Thank you

user 3.14159 replying to topnush

I think that now I finally understood what you want to achieve. Please have a look.

topnush

similarly S should be {1}, {1,2} etc.

topnush replying to user 3.14159

it (at least the animation) still isn't right. s/t should be 1,1,1, 0.75, 0.6, 0.5,0.429,0.375,0.333 but in your animation it is 0.6, 0.5, 0.429, 0.375, 0.333 then blank

user 3.14159 replying to topnush

Better now? (I am wondering what the purpose of `\uncover<1->{...}` is, naively it seems not to have an effect.)

topnush replying to user 3.14159

it still seems to be 1-9 where it should be 1-

user 3.14159

@topnush Done.

samcarter replying to topnush

I wonder who said this :) Yes, it is true that `\uncover` is more powerful than `\pause` - but that does not mean that you should use both, one of them is enough :)

topnush

but having both is clearly silly here

topnush replying to samcarter

just deleted the (first) uncover :)  I feel someone said that \pause wasn't as good as \uncover that's the only reason

samcarter replying to topnush

Why are you using the `\uncover` anyway? The `\pause` before takes care of it not appearing earlier.

topnush replying to samcarter

thank you

topnush replying to user 3.14159

this actually works perfectly. Thank you. You just need to change it to uncover<1->.  Could you include it in your answer please?

samcarter replying to topnush

The reason not all your values are shown is that with `\uncover<1-9>{}` you stop it after half the list. You need at least overlays until 13 and not 9 (but the `\uncover` seems to be superfluous anyway, so I would just remove it)

samcarter

For the first part, you can avoid all the problems with counters and offsets if you let beamer create the overlays with   
```  
\alt<+>{\underaccent{\bar}{\x}}{\x}  
```  
  
(for the second part, I think @marmots `count= ... from` is the easiest solution)  
```  
\documentclass{beamer}

\usepackage{accents}

\usepackage{tikz}

\begin{document}


\begin{frame}[fragile]{}

\begin{itemize}[<+(1)->]
\item one
\item two
\item three
\end{itemize}
\pause
First line:

\[
\foreach \x  in {1,2,3,4,5,6,7,8,9}{
\alt<+>{\underaccent{\bar}{\x}}{\x}
}
\]


\vskip 0.5cm
\edef\myoffset{\number\value{beamerpauses}}
\hspace*{1cm}
 $S$ = \foreach [count=\y from \myoffset] \x in {{\{1\}},{\{1,2\}},{\{1,2,3\}},{\{1,4,3\}},{\{5,4,3\}},{\{5,6,3\}},{\{5,6,3\}},{\{5,6,3\}},{\{5,6,9\}}}{\only<\y>{\x}}\\
\hspace*{1.25cm}$\frac{s}{t}$ = \foreach [count=\y from \myoffset] \x in  {1,1,1,0.75,0.6,0.5,0.429,0.375,0.333}{\only<\y>{\x}}



\end{frame}
\end{document}  
```

topnush

instead S and s/t should go through their lists (as your new example does for the first 5) for all 9

topnush

Your animation hasn't been updated but you can see that S and s/t disappear once 6 is underlined which shouldn't happen.

topnush replying to user 3.14159

that is almost it! It is all perfect until I get S = {5,4,3}, s/t = 0.6 and then S and s/t are blank after that. I guess the counter was reset but it still thinks there are only 5 of them instead of 9?

user 3.14159 replying to topnush

I'd say yes but I do not know what the expected result is.  
```  
\documentclass{beamer}
\usepackage{accents}
\usepackage{pgffor}
\begin{document}
\begin{frame}[fragile]{}

\begin{itemize}[<+(1)->]
\item one
\item two
\item three
\end{itemize}
\pause
\uncover<4->{
First line:

\[\edef\myoffset{\number\value{beamerpauses}}
\foreach \x [count=\y from \myoffset] in {1,2,3,4,5,6,7,8,9}{
\alt<\y>{\underaccent{\bar}{\x}}{\x}
}
\]


\vskip 0.5cm

\hspace*{1cm}
\uncover<1-9>
{\edef\myoffsetB{\number\value{beamerpauses}}
 $S$ = \foreach [count=\y from \myoffsetB] \x in {{\{1\}},{\{1,2\}},{\{1,2,3\}},{\{1,4,3\}},{\{5,4,3\}},{\{5,6,3\}},{\{5,6,3\}},{\{5,6,3\}},{\{5,6,9\}}}{\only<\y>{\x}}\\
\hspace*{1.25cm}$\frac{s}{t}$ = \foreach [count=\y from \myoffsetB] \x in 
 {1,1,1,0.75,0.6,0.5,0.429,0.375,0.333}{\only<\y>{\x}}
}
}
\end{frame}
\end{document}  
```

topnush

now it is starting at index 5

topnush

that part should also start from index 1

topnush replying to user 3.14159

```  
\uncover<1-9>
{
 $S$ = \foreach [count=\y] \x in {{\{1\}},{\{1,2\}},{\{1,2,3\}},{\{1,4,3\}},{\{5,4,3\}},{\{5,6,3\}},{\{5,6,3\}},{\{5,6,3\}},{\{5,6,9\}}}{\only<\y>{\x}}\\
\hspace*{1.25cm}$\frac{s}{t}$ = \foreach [count=\y] \x in  {1,1,1,0.75,0.6,0.5,0.429,0.375,0.333}{\only<\y>{\x}}
}  
```

user 3.14159 replying to topnush

Sorry, can you be a bit more specific? What are the two lines below?

topnush

@marmot, re: [your answer](#a1997), will the same trick work if you include the two lines below as well?

1 Answer