How to reset the counter back to one - TeX

topnush replying to user 3.14159 — Wednesday, 5th May 2021 17:36

the new version is perfect. Thank you

user 3.14159 replying to topnush — Wednesday, 5th May 2021 17:36

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

topnush — Wednesday, 5th May 2021 17:13

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

topnush replying to user 3.14159 — Wednesday, 5th May 2021 17:13

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 — Wednesday, 5th May 2021 17:09

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 — Wednesday, 5th May 2021 17:07

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

user 3.14159 — Wednesday, 5th May 2021 16:58

@topnush Done.

samcarter replying to topnush — Wednesday, 5th May 2021 10:33

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 — Wednesday, 5th May 2021 10:14

but having both is clearly silly here

topnush replying to samcarter — Wednesday, 5th May 2021 10:14

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 — Wednesday, 5th May 2021 10:11

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

topnush replying to samcarter — Wednesday, 5th May 2021 09:56

thank you

topnush replying to user 3.14159 — Wednesday, 5th May 2021 09:56

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 — Tuesday, 4th May 2021 17:50

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 — Tuesday, 4th May 2021 17:45

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}

xxxxxxxxxx
 
\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}

xxxxxxxxxx
 
\documentclass{beamer}​\usepackage{accents}​\usepackage{tikz}​\begin{document}​​\begin{frame}[fragile]{}​\begin{itemize}[<+(1)->]\item one\item two\item three\end{itemize}\pauseFirst 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}  
show all 36 lines

topnush — Tuesday, 4th May 2021 16:51

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

topnush — Tuesday, 4th May 2021 16:49

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 — Tuesday, 4th May 2021 16:48

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 — Tuesday, 4th May 2021 16:45

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}

xxxxxxxxxx
 
\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}  
show all 34 lines

topnush — Tuesday, 4th May 2021 16:41

now it is starting at index 5

topnush — Tuesday, 4th May 2021 16:40

that part should also start from index 1

topnush replying to user 3.14159 — Tuesday, 4th May 2021 16:40

\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}}
}

xxxxxxxxxx
 
\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 — Tuesday, 4th May 2021 16:40

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

topnush — Tuesday, 4th May 2021 16:38

@marmot, re: your answer, will the same trick work if you include the two lines below as well?

1 Answer