Precision problems when not using the parser with luamath

[hmenke]

Probably the PGF math parser evaluates subexpressions during parsing which leads to a massive loss of precision.

\documentclass[a5paper]{article}
\pagestyle{empty}
\usepackage{tikz}
\usetikzlibrary{luamath}
\begin{document}
\pgfkeys{/pgf/trig format/rad}

\pgfkeys{/pgf/luamath=only computation}
\pgfmathparse{2*11000*(1 - 1.40576 - cos(0.02) + sqrt(1.40576^2 - sin(0.02)^2))}
only computation:\hfill\pgfmathresult

\pgfkeys{/pgf/luamath=parser and computation}
\pgfmathparse{2*11000*(1 - 1.40576 - cos(0.02) + sqrt(1.40576^2 - sin(0.02)^2))}
parser and computation:\hfill\pgfmathresult

Lua:\hfill\directlua{
    local print = tex.print
    local _ENV = math
    print(2*11000*(1 - 1.40576 - cos(0.02) + sqrt(1.40576^2 - sin(0.02)^2)))
}
\end{document}

[Mo-Gul]

@hmenke, in https://github.com/pgf-tikz/pgf/issues/861#issuecomment-633431073 you stated that LUA is broken. Could you elaborate on this? What do you think the "right" result is of the above expression?

(When I calculate it using Excel I get 1.270132770125 as well.)

[hmenke]

As you can see from the image in the first post, the only computation mode gives a completely wrong result.

[Mo-Gul]

Aah, now I see. I thought "only computation" was the result not using Lua (at all).

I had a closer look and it seems that the (large) difference in the output comes from 1.40576^2 and the rest is just a subsequent error. See the red marked entry in the following image. When I replace 1.40576^2 in the expression with the result of the computation from Lua, i.e. 1.9761611776, then the results are almost identical (the last line in the table marked in green).

When I am not mistaken the "large" difference comes from https://github.com/pgf-tikz/pgf/blob/e0b798ba19471877f80633a727c45e09439344b3/tex/generic/pgf/libraries/luamath/pgflibraryluamath.code.tex#L237-L242

All in all I wouldn't say that "Lua is broken". "Just" only computation is a bit inaccurate which seems to be explainable from the above comment in the code. 1.40576^2 gives the same result using LaTeX/pgf calculation (only) or using only computation.

Evaluating the whole expression not using Lua or xfp results in a Dimension too large error. And using fpu then gives a result that is even more off than the only computation result (which seems to be originated from the sqrt(a) computation result).

So when it is known that only computation is "inaccurate", why not (simply) use parser and computation instead?

Code to generate above result

```latex \documentclass{article} \usepackage[a4paper,margin=1in,landscape]{geometry} \usepackage{mathtools} \usepackage{xparse} \usepackage{xfp} \usepackage{tikz} \usetikzlibrary{ fpu, luamath, matrix, } % from \def\sqrtexplained#1{% \begingroup \sbox0{$#1$} \def\underbrace##1_##2{##1} \sbox2{$#1$} \dimen0=\wd0 \advance\dimen0-\wd2 \mathrlap{\sqrt{\phantom{\displaystyle#1}\kern\dimen0 }} \hphantom{\sqrt{\vphantom{\displaystyle#1}}} \endgroup #1% } \NewDocumentCommand{\mylua}{m}{ \directlua{ local print = tex.print; local _ENV = math; print(#1) } } \pgfkeys{ /pgf/trig format/rad, /pgf/number format/.cd, sci, precision=12, } \NewDocumentCommand{\myrow}{sm}{% % 2nd column \IfBooleanTF{#1}{ (error) }{ \pgfmathparse{#2}\pgfmathresult% % \pgfmathparse{#2}\pgfmathprintnumber{\pgfmathresult}% } \pgfmatrixnextcell % 3rd column \pgfkeys{ /pgf/fpu=true, % /pgf/fpu/output format=sci, } \pgfmathparse{#2}\pgfmathresult% % \pgfmathparse{#2}\pgfmathprintnumber{\pgfmathresult}% \pgfkeys{/pgf/fpu=false}% \pgfmatrixnextcell \pgfkeys{/pgf/luamath=only computation} \pgfmathparse{#2}\pgfmathresult% % \pgfmathparse{#2}\pgfmathprintnumber{\pgfmathresult}% \pgfmatrixnextcell \pgfkeys{/pgf/luamath=parser and computation} \pgfmathparse{#2}\pgfmathresult% % \pgfmathparse{#2}\pgfmathprintnumber{\pgfmathresult}% \pgfmatrixnextcell \mylua{#2} % \pgfmathprintnumber{\mylua{#2}} \pgfmatrixnextcell \fpeval{#2} % \pgfmathprintnumber{\fpeval{#2}} } \begin{document} \begin{equation*} 2 \cdot 11000 \cdot \bigg( \underbrace{1 - 1.40576 - \cos(0.02) + \sqrtexplained{\underbrace{1.40576^2 - \sin(0.02)^2}_{a}}}_{b} \bigg) \end{equation*} \begin{tikzpicture}[ row 2 column 4/.style=red, row 10/.style=green!50!black, ] \matrix (a) [ matrix of nodes, anchor=west, ] { expression & off & fpu & only computation & parser and computation & Lua & xfp \\ $1.40576^2$ & \myrow{1.40576^2} \\ $\sin(0.02)^2$ & \myrow{sin(0.02)^2} \\ $a$ & \myrow{1.40576^2 - sin(0.02)^2} \\ $\sqrt{a}$ & \myrow{sqrt(1.40576^2 - sin(0.02)^2)} \\ $\cos(0.02)$ & \myrow{cos(0.02)} \\ $1 - 1.40576 - \cos(0.02)$ & \myrow{1 - 1.40576 - cos(0.02)} \\ $b$ & \myrow*{1 - 1.40576 - cos(0.02) + sqrt(1.40576^2 - sin(0.02)^2)} \\ $2 \cdot 11000 \cdot b$ & \myrow*{2*11000*(1 - 1.40576 - cos(0.02) + sqrt(1.40576^2 - sin(0.02)^2))} \\ using $1.40576^2$ = Lua result & \myrow*{2*11000*(1 - 1.40576 - cos(0.02) + sqrt(1.9761611776 - sin(0.02)^2))} \\ }; \draw (a-1-1.south -| a.west) -- (a-1-7.south -| a.east); \end{tikzpicture} \end{document} ```

[davidcarlisle]

@Mo-Gul you confused me for a while as 1.460576^2 is 2.133 not 1.976 you intended 1.40576^2

[Mo-Gul]

@davidcarlisle, I am so sorry. Don't know how this typo could happen. I edited my previous post accordingly so that nobody else gets confused too.

[hmenke]

@Mo-Gul The comment you linked applies to the line below https://github.com/pgf-tikz/pgf/blob/e0b798ba19471877f80633a727c45e09439344b3/tex/generic/pgf/libraries/luamath/pgflibraryluamath.code.tex#L239-L243 Thanks for debugging this. I'll have a closer look.

[hmenke]

I have found part of the problem. pow is simply not implemented. https://github.com/pgf-tikz/pgf/blob/e0b798ba19471877f80633a727c45e09439344b3/tex/generic/pgf/libraries/luamath/pgflibraryluamath.code.tex#L307 However, the corresponding Lua code is there. Same for other functions. https://github.com/pgf-tikz/pgf/blob/e0b798ba19471877f80633a727c45e09439344b3/tex/generic/pgf/libraries/luamath/pgf/luamath/functions.lua#L105-L108 I have to email Christian anyway, so I will ask about that.