On a conditional command: more powerful than \ifnum

Question

According to the TEX in a Nutshell, Petr Olšák, page 16/29, we get:

\ifnum ⟨number 1⟩ ⟨relation⟩ ⟨number 2⟩ . The ⟨relation⟩ could be < or = or >. It returns true if the comparison of the two numbers is true.

I am looking for a more powerful conditional command to include (~=) (not equal to) relation.

• Do we have such a conditional command?

The reason I asked is related to page 1005/1318 TikZ manual. Which I modified the code a little bit to generate an array of elliptical objects as below:

I rotated all ellipses 45 degree clockwise. What I want to do is not to rotate the highlighted ellipses in yellow so the rot=0 for them.

• How to do it with a conditional command?

Below is my code:

 \documentclass{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
 \foreach \x in {1,...,4}
\foreach \y in {1,...,4}

{
\fill[red!50] (\x,\y) ellipse [x radius=3pt, y radius=6pt, rotate=-45];
\ifnum {\x<\y} & {\x>1}
\breakforeach
\fi
}
\draw [|-|] (.895,1) -- ++(0.211,0);
\end{tikzpicture}
\end{document}

It seems ifthen package is the answer. I used the code below using \ifthenelse, but got an error:

\documentclass{standalone}
\usepackage{tikz,ifthen}
\begin{document}
\ifthenelse{1>2 \AND 3=3}{yes}{no}
\begin{tikzpicture}
\foreach \x in {1,...,4}
\foreach \y in {1,...,4}
{
\newcommand{\first}{\(\x=1 \and \y=1\)}
\newcommand{\second}{\(\x=2 \and \y=1\) }
\ifthenelse{\(\first\) \or \(\second\)}
 {\fill[red!50] (\x,\y) ellipse [x radius=3pt , y radius= 6pt, rotate=0];}
{\fill[red!50] (\x,\y) ellipse [x radius=3pt , y radius= 6pt, rotate=-45];}
\ifnum \x<\y
\breakforeach
\fi
}
\draw [|-|] (.895,1) -- ++(0.211,0);
\end{tikzpicture}
\end{document}

Error: ! Extra \or. ...=1) } \ifthenelse {(\first ) \or (\second )} {\fill [red!... Do you know how to debug this code to make it work?

Answer 1

In order to reproduce the picture, the test should be 1 ≤ x < y.

You can parametrize the rotation angle and test for the two special cases.

\documentclass{standalone}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
\foreach \x in {1,...,4} {
  \foreach \y in {1,...,4} {
    \def\rotation{-45}
    \ifnum\y=1
      \ifnum\x=1 \def\rotation{0} \fi
      \ifnum\x=2 \def\rotation{0} \fi
    \fi
    \fill[red!50] (\x,\y) ellipse [x radius=3pt, y radius=6pt, rotate=\rotation];
    \ifnum \x<\y \unless\ifnum \x<1 \breakforeach \fi\fi
  }
}
\draw [|-|] (.895,1) -- ++(0.211,0);
\end{tikzpicture}
\end{document}

You can also use \ifthenelse, of course.

\documentclass{standalone}
\usepackage{tikz}
\usepackage{xifthen}

\begin{document}
\begin{tikzpicture}
\foreach \x in {1,...,4} {
  \foreach \y in {1,...,4} {
    \ifthenelse{\y=1 \AND \(\x=1 \OR \x=2\)}{\def\rotation{0}}{\def\rotation{-45}}
    \fill[red!50] (\x,\y) ellipse [x radius=3pt, y radius=6pt, rotate=\rotation];
    \ifthenelse{ \x<\y \AND \NOT\(\x<1\) }{\breakforeach}{}
  }
}
\draw [|-|] (.895,1) -- ++(0.211,0);
\end{tikzpicture}
\end{document}

With a somewhat easier syntax, see https://tex.stackexchange.com/a/467527/4427

\documentclass[border=4]{standalone}
\usepackage{tikz}
\usepackage{xparse}

\ExplSyntaxOn
\NewExpandableDocumentCommand{\xifthenelse}{mmm}
 {
  \bool_if:nTF { #1 } { #2 } { #3 }
 }

\cs_new_eq:NN \numtest     \int_compare_p:n
\cs_new_eq:NN \oddtest     \int_if_odd_p:n
\cs_new_eq:NN \fptest      \fp_compare_p:n
\cs_new_eq:NN \dimtest     \dim_compare_p:n
\cs_new_eq:NN \deftest     \cs_if_exist_p:N
\cs_new_eq:NN \namedeftest \cs_if_exist_p:c
\cs_new_eq:NN \eqdeftest   \token_if_eq_meaning_p:NN
\cs_new_eq:NN \streqtest   \str_if_eq_p:ee
\cs_new_eq:NN \emptytest   \tl_if_blank_p:n
\prg_new_conditional:Nnn \xxifthen_legacy_conditional:n { p,T,F,TF }
 {
  \use:c { if#1 } \prg_return_true: \else: \prg_return_false: \fi:
 }
\cs_new_eq:NN \boolean \xxifthen_legacy_conditional_p:n
\ExplSyntaxOff

\begin{document}
\begin{tikzpicture}
\foreach \x in {1,...,4} {
  \foreach \y in {1,...,4} {
    \xifthenelse{\numtest{\y=1} && (\numtest{\x=1} || \numtest{\x=2})}
      {\def\rotation{0}}
      {\def\rotation{-45}}
    \fill[red!50] (\x,\y) ellipse [x radius=3pt, y radius=6pt, rotate=\rotation];
    \xifthenelse{ \numtest{1<=\x<\y} }{\breakforeach}{}
  }
}
\draw [|-|] (.895,1) -- ++(0.211,0);
\end{tikzpicture}
\end{document}

Answer 2

LaTeX's ifthen package has some facility for combining conditionals with \and and \or, \not and parentheses. But your case is easy to do with \ifnum:

\ifnum \x<\y \ifnum \x>1
  \breakforeach
\fi\fi

Answer 3

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{math}
\begin{document}
\begin{tikzpicture}
\foreach \x in {1,...,4} {
\tikzmath {\xEnd=\x+1;}
\foreach \y in {1,...,\xEnd} {
\fill[red!50] (\x,\y) 
ellipse [x radius=3pt, y radius=6pt, rotate={ifthenelse(\x==1 || \x==2 && \y==1,0,-45)}];
}}
\draw [|-|] (.895,1) -- ++(0.211,0);
\end{tikzpicture}
\end{document}

Answer 4

Found a much simpler solution. Below are the key ideas:

• \ifthenelse from ifthen package in CTAN is that powerful conditional command.

• \ifthenelse{<test>}{<then clause>}{<else clause>}is the command format wherein <test> is a boolean expression using the infix connectives, \and, \or, the unary \not and parentheses \( \).

• Also \OR and \AND are safer to be used with \ifthenelse rather than using \or and \andsince it can’t be misinterpreted when appearing inside a TEX-conditional in which \or has a different meaning. So here is the code:

  \documentclass{standalone}
  \usepackage{tikz,ifthen}
  \begin{document}
  \begin{tikzpicture}
  \foreach \x in {1,...,4}
  \foreach \y in {1,...,4}
  {
  \ifthenelse{\(\x=1 \AND \y=1\) \OR \(\x=2 \AND \y=1\) }
  {\fill[red!50] (\x,\y) ellipse [x radius=3pt , y radius= 6pt, rotate=0];}
  {\fill[red!50] (\x,\y) ellipse [x radius=3pt , y radius= 6pt, rotate=-45];}
  \ifnum \x<\y
  \breakforeach
  \fi
  }
   \draw [|-|] (.895,1) -- ++(0.211,0);
  \end{tikzpicture}
  \end{document}