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?

conditionals foreach ifthenelse tikz pgf

egreg · Accepted Answer

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} {
    defrotation{-45}
    ifnumy=1
      ifnumx=1 defrotation{0} fi
      ifnumx=2 defrotation{0} fi
    fi
    fill[red!50] (x,y) ellipse [x radius=3pt, y radius=6pt, rotate=rotation];
    ifnum x<y unlessifnum x<1 breakforeach fifi
  }
}
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)}{defrotation{0}}{defrotation{-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})}
      {defrotation{0}}
      {defrotation{-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}

Donald Arseneau · Answer

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
fifi

Aria · Answer

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}

vi pa · Answer

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}

On a conditional command: more powerful than ifnum

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