Desafio

Escreva um código que produza o código de equação matemática TeX (LaTeX) (fornecido abaixo) que digitará o Fractal do Triângulo de Sierpinski de 5 níveis. O menor código vence .

Detalhes

O TeX (e amigos como o LaTeX, etc.) é um sofisticado sistema de composição tipográfica. Ele pode renderizar expressões complexas aninhadas arbitrárias para fórmulas matemáticas. Coincidentemente, esse "complexo aninhado" também é descritivo de fractais. O seguinte é renderizado com MathJaX

$${{{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}^{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}_{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}}^{{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}^{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}_{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}}_{{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}^{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}_{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}}}$$

by the following plain-text math-equation code consisting of nested super- and sub-scripts:

{{{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}^{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}_{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}}^{{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}^{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}_{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}}_{{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}^{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}_{{{x^x_x}^{x^x_x}_{x^x_x}}^{{x^x_x}^{x^x_x}_{x^x_x}}_{{x^x_x}^{x^x_x}_{x^x_x}}}}}

Note this is just a 5-level nesting. You do not need to generate $...$ or $$...$$ or other markup required to start/end a math equation in TeX & Co. You can preview generated TeX in many online editors, for instance: http://www.hostmath.com but you can find many others too. This question was inspired by a discussion with friends.

Update

There is a similar question but it much more general and will produce different solutions. I wanted to see really kolmogorov-complexity for a very fixed simple code that in one system (TeX) is completely explicit while in another compressed. This also address the n instead of 5 levels comment.

SOGL V0.12, 16 12 bytes

 x5{"{^_}”;∑

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Port of Erik The Outgolfer's Python 2 answer

Python 2, 32 bytes

exec"print'x'"+".join('{^_}')"*5

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plain TeX, 29 bytes

\def~#1x{{#1x_#1x^#1x}}~~~~~x

That outputs what others have output. But if we need the code to be compilable it would be 6 bytes more

\def~#1x{{#1x_#1x^#1x}}$~~~~~x$\bye

Explanation

~ is an active character in TeX, so we can give it a (new) definition.

\def~#1x{{#1x_#1x^#1x}} defines ~ as a macro, so that when TeX sees ~, it does the following:

• Read everything up to the next x, and call that #1 (pattern-matching).
• Replace the whole thing with {#1x_#1x^#1x}

For example, ~ABCx would get replaced with {ABCx_ABCx^ABCx}.

When ~~~~~x is used, #1 is ~~~~, so the whole thing gets replaced with {~~~~x_~~~~x^~~~~x}. And so on.

Once we have the long string, we can print it out to terminal with \message (and ending with a \bye so TeX stops), so \message{~~~~~x}\bye. Or typeset the resulting expression (as a mathematical formula), by surrounding it in $s : so $~~~~~x$\bye.

05AB1E, 17 bytes

'x5F'x¡"{x^x_x}"ý

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Explanation

'x                  # push "x"
  5F                # 5 times do
    'x¡             # split on "x"
       "{x^x_x}"ý   # join on "{x^x_x}"

Other programs at the same byte-count include

"{x^x_x}"©4F'x¡®ý
'x5F'x"{x^x_x}".:

PowerShell,  44  35 bytes

"'x'"+"-replace'x','{x^x_x}'"*5|iex

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Uses string multiplication to repeatedly -replace xes with the sub- and super-scripts, then output.

Saved 9 bytes thanks to Joey.

MATL, 21 20 bytes

'x'XJ5:"J'{x^x_x}'Zt

-1 byte thanks to Giuseppe

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JavaScript (ES6), 45 42 37 bytes

f=n=>n>4?'x':[...'{^_}'].join(f(-~n))

Edit: Saved 3 2 bytes thanks to @Arnauld. Specifying 5 still costs me 2 bytes; this 41 40 35-byte version takes a parameter instead:

f=n=>n?[...'{^_}'].join(f(n-1)):'x'

05AB1E, 13 bytes

'x5F"{^_}"Ssý

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Port of my Python 2 answer.

Jelly, 12 bytes

”x“{^_}”j$5¡

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Port of my Python 2 answer.

Japt, 21 20 18 bytes

5Æ="\{^_}"¬qUª'xÃÌ

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---

Explanation

5Æ             Ã

Generate an array of length 5 and map over it.

"\{^_}"¬

Split a string to an array of characters

qUª'x

Rejoin (q) to a string using the current value of U or (ª) "x".

=

Assign the result of that to U.

Ì

Get the last element in the array.

---

Alternatives, 18 bytes

Same as above but reducing the array after it's been created.

5o r@"\{^_}"¬qX}'x

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The recursive option.

>4©'xª"\{^_}"¬qßUÄ

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Java (OpenJDK 8), 179 167 bytes

@Neil port

interface Y{static void main(String[]a){System.out.println(t.apply(1));}java.util.function.Function<Integer,String>t=N->N>0?Y.t.apply(N-1).replace("x","{x^x_x}"):"x";}

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Wolfram Language (Mathematica) - 40 characters

Summarizing 3 best answers here:

40 bytes:

Nest["{"<>#<>"_"<>#<>"^"<>#<>"}"&,"x",5]

41 bytes:

Nest[StringReplace["x"->"{x^x_x}"],"x",5]

44 bytes:

Last@SubstitutionSystem["x"->"{x^x_x}","x",5]

C (gcc), 82 bytes

O(o){o--?_(125,O(o,_(95,O(o,_(94,O(o,_(123))))))):_(120);}main(){O(5);}

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Pyth, 17 16 13 bytes

jF+\x*5]"{^_}

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Python 3 translation:

from functools import*;print(reduce(lambda x,y:x.join(y),["x"]+5*["{^_}"]))