Meydan okuma

5 seviyeli Sierpinski Üçgen Fraktalını dizecek TeX (LaTeX) matematik denklem kodunu (aşağıda verilmiştir) veren kodu yazın. En kısa kod kazanır .

ayrıntılar

TeX (ve LaTeX gibi arkadaşlar) sofistike bir dizgi sistemidir. Matematiksel formüller için rasgele iç içe geçmiş karmaşık ifadeler yapabilir. Tesadüfen bu "iç içe geçmiş kompleks" aynı zamanda fraktalların tanımlayıcısıdır. Aşağıdaki MathJaX ile işlenir

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

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

Test it

---

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

Test it

The recursive option.

>4©'xª"\{^_}"¬qßUÄ

Test it

Java (OpenJDK 8) , 179 167 bayt

@Neil limanı

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 Dili ( Mathematica ) - 40 karakter

Burada en iyi 3 cevabı özetliyoruz :

40 bayt:

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

41 bayt:

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

44 bayt:

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

C (gcc) , 82 bayt

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 bayt

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

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Python 3 çeviri:

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