Prosta złożoność

17

Wynik

                  # # # # # ##### ## #         
                  ## #### ####### # # ##        
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                ####### ### # ### # ## ####        
                ## # # ### # ####### ##          
                 # ## ##### ##### # # #          
                ## # # # # #####  
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    # # ##### ## # #                     
    ####### # # ## ##                    
    ## # ### ####### ####                    
     # ##### ### # # # #                    
 ##### # # # # # ###                
### # ## #### #######                 
  # ####### ## ## #                 
  ##### # # # # ##                
          # # # ##### ##### ## #                 
          ## ####### # ### # # ##                
        #### ## # ### # ### #######                
        # # # ##### ##### ### # #                
        # # ### ##### # # # #                
        ####### ### # ## ## ####                
        ## # # ####### #### ##                  
         # ## ##### # # # # #                  
                                          #                     
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                                        # # ###                
                                        #######                 
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                                 ##### ## #                 
                                ### # # ##                
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                                #### ##                  
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                                  # # # # # #####  
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                                ####### ### # ### # ##    
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                                ## # #     
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                                ### # # # #    
                                    # # # # ###
                                    #### ####### 
                                    ## ## # 
                                     # # ##

Dozwolone spacje końcowe. Najkrótsze rozwiązanie wygrywa.

wskazówka 0

wskazówka 1:

hint1

dzięki @Tahg za poprawienie podpowiedzi 1

code-golf ascii-art kolmogorov-complexity ngn
źródło
8
@ngn wyrzuca tylko wyzwania bez wyjaśnienia, w jaki sposób zostały utworzone dane, co jest zwykle odrzucane (patrz codegolf.stackexchange.com/q/126037 ), ponieważ pierwsza odpowiedź ma tendencję do zdmuchiwania „tajemniczej” części
Uriel
16
Osobiście nie lubię takich wyzwań, w których ustalenie reguły / przepisu jest częścią zadania. Poza tym, gdy ktoś go znajdzie, wszyscy inni mogą po prostu go śledzić, więc dlaczego nie napisać w pierwszej kolejności?
Luis Mendo,
11
@LuisMendo To interesujący argument. Code golf w najczystszej formie polega na „wdrożeniu znanego rozwiązania”. Zagadki w najczystszej formie polegają na znalezieniu rozwiązania - wdrożenie nie jest istotne lub jest uważane za zajęty. Jedną z opcji byłoby opublikowanie „sztuczki” w spoilerach. W ten sposób golfiści z czystym kodem zaatakują problem jako wyzwanie dla golfistów, a ci, którzy lubią zarówno golfa, jak i puzzle, mogą uniknąć patrzenia i rozwiązania obu.
Jonasz
5
Myślę, że „tajemnicza” część jest czymś, co zdecydowanie mogę zobaczyć więcej. Chociaż jedna odpowiedź może rozwiązać ten problem, reszta nadal stanowi normalne wyzwanie dla golfa ... I stanowi dobre wyzwanie dla tych, którzy chcą znaleźć wzór na początku.
całkowicie ludzki,
3
@ H.PWiz, nie jestem pewien, czy + 26 / -7 jest naprawdę „dobrze przyjęty”. Wygląda bardziej jak „słabo odebrany, ale udało mu się trafić w HNQ wystarczająco szybko, aby uzyskać wypaczony wynik”.
Peter Taylor,

Odpowiedzi:

11

SOGL V0.12 , 17 16 14 bajtów

 #6{³IIč▓┼;I+§

Wypróbuj tutaj!

W późniejszej aktualizacji č▓można usunąć dla 12 bajtów - które konwertują ToS z tablicy tablic znaków, ciąg wielowierszowy na tablicę ciągów - [["#","#"],[" ","#"]] -> ["##"," #"]-, ponieważ ┼- dołączenie poziome - nie radzi sobie dobrze z tablicami tablic znaków - które Itworzy, ponieważ służy również do obracania tablicy. W SOGL tablica tablic znaków powinna być = tablica ciągów, ale wiele rzeczy jeszcze nie obsługuje…

Wyjaśnienie:

 #            push "#"
  6{          6 times do
    ³           create 3 total copies of ToS
     II         rotate clockwise twice
       č▓       normalize as explained above
         ┼      append horizontally
          ;     get the 3rd copy ontop
           I    rotate clockwise
            +   append vertically
             §  reverse horizontally
dzaima
źródło
1
poczekaj, jak to działa
Conor O'Brien
@ ConorO'Brien właśnie dodał wyjaśnienie: p
dzaima
och fajnie `` - ``
Conor O'Brien
22

JavaScript (ES6), 233 217 213 198 182 170 163 122 bajtów

f=_=>[...Array(64)].map((_,x,a)=>a.map(g=(i=(n=64,x),j)=>(n>>=1)?i&n?j&n?g(j,i):` `:j&n?g(i,~j):g(~i,j):`#`).join``).join`
`
document.write(`<pre>${f()}</pre>`)
Rozwiń fragment kodu

Edycja: Zapisano 14 18 bajtów dzięki @Shaggy. Zaoszczędzono 3 bajty dzięki @ngn. Oszczędność kolejnych 12 bajtów dzięki dwóm pracującym razem. Zaoszczędzono 41 bajtów, kradnąc obserwacje @ user202729, że ćwiartki używają odbić, a nie rotacji. Nie golfowany:

function f() {
    var s = '';
    for (var i = 0; i < 64; i++) {
        for (var j = 0; j < 64; j++) {
            var x = i, y = j, c = '#'; // Default to #
            // Each non-blank quadrant maps to to the original
            // image by doubling and a reflection. Repeat
            // this six times unless we hit the blank quadrant.
            for (var n = 0; n < 6; n++) {
                if (x >= 32) {
                    if (y >= 32) {
                        // Bottom right quarter is a diagonal reflection
                        var t = x - 32;
                        x = y - 32;
                        y = t;
                    } else {
                        // Bottom left quarter is blank
                        c = ' ';
                        break;
                    }
                } else {
                    if (y >= 32) {
                       // Top right corner is a horizontal reflection
                       y = 63 - y;
                    } else {
                       // Top left corner is a vertical reflection
                       x = 31 - x;
                    }
                }
                x *= 2;
                y *= 2;
            }
            s += c;
        }
        s += '\n';
    }
    return s;
}
Neil
źródło
Czy musisz f=tutaj liczyć ? Nawiasem mówiąc, ładnie zrobione.
Shaggy
217 bajtów
Kudłaty
1
Nie potrzebujesz f=liczenia bajtów, ale fragment kodu nie działa bez niego.
1
@Neil Czy jest szansa, że ​​możesz dodać wyjaśnienie lub wersję bez golfa?
Jonasz
1
@Jonah Mam nadzieję, że okaże się to pomocne.
Neil
11

LOGO, 375 341 297 295 278 + 3 bajty

Dodaj 3 bajty ze względu na -pflagę, która domyślnie włącza tryb perspektywy, więc nie trzeba uruchamiać perspectivepolecenia, oszczędza ogółem 9 bajtów.

Użyj FMSLogo w systemie Windows z formatem nowej linii Unix (LF) (FMSLogo ma problem z analizowaniem formatu nowej linii CR)

to R
rt 90
end
to g :w :l
R fd 2*:l R bk :l up 180
run :w
R run :w
fd 2*:l R bk :l run :w
fd 2*:l up 180
end
to h
pu
ask -1[setxyz 0 0 870]g[g[g[g[g[g[rt 45 fd .7 pd fd 0 pu bk .7 lt 45]1]2]4]8]16]32
repeat 64[sety 64-# repeat 64[setx #-1 type if pixel=[0 0 0]""#["\ ]](pr)]
end

Niestety nie ma „Wypróbuj online!” link, ponieważ nie mogę znaleźć trybu perspektywy wsparcia tłumacza online.

Pomysł: narysuj obraz obrazu, a następnie pobierz piksele z obrazu i wydrukuj jako dane wyjściowe.

Podział obrazu na proste powtarzające się części:

.

Skorzystaj z podpowiedzi powyżej. Ponieważ jednak LOGO nie obsługuje odbicia, możemy to tylko zasymulować, wchodząc w perspectivetryb 3D ( ) i obracając żółwia o 180 stopni wokół osi równoległej do ekranu komputera.

This defines a helper function g, which given 2 parameters l (image side length) and w (procedure used to draw the image), draw 3 copies of its reflected image. (see hint in the question) The procedure h perform the main procedure.

user202729
źródło
You can probably omit some spaces like 90 fd -> 90fd and save a few bytes.
Jonathan Frech
@JonathanFrech Unfortunately, FMSLogo doesn't support that.
user202729
Why does papert support it then?
Jonathan Frech
2
Is this the only answer that actually exploits the pattern rather than bit packing?
Jonah
2
@Jonah The JS solution was first to exploit the pattern, though it does it in an interesting way - with modular arithmetic and bit operations instead of concatenating matrices. This LOGO solution is also interesting and unique with its pixel-reading approach. I don't know enough LOGO to understand the code but judging by the description it repeats certain patterns without going all the way down the fractal's structure (which I already gave away in hint1).
ngn
10

05AB1E, 30 26 bytes

'#)6F©R®í®'#ð:®ζ)2ä€ø€J˜}»

Try it online!

Emigna
źródło
8

Python 2, 205 195 ... 145 144 142 144 bytes

g=lambda n,i,j,*_:g(n/2,*[~i,j,i,~j][2*(j&n>0)-(i&n>0):])if n*(i&n<=j&n)else'# '[i&n>0]
r=range(64)
for j in r:print''.join(g(32,j,i)for i in r)

Try it online!

Inspired by Neil's JS answer.

TFeld
źródło
n>>1 is n/=2.
Jonathan Frech
a is defined and then used once. Replacing its reference with its value could save four bytes.
Jonathan Frech
I overlooked a useful simplification - rather than doing everything using rotations, I should have been using reflections instead. It saved me 41 bytes on my answer!
Neil
I think you can save two bytes by simplifying your [...[x],...[x]][z] selection.
Jonathan Frech
1
@totallyhuman Fair enough
TFeld
7

Haskell, 126 125 113 106 103 bytes

e=[]:e
z=zipWith
r=reverse
m s=z(++)(r s++map(' '<$)s)$map r s++foldr(z(:))e s
unlines$m$m$m$m$m$m["#"]

A direct implementation of hint1 in the spoiler.

Function m generates the next iteration. The main functions applies m 6 times. Details to m:

m s=                          -- s is the current pattern
     zipWith(++)              -- concatenate pairwise the lines of
                (r s)++       --    the lines of s in reverse order, followed by
                map(' '<$)s   --    each char in s replaced by a space
                              -- and
                map r s++     --    each line of s reversed, followed by
                foldr ... s   --    the transposition of s


e=[]:e;foldr(zipWith(:))e     -- this right fold transposes a matrix

Edit: @ngn saved a byte and @Laikoni another 3. Thanks!

nimi
źródło
n$n$n$n$n$n["#"] is shorter than iterate n["#"]!!6 :)
ngn
@ngn: well spotted. Thanks a lot!
nimi
You can use e=[]:e;foldr(zipWith(:))e instead of import Data.List;transpose and then shorten zipWith.
Laikoni
6

Java 7, 259 238 237 200 bytes

void f(){char[][]o={{35}},n;for(int s=1,x,y;s<64;s*=2,o=n)for(n=new char[64][64],x=s;x-->0;)for(y=0;y<s;n[s+y++][x]=32)n[s+~y][x]=n[y][2*s+~x]=n[s+x][s+y]=o[y][x];for(char[]b:o)System.out.println(b);}

Saved 2 bytes by removing the {} on the x loop, thanks ngn
Saved 19 bytes from various assignment changes, thanks Jonathan
Saved 24 bytes for print vs return (didn't know this was allowed), and
Saved 13 bytes for loop changes, thanks Nevay

Try it online!

My first challenge, and I think respectable for Java. Uses hint 1, (which is wrong btw, but I can't write comments). It can probably be golfed further, this pass was basically as is, without pretty printing.

Tahg
źródło
I am not sure if it might be because of a version difference, but I tested your code on TIO using Java 8, and it prints null characters instead of spaces.
Jonathan Frech
It sure did, thanks for noticing. Apparently, Java is quite happy with displaying nulls as spaces in its Strings.
Tahg
You can save a byte by removing s=1 from your for loop and replacing int s,t,x,y; with int s=1,t,x,y;.
Jonathan Frech
1
n[s-1-y][x]=o[y][x];n[y][t-1-x]=o[y][x]; -> n[s-1-y][x]=n[y][t-1-x]=o[y][x];.
Jonathan Frech
2
the proper transformation matrix is [[V,H],[_,T]] (Either that or there's some subtle difference in your algorithm, but that's what I needed)
Tahg
3

Python 2, 586 bytes

import zlib,base64 as b
print zlib.decompress(b.b64decode('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'))

Try it online.

Python 2, 1032 1025 bytes

I like this one more. But it is much longer. It could be golfed down but there is no need for that. officialaimm's approach is much shorter.

-7 thanks to Jonathan Frech

print''.join(' '*int(x)if'/'<x<'@'else("#"*(ord(x)-64),"\n")[x<"#"]for x in'99A9A2A4A2A1E2B4A9!99B8D4G2A3A4B8!88D8B6B4A2C1G8!88A2A9A7A4E1C1A2A8!88A2A1C1E3E4A9A2A8!88G1C2A2C2A4B8D8!88B4A3A2C2A2G8B55!98A4B2E3E1A2A9A55!88B4A559A9A2A1E2!98A4B99B8G2A2!98G88D8B4A2C!88C1A2A88A2A9A4E1!992A2A88A2A1C1E4A5!992D88G1C2A4B4!992B99B4A3A2G4!777A559A4B2E1A2A4!4A2A1E2B4A559A777!4G2A3A4B99B992!4B4A2C1G88D992!5A4E1C1A2A88A2A569!1E4A9A2A88A2A1C88!C2A4B8D88G98!2A2G8B99B4A89!2E1A2A9A559A4B88!55A9A2A1E3E2B4A98!55B8G2A2C2A3A4B88!8D8B4A2C2A2C1G88!8A2A9A4E3E1C1A2A88!8A2A1C1E4A7A9A2A88!8G1C2A4B6B8D88!8B4A3A2G4D8B99!9A4B2E1A2A4A2A9A99!99996A777!99996B992!99994D299!99994A2A992!99994A2A1C88!99994G98!99994B4A98!99995A4B88!9996E2B4A89!9995C2A3A4B88!9997A2C1G88!9997E1C1A2A88!9997A9A2A88!9997B8D88!9995D8B99!9995A2A9A99!9997A9A2A4A2A1E2!9997B8D4G2A2!9995D8B6B4A2C!9995A2A9A7A4E1!9995A2A1C1E3E4A5!9995G1C2A2C2A4B4!9995B4A3A2C2A2G4!9996A4B2E3E1A2A4!9995B4A559A5!9996A4B99B4!9996G88D4!9995C1A2A88A2A4!9999A2A88A2A1C!9999D88G1!9999B99B4A1!99991A559A4B')

Try it online.

Simon
źródło
The question allows [t]railing spaces, though your solution omits some spaces.
Jonathan Frech
allowing trailing spaces means, that I dont have to output them. Is that correct? It would save bytes on my second approach. On the first one it is already done because I deleted the trailing spaces while saving my file...
Simon
I do not know for sure, but I interpreted it as allowing additional spaces, not allowing missing spaces.
Jonathan Frech
x=='!' -> x<'#'
Jonathan Frech
if x.isdigit() -> if"/"<x<"@"
Jonathan Frech
3

Mathematica, 112 90 bytes

Thanks to Jonathan Frech for help saving 2 bytes!

""<>Nest[{{(r=Reverse)@#,r/@#},{" "+0#,#}}~Flatten~{{1,3},{2,4}}&,{{"#"}},6]~Riffle~"
"

Try it online! (Mathics)

For some reasons, Mathics prints leading spaces in all lines except the first one when print multiline string. Also Mathics doesn't support operator.

Explanation:

  • {{Reverse@#,Reverse/@#},{" "+0#,#}} : Represent the reverse-horizontally, reverse-vertically, replace-all-by-" ", and transpose ( is transpose operator in Mathematica), corresponding to different ways to reflect or rotate the image.
  • ~Flatten~{{1,3},{2,4}} : Flatten in particular dimensions.
  • Nest[ ... ,{{"#"}},6] : Apply the function inside to {{"#"}} 6 times.
  • ~Riffle~"<literal newline character>" : Riffle a newline character between each line.
  • ""<> : Join all strings together.
user202729
źródło
You can define replace Reverse with R and define R=Reverse; to save two bytes.
Jonathan Frech
2

C# (.NET Core), 1016 1002 980 955 bytes

Saved 14 bytes thanks to Kevin Cruijssen!
Saved 47 bytes thanks to Jonathan Frech!

o=>{foreach(var l in new[]{35223185965568,52841249260288,0xf00c0c277f00,0x9004043ee900,0x977c7c200900,0xfee4e4300f00,0xc22727f00c00,73934616658944,0xc2000020097c,73667282210788,0x7f0000f00c27,0xe9000090043e,9895614577696,0xf0000fee430,0xc0000c227f0,4398050918032,325982355457<<21,1092065689603<<20,835235872783<<20,291765223433<<20,0x7c20090000970000,-15289957744513<<17,21955973480545<<17,68788263321667<<16,68799053409<<17,206425089091<<16,0xf00c27277f0000,618546478825<<16,650622541833<<16,0xfee430300f0000,208473439235<<18,72203117569<<18,1<<21,3<<20,15<<20,9<<20,9895936,127<<17,97<<17,67<<16,15969<<17,3829596160,662634496,16105<<16,8201<<16,806289408,4027318272,9217L<<18,537463164,806293476,4027321383,2416182334,2541517856,4276413488,3257346032,1128152720,3254779936,1124073520,2130706672,3909091472,150995095,251658494,201326786,67108931})System.Console.WriteLine(System.Convert.ToString(l,2).PadLeft(64,'0').Replace('0',' ').Replace('1','#'));}

Try it online!

Explanation

The format of the output is stored inside an array of signed 64-bit numbers, which fits perfectly since each line is 64 characters long. Empty spaces are represented by 0 and # is represented by 1.

The numbers are then converted to their binary string, zeros are padded left until the string is 64 characters wide and then the 0 and 1 characters are replaced with   and #.

The code is stored inside a lamba function, more specifically a System.Func<string>.

Some constants in the long[] are shortened by performing some bit-shifting.

Ian H.
źródło
Is there not an unnecessary space in long[] n=new[]?
Jonathan Frech
@JonathanFrech Ah you are right, apparently it snuck in there somehow.
Ian H.
1
You would save 4 bytes by not having to specify the padding character.
Neil
1
Well, if the trailing l is not needed, you can probably save even more bytes.
Jonathan Frech
1
Looks like the bottom of this is bugged.
Shaggy
2

Charcoal, 42 bytes

#↓##ＦＥ⁵Ｘ²ι«‖↑Ｊ⁻×³ι¹±⁰ψ⟲ＯＯ⁴ײι⟲ＣＪ⁰±φＴ×⁴ι×⁴ι

Try it online! Link is to verbose version of code. Would be 25 bytes if this worked:

#ＦＥ⁷Ｘ²ι«‖↑⟲Ｃ→⁴⟲ＣＪ⁰±φＴιι

Explanation:

#↓##

Manually generate the first recursion, since it's not possible to rotate around (1, 0.5). (Such a rotation would only make sense if it was 180°.)

ＦＥ⁵Ｘ²ι«

Loop over the first five powers of 2 (1, 2, 4, 8, 16).

‖↑

Reflect the canvas vertically. This completes the top left quarter of the result.

Ｊ⁻×³ι¹±⁰ψ⟲ＯＯ⁴ײι

Rotate the canvas 180° around a point half-way up the right-hand side. The copy ends up in the correct position for the top right quarter of the result.

⟲Ｃ

Rotate the canvas 90° around the bottom right corner. The copy of the top right corner ends up in the correct position for the bottom right corner of the result. The copy of the top left corner is extraneous.

Ｊ⁰±φＴ×⁴ι×⁴ι

Trim the canvas to the size we want. The trim starts at the cursor or the top left of the canvas, whichever is bottom rightmost. The cursor is therefore sent to (0, -1000) to ensure that it doesn't interfere with the trim.

Neil
źródło
1

Python 2, 715 711 bytes

  • -4 bytes thanks to @Jonathan Frech
for k in range(64):print"".join(ord(i)*" #"[j%2]for j,i in enumerate('												<?<<<89:11222	202	4	$" 	!%$%$&%$($%&'))[k<<6:][:64]

Try it online!

officialaimm
źródło
1
711 bytes.
Jonathan Frech
@JonathanFrech Thank you, I didn't think there were no new-lines. :D
officialaimm
1

Perl 5, 452 bytes

451 bytes code + 1 for -p.

Packs all the data into the string, rather than stealing @Neil's correct answer.

$_=unpack"B*","\x00\x00\x20\x09\x09\x7c\xc2\x33\x00\x30\x0f\x0f\xe4\x43\x33\x00\xf0\x0c\x0c\x27\x7f\x33\x00\x90\x04\x04\x3e\xe9\x33\x00\x97\x7c\x7c\x20\x09\x33\x00\xfe\xe4\xe4\x30\x0f\x33\x00\xc2\x27\x27\xf0\x0c\x33\x00\x43\x3e\x3e\x90\x04\x33\x00\xc2\x00\x00\x20\x09\x7c\x00\x00\x43\x00\x00\x30\x0f\xe4\x00\x00\x7f\x00\x00\xf0\x0c\x27\x00\x00\xe9\x00\x00\x90\x04\x3e\x00\x00\x09\x00\x00\x97\x7c\x20\x00\x00\x0f\x00\x00\xfe\xe4\x30\x00\x00\x0c\x00\x00\xc2\x27\xf0\x00\x00\x04\x00\x00\x43\x3e\x90\x09\x7c\xc2\x00\x00\x20\x00\x00\x0f\xe4\x43\x00\x00\x30\x00\x00\x0c\x27\x7f\x00\x00\xf0\x00\x00\x04\x3e\xe9\x00\x00\x90\x00\x00\x7c\x20\x09\x00\x00\x97\x00\x00\xe4\x30\x0f\x00\x00\xfe\x00\x00\x27\xf0\x0c\x00\x00\xc2\x00\x00\x3e\x90\x04\x00\x00\x43\x33\x00\x20\x09\x7c\x7c\xc2\x33\x00\x30\x0f\xe4\xe4\x43\x33\x00\xf0\x0c\x27\x27\x7f\x33\x00\x90\x04\x3e\x3e\xe9\x33\x00\x97\x7c\x20\x20\x09\x33\x00\xfe\xe4\x30\x30\x0f\x33\x00\xc2\x27\xf0\xf0\x0c\x33\x00\x43\x3e\x90\x90\x04\x37\x00\x20\x37\x00\x30\x37\x00\xf0\x37\x00\x90\x37\x00\x97\x37\x00\xfe\x37\x00\xc2\x37\x00\x43\x36\x00\x7c\xc2\x36\x00\xe4\x43\x36\x00\x27\x7f\x36\x00\x3e\xe9\x36\x00\x20\x09\x36\x00\x30\x0f\x36\x00\xf0\x0c\x36\x00\x90\x04\x36\x00\x20\x09\x09\x7c\x34\x00\x30\x0f\x0f\xe4\x34\x00\xf0\x0c\x0c\x27\x34\x00\x90\x04\x04\x3e\x34\x00\x97\x7c\x7c\x20\x34\x00\xfe\xe4\xe4\x30\x34\x00\xc2\x27\x27\xf0\x34\x00\x43\x3e\x3e\x90\x34\x00\xc2\x00\x00\x20\x34\x00\x43\x00\x00\x30\x34\x00\x7f\x00\x00\xf0\x34\x00\xe9\x00\x00\x90\x34\x00\x09\x00\x00\x97\x34\x00\x0f\x00\x00\xfe\x34\x00\x0c\x00\x00\xc2\x34\x00\x04\x00\x00\x43"=~s/[3-9](.)/$1x$&/ger;y/01/ #/;s/.{64}/$&
/g

Reversible output from xxd for 451 byte file:

00000000: 245f 3d75 6e70 6163 6b22 422a 222c 2200  $_=unpack"B*",".
00000010: 0020 0909 7cc2 3300 300f 0fe4 4333 00f0  . ..|.3.0...C3..
00000020: 0c0c 277f 3300 9004 043e e933 0097 7c7c  ..'.3....>.3..||
00000030: 2009 3300 fee4 e430 0f33 00c2 2727 f00c   .3....0.3..''..
00000040: 3300 433e 3e90 0433 00c2 0000 2009 7c00  3.C>>..3.... .|.
00000050: 0043 0000 300f e400 007f 0000 f00c 2700  .C..0.........'.
00000060: 00e9 0000 9004 3e00 0009 0000 977c 2000  ......>......| .
00000070: 000f 0000 fee4 3000 000c 0000 c227 f000  ......0......'..
00000080: 0004 0000 433e 9009 7cc2 0000 2000 000f  ....C>..|... ...
00000090: e443 0000 3000 000c 277f 0000 f000 0004  .C..0...'.......
000000a0: 3ee9 0000 9000 007c 2009 0000 9700 00e4  >......| .......
000000b0: 300f 0000 fe00 0027 f00c 0000 c200 003e  0......'.......>
000000c0: 9004 0000 4333 0020 097c 7cc2 3300 300f  ....C3. .||.3.0.
000000d0: e4e4 4333 00f0 0c27 277f 3300 9004 3e3e  ..C3...''.3...>>
000000e0: e933 0097 7c20 2009 3300 fee4 3030 0f33  .3..|  .3...00.3
000000f0: 00c2 27f0 f00c 3300 433e 9090 0437 0020  ..'...3.C>...7. 
00000100: 3700 3037 00f0 3700 9037 0097 3700 fe37  7.07..7..7..7..7
00000110: 00c2 3700 4336 007c c236 00e4 4336 0027  ..7.C6.|.6..C6.'
00000120: 7f36 003e e936 0020 0936 0030 0f36 00f0  .6.>.6. .6.0.6..
00000130: 0c36 0090 0436 0020 0909 7c34 0030 0f0f  .6...6. ..|4.0..
00000140: e434 00f0 0c0c 2734 0090 0404 3e34 0097  .4....'4....>4..
00000150: 7c7c 2034 00fe e4e4 3034 00c2 2727 f034  || 4....04..''.4
00000160: 0043 3e3e 9034 00c2 0000 2034 0043 0000  .C>>.4.... 4.C..
00000170: 3034 007f 0000 f034 00e9 0000 9034 0009  04.....4.....4..
00000180: 0000 9734 000f 0000 fe34 000c 0000 c234  ...4.....4.....4
00000190: 0004 0000 4322 3d7e 732f 5b33 2d39 5d28  ....C"=~s/[3-9](
000001a0: 2e29 2f24 3178 2426 2f67 6572 3b79 2f30  .)/$1x$&/ger;y/0
000001b0: 312f 2023 2f3b 732f 2e7b 3634 7d2f 2426  1/ #/;s/.{64}/$&
000001c0: 0a2f 67                                  ./g

Try it online!

Dom Hastings
źródło
1

Jq 1.5, 538 535 488 476 bytes

This is a straightforward representation. I haven't attempted any fancy encodings yet. Replaced 0,0 pairs with Z function.

Thanks again to Jonathan Frech for helping eliminate 3 bytes!

def B:recurse(if.>0then./2|floor else empty end)|.%2;def S:256+.|[B]|reverse[2:]|map(if.>0then"#"else" "end)|join("");def Z:0,0;[[Z,3,1,1,4,5,2],[1,4,5,2,2,3,Z],[Z,Z,2,3,Z],[Z,Z,3,1,1,4]]as$m|[[],[9,15,12,4,124,228,39,62],[Z,Z,Z,Z,124,228,39,62,32,48,240,144],[32,48,240,144,151,254,194,67,194,67,127,233,9,15,12,4],[124,228,39,62,32,48,240,144,32,48,240,144,151,254,194,67],[194,67,127,233,9,15,12,4,9,15,12,4,124,228,39,62]]as$c|$m[]|range(16)as$l|map($c[.][$l]|S)|join("")

Character count

$ wc -c picture.jq
 476 picture.jq

Sample run

$ jq -Mnr -f picture.jq
                  #         #  #    #  # #####  ##    #         
                  ##        ####    #######  #   #    ##        
                ####        ##      ##    #  ### #######        
                #  #         #       #    ##### ### #  #        
                #  # ### #####   #####    #         #  #        
                ####### ###  #  ###  #    ##        ####        
                ##    #   #  ###  #  #######        ##          
                 #    ##  #####   ##### #  #         #          
                ##    #                   #         #  # #####  
                 #    ##                  ##        #######  #  
                 #######                ####        ##    #  ###
                ### #  #                #  #         #    ##### 
                    #  #                #  # ### #####    #     
                    ####                ####### ###  #    ##    
                    ##                  ##    #   #  #######    
                     #                   #    ##  ##### #  #    
    #  # #####  ##    #                   #                     
    #######  #   #    ##                  ##                    
    ##    #  ### #######                ####                    
     #    ##### ### #  #                #  #                    
 #####    #         #  #                #  # ###                
###  #    ##        ####                #######                 
  #  #######        ##                  ##    #                 
  ##### #  #         #                   #    ##                
          #         #  # #####   #####  ##    #                 
          ##        #######  #  ###  #   #    ##                
        ####        ##    #  ###  #  ### #######                
        #  #         #    #####   ##### ### #  #                
        #  # ### #####    #       #         #  #                
        ####### ###  #    ##      ##        ####                
        ##    #   #  #######    ####        ##                  
         #    ##  ##### #  #    #  #         #                  
                                          #                     
                                          ##                    
                                        ####                    
                                        #  #                    
                                        #  # ###                
                                        #######                 
                                        ##    #                 
                                         #    ##                
                                 #####  ##    #                 
                                ###  #   #    ##                
                                  #  ### #######                
                                  ##### ### #  #                
                                  #         #  #                
                                  ##        ####                
                                ####        ##                  
                                #  #         #                  
                                  #         #  #    #  # #####  
                                  ##        ####    #######  #  
                                ####        ##      ##    #  ###
                                #  #         #       #    ##### 
                                #  # ### #####   #####    #     
                                ####### ###  #  ###  #    ##    
                                ##    #   #  ###  #  #######    
                                 #    ##  #####   ##### #  #    
                                ##    #                   #     
                                 #    ##                  ##    
                                 #######                ####    
                                ### #  #                #  #    
                                    #  #                #  # ###
                                    ####                ####### 
                                    ##                  ##    # 
                                     #                   #    ##
jq170727
źródło
1
Three omittable spaces in B: rec and 16) as $l|.
Jonathan Frech
1
if.==0then" "else"#"end -> if.>0then"#"else" "end.
Jonathan Frech
1

Python 2, 174 164 154 bytes (based on matrix operations)

  • -10 bytes thanks to ngn.
  • -10 bytes thanks to Jonathan Frech.
from numpy import*
s=[['#']]
exec'b=fliplr(s);s=vstack((hstack((flipud(s),b)),hstack((eye(len(s)),rot90(b)))));'*6
s[s>'#']=' '
for l in s:print''.join(l)

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mdahmoune
źródło
import * -> import*; in s.tolist() -> in s
ngn
1
154 bytes.
Jonathan Frech
0

JavaScript (Node.js), 1233 bytes

_=>` 18# 9# 2# 4# 2# #5 2#2 4#
 18#2 8#4 4#7 2# 3# 4#2
 16#4 8#2 6#2 4# 2#3 #7
 16# 2# 9# 7# 4#5 #3 # 2#
 16# 2# #3 #5 3#5 4# 9# 2#
 16#7 #3 2# 2#3 2# 4#2 8#4
 16#2 4# 3# 2#3 2# 2#7 8#2
 17# 4#2 2#5 3#5 # 2# 9#
 16#2 4# 19# 9# 2# #5
 17# 4#2 18#2 8#7 2#
 17#7 16#4 8#2 4# 2#3
 16#3 # 2# 16# 2# 9# 4#5
 20# 2# 16# 2# #3 #5 4#
 20#4 16#7 #3 2# 4#2
 20#2 18#2 4# 3# 2#7
 21# 19# 4#2 2#5 # 2#
 4# 2# #5 2#2 4# 19#
 4#7 2# 3# 4#2 18#2
 4#2 4# 2#3 #7 16#4
 5# 4#5 #3 # 2# 16# 2#
 #5 4# 9# 2# 16# 2# #3
#3 2# 4#2 8#4 16#7
 2# 2#7 8#2 18#2 4#
 2#5 # 2# 9# 19# 4#2
 10# 9# 2# #5 3#5 2#2 4#
 10#2 8#7 2# 2#3 2# 3# 4#2
 8#4 8#2 4# 2#3 2# 2#3 #7
 8# 2# 9# 4#5 3#5 #3 # 2#
 8# 2# #3 #5 4# 7# 9# 2#
 8#7 #3 2# 4#2 6#2 8#4
 8#2 4# 3# 2#7 4#4 8#2
 9# 4#2 2#5 # 2# 4# 2# 9#
 42#
 42#2
 40#4
 40# 2#
 40# 2# #3
 40#7
 40#2 4#
 41# 4#2
 33#5 2#2 4#
 32#3 2# 3# 4#2
 34# 2#3 #7
 34#5 #3 # 2#
 34# 9# 2#
 34#2 8#4
 32#4 8#2
 32# 2# 9#
 34# 9# 2# 4# 2# #5
 34#2 8#4 4#7 2#
 32#4 8#2 6#2 4# 2#3
 32# 2# 9# 7# 4#5
 32# 2# #3 #5 3#5 4#
 32#7 #3 2# 2#3 2# 4#2
 32#2 4# 3# 2#3 2# 2#7
 33# 4#2 2#5 3#5 # 2#
 32#2 4# 19#
 33# 4#2 18#2
 33#7 16#4
 32#3 # 2# 16# 2#
 36# 2# 16# 2# #3
 36#4 16#7
 36#2 18#2 4#
 37# 19# 4#2`.replace(/(.)(\d+)/g,(_,c,n)=>c.repeat(n))

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totallyhuman
źródło
Ooh, what's the problem now? Erm... Am I missing something?
totallyhuman
1
I didn't downvote, but, fyi, it can be compressed much more using this.
I didn't either, but this isn't in the spirit of the question. The pattern was generated, presumably, by some simple recursive procedure. The challenge is determine the rule, at which point your solution will be extremely short.
Jonah
1
Convert to base 36 to save ~50 bytes.
Shaggy
2
I am completely aware that this is not the optimal solution or the shortest solution or a clever one. I simply tried using an algorithm and it ended up being this. This is, however, a perfectly valid solution and hence doesn't deserve downvotes. Boring solutions should not be upvoted but they shouldn't be downvoted either.
totallyhuman
0

C# (.NET Core), 976 969 bytes

o=>{foreach(var l in new[]{35223185965568,52841249260288,0xf00c0c277f00,0x9004043ee900,0x977c7c200900,0xfee4e4300f00,0xc22727f00c00,73934616658944,0xc2000020097c,73667282210788,0x7f0000f00c27,0xe9000090043e,9895614577696,0xf0000fee430,0xc0000c227f0,4398050918032,0x97cc20000200000,0xfe44300003<<20,0xc277f0000f<<20,0x43ee900009<<20,0x7c20090000970000,-0x1bcff0ffff020000,0x27f00c0000c20000,0x3e90040000430000,9017629528424448,0x300fe4e4430000,0xf00c27277f0000,0x90043e3ee90000,0x977c2020090000,0xfee430300f0000,0xc227f0f00c0000,0x433e9090040000,2097152,3145728,15728640,9437184,9895936,16646144,12713984,4390912,2093088768,3829596160,662634496,1055457280,537460736,806289408,4027318272,2416181248,537463164,806293476,4027321383,2416182334,2541517856,4276413488,3257346032,1128152720,3254779936,1124073520,2130706672,3909091472,150995095,251658494,201326786,67108931})System.Console.WriteLine(System.Convert.ToString(l,2).PadLeft(64).Replace('0',' ').Replace('1','#'));}

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my pronoun is monicareinstate
źródło
Hi, welcome to PPCG! This looks almost exactly the same as @IanH's C# .NET answer.. So if you have any improvements for his, make a comment instead of creating a new answer that is almost exactly the same. As for your question, you can use <s>969</s> to cross out the previous bytes.
Kevin Cruijssen
4
I do not have the reputation needed to comment.
my pronoun is monicareinstate
Please don't try to circumvent the rep requirements.
Shaggy
2
Does this mean that in PPCG I cannot try to help others if my reputation is too low?
my pronoun is monicareinstate
3
@Shaggy to be fair, stackexchange is a bit harsh on newcomers, let's not make it harder for them by assuming malice too quickly
ngn
0

C# (.NET Core), 739 bytes

_=>{var r="";for(int i=0,j,k=0;i<626;i++)for(j=0;j++<@"4#+#$#&#$##'$$&#=$*&&)$#%#&$:&*$($&#$%#):#$#+#)#&'#%##$#:#$##%#'%'&#+#$#:)#%$#$%$#&$*&:$&#%#$%$#$)*$=#&$$'%'##$#+#<$&#5#+#$##'5#&$4$*)$#5)2&*$&#$%2%##$#2#$#+#&'7#$#2#$##%#'&#;&2)#%$#&$:$4$&#%#$);#5#&$$'##$#*#$##'$$&#5#;)$#%#&$4$:$&#$%#)2&;#&'#%##$#2#$#7'&#+#$#2#$##%2%$#&$*&2)5#$)*$4$&#5'##$#+#5#&$<#+#$##'%'$$&#=$*)$#$%$#%#&$:&*$&#$%$#$%#):#$#+#&'%'#%##$#:#$##%#'&#)#+#$#:)#%$#&$($*&:$&#%#$)&&*$=#&$$'##$#&#$#+#^#a$^&^#$#^#$##%Z)[$&#\#&$S'$$&#S%$#%#&$T#$%#)T'#%##$#T#+#$#T$*&R&*$T#$#+#V#+#$#&#$##'F$*&&)$#D&*$($&#$%B#$#+#)#&'C#$##%#'%'&#G)#%$#$%$#&$F$&#%#$%$#$)G#&$$'%'##$#F$&#5#H#&$4$G)2&F%##$#2#$#J#$#2#$##%F&2)G$4$&#H#5#&$"[i]-34;){r+=i%2<1?' ':'#';if(++k%64<1)r+='\n';}return r;}

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Uses the same approach is this other answer.

Charlie
źródło
0

K (ngn/k), 32 31 27 bytes

6{,[|x;""x],'(+|+x),+x}/"#"

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6{ }/ 6 times do

+x transpose

|x reverse vertically

+|+x reverse horizontally

, concatenate vertically

,' concatenate horizontally

,[A;B] is the same as A,B. it helps avoid parentheses around A and around the whole expression

""x use the elements of x as indices in the empty string. out-of-bounds indexing produces spaces, so this expression will return an all-spaces matrix, same size as x

ngn
źródło