# Output the Arecibo Message

The Arecibo message is a 1974 interstellar radio message carrying basic information about humanity and Earth sent to globular star cluster M13 in the hope that extraterrestrial intelligence might receive and decipher it... The message consisted of 1,679 binary digits, approximately 210 bytes...

The number 1,679 was chosen because it is a semiprime (the product of two prime numbers), to be arranged rectangularly as 73 rows by 23 columns. The alternative arrangement, 23 rows by 73 columns, produces an unintelligible set of characters (as do all other X/Y formats).

This is the message with color added to highlight its separate parts. The actual binary transmission carried no color information.

Source: Wikipedia

Your task is to output the Arecibo Message in the exact 23x73 arrangement shown in the image. Any of these output formats is acceptable:

• Text, using one character for ones and another for zeros (using the usual rules for row separation)
• A 2D array of two distinct values
• A 23x73 image with two distinct colors
• Aan uninterrupted stream of 1679 items of two distinct values (i.e. any of the above formats, but flat.)
• A 1679-bit integer. Indicate bit and byte order (endianness) in your solution.

For your convenience, here is a copy-pastable version (also an example output in text format):

00000010101010000000000
00101000001010000000100
10001000100010010110010
10101010101010100100100
00000000000000000000000
00000000000011000000000
00000000001101000000000
00000000001101000000000
00000000010101000000000
00000000011111000000000
00000000000000000000000
11000011100011000011000
10000000000000110010000
11010001100011000011010
11111011111011111011111
00000000000000000000000
00010000000000000000010
00000000000000000000000
00001000000000000000001
11111000000000000011111
00000000000000000000000
11000011000011100011000
10000000100000000010000
11010000110001110011010
11111011111011111011111
00000000000000000000000
00010000001100000000010
00000000001100000000000
00001000001100000000001
11111000001100000011111
00000000001100000000000
00100000000100000000100
00010000001100000001000
00001100001100000010000
00000011000100001100000
00000000001100110000000
00000011000100001100000
00001100001100000010000
00010000001000000001000
00100000001100000000100
01000000001100000000100
01000000000100000001000
00100000001000000010000
00010000000000001100000
00001100000000110000000
00100011101011000000000
00100000001000000000000
00100000111110000000000
00100001011101001011011
00000010011100100111111
10111000011100000110111
00000000010100000111011
00100000010100000111111
00100000010100000110000
00100000110110000000000
00000000000000000000000
00111000001000000000000
00111010100010101010101
00111000000000101010100
00000000000000101000000
00000000111110000000000
00000011111111100000000
00001110000000111000000
00011000000000001100000
00110100000000010110000
01100110000000110011000
01000101000001010001000
01000100100010010001000
00000100010100010000000
00000100001000010000000
00000100000000010000000
00000001001010000000000
01111001111101001111000


If your language, for some reason, has a builtin for the Arecibo Message, you may not use that builtin.

Good Luck!

UPDATE: I accepted the 05AB1E answer since it was the first one to be shorter than the original message. Don't let that dissuade you from new solutions.

UPDATE 2019-09-09: Accepted answer moved to a new 05AB1E answer, as it obsoletes the previous 05AB1E answer. Same point goes as previous update; new solutions still welcome.

• If a language has a built in for Arecibo message, I can tell, with all certainty, that I've seen everything in this world c: Commented Apr 9, 2019 at 18:04
• Mathematica (IIRC) has an image builtin for Lena, so it wouldn't surprise me if it also has an Aricebo message as well. Commented Apr 9, 2019 at 18:06
• @RobertS. no because there are other valid formats besides text. Commented Apr 9, 2019 at 18:27
• In the spirit of the original format, a flat result/output should be allowed. The whole point of 1679 bits is exactly that the proper number of rows and columns can be inferred from the signal length.
Commented Apr 9, 2019 at 19:10
• @LuisfelipeDejesusMunoz Mathematica has a builtin to determine goats in a picture, so an Arecibo message builtin wouldn't really surprise me either.. That language is built on builtins of the universe.. >.> Commented Apr 10, 2019 at 6:47

# 05AB1E, 215210 200 bytes

Saved 15 bytes thanks to Magic Octopus Urn

•cOž¤4é57ñΛ\Ö₃BαöĀíL½₅üBdoÙRθLγ¨G×Tćú$G(˜ƒ¦!€R»SDrµCnJ†d∊ζ·<8‡T@|‹ï=BζćósxG\ÙÎ$¿o₁5/ÔŸÇBûXé-”a::Ž]°∊y;ζ]MÜβ‘иL”β{üÃÇíäc€÷›ÎU=}¨иaŸdY»¾ÚUβ:ô©¦β†₅DGŠβ3Jêθ,äá!ícqšVÖ›lÈΣ¯pε €êÊƒDpÙ/¬Žλ8:ãÿ3=€.Þć•3BY¾4×:


Base-255 encoded trinary string with occurrences of 0000 replaced by 2.

• @MagicOctopusUrn: Thanks! That makes it 210 even :) Commented Apr 10, 2019 at 14:15
• It's actually even better if you replace 0000 with 2 by 9 more bytes. - pastebin.com/aZ6tHxjx for 201 Commented Apr 10, 2019 at 14:19
• @MagicOctopusUrn: Yes, I found that as well and was just about to post it :) Commented Apr 10, 2019 at 14:24
• Cool! Since the Arecibo messages is 210 bytes (23 * 73 / 8 = 209.875), your solution (currently at 200 bytes) is shorter than the message itself!
– J-L
Commented Apr 11, 2019 at 16:51
• I went ahead and made this the accepted answer since it was the first one to be shorter than the message itself. Commented Apr 11, 2019 at 17:42

# Java, 688 678 590 379 361 bytes

Returns a string.

n->new java.math.BigInteger("in95mzupnpa2r0khpoepyql6ioqyn413avucdtfay6indx4wh9dehe3sn18klobtf4z9g9q17umqmwpegr2khb5eqinn7azl4jpfp2a8eui0xfrx5qwrou6gd65jh4ge3ls14k5lu7qrvmg6942ms29u5rb8fa6yrdhfoh5zoi9bdi7uh5ig0u0ff9kounth8sh357x7qox4m3oqviqsbrvakonbka4ahp21bgzi5v1akzzuqoncszhpabbru9q1uo2g11zr73iuyiqr5ikr69zn7cdv7e1lhd6ese9",36).toString(3).replace("2","0000")


-10 bytes by returning the raw stream (old answer)
-88 bytes by using base 10 numerics (thanks @ceilingcat!)
-211 bytes (I knew it could be golfed!) by using a base-36 encoded BigInteger (thanks @JollyJoker!)
-18 bytes by using a different encoded integer (thanks again @JollyJoker)

Try it online!

Explanation:

n->new java.math.BigInteger("base36 string",36) // Decode the base-36 integer.
.toString(3)                                 // Re-encode as ternary
.replace("2","0000")                         // Replace 2 with "0000"
// Implicit return

• Comments are not for extended discussion; this conversation has been moved to chat. Commented Apr 11, 2019 at 22:37

# Jelly, 213 bytes

“H²ɓ¶Ṡḷ€ẹ]ƒf*ḳḢ&ƁṇOḥ{ḄṫwỊ+oLạʋß¢H9¢¹÷ỴɗÇ⁶ƲƙæḊẋ3³=1!VƇƁ'D⁺3Ỵɱ©⁵%fȯez#ƈjƒżṆo.ZF⁶ċṢ⁶ọṛb9Ȯƒd?ƁUĠt4ẇ,ḞġƒµƭfʠƁP§÷øȤŻPɲẋ(¢ß¢(⁽3¶ṙėɗy@ṁYȮL~e⁷ƤĊ§nỊṅµṠ°@7ẠB>Ġ⁻İ}uy¡½:esOpḢt}qS©HÞṬĖṛṇḣ9÷;ESḢ,Ẉ^ṙpƲ©tṃwçnẒṆ¡⁻Jıƒị£-&Ɱ*ẋʂżoȯÑḢɼ’


Try it online!

I played around with Huffman coding, but the improvements in data size were outweighed by the extra code. As such, this is simply a base-250 encoded version of the desired output. Output consists of an integer that when decoded as bijective base 2 will yield the 1D list of 1s and 2s. Thanks @Emigna for pointing out the change in rules.

Try it online - with further decoding to demonstrate output!

If a more conventional binary encoding is preferred, here is one that encodes an integer representation of the inverted binary message. The most significant bit of the integer represents the beginning of the message.

# Brainfuck, 236020081938 1902 bytes

-[>+<-----]>---......+.-.+.-.+.-.+.-............>++[-<+.-.+.-.....>]<..+.-.>++[-<.+.-...+.-..>]<+.-.+..-..>+++[-<+.-.+.-.+.-.>]<.+.-..+.-.>+++[-<............>]<+..-..................>++[-<.+..-.+.-..................>]<+.-.+.-.+.-..................+.....-..>+++[-<..........>]<+..-....+...-...+..-....+..-...+.-.............+..-..+.-....+..-.+.-...+..-...+..-....+..-.+.-.>++[-<+.....-.+.....-.>]++[-<............>]<.+.-......>++[-<...........+.-.................>]<+......-.............+.....-.>++[-<...........>]<+..-....+..-....+...-...+..-...>++[-<+.-.......>]<..>++[-<+.-....+..-.>]<..+...-..+..-.+.-.>++[-<+.....-.+.....-.>]++[-<............>]<.+.-......+..-.........+.-...........+..-...............+.-.....+..-..........+......-.....+..-......+.....-..........+..-..........>+++[-<...+.-.....>]<+.-......+..>++[-<-.......+.>]<.-....+..-......+.-..........+..-...+.-....+..-...............+..-..+..-.............+..-...+.-....+..-.........+..-....+..-..>++[-<....+.-...>]++[-<...+.-.....>]<+.-.......>++[-<+..-........+.-...+.-........>]<.+.-.......+.-.....>++++[-<+.-.......>]++[-<.....+..-....>]<....+..-.........+.-...+...-.+.-.+..-...........>++[-<+.-.......>]<.......+.-.....+.....-............+.-....+.-.+...-.+.-..+.-.+..-.+..-......+.-..+...-..+.-..+.......-.+...-....+...-.....+..-.+...-.........+.-.+.-.....+...-.+..-..+.-......+.-.+.-.....+......-..+.-......+.-.+.-.....+..-......+.-.....+..-.+..-..>+++[-<...........>]<+...-.....+.-..............+...-.+.-.+.-...>++[-<+.-.+.-.+.-.>]<.+...-.........+.-.+.-.+.-.+.-................+.-.+.-..............+.....-................+.........-..........>++[-<..+...-.....>]++[-<....+..-.......>]<+..-.+.-.........+.-.+..-..>++[-<...+..-..+..-....>]<+.-.>++[-<..+.-.+.-...>]<+.-....>+++[-<+.-...+.-..>]<......>++[-<+.-...+.-.>]<...........+.-....+.-.>++[-<...+.-.........>]<+.-..............+.-..+.-.+.-...........+....-..+.....-.+.-..+....-...


Try it online!

My idea was to first generate a simple solution consisting of setting up an ASCII 0 followed only by +-. instructions to produce the output. Repeating sections can then be shortened using loops.

Choosing the optimal set of non-overlapping sections for looping basically comes down to solving a maximum-weight independent set problem. Currently, I just use a greedy algorithm which takes the section that can be optimized the most and then removes all conflicting sections. This is then repeated until no sections are left.

Using a more sophisticated algorithm one can surely achieve better results.

# Piet, 1763 codels

Outputs a stream of 0s and 1s (no line breaks).

Codel size 1:

Codel size 4, for easier viewing:

## Explanation

• First, push a sentinel value of -1 onto the stack.
• Then push the Arecibo message, in reverse order (because it's a stack), using run-length encoding.
• Finally, alternate between two loops, one printing zeroes and the other printing ones.
• The loop counter is the current stack value, decremented until it hits zero, at which point it is discarded and we switch to the other loop.
• Between the zeroes-loop and the ones-loop, check for the sentinel value, exiting if it is found.

## Notes

The program follows a spiral path, clockwise from top left into the centre. The scattered black blocks that roughly follow the diagonals are the flow control. Here's the trace from NPiet.

I've been working on this since the day this challenge went up, but it took a little bit of time to get the message "written" into the picture! I wrote the final loops and the sentinel value first, and then built up the message from the centre outwards. (Since Piet always starts execution from the top left, I expected to have to shuffle and rotate the image around to avoid excess whitespace, but it fit perfectly!)

Fun fact: Run-length encoding in Piet doesn't (by itself) save any space. It takes n codels of one colour to push the value n onto the stack, or n codels of different colours to push that many 1s onto the stack. So it's the same number of codels either way. But the bigger numbers that RLE gives you mean you can use arithmetic tricks (e.g. instead of pushing 9, you can push 3, duplicate, and multiply) to reduce the number of codels, and funny-shaped blocks to fill in available whitespace.

I wasn't sure about how to count score for Piet entries. I found some that seem to count all codels, and others that explicitly only count those actively used. I just counted them all; ignoring white codels (even those that the program never moves through) seems akin to ignoring whitespace in a more typical programming language.

Oh, and I've just now (two hours after posting) realised that I wasted the last bit of time working on this. I wanted to trim off the almost-completely-white last row and column, so I shuffled things around... including the black flow-control blocks. But the edges of the image work the same as black! If I'd just remembered that, I wouldn't have needed to spend so much time puzzling over the intricacies of DPs and CCs...

# C# (Visual C# Interactive Compiler), 366332329 319 bytes

int i;foreach(var g in"*ЀʂЄ࢈ҲપԤ␀␀␀؀ȀȀȀ؀␀␀సؘࠀƐഘؚ྾ߟ␀␀Ā␀␀ྀ␀␀రܘࠈഌΚ྾ߟ␀␀ă␀ྃ␀ȁăÃ1ƀ1ÃĂȃЃЁȂĀÀƀȺ؀Ȃ␀ȏЀȗɛ'Ŀஇ7;ȅ?ȅ0ȍЀ␀␀΂␀ΨՕ΀Ŕ␀ŀЀ?܀àǀƀ̀°٠Ƙѐʈш҈EB@Ѐޟɸ")Write(Convert.ToString(g,2).PadLeft(12-i++%2,'0'));


Replace all instances of ␀ with \0 to test.

Try it online!

# C# (Visual C# Interactive Compiler), 305 bytes, 210 chars

_=>"*ЀʂЄ࢈ҲપԤ␀␀␀؀ȀȀȀ؀␀␀సؘࠀƐഘؚ྾ߟ␀␀Ā␀␀ྀ␀␀రܘࠈഌΚ྾ߟ␀␀ă␀ྃ␀ȁăÃ1ƀ1ÃĂȃЃЁȂĀÀƀȺ؀Ȃ␀ȏЀȗɛ'Ŀஇ7;ȅ?ȅ0ȍЀ␀␀΂␀ΨՕ΀Ŕ␀ŀЀ?܀àǀƀ̀°٠Ƙѐʈш҈EB@Ѐޟɸ".Select((g,i)=>Convert.ToString(g,2).PadLeft(12-i%2,'0'))


Same with above, replace with ␀ with \0 to test. Output as IEnumerable<string>.

Try it online!(Courtesy of Jo King)

• TIO link with NUL bytes
– Jo King
Commented Apr 10, 2019 at 6:47
• I believe the ++ in 12-i++%2 is a nop (at least, it worked for me when I removed it) Commented Apr 10, 2019 at 12:24
• @someone They are a copy-paste from the old answer I forgot to remove Commented Apr 10, 2019 at 14:56
• Your 319 byte version with actual NUL characters: Try it online! Commented Apr 7, 2021 at 19:20

# 05AB1E, 182 bytes

•sv¯ö¨₁ÿ.ÛïžôΔ¨γ_Ígv…=Bм„Ð.(Ü¦i´…ε±G½0^/₃öRÛž¼¤"āêL!ˆ6‘GāÜ‡ðв₁÷Ã7€₂δ¬‚Cć¨g¾†@÷[_-68¯a∍iG*6ÆîÆ;>éjζãÎÂ+ºžnî¼ć'(ÝÞΔ‹∞Ã‰Ý¹Õ5λ₆*a|§oÄmôæ¨;Â—:hž¥ð¢ocË'¨%¡4Ćáß©ìća;FÁ?iˆèεƒʒ•Ž6–FD4‰3ÊD?i-


Try it online! (uses 1 for 0 and 0 for 1, as allowed by the question).

Try it online! (5 bytes longer, 0 for 0 and 1 for 1, added newlines for readability).

Most of the code is a base-255 integer constant N, the rest is an Asymmetric Numeral System decoder, using hardcoded probabilities of 75% / 25% (the actual frequency of 0 is 76.35%, which is so close to 75% that it would only save 1.2 bits in the payload, while the nice and round 75% lets us save several bytes in the decoder).

Ž6–F                  # repeat the following 1679 times:
D                 #  duplicate N
4‰              #  divmod 4: pushes N / 4, N % 4 on the stack
3Ê            #  is N % 4 != 3 ? (boolean 1 or 0)
D?          #  print a copy
i-        #  if it's 1, subtract: N = N - (N / 4)
#  (otherwise, N = N / 4, since that's the top of the stack)


Here's the ANS encoder that generated the constant: Try it online!

Î                          # start from N = 0
Rv         ]              # for each bit in the reversed input:
4*                      #  N *= 4
yi                    #  if the bit is 1:
3+                  #   N += 3
ë                 #  else:
3÷               #   N /= 3 (integer division)
₅B'•.ø        # compress N as base-255

• Good work at obsoleting the previous 05AB1E answer! Commented Sep 9, 2019 at 21:11

# [Python 2], 345 bytes

s=''
for c in")pG/K(K*j$h%kk$ppjGE&I6S6S5[5eCv~vw0x&z$wgqcde$e=G4G?G4eG0e:vv~w*G,gn$wy$uuuuG=G)I,G.I2G(I-eG(I)e-I0G+G+G(G)I*G*vI)G-w'I2y0w'I,vI)G*G)G+G(G*I+W+I+W,G*G(G*G*G*G/I,I+I,iq.G*G1G(e/g$c%sG)m%md~$M(},K(cO)K(eO)K(I)G(aE$M(G1c$hpoI,pG3K1e3eU/M*M,I.I*S,Q(y*y'hG(ng&j$j$G+hW/g'G/G,G1k.d$e$mN":c=ord(c)-35;s+=[bin(c-35)[2:],'0'*c][c<35]
print s


I encoded the length of strings of 0s as a byte starting at chr(31). Then I encoded the remaining 10101 as binary numbers starting at chr(70) up to chr(126). Binary strings that didn't fit were split up into smaller chunks.

Edit: Reduced to 326 bytes. Thanks Jo King

Edit: Fixed a bug in the code generator program

Edit: Final Edit

• You don't need to save the big string in o to a variable.
– xnor
Commented Apr 12, 2019 at 1:16

# Perl 6, 368 bytes

.say for :36('FJXOE0PDDNF5Y5EHGB8M9SWMXQOXIKIT9F6ZKWWDEACHCBGXL1N2H60CN0CJ4EMKF7D6MODSKYJVNR4SFTDR6NSM421LQ67B6MWF0G5BQATFOJJJBQ0UFQM64T0MWSQN41C4S5D1QR5KJM2L9UTYMMKUBBQWY45YCMRGO8ZRGTQH7LXMZBUASLCTKX30IH0AYKYEPHO8HFHX8GAY5WM38YOSUX0HABYSH2PPBLRDRZIN5ANAQ3V8PLOZ6EHC0UI95EVJVYD1820T6J14HGX85NWFQET2NWOMSNUT0JW4LHMY90X094TEE9KXJXSNN6YPERFQW').base(2).substr(1).comb(23)


Try it online!

The long string is the message as a single base-36 number (with a single prefixed 1 bit to preserve the leading zeroes) which is then converted back to binary and printed 23 bits at a time.

• You can use >>.say and &{S/.//} to save bytes. Have you thought about using a different base instead?
– Jo King
Commented Apr 9, 2019 at 22:08
• Yes, I tried using base 65536/Unicode, but the message runs into some forbidden surrogate codepoints. I was able to avoid that by adding a fixed offset to each codepoint, but astonishingly, some of the generated strings crashed Emacs a few times. Resolving the issues started taking more time than I could afford to commit. I'll probably revisit the problem later.
– Sean
Commented Apr 9, 2019 at 22:36
• There's no point going into multiple byte characters, since that increases your byte count. 289 bytes by implementing my tips from above and using base 122 (to avoid carriage returns)
– Jo King
Commented Apr 10, 2019 at 0:58

# Wolfram Language (Mathematica), 383 bytes

StringPartition[Uncompress@"1:eJylVNsRwjAM44s92KBaBTZgAUZhZXqtHUuOWrgjfSRRHFlO4tyer/vjfb1clq0gHirZLRjby986hppcT5p+L3BmgJ3t4Ul4GsNyG++7YbaXLh0ZTPhXa4Sn+X/s9Qfk3Hx2cOaSIuNYaVu5laschvgzSqAjHeZBhilKgKBDEhw0upJRg+HOK4MyNC29sfbc3RV0VPDqeLiRTsG1ulExq1IitpunOa7asnYM9siDZ6eidUCkEzBOUbCkGIig4aTyUGBYWAX6W6aXIWGGI/HlhmsqzSU0QTZjkMVpaX5sBsm1OGKVg1qdjKP0EdyqZBRLhukn8DLBQav6kccgz8OKfgBzjj6Z",23]


Try it online!

# Node.js, 333 bytes

Returns a binary string of 1,679 characters.

_=>Buffer(")SI)=.);1K?>>>2333A3,93/I3>3)g33)AEAAI)5JQZJTddda3)*3*33+3,e)*e3)//0/1+)1C/7Cgggg3395)9)A3IY)h*IH),39+)995*)AA-)59Y)*O3Z,)//*)91**)A*-)Y+1)I11+)I1)/)5)Y*0?)+)I)-0Y)1@;_*7<gaE/a)Q7[*9HM+IY16I33)a1)*^.><I+S3.38I)*hY)7)a)Y)A,9*A5Y/A:9=9K:1I-=9)19I)9*329)GH<").map(x=>s+=(x-51&&x-41).toString(2).padStart(x-51?6:12,0),s='')&&s


Try it online! (with formatted output)

# JavaScript (ES8), 413 bytes

Returns a binary string of 1,679 characters.

_=>atob('AsKoAFBQEiIlwpVVUk!!ABwo!DQ!Gg!V!Aw7g!!GHGHCi!yGjHChsK+w7vDr8K!!Q!Q!!g!/DgAHDs!AGGHDhiAgCGhjwprDu8Ovwr4!ABAw4AQAcK!MKDAD8GB8OAD!QCAQQMBAYYEAMQwABwpgAMQwBwoYEBAgEEBgEQDAIwoAgIMKAwoDCgMKABgDDgDAEdcKACAgAEHwAIXTCtgTDpMO+w6HDgcK4AsKDwrIFB8OkCgwINg!!A4IAB1FVTDoArCo!U!/CgAB/w4ADwoDDoAwAYDQBYMOMBmEUFEIkSMKARR!woQgAQB!MKUAHnDtA'.split!.joinAA).replace(/[\s\S]/g,c=>c.charCodeAt().toString(2).padStart(8,0))


Try it online! (with formatted output)

# Bubblegum, 275 236 bytes

00000000: e006 8e00 e45d 0018 6988 6507 a228 f86f  .....]..i.e..(.o
00000010: f042 c62f d4d7 b99e 38bc 56c4 52e8 2630  .B./....8.V.R.&0
00000020: 8aaa 7252 d47d 5ef4 c96a 511f 6842 423f  ..rR.}^..jQ.hBB?
00000030: 4532 ca9f 22d3 1633 e0c4 665a d5dc 4e68  E2.."..3..fZ..Nh
00000040: 7b09 76ae 3c7e f9d4 fa4a 05e0 4163 c580  {.v.<~...J..Ac..
00000050: c585 a383 2396 4ca9 1f48 a4b9 744e 37c8  ....#.L..H..tN7.
00000060: 68c5 af23 645d 59a7 542a e6d1 23b9 3aba  h..#d]Y.T*..#.:.
00000070: f0e6 2738 dfd5 b0a3 c6a3 60bf c5b6 5ae6  ..'8.........Z.
00000080: 7893 30a8 ae04 edf9 298b b777 4d56 285b  x.0.....)..wMV([
00000090: cb74 07cc 7a7b a399 3dc7 c6e7 b693 e715  .t..z{..=.......
000000a0: d908 876e 001f 7408 3c6a 5fcd 37cb 02c4  ...n..t.<j_.7...
000000b0: 93de 33c2 a11e 5bac cd12 d99a fac3 e0fa  ..3...[.........
000000c0: 5268 94f7 d640 0f73 cede f79d 821f 39d1  [email protected].
000000d0: dc49 ff06 6962 6c31 dc29 a077 01c3 7690  .I..ibl1.).w..v.
000000e0: 85ef bbec 31d7 5c7f f9fc 8c00            ....1.\.....


Try it online!

# bash + GNU tools, 351 bytes

base64 -d<<<H4sIAPnNrVwCA6WUCRLDIAwDv8T+/3NNG4wvkTBTcisGSyA8xrcxj9Ds02F+Z7yuf3hnPyz0vYEGz+FG3IKBs+x3oL2PSh0TM/PnaGamft9nPUCew3uCp5RBWdRKGz+qNJn8qRKkkNaTBgeVIFXWOdi8VCaIeUnsfHo6TXpaFa3H5olf6J5MuIHLoEi0uKcRFCvEXG4xseglKzZg7kpYJSLMA3M7wXKR+/L2WiK0kvg+TDASLp6Co1KEVIlVmFzhCktRhBBDdSZYU1xKHrmDUllcWpNR/YNW2QNcHtF0rSySb0MXk/SDUgEwG5gfLvQDxuEdDo8GAAA=|gunzip


TIO

• didn't see your bash post before I posted mine - so much smaller! Commented Apr 10, 2019 at 13:29

# MathGolf, 223 220 bytes

'06*"x·<×J× Ç×►B×ê\"@$↕!◙è0♥f░×→0×♠p└☺α?×└•×◙×P♣p¬è:×►∟××××←♦♠♣≥â☻┘A☺▄αp⌂r☻[║►×>◘×♦♦└:☻↑×♥@@@@►►@♦↑ ☻♀◘☻♦☻☻├×å↑×╠×Ç!♠ 0♀◘↑◘☻♦◘×♠α♥â▼ÇA×└×◘ ×××▀≈}═14♦►►π0♀××°×α•×►×××☻×◘××└≈}_├↑♪↓×─0♫♥×××|××*××♪×Ç♠×└×××× ⌐¬╩↕◄○((×T☺"$à+


Try it online!

## Explanation

'0                push single character "0"
6*              repeat 6 times
"..."         push the magic string
$convert to ordinal à convert to binary string + pop a, b : push(a+b) (add the 6 zeroes) below is the footer L/ divide into groups of 23 characters n join array with newlines into string  • You can move the L/n to the footer, so it's actually 220 bytes. Can more bytes be saved by porting the 05AB1E / Java answers (using this compressed integer, convert it to base-3, and replace all 2s with 0000s)? Commented Apr 12, 2019 at 12:11 • @KevinCruijssen nice catch! I wish I had more string replacement functions, but that's not part of MathGolf right now. – maxb Commented Apr 12, 2019 at 16:18 • How about a map per digit, which maps 2 to ♫░╞? EDIT: Never mind. I see you don't have a Base Conversion builtin (except for binary/hexadecimal) to convert to base-3? Commented Apr 12, 2019 at 16:19 • @KevinCruijssen I have thought a bit on how I should change the base conversion. I certainly don't need one command for converting binary string to decimal, and another for converting a binary array to decimal. That way, I can fit another base conversion command in (and the idea has always been to have general base conversion in the language). – maxb Commented Apr 14, 2019 at 13:51 • Surely you can move the + to the footer as well – Jo King Commented Apr 20, 2019 at 22:30 # Python 3, 331 bytes exit(''.join(bin(i)[3:]for i in b'UP@JBIDQKJjjd@@@@@L@@Ah@@CP@@J@@_@@@@@LNLLP@FPtXpu}}}|@@@@@@@@@A@@A~@@~@@@CCCcDA@DMCGM____@@@@HF@H@L@@PX@_pOA@HA@HHF@LLB@FHX@@s@@XaCC@HD@L@b@XAD@PDDA@PD@C@F@X@ck@A@P@BCx@DKi[@gI\x7f\\NC\\@TGY@hOrAPXDFp@@@@@\\D@@zbjipAU@@B@Gp@@\x7fx@G@\\@X@LAh@lFXCLHhJHQHdPBJH@DHP@H@@Dh@OOix')[1:])  Try it online! # Perl 5, 460 bytes printf"%023b",oct"0x$_"for qw/15400 141404 4444b2 555524 0 600 1a00 1a00 2a00 3e00 0 61c618 400190 68c61a 7df7df 0 80002 0 40001 7c001f 0 618718 404010 68639a 7df7df 0 81802 1800 41801 7c181f 1800 100804 81808 61810 18860 1980 18860 61810 81008 101804 201804 200808 101010 80060 60180 11d600 101000 107c00 10ba5b 1393f 5c3837 283b 10283f 102830 106c00 0 1c1000 1d4555 1c0154 140 7c00 1ff00 701c0 c0060 1a00b0 330198 228288 224488 22880 21080 20080 9400 3cfa78/


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# Python 2, 336 bytes

print bin(int('gVQAKCgJERLKqqkgAAAAAMAABoAADQAAKgAAfAAAAAMOMMQAGQ0Yw19998AAAAgAAgAAABAAB+AA+AAADDDjEBAENDHNffffAAAAIGAIAMAAQYAfgwPgBgAIBAIIGAgMMCAGIYAAzAAYhgDDAgIEAggMAiAYBEAQEEBAQEADAGAYAjrABAQACD4AELpbAnJ/cODcAUHZAoPyBQYEGwAAAAAcEAA6iqpwBVAACgAHwAA/4AHAcAYAMBoAsGYDMIoKIRIkQCKIAEIQAIAgAEoAPPp4'.decode('base64').encode('hex'),16))[3:]


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Prints a string of bytes

# Perl 6, 276 bytes

:122[q{3tD!gLF['u#.K>'uCG8cvJZ[on1Z<D!	Fhqq-_V'"qQn+n0h\:b,vXjo&1TMFaW;wvX;eUS
(ueelNSu,y93\kjGI&?UU/38 :ndi4Y5cyC+ME\g7LWaS;QLjtD^L+aVd(XW%gy\8'Eqk-+f72S5J?(r5!m^5px T[Z'3&jd0lZ'/x%#(}.ords].base(2)~~S/.//.say


Try it online!

Outputs as a series of 1679 0s and 1s. You can have it on different lines by adding .comb(23)>> before the say.

### Explanation:

:122["...".ords]                     # Convert the string from base 122
.base(2)             # To base 2
~~S/.//      # Remove the leading 1
.say  # And print


I can probably save bytes by using the output as a 1679 bit integer instead, or reversing the bit representation.

# Zsh (+coreutils), 320 577 bytes

It runs on my Mac, but try it online doesn't work because TIO doesn't have basenc :(
It's a one-line program, I've added newlines here for readability. Data compressed to z85 encoding (similar to Ascii85).

basenc -d --z85<<<'0&M8}p.B@#L#QZE00001Fb*2U02+?qq#={[000069asT6g8.)4ZwEwH00
0Mg1onA4aoh}/0Q{6?7?7Abap8T<N!#M/0000:ZZ{J%0d$no1{.{L01Yzs5hjUju{Cb/00eC!f:[ 8q1oX#k5eUKnfBigKaB[#)FcD1?fA=JG2NyzwD&LsiWGLW<&LA29Gr+T=<o=q3huiZS05#83BRN) l3zmwk01TOWF4#ep&0YP[gY.1d24nhWll]9LmgYDLaoqEc0f^Fe]ZR$Y'|basenc --base2msbf
-w0|cut -c-1679


Encoding steps
. Convert Arecibo message to a digestible format → bitstr.8
. Convert bitstr.8 into binary, using script demo16.sh**bits.16.again
. Convert binary to z85basenc --z85 bits.16.again >bits.85
. Use the z85-encoded string in bits.85 for a much shorter code golf!

Decoding steps -- i.e. what the main script does
. basenc -d --z85<<< decodes the z85 string to binary
. |basenc --base2msbf -w0 decodes binary to Ascii 1s and 0s
. |cut -c-1679 trims extra data

#### Original 577-byte solution

try it online!!

S=fxxxxibxxdxxfyzzzyx1yxxxxxxxxyyywl2ij1xxhj1xxhixxxhi5iw2d3c2d2cxl2bxc2az2c2d2ax4x4x4x5wcxpxwdxp15m5w2d2d3c2cxfxhxc1xxc2c2y1xx4x4x4x5wc1f2ixj2kd1e2j15e2f5j2kb1h1hyc1f2gzd2d2fzaf2c1d2ej2b2gf2c1d2ed2d2f1dc1f1hzb1g2hya1h2hya1i1gzb1g1g1dc1l2ed2h2gbz2xx2ibzd1lbzb5jb1dx2xyx1x2fy2yy6x3d3e2a3ixxd2x2b1fxxd6b1fxxd2db1e2a2jwb3e1lb2xxzxxxxx1b3ixxxynxxeh5jf9hd3g3fc2k2eb2a1ix2da2b2g2b2cazxxdxzzazyzyzzezxz1ge1d1d1ge1i1ggyxxia3y4xy3z
for X in ${(s::)S};{case$X in
[a-w])V=$[##$X-96];printf 0%.s {1..$V};;[1-9])V=$[##$X-48];printf 1%.s {1..$V};;[x-z])V=$[##$X-119];printf $[10**$V];;esac}


Used custom encoding logic. The string S is 421 characters, could possibly be compressed a bit more. Letters a-w represent repeated 0s. Numbers 1-9 represent repeated 1s. Letters x y z represent 10 100 1000 respectively.

Maybe I should have tried byte-pair encoding or Ascii85 --> Done. ✅

# Bash, 702 697 bytes

xxd -r -p<<X|xz -dc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X


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# Ruby, 362 bytes

puts"5r0afnfm8wyke8tfy1pwt7xnuaxyh3wodfp7bhsdufyw0xbdp1pumrz2xir652tuc0ss9oec8yad9vefivd66j126wybhefgk2lv38uqqiur11u26q275jk3h2ucithd59awpaenqpqi1pszh52179zw0ddqtbrvo6kyrrgv8c34pqrp83j8estjp63v29t4hqp9yg2hhzjlq1e9zqx6gh20n9lsttimz3nbq060ritrphxaru7quwmv3oujhd9xjddpbacq4bnpf270znhgto59yn0980itylf95pxw9x7rvkvi7mfql1sx46puo8rg4dq0".to_i(36).to_s(2).rjust(1679,?0)


Integer written in base 36. There's surely a more efficient way to compress the integer, e.g. with zlib or base64.

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• 350 bytes if you: (1) use "%b" for formatting instead of to_s, (2) prepend ?0*6 instead of calling rjust
– G B
Commented Sep 25, 2019 at 7:19

# [C++ (VC++)(but tested with gcc as well)], 585 bytes

#define l(x,y)for(int x=0;x<y;x++)
void f(){const char*a="02A800505012Y595Y240U180YD0Y1A0Y540YF80V61C618800321A3186BEFBEF80X10Y40W20YFC001F0X1861C620200868639AFBEFBE0W40C0100180Z83003F0607C00C001008041030101860400C430Z19800310C018604040804101804403008802020808080800600C030047580080800107C002174B604E4FEE1C1B80283B20507E40A0C08360U3820Z751554E00AA0Z140ZF80Z7FC00380E00C0060340160CC06611414422448804510Z8420010040Z940079F4F0";int x=1679;l(i,365){int d=a[i],c=0;d-=(d>47&d<58)?48:((d>64&d<71)?55:0);if(d>70&d<91)c=91-d,d=a[i-1];for(c;c>=0;c--)l(j,4){if(x--)cout<<(int)((d&(8>>j))>0);}}}



Try it online!

ungolfed Version (lacks the break after the 1679th element though and goes until the 1680th):

#include <stdio.h>
#include <iostream>
using namespace std;
int main()
{
const char arecibo[]="02A800505012Y595Y240U180YD0Y1A0Y540YF80V61C618800321A3186BEFBEF80X10Y40W20YFC001F0X1861C620200868639AFBEFBE0W40C0100180Z83003F0607C00C001008041030101860400C430Z19800310C018604040804101804403008802020808080800600C030047580080800107C002174B604E4FEE1C1B80283B20507E40A0C08360U3820Z751554E00AA0Z140ZF80Z7FC00380E00C0060340160CC06611414422448804510Z8420010040Z940079F4F0";
int i = 0,j;
while (i<sizeof(arecibo)-1)
{
char digit = arecibo[i];
int count=0;
if (digit >= '0' & digit <= '9')
{
digit -= '0';
}
else if (digit>='A'& digit<='F')
{
digit -=  'A'-10;
}
else if (digit > 'F'&digit<='Z')
{
//digit does not contain any valid hex digit in this case
count = 'Z' - digit+1; //digit = Z means repeat 2 more times...
digit = arecibo[i - 1];
}
for (count; count >= 0; count--)
{
for (j = 0; j<4; j++)
{
cout << (int)(((digit)&(8 >> j))>0);
}
}
i++;
}
return 0;
}


as an Explanation: i concatenated the 73 lines of sample output given to one long line. i encoded them in hexadecimal where the bit order is msbfirst (using this program https://github.com/Marc-Bender/longBinaryStreamToHex/releases/download/addedErrorCode-4/longBinaryStreamToHex.exe) i shortened the Output of that by about 70 hexadecimal Digits by using the letters 'G'-'Z' as a sign to repeat the last Digit for a certain amount of times (Z = 2more times, Y = 3more times …) the rest should be relatively self explainatory for Code-Golfers. abusing the preprocessor to shorten loops, abusing the , Operator and the like.

Output Format is uninterrupted stream of 1679 0/1-values.

• before anybody asks the const is for compatibility with VC++ (what the main development work was done with) Commented Apr 10, 2019 at 21:41
• @ceilingcat i think you can even short that more by pitting your include in the header section... Commented Apr 12, 2019 at 21:20
• 554 bytes Commented Jul 28, 2019 at 20:37

# Perl 6, 348 bytes

{"000000"~:36<5r0afnfm8wyke8tfy1pwt7xnuaxyh3wodfp7bhsdufyw0xbdp1pumrz2xir652tuc0ss9oec8yad9vefivd66j126wybhefgk2lv38uqqiur11u26q275jk3h2ucithd59awpaenqpqi1pszh52179zw0ddqtbrvo6kyrrgv8c34pqrp83j8estjp63v29t4hqp9yg2hhzjlq1e9zqx6gh20n9lsttimz3nbq060ritrphxaru7quwmv3oujhd9xjddpbacq4bnpf270znhgto59yn0980itylf95pxw9x7rvkvi7mfql1sx46puo8rg4dq0>.base(2)}


Based on Benjamin Urquhart's Java solution.

Uses a straight stream of 0 and 1 characters. The link below has some code to prettify the output.

Try it online!

# Tcl, 366 bytes

binary scan [binary decode base64 QBUACgpIRKSpqkoCAAAAgAEAsAAAWAAAKgAAHwAAAGA4hhEATFiMYX3f9wEAAAgAIAAAAAQA8AOADwAAGIZjBAQQFsZZ3/d9AAAAAgMIgAEAwQD8YOADMAAIECAIDAgYBgIwwgCAGQCMMIBhICAQIAgYIAIMEAEEBAEBAQFgAAMMIK4BEBAACD4AhC5tICd/h4MdQMFNoOAnUDAQbAAAAAAcBACuqCoHUAUAKADwAQD+A8ABBzAABiyABjNghigoQiQSAaIIACEEgAACACkAni8P] b* z
puts [join [regexp -all -inline .{23} \$z] \n]


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# C++ (with Gnu multi-precision library), 359 bytes

This outputs the string as one line. It uses '1' for 0, and '0' for 1 :/

It simply reads the embedded string as base 62, and prints it as base 2.

Use g++ -g arecibo.cpp -lgmp -lgmpxx to compile and link

#include<gmpxx.h>
main(){mpz_out_str(stdout,2,class_mpz("vuXXKBYAu1hPsJPbFSf49akyFd0bjJbMIV3upYndU8kYFPsXcpRUK6c9qnzLfORxkxGL7ZfoSwgSxFbCrydjHL603QcxexopAzoYAcAyEiENJJU2vQA2zM8NDbeY6nHgL8rfFoPJaclwxx6TeKeOEknEQHzp7C4la3o8xijBQLOVweFZ1CI9dXy2VQhgnuwF5FeW5mQ909pRMxhn6G3RJ1QEtkS7oGMZYHM03fh4fj391IoYLGVv3iUVuTrV2ghz9CUa9hfGYrdhGoVE2w1inYalYl",62).get_mpz_t());}

• I couldn't get this to work until I replaced class_mpz with mpz_class Commented Apr 23, 2019 at 6:41

# C++ (gcc), 748 bytes

#define l(X,Y)if(z[i]==#X[0])z.replace(i,1,#Y);
void f(){std::string z = {"ab1eeedala1ebeeheal1mmma1fa1g1eeeeeeea1a1alddhgdbcdgdacedgdacedgdeeedgdndgddhgqiafbcag1dbfa1blceafafbcegcinnnlddhgmddegddhgb1ddelcidbnlddhgqqiafag1hedeblcebcaf1acegcinnnlddhgmhcdegdacdagb1bfda1lcibfhcildacdaga1d1d1almhcheagbqch1blhcmbqgdacachghcmbqgbqch1blmh1d1aga1hfd1aledcd1aledeheaga1heheblmdbqgbcdchga1af1efdga1hedbla1bndala1b1f1ea1fflh1aia1acccl1f1bibff1ldeebf1fla1h1ebfccla1h1ebfbla1bffdalddhgaibedblaieemeeeeelaideeeealdh1ehldcidalhcccidlbihf1hlafdafbgacedefblfachfacagemebeemagema1ma1magbememhgbeb1b1hgbedehghea1edalfcacieacca0"};for(int i=0;i<1751;i++){l(q,cb)l(n,fi)l(m,ae)l(i,c1)l(h,ba)l(g,0l)l(f,0c)l(e,01)l(d,bb)l(c,11)l(b,aa)l(a,00)l(l,\n)printf("%c",z[i]);}}
`

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By replacing the most used substring with a new character until it's not worth it anymore

• 735 bytes Commented Jun 1, 2019 at 23:26