Tribute to Stan Lee
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Unfortunately one of the greatest comic book writers passed away yesterday afternoon. Lots of Hollywood stars, musicians, actors, and many other people are paying tribute to this awesome writer, so we must do something as well.
Challenge
Print The Avengers Logo
Note: You may use any other character in place of # other than a space character; while you must use a space character for the space
In ASCII-art
######
###############
##### ##########
#### ####### ####
### #### ### ###
### #### ### ###
### #### ### ###
### #### ### ###
### #### ### ###
### #### ### ###
### ########### ###
### ########### ###
### #### ### ###
### ### ### ###
#### #### ####
######## ######
#################
###
Optional
Who was (in your opinion) his greatest hero and villain in the entire marvel universe?
Standard code-golf rules apply
code-golf ascii-art kolmogorov-complexity
asked Nov 13 '18 at 17:46
Luis felipe De jesus MunozLuis felipe De jesus Munoz
4,16421255
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show 1 more comment
39
$begingroup$
Unfortunately one of the greatest comic book writers passed away yesterday afternoon. Lots of Hollywood stars, musicians, actors, and many other people are paying tribute to this awesome writer, so we must do something as well.
Challenge
Print The Avengers Logo
Note: You may use any other character in place of # other than a space character; while you must use a space character for the space
In ASCII-art
######
###############
##### ##########
#### ####### ####
### #### ### ###
### #### ### ###
### #### ### ###
### #### ### ###
### #### ### ###
### #### ### ###
### ########### ###
### ########### ###
### #### ### ###
### ### ### ###
#### #### ####
######## ######
#################
###
Optional
Who was (in your opinion) his greatest hero and villain in the entire marvel universe?
Standard code-golf rules apply
code-golf ascii-art kolmogorov-complexity
asked Nov 13 '18 at 17:46
Luis felipe De jesus MunozLuis felipe De jesus Munoz
4,16421255
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3
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What happened to the → in Ⓐ?
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– Adám
Nov 13 '18 at 20:00
3
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@Adám 2 Reasons. I believe that the arrow means the continuation of the series (in this case the comics about avengers). Since Stan passed away there wont be a continuation of it (In my opinion, it is not the same without Stan). That is one of the reasons
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– Luis felipe De jesus Munoz
Nov 13 '18 at 20:05
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The second one is that I didnt know how to draw it :c
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– Luis felipe De jesus Munoz
Nov 13 '18 at 20:05
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Are the trailing spaces required?
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– Scoots
Nov 14 '18 at 12:40
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@Scoots no, it is not required as long as your output looks the same
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– Luis felipe De jesus Munoz
Nov 14 '18 at 12:47
|
show 1 more comment
39
39
39
8
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Unfortunately one of the greatest comic book writers passed away yesterday afternoon. Lots of Hollywood stars, musicians, actors, and many other people are paying tribute to this awesome writer, so we must do something as well.
Challenge
Print The Avengers Logo
Note: You may use any other character in place of # other than a space character; while you must use a space character for the space
In ASCII-art
######
###############
##### ##########
#### ####### ####
### #### ### ###
### #### ### ###
### #### ### ###
### #### ### ###
### #### ### ###
### #### ### ###
### ########### ###
### ########### ###
### #### ### ###
### ### ### ###
#### #### ####
######## ######
#################
###
Optional
Who was (in your opinion) his greatest hero and villain in the entire marvel universe?
Standard code-golf rules apply
code-golf ascii-art kolmogorov-complexity
asked Nov 13 '18 at 17:46
Luis felipe De jesus MunozLuis felipe De jesus Munoz
4,16421255
$endgroup$
Unfortunately one of the greatest comic book writers passed away yesterday afternoon. Lots of Hollywood stars, musicians, actors, and many other people are paying tribute to this awesome writer, so we must do something as well.
Challenge
Print The Avengers Logo
Note: You may use any other character in place of # other than a space character; while you must use a space character for the space
In ASCII-art
######
###############
##### ##########
#### ####### ####
### #### ### ###
### #### ### ###
### #### ### ###
### #### ### ###
### #### ### ###
### #### ### ###
### ########### ###
### ########### ###
### #### ### ###
### ### ### ###
#### #### ####
######## ######
#################
###
Optional
Who was (in your opinion) his greatest hero and villain in the entire marvel universe?
Standard code-golf rules apply
code-golf ascii-art kolmogorov-complexity
code-golf ascii-art kolmogorov-complexity
asked Nov 13 '18 at 17:46
Luis felipe De jesus MunozLuis felipe De jesus Munoz
4,16421255
asked Nov 13 '18 at 17:46
Luis felipe De jesus MunozLuis felipe De jesus Munoz
4,16421255
edited Nov 14 '18 at 20:35
Luis felipe De jesus Munoz
asked Nov 13 '18 at 17:46
Luis felipe De jesus MunozLuis felipe De jesus Munoz
4,16421255
asked Nov 13 '18 at 17:46
Luis felipe De jesus MunozLuis felipe De jesus Munoz
4,16421255
asked Nov 13 '18 at 17:46
Luis felipe De jesus MunozLuis felipe De jesus Munoz
4,16421255
4,16421255
3
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What happened to the → in Ⓐ?
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– Adám
Nov 13 '18 at 20:00
3
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@Adám 2 Reasons. I believe that the arrow means the continuation of the series (in this case the comics about avengers). Since Stan passed away there wont be a continuation of it (In my opinion, it is not the same without Stan). That is one of the reasons
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– Luis felipe De jesus Munoz
Nov 13 '18 at 20:05
43
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The second one is that I didnt know how to draw it :c
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– Luis felipe De jesus Munoz
Nov 13 '18 at 20:05
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Are the trailing spaces required?
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– Scoots
Nov 14 '18 at 12:40
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@Scoots no, it is not required as long as your output looks the same
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– Luis felipe De jesus Munoz
Nov 14 '18 at 12:47
|
show 1 more comment
3
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What happened to the → in Ⓐ?
$endgroup$
– Adám
Nov 13 '18 at 20:00
3
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@Adám 2 Reasons. I believe that the arrow means the continuation of the series (in this case the comics about avengers). Since Stan passed away there wont be a continuation of it (In my opinion, it is not the same without Stan). That is one of the reasons
$endgroup$
– Luis felipe De jesus Munoz
Nov 13 '18 at 20:05
43
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The second one is that I didnt know how to draw it :c
$endgroup$
– Luis felipe De jesus Munoz
Nov 13 '18 at 20:05
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Are the trailing spaces required?
$endgroup$
– Scoots
Nov 14 '18 at 12:40
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@Scoots no, it is not required as long as your output looks the same
$endgroup$
– Luis felipe De jesus Munoz
Nov 14 '18 at 12:47
3
3
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What happened to the → in Ⓐ?
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– Adám
Nov 13 '18 at 20:00
$begingroup$
What happened to the → in Ⓐ?
$endgroup$
– Adám
Nov 13 '18 at 20:00
3
3
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@Adám 2 Reasons. I believe that the arrow means the continuation of the series (in this case the comics about avengers). Since Stan passed away there wont be a continuation of it (In my opinion, it is not the same without Stan). That is one of the reasons
$endgroup$
– Luis felipe De jesus Munoz
Nov 13 '18 at 20:05
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@Adám 2 Reasons. I believe that the arrow means the continuation of the series (in this case the comics about avengers). Since Stan passed away there wont be a continuation of it (In my opinion, it is not the same without Stan). That is one of the reasons
$endgroup$
– Luis felipe De jesus Munoz
Nov 13 '18 at 20:05
43
43
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The second one is that I didnt know how to draw it :c
$endgroup$
– Luis felipe De jesus Munoz
Nov 13 '18 at 20:05
$begingroup$
The second one is that I didnt know how to draw it :c
$endgroup$
– Luis felipe De jesus Munoz
Nov 13 '18 at 20:05
$begingroup$
Are the trailing spaces required?
$endgroup$
– Scoots
Nov 14 '18 at 12:40
$begingroup$
Are the trailing spaces required?
$endgroup$
– Scoots
Nov 14 '18 at 12:40
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@Scoots no, it is not required as long as your output looks the same
$endgroup$
– Luis felipe De jesus Munoz
Nov 14 '18 at 12:47
$begingroup$
@Scoots no, it is not required as long as your output looks the same
$endgroup$
– Luis felipe De jesus Munoz
Nov 14 '18 at 12:47
|
show 1 more comment
22 Answers
22
active
oldest
votes
25
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Jelly, 62 59 58 bytes
“ḢE{ɠs{Ƒ0Ṇṁỵ8ỊṢṂƊeLṫfIWẈḞ'ʠJ£ṗɱçoȧ?ƒnØẆƥṂ⁷ɱ’b12ĖŒṙḂa⁶s27Y
Thanks to @JonathanAllan for golfing off 1 byte!
Try it online!
How it works
“ḢE{ɠs{Ƒ0Ṇṁỵ8ỊṢṂƊeLṫfIWẈḞ'ʠJ£ṗɱçoȧ?ƒnØẆƥṂ⁷ɱ’
is a bijective base-250 integer literal, which uses the characters in Jelly's code page as digits.
The literal encodes the integer $scriptsize 10250938842396786963034911279002199266186481794751691439873548591280332406943905758890346943197901909163$, which b12 converts to duodecimal, yielding $scriptsize 30b620b40ba54a64771433841333139413468423468423467433467433466b466b465453465363431424b43896860ba3_{12}$.
There are some zeroes, because 14 spaces (e.g.) are encoded as 3 spaces, 0 hashes, and 11 spaces. This maintains the base small (the largest run consists of 17 hashes), without adding any additional logic to the decoder.
Ė (enumerate) prefixes every base-12 digit by its 1-based index, then Œṙ performs run-length decoding.
Now, for each integer in the resulting array, Ḃ extracts the least significant bit, then a⁶ (AND space) replaces ones with spaces, while leaving zeroes unchanged.
Finally, s27 splits the unformatted character array into chunks of length 27, which Y separates by linefeeds.
answered Nov 13 '18 at 21:22
Dennis♦Dennis
187k32297737
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add a comment |
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R, 198, 173, 163, 157, 144, 142 bytes
write(rep(rep(c(' ',4),44),utf8ToInt("NFMOJEDJFDGGADCCHDACCCACIDACDFHDBCDFHDBCDFGDCCDFGDCCDFFKDFFKDFEDECDFECFCDCADBDKDCHIFHQJC")-64),1,27,,'')
Try it online!
Also R honors to the great Stan.
Notes :
- instead of the character
#we've used the character4because saves 2 bytes and :
- it's a reference to the Fantastic Four and/or the next Avengers 4 movie
- it like we're repeating the logo inside the logo (since
4seems a mini "slanted"A)
- -15 bytes thanks to @Giuseppe
Explanation :
Applying a Run Length Encoding (rle function) on the characters of the string (excluding 'n') returns 88 repetitions of the alternating <space>,<hash>,<space>,<hash>,<space>,<hash>....
The 88 repetitions are all values in the range [1,17] hence, summing 64 we get the code points of the letters [A...Q] obtaining the string : "NFMOJEDJFDGGADCCHDACCCACIDACDFHDBCDFHDBCDFGDCCDFGDCCDFFKDFFKDFEDECDFECFCDCADBDKDCHIFHQJC"
Hence the code basically does the opposite operations :
v <- utf8ToInt("NFMOJEDJFDGGADC...")-64 # convert the string back to [1...17] vector
e <- rep(rep(c(' ',4),44),v) # perform the inverse of rle function obtaining an
# expanded vector of <space>'s and <four>'s
write(e,1,27,,'') # write the vector to stdout arranging it
# in rows of maximum 27 characters
answered Nov 13 '18 at 18:48
digEmAlldigEmAll
3,029414
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Your solution is nicer, but your own gzip 2-parter gives 89+47 = 136 bytes.
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– J.Doe
Nov 14 '18 at 19:43
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@J.Doe : ahah I even forgot to have used that approach! please post it as your own answer :)
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– digEmAll
Nov 14 '18 at 19:51
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Since we can use any character for#, using0in place of"#"should save 2 more bytes. @J.Doe that may work better for your gzip as well.
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– Giuseppe
Nov 14 '18 at 20:50
add a comment |
12
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Python 2, 129 bytes
00000000: 2363 6f64 696e 673a 4c31 0a70 7269 6e74 #coding:L1.print
00000010: 2778 da95 9141 0e5c 3021 0803 ef7d 4593 'x...A.!...}E.
00000020: feff 8f9b 5d14 5c6e f1b0 f622 7422 2890 ....].n..."t"(.
00000030: 2e7d 5a09 dc4b 19cb bc53 84d1 4a6a 5960 .}Z..K...S..JjY`
00000040: 116e 3f42 c290 b3f7 c0bc 76cf 549d 6ed8 .n?B......v.T.n.
00000050: f8fa 5f26 0b0e 8c93 d5cb 35f6 b1e7 a939 .._&......5....9
00000060: 9e98 e769 47b9 87d6 cdf5 5c30 3030 32c0 ...iG.....002.
00000070: 4029 272e 6465 636f 6465 2827 7a69 7027 @)'.decode('zip'
00000080: 29 )
Try it online!
answered Nov 13 '18 at 23:16
LynnLynn
49.5k794227
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129 bytes, I removed the trailingnfrom the compressed string.
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– ovs
Nov 14 '18 at 20:46
add a comment |
12
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Bubblegum, 71 66 65 64 bytes
0000000: 95 91 21 12 00 20 08 c0 3a af 90 ff 7f 52 4e 58 ..!.. ..:....RNX
0000010: 58 93 25 d8 8a 27 47 e4 23 66 01 2c ce b1 3c 12 X.%..'G.#f.,..<.
0000020: 5f 5b 17 7c 81 ed 31 28 4e 6e a4 6d 23 ad 5b c2 _[.|..1(Nn.m#.[.
0000030: ae 25 83 9a 94 1a 2f 6f 05 fc e5 f0 73 13 f9 0b .%..../o....s...
Thanks to @ovs for golfing off 1 byte!
Try it online!
answered Nov 14 '18 at 1:53
Dennis♦Dennis
187k32297737
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How exactly do you golf bytes off of Bubblegum? I assumed the strategy would be to try each compression algorithm and pick the one that compresses your input the most. Where do you go from there?
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– Cowabunghole
Nov 16 '18 at 18:54
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@Cowabunghole The first golf removed the trailing spaces, the second one increased the number of zopfli iterations (ovs's idea), and the last one switched from#to!.
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– Dennis♦
Nov 16 '18 at 21:50
add a comment |
12
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Charcoal, 71 68 67 bytes
←⁸↙Ｐ⁵↘←⁴↓Ｐ³↙Ｆ⁹⟦³⟧→Ｐ⁴↘Ｐ⁶↘‖ＯＭ←←¹⁷↘Ｆ⁹«Ｐ⁺³›ι³↑Ｐ⁴↗¿⁼ι²Ｂ⁸±²»↘→ＵＯ³¦¹⁴ＵＭＫＡ#
Try it online! Link is to verbose version of code. Explanation:
←⁸↙Ｐ⁵↘←⁴↓Ｐ³↙Ｆ⁹⟦³⟧→Ｐ⁴↘Ｐ⁶
Output half of the circle.
↘‖ＯＭ←←¹⁷↘
Reflect it and draw in the bottom.
Ｆ⁹«Ｐ⁺³›ι³↑Ｐ⁴↗¿⁼ι²Ｂ⁸±²»
Draw the left arm and crossbar of the A.
↘→ＵＯ³¦¹⁴
Draw the right arm of the A.
ＵＭＫＡ#
Change all the characters to #s.
answered Nov 14 '18 at 1:34
NeilNeil
79.9k744178
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C (gcc), 244 243 242 bytes
#define z <<10|117441415
x={8064,2097088,8142832,31473596,58751886,30 z,60 z,60 z,120 z,120 z,255 z,255 z,480 z,448 z,63897630,33423612,2097136,1835008};b(_,i){i&&b(_/2,i-1),putchar(32+_%2*3);}main(_){_<19&&b(x[_-1],27)&puts("")&main(_+1);}
Try it online!
-1 Thanks to Peter Cordes
-1 Thanks to ceilingcat
Uses bit compression.
Explanation:
The goal is to print a set of numbers in binary using space and # as digits which represent the logo. A bit of bash magic converts the logo to binary masks:
echo "ibase=2;$(<code which echoes the logo [see my bash solution for example]> | tr ' #' 01)" | bc
This results in the binary 'numbers' being:
8064,2097088,8142832,31473596,58751886,117472135,117502855,117502855,117564295,117564295,117702535,117702535,117932935,117900167,63897630,33423612,2097136,1835008
There is an obvious pattern in the middle where every line contains ### ### ###
We can save some space by compressing that middle section based on saving that pattern and OR-ing against it. In addition, all of those lines merely add some stuff to the left of the middle section, so we make the z macro which takes ?????????????? and converts it into ###??????????????### ###. This involves bitshifting left by 10 and OR-ing with the binary of that pattern, which is 117441415.
Now we can more easily understand the code:
#define z <<10|117441415 // Define z to be the compression defined above
x={ // Define x to be an array storing each line's number
8064,2097088,8142832, // The first 5 lines are uncompressed
31473596,58751886,
30 z,60 z,60 z,120 z, // The middle 9 lines are z-compressed
120 z,255 z,255 z,480 z,
448 z,
63897630,33423612, // The last 4 lines are uncompressed
2097136,1835008};
b(_,i){ // we also need a function to print n-bit binary numbers
i&& // i is the bit counter, we recurse until its zero
b(_/2,i-1), // each recursive call halves the input and decrements i
putchar(" #"[_%2]);} // this just prints the correct character
main(_){ // this is the main function, called as ./? will have 1 in _ (argc)
_<19? // if _ is less than 19 then...
b(x[_-1],27), // print the binary expansion of x[_-1]
puts(""), // print a new line
main(_+1) // recurse with _++
:0;}
answered Nov 13 '18 at 20:58
LambdaBetaLambdaBeta
2,109418
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Canx?y:0be replaced withx&&y, or does operator precedence not work? Maybe with&instead of,separating the 3 function calls because&has higher precedence than&&or,(en.cppreference.com/w/c/language/operator_precedence). Or not because only,provides a sequence point for the printing vs. recursion to avoid undefined behaviour.
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– Peter Cordes
Nov 13 '18 at 21:18
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It doesn't work with&&indeed because of the precedence, while&doesn't short-circuit. As of yet I haven't found a shorter alternative, though I wouldn't rule one out.
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– LambdaBeta
Nov 13 '18 at 21:51
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I meant with&or|or^replacing,(because they don't short-circuit) and&&replacing?(because it does). Like_<19 && b() & puts() & main(_+1);(spaces added for readability). The order of evaluation is undefined, which may actually undefined behaviour because of unsequenced side-effects onstdouten.cppreference.com/w/c/language/eval_order, but in practice any given compiler will pick some order for a given set of target + options.
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– Peter Cordes
Nov 13 '18 at 21:56
1
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Ahh, I see what you mean. You're right - that does work! -1 byte.
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– LambdaBeta
Nov 13 '18 at 21:59
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You don't need the space between the macro and the definition.
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– Zacharý
Nov 16 '18 at 17:07
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T-SQL, 341 338 bytes
DECLARE @ CHAR(2000)=REPLACE(REPLACE(REPLACE(
'PRINT SPACE(14*6&$6*15&$4*5$4*10&$2*4$7*7$1*4&$1*3$8*4$1*3$3*3&*3$9*4$1*3$4
*3&*3$8*4$2*3$4*3&*3$8*4$2*3$4*3&*3$7*4$3*3$4*3&*3$7*4$3*3$4*3&*3$6*11$4*3&
*3$6*11$4*3&*3$5*4$5*3$4*3&*3$5*3$6*3$4*3&$1*4$2*4$11*4&$2*8$9*6&$6*17&$6*3)'
,'*',')+REPLICATE(''#'','),'$',')+SPACE('),'&',')+(''
''')EXEC(@)
The first 4 line breaks are for readability only, the final line break is part of a string literal.
Similar to my Adam West tribute, I've manually encoded a long string, and made the following replacements:
*7gets replaced by+REPLICATE('#',7)
$4gets replaced by+SPACE(4)
&gets replaced by a line break inside quotes
This results in a massive SQL command string:
PRINT SPACE(14)+REPLICATE('#',6)+('
')+SPACE(6)+REPLICATE('#',15)+('
')+SPACE(4)+REPLICATE('#',5)+SPACE(4)+REPLICATE('#',10)+('
')+SPACE(2)+REPLICATE('#',4)+SPACE(7)+REPLICATE('#',7)+SPACE(1)+REPLICATE('#',4)+('
')+SPACE(1)+REPLICATE('#',3)+SPACE(8)+REPLICATE('#',4)+SPACE(1)+REPLICATE('#',3)+SPACE(3)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(9)+REPLICATE('#',4)+SPACE(1)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(8)+REPLICATE('#',4)+SPACE(2)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(8)+REPLICATE('#',4)+SPACE(2)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(7)+REPLICATE('#',4)+SPACE(3)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(7)+REPLICATE('#',4)+SPACE(3)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(6)+REPLICATE('#',11)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(6)+REPLICATE('#',11)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(5)+REPLICATE('#',4)+SPACE(5)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(5)+REPLICATE('#',3)+SPACE(6)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+SPACE(1)+REPLICATE('#',4)+SPACE(2)+REPLICATE('#',4)+SPACE(11)+REPLICATE('#',4)+('
')+SPACE(2)+REPLICATE('#',8)+SPACE(9)+REPLICATE('#',6)+('
')+SPACE(6)+REPLICATE('#',17)+('
')+SPACE(6)+REPLICATE('#',3)
Which, when run, produces the necessary output.
Long, but still better than my best set-based solution (463 bytes):
SELECT SPACE(a)+REPLICATE('#',b)+SPACE(c)+REPLICATE('#',d)
+SPACE(e)+REPLICATE('#',f)+SPACE(g)+REPLICATE('#',h)
FROM(VALUES(7,0,7,6,0,0,0,0),(6,8,0,7,0,0,0,0),(4,5,4,5,0,5,0,0),(2,4,7,7,1,4,0,0),
(1,3,8,4,1,3,3,3),(0,3,9,4,1,3,4,3),(0,3,8,4,2,3,4,3),(0,3,8,4,2,3,4,3),
(0,3,7,4,3,3,4,3),(0,3,7,4,3,3,4,3),(0,3,6,6,0,5,4,3),(0,3,6,6,0,5,4,3),
(0,3,5,4,5,3,4,3),(0,3,5,3,6,3,4,3),(1,4,2,4,5,0,6,4),(2,8,9,6,0,0,0,0),
(6,9,0,8,0,0,0,0),(6,3,0,0,0,0,0,0))t(a,b,c,d,e,f,g,h)
answered Nov 13 '18 at 23:03
BradCBradC
3,739523
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can drop 2 chars usingCHAR(8000)instead ofVARCHAR(max)(the huge command is 1453 bytes)
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– Shameen
Nov 14 '18 at 13:21
1
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Thanks, @Shameen. I tried to save one more and useCHAR(2E3)but it didn't seem to like that. I also saved a byte by using a line break literal instead ofCHAR(13).
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– BradC
Nov 14 '18 at 15:14
add a comment |
10
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Canvas, 74 73 71 bytes
ｑｃ２－Ａｚ↓n⁴╫ｍ┬ｆｆ╷＼_↘Tｔ)%⁵6＞Ｙy7pＱ∔Ｉ％SIŗ＋T＾≤?↔↖¶８^`O‾+Ｏ│≤n≡j↶82„２┬｛ #＠］∑‾＋ｎ
Try it here!
answered Nov 13 '18 at 18:27
dzaimadzaima
14.6k21755
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im just curious, how u guys type those chars? u have special keyboard or something?
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– Nikko Khresna
Nov 16 '18 at 14:24
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@NikkoKhresna 85% of my code is the data generated by a generator, but the last 11 bytes doing all the logic were typed in the Canvas dev mode and spamming of tab after every char. That's only the case for me & Canvas though, e.g. AFAIK all of Jellys chars are typable with linux compose sequences.
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– dzaima
Nov 16 '18 at 14:44
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@NikkoKhresna yeah, a lot of languages have different ways of typing the chars. some are typable with US keyboard (Jelly), some require pasting, and some (Charcoal, Actually and Sesos afaik) have a sort of verbose mode which only requires you to know the command names
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– ASCII-only
Nov 18 '18 at 1:37
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@NikkoKhresna Charcoal also has (a very, very outdated) Linux keyboard, and probably some other languages too
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– ASCII-only
Nov 18 '18 at 1:39
add a comment |
10
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05AB1E, 101 88 64 bytes
„# •∍ΔÎë,½=bOÅ.âαUΔ'òõƶαÔγλ#xÆ]~”FbćÁ˜Ð”wнQ_wā©•12вεN>yи}˜èJ27ô»
-24 bytes by creating a port of @Dennis♦' Jelly answer.
Try it online.
Explanation:
„# # Push string "# "
•∍ΔÎë,½=bOÅ.âαUΔ'òõƶαÔγλ#xÆ]~”FbćÁ˜Ð”wнQ_wā©•
'# Push compressed integer 10250938842396786963034911279002199266186481794751691439873548591280332406943905758890346943197901909163
12в # Convert to Base-12 as list: [3,0,11,6,2,0,11,4,0,11,10,5,4,10,6,4,7,7,1,4,3,3,8,4,1,3,3,3,1,3,9,4,1,3,4,6,8,4,2,3,4,6,8,4,2,3,4,6,7,4,3,3,4,6,7,4,3,3,4,6,6,11,4,6,6,11,4,6,5,4,5,3,4,6,5,3,6,3,4,3,1,4,2,4,11,4,3,8,9,6,8,6,0,11,10,3]
ε } # Map each `y` to:
N> # The index+1
yи # Repeated `y` amount of times
˜ # Flatten the list
è # Index each in the string "# " (with automatic wraparound)
J # Join everything together
27ô # Split into parts of length 27
» # And join by newlines
See this 05AB1E tip of mine (sections How to compress large integers? and How to compress integer lists?) to understand how the compression of the integer and list works.
answered Nov 14 '18 at 7:36
Kevin CruijssenKevin Cruijssen
36.6k555192
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add a comment |
9
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JavaScript (ES6), 173 170 bytes
_=>`tc
du
9a9k
58fe38
36h83676
6j83!h85!h85!f87!f87!dm96
6dm96
6b8b!b6d696
3858n8
5gjc
dy
d6`.split`!`.join`696
6`.replace(/./g,c=>'# '[(k=parseInt(c,36))&1].repeat(k/2))
Try it online!
Node.js, 163 bytes
Provided that an array of strings is a valid output:
_=>Buffer(`.&-/*%$*&$''!$##($!###!#)$!#$&($"#$&($"#$&'$##$&'$##$&&+$&&+$&%$%#$&%#&#$#!$"$+$#()&(1*#2`).map(c=>s+='# '[x^=1].repeat(c-32),s=x='')&&s.match(/.{27}/g)
Try it online!
answered Nov 13 '18 at 18:05
ArnauldArnauld
73.7k689309
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add a comment |
8
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///, 170 bytes
/+/////*/ +)/'$# +(/"$!+'/" +&/!#+%/"!
!+$/*!+"/**+!/###/""('*
(!!!'
"&#"!!&"
$#'$& &*
!""& ! $
!"'& !%""&$%""&$%)$%)$%(&#%(&#%'&'!%'!"$"!
&$#")
$&#"'!!*
(!!&#"
"$
Try it online!
answered Nov 13 '18 at 19:02
Conor O'BrienConor O'Brien
29.2k263162
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add a comment |
8
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C (gcc), 174 168 164 bytes
-6 bytes thanks to Dennis.
i,j=88;main(l){while(j--)for(i="CJQHFIHCDKDBDACDCFCEFDCEDEFDKFFDKFFDCCDGFDCCDGFDCBDHFDCBDHFDCADICACCCADHCCDAGGDFJDEJOMFN"[j]-64;i;)putchar(l++%28?--i,35-j%2*3:10);}
Try it online!
answered Nov 13 '18 at 22:10
gastropnergastropner
2,0501411
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add a comment |
7
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Bash, 192 176 bytes
dc<<<"16i2o81C0000 81FFFF0 9FE00FC BCF001E F070387 F078387 F03FF87 F03FF87 F01E387 F01E387 F00F387 F00F387 F007B87 B807B8E 9E03FBC 87C3FF0 81FFFC0 8001F80f"|tr 01 ' #'|cut -c2-
Try it online!
-16 thanks to manatwork
This is similar to my C answer, except it just uses a raw base-16 compression and passes it through bc, then uses tr to convert 1 to # and 0 to space. Each row has 1 appended to it and stripped off of it to maintain alignment.
Unfortunately dc is shorter than bc.
answered Nov 13 '18 at 21:00
LambdaBetaLambdaBeta
2,109418
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No need for^. But even better, usecut -c2-instead of thesedpart.
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– manatwork
Nov 13 '18 at 22:59
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And also shorter with here-string instead ofechoand even shorter withdcinstead ofbc: Try it online!
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– manatwork
Nov 14 '18 at 9:07
2
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I'll give you theecho, but come on -dcfor Stan Lee... you must be joking ;)
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– LambdaBeta
Nov 14 '18 at 15:31
1
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134 bytes
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– Dennis♦
Nov 14 '18 at 18:55
add a comment |
6
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Haskell, 170 163 bytes
Edit: -7 bytes thanks to @Ørjan Johansen
m=<<"Mc Ul WeWg YdTgZa ZcSdZcX cRdZcW cSdYcW cSdYcW cTdXcW cTdXcW cUkW cUkW cVdVcW cVcUcW ZdYdPa YhRc Un U "
m ' '="###n"
m c=(' '<$['Z','Y'..c])++('#'<$['a'..c])
Try it online!
Spaces are encoded as uppercase characters (length: Z down to char), hash signs as lowercase characters (length: a to char) and the last three # of each line plus newline as a space. Function m decodes it.
answered Nov 14 '18 at 6:20
niminimi
31.5k32185
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1
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I think you can save some by letting space encode"###n".
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– Ørjan Johansen
Nov 14 '18 at 6:36
add a comment |
6
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Perl 6, 136 112 bytes
:128['?@~p<?x?aB~<|xpqpac`?@~x>q|<!Ca`||?x'.ords].base(2)~~TR/1/ /.comb(27)>>.say
Try it online!
Parses a base 128 number from the ordinal values of the string, converts to base 2 and replaces the 1s with spaces. This uses zeroes as the main character. The string has like 9 null bytes, which was a bitch to type up in TIO. I would have used base 127 for the same amount of bytes, but no nulls, but that has a carriage return, which seems to be impossible to type :(
Explanation:
:128['...'.ords] # Convert the bytes of the string to base 127
.base(2) # Convert to base 2
~~TR/1/ / # Translate 1s to spaces
# Alternatively, this could be .split(1)
.comb(27) # Split to strings of length 27
>>.say # Print each on a newline
answered Nov 14 '18 at 3:13
Jo KingJo King
21.4k248110
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add a comment |
5
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J, 130 128 bytes
echo' #'{~18 27$;(_243{.2#.inv 92x#._32+a.i.])&>'!TYPW.ajz i8hIhXl''3lOH8GvV.C2Z{r/=,G';'"a*2ZDxRplkh2tzRakz.?ZwVmeOT6L^lFB^eyT'
Try it online!
Initial solution
J, 164 bytes
echo' #'{~18 27$,#:849239965469633263905532594449192007713271791872263657753301928240007 12380965417202148347902847903517734495157419855048834759608223758433386496x
Try it online!
answered Nov 13 '18 at 20:04
Galen IvanovGalen Ivanov
6,56711032
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add a comment |
4
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Perl 5, 181 bytes
say+(sprintf"%28b",oct"0x$_")=~y/10/# /r for"0003f0003fff800f87fe03c07f78700f71ce00f70ee01e70ee01e70ee03c70ee03c70ee07ff0ee07ff0ee0f070ee0e070e79e003c3fc01f803fffe00380000"=~/.{7}/g
Try it online!
answered Nov 13 '18 at 22:14
XcaliXcali
5,238520
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add a comment |
4
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MATLAB : 144 Bytes
reshape(repelem([repmat(' #',1,44),' '],'NFMOJEDJFDGGADCCHDACCCACIDACDFHDBCDFHDBCDFGDCCDFGDCCDFFKDFFKDFEDECDFECFCDCADBDKDCHIFHQJCR'-64),27,)'
Try it online! (Technically in Octave)
Explanation:
This uses the same strategy as digEmAll in R, just with MATLAB syntax. The main difference is that MATLAB has automatic conversion from characters to integers.
answered Nov 14 '18 at 16:36
Nicky MattssonNicky Mattsson
1414
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add a comment |
4
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PHP, 286 212 209 bytes
<?php foreach(str_split('000680018y4g04uj1s0iql6k0yz98u1xxu2v1xyhs71xyhs71xzt6v1xzt6v1y2ruv1y2ruv1y7pmv1y70cn121jq60jwdto018y5s013bwg',6)as $i)echo str_replace(0,' ',str_pad(base_convert($i,36,2),27,0,0))."
";
Try it online!
answered Nov 14 '18 at 13:56
ScootsScoots
404311
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add a comment |
3
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C# (.NET Core), 199 bytes
_=>{var r="";for(int i=0,j,k=0;i<88;i++)for(j=0;j++<"0(/1,'&,(&))#&%%*&#%%%#%+&#%&(*&$%&(*&$%&()&%%&()&%%&((-&((-&('&'%&('%(%&%#&$&-&%*+(*3,%"[i]-34;){r+=i%2<1?' ':'#';if(++k%27<1)r+='n';}return r;}
Try it online!
Uses the same approach as my solution to the tribute to Adam West.
answered Nov 14 '18 at 11:37
CharlieCharlie
7,3162389
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add a comment |
1
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deflate, 79 bytes
eJyVkcsJADAIQ++ZIpD9dyxUqrGUgjlpHvglXdqCJymk1yEiaRXEIOXzuBHCiKObReUxUzYaMdt2wmTBg/FmNXndgLRbNvL7ifsLfMw6iQ==
answered Nov 14 '18 at 6:26
WhaleWhale
211
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6
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You appear to have posted two answers in the same language. If you make an improvement to your submission it is better to edit your existing answer. Additionally, could you provide a link to thedeflatelanguage, as this just looks like a base64 string
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– Jo King
Nov 14 '18 at 6:28
3
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I don't think deflate counts as a programming language (although I could be wrong as I don't participate on PPCG).
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– NobodyNada
Nov 14 '18 at 7:28
2
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This is not DEFLATE, but zlib, which is DEFLATE plus a 2 byte header and a 4 byte checksum. By stripping those, you can use the Bubblegum interpreter.
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– Dennis♦
Nov 14 '18 at 14:24
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@NobodyNada deflate is not a programming language but we do not require that submissions be in programming languages.
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– Wît Wisarhd
Nov 14 '18 at 23:06
add a comment |
0
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Turing Machine But Way Worse - 16913 bytes
0 0 0 1 1 0 0
0 1 0 1 2 0 0
0 2 1 0 3 1 0
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1 0 1 1 4 1 0
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1 246 1 0 716 1 0
0 716 0 0 717 1 1
Made with ASCII_only's program generator
answered Dec 27 '18 at 3:36
MilkyWay90MilkyWay90
1799
$endgroup$
add a comment |
Your Answer
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25
$begingroup$
Jelly, 62 59 58 bytes
“ḢE{ɠs{Ƒ0Ṇṁỵ8ỊṢṂƊeLṫfIWẈḞ'ʠJ£ṗɱçoȧ?ƒnØẆƥṂ⁷ɱ’b12ĖŒṙḂa⁶s27Y
Thanks to @JonathanAllan for golfing off 1 byte!
Try it online!
How it works
“ḢE{ɠs{Ƒ0Ṇṁỵ8ỊṢṂƊeLṫfIWẈḞ'ʠJ£ṗɱçoȧ?ƒnØẆƥṂ⁷ɱ’
is a bijective base-250 integer literal, which uses the characters in Jelly's code page as digits.
The literal encodes the integer $scriptsize 10250938842396786963034911279002199266186481794751691439873548591280332406943905758890346943197901909163$, which b12 converts to duodecimal, yielding $scriptsize 30b620b40ba54a64771433841333139413468423468423467433467433466b466b465453465363431424b43896860ba3_{12}$.
There are some zeroes, because 14 spaces (e.g.) are encoded as 3 spaces, 0 hashes, and 11 spaces. This maintains the base small (the largest run consists of 17 hashes), without adding any additional logic to the decoder.
Ė (enumerate) prefixes every base-12 digit by its 1-based index, then Œṙ performs run-length decoding.
Now, for each integer in the resulting array, Ḃ extracts the least significant bit, then a⁶ (AND space) replaces ones with spaces, while leaving zeroes unchanged.
Finally, s27 splits the unformatted character array into chunks of length 27, which Y separates by linefeeds.
answered Nov 13 '18 at 21:22
Dennis♦Dennis
187k32297737
$endgroup$
add a comment |
25
$begingroup$
Jelly, 62 59 58 bytes
“ḢE{ɠs{Ƒ0Ṇṁỵ8ỊṢṂƊeLṫfIWẈḞ'ʠJ£ṗɱçoȧ?ƒnØẆƥṂ⁷ɱ’b12ĖŒṙḂa⁶s27Y
Thanks to @JonathanAllan for golfing off 1 byte!
Try it online!
How it works
“ḢE{ɠs{Ƒ0Ṇṁỵ8ỊṢṂƊeLṫfIWẈḞ'ʠJ£ṗɱçoȧ?ƒnØẆƥṂ⁷ɱ’
is a bijective base-250 integer literal, which uses the characters in Jelly's code page as digits.
The literal encodes the integer $scriptsize 10250938842396786963034911279002199266186481794751691439873548591280332406943905758890346943197901909163$, which b12 converts to duodecimal, yielding $scriptsize 30b620b40ba54a64771433841333139413468423468423467433467433466b466b465453465363431424b43896860ba3_{12}$.
There are some zeroes, because 14 spaces (e.g.) are encoded as 3 spaces, 0 hashes, and 11 spaces. This maintains the base small (the largest run consists of 17 hashes), without adding any additional logic to the decoder.
Ė (enumerate) prefixes every base-12 digit by its 1-based index, then Œṙ performs run-length decoding.
Now, for each integer in the resulting array, Ḃ extracts the least significant bit, then a⁶ (AND space) replaces ones with spaces, while leaving zeroes unchanged.
Finally, s27 splits the unformatted character array into chunks of length 27, which Y separates by linefeeds.
answered Nov 13 '18 at 21:22
Dennis♦Dennis
187k32297737
$endgroup$
add a comment |
25
25
25
$begingroup$
Jelly, 62 59 58 bytes
“ḢE{ɠs{Ƒ0Ṇṁỵ8ỊṢṂƊeLṫfIWẈḞ'ʠJ£ṗɱçoȧ?ƒnØẆƥṂ⁷ɱ’b12ĖŒṙḂa⁶s27Y
Thanks to @JonathanAllan for golfing off 1 byte!
Try it online!
How it works
“ḢE{ɠs{Ƒ0Ṇṁỵ8ỊṢṂƊeLṫfIWẈḞ'ʠJ£ṗɱçoȧ?ƒnØẆƥṂ⁷ɱ’
is a bijective base-250 integer literal, which uses the characters in Jelly's code page as digits.
The literal encodes the integer $scriptsize 10250938842396786963034911279002199266186481794751691439873548591280332406943905758890346943197901909163$, which b12 converts to duodecimal, yielding $scriptsize 30b620b40ba54a64771433841333139413468423468423467433467433466b466b465453465363431424b43896860ba3_{12}$.
There are some zeroes, because 14 spaces (e.g.) are encoded as 3 spaces, 0 hashes, and 11 spaces. This maintains the base small (the largest run consists of 17 hashes), without adding any additional logic to the decoder.
Ė (enumerate) prefixes every base-12 digit by its 1-based index, then Œṙ performs run-length decoding.
Now, for each integer in the resulting array, Ḃ extracts the least significant bit, then a⁶ (AND space) replaces ones with spaces, while leaving zeroes unchanged.
Finally, s27 splits the unformatted character array into chunks of length 27, which Y separates by linefeeds.
answered Nov 13 '18 at 21:22
Dennis♦Dennis
187k32297737
$endgroup$
Jelly, 62 59 58 bytes
“ḢE{ɠs{Ƒ0Ṇṁỵ8ỊṢṂƊeLṫfIWẈḞ'ʠJ£ṗɱçoȧ?ƒnØẆƥṂ⁷ɱ’b12ĖŒṙḂa⁶s27Y
Thanks to @JonathanAllan for golfing off 1 byte!
Try it online!
How it works
“ḢE{ɠs{Ƒ0Ṇṁỵ8ỊṢṂƊeLṫfIWẈḞ'ʠJ£ṗɱçoȧ?ƒnØẆƥṂ⁷ɱ’
is a bijective base-250 integer literal, which uses the characters in Jelly's code page as digits.
The literal encodes the integer $scriptsize 10250938842396786963034911279002199266186481794751691439873548591280332406943905758890346943197901909163$, which b12 converts to duodecimal, yielding $scriptsize 30b620b40ba54a64771433841333139413468423468423467433467433466b466b465453465363431424b43896860ba3_{12}$.
There are some zeroes, because 14 spaces (e.g.) are encoded as 3 spaces, 0 hashes, and 11 spaces. This maintains the base small (the largest run consists of 17 hashes), without adding any additional logic to the decoder.
Ė (enumerate) prefixes every base-12 digit by its 1-based index, then Œṙ performs run-length decoding.
Now, for each integer in the resulting array, Ḃ extracts the least significant bit, then a⁶ (AND space) replaces ones with spaces, while leaving zeroes unchanged.
Finally, s27 splits the unformatted character array into chunks of length 27, which Y separates by linefeeds.
answered Nov 13 '18 at 21:22
Dennis♦Dennis
187k32297737
edited Nov 14 '18 at 20:19
answered Nov 13 '18 at 21:22
Dennis♦Dennis
187k32297737
answered Nov 13 '18 at 21:22
Dennis♦Dennis
187k32297737
answered Nov 13 '18 at 21:22
Dennis♦Dennis
187k32297737
187k32297737
add a comment |
add a comment |
17
$begingroup$
R, 198, 173, 163, 157, 144, 142 bytes
write(rep(rep(c(' ',4),44),utf8ToInt("NFMOJEDJFDGGADCCHDACCCACIDACDFHDBCDFHDBCDFGDCCDFGDCCDFFKDFFKDFEDECDFECFCDCADBDKDCHIFHQJC")-64),1,27,,'')
Try it online!
Also R honors to the great Stan.
Notes :
- instead of the character
#we've used the character4because saves 2 bytes and :
- it's a reference to the Fantastic Four and/or the next Avengers 4 movie
- it like we're repeating the logo inside the logo (since
4seems a mini "slanted"A)
- -15 bytes thanks to @Giuseppe
Explanation :
Applying a Run Length Encoding (rle function) on the characters of the string (excluding 'n') returns 88 repetitions of the alternating <space>,<hash>,<space>,<hash>,<space>,<hash>....
The 88 repetitions are all values in the range [1,17] hence, summing 64 we get the code points of the letters [A...Q] obtaining the string : "NFMOJEDJFDGGADCCHDACCCACIDACDFHDBCDFHDBCDFGDCCDFGDCCDFFKDFFKDFEDECDFECFCDCADBDKDCHIFHQJC"
Hence the code basically does the opposite operations :
v <- utf8ToInt("NFMOJEDJFDGGADC...")-64 # convert the string back to [1...17] vector
e <- rep(rep(c(' ',4),44),v) # perform the inverse of rle function obtaining an
# expanded vector of <space>'s and <four>'s
write(e,1,27,,'') # write the vector to stdout arranging it
# in rows of maximum 27 characters
answered Nov 13 '18 at 18:48
digEmAlldigEmAll
3,029414
$endgroup$
1
$begingroup$
Your solution is nicer, but your own gzip 2-parter gives 89+47 = 136 bytes.
$endgroup$
– J.Doe
Nov 14 '18 at 19:43
$begingroup$
@J.Doe : ahah I even forgot to have used that approach! please post it as your own answer :)
$endgroup$
– digEmAll
Nov 14 '18 at 19:51
$begingroup$
Since we can use any character for#, using0in place of"#"should save 2 more bytes. @J.Doe that may work better for your gzip as well.
$endgroup$
– Giuseppe
Nov 14 '18 at 20:50
add a comment |
17
$begingroup$
R, 198, 173, 163, 157, 144, 142 bytes
write(rep(rep(c(' ',4),44),utf8ToInt("NFMOJEDJFDGGADCCHDACCCACIDACDFHDBCDFHDBCDFGDCCDFGDCCDFFKDFFKDFEDECDFECFCDCADBDKDCHIFHQJC")-64),1,27,,'')
Try it online!
Also R honors to the great Stan.
Notes :
- instead of the character
#we've used the character4because saves 2 bytes and :
- it's a reference to the Fantastic Four and/or the next Avengers 4 movie
- it like we're repeating the logo inside the logo (since
4seems a mini "slanted"A)
- -15 bytes thanks to @Giuseppe
Explanation :
Applying a Run Length Encoding (rle function) on the characters of the string (excluding 'n') returns 88 repetitions of the alternating <space>,<hash>,<space>,<hash>,<space>,<hash>....
The 88 repetitions are all values in the range [1,17] hence, summing 64 we get the code points of the letters [A...Q] obtaining the string : "NFMOJEDJFDGGADCCHDACCCACIDACDFHDBCDFHDBCDFGDCCDFGDCCDFFKDFFKDFEDECDFECFCDCADBDKDCHIFHQJC"
Hence the code basically does the opposite operations :
v <- utf8ToInt("NFMOJEDJFDGGADC...")-64 # convert the string back to [1...17] vector
e <- rep(rep(c(' ',4),44),v) # perform the inverse of rle function obtaining an
# expanded vector of <space>'s and <four>'s
write(e,1,27,,'') # write the vector to stdout arranging it
# in rows of maximum 27 characters
answered Nov 13 '18 at 18:48
digEmAlldigEmAll
3,029414
$endgroup$
1
$begingroup$
Your solution is nicer, but your own gzip 2-parter gives 89+47 = 136 bytes.
$endgroup$
– J.Doe
Nov 14 '18 at 19:43
$begingroup$
@J.Doe : ahah I even forgot to have used that approach! please post it as your own answer :)
$endgroup$
– digEmAll
Nov 14 '18 at 19:51
$begingroup$
Since we can use any character for#, using0in place of"#"should save 2 more bytes. @J.Doe that may work better for your gzip as well.
$endgroup$
– Giuseppe
Nov 14 '18 at 20:50
add a comment |
17
17
17
$begingroup$
R, 198, 173, 163, 157, 144, 142 bytes
write(rep(rep(c(' ',4),44),utf8ToInt("NFMOJEDJFDGGADCCHDACCCACIDACDFHDBCDFHDBCDFGDCCDFGDCCDFFKDFFKDFEDECDFECFCDCADBDKDCHIFHQJC")-64),1,27,,'')
Try it online!
Also R honors to the great Stan.
Notes :
- instead of the character
#we've used the character4because saves 2 bytes and :
- it's a reference to the Fantastic Four and/or the next Avengers 4 movie
- it like we're repeating the logo inside the logo (since
4seems a mini "slanted"A)
- -15 bytes thanks to @Giuseppe
Explanation :
Applying a Run Length Encoding (rle function) on the characters of the string (excluding 'n') returns 88 repetitions of the alternating <space>,<hash>,<space>,<hash>,<space>,<hash>....
The 88 repetitions are all values in the range [1,17] hence, summing 64 we get the code points of the letters [A...Q] obtaining the string : "NFMOJEDJFDGGADCCHDACCCACIDACDFHDBCDFHDBCDFGDCCDFGDCCDFFKDFFKDFEDECDFECFCDCADBDKDCHIFHQJC"
Hence the code basically does the opposite operations :
v <- utf8ToInt("NFMOJEDJFDGGADC...")-64 # convert the string back to [1...17] vector
e <- rep(rep(c(' ',4),44),v) # perform the inverse of rle function obtaining an
# expanded vector of <space>'s and <four>'s
write(e,1,27,,'') # write the vector to stdout arranging it
# in rows of maximum 27 characters
answered Nov 13 '18 at 18:48
digEmAlldigEmAll
3,029414
$endgroup$
R, 198, 173, 163, 157, 144, 142 bytes
write(rep(rep(c(' ',4),44),utf8ToInt("NFMOJEDJFDGGADCCHDACCCACIDACDFHDBCDFHDBCDFGDCCDFGDCCDFFKDFFKDFEDECDFECFCDCADBDKDCHIFHQJC")-64),1,27,,'')
Try it online!
Also R honors to the great Stan.
Notes :
- instead of the character
#we've used the character4because saves 2 bytes and :
- it's a reference to the Fantastic Four and/or the next Avengers 4 movie
- it like we're repeating the logo inside the logo (since
4seems a mini "slanted"A)
- -15 bytes thanks to @Giuseppe
Explanation :
Applying a Run Length Encoding (rle function) on the characters of the string (excluding 'n') returns 88 repetitions of the alternating <space>,<hash>,<space>,<hash>,<space>,<hash>....
The 88 repetitions are all values in the range [1,17] hence, summing 64 we get the code points of the letters [A...Q] obtaining the string : "NFMOJEDJFDGGADCCHDACCCACIDACDFHDBCDFHDBCDFGDCCDFGDCCDFFKDFFKDFEDECDFECFCDCADBDKDCHIFHQJC"
Hence the code basically does the opposite operations :
v <- utf8ToInt("NFMOJEDJFDGGADC...")-64 # convert the string back to [1...17] vector
e <- rep(rep(c(' ',4),44),v) # perform the inverse of rle function obtaining an
# expanded vector of <space>'s and <four>'s
write(e,1,27,,'') # write the vector to stdout arranging it
# in rows of maximum 27 characters
answered Nov 13 '18 at 18:48
digEmAlldigEmAll
3,029414
edited Nov 15 '18 at 7:18
answered Nov 13 '18 at 18:48
digEmAlldigEmAll
3,029414
answered Nov 13 '18 at 18:48
digEmAlldigEmAll
3,029414
answered Nov 13 '18 at 18:48
digEmAlldigEmAll
3,029414
3,029414
1
$begingroup$
Your solution is nicer, but your own gzip 2-parter gives 89+47 = 136 bytes.
$endgroup$
– J.Doe
Nov 14 '18 at 19:43
$begingroup$
@J.Doe : ahah I even forgot to have used that approach! please post it as your own answer :)
$endgroup$
– digEmAll
Nov 14 '18 at 19:51
$begingroup$
Since we can use any character for#, using0in place of"#"should save 2 more bytes. @J.Doe that may work better for your gzip as well.
$endgroup$
– Giuseppe
Nov 14 '18 at 20:50
add a comment |
1
$begingroup$
Your solution is nicer, but your own gzip 2-parter gives 89+47 = 136 bytes.
$endgroup$
– J.Doe
Nov 14 '18 at 19:43
$begingroup$
@J.Doe : ahah I even forgot to have used that approach! please post it as your own answer :)
$endgroup$
– digEmAll
Nov 14 '18 at 19:51
$begingroup$
Since we can use any character for#, using0in place of"#"should save 2 more bytes. @J.Doe that may work better for your gzip as well.
$endgroup$
– Giuseppe
Nov 14 '18 at 20:50
1
1
$begingroup$
Your solution is nicer, but your own gzip 2-parter gives 89+47 = 136 bytes.
$endgroup$
– J.Doe
Nov 14 '18 at 19:43
$begingroup$
Your solution is nicer, but your own gzip 2-parter gives 89+47 = 136 bytes.
$endgroup$
– J.Doe
Nov 14 '18 at 19:43
$begingroup$
@J.Doe : ahah I even forgot to have used that approach! please post it as your own answer :)
$endgroup$
– digEmAll
Nov 14 '18 at 19:51
$begingroup$
@J.Doe : ahah I even forgot to have used that approach! please post it as your own answer :)
$endgroup$
– digEmAll
Nov 14 '18 at 19:51
$begingroup$
Since we can use any character for
#, using 0 in place of "#" should save 2 more bytes. @J.Doe that may work better for your gzip as well.$endgroup$
– Giuseppe
Nov 14 '18 at 20:50
$begingroup$
Since we can use any character for
#, using 0 in place of "#" should save 2 more bytes. @J.Doe that may work better for your gzip as well.$endgroup$
– Giuseppe
Nov 14 '18 at 20:50
add a comment |
12
$begingroup$
Python 2, 129 bytes
00000000: 2363 6f64 696e 673a 4c31 0a70 7269 6e74 #coding:L1.print
00000010: 2778 da95 9141 0e5c 3021 0803 ef7d 4593 'x...A.!...}E.
00000020: feff 8f9b 5d14 5c6e f1b0 f622 7422 2890 ....].n..."t"(.
00000030: 2e7d 5a09 dc4b 19cb bc53 84d1 4a6a 5960 .}Z..K...S..JjY`
00000040: 116e 3f42 c290 b3f7 c0bc 76cf 549d 6ed8 .n?B......v.T.n.
00000050: f8fa 5f26 0b0e 8c93 d5cb 35f6 b1e7 a939 .._&......5....9
00000060: 9e98 e769 47b9 87d6 cdf5 5c30 3030 32c0 ...iG.....002.
00000070: 4029 272e 6465 636f 6465 2827 7a69 7027 @)'.decode('zip'
00000080: 29 )
Try it online!
answered Nov 13 '18 at 23:16
LynnLynn
49.5k794227
$endgroup$
$begingroup$
129 bytes, I removed the trailingnfrom the compressed string.
$endgroup$
– ovs
Nov 14 '18 at 20:46
add a comment |
12
$begingroup$
Python 2, 129 bytes
00000000: 2363 6f64 696e 673a 4c31 0a70 7269 6e74 #coding:L1.print
00000010: 2778 da95 9141 0e5c 3021 0803 ef7d 4593 'x...A.!...}E.
00000020: feff 8f9b 5d14 5c6e f1b0 f622 7422 2890 ....].n..."t"(.
00000030: 2e7d 5a09 dc4b 19cb bc53 84d1 4a6a 5960 .}Z..K...S..JjY`
00000040: 116e 3f42 c290 b3f7 c0bc 76cf 549d 6ed8 .n?B......v.T.n.
00000050: f8fa 5f26 0b0e 8c93 d5cb 35f6 b1e7 a939 .._&......5....9
00000060: 9e98 e769 47b9 87d6 cdf5 5c30 3030 32c0 ...iG.....002.
00000070: 4029 272e 6465 636f 6465 2827 7a69 7027 @)'.decode('zip'
00000080: 29 )
Try it online!
answered Nov 13 '18 at 23:16
LynnLynn
49.5k794227
$endgroup$
$begingroup$
129 bytes, I removed the trailingnfrom the compressed string.
$endgroup$
– ovs
Nov 14 '18 at 20:46
add a comment |
12
12
12
$begingroup$
Python 2, 129 bytes
00000000: 2363 6f64 696e 673a 4c31 0a70 7269 6e74 #coding:L1.print
00000010: 2778 da95 9141 0e5c 3021 0803 ef7d 4593 'x...A.!...}E.
00000020: feff 8f9b 5d14 5c6e f1b0 f622 7422 2890 ....].n..."t"(.
00000030: 2e7d 5a09 dc4b 19cb bc53 84d1 4a6a 5960 .}Z..K...S..JjY`
00000040: 116e 3f42 c290 b3f7 c0bc 76cf 549d 6ed8 .n?B......v.T.n.
00000050: f8fa 5f26 0b0e 8c93 d5cb 35f6 b1e7 a939 .._&......5....9
00000060: 9e98 e769 47b9 87d6 cdf5 5c30 3030 32c0 ...iG.....002.
00000070: 4029 272e 6465 636f 6465 2827 7a69 7027 @)'.decode('zip'
00000080: 29 )
Try it online!
answered Nov 13 '18 at 23:16
LynnLynn
49.5k794227
$endgroup$
Python 2, 129 bytes
00000000: 2363 6f64 696e 673a 4c31 0a70 7269 6e74 #coding:L1.print
00000010: 2778 da95 9141 0e5c 3021 0803 ef7d 4593 'x...A.!...}E.
00000020: feff 8f9b 5d14 5c6e f1b0 f622 7422 2890 ....].n..."t"(.
00000030: 2e7d 5a09 dc4b 19cb bc53 84d1 4a6a 5960 .}Z..K...S..JjY`
00000040: 116e 3f42 c290 b3f7 c0bc 76cf 549d 6ed8 .n?B......v.T.n.
00000050: f8fa 5f26 0b0e 8c93 d5cb 35f6 b1e7 a939 .._&......5....9
00000060: 9e98 e769 47b9 87d6 cdf5 5c30 3030 32c0 ...iG.....002.
00000070: 4029 272e 6465 636f 6465 2827 7a69 7027 @)'.decode('zip'
00000080: 29 )
Try it online!
answered Nov 13 '18 at 23:16
LynnLynn
49.5k794227
edited Nov 14 '18 at 22:39
answered Nov 13 '18 at 23:16
LynnLynn
49.5k794227
answered Nov 13 '18 at 23:16
LynnLynn
49.5k794227
answered Nov 13 '18 at 23:16
LynnLynn
49.5k794227
49.5k794227
$begingroup$
129 bytes, I removed the trailingnfrom the compressed string.
$endgroup$
– ovs
Nov 14 '18 at 20:46
add a comment |
$begingroup$
129 bytes, I removed the trailingnfrom the compressed string.
$endgroup$
– ovs
Nov 14 '18 at 20:46
$begingroup$
129 bytes, I removed the trailing
n from the compressed string.$endgroup$
– ovs
Nov 14 '18 at 20:46
$begingroup$
129 bytes, I removed the trailing
n from the compressed string.$endgroup$
– ovs
Nov 14 '18 at 20:46
add a comment |
12
$begingroup$
Bubblegum, 71 66 65 64 bytes
0000000: 95 91 21 12 00 20 08 c0 3a af 90 ff 7f 52 4e 58 ..!.. ..:....RNX
0000010: 58 93 25 d8 8a 27 47 e4 23 66 01 2c ce b1 3c 12 X.%..'G.#f.,..<.
0000020: 5f 5b 17 7c 81 ed 31 28 4e 6e a4 6d 23 ad 5b c2 _[.|..1(Nn.m#.[.
0000030: ae 25 83 9a 94 1a 2f 6f 05 fc e5 f0 73 13 f9 0b .%..../o....s...
Thanks to @ovs for golfing off 1 byte!
Try it online!
answered Nov 14 '18 at 1:53
Dennis♦Dennis
187k32297737
$endgroup$
$begingroup$
How exactly do you golf bytes off of Bubblegum? I assumed the strategy would be to try each compression algorithm and pick the one that compresses your input the most. Where do you go from there?
$endgroup$
– Cowabunghole
Nov 16 '18 at 18:54
1
$begingroup$
@Cowabunghole The first golf removed the trailing spaces, the second one increased the number of zopfli iterations (ovs's idea), and the last one switched from#to!.
$endgroup$
– Dennis♦
Nov 16 '18 at 21:50
add a comment |
12
$begingroup$
Bubblegum, 71 66 65 64 bytes
0000000: 95 91 21 12 00 20 08 c0 3a af 90 ff 7f 52 4e 58 ..!.. ..:....RNX
0000010: 58 93 25 d8 8a 27 47 e4 23 66 01 2c ce b1 3c 12 X.%..'G.#f.,..<.
0000020: 5f 5b 17 7c 81 ed 31 28 4e 6e a4 6d 23 ad 5b c2 _[.|..1(Nn.m#.[.
0000030: ae 25 83 9a 94 1a 2f 6f 05 fc e5 f0 73 13 f9 0b .%..../o....s...
Thanks to @ovs for golfing off 1 byte!
Try it online!
answered Nov 14 '18 at 1:53
Dennis♦Dennis
187k32297737
$endgroup$
$begingroup$
How exactly do you golf bytes off of Bubblegum? I assumed the strategy would be to try each compression algorithm and pick the one that compresses your input the most. Where do you go from there?
$endgroup$
– Cowabunghole
Nov 16 '18 at 18:54
1
$begingroup$
@Cowabunghole The first golf removed the trailing spaces, the second one increased the number of zopfli iterations (ovs's idea), and the last one switched from#to!.
$endgroup$
– Dennis♦
Nov 16 '18 at 21:50
add a comment |
12
12
12
$begingroup$
Bubblegum, 71 66 65 64 bytes
0000000: 95 91 21 12 00 20 08 c0 3a af 90 ff 7f 52 4e 58 ..!.. ..:....RNX
0000010: 58 93 25 d8 8a 27 47 e4 23 66 01 2c ce b1 3c 12 X.%..'G.#f.,..<.
0000020: 5f 5b 17 7c 81 ed 31 28 4e 6e a4 6d 23 ad 5b c2 _[.|..1(Nn.m#.[.
0000030: ae 25 83 9a 94 1a 2f 6f 05 fc e5 f0 73 13 f9 0b .%..../o....s...
Thanks to @ovs for golfing off 1 byte!
Try it online!
answered Nov 14 '18 at 1:53
Dennis♦Dennis
187k32297737
$endgroup$
Bubblegum, 71 66 65 64 bytes
0000000: 95 91 21 12 00 20 08 c0 3a af 90 ff 7f 52 4e 58 ..!.. ..:....RNX
0000010: 58 93 25 d8 8a 27 47 e4 23 66 01 2c ce b1 3c 12 X.%..'G.#f.,..<.
0000020: 5f 5b 17 7c 81 ed 31 28 4e 6e a4 6d 23 ad 5b c2 _[.|..1(Nn.m#.[.
0000030: ae 25 83 9a 94 1a 2f 6f 05 fc e5 f0 73 13 f9 0b .%..../o....s...
Thanks to @ovs for golfing off 1 byte!
Try it online!
answered Nov 14 '18 at 1:53
Dennis♦Dennis
187k32297737
edited Nov 14 '18 at 22:58
answered Nov 14 '18 at 1:53
Dennis♦Dennis
187k32297737
answered Nov 14 '18 at 1:53
Dennis♦Dennis
187k32297737
answered Nov 14 '18 at 1:53
Dennis♦Dennis
187k32297737
187k32297737
$begingroup$
How exactly do you golf bytes off of Bubblegum? I assumed the strategy would be to try each compression algorithm and pick the one that compresses your input the most. Where do you go from there?
$endgroup$
– Cowabunghole
Nov 16 '18 at 18:54
1
$begingroup$
@Cowabunghole The first golf removed the trailing spaces, the second one increased the number of zopfli iterations (ovs's idea), and the last one switched from#to!.
$endgroup$
– Dennis♦
Nov 16 '18 at 21:50
add a comment |
$begingroup$
How exactly do you golf bytes off of Bubblegum? I assumed the strategy would be to try each compression algorithm and pick the one that compresses your input the most. Where do you go from there?
$endgroup$
– Cowabunghole
Nov 16 '18 at 18:54
1
$begingroup$
@Cowabunghole The first golf removed the trailing spaces, the second one increased the number of zopfli iterations (ovs's idea), and the last one switched from#to!.
$endgroup$
– Dennis♦
Nov 16 '18 at 21:50
$begingroup$
How exactly do you golf bytes off of Bubblegum? I assumed the strategy would be to try each compression algorithm and pick the one that compresses your input the most. Where do you go from there?
$endgroup$
– Cowabunghole
Nov 16 '18 at 18:54
$begingroup$
How exactly do you golf bytes off of Bubblegum? I assumed the strategy would be to try each compression algorithm and pick the one that compresses your input the most. Where do you go from there?
$endgroup$
– Cowabunghole
Nov 16 '18 at 18:54
1
1
$begingroup$
@Cowabunghole The first golf removed the trailing spaces, the second one increased the number of zopfli iterations (ovs's idea), and the last one switched from
# to !.$endgroup$
– Dennis♦
Nov 16 '18 at 21:50
$begingroup$
@Cowabunghole The first golf removed the trailing spaces, the second one increased the number of zopfli iterations (ovs's idea), and the last one switched from
# to !.$endgroup$
– Dennis♦
Nov 16 '18 at 21:50
add a comment |
12
$begingroup$
Charcoal, 71 68 67 bytes
←⁸↙Ｐ⁵↘←⁴↓Ｐ³↙Ｆ⁹⟦³⟧→Ｐ⁴↘Ｐ⁶↘‖ＯＭ←←¹⁷↘Ｆ⁹«Ｐ⁺³›ι³↑Ｐ⁴↗¿⁼ι²Ｂ⁸±²»↘→ＵＯ³¦¹⁴ＵＭＫＡ#
Try it online! Link is to verbose version of code. Explanation:
←⁸↙Ｐ⁵↘←⁴↓Ｐ³↙Ｆ⁹⟦³⟧→Ｐ⁴↘Ｐ⁶
Output half of the circle.
↘‖ＯＭ←←¹⁷↘
Reflect it and draw in the bottom.
Ｆ⁹«Ｐ⁺³›ι³↑Ｐ⁴↗¿⁼ι²Ｂ⁸±²»
Draw the left arm and crossbar of the A.
↘→ＵＯ³¦¹⁴
Draw the right arm of the A.
ＵＭＫＡ#
Change all the characters to #s.
answered Nov 14 '18 at 1:34
NeilNeil
79.9k744178
$endgroup$
add a comment |
12
$begingroup$
Charcoal, 71 68 67 bytes
←⁸↙Ｐ⁵↘←⁴↓Ｐ³↙Ｆ⁹⟦³⟧→Ｐ⁴↘Ｐ⁶↘‖ＯＭ←←¹⁷↘Ｆ⁹«Ｐ⁺³›ι³↑Ｐ⁴↗¿⁼ι²Ｂ⁸±²»↘→ＵＯ³¦¹⁴ＵＭＫＡ#
Try it online! Link is to verbose version of code. Explanation:
←⁸↙Ｐ⁵↘←⁴↓Ｐ³↙Ｆ⁹⟦³⟧→Ｐ⁴↘Ｐ⁶
Output half of the circle.
↘‖ＯＭ←←¹⁷↘
Reflect it and draw in the bottom.
Ｆ⁹«Ｐ⁺³›ι³↑Ｐ⁴↗¿⁼ι²Ｂ⁸±²»
Draw the left arm and crossbar of the A.
↘→ＵＯ³¦¹⁴
Draw the right arm of the A.
ＵＭＫＡ#
Change all the characters to #s.
answered Nov 14 '18 at 1:34
NeilNeil
79.9k744178
$endgroup$
add a comment |
12
12
12
$begingroup$
Charcoal, 71 68 67 bytes
←⁸↙Ｐ⁵↘←⁴↓Ｐ³↙Ｆ⁹⟦³⟧→Ｐ⁴↘Ｐ⁶↘‖ＯＭ←←¹⁷↘Ｆ⁹«Ｐ⁺³›ι³↑Ｐ⁴↗¿⁼ι²Ｂ⁸±²»↘→ＵＯ³¦¹⁴ＵＭＫＡ#
Try it online! Link is to verbose version of code. Explanation:
←⁸↙Ｐ⁵↘←⁴↓Ｐ³↙Ｆ⁹⟦³⟧→Ｐ⁴↘Ｐ⁶
Output half of the circle.
↘‖ＯＭ←←¹⁷↘
Reflect it and draw in the bottom.
Ｆ⁹«Ｐ⁺³›ι³↑Ｐ⁴↗¿⁼ι²Ｂ⁸±²»
Draw the left arm and crossbar of the A.
↘→ＵＯ³¦¹⁴
Draw the right arm of the A.
ＵＭＫＡ#
Change all the characters to #s.
answered Nov 14 '18 at 1:34
NeilNeil
79.9k744178
$endgroup$
Charcoal, 71 68 67 bytes
←⁸↙Ｐ⁵↘←⁴↓Ｐ³↙Ｆ⁹⟦³⟧→Ｐ⁴↘Ｐ⁶↘‖ＯＭ←←¹⁷↘Ｆ⁹«Ｐ⁺³›ι³↑Ｐ⁴↗¿⁼ι²Ｂ⁸±²»↘→ＵＯ³¦¹⁴ＵＭＫＡ#
Try it online! Link is to verbose version of code. Explanation:
←⁸↙Ｐ⁵↘←⁴↓Ｐ³↙Ｆ⁹⟦³⟧→Ｐ⁴↘Ｐ⁶
Output half of the circle.
↘‖ＯＭ←←¹⁷↘
Reflect it and draw in the bottom.
Ｆ⁹«Ｐ⁺³›ι³↑Ｐ⁴↗¿⁼ι²Ｂ⁸±²»
Draw the left arm and crossbar of the A.
↘→ＵＯ³¦¹⁴
Draw the right arm of the A.
ＵＭＫＡ#
Change all the characters to #s.
answered Nov 14 '18 at 1:34
NeilNeil
79.9k744178
edited Nov 17 '18 at 11:08
answered Nov 14 '18 at 1:34
NeilNeil
79.9k744178
answered Nov 14 '18 at 1:34
NeilNeil
79.9k744178
answered Nov 14 '18 at 1:34
NeilNeil
79.9k744178
79.9k744178
add a comment |
add a comment |
12
$begingroup$
C (gcc), 244 243 242 bytes
#define z <<10|117441415
x={8064,2097088,8142832,31473596,58751886,30 z,60 z,60 z,120 z,120 z,255 z,255 z,480 z,448 z,63897630,33423612,2097136,1835008};b(_,i){i&&b(_/2,i-1),putchar(32+_%2*3);}main(_){_<19&&b(x[_-1],27)&puts("")&main(_+1);}
Try it online!
-1 Thanks to Peter Cordes
-1 Thanks to ceilingcat
Uses bit compression.
Explanation:
The goal is to print a set of numbers in binary using space and # as digits which represent the logo. A bit of bash magic converts the logo to binary masks:
echo "ibase=2;$(<code which echoes the logo [see my bash solution for example]> | tr ' #' 01)" | bc
This results in the binary 'numbers' being:
8064,2097088,8142832,31473596,58751886,117472135,117502855,117502855,117564295,117564295,117702535,117702535,117932935,117900167,63897630,33423612,2097136,1835008
There is an obvious pattern in the middle where every line contains ### ### ###
We can save some space by compressing that middle section based on saving that pattern and OR-ing against it. In addition, all of those lines merely add some stuff to the left of the middle section, so we make the z macro which takes ?????????????? and converts it into ###??????????????### ###. This involves bitshifting left by 10 and OR-ing with the binary of that pattern, which is 117441415.
Now we can more easily understand the code:
#define z <<10|117441415 // Define z to be the compression defined above
x={ // Define x to be an array storing each line's number
8064,2097088,8142832, // The first 5 lines are uncompressed
31473596,58751886,
30 z,60 z,60 z,120 z, // The middle 9 lines are z-compressed
120 z,255 z,255 z,480 z,
448 z,
63897630,33423612, // The last 4 lines are uncompressed
2097136,1835008};
b(_,i){ // we also need a function to print n-bit binary numbers
i&& // i is the bit counter, we recurse until its zero
b(_/2,i-1), // each recursive call halves the input and decrements i
putchar(" #"[_%2]);} // this just prints the correct character
main(_){ // this is the main function, called as ./? will have 1 in _ (argc)
_<19? // if _ is less than 19 then...
b(x[_-1],27), // print the binary expansion of x[_-1]
puts(""), // print a new line
main(_+1) // recurse with _++
:0;}
answered Nov 13 '18 at 20:58
LambdaBetaLambdaBeta
2,109418
$endgroup$
$begingroup$
Canx?y:0be replaced withx&&y, or does operator precedence not work? Maybe with&instead of,separating the 3 function calls because&has higher precedence than&&or,(en.cppreference.com/w/c/language/operator_precedence). Or not because only,provides a sequence point for the printing vs. recursion to avoid undefined behaviour.
$endgroup$
– Peter Cordes
Nov 13 '18 at 21:18
$begingroup$
It doesn't work with&&indeed because of the precedence, while&doesn't short-circuit. As of yet I haven't found a shorter alternative, though I wouldn't rule one out.
$endgroup$
– LambdaBeta
Nov 13 '18 at 21:51
$begingroup$
I meant with&or|or^replacing,(because they don't short-circuit) and&&replacing?(because it does). Like_<19 && b() & puts() & main(_+1);(spaces added for readability). The order of evaluation is undefined, which may actually undefined behaviour because of unsequenced side-effects onstdouten.cppreference.com/w/c/language/eval_order, but in practice any given compiler will pick some order for a given set of target + options.
$endgroup$
– Peter Cordes
Nov 13 '18 at 21:56
1
$begingroup$
Ahh, I see what you mean. You're right - that does work! -1 byte.
$endgroup$
– LambdaBeta
Nov 13 '18 at 21:59
$begingroup$
You don't need the space between the macro and the definition.
$endgroup$
– Zacharý
Nov 16 '18 at 17:07
|
show 1 more comment
12
$begingroup$
C (gcc), 244 243 242 bytes
#define z <<10|117441415
x={8064,2097088,8142832,31473596,58751886,30 z,60 z,60 z,120 z,120 z,255 z,255 z,480 z,448 z,63897630,33423612,2097136,1835008};b(_,i){i&&b(_/2,i-1),putchar(32+_%2*3);}main(_){_<19&&b(x[_-1],27)&puts("")&main(_+1);}
Try it online!
-1 Thanks to Peter Cordes
-1 Thanks to ceilingcat
Uses bit compression.
Explanation:
The goal is to print a set of numbers in binary using space and # as digits which represent the logo. A bit of bash magic converts the logo to binary masks:
echo "ibase=2;$(<code which echoes the logo [see my bash solution for example]> | tr ' #' 01)" | bc
This results in the binary 'numbers' being:
8064,2097088,8142832,31473596,58751886,117472135,117502855,117502855,117564295,117564295,117702535,117702535,117932935,117900167,63897630,33423612,2097136,1835008
There is an obvious pattern in the middle where every line contains ### ### ###
We can save some space by compressing that middle section based on saving that pattern and OR-ing against it. In addition, all of those lines merely add some stuff to the left of the middle section, so we make the z macro which takes ?????????????? and converts it into ###??????????????### ###. This involves bitshifting left by 10 and OR-ing with the binary of that pattern, which is 117441415.
Now we can more easily understand the code:
#define z <<10|117441415 // Define z to be the compression defined above
x={ // Define x to be an array storing each line's number
8064,2097088,8142832, // The first 5 lines are uncompressed
31473596,58751886,
30 z,60 z,60 z,120 z, // The middle 9 lines are z-compressed
120 z,255 z,255 z,480 z,
448 z,
63897630,33423612, // The last 4 lines are uncompressed
2097136,1835008};
b(_,i){ // we also need a function to print n-bit binary numbers
i&& // i is the bit counter, we recurse until its zero
b(_/2,i-1), // each recursive call halves the input and decrements i
putchar(" #"[_%2]);} // this just prints the correct character
main(_){ // this is the main function, called as ./? will have 1 in _ (argc)
_<19? // if _ is less than 19 then...
b(x[_-1],27), // print the binary expansion of x[_-1]
puts(""), // print a new line
main(_+1) // recurse with _++
:0;}
answered Nov 13 '18 at 20:58
LambdaBetaLambdaBeta
2,109418
$endgroup$
$begingroup$
Canx?y:0be replaced withx&&y, or does operator precedence not work? Maybe with&instead of,separating the 3 function calls because&has higher precedence than&&or,(en.cppreference.com/w/c/language/operator_precedence). Or not because only,provides a sequence point for the printing vs. recursion to avoid undefined behaviour.
$endgroup$
– Peter Cordes
Nov 13 '18 at 21:18
$begingroup$
It doesn't work with&&indeed because of the precedence, while&doesn't short-circuit. As of yet I haven't found a shorter alternative, though I wouldn't rule one out.
$endgroup$
– LambdaBeta
Nov 13 '18 at 21:51
$begingroup$
I meant with&or|or^replacing,(because they don't short-circuit) and&&replacing?(because it does). Like_<19 && b() & puts() & main(_+1);(spaces added for readability). The order of evaluation is undefined, which may actually undefined behaviour because of unsequenced side-effects onstdouten.cppreference.com/w/c/language/eval_order, but in practice any given compiler will pick some order for a given set of target + options.
$endgroup$
– Peter Cordes
Nov 13 '18 at 21:56
1
$begingroup$
Ahh, I see what you mean. You're right - that does work! -1 byte.
$endgroup$
– LambdaBeta
Nov 13 '18 at 21:59
$begingroup$
You don't need the space between the macro and the definition.
$endgroup$
– Zacharý
Nov 16 '18 at 17:07
|
show 1 more comment
12
12
12
$begingroup$
C (gcc), 244 243 242 bytes
#define z <<10|117441415
x={8064,2097088,8142832,31473596,58751886,30 z,60 z,60 z,120 z,120 z,255 z,255 z,480 z,448 z,63897630,33423612,2097136,1835008};b(_,i){i&&b(_/2,i-1),putchar(32+_%2*3);}main(_){_<19&&b(x[_-1],27)&puts("")&main(_+1);}
Try it online!
-1 Thanks to Peter Cordes
-1 Thanks to ceilingcat
Uses bit compression.
Explanation:
The goal is to print a set of numbers in binary using space and # as digits which represent the logo. A bit of bash magic converts the logo to binary masks:
echo "ibase=2;$(<code which echoes the logo [see my bash solution for example]> | tr ' #' 01)" | bc
This results in the binary 'numbers' being:
8064,2097088,8142832,31473596,58751886,117472135,117502855,117502855,117564295,117564295,117702535,117702535,117932935,117900167,63897630,33423612,2097136,1835008
There is an obvious pattern in the middle where every line contains ### ### ###
We can save some space by compressing that middle section based on saving that pattern and OR-ing against it. In addition, all of those lines merely add some stuff to the left of the middle section, so we make the z macro which takes ?????????????? and converts it into ###??????????????### ###. This involves bitshifting left by 10 and OR-ing with the binary of that pattern, which is 117441415.
Now we can more easily understand the code:
#define z <<10|117441415 // Define z to be the compression defined above
x={ // Define x to be an array storing each line's number
8064,2097088,8142832, // The first 5 lines are uncompressed
31473596,58751886,
30 z,60 z,60 z,120 z, // The middle 9 lines are z-compressed
120 z,255 z,255 z,480 z,
448 z,
63897630,33423612, // The last 4 lines are uncompressed
2097136,1835008};
b(_,i){ // we also need a function to print n-bit binary numbers
i&& // i is the bit counter, we recurse until its zero
b(_/2,i-1), // each recursive call halves the input and decrements i
putchar(" #"[_%2]);} // this just prints the correct character
main(_){ // this is the main function, called as ./? will have 1 in _ (argc)
_<19? // if _ is less than 19 then...
b(x[_-1],27), // print the binary expansion of x[_-1]
puts(""), // print a new line
main(_+1) // recurse with _++
:0;}
answered Nov 13 '18 at 20:58
LambdaBetaLambdaBeta
2,109418
$endgroup$
C (gcc), 244 243 242 bytes
#define z <<10|117441415
x={8064,2097088,8142832,31473596,58751886,30 z,60 z,60 z,120 z,120 z,255 z,255 z,480 z,448 z,63897630,33423612,2097136,1835008};b(_,i){i&&b(_/2,i-1),putchar(32+_%2*3);}main(_){_<19&&b(x[_-1],27)&puts("")&main(_+1);}
Try it online!
-1 Thanks to Peter Cordes
-1 Thanks to ceilingcat
Uses bit compression.
Explanation:
The goal is to print a set of numbers in binary using space and # as digits which represent the logo. A bit of bash magic converts the logo to binary masks:
echo "ibase=2;$(<code which echoes the logo [see my bash solution for example]> | tr ' #' 01)" | bc
This results in the binary 'numbers' being:
8064,2097088,8142832,31473596,58751886,117472135,117502855,117502855,117564295,117564295,117702535,117702535,117932935,117900167,63897630,33423612,2097136,1835008
There is an obvious pattern in the middle where every line contains ### ### ###
We can save some space by compressing that middle section based on saving that pattern and OR-ing against it. In addition, all of those lines merely add some stuff to the left of the middle section, so we make the z macro which takes ?????????????? and converts it into ###??????????????### ###. This involves bitshifting left by 10 and OR-ing with the binary of that pattern, which is 117441415.
Now we can more easily understand the code:
#define z <<10|117441415 // Define z to be the compression defined above
x={ // Define x to be an array storing each line's number
8064,2097088,8142832, // The first 5 lines are uncompressed
31473596,58751886,
30 z,60 z,60 z,120 z, // The middle 9 lines are z-compressed
120 z,255 z,255 z,480 z,
448 z,
63897630,33423612, // The last 4 lines are uncompressed
2097136,1835008};
b(_,i){ // we also need a function to print n-bit binary numbers
i&& // i is the bit counter, we recurse until its zero
b(_/2,i-1), // each recursive call halves the input and decrements i
putchar(" #"[_%2]);} // this just prints the correct character
main(_){ // this is the main function, called as ./? will have 1 in _ (argc)
_<19? // if _ is less than 19 then...
b(x[_-1],27), // print the binary expansion of x[_-1]
puts(""), // print a new line
main(_+1) // recurse with _++
:0;}
answered Nov 13 '18 at 20:58
LambdaBetaLambdaBeta
2,109418
edited Nov 21 '18 at 15:42
answered Nov 13 '18 at 20:58
LambdaBetaLambdaBeta
2,109418
answered Nov 13 '18 at 20:58
LambdaBetaLambdaBeta
2,109418
answered Nov 13 '18 at 20:58
LambdaBetaLambdaBeta
2,109418
2,109418
$begingroup$
Canx?y:0be replaced withx&&y, or does operator precedence not work? Maybe with&instead of,separating the 3 function calls because&has higher precedence than&&or,(en.cppreference.com/w/c/language/operator_precedence). Or not because only,provides a sequence point for the printing vs. recursion to avoid undefined behaviour.
$endgroup$
– Peter Cordes
Nov 13 '18 at 21:18
$begingroup$
It doesn't work with&&indeed because of the precedence, while&doesn't short-circuit. As of yet I haven't found a shorter alternative, though I wouldn't rule one out.
$endgroup$
– LambdaBeta
Nov 13 '18 at 21:51
$begingroup$
I meant with&or|or^replacing,(because they don't short-circuit) and&&replacing?(because it does). Like_<19 && b() & puts() & main(_+1);(spaces added for readability). The order of evaluation is undefined, which may actually undefined behaviour because of unsequenced side-effects onstdouten.cppreference.com/w/c/language/eval_order, but in practice any given compiler will pick some order for a given set of target + options.
$endgroup$
– Peter Cordes
Nov 13 '18 at 21:56
1
$begingroup$
Ahh, I see what you mean. You're right - that does work! -1 byte.
$endgroup$
– LambdaBeta
Nov 13 '18 at 21:59
$begingroup$
You don't need the space between the macro and the definition.
$endgroup$
– Zacharý
Nov 16 '18 at 17:07
|
show 1 more comment
$begingroup$
Canx?y:0be replaced withx&&y, or does operator precedence not work? Maybe with&instead of,separating the 3 function calls because&has higher precedence than&&or,(en.cppreference.com/w/c/language/operator_precedence). Or not because only,provides a sequence point for the printing vs. recursion to avoid undefined behaviour.
$endgroup$
– Peter Cordes
Nov 13 '18 at 21:18
$begingroup$
It doesn't work with&&indeed because of the precedence, while&doesn't short-circuit. As of yet I haven't found a shorter alternative, though I wouldn't rule one out.
$endgroup$
– LambdaBeta
Nov 13 '18 at 21:51
$begingroup$
I meant with&or|or^replacing,(because they don't short-circuit) and&&replacing?(because it does). Like_<19 && b() & puts() & main(_+1);(spaces added for readability). The order of evaluation is undefined, which may actually undefined behaviour because of unsequenced side-effects onstdouten.cppreference.com/w/c/language/eval_order, but in practice any given compiler will pick some order for a given set of target + options.
$endgroup$
– Peter Cordes
Nov 13 '18 at 21:56
1
$begingroup$
Ahh, I see what you mean. You're right - that does work! -1 byte.
$endgroup$
– LambdaBeta
Nov 13 '18 at 21:59
$begingroup$
You don't need the space between the macro and the definition.
$endgroup$
– Zacharý
Nov 16 '18 at 17:07
$begingroup$
Can
x?y:0 be replaced with x&&y, or does operator precedence not work? Maybe with & instead of , separating the 3 function calls because & has higher precedence than && or , (en.cppreference.com/w/c/language/operator_precedence). Or not because only , provides a sequence point for the printing vs. recursion to avoid undefined behaviour.$endgroup$
– Peter Cordes
Nov 13 '18 at 21:18
$begingroup$
Can
x?y:0 be replaced with x&&y, or does operator precedence not work? Maybe with & instead of , separating the 3 function calls because & has higher precedence than && or , (en.cppreference.com/w/c/language/operator_precedence). Or not because only , provides a sequence point for the printing vs. recursion to avoid undefined behaviour.$endgroup$
– Peter Cordes
Nov 13 '18 at 21:18
$begingroup$
It doesn't work with
&& indeed because of the precedence, while & doesn't short-circuit. As of yet I haven't found a shorter alternative, though I wouldn't rule one out.$endgroup$
– LambdaBeta
Nov 13 '18 at 21:51
$begingroup$
It doesn't work with
&& indeed because of the precedence, while & doesn't short-circuit. As of yet I haven't found a shorter alternative, though I wouldn't rule one out.$endgroup$
– LambdaBeta
Nov 13 '18 at 21:51
$begingroup$
I meant with
& or | or ^ replacing , (because they don't short-circuit) and && replacing ? (because it does). Like _<19 && b() & puts() & main(_+1); (spaces added for readability). The order of evaluation is undefined, which may actually undefined behaviour because of unsequenced side-effects on stdout en.cppreference.com/w/c/language/eval_order, but in practice any given compiler will pick some order for a given set of target + options.$endgroup$
– Peter Cordes
Nov 13 '18 at 21:56
$begingroup$
I meant with
& or | or ^ replacing , (because they don't short-circuit) and && replacing ? (because it does). Like _<19 && b() & puts() & main(_+1); (spaces added for readability). The order of evaluation is undefined, which may actually undefined behaviour because of unsequenced side-effects on stdout en.cppreference.com/w/c/language/eval_order, but in practice any given compiler will pick some order for a given set of target + options.$endgroup$
– Peter Cordes
Nov 13 '18 at 21:56
1
1
$begingroup$
Ahh, I see what you mean. You're right - that does work! -1 byte.
$endgroup$
– LambdaBeta
Nov 13 '18 at 21:59
$begingroup$
Ahh, I see what you mean. You're right - that does work! -1 byte.
$endgroup$
– LambdaBeta
Nov 13 '18 at 21:59
$begingroup$
You don't need the space between the macro and the definition.
$endgroup$
– Zacharý
Nov 16 '18 at 17:07
$begingroup$
You don't need the space between the macro and the definition.
$endgroup$
– Zacharý
Nov 16 '18 at 17:07
|
show 1 more comment
11
$begingroup$
T-SQL, 341 338 bytes
DECLARE @ CHAR(2000)=REPLACE(REPLACE(REPLACE(
'PRINT SPACE(14*6&$6*15&$4*5$4*10&$2*4$7*7$1*4&$1*3$8*4$1*3$3*3&*3$9*4$1*3$4
*3&*3$8*4$2*3$4*3&*3$8*4$2*3$4*3&*3$7*4$3*3$4*3&*3$7*4$3*3$4*3&*3$6*11$4*3&
*3$6*11$4*3&*3$5*4$5*3$4*3&*3$5*3$6*3$4*3&$1*4$2*4$11*4&$2*8$9*6&$6*17&$6*3)'
,'*',')+REPLICATE(''#'','),'$',')+SPACE('),'&',')+(''
''')EXEC(@)
The first 4 line breaks are for readability only, the final line break is part of a string literal.
Similar to my Adam West tribute, I've manually encoded a long string, and made the following replacements:
*7gets replaced by+REPLICATE('#',7)
$4gets replaced by+SPACE(4)
&gets replaced by a line break inside quotes
This results in a massive SQL command string:
PRINT SPACE(14)+REPLICATE('#',6)+('
')+SPACE(6)+REPLICATE('#',15)+('
')+SPACE(4)+REPLICATE('#',5)+SPACE(4)+REPLICATE('#',10)+('
')+SPACE(2)+REPLICATE('#',4)+SPACE(7)+REPLICATE('#',7)+SPACE(1)+REPLICATE('#',4)+('
')+SPACE(1)+REPLICATE('#',3)+SPACE(8)+REPLICATE('#',4)+SPACE(1)+REPLICATE('#',3)+SPACE(3)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(9)+REPLICATE('#',4)+SPACE(1)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(8)+REPLICATE('#',4)+SPACE(2)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(8)+REPLICATE('#',4)+SPACE(2)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(7)+REPLICATE('#',4)+SPACE(3)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(7)+REPLICATE('#',4)+SPACE(3)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(6)+REPLICATE('#',11)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(6)+REPLICATE('#',11)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(5)+REPLICATE('#',4)+SPACE(5)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(5)+REPLICATE('#',3)+SPACE(6)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+SPACE(1)+REPLICATE('#',4)+SPACE(2)+REPLICATE('#',4)+SPACE(11)+REPLICATE('#',4)+('
')+SPACE(2)+REPLICATE('#',8)+SPACE(9)+REPLICATE('#',6)+('
')+SPACE(6)+REPLICATE('#',17)+('
')+SPACE(6)+REPLICATE('#',3)
Which, when run, produces the necessary output.
Long, but still better than my best set-based solution (463 bytes):
SELECT SPACE(a)+REPLICATE('#',b)+SPACE(c)+REPLICATE('#',d)
+SPACE(e)+REPLICATE('#',f)+SPACE(g)+REPLICATE('#',h)
FROM(VALUES(7,0,7,6,0,0,0,0),(6,8,0,7,0,0,0,0),(4,5,4,5,0,5,0,0),(2,4,7,7,1,4,0,0),
(1,3,8,4,1,3,3,3),(0,3,9,4,1,3,4,3),(0,3,8,4,2,3,4,3),(0,3,8,4,2,3,4,3),
(0,3,7,4,3,3,4,3),(0,3,7,4,3,3,4,3),(0,3,6,6,0,5,4,3),(0,3,6,6,0,5,4,3),
(0,3,5,4,5,3,4,3),(0,3,5,3,6,3,4,3),(1,4,2,4,5,0,6,4),(2,8,9,6,0,0,0,0),
(6,9,0,8,0,0,0,0),(6,3,0,0,0,0,0,0))t(a,b,c,d,e,f,g,h)
answered Nov 13 '18 at 23:03
BradCBradC
3,739523
$endgroup$
$begingroup$
can drop 2 chars usingCHAR(8000)instead ofVARCHAR(max)(the huge command is 1453 bytes)
$endgroup$
– Shameen
Nov 14 '18 at 13:21
1
$begingroup$
Thanks, @Shameen. I tried to save one more and useCHAR(2E3)but it didn't seem to like that. I also saved a byte by using a line break literal instead ofCHAR(13).
$endgroup$
– BradC
Nov 14 '18 at 15:14
add a comment |
11
$begingroup$
T-SQL, 341 338 bytes
DECLARE @ CHAR(2000)=REPLACE(REPLACE(REPLACE(
'PRINT SPACE(14*6&$6*15&$4*5$4*10&$2*4$7*7$1*4&$1*3$8*4$1*3$3*3&*3$9*4$1*3$4
*3&*3$8*4$2*3$4*3&*3$8*4$2*3$4*3&*3$7*4$3*3$4*3&*3$7*4$3*3$4*3&*3$6*11$4*3&
*3$6*11$4*3&*3$5*4$5*3$4*3&*3$5*3$6*3$4*3&$1*4$2*4$11*4&$2*8$9*6&$6*17&$6*3)'
,'*',')+REPLICATE(''#'','),'$',')+SPACE('),'&',')+(''
''')EXEC(@)
The first 4 line breaks are for readability only, the final line break is part of a string literal.
Similar to my Adam West tribute, I've manually encoded a long string, and made the following replacements:
*7gets replaced by+REPLICATE('#',7)
$4gets replaced by+SPACE(4)
&gets replaced by a line break inside quotes
This results in a massive SQL command string:
PRINT SPACE(14)+REPLICATE('#',6)+('
')+SPACE(6)+REPLICATE('#',15)+('
')+SPACE(4)+REPLICATE('#',5)+SPACE(4)+REPLICATE('#',10)+('
')+SPACE(2)+REPLICATE('#',4)+SPACE(7)+REPLICATE('#',7)+SPACE(1)+REPLICATE('#',4)+('
')+SPACE(1)+REPLICATE('#',3)+SPACE(8)+REPLICATE('#',4)+SPACE(1)+REPLICATE('#',3)+SPACE(3)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(9)+REPLICATE('#',4)+SPACE(1)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(8)+REPLICATE('#',4)+SPACE(2)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(8)+REPLICATE('#',4)+SPACE(2)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(7)+REPLICATE('#',4)+SPACE(3)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(7)+REPLICATE('#',4)+SPACE(3)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(6)+REPLICATE('#',11)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(6)+REPLICATE('#',11)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(5)+REPLICATE('#',4)+SPACE(5)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(5)+REPLICATE('#',3)+SPACE(6)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+SPACE(1)+REPLICATE('#',4)+SPACE(2)+REPLICATE('#',4)+SPACE(11)+REPLICATE('#',4)+('
')+SPACE(2)+REPLICATE('#',8)+SPACE(9)+REPLICATE('#',6)+('
')+SPACE(6)+REPLICATE('#',17)+('
')+SPACE(6)+REPLICATE('#',3)
Which, when run, produces the necessary output.
Long, but still better than my best set-based solution (463 bytes):
SELECT SPACE(a)+REPLICATE('#',b)+SPACE(c)+REPLICATE('#',d)
+SPACE(e)+REPLICATE('#',f)+SPACE(g)+REPLICATE('#',h)
FROM(VALUES(7,0,7,6,0,0,0,0),(6,8,0,7,0,0,0,0),(4,5,4,5,0,5,0,0),(2,4,7,7,1,4,0,0),
(1,3,8,4,1,3,3,3),(0,3,9,4,1,3,4,3),(0,3,8,4,2,3,4,3),(0,3,8,4,2,3,4,3),
(0,3,7,4,3,3,4,3),(0,3,7,4,3,3,4,3),(0,3,6,6,0,5,4,3),(0,3,6,6,0,5,4,3),
(0,3,5,4,5,3,4,3),(0,3,5,3,6,3,4,3),(1,4,2,4,5,0,6,4),(2,8,9,6,0,0,0,0),
(6,9,0,8,0,0,0,0),(6,3,0,0,0,0,0,0))t(a,b,c,d,e,f,g,h)
answered Nov 13 '18 at 23:03
BradCBradC
3,739523
$endgroup$
$begingroup$
can drop 2 chars usingCHAR(8000)instead ofVARCHAR(max)(the huge command is 1453 bytes)
$endgroup$
– Shameen
Nov 14 '18 at 13:21
1
$begingroup$
Thanks, @Shameen. I tried to save one more and useCHAR(2E3)but it didn't seem to like that. I also saved a byte by using a line break literal instead ofCHAR(13).
$endgroup$
– BradC
Nov 14 '18 at 15:14
add a comment |
11
11
11
$begingroup$
T-SQL, 341 338 bytes
DECLARE @ CHAR(2000)=REPLACE(REPLACE(REPLACE(
'PRINT SPACE(14*6&$6*15&$4*5$4*10&$2*4$7*7$1*4&$1*3$8*4$1*3$3*3&*3$9*4$1*3$4
*3&*3$8*4$2*3$4*3&*3$8*4$2*3$4*3&*3$7*4$3*3$4*3&*3$7*4$3*3$4*3&*3$6*11$4*3&
*3$6*11$4*3&*3$5*4$5*3$4*3&*3$5*3$6*3$4*3&$1*4$2*4$11*4&$2*8$9*6&$6*17&$6*3)'
,'*',')+REPLICATE(''#'','),'$',')+SPACE('),'&',')+(''
''')EXEC(@)
The first 4 line breaks are for readability only, the final line break is part of a string literal.
Similar to my Adam West tribute, I've manually encoded a long string, and made the following replacements:
*7gets replaced by+REPLICATE('#',7)
$4gets replaced by+SPACE(4)
&gets replaced by a line break inside quotes
This results in a massive SQL command string:
PRINT SPACE(14)+REPLICATE('#',6)+('
')+SPACE(6)+REPLICATE('#',15)+('
')+SPACE(4)+REPLICATE('#',5)+SPACE(4)+REPLICATE('#',10)+('
')+SPACE(2)+REPLICATE('#',4)+SPACE(7)+REPLICATE('#',7)+SPACE(1)+REPLICATE('#',4)+('
')+SPACE(1)+REPLICATE('#',3)+SPACE(8)+REPLICATE('#',4)+SPACE(1)+REPLICATE('#',3)+SPACE(3)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(9)+REPLICATE('#',4)+SPACE(1)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(8)+REPLICATE('#',4)+SPACE(2)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(8)+REPLICATE('#',4)+SPACE(2)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(7)+REPLICATE('#',4)+SPACE(3)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(7)+REPLICATE('#',4)+SPACE(3)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(6)+REPLICATE('#',11)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(6)+REPLICATE('#',11)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(5)+REPLICATE('#',4)+SPACE(5)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(5)+REPLICATE('#',3)+SPACE(6)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+SPACE(1)+REPLICATE('#',4)+SPACE(2)+REPLICATE('#',4)+SPACE(11)+REPLICATE('#',4)+('
')+SPACE(2)+REPLICATE('#',8)+SPACE(9)+REPLICATE('#',6)+('
')+SPACE(6)+REPLICATE('#',17)+('
')+SPACE(6)+REPLICATE('#',3)
Which, when run, produces the necessary output.
Long, but still better than my best set-based solution (463 bytes):
SELECT SPACE(a)+REPLICATE('#',b)+SPACE(c)+REPLICATE('#',d)
+SPACE(e)+REPLICATE('#',f)+SPACE(g)+REPLICATE('#',h)
FROM(VALUES(7,0,7,6,0,0,0,0),(6,8,0,7,0,0,0,0),(4,5,4,5,0,5,0,0),(2,4,7,7,1,4,0,0),
(1,3,8,4,1,3,3,3),(0,3,9,4,1,3,4,3),(0,3,8,4,2,3,4,3),(0,3,8,4,2,3,4,3),
(0,3,7,4,3,3,4,3),(0,3,7,4,3,3,4,3),(0,3,6,6,0,5,4,3),(0,3,6,6,0,5,4,3),
(0,3,5,4,5,3,4,3),(0,3,5,3,6,3,4,3),(1,4,2,4,5,0,6,4),(2,8,9,6,0,0,0,0),
(6,9,0,8,0,0,0,0),(6,3,0,0,0,0,0,0))t(a,b,c,d,e,f,g,h)
answered Nov 13 '18 at 23:03
BradCBradC
3,739523
$endgroup$
T-SQL, 341 338 bytes
DECLARE @ CHAR(2000)=REPLACE(REPLACE(REPLACE(
'PRINT SPACE(14*6&$6*15&$4*5$4*10&$2*4$7*7$1*4&$1*3$8*4$1*3$3*3&*3$9*4$1*3$4
*3&*3$8*4$2*3$4*3&*3$8*4$2*3$4*3&*3$7*4$3*3$4*3&*3$7*4$3*3$4*3&*3$6*11$4*3&
*3$6*11$4*3&*3$5*4$5*3$4*3&*3$5*3$6*3$4*3&$1*4$2*4$11*4&$2*8$9*6&$6*17&$6*3)'
,'*',')+REPLICATE(''#'','),'$',')+SPACE('),'&',')+(''
''')EXEC(@)
The first 4 line breaks are for readability only, the final line break is part of a string literal.
Similar to my Adam West tribute, I've manually encoded a long string, and made the following replacements:
*7gets replaced by+REPLICATE('#',7)
$4gets replaced by+SPACE(4)
&gets replaced by a line break inside quotes
This results in a massive SQL command string:
PRINT SPACE(14)+REPLICATE('#',6)+('
')+SPACE(6)+REPLICATE('#',15)+('
')+SPACE(4)+REPLICATE('#',5)+SPACE(4)+REPLICATE('#',10)+('
')+SPACE(2)+REPLICATE('#',4)+SPACE(7)+REPLICATE('#',7)+SPACE(1)+REPLICATE('#',4)+('
')+SPACE(1)+REPLICATE('#',3)+SPACE(8)+REPLICATE('#',4)+SPACE(1)+REPLICATE('#',3)+SPACE(3)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(9)+REPLICATE('#',4)+SPACE(1)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(8)+REPLICATE('#',4)+SPACE(2)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(8)+REPLICATE('#',4)+SPACE(2)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(7)+REPLICATE('#',4)+SPACE(3)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(7)+REPLICATE('#',4)+SPACE(3)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(6)+REPLICATE('#',11)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(6)+REPLICATE('#',11)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(5)+REPLICATE('#',4)+SPACE(5)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+REPLICATE('#',3)+SPACE(5)+REPLICATE('#',3)+SPACE(6)+REPLICATE('#',3)+SPACE(4)+REPLICATE('#',3)+('
')+SPACE(1)+REPLICATE('#',4)+SPACE(2)+REPLICATE('#',4)+SPACE(11)+REPLICATE('#',4)+('
')+SPACE(2)+REPLICATE('#',8)+SPACE(9)+REPLICATE('#',6)+('
')+SPACE(6)+REPLICATE('#',17)+('
')+SPACE(6)+REPLICATE('#',3)
Which, when run, produces the necessary output.
Long, but still better than my best set-based solution (463 bytes):
SELECT SPACE(a)+REPLICATE('#',b)+SPACE(c)+REPLICATE('#',d)
+SPACE(e)+REPLICATE('#',f)+SPACE(g)+REPLICATE('#',h)
FROM(VALUES(7,0,7,6,0,0,0,0),(6,8,0,7,0,0,0,0),(4,5,4,5,0,5,0,0),(2,4,7,7,1,4,0,0),
(1,3,8,4,1,3,3,3),(0,3,9,4,1,3,4,3),(0,3,8,4,2,3,4,3),(0,3,8,4,2,3,4,3),
(0,3,7,4,3,3,4,3),(0,3,7,4,3,3,4,3),(0,3,6,6,0,5,4,3),(0,3,6,6,0,5,4,3),
(0,3,5,4,5,3,4,3),(0,3,5,3,6,3,4,3),(1,4,2,4,5,0,6,4),(2,8,9,6,0,0,0,0),
(6,9,0,8,0,0,0,0),(6,3,0,0,0,0,0,0))t(a,b,c,d,e,f,g,h)
answered Nov 13 '18 at 23:03
BradCBradC
3,739523
edited Nov 14 '18 at 16:21
answered Nov 13 '18 at 23:03
BradCBradC
3,739523
answered Nov 13 '18 at 23:03
BradCBradC
3,739523
answered Nov 13 '18 at 23:03
BradCBradC
3,739523
3,739523
$begingroup$
can drop 2 chars usingCHAR(8000)instead ofVARCHAR(max)(the huge command is 1453 bytes)
$endgroup$
– Shameen
Nov 14 '18 at 13:21
1
$begingroup$
Thanks, @Shameen. I tried to save one more and useCHAR(2E3)but it didn't seem to like that. I also saved a byte by using a line break literal instead ofCHAR(13).
$endgroup$
– BradC
Nov 14 '18 at 15:14
add a comment |
$begingroup$
can drop 2 chars usingCHAR(8000)instead ofVARCHAR(max)(the huge command is 1453 bytes)
$endgroup$
– Shameen
Nov 14 '18 at 13:21
1
$begingroup$
Thanks, @Shameen. I tried to save one more and useCHAR(2E3)but it didn't seem to like that. I also saved a byte by using a line break literal instead ofCHAR(13).
$endgroup$
– BradC
Nov 14 '18 at 15:14
$begingroup$
can drop 2 chars using
CHAR(8000) instead of VARCHAR(max) (the huge command is 1453 bytes)$endgroup$
– Shameen
Nov 14 '18 at 13:21
$begingroup$
can drop 2 chars using
CHAR(8000) instead of VARCHAR(max) (the huge command is 1453 bytes)$endgroup$
– Shameen
Nov 14 '18 at 13:21
1
1
$begingroup$
Thanks, @Shameen. I tried to save one more and use
CHAR(2E3) but it didn't seem to like that. I also saved a byte by using a line break literal instead of CHAR(13).$endgroup$
– BradC
Nov 14 '18 at 15:14
$begingroup$
Thanks, @Shameen. I tried to save one more and use
CHAR(2E3) but it didn't seem to like that. I also saved a byte by using a line break literal instead of CHAR(13).$endgroup$
– BradC
Nov 14 '18 at 15:14
add a comment |
10
$begingroup$
Canvas, 74 73 71 bytes
ｑｃ２－Ａｚ↓n⁴╫ｍ┬ｆｆ╷＼_↘Tｔ)%⁵6＞Ｙy7pＱ∔Ｉ％SIŗ＋T＾≤?↔↖¶８^`O‾+Ｏ│≤n≡j↶82„２┬｛ #＠］∑‾＋ｎ
Try it here!
answered Nov 13 '18 at 18:27
dzaimadzaima
14.6k21755
$endgroup$
$begingroup$
im just curious, how u guys type those chars? u have special keyboard or something?
$endgroup$
– Nikko Khresna
Nov 16 '18 at 14:24
$begingroup$
@NikkoKhresna 85% of my code is the data generated by a generator, but the last 11 bytes doing all the logic were typed in the Canvas dev mode and spamming of tab after every char. That's only the case for me & Canvas though, e.g. AFAIK all of Jellys chars are typable with linux compose sequences.
$endgroup$
– dzaima
Nov 16 '18 at 14:44
$begingroup$
@NikkoKhresna yeah, a lot of languages have different ways of typing the chars. some are typable with US keyboard (Jelly), some require pasting, and some (Charcoal, Actually and Sesos afaik) have a sort of verbose mode which only requires you to know the command names
$endgroup$
– ASCII-only
Nov 18 '18 at 1:37
$begingroup$
@NikkoKhresna Charcoal also has (a very, very outdated) Linux keyboard, and probably some other languages too
$endgroup$
– ASCII-only
Nov 18 '18 at 1:39
add a comment |
10
$begingroup$
Canvas, 74 73 71 bytes
ｑｃ２－Ａｚ↓n⁴╫ｍ┬ｆｆ╷＼_↘Tｔ)%⁵6＞Ｙy7pＱ∔Ｉ％SIŗ＋T＾≤?↔↖¶８^`O‾+Ｏ│≤n≡j↶82„２┬｛ #＠］∑‾＋ｎ
Try it here!
answered Nov 13 '18 at 18:27
dzaimadzaima
14.6k21755
$endgroup$
$begingroup$
im just curious, how u guys type those chars? u have special keyboard or something?
$endgroup$
– Nikko Khresna
Nov 16 '18 at 14:24
$begingroup$
@NikkoKhresna 85% of my code is the data generated by a generator, but the last 11 bytes doing all the logic were typed in the Canvas dev mode and spamming of tab after every char. That's only the case for me & Canvas though, e.g. AFAIK all of Jellys chars are typable with linux compose sequences.
$endgroup$
– dzaima
Nov 16 '18 at 14:44
$begingroup$
@NikkoKhresna yeah, a lot of languages have different ways of typing the chars. some are typable with US keyboard (Jelly), some require pasting, and some (Charcoal, Actually and Sesos afaik) have a sort of verbose mode which only requires you to know the command names
$endgroup$
– ASCII-only
Nov 18 '18 at 1:37
$begingroup$
@NikkoKhresna Charcoal also has (a very, very outdated) Linux keyboard, and probably some other languages too
$endgroup$
– ASCII-only
Nov 18 '18 at 1:39
add a comment |
10
10
10
$begingroup$
Canvas, 74 73 71 bytes
ｑｃ２－Ａｚ↓n⁴╫ｍ┬ｆｆ╷＼_↘Tｔ)%⁵6＞Ｙy7pＱ∔Ｉ％SIŗ＋T＾≤?↔↖¶８^`O‾+Ｏ│≤n≡j↶82„２┬｛ #＠］∑‾＋ｎ
Try it here!
answered Nov 13 '18 at 18:27
dzaimadzaima
14.6k21755
$endgroup$
Canvas, 74 73 71 bytes
ｑｃ２－Ａｚ↓n⁴╫ｍ┬ｆｆ╷＼_↘Tｔ)%⁵6＞Ｙy7pＱ∔Ｉ％SIŗ＋T＾≤?↔↖¶８^`O‾+Ｏ│≤n≡j↶82„２┬｛ #＠］∑‾＋ｎ
Try it here!
answered Nov 13 '18 at 18:27
dzaimadzaima
14.6k21755
edited Nov 13 '18 at 18:35
answered Nov 13 '18 at 18:27
dzaimadzaima
14.6k21755
answered Nov 13 '18 at 18:27
dzaimadzaima
14.6k21755
answered Nov 13 '18 at 18:27
dzaimadzaima
14.6k21755
14.6k21755
$begingroup$
im just curious, how u guys type those chars? u have special keyboard or something?
$endgroup$
– Nikko Khresna
Nov 16 '18 at 14:24
$begingroup$
@NikkoKhresna 85% of my code is the data generated by a generator, but the last 11 bytes doing all the logic were typed in the Canvas dev mode and spamming of tab after every char. That's only the case for me & Canvas though, e.g. AFAIK all of Jellys chars are typable with linux compose sequences.
$endgroup$
– dzaima
Nov 16 '18 at 14:44
$begingroup$
@NikkoKhresna yeah, a lot of languages have different ways of typing the chars. some are typable with US keyboard (Jelly), some require pasting, and some (Charcoal, Actually and Sesos afaik) have a sort of verbose mode which only requires you to know the command names
$endgroup$
– ASCII-only
Nov 18 '18 at 1:37
$begingroup$
@NikkoKhresna Charcoal also has (a very, very outdated) Linux keyboard, and probably some other languages too
$endgroup$
– ASCII-only
Nov 18 '18 at 1:39
add a comment |
$begingroup$
im just curious, how u guys type those chars? u have special keyboard or something?
$endgroup$
– Nikko Khresna
Nov 16 '18 at 14:24
$begingroup$
@NikkoKhresna 85% of my code is the data generated by a generator, but the last 11 bytes doing all the logic were typed in the Canvas dev mode and spamming of tab after every char. That's only the case for me & Canvas though, e.g. AFAIK all of Jellys chars are typable with linux compose sequences.
$endgroup$
– dzaima
Nov 16 '18 at 14:44
$begingroup$
@NikkoKhresna yeah, a lot of languages have different ways of typing the chars. some are typable with US keyboard (Jelly), some require pasting, and some (Charcoal, Actually and Sesos afaik) have a sort of verbose mode which only requires you to know the command names
$endgroup$
– ASCII-only
Nov 18 '18 at 1:37
$begingroup$
@NikkoKhresna Charcoal also has (a very, very outdated) Linux keyboard, and probably some other languages too
$endgroup$
– ASCII-only
Nov 18 '18 at 1:39
$begingroup$
im just curious, how u guys type those chars? u have special keyboard or something?
$endgroup$
– Nikko Khresna
Nov 16 '18 at 14:24
$begingroup$
im just curious, how u guys type those chars? u have special keyboard or something?
$endgroup$
– Nikko Khresna
Nov 16 '18 at 14:24
$begingroup$
@NikkoKhresna 85% of my code is the data generated by a generator, but the last 11 bytes doing all the logic were typed in the Canvas dev mode and spamming of tab after every char. That's only the case for me & Canvas though, e.g. AFAIK all of Jellys chars are typable with linux compose sequences.
$endgroup$
– dzaima
Nov 16 '18 at 14:44
$begingroup$
@NikkoKhresna 85% of my code is the data generated by a generator, but the last 11 bytes doing all the logic were typed in the Canvas dev mode and spamming of tab after every char. That's only the case for me & Canvas though, e.g. AFAIK all of Jellys chars are typable with linux compose sequences.
$endgroup$
– dzaima
Nov 16 '18 at 14:44
$begingroup$
@NikkoKhresna yeah, a lot of languages have different ways of typing the chars. some are typable with US keyboard (Jelly), some require pasting, and some (Charcoal, Actually and Sesos afaik) have a sort of verbose mode which only requires you to know the command names
$endgroup$
– ASCII-only
Nov 18 '18 at 1:37
$begingroup$
@NikkoKhresna yeah, a lot of languages have different ways of typing the chars. some are typable with US keyboard (Jelly), some require pasting, and some (Charcoal, Actually and Sesos afaik) have a sort of verbose mode which only requires you to know the command names
$endgroup$
– ASCII-only
Nov 18 '18 at 1:37
$begingroup$
@NikkoKhresna Charcoal also has (a very, very outdated) Linux keyboard, and probably some other languages too
$endgroup$
– ASCII-only
Nov 18 '18 at 1:39
$begingroup$
@NikkoKhresna Charcoal also has (a very, very outdated) Linux keyboard, and probably some other languages too
$endgroup$
– ASCII-only
Nov 18 '18 at 1:39
add a comment |
10
$begingroup$
05AB1E, 101 88 64 bytes
„# •∍ΔÎë,½=bOÅ.âαUΔ'òõƶαÔγλ#xÆ]~”FbćÁ˜Ð”wнQ_wā©•12вεN>yи}˜èJ27ô»
-24 bytes by creating a port of @Dennis♦' Jelly answer.
Try it online.
Explanation:
„# # Push string "# "
•∍ΔÎë,½=bOÅ.âαUΔ'òõƶαÔγλ#xÆ]~”FbćÁ˜Ð”wнQ_wā©•
'# Push compressed integer 10250938842396786963034911279002199266186481794751691439873548591280332406943905758890346943197901909163
12в # Convert to Base-12 as list: [3,0,11,6,2,0,11,4,0,11,10,5,4,10,6,4,7,7,1,4,3,3,8,4,1,3,3,3,1,3,9,4,1,3,4,6,8,4,2,3,4,6,8,4,2,3,4,6,7,4,3,3,4,6,7,4,3,3,4,6,6,11,4,6,6,11,4,6,5,4,5,3,4,6,5,3,6,3,4,3,1,4,2,4,11,4,3,8,9,6,8,6,0,11,10,3]
ε } # Map each `y` to:
N> # The index+1
yи # Repeated `y` amount of times
˜ # Flatten the list
è # Index each in the string "# " (with automatic wraparound)
J # Join everything together
27ô # Split into parts of length 27
» # And join by newlines
See this 05AB1E tip of mine (sections How to compress large integers? and How to compress integer lists?) to understand how the compression of the integer and list works.
answered Nov 14 '18 at 7:36
Kevin CruijssenKevin Cruijssen
36.6k555192
$endgroup$
add a comment |
10
$begingroup$
05AB1E, 101 88 64 bytes
„# •∍ΔÎë,½=bOÅ.âαUΔ'òõƶαÔγλ#xÆ]~”FbćÁ˜Ð”wнQ_wā©•12вεN>yи}˜èJ27ô»
-24 bytes by creating a port of @Dennis♦' Jelly answer.
Try it online.
Explanation:
„# # Push string "# "
•∍ΔÎë,½=bOÅ.âαUΔ'òõƶαÔγλ#xÆ]~”FbćÁ˜Ð”wнQ_wā©•
'# Push compressed integer 10250938842396786963034911279002199266186481794751691439873548591280332406943905758890346943197901909163
12в # Convert to Base-12 as list: [3,0,11,6,2,0,11,4,0,11,10,5,4,10,6,4,7,7,1,4,3,3,8,4,1,3,3,3,1,3,9,4,1,3,4,6,8,4,2,3,4,6,8,4,2,3,4,6,7,4,3,3,4,6,7,4,3,3,4,6,6,11,4,6,6,11,4,6,5,4,5,3,4,6,5,3,6,3,4,3,1,4,2,4,11,4,3,8,9,6,8,6,0,11,10,3]
ε } # Map each `y` to:
N> # The index+1
yи # Repeated `y` amount of times
˜ # Flatten the list
è # Index each in the string "# " (with automatic wraparound)
J # Join everything together
27ô # Split into parts of length 27
» # And join by newlines
See this 05AB1E tip of mine (sections How to compress large integers? and How to compress integer lists?) to understand how the compression of the integer and list works.
answered Nov 14 '18 at 7:36
Kevin CruijssenKevin Cruijssen
36.6k555192
$endgroup$
add a comment |
10
10
10
$begingroup$
05AB1E, 101 88 64 bytes
„# •∍ΔÎë,½=bOÅ.âαUΔ'òõƶαÔγλ#xÆ]~”FbćÁ˜Ð”wнQ_wā©•12вεN>yи}˜èJ27ô»
-24 bytes by creating a port of @Dennis♦' Jelly answer.
Try it online.
Explanation:
„# # Push string "# "
•∍ΔÎë,½=bOÅ.âαUΔ'òõƶαÔγλ#xÆ]~”FbćÁ˜Ð”wнQ_wā©•
'# Push compressed integer 10250938842396786963034911279002199266186481794751691439873548591280332406943905758890346943197901909163
12в # Convert to Base-12 as list: [3,0,11,6,2,0,11,4,0,11,10,5,4,10,6,4,7,7,1,4,3,3,8,4,1,3,3,3,1,3,9,4,1,3,4,6,8,4,2,3,4,6,8,4,2,3,4,6,7,4,3,3,4,6,7,4,3,3,4,6,6,11,4,6,6,11,4,6,5,4,5,3,4,6,5,3,6,3,4,3,1,4,2,4,11,4,3,8,9,6,8,6,0,11,10,3]
ε } # Map each `y` to:
N> # The index+1
yи # Repeated `y` amount of times
˜ # Flatten the list
è # Index each in the string "# " (with automatic wraparound)
J # Join everything together
27ô # Split into parts of length 27
» # And join by newlines
See this 05AB1E tip of mine (sections How to compress large integers? and How to compress integer lists?) to understand how the compression of the integer and list works.
answered Nov 14 '18 at 7:36
Kevin CruijssenKevin Cruijssen
36.6k555192
$endgroup$
05AB1E, 101 88 64 bytes
„# •∍ΔÎë,½=bOÅ.âαUΔ'òõƶαÔγλ#xÆ]~”FbćÁ˜Ð”wнQ_wā©•12вεN>yи}˜èJ27ô»
-24 bytes by creating a port of @Dennis♦' Jelly answer.
Try it online.
Explanation:
„# # Push string "# "
•∍ΔÎë,½=bOÅ.âαUΔ'òõƶαÔγλ#xÆ]~”FbćÁ˜Ð”wнQ_wā©•
'# Push compressed integer 10250938842396786963034911279002199266186481794751691439873548591280332406943905758890346943197901909163
12в # Convert to Base-12 as list: [3,0,11,6,2,0,11,4,0,11,10,5,4,10,6,4,7,7,1,4,3,3,8,4,1,3,3,3,1,3,9,4,1,3,4,6,8,4,2,3,4,6,8,4,2,3,4,6,7,4,3,3,4,6,7,4,3,3,4,6,6,11,4,6,6,11,4,6,5,4,5,3,4,6,5,3,6,3,4,3,1,4,2,4,11,4,3,8,9,6,8,6,0,11,10,3]
ε } # Map each `y` to:
N> # The index+1
yи # Repeated `y` amount of times
˜ # Flatten the list
è # Index each in the string "# " (with automatic wraparound)
J # Join everything together
27ô # Split into parts of length 27
» # And join by newlines
See this 05AB1E tip of mine (sections How to compress large integers? and How to compress integer lists?) to understand how the compression of the integer and list works.
answered Nov 14 '18 at 7:36
Kevin CruijssenKevin Cruijssen
36.6k555192
edited Nov 14 '18 at 13:01
answered Nov 14 '18 at 7:36
Kevin CruijssenKevin Cruijssen
36.6k555192
answered Nov 14 '18 at 7:36
Kevin CruijssenKevin Cruijssen
36.6k555192
answered Nov 14 '18 at 7:36
Kevin CruijssenKevin Cruijssen
36.6k555192
36.6k555192
add a comment |
add a comment |
9
$begingroup$
JavaScript (ES6), 173 170 bytes
_=>`tc
du
9a9k
58fe38
36h83676
6j83!h85!h85!f87!f87!dm96
6dm96
6b8b!b6d696
3858n8
5gjc
dy
d6`.split`!`.join`696
6`.replace(/./g,c=>'# '[(k=parseInt(c,36))&1].repeat(k/2))
Try it online!
Node.js, 163 bytes
Provided that an array of strings is a valid output:
_=>Buffer(`.&-/*%$*&$''!$##($!###!#)$!#$&($"#$&($"#$&'$##$&'$##$&&+$&&+$&%$%#$&%#&#$#!$"$+$#()&(1*#2`).map(c=>s+='# '[x^=1].repeat(c-32),s=x='')&&s.match(/.{27}/g)
Try it online!
answered Nov 13 '18 at 18:05
ArnauldArnauld
73.7k689309
$endgroup$
add a comment |
9
$begingroup$
JavaScript (ES6), 173 170 bytes
_=>`tc
du
9a9k
58fe38
36h83676
6j83!h85!h85!f87!f87!dm96
6dm96
6b8b!b6d696
3858n8
5gjc
dy
d6`.split`!`.join`696
6`.replace(/./g,c=>'# '[(k=parseInt(c,36))&1].repeat(k/2))
Try it online!
Node.js, 163 bytes
Provided that an array of strings is a valid output:
_=>Buffer(`.&-/*%$*&$''!$##($!###!#)$!#$&($"#$&($"#$&'$##$&'$##$&&+$&&+$&%$%#$&%#&#$#!$"$+$#()&(1*#2`).map(c=>s+='# '[x^=1].repeat(c-32),s=x='')&&s.match(/.{27}/g)
Try it online!
answered Nov 13 '18 at 18:05
ArnauldArnauld
73.7k689309
$endgroup$
add a comment |
9
9
9
$begingroup$
JavaScript (ES6), 173 170 bytes
_=>`tc
du
9a9k
58fe38
36h83676
6j83!h85!h85!f87!f87!dm96
6dm96
6b8b!b6d696
3858n8
5gjc
dy
d6`.split`!`.join`696
6`.replace(/./g,c=>'# '[(k=parseInt(c,36))&1].repeat(k/2))
Try it online!
Node.js, 163 bytes
Provided that an array of strings is a valid output:
_=>Buffer(`.&-/*%$*&$''!$##($!###!#)$!#$&($"#$&($"#$&'$##$&'$##$&&+$&&+$&%$%#$&%#&#$#!$"$+$#()&(1*#2`).map(c=>s+='# '[x^=1].repeat(c-32),s=x='')&&s.match(/.{27}/g)
Try it online!
answered Nov 13 '18 at 18:05
ArnauldArnauld
73.7k689309
$endgroup$
JavaScript (ES6), 173 170 bytes
_=>`tc
du
9a9k
58fe38
36h83676
6j83!h85!h85!f87!f87!dm96
6dm96
6b8b!b6d696
3858n8
5gjc
dy
d6`.split`!`.join`696
6`.replace(/./g,c=>'# '[(k=parseInt(c,36))&1].repeat(k/2))
Try it online!
Node.js, 163 bytes
Provided that an array of strings is a valid output:
_=>Buffer(`.&-/*%$*&$''!$##($!###!#)$!#$&($"#$&($"#$&'$##$&'$##$&&+$&&+$&%$%#$&%#&#$#!$"$+$#()&(1*#2`).map(c=>s+='# '[x^=1].repeat(c-32),s=x='')&&s.match(/.{27}/g)
Try it online!
answered Nov 13 '18 at 18:05
ArnauldArnauld
73.7k689309
edited Nov 13 '18 at 19:34
answered Nov 13 '18 at 18:05
ArnauldArnauld
73.7k689309
answered Nov 13 '18 at 18:05
ArnauldArnauld
73.7k689309
answered Nov 13 '18 at 18:05
ArnauldArnauld
73.7k689309
73.7k689309
add a comment |
add a comment |
8
$begingroup$
///, 170 bytes
/+/////*/ +)/'$# +(/"$!+'/" +&/!#+%/"!
!+$/*!+"/**+!/###/""('*
(!!!'
"&#"!!&"
$#'$& &*
!""& ! $
!"'& !%""&$%""&$%)$%)$%(&#%(&#%'&'!%'!"$"!
&$#")
$&#"'!!*
(!!&#"
"$
Try it online!
answered Nov 13 '18 at 19:02
Conor O'BrienConor O'Brien
29.2k263162
$endgroup$
add a comment |
8
$begingroup$
///, 170 bytes
/+/////*/ +)/'$# +(/"$!+'/" +&/!#+%/"!
!+$/*!+"/**+!/###/""('*
(!!!'
"&#"!!&"
$#'$& &*
!""& ! $
!"'& !%""&$%""&$%)$%)$%(&#%(&#%'&'!%'!"$"!
&$#")
$&#"'!!*
(!!&#"
"$
Try it online!
answered Nov 13 '18 at 19:02
Conor O'BrienConor O'Brien
29.2k263162
$endgroup$
add a comment |
8
8
8
$begingroup$
///, 170 bytes
/+/////*/ +)/'$# +(/"$!+'/" +&/!#+%/"!
!+$/*!+"/**+!/###/""('*
(!!!'
"&#"!!&"
$#'$& &*
!""& ! $
!"'& !%""&$%""&$%)$%)$%(&#%(&#%'&'!%'!"$"!
&$#")
$&#"'!!*
(!!&#"
"$
Try it online!
answered Nov 13 '18 at 19:02
Conor O'BrienConor O'Brien
29.2k263162
$endgroup$
///, 170 bytes
/+/////*/ +)/'$# +(/"$!+'/" +&/!#+%/"!
!+$/*!+"/**+!/###/""('*
(!!!'
"&#"!!&"
$#'$& &*
!""& ! $
!"'& !%""&$%""&$%)$%)$%(&#%(&#%'&'!%'!"$"!
&$#")
$&#"'!!*
(!!&#"
"$
Try it online!
answered Nov 13 '18 at 19:02
Conor O'BrienConor O'Brien
29.2k263162
answered Nov 13 '18 at 19:02
Conor O'BrienConor O'Brien
29.2k263162
answered Nov 13 '18 at 19:02
Conor O'BrienConor O'Brien
29.2k263162
answered Nov 13 '18 at 19:02
Conor O'BrienConor O'Brien
29.2k263162
29.2k263162
add a comment |
add a comment |
8
$begingroup$
C (gcc), 174 168 164 bytes
-6 bytes thanks to Dennis.
i,j=88;main(l){while(j--)for(i="CJQHFIHCDKDBDACDCFCEFDCEDEFDKFFDKFFDCCDGFDCCDGFDCBDHFDCBDHFDCADICACCCADHCCDAGGDFJDEJOMFN"[j]-64;i;)putchar(l++%28?--i,35-j%2*3:10);}
Try it online!
answered Nov 13 '18 at 22:10
gastropnergastropner
2,0501411
$endgroup$
add a comment |
8
$begingroup$
C (gcc), 174 168 164 bytes
-6 bytes thanks to Dennis.
i,j=88;main(l){while(j--)for(i="CJQHFIHCDKDBDACDCFCEFDCEDEFDKFFDKFFDCCDGFDCCDGFDCBDHFDCBDHFDCADICACCCADHCCDAGGDFJDEJOMFN"[j]-64;i;)putchar(l++%28?--i,35-j%2*3:10);}
Try it online!
answered Nov 13 '18 at 22:10
gastropnergastropner
2,0501411
$endgroup$
add a comment |
8
8
8
$begingroup$
C (gcc), 174 168 164 bytes
-6 bytes thanks to Dennis.
i,j=88;main(l){while(j--)for(i="CJQHFIHCDKDBDACDCFCEFDCEDEFDKFFDKFFDCCDGFDCCDGFDCBDHFDCBDHFDCADICACCCADHCCDAGGDFJDEJOMFN"[j]-64;i;)putchar(l++%28?--i,35-j%2*3:10);}
Try it online!
answered Nov 13 '18 at 22:10
gastropnergastropner
2,0501411
$endgroup$
C (gcc), 174 168 164 bytes
-6 bytes thanks to Dennis.
i,j=88;main(l){while(j--)for(i="CJQHFIHCDKDBDACDCFCEFDCEDEFDKFFDKFFDCCDGFDCCDGFDCBDHFDCBDHFDCADICACCCADHCCDAGGDFJDEJOMFN"[j]-64;i;)putchar(l++%28?--i,35-j%2*3:10);}
Try it online!
answered Nov 13 '18 at 22:10
gastropnergastropner
2,0501411
edited Nov 14 '18 at 6:29
answered Nov 13 '18 at 22:10
gastropnergastropner
2,0501411
answered Nov 13 '18 at 22:10
gastropnergastropner
2,0501411
answered Nov 13 '18 at 22:10
gastropnergastropner
2,0501411
2,0501411
add a comment |
add a comment |
7
$begingroup$
Bash, 192 176 bytes
dc<<<"16i2o81C0000 81FFFF0 9FE00FC BCF001E F070387 F078387 F03FF87 F03FF87 F01E387 F01E387 F00F387 F00F387 F007B87 B807B8E 9E03FBC 87C3FF0 81FFFC0 8001F80f"|tr 01 ' #'|cut -c2-
Try it online!
-16 thanks to manatwork
This is similar to my C answer, except it just uses a raw base-16 compression and passes it through bc, then uses tr to convert 1 to # and 0 to space. Each row has 1 appended to it and stripped off of it to maintain alignment.
Unfortunately dc is shorter than bc.
answered Nov 13 '18 at 21:00
LambdaBetaLambdaBeta
2,109418
$endgroup$
$begingroup$
No need for^. But even better, usecut -c2-instead of thesedpart.
$endgroup$
– manatwork
Nov 13 '18 at 22:59
$begingroup$
And also shorter with here-string instead ofechoand even shorter withdcinstead ofbc: Try it online!
$endgroup$
– manatwork
Nov 14 '18 at 9:07
2
$begingroup$
I'll give you theecho, but come on -dcfor Stan Lee... you must be joking ;)
$endgroup$
– LambdaBeta
Nov 14 '18 at 15:31
1
$begingroup$
134 bytes
$endgroup$
– Dennis♦
Nov 14 '18 at 18:55
add a comment |
7
$begingroup$
Bash, 192 176 bytes
dc<<<"16i2o81C0000 81FFFF0 9FE00FC BCF001E F070387 F078387 F03FF87 F03FF87 F01E387 F01E387 F00F387 F00F387 F007B87 B807B8E 9E03FBC 87C3FF0 81FFFC0 8001F80f"|tr 01 ' #'|cut -c2-
Try it online!
-16 thanks to manatwork
This is similar to my C answer, except it just uses a raw base-16 compression and passes it through bc, then uses tr to convert 1 to # and 0 to space. Each row has 1 appended to it and stripped off of it to maintain alignment.
Unfortunately dc is shorter than bc.
answered Nov 13 '18 at 21:00
LambdaBetaLambdaBeta
2,109418
$endgroup$
$begingroup$
No need for^. But even better, usecut -c2-instead of thesedpart.
$endgroup$
– manatwork
Nov 13 '18 at 22:59
$begingroup$
And also shorter with here-string instead ofechoand even shorter withdcinstead ofbc: Try it online!
$endgroup$
– manatwork
Nov 14 '18 at 9:07
2
$begingroup$
I'll give you theecho, but come on -dcfor Stan Lee... you must be joking ;)
$endgroup$
– LambdaBeta
Nov 14 '18 at 15:31
1
$begingroup$
134 bytes
$endgroup$
– Dennis♦
Nov 14 '18 at 18:55
add a comment |
7
7
7
$begingroup$
Bash, 192 176 bytes
dc<<<"16i2o81C0000 81FFFF0 9FE00FC BCF001E F070387 F078387 F03FF87 F03FF87 F01E387 F01E387 F00F387 F00F387 F007B87 B807B8E 9E03FBC 87C3FF0 81FFFC0 8001F80f"|tr 01 ' #'|cut -c2-
Try it online!
-16 thanks to manatwork
This is similar to my C answer, except it just uses a raw base-16 compression and passes it through bc, then uses tr to convert 1 to # and 0 to space. Each row has 1 appended to it and stripped off of it to maintain alignment.
Unfortunately dc is shorter than bc.
answered Nov 13 '18 at 21:00
LambdaBetaLambdaBeta
2,109418
$endgroup$
Bash, 192 176 bytes
dc<<<"16i2o81C0000 81FFFF0 9FE00FC BCF001E F070387 F078387 F03FF87 F03FF87 F01E387 F01E387 F00F387 F00F387 F007B87 B807B8E 9E03FBC 87C3FF0 81FFFC0 8001F80f"|tr 01 ' #'|cut -c2-
Try it online!
-16 thanks to manatwork
This is similar to my C answer, except it just uses a raw base-16 compression and passes it through bc, then uses tr to convert 1 to # and 0 to space. Each row has 1 appended to it and stripped off of it to maintain alignment.
Unfortunately dc is shorter than bc.
answered Nov 13 '18 at 21:00
LambdaBetaLambdaBeta
2,109418
edited Nov 14 '18 at 15:32
answered Nov 13 '18 at 21:00
LambdaBetaLambdaBeta
2,109418
answered Nov 13 '18 at 21:00
LambdaBetaLambdaBeta
2,109418
answered Nov 13 '18 at 21:00
LambdaBetaLambdaBeta
2,109418
2,109418
$begingroup$
No need for^. But even better, usecut -c2-instead of thesedpart.
$endgroup$
– manatwork
Nov 13 '18 at 22:59
$begingroup$
And also shorter with here-string instead ofechoand even shorter withdcinstead ofbc: Try it online!
$endgroup$
– manatwork
Nov 14 '18 at 9:07
2
$begingroup$
I'll give you theecho, but come on -dcfor Stan Lee... you must be joking ;)
$endgroup$
– LambdaBeta
Nov 14 '18 at 15:31
1
$begingroup$
134 bytes
$endgroup$
– Dennis♦
Nov 14 '18 at 18:55
add a comment |
$begingroup$
No need for^. But even better, usecut -c2-instead of thesedpart.
$endgroup$
– manatwork
Nov 13 '18 at 22:59
$begingroup$
And also shorter with here-string instead ofechoand even shorter withdcinstead ofbc: Try it online!
$endgroup$
– manatwork
Nov 14 '18 at 9:07
2
$begingroup$
I'll give you theecho, but come on -dcfor Stan Lee... you must be joking ;)
$endgroup$
– LambdaBeta
Nov 14 '18 at 15:31
1
$begingroup$
134 bytes
$endgroup$
– Dennis♦
Nov 14 '18 at 18:55
$begingroup$
No need for
^. But even better, use cut -c2- instead of the sed part.$endgroup$
– manatwork
Nov 13 '18 at 22:59
$begingroup$
No need for
^. But even better, use cut -c2- instead of the sed part.$endgroup$
– manatwork
Nov 13 '18 at 22:59
$begingroup$
And also shorter with here-string instead of
echo and even shorter with dc instead of bc: Try it online!$endgroup$
– manatwork
Nov 14 '18 at 9:07
$begingroup$
And also shorter with here-string instead of
echo and even shorter with dc instead of bc: Try it online!$endgroup$
– manatwork
Nov 14 '18 at 9:07
2
2
$begingroup$
I'll give you the
echo, but come on - dc for Stan Lee... you must be joking ;)$endgroup$
– LambdaBeta
Nov 14 '18 at 15:31
$begingroup$
I'll give you the
echo, but come on - dc for Stan Lee... you must be joking ;)$endgroup$
– LambdaBeta
Nov 14 '18 at 15:31
1
1
$begingroup$
134 bytes
$endgroup$
– Dennis♦
Nov 14 '18 at 18:55
$begingroup$
134 bytes
$endgroup$
– Dennis♦
Nov 14 '18 at 18:55
add a comment |
6
$begingroup$
Haskell, 170 163 bytes
Edit: -7 bytes thanks to @Ørjan Johansen
m=<<"Mc Ul WeWg YdTgZa ZcSdZcX cRdZcW cSdYcW cSdYcW cTdXcW cTdXcW cUkW cUkW cVdVcW cVcUcW ZdYdPa YhRc Un U "
m ' '="###n"
m c=(' '<$['Z','Y'..c])++('#'<$['a'..c])
Try it online!
Spaces are encoded as uppercase characters (length: Z down to char), hash signs as lowercase characters (length: a to char) and the last three # of each line plus newline as a space. Function m decodes it.
answered Nov 14 '18 at 6:20
niminimi
31.5k32185
$endgroup$
1
$begingroup$
I think you can save some by letting space encode"###n".
$endgroup$
– Ørjan Johansen
Nov 14 '18 at 6:36
add a comment |
6
$begingroup$
Haskell, 170 163 bytes
Edit: -7 bytes thanks to @Ørjan Johansen
m=<<"Mc Ul WeWg YdTgZa ZcSdZcX cRdZcW cSdYcW cSdYcW cTdXcW cTdXcW cUkW cUkW cVdVcW cVcUcW ZdYdPa YhRc Un U "
m ' '="###n"
m c=(' '<$['Z','Y'..c])++('#'<$['a'..c])
Try it online!
Spaces are encoded as uppercase characters (length: Z down to char), hash signs as lowercase characters (length: a to char) and the last three # of each line plus newline as a space. Function m decodes it.
answered Nov 14 '18 at 6:20
niminimi
31.5k32185
$endgroup$
1
$begingroup$
I think you can save some by letting space encode"###n".
$endgroup$
– Ørjan Johansen
Nov 14 '18 at 6:36
add a comment |
6
6
6
$begingroup$
Haskell, 170 163 bytes
Edit: -7 bytes thanks to @Ørjan Johansen
m=<<"Mc Ul WeWg YdTgZa ZcSdZcX cRdZcW cSdYcW cSdYcW cTdXcW cTdXcW cUkW cUkW cVdVcW cVcUcW ZdYdPa YhRc Un U "
m ' '="###n"
m c=(' '<$['Z','Y'..c])++('#'<$['a'..c])
Try it online!
Spaces are encoded as uppercase characters (length: Z down to char), hash signs as lowercase characters (length: a to char) and the last three # of each line plus newline as a space. Function m decodes it.
answered Nov 14 '18 at 6:20
niminimi
31.5k32185
$endgroup$
Haskell, 170 163 bytes
Edit: -7 bytes thanks to @Ørjan Johansen
m=<<"Mc Ul WeWg YdTgZa ZcSdZcX cRdZcW cSdYcW cSdYcW cTdXcW cTdXcW cUkW cUkW cVdVcW cVcUcW ZdYdPa YhRc Un U "
m ' '="###n"
m c=(' '<$['Z','Y'..c])++('#'<$['a'..c])
Try it online!
Spaces are encoded as uppercase characters (length: Z down to char), hash signs as lowercase characters (length: a to char) and the last three # of each line plus newline as a space. Function m decodes it.
answered Nov 14 '18 at 6:20
niminimi
31.5k32185
edited Nov 14 '18 at 6:50
answered Nov 14 '18 at 6:20
niminimi
31.5k32185
answered Nov 14 '18 at 6:20
niminimi
31.5k32185
answered Nov 14 '18 at 6:20
niminimi
31.5k32185
31.5k32185
1
$begingroup$
I think you can save some by letting space encode"###n".
$endgroup$
– Ørjan Johansen
Nov 14 '18 at 6:36
add a comment |
1
$begingroup$
I think you can save some by letting space encode"###n".
$endgroup$
– Ørjan Johansen
Nov 14 '18 at 6:36
1
1
$begingroup$
I think you can save some by letting space encode
"###n".$endgroup$
– Ørjan Johansen
Nov 14 '18 at 6:36
$begingroup$
I think you can save some by letting space encode
"###n".$endgroup$
– Ørjan Johansen
Nov 14 '18 at 6:36
add a comment |
6
$begingroup$
Perl 6, 136 112 bytes
:128['?@~p<?x?aB~<|xpqpac`?@~x>q|<!Ca`||?x'.ords].base(2)~~TR/1/ /.comb(27)>>.say
Try it online!
Parses a base 128 number from the ordinal values of the string, converts to base 2 and replaces the 1s with spaces. This uses zeroes as the main character. The string has like 9 null bytes, which was a bitch to type up in TIO. I would have used base 127 for the same amount of bytes, but no nulls, but that has a carriage return, which seems to be impossible to type :(
Explanation:
:128['...'.ords] # Convert the bytes of the string to base 127
.base(2) # Convert to base 2
~~TR/1/ / # Translate 1s to spaces
# Alternatively, this could be .split(1)
.comb(27) # Split to strings of length 27
>>.say # Print each on a newline
answered Nov 14 '18 at 3:13
Jo KingJo King
21.4k248110
$endgroup$
add a comment |
6
$begingroup$
Perl 6, 136 112 bytes
:128['?@~p<?x?aB~<|xpqpac`?@~x>q|<!Ca`||?x'.ords].base(2)~~TR/1/ /.comb(27)>>.say
Try it online!
Parses a base 128 number from the ordinal values of the string, converts to base 2 and replaces the 1s with spaces. This uses zeroes as the main character. The string has like 9 null bytes, which was a bitch to type up in TIO. I would have used base 127 for the same amount of bytes, but no nulls, but that has a carriage return, which seems to be impossible to type :(
Explanation:
:128['...'.ords] # Convert the bytes of the string to base 127
.base(2) # Convert to base 2
~~TR/1/ / # Translate 1s to spaces
# Alternatively, this could be .split(1)
.comb(27) # Split to strings of length 27
>>.say # Print each on a newline
answered Nov 14 '18 at 3:13
Jo KingJo King
21.4k248110
$endgroup$
add a comment |
6
6
6
$begingroup$
Perl 6, 136 112 bytes
:128['?@~p<?x?aB~<|xpqpac`?@~x>q|<!Ca`||?x'.ords].base(2)~~TR/1/ /.comb(27)>>.say
Try it online!
Parses a base 128 number from the ordinal values of the string, converts to base 2 and replaces the 1s with spaces. This uses zeroes as the main character. The string has like 9 null bytes, which was a bitch to type up in TIO. I would have used base 127 for the same amount of bytes, but no nulls, but that has a carriage return, which seems to be impossible to type :(
Explanation:
:128['...'.ords] # Convert the bytes of the string to base 127
.base(2) # Convert to base 2
~~TR/1/ / # Translate 1s to spaces
# Alternatively, this could be .split(1)
.comb(27) # Split to strings of length 27
>>.say # Print each on a newline
answered Nov 14 '18 at 3:13
Jo KingJo King
21.4k248110
$endgroup$
Perl 6, 136 112 bytes
:128['?@~p<?x?aB~<|xpqpac`?@~x>q|<!Ca`||?x'.ords].base(2)~~TR/1/ /.comb(27)>>.say
Try it online!
Parses a base 128 number from the ordinal values of the string, converts to base 2 and replaces the 1s with spaces. This uses zeroes as the main character. The string has like 9 null bytes, which was a bitch to type up in TIO. I would have used base 127 for the same amount of bytes, but no nulls, but that has a carriage return, which seems to be impossible to type :(
Explanation:
:128['...'.ords] # Convert the bytes of the string to base 127
.base(2) # Convert to base 2
~~TR/1/ / # Translate 1s to spaces
# Alternatively, this could be .split(1)
.comb(27) # Split to strings of length 27
>>.say # Print each on a newline
answered Nov 14 '18 at 3:13
Jo KingJo King
21.4k248110
edited Nov 17 '18 at 11:46
answered Nov 14 '18 at 3:13
Jo KingJo King
21.4k248110
answered Nov 14 '18 at 3:13
Jo KingJo King
21.4k248110
answered Nov 14 '18 at 3:13
Jo KingJo King
21.4k248110
21.4k248110
add a comment |
add a comment |
5
$begingroup$
J, 130 128 bytes
echo' #'{~18 27$;(_243{.2#.inv 92x#._32+a.i.])&>'!TYPW.ajz i8hIhXl''3lOH8GvV.C2Z{r/=,G';'"a*2ZDxRplkh2tzRakz.?ZwVmeOT6L^lFB^eyT'
Try it online!
Initial solution
J, 164 bytes
echo' #'{~18 27$,#:849239965469633263905532594449192007713271791872263657753301928240007 12380965417202148347902847903517734495157419855048834759608223758433386496x
Try it online!
answered Nov 13 '18 at 20:04
Galen IvanovGalen Ivanov
6,56711032
$endgroup$
add a comment |
5
$begingroup$
J, 130 128 bytes
echo' #'{~18 27$;(_243{.2#.inv 92x#._32+a.i.])&>'!TYPW.ajz i8hIhXl''3lOH8GvV.C2Z{r/=,G';'"a*2ZDxRplkh2tzRakz.?ZwVmeOT6L^lFB^eyT'
Try it online!
Initial solution
J, 164 bytes
echo' #'{~18 27$,#:849239965469633263905532594449192007713271791872263657753301928240007 12380965417202148347902847903517734495157419855048834759608223758433386496x
Try it online!
answered Nov 13 '18 at 20:04
Galen IvanovGalen Ivanov
6,56711032
$endgroup$
add a comment |
5
5
5
$begingroup$
J, 130 128 bytes
echo' #'{~18 27$;(_243{.2#.inv 92x#._32+a.i.])&>'!TYPW.ajz i8hIhXl''3lOH8GvV.C2Z{r/=,G';'"a*2ZDxRplkh2tzRakz.?ZwVmeOT6L^lFB^eyT'
Try it online!
Initial solution
J, 164 bytes
echo' #'{~18 27$,#:849239965469633263905532594449192007713271791872263657753301928240007 12380965417202148347902847903517734495157419855048834759608223758433386496x
Try it online!
answered Nov 13 '18 at 20:04
Galen IvanovGalen Ivanov
6,56711032
$endgroup$
J, 130 128 bytes
echo' #'{~18 27$;(_243{.2#.inv 92x#._32+a.i.])&>'!TYPW.ajz i8hIhXl''3lOH8GvV.C2Z{r/=,G';'"a*2ZDxRplkh2tzRakz.?ZwVmeOT6L^lFB^eyT'
Try it online!
Initial solution
J, 164 bytes
echo' #'{~18 27$,#:849239965469633263905532594449192007713271791872263657753301928240007 12380965417202148347902847903517734495157419855048834759608223758433386496x
Try it online!
answered Nov 13 '18 at 20:04
Galen IvanovGalen Ivanov
6,56711032
edited Nov 14 '18 at 11:15
answered Nov 13 '18 at 20:04
Galen IvanovGalen Ivanov
6,56711032
answered Nov 13 '18 at 20:04
Galen IvanovGalen Ivanov
6,56711032
answered Nov 13 '18 at 20:04
Galen IvanovGalen Ivanov
6,56711032
6,56711032
add a comment |
add a comment |
4
$begingroup$
Perl 5, 181 bytes
say+(sprintf"%28b",oct"0x$_")=~y/10/# /r for"0003f0003fff800f87fe03c07f78700f71ce00f70ee01e70ee01e70ee03c70ee03c70ee07ff0ee07ff0ee0f070ee0e070e79e003c3fc01f803fffe00380000"=~/.{7}/g
Try it online!
answered Nov 13 '18 at 22:14
XcaliXcali
5,238520
$endgroup$
add a comment |
4
$begingroup$
Perl 5, 181 bytes
say+(sprintf"%28b",oct"0x$_")=~y/10/# /r for"0003f0003fff800f87fe03c07f78700f71ce00f70ee01e70ee01e70ee03c70ee03c70ee07ff0ee07ff0ee0f070ee0e070e79e003c3fc01f803fffe00380000"=~/.{7}/g
Try it online!
answered Nov 13 '18 at 22:14
XcaliXcali
5,238520
$endgroup$
add a comment |
4
4
4
$begingroup$
Perl 5, 181 bytes
say+(sprintf"%28b",oct"0x$_")=~y/10/# /r for"0003f0003fff800f87fe03c07f78700f71ce00f70ee01e70ee01e70ee03c70ee03c70ee07ff0ee07ff0ee0f070ee0e070e79e003c3fc01f803fffe00380000"=~/.{7}/g
Try it online!
answered Nov 13 '18 at 22:14
XcaliXcali
5,238520
$endgroup$
Perl 5, 181 bytes
say+(sprintf"%28b",oct"0x$_")=~y/10/# /r for"0003f0003fff800f87fe03c07f78700f71ce00f70ee01e70ee01e70ee03c70ee03c70ee07ff0ee07ff0ee0f070ee0e070e79e003c3fc01f803fffe00380000"=~/.{7}/g
Try it online!
answered Nov 13 '18 at 22:14
XcaliXcali
5,238520
answered Nov 13 '18 at 22:14
XcaliXcali
5,238520
answered Nov 13 '18 at 22:14
XcaliXcali
5,238520
answered Nov 13 '18 at 22:14
XcaliXcali
5,238520
5,238520
add a comment |
add a comment |
4
$begingroup$
MATLAB : 144 Bytes
reshape(repelem([repmat(' #',1,44),' '],'NFMOJEDJFDGGADCCHDACCCACIDACDFHDBCDFHDBCDFGDCCDFGDCCDFFKDFFKDFEDECDFECFCDCADBDKDCHIFHQJCR'-64),27,)'
Try it online! (Technically in Octave)
Explanation:
This uses the same strategy as digEmAll in R, just with MATLAB syntax. The main difference is that MATLAB has automatic conversion from characters to integers.
answered Nov 14 '18 at 16:36
Nicky MattssonNicky Mattsson
1414
$endgroup$
add a comment |
4
$begingroup$
MATLAB : 144 Bytes
reshape(repelem([repmat(' #',1,44),' '],'NFMOJEDJFDGGADCCHDACCCACIDACDFHDBCDFHDBCDFGDCCDFGDCCDFFKDFFKDFEDECDFECFCDCADBDKDCHIFHQJCR'-64),27,)'
Try it online! (Technically in Octave)
Explanation:
This uses the same strategy as digEmAll in R, just with MATLAB syntax. The main difference is that MATLAB has automatic conversion from characters to integers.
answered Nov 14 '18 at 16:36
Nicky MattssonNicky Mattsson
1414
$endgroup$
add a comment |
4
4
4
$begingroup$
MATLAB : 144 Bytes
reshape(repelem([repmat(' #',1,44),' '],'NFMOJEDJFDGGADCCHDACCCACIDACDFHDBCDFHDBCDFGDCCDFGDCCDFFKDFFKDFEDECDFECFCDCADBDKDCHIFHQJCR'-64),27,)'
Try it online! (Technically in Octave)
Explanation:
This uses the same strategy as digEmAll in R, just with MATLAB syntax. The main difference is that MATLAB has automatic conversion from characters to integers.
answered Nov 14 '18 at 16:36
Nicky MattssonNicky Mattsson
1414
$endgroup$
MATLAB : 144 Bytes
reshape(repelem([repmat(' #',1,44),' '],'NFMOJEDJFDGGADCCHDACCCACIDACDFHDBCDFHDBCDFGDCCDFGDCCDFFKDFFKDFEDECDFECFCDCADBDKDCHIFHQJCR'-64),27,)'
Try it online! (Technically in Octave)
Explanation:
This uses the same strategy as digEmAll in R, just with MATLAB syntax. The main difference is that MATLAB has automatic conversion from characters to integers.
answered Nov 14 '18 at 16:36
Nicky MattssonNicky Mattsson
1414
edited Nov 14 '18 at 19:35
answered Nov 14 '18 at 16:36
Nicky MattssonNicky Mattsson
1414
answered Nov 14 '18 at 16:36
Nicky MattssonNicky Mattsson
1414
answered Nov 14 '18 at 16:36
Nicky MattssonNicky Mattsson
1414
1414
add a comment |
add a comment |
4
$begingroup$
PHP, 286 212 209 bytes
<?php foreach(str_split('000680018y4g04uj1s0iql6k0yz98u1xxu2v1xyhs71xyhs71xzt6v1xzt6v1y2ruv1y2ruv1y7pmv1y70cn121jq60jwdto018y5s013bwg',6)as $i)echo str_replace(0,' ',str_pad(base_convert($i,36,2),27,0,0))."
";
Try it online!
answered Nov 14 '18 at 13:56
ScootsScoots
404311
$endgroup$
add a comment |
4
$begingroup$
PHP, 286 212 209 bytes
<?php foreach(str_split('000680018y4g04uj1s0iql6k0yz98u1xxu2v1xyhs71xyhs71xzt6v1xzt6v1y2ruv1y2ruv1y7pmv1y70cn121jq60jwdto018y5s013bwg',6)as $i)echo str_replace(0,' ',str_pad(base_convert($i,36,2),27,0,0))."
";
Try it online!
answered Nov 14 '18 at 13:56
ScootsScoots
404311
$endgroup$
add a comment |
4
4
4
$begingroup$
PHP, 286 212 209 bytes
<?php foreach(str_split('000680018y4g04uj1s0iql6k0yz98u1xxu2v1xyhs71xyhs71xzt6v1xzt6v1y2ruv1y2ruv1y7pmv1y70cn121jq60jwdto018y5s013bwg',6)as $i)echo str_replace(0,' ',str_pad(base_convert($i,36,2),27,0,0))."
";
Try it online!
answered Nov 14 '18 at 13:56
ScootsScoots
404311
$endgroup$
PHP, 286 212 209 bytes
<?php foreach(str_split('000680018y4g04uj1s0iql6k0yz98u1xxu2v1xyhs71xyhs71xzt6v1xzt6v1y2ruv1y2ruv1y7pmv1y70cn121jq60jwdto018y5s013bwg',6)as $i)echo str_replace(0,' ',str_pad(base_convert($i,36,2),27,0,0))."
";
Try it online!
answered Nov 14 '18 at 13:56
ScootsScoots
404311
edited Nov 16 '18 at 13:11
answered Nov 14 '18 at 13:56
ScootsScoots
404311
answered Nov 14 '18 at 13:56
ScootsScoots
404311
answered Nov 14 '18 at 13:56
ScootsScoots
404311
404311
add a comment |
add a comment |
3
$begingroup$
C# (.NET Core), 199 bytes
_=>{var r="";for(int i=0,j,k=0;i<88;i++)for(j=0;j++<"0(/1,'&,(&))#&%%*&#%%%#%+&#%&(*&$%&(*&$%&()&%%&()&%%&((-&((-&('&'%&('%(%&%#&$&-&%*+(*3,%"[i]-34;){r+=i%2<1?' ':'#';if(++k%27<1)r+='n';}return r;}
Try it online!
Uses the same approach as my solution to the tribute to Adam West.
answered Nov 14 '18 at 11:37
CharlieCharlie
7,3162389
$endgroup$
add a comment |
3
$begingroup$
C# (.NET Core), 199 bytes
_=>{var r="";for(int i=0,j,k=0;i<88;i++)for(j=0;j++<"0(/1,'&,(&))#&%%*&#%%%#%+&#%&(*&$%&(*&$%&()&%%&()&%%&((-&((-&('&'%&('%(%&%#&$&-&%*+(*3,%"[i]-34;){r+=i%2<1?' ':'#';if(++k%27<1)r+='n';}return r;}
Try it online!
Uses the same approach as my solution to the tribute to Adam West.
answered Nov 14 '18 at 11:37
CharlieCharlie
7,3162389
$endgroup$
add a comment |
3
3
3
$begingroup$
C# (.NET Core), 199 bytes
_=>{var r="";for(int i=0,j,k=0;i<88;i++)for(j=0;j++<"0(/1,'&,(&))#&%%*&#%%%#%+&#%&(*&$%&(*&$%&()&%%&()&%%&((-&((-&('&'%&('%(%&%#&$&-&%*+(*3,%"[i]-34;){r+=i%2<1?' ':'#';if(++k%27<1)r+='n';}return r;}
Try it online!
Uses the same approach as my solution to the tribute to Adam West.
answered Nov 14 '18 at 11:37
CharlieCharlie
7,3162389
$endgroup$
C# (.NET Core), 199 bytes
_=>{var r="";for(int i=0,j,k=0;i<88;i++)for(j=0;j++<"0(/1,'&,(&))#&%%*&#%%%#%+&#%&(*&$%&(*&$%&()&%%&()&%%&((-&((-&('&'%&('%(%&%#&$&-&%*+(*3,%"[i]-34;){r+=i%2<1?' ':'#';if(++k%27<1)r+='n';}return r;}
Try it online!
Uses the same approach as my solution to the tribute to Adam West.
answered Nov 14 '18 at 11:37
CharlieCharlie
7,3162389
answered Nov 14 '18 at 11:37
CharlieCharlie
7,3162389
answered Nov 14 '18 at 11:37
CharlieCharlie
7,3162389
answered Nov 14 '18 at 11:37
CharlieCharlie
7,3162389
7,3162389
add a comment |
add a comment |
1
$begingroup$
deflate, 79 bytes
eJyVkcsJADAIQ++ZIpD9dyxUqrGUgjlpHvglXdqCJymk1yEiaRXEIOXzuBHCiKObReUxUzYaMdt2wmTBg/FmNXndgLRbNvL7ifsLfMw6iQ==
answered Nov 14 '18 at 6:26
WhaleWhale
211
$endgroup$
6
$begingroup$
You appear to have posted two answers in the same language. If you make an improvement to your submission it is better to edit your existing answer. Additionally, could you provide a link to thedeflatelanguage, as this just looks like a base64 string
$endgroup$
– Jo King
Nov 14 '18 at 6:28
3
$begingroup$
I don't think deflate counts as a programming language (although I could be wrong as I don't participate on PPCG).
$endgroup$
– NobodyNada
Nov 14 '18 at 7:28
2
$begingroup$
This is not DEFLATE, but zlib, which is DEFLATE plus a 2 byte header and a 4 byte checksum. By stripping those, you can use the Bubblegum interpreter.
$endgroup$
– Dennis♦
Nov 14 '18 at 14:24
$begingroup$
@NobodyNada deflate is not a programming language but we do not require that submissions be in programming languages.
$endgroup$
– Wît Wisarhd
Nov 14 '18 at 23:06
add a comment |
1
$begingroup$
deflate, 79 bytes
eJyVkcsJADAIQ++ZIpD9dyxUqrGUgjlpHvglXdqCJymk1yEiaRXEIOXzuBHCiKObReUxUzYaMdt2wmTBg/FmNXndgLRbNvL7ifsLfMw6iQ==
answered Nov 14 '18 at 6:26
WhaleWhale
211
$endgroup$
6
$begingroup$
You appear to have posted two answers in the same language. If you make an improvement to your submission it is better to edit your existing answer. Additionally, could you provide a link to thedeflatelanguage, as this just looks like a base64 string
$endgroup$
– Jo King
Nov 14 '18 at 6:28
3
$begingroup$
I don't think deflate counts as a programming language (although I could be wrong as I don't participate on PPCG).
$endgroup$
– NobodyNada
Nov 14 '18 at 7:28
2
$begingroup$
This is not DEFLATE, but zlib, which is DEFLATE plus a 2 byte header and a 4 byte checksum. By stripping those, you can use the Bubblegum interpreter.
$endgroup$
– Dennis♦
Nov 14 '18 at 14:24
$begingroup$
@NobodyNada deflate is not a programming language but we do not require that submissions be in programming languages.
$endgroup$
– Wît Wisarhd
Nov 14 '18 at 23:06
add a comment |
1
1
1
$begingroup$
deflate, 79 bytes
eJyVkcsJADAIQ++ZIpD9dyxUqrGUgjlpHvglXdqCJymk1yEiaRXEIOXzuBHCiKObReUxUzYaMdt2wmTBg/FmNXndgLRbNvL7ifsLfMw6iQ==
answered Nov 14 '18 at 6:26
WhaleWhale
211
$endgroup$
deflate, 79 bytes
eJyVkcsJADAIQ++ZIpD9dyxUqrGUgjlpHvglXdqCJymk1yEiaRXEIOXzuBHCiKObReUxUzYaMdt2wmTBg/FmNXndgLRbNvL7ifsLfMw6iQ==
answered Nov 14 '18 at 6:26
WhaleWhale
211
answered Nov 14 '18 at 6:26
WhaleWhale
211
answered Nov 14 '18 at 6:26
WhaleWhale
211
answered Nov 14 '18 at 6:26
WhaleWhale
211
211
6
$begingroup$
You appear to have posted two answers in the same language. If you make an improvement to your submission it is better to edit your existing answer. Additionally, could you provide a link to thedeflatelanguage, as this just looks like a base64 string
$endgroup$
– Jo King
Nov 14 '18 at 6:28
3
$begingroup$
I don't think deflate counts as a programming language (although I could be wrong as I don't participate on PPCG).
$endgroup$
– NobodyNada
Nov 14 '18 at 7:28
2
$begingroup$
This is not DEFLATE, but zlib, which is DEFLATE plus a 2 byte header and a 4 byte checksum. By stripping those, you can use the Bubblegum interpreter.
$endgroup$
– Dennis♦
Nov 14 '18 at 14:24
$begingroup$
@NobodyNada deflate is not a programming language but we do not require that submissions be in programming languages.
$endgroup$
– Wît Wisarhd
Nov 14 '18 at 23:06
add a comment |
6
$begingroup$
You appear to have posted two answers in the same language. If you make an improvement to your submission it is better to edit your existing answer. Additionally, could you provide a link to thedeflatelanguage, as this just looks like a base64 string
$endgroup$
– Jo King
Nov 14 '18 at 6:28
3
$begingroup$
I don't think deflate counts as a programming language (although I could be wrong as I don't participate on PPCG).
$endgroup$
– NobodyNada
Nov 14 '18 at 7:28
2
$begingroup$
This is not DEFLATE, but zlib, which is DEFLATE plus a 2 byte header and a 4 byte checksum. By stripping those, you can use the Bubblegum interpreter.
$endgroup$
– Dennis♦
Nov 14 '18 at 14:24
$begingroup$
@NobodyNada deflate is not a programming language but we do not require that submissions be in programming languages.
$endgroup$
– Wît Wisarhd
Nov 14 '18 at 23:06
6
6
$begingroup$
You appear to have posted two answers in the same language. If you make an improvement to your submission it is better to edit your existing answer. Additionally, could you provide a link to the
deflate language, as this just looks like a base64 string$endgroup$
– Jo King
Nov 14 '18 at 6:28
$begingroup$
You appear to have posted two answers in the same language. If you make an improvement to your submission it is better to edit your existing answer. Additionally, could you provide a link to the
deflate language, as this just looks like a base64 string$endgroup$
– Jo King
Nov 14 '18 at 6:28
3
3
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I don't think deflate counts as a programming language (although I could be wrong as I don't participate on PPCG).
$endgroup$
– NobodyNada
Nov 14 '18 at 7:28
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I don't think deflate counts as a programming language (although I could be wrong as I don't participate on PPCG).
$endgroup$
– NobodyNada
Nov 14 '18 at 7:28
2
2
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This is not DEFLATE, but zlib, which is DEFLATE plus a 2 byte header and a 4 byte checksum. By stripping those, you can use the Bubblegum interpreter.
$endgroup$
– Dennis♦
Nov 14 '18 at 14:24
$begingroup$
This is not DEFLATE, but zlib, which is DEFLATE plus a 2 byte header and a 4 byte checksum. By stripping those, you can use the Bubblegum interpreter.
$endgroup$
– Dennis♦
Nov 14 '18 at 14:24
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@NobodyNada deflate is not a programming language but we do not require that submissions be in programming languages.
$endgroup$
– Wît Wisarhd
Nov 14 '18 at 23:06
$begingroup$
@NobodyNada deflate is not a programming language but we do not require that submissions be in programming languages.
$endgroup$
– Wît Wisarhd
Nov 14 '18 at 23:06
add a comment |
0
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Turing Machine But Way Worse - 16913 bytes
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Made with ASCII_only's program generator
answered Dec 27 '18 at 3:36
MilkyWay90MilkyWay90
1799
$endgroup$
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0
$begingroup$
Turing Machine But Way Worse - 16913 bytes
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Made with ASCII_only's program generator
answered Dec 27 '18 at 3:36
MilkyWay90MilkyWay90
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Turing Machine But Way Worse - 16913 bytes
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Made with ASCII_only's program generator
answered Dec 27 '18 at 3:36
MilkyWay90MilkyWay90
1799
$endgroup$
Turing Machine But Way Worse - 16913 bytes
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Made with ASCII_only's program generator
answered Dec 27 '18 at 3:36
MilkyWay90MilkyWay90
1799
answered Dec 27 '18 at 3:36
MilkyWay90MilkyWay90
1799
answered Dec 27 '18 at 3:36
MilkyWay90MilkyWay90
1799
answered Dec 27 '18 at 3:36
MilkyWay90MilkyWay90
1799
1799
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3
$begingroup$
What happened to the → in Ⓐ?
$endgroup$
– Adám
Nov 13 '18 at 20:00
3
$begingroup$
@Adám 2 Reasons. I believe that the arrow means the continuation of the series (in this case the comics about avengers). Since Stan passed away there wont be a continuation of it (In my opinion, it is not the same without Stan). That is one of the reasons
$endgroup$
– Luis felipe De jesus Munoz
Nov 13 '18 at 20:05
43
$begingroup$
The second one is that I didnt know how to draw it :c
$endgroup$
– Luis felipe De jesus Munoz
Nov 13 '18 at 20:05
$begingroup$
Are the trailing spaces required?
$endgroup$
– Scoots
Nov 14 '18 at 12:40
$begingroup$
@Scoots no, it is not required as long as your output looks the same
$endgroup$
– Luis felipe De jesus Munoz
Nov 14 '18 at 12:47