Tribute to Stan Lee
up vote
37
down vote
favorite
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
|
show 1 more comment
up vote
37
down vote
favorite
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
3
What happened to the → in Ⓐ?
– Adám
Nov 13 at 20:00
3
@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
– Luis felipe De jesus Munoz
Nov 13 at 20:05
38
The second one is that I didnt know how to draw it :c
– Luis felipe De jesus Munoz
Nov 13 at 20:05
Are the trailing spaces required?
– Scoots
2 days ago
@Scoots no, it is not required as long as your output looks the same
– Luis felipe De jesus Munoz
2 days ago
|
show 1 more comment
up vote
37
down vote
favorite
up vote
37
down vote
favorite
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
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
edited 2 days ago
asked Nov 13 at 17:46
Luis felipe De jesus Munoz
3,98911253
3,98911253
3
What happened to the → in Ⓐ?
– Adám
Nov 13 at 20:00
3
@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
– Luis felipe De jesus Munoz
Nov 13 at 20:05
38
The second one is that I didnt know how to draw it :c
– Luis felipe De jesus Munoz
Nov 13 at 20:05
Are the trailing spaces required?
– Scoots
2 days ago
@Scoots no, it is not required as long as your output looks the same
– Luis felipe De jesus Munoz
2 days ago
|
show 1 more comment
3
What happened to the → in Ⓐ?
– Adám
Nov 13 at 20:00
3
@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
– Luis felipe De jesus Munoz
Nov 13 at 20:05
38
The second one is that I didnt know how to draw it :c
– Luis felipe De jesus Munoz
Nov 13 at 20:05
Are the trailing spaces required?
– Scoots
2 days ago
@Scoots no, it is not required as long as your output looks the same
– Luis felipe De jesus Munoz
2 days ago
3
3
What happened to the → in Ⓐ?
– Adám
Nov 13 at 20:00
What happened to the → in Ⓐ?
– Adám
Nov 13 at 20:00
3
3
@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
– Luis felipe De jesus Munoz
Nov 13 at 20:05
@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
– Luis felipe De jesus Munoz
Nov 13 at 20:05
38
38
The second one is that I didnt know how to draw it :c
– Luis felipe De jesus Munoz
Nov 13 at 20:05
The second one is that I didnt know how to draw it :c
– Luis felipe De jesus Munoz
Nov 13 at 20:05
Are the trailing spaces required?
– Scoots
2 days ago
Are the trailing spaces required?
– Scoots
2 days ago
@Scoots no, it is not required as long as your output looks the same
– Luis felipe De jesus Munoz
2 days ago
@Scoots no, it is not required as long as your output looks the same
– Luis felipe De jesus Munoz
2 days ago
|
show 1 more comment
21 Answers
21
active
oldest
votes
up vote
22
down vote
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.
add a comment |
up vote
15
down vote
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 character4
because 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
4
seems 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
1
Your solution is nicer, but your own gzip 2-parter gives 89+47 = 136 bytes.
– J.Doe
2 days ago
@J.Doe : ahah I even forgot to have used that approach! please post it as your own answer :)
– digEmAll
2 days ago
Since we can use any character for#
, using0
in place of"#"
should save 2 more bytes. @J.Doe that may work better for your gzip as well.
– Giuseppe
2 days ago
add a comment |
up vote
10
down vote
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!
im just curious, how u guys type those chars? u have special keyboard or something?
– Nikko Khresna
8 hours ago
@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.
– dzaima
8 hours ago
add a comment |
up vote
10
down vote
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!
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?
– Cowabunghole
4 hours ago
@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!
.
– Dennis♦
1 hour ago
add a comment |
up vote
9
down vote
C (gcc), 244 243 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(" #"[_%2]);}main(_){_<19&&b(x[_-1],27)&puts("")&main(_+1);}
Try it online!
-1 Thanks to Peter Cordes
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;}
Canx?y:0
be 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.
– Peter Cordes
Nov 13 at 21:18
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.
– LambdaBeta
Nov 13 at 21:51
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 onstdout
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.
– Peter Cordes
Nov 13 at 21:56
1
Ahh, I see what you mean. You're right - that does work! -1 byte.
– LambdaBeta
Nov 13 at 21:59
You don't need the space between the macro and the definition.
– Zacharý
5 hours ago
add a comment |
up vote
9
down vote
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:
*7
gets replaced by+REPLICATE('#',7)
$4
gets 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)
can drop 2 chars usingCHAR(8000)
instead ofVARCHAR(max)
(the huge command is 1453 bytes)
– Shameen
2 days ago
1
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)
.
– BradC
2 days ago
add a comment |
up vote
9
down vote
Charcoal, 71 68 bytes
P⁸↓P-³↙←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:
P⁸↓P-³↙←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.
add a comment |
up vote
8
down vote
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(`.&-/*%$*&$''!$##($!###!#)$!#$&($"#$&($"#$&'$##$&'$##$&&+$&&+$&%$%#$&%#&#$#!$"$+$#()&(1*#2`).map(c=>s+='# '[x^=1].repeat(c-32),s=x='')&&s.match(/.{27}/g)
Try it online!
add a comment |
up vote
8
down vote
Python 2, 129 bytes
00000000: 2363 6f64 696e 673a 4c31 0a70 7269 6e74 #coding:L1.print
00000010: 2778 da95 9141 0e5c 3021 0803 ef7d 4593 'x...A.!...}E.
00000020: feff 8f9b 5d14 5c6e f1b0 f622 7422 2890 ....].n..."t"(.
00000030: 2e7d 5a09 dc4b 19cb bc53 84d1 4a6a 5960 .}Z..K...S..JjY`
00000040: 116e 3f42 c290 b3f7 c0bc 76cf 549d 6ed8 .n?B......v.T.n.
00000050: f8fa 5f26 0b0e 8c93 d5cb 35f6 b1e7 a939 .._&......5....9
00000060: 9e98 e769 47b9 87d6 cdf5 5c30 3030 32c0 ...iG.....002.
00000070: 4029 272e 6465 636f 6465 2827 7a69 7027 @)'.decode('zip'
00000080: 29 )
Try it online!
129 bytes, I removed the trailingn
from the compressed string.
– ovs
2 days ago
add a comment |
up vote
7
down vote
///, 170 bytes
/+/////*/ +)/'$# +(/"$!+'/" +&/!#+%/"!
!+$/*!+"/**+!/###/""('*
(!!!'
"&#"!!&"
$#'$& &*
!""& ! $
!"'& !%""&$%""&$%)$%)$%(&#%(&#%'&'!%'!"$"!
&$#")
$&#"'!!*
(!!&#"
"$
Try it online!
add a comment |
up vote
7
down vote
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!
add a comment |
up vote
7
down vote
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.
add a comment |
up vote
6
down vote
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.
1
I think you can save some by letting space encode"###n"
.
– Ørjan Johansen
2 days ago
add a comment |
up vote
6
down vote
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
.
No need for^
. But even better, usecut -c2-
instead of thesed
part.
– manatwork
Nov 13 at 22:59
And also shorter with here-string instead ofecho
and even shorter withdc
instead ofbc
: Try it online!
– manatwork
2 days ago
2
I'll give you theecho
, but come on -dc
for Stan Lee... you must be joking ;)
– LambdaBeta
2 days ago
1
134 bytes
– Dennis♦
2 days ago
add a comment |
up vote
5
down vote
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!
add a comment |
up vote
4
down vote
Perl 5, 181 bytes
say+(sprintf"%28b",oct"0x$_")=~y/10/# /r for"0003f0003fff800f87fe03c07f78700f71ce00f70ee01e70ee01e70ee03c70ee03c70ee07ff0ee07ff0ee0f070ee0e070e79e003c3fc01f803fffe00380000"=~/.{7}/g
Try it online!
add a comment |
up vote
4
down vote
Perl 6, 136 116 bytes
:100['> b3V$^%&!U3X/_9-D>Z(+, OD_=D,-:&3S>#OL
8C"> c H'.ords].base(2)~~TR/1/ /.comb(27)>>.say
Try it online!
Parses a base 100 number from the ordinal values of the string, converts to base 2 and replaces the 1
s with spaces. This uses zeroes as the main character.
Explanation:
:100['...'.ords] # Convert the bytes of the string to base 100
.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
add a comment |
up vote
3
down vote
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.
New contributor
add a comment |
up vote
3
down vote
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!
add a comment |
up vote
2
down vote
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.
add a comment |
up vote
0
down vote
deflate, 79 bytes
eJyVkcsJADAIQ++ZIpD9dyxUqrGUgjlpHvglXdqCJymk1yEiaRXEIOXzuBHCiKObReUxUzYaMdt2wmTBg/FmNXndgLRbNvL7ifsLfMw6iQ==
New contributor
6
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 thedeflate
language, as this just looks like a base64 string
– Jo King
2 days ago
3
I don't think deflate counts as a programming language (although I could be wrong as I don't participate on PPCG).
– NobodyNada
2 days ago
2
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.
– Dennis♦
2 days ago
@NobodyNada deflate is not a programming language but we do not require that submissions be in programming languages.
– Post Left Garf Hunter
yesterday
add a comment |
21 Answers
21
active
oldest
votes
21 Answers
21
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
22
down vote
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.
add a comment |
up vote
22
down vote
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.
add a comment |
up vote
22
down vote
up vote
22
down vote
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.
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.
edited 2 days ago
answered Nov 13 at 21:22
Dennis♦
184k32293729
184k32293729
add a comment |
add a comment |
up vote
15
down vote
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 character4
because 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
4
seems 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
1
Your solution is nicer, but your own gzip 2-parter gives 89+47 = 136 bytes.
– J.Doe
2 days ago
@J.Doe : ahah I even forgot to have used that approach! please post it as your own answer :)
– digEmAll
2 days ago
Since we can use any character for#
, using0
in place of"#"
should save 2 more bytes. @J.Doe that may work better for your gzip as well.
– Giuseppe
2 days ago
add a comment |
up vote
15
down vote
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 character4
because 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
4
seems 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
1
Your solution is nicer, but your own gzip 2-parter gives 89+47 = 136 bytes.
– J.Doe
2 days ago
@J.Doe : ahah I even forgot to have used that approach! please post it as your own answer :)
– digEmAll
2 days ago
Since we can use any character for#
, using0
in place of"#"
should save 2 more bytes. @J.Doe that may work better for your gzip as well.
– Giuseppe
2 days ago
add a comment |
up vote
15
down vote
up vote
15
down vote
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 character4
because 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
4
seems 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
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 character4
because 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
4
seems 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
edited yesterday
answered Nov 13 at 18:48
digEmAll
2,20148
2,20148
1
Your solution is nicer, but your own gzip 2-parter gives 89+47 = 136 bytes.
– J.Doe
2 days ago
@J.Doe : ahah I even forgot to have used that approach! please post it as your own answer :)
– digEmAll
2 days ago
Since we can use any character for#
, using0
in place of"#"
should save 2 more bytes. @J.Doe that may work better for your gzip as well.
– Giuseppe
2 days ago
add a comment |
1
Your solution is nicer, but your own gzip 2-parter gives 89+47 = 136 bytes.
– J.Doe
2 days ago
@J.Doe : ahah I even forgot to have used that approach! please post it as your own answer :)
– digEmAll
2 days ago
Since we can use any character for#
, using0
in place of"#"
should save 2 more bytes. @J.Doe that may work better for your gzip as well.
– Giuseppe
2 days ago
1
1
Your solution is nicer, but your own gzip 2-parter gives 89+47 = 136 bytes.
– J.Doe
2 days ago
Your solution is nicer, but your own gzip 2-parter gives 89+47 = 136 bytes.
– J.Doe
2 days ago
@J.Doe : ahah I even forgot to have used that approach! please post it as your own answer :)
– digEmAll
2 days ago
@J.Doe : ahah I even forgot to have used that approach! please post it as your own answer :)
– digEmAll
2 days ago
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.– Giuseppe
2 days ago
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.– Giuseppe
2 days ago
add a comment |
up vote
10
down vote
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!
im just curious, how u guys type those chars? u have special keyboard or something?
– Nikko Khresna
8 hours ago
@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.
– dzaima
8 hours ago
add a comment |
up vote
10
down vote
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!
im just curious, how u guys type those chars? u have special keyboard or something?
– Nikko Khresna
8 hours ago
@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.
– dzaima
8 hours ago
add a comment |
up vote
10
down vote
up vote
10
down vote
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!
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!
edited Nov 13 at 18:35
answered Nov 13 at 18:27
dzaima
14k21654
14k21654
im just curious, how u guys type those chars? u have special keyboard or something?
– Nikko Khresna
8 hours ago
@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.
– dzaima
8 hours ago
add a comment |
im just curious, how u guys type those chars? u have special keyboard or something?
– Nikko Khresna
8 hours ago
@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.
– dzaima
8 hours ago
im just curious, how u guys type those chars? u have special keyboard or something?
– Nikko Khresna
8 hours ago
im just curious, how u guys type those chars? u have special keyboard or something?
– Nikko Khresna
8 hours ago
@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.
– dzaima
8 hours ago
@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.
– dzaima
8 hours ago
add a comment |
up vote
10
down vote
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!
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?
– Cowabunghole
4 hours ago
@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!
.
– Dennis♦
1 hour ago
add a comment |
up vote
10
down vote
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!
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?
– Cowabunghole
4 hours ago
@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!
.
– Dennis♦
1 hour ago
add a comment |
up vote
10
down vote
up vote
10
down vote
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!
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!
edited 2 days ago
answered 2 days ago
Dennis♦
184k32293729
184k32293729
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?
– Cowabunghole
4 hours ago
@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!
.
– Dennis♦
1 hour ago
add a comment |
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?
– Cowabunghole
4 hours ago
@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!
.
– Dennis♦
1 hour ago
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?
– Cowabunghole
4 hours ago
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?
– Cowabunghole
4 hours ago
@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 !
.– Dennis♦
1 hour ago
@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 !
.– Dennis♦
1 hour ago
add a comment |
up vote
9
down vote
C (gcc), 244 243 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(" #"[_%2]);}main(_){_<19&&b(x[_-1],27)&puts("")&main(_+1);}
Try it online!
-1 Thanks to Peter Cordes
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;}
Canx?y:0
be 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.
– Peter Cordes
Nov 13 at 21:18
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.
– LambdaBeta
Nov 13 at 21:51
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 onstdout
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.
– Peter Cordes
Nov 13 at 21:56
1
Ahh, I see what you mean. You're right - that does work! -1 byte.
– LambdaBeta
Nov 13 at 21:59
You don't need the space between the macro and the definition.
– Zacharý
5 hours ago
add a comment |
up vote
9
down vote
C (gcc), 244 243 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(" #"[_%2]);}main(_){_<19&&b(x[_-1],27)&puts("")&main(_+1);}
Try it online!
-1 Thanks to Peter Cordes
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;}
Canx?y:0
be 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.
– Peter Cordes
Nov 13 at 21:18
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.
– LambdaBeta
Nov 13 at 21:51
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 onstdout
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.
– Peter Cordes
Nov 13 at 21:56
1
Ahh, I see what you mean. You're right - that does work! -1 byte.
– LambdaBeta
Nov 13 at 21:59
You don't need the space between the macro and the definition.
– Zacharý
5 hours ago
add a comment |
up vote
9
down vote
up vote
9
down vote
C (gcc), 244 243 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(" #"[_%2]);}main(_){_<19&&b(x[_-1],27)&puts("")&main(_+1);}
Try it online!
-1 Thanks to Peter Cordes
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;}
C (gcc), 244 243 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(" #"[_%2]);}main(_){_<19&&b(x[_-1],27)&puts("")&main(_+1);}
Try it online!
-1 Thanks to Peter Cordes
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;}
edited Nov 13 at 21:59
answered Nov 13 at 20:58
LambdaBeta
1,989416
1,989416
Canx?y:0
be 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.
– Peter Cordes
Nov 13 at 21:18
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.
– LambdaBeta
Nov 13 at 21:51
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 onstdout
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.
– Peter Cordes
Nov 13 at 21:56
1
Ahh, I see what you mean. You're right - that does work! -1 byte.
– LambdaBeta
Nov 13 at 21:59
You don't need the space between the macro and the definition.
– Zacharý
5 hours ago
add a comment |
Canx?y:0
be 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.
– Peter Cordes
Nov 13 at 21:18
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.
– LambdaBeta
Nov 13 at 21:51
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 onstdout
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.
– Peter Cordes
Nov 13 at 21:56
1
Ahh, I see what you mean. You're right - that does work! -1 byte.
– LambdaBeta
Nov 13 at 21:59
You don't need the space between the macro and the definition.
– Zacharý
5 hours ago
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.– Peter Cordes
Nov 13 at 21:18
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.– Peter Cordes
Nov 13 at 21:18
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.– LambdaBeta
Nov 13 at 21:51
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.– LambdaBeta
Nov 13 at 21:51
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.– Peter Cordes
Nov 13 at 21:56
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.– Peter Cordes
Nov 13 at 21:56
1
1
Ahh, I see what you mean. You're right - that does work! -1 byte.
– LambdaBeta
Nov 13 at 21:59
Ahh, I see what you mean. You're right - that does work! -1 byte.
– LambdaBeta
Nov 13 at 21:59
You don't need the space between the macro and the definition.
– Zacharý
5 hours ago
You don't need the space between the macro and the definition.
– Zacharý
5 hours ago
add a comment |
up vote
9
down vote
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:
*7
gets replaced by+REPLICATE('#',7)
$4
gets 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)
can drop 2 chars usingCHAR(8000)
instead ofVARCHAR(max)
(the huge command is 1453 bytes)
– Shameen
2 days ago
1
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)
.
– BradC
2 days ago
add a comment |
up vote
9
down vote
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:
*7
gets replaced by+REPLICATE('#',7)
$4
gets 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)
can drop 2 chars usingCHAR(8000)
instead ofVARCHAR(max)
(the huge command is 1453 bytes)
– Shameen
2 days ago
1
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)
.
– BradC
2 days ago
add a comment |
up vote
9
down vote
up vote
9
down vote
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:
*7
gets replaced by+REPLICATE('#',7)
$4
gets 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)
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:
*7
gets replaced by+REPLICATE('#',7)
$4
gets 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)
edited 2 days ago
answered 2 days ago
BradC
3,544521
3,544521
can drop 2 chars usingCHAR(8000)
instead ofVARCHAR(max)
(the huge command is 1453 bytes)
– Shameen
2 days ago
1
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)
.
– BradC
2 days ago
add a comment |
can drop 2 chars usingCHAR(8000)
instead ofVARCHAR(max)
(the huge command is 1453 bytes)
– Shameen
2 days ago
1
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)
.
– BradC
2 days ago
can drop 2 chars using
CHAR(8000)
instead of VARCHAR(max)
(the huge command is 1453 bytes)– Shameen
2 days ago
can drop 2 chars using
CHAR(8000)
instead of VARCHAR(max)
(the huge command is 1453 bytes)– Shameen
2 days ago
1
1
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)
.– BradC
2 days ago
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)
.– BradC
2 days ago
add a comment |
up vote
9
down vote
Charcoal, 71 68 bytes
P⁸↓P-³↙←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:
P⁸↓P-³↙←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.
add a comment |
up vote
9
down vote
Charcoal, 71 68 bytes
P⁸↓P-³↙←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:
P⁸↓P-³↙←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.
add a comment |
up vote
9
down vote
up vote
9
down vote
Charcoal, 71 68 bytes
P⁸↓P-³↙←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:
P⁸↓P-³↙←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.
Charcoal, 71 68 bytes
P⁸↓P-³↙←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:
P⁸↓P-³↙←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.
edited 10 hours ago
answered 2 days ago
Neil
77.9k744174
77.9k744174
add a comment |
add a comment |
up vote
8
down vote
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(`.&-/*%$*&$''!$##($!###!#)$!#$&($"#$&($"#$&'$##$&'$##$&&+$&&+$&%$%#$&%#&#$#!$"$+$#()&(1*#2`).map(c=>s+='# '[x^=1].repeat(c-32),s=x='')&&s.match(/.{27}/g)
Try it online!
add a comment |
up vote
8
down vote
JavaScript (ES6), 173 170 bytes
_=>`tc
du
9a9k
58fe38
36h83676
6j83!h85!h85!f87!f87!dm96
6dm96
6b8b!b6d696
3858n8
5gjc
dy
d6`.split`!`.join`696
6`.replace(/./g,c=>'# '[(k=parseInt(c,36))&1].repeat(k/2))
Try it online!
Node.js, 163 bytes
Provided that an array of strings is a valid output:
_=>Buffer(`.&-/*%$*&$''!$##($!###!#)$!#$&($"#$&($"#$&'$##$&'$##$&&+$&&+$&%$%#$&%#&#$#!$"$+$#()&(1*#2`).map(c=>s+='# '[x^=1].repeat(c-32),s=x='')&&s.match(/.{27}/g)
Try it online!
add a comment |
up vote
8
down vote
up vote
8
down vote
JavaScript (ES6), 173 170 bytes
_=>`tc
du
9a9k
58fe38
36h83676
6j83!h85!h85!f87!f87!dm96
6dm96
6b8b!b6d696
3858n8
5gjc
dy
d6`.split`!`.join`696
6`.replace(/./g,c=>'# '[(k=parseInt(c,36))&1].repeat(k/2))
Try it online!
Node.js, 163 bytes
Provided that an array of strings is a valid output:
_=>Buffer(`.&-/*%$*&$''!$##($!###!#)$!#$&($"#$&($"#$&'$##$&'$##$&&+$&&+$&%$%#$&%#&#$#!$"$+$#()&(1*#2`).map(c=>s+='# '[x^=1].repeat(c-32),s=x='')&&s.match(/.{27}/g)
Try it online!
JavaScript (ES6), 173 170 bytes
_=>`tc
du
9a9k
58fe38
36h83676
6j83!h85!h85!f87!f87!dm96
6dm96
6b8b!b6d696
3858n8
5gjc
dy
d6`.split`!`.join`696
6`.replace(/./g,c=>'# '[(k=parseInt(c,36))&1].repeat(k/2))
Try it online!
Node.js, 163 bytes
Provided that an array of strings is a valid output:
_=>Buffer(`.&-/*%$*&$''!$##($!###!#)$!#$&($"#$&($"#$&'$##$&'$##$&&+$&&+$&%$%#$&%#&#$#!$"$+$#()&(1*#2`).map(c=>s+='# '[x^=1].repeat(c-32),s=x='')&&s.match(/.{27}/g)
Try it online!
edited Nov 13 at 19:34
answered Nov 13 at 18:05
Arnauld
68.6k584289
68.6k584289
add a comment |
add a comment |
up vote
8
down vote
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!
129 bytes, I removed the trailingn
from the compressed string.
– ovs
2 days ago
add a comment |
up vote
8
down vote
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!
129 bytes, I removed the trailingn
from the compressed string.
– ovs
2 days ago
add a comment |
up vote
8
down vote
up vote
8
down vote
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!
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!
edited 2 days ago
answered 2 days ago
Lynn
49.1k694223
49.1k694223
129 bytes, I removed the trailingn
from the compressed string.
– ovs
2 days ago
add a comment |
129 bytes, I removed the trailingn
from the compressed string.
– ovs
2 days ago
129 bytes, I removed the trailing
n
from the compressed string.– ovs
2 days ago
129 bytes, I removed the trailing
n
from the compressed string.– ovs
2 days ago
add a comment |
up vote
7
down vote
///, 170 bytes
/+/////*/ +)/'$# +(/"$!+'/" +&/!#+%/"!
!+$/*!+"/**+!/###/""('*
(!!!'
"&#"!!&"
$#'$& &*
!""& ! $
!"'& !%""&$%""&$%)$%)$%(&#%(&#%'&'!%'!"$"!
&$#")
$&#"'!!*
(!!&#"
"$
Try it online!
add a comment |
up vote
7
down vote
///, 170 bytes
/+/////*/ +)/'$# +(/"$!+'/" +&/!#+%/"!
!+$/*!+"/**+!/###/""('*
(!!!'
"&#"!!&"
$#'$& &*
!""& ! $
!"'& !%""&$%""&$%)$%)$%(&#%(&#%'&'!%'!"$"!
&$#")
$&#"'!!*
(!!&#"
"$
Try it online!
add a comment |
up vote
7
down vote
up vote
7
down vote
///, 170 bytes
/+/////*/ +)/'$# +(/"$!+'/" +&/!#+%/"!
!+$/*!+"/**+!/###/""('*
(!!!'
"&#"!!&"
$#'$& &*
!""& ! $
!"'& !%""&$%""&$%)$%)$%(&#%(&#%'&'!%'!"$"!
&$#")
$&#"'!!*
(!!&#"
"$
Try it online!
///, 170 bytes
/+/////*/ +)/'$# +(/"$!+'/" +&/!#+%/"!
!+$/*!+"/**+!/###/""('*
(!!!'
"&#"!!&"
$#'$& &*
!""& ! $
!"'& !%""&$%""&$%)$%)$%(&#%(&#%'&'!%'!"$"!
&$#")
$&#"'!!*
(!!&#"
"$
Try it online!
answered Nov 13 at 19:02
Conor O'Brien
28.8k263160
28.8k263160
add a comment |
add a comment |
up vote
7
down vote
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!
add a comment |
up vote
7
down vote
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!
add a comment |
up vote
7
down vote
up vote
7
down vote
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!
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!
edited 2 days ago
answered Nov 13 at 22:10
gastropner
1,8001410
1,8001410
add a comment |
add a comment |
up vote
7
down vote
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.
add a comment |
up vote
7
down vote
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.
add a comment |
up vote
7
down vote
up vote
7
down vote
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.
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.
edited 2 days ago
answered 2 days ago
Kevin Cruijssen
34k554181
34k554181
add a comment |
add a comment |
up vote
6
down vote
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.
1
I think you can save some by letting space encode"###n"
.
– Ørjan Johansen
2 days ago
add a comment |
up vote
6
down vote
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.
1
I think you can save some by letting space encode"###n"
.
– Ørjan Johansen
2 days ago
add a comment |
up vote
6
down vote
up vote
6
down vote
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.
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.
edited 2 days ago
answered 2 days ago
nimi
30.7k31985
30.7k31985
1
I think you can save some by letting space encode"###n"
.
– Ørjan Johansen
2 days ago
add a comment |
1
I think you can save some by letting space encode"###n"
.
– Ørjan Johansen
2 days ago
1
1
I think you can save some by letting space encode
"###n"
.– Ørjan Johansen
2 days ago
I think you can save some by letting space encode
"###n"
.– Ørjan Johansen
2 days ago
add a comment |
up vote
6
down vote
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
.
No need for^
. But even better, usecut -c2-
instead of thesed
part.
– manatwork
Nov 13 at 22:59
And also shorter with here-string instead ofecho
and even shorter withdc
instead ofbc
: Try it online!
– manatwork
2 days ago
2
I'll give you theecho
, but come on -dc
for Stan Lee... you must be joking ;)
– LambdaBeta
2 days ago
1
134 bytes
– Dennis♦
2 days ago
add a comment |
up vote
6
down vote
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
.
No need for^
. But even better, usecut -c2-
instead of thesed
part.
– manatwork
Nov 13 at 22:59
And also shorter with here-string instead ofecho
and even shorter withdc
instead ofbc
: Try it online!
– manatwork
2 days ago
2
I'll give you theecho
, but come on -dc
for Stan Lee... you must be joking ;)
– LambdaBeta
2 days ago
1
134 bytes
– Dennis♦
2 days ago
add a comment |
up vote
6
down vote
up vote
6
down vote
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
.
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
.
edited 2 days ago
answered Nov 13 at 21:00
LambdaBeta
1,989416
1,989416
No need for^
. But even better, usecut -c2-
instead of thesed
part.
– manatwork
Nov 13 at 22:59
And also shorter with here-string instead ofecho
and even shorter withdc
instead ofbc
: Try it online!
– manatwork
2 days ago
2
I'll give you theecho
, but come on -dc
for Stan Lee... you must be joking ;)
– LambdaBeta
2 days ago
1
134 bytes
– Dennis♦
2 days ago
add a comment |
No need for^
. But even better, usecut -c2-
instead of thesed
part.
– manatwork
Nov 13 at 22:59
And also shorter with here-string instead ofecho
and even shorter withdc
instead ofbc
: Try it online!
– manatwork
2 days ago
2
I'll give you theecho
, but come on -dc
for Stan Lee... you must be joking ;)
– LambdaBeta
2 days ago
1
134 bytes
– Dennis♦
2 days ago
No need for
^
. But even better, use cut -c2-
instead of the sed
part.– manatwork
Nov 13 at 22:59
No need for
^
. But even better, use cut -c2-
instead of the sed
part.– manatwork
Nov 13 at 22:59
And also shorter with here-string instead of
echo
and even shorter with dc
instead of bc
: Try it online!– manatwork
2 days ago
And also shorter with here-string instead of
echo
and even shorter with dc
instead of bc
: Try it online!– manatwork
2 days ago
2
2
I'll give you the
echo
, but come on - dc
for Stan Lee... you must be joking ;)– LambdaBeta
2 days ago
I'll give you the
echo
, but come on - dc
for Stan Lee... you must be joking ;)– LambdaBeta
2 days ago
1
1
134 bytes
– Dennis♦
2 days ago
134 bytes
– Dennis♦
2 days ago
add a comment |
up vote
5
down vote
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!
add a comment |
up vote
5
down vote
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!
add a comment |
up vote
5
down vote
up vote
5
down vote
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!
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!
edited 2 days ago
answered Nov 13 at 20:04
Galen Ivanov
5,88711032
5,88711032
add a comment |
add a comment |
up vote
4
down vote
Perl 5, 181 bytes
say+(sprintf"%28b",oct"0x$_")=~y/10/# /r for"0003f0003fff800f87fe03c07f78700f71ce00f70ee01e70ee01e70ee03c70ee03c70ee07ff0ee07ff0ee0f070ee0e070e79e003c3fc01f803fffe00380000"=~/.{7}/g
Try it online!
add a comment |
up vote
4
down vote
Perl 5, 181 bytes
say+(sprintf"%28b",oct"0x$_")=~y/10/# /r for"0003f0003fff800f87fe03c07f78700f71ce00f70ee01e70ee01e70ee03c70ee03c70ee07ff0ee07ff0ee0f070ee0e070e79e003c3fc01f803fffe00380000"=~/.{7}/g
Try it online!
add a comment |
up vote
4
down vote
up vote
4
down vote
Perl 5, 181 bytes
say+(sprintf"%28b",oct"0x$_")=~y/10/# /r for"0003f0003fff800f87fe03c07f78700f71ce00f70ee01e70ee01e70ee03c70ee03c70ee07ff0ee07ff0ee0f070ee0e070e79e003c3fc01f803fffe00380000"=~/.{7}/g
Try it online!
Perl 5, 181 bytes
say+(sprintf"%28b",oct"0x$_")=~y/10/# /r for"0003f0003fff800f87fe03c07f78700f71ce00f70ee01e70ee01e70ee03c70ee03c70ee07ff0ee07ff0ee0f070ee0e070e79e003c3fc01f803fffe00380000"=~/.{7}/g
Try it online!
answered Nov 13 at 22:14
Xcali
4,970520
4,970520
add a comment |
add a comment |
up vote
4
down vote
Perl 6, 136 116 bytes
:100['> b3V$^%&!U3X/_9-D>Z(+, OD_=D,-:&3S>#OL
8C"> c H'.ords].base(2)~~TR/1/ /.comb(27)>>.say
Try it online!
Parses a base 100 number from the ordinal values of the string, converts to base 2 and replaces the 1
s with spaces. This uses zeroes as the main character.
Explanation:
:100['...'.ords] # Convert the bytes of the string to base 100
.base(2) # Convert to base 2
~~TR/1/ / # Translate 1s to spaces
# Alternatively, this could be .split(1)
.comb(27) # Split to strings of length 27
>>.say # Print each on a newline
add a comment |
up vote
4
down vote
Perl 6, 136 116 bytes
:100['> b3V$^%&!U3X/_9-D>Z(+, OD_=D,-:&3S>#OL
8C"> c H'.ords].base(2)~~TR/1/ /.comb(27)>>.say
Try it online!
Parses a base 100 number from the ordinal values of the string, converts to base 2 and replaces the 1
s with spaces. This uses zeroes as the main character.
Explanation:
:100['...'.ords] # Convert the bytes of the string to base 100
.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
add a comment |
up vote
4
down vote
up vote
4
down vote
Perl 6, 136 116 bytes
:100['> b3V$^%&!U3X/_9-D>Z(+, OD_=D,-:&3S>#OL
8C"> c H'.ords].base(2)~~TR/1/ /.comb(27)>>.say
Try it online!
Parses a base 100 number from the ordinal values of the string, converts to base 2 and replaces the 1
s with spaces. This uses zeroes as the main character.
Explanation:
:100['...'.ords] # Convert the bytes of the string to base 100
.base(2) # Convert to base 2
~~TR/1/ / # Translate 1s to spaces
# Alternatively, this could be .split(1)
.comb(27) # Split to strings of length 27
>>.say # Print each on a newline
Perl 6, 136 116 bytes
:100['> b3V$^%&!U3X/_9-D>Z(+, OD_=D,-:&3S>#OL
8C"> c H'.ords].base(2)~~TR/1/ /.comb(27)>>.say
Try it online!
Parses a base 100 number from the ordinal values of the string, converts to base 2 and replaces the 1
s with spaces. This uses zeroes as the main character.
Explanation:
:100['...'.ords] # Convert the bytes of the string to base 100
.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
edited 2 days ago
answered 2 days ago
Jo King
19.1k242102
19.1k242102
add a comment |
add a comment |
up vote
3
down vote
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.
New contributor
add a comment |
up vote
3
down vote
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.
New contributor
add a comment |
up vote
3
down vote
up vote
3
down vote
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.
New contributor
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.
New contributor
edited 2 days ago
New contributor
answered 2 days ago
Nicky Mattsson
1314
1314
New contributor
New contributor
add a comment |
add a comment |
up vote
3
down vote
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!
add a comment |
up vote
3
down vote
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!
add a comment |
up vote
3
down vote
up vote
3
down vote
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!
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!
edited 9 hours ago
answered 2 days ago
Scoots
394311
394311
add a comment |
add a comment |
up vote
2
down vote
C# (.NET Core), 199 bytes
_=>{var r="";for(int i=0,j,k=0;i<88;i++)for(j=0;j++<"0(/1,'&,(&))#&%%*&#%%%#%+&#%&(*&$%&(*&$%&()&%%&()&%%&((-&((-&('&'%&('%(%&%#&$&-&%*+(*3,%"[i]-34;){r+=i%2<1?' ':'#';if(++k%27<1)r+='n';}return r;}
Try it online!
Uses the same approach as my solution to the tribute to Adam West.
add a comment |
up vote
2
down vote
C# (.NET Core), 199 bytes
_=>{var r="";for(int i=0,j,k=0;i<88;i++)for(j=0;j++<"0(/1,'&,(&))#&%%*&#%%%#%+&#%&(*&$%&(*&$%&()&%%&()&%%&((-&((-&('&'%&('%(%&%#&$&-&%*+(*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.
add a comment |
up vote
2
down vote
up vote
2
down vote
C# (.NET Core), 199 bytes
_=>{var r="";for(int i=0,j,k=0;i<88;i++)for(j=0;j++<"0(/1,'&,(&))#&%%*&#%%%#%+&#%&(*&$%&(*&$%&()&%%&()&%%&((-&((-&('&'%&('%(%&%#&$&-&%*+(*3,%"[i]-34;){r+=i%2<1?' ':'#';if(++k%27<1)r+='n';}return r;}
Try it online!
Uses the same approach as my solution to the tribute to Adam West.
C# (.NET Core), 199 bytes
_=>{var r="";for(int i=0,j,k=0;i<88;i++)for(j=0;j++<"0(/1,'&,(&))#&%%*&#%%%#%+&#%&(*&$%&(*&$%&()&%%&()&%%&((-&((-&('&'%&('%(%&%#&$&-&%*+(*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 2 days ago
Charlie
7,2832388
7,2832388
add a comment |
add a comment |
up vote
0
down vote
deflate, 79 bytes
eJyVkcsJADAIQ++ZIpD9dyxUqrGUgjlpHvglXdqCJymk1yEiaRXEIOXzuBHCiKObReUxUzYaMdt2wmTBg/FmNXndgLRbNvL7ifsLfMw6iQ==
New contributor
6
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 thedeflate
language, as this just looks like a base64 string
– Jo King
2 days ago
3
I don't think deflate counts as a programming language (although I could be wrong as I don't participate on PPCG).
– NobodyNada
2 days ago
2
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.
– Dennis♦
2 days ago
@NobodyNada deflate is not a programming language but we do not require that submissions be in programming languages.
– Post Left Garf Hunter
yesterday
add a comment |
up vote
0
down vote
deflate, 79 bytes
eJyVkcsJADAIQ++ZIpD9dyxUqrGUgjlpHvglXdqCJymk1yEiaRXEIOXzuBHCiKObReUxUzYaMdt2wmTBg/FmNXndgLRbNvL7ifsLfMw6iQ==
New contributor
6
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 thedeflate
language, as this just looks like a base64 string
– Jo King
2 days ago
3
I don't think deflate counts as a programming language (although I could be wrong as I don't participate on PPCG).
– NobodyNada
2 days ago
2
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.
– Dennis♦
2 days ago
@NobodyNada deflate is not a programming language but we do not require that submissions be in programming languages.
– Post Left Garf Hunter
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
deflate, 79 bytes
eJyVkcsJADAIQ++ZIpD9dyxUqrGUgjlpHvglXdqCJymk1yEiaRXEIOXzuBHCiKObReUxUzYaMdt2wmTBg/FmNXndgLRbNvL7ifsLfMw6iQ==
New contributor
deflate, 79 bytes
eJyVkcsJADAIQ++ZIpD9dyxUqrGUgjlpHvglXdqCJymk1yEiaRXEIOXzuBHCiKObReUxUzYaMdt2wmTBg/FmNXndgLRbNvL7ifsLfMw6iQ==
New contributor
New contributor
answered 2 days ago
Whale
11
11
New contributor
New contributor
6
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 thedeflate
language, as this just looks like a base64 string
– Jo King
2 days ago
3
I don't think deflate counts as a programming language (although I could be wrong as I don't participate on PPCG).
– NobodyNada
2 days ago
2
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.
– Dennis♦
2 days ago
@NobodyNada deflate is not a programming language but we do not require that submissions be in programming languages.
– Post Left Garf Hunter
yesterday
add a comment |
6
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 thedeflate
language, as this just looks like a base64 string
– Jo King
2 days ago
3
I don't think deflate counts as a programming language (although I could be wrong as I don't participate on PPCG).
– NobodyNada
2 days ago
2
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.
– Dennis♦
2 days ago
@NobodyNada deflate is not a programming language but we do not require that submissions be in programming languages.
– Post Left Garf Hunter
yesterday
6
6
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– Jo King
2 days ago
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– Jo King
2 days ago
3
3
I don't think deflate counts as a programming language (although I could be wrong as I don't participate on PPCG).
– NobodyNada
2 days ago
I don't think deflate counts as a programming language (although I could be wrong as I don't participate on PPCG).
– NobodyNada
2 days ago
2
2
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.
– Dennis♦
2 days ago
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.
– Dennis♦
2 days ago
@NobodyNada deflate is not a programming language but we do not require that submissions be in programming languages.
– Post Left Garf Hunter
yesterday
@NobodyNada deflate is not a programming language but we do not require that submissions be in programming languages.
– Post Left Garf Hunter
yesterday
add a comment |
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3
What happened to the → in Ⓐ?
– Adám
Nov 13 at 20:00
3
@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
– Luis felipe De jesus Munoz
Nov 13 at 20:05
38
The second one is that I didnt know how to draw it :c
– Luis felipe De jesus Munoz
Nov 13 at 20:05
Are the trailing spaces required?
– Scoots
2 days ago
@Scoots no, it is not required as long as your output looks the same
– Luis felipe De jesus Munoz
2 days ago