Output the ŋarâþ crîþ alphabet song without using (m)any lettersProduce a palindrome from this textAlphanumeric balanceFinding prime numbers without using “prime characters”Holy Hole In A Donut, Batman!Shoot the ASCII MoonCreate an Alphabet SongCompressing the Atomic Ionization Energies
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Output the ŋarâþ crîþ alphabet song without using (m)any letters
Produce a palindrome from this textAlphanumeric balanceFinding prime numbers without using “prime characters”Holy Hole In A Donut, Batman!Shoot the ASCII MoonCreate an Alphabet SongCompressing the Atomic Ionization Energies
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Your goal is to write a program that takes no input and outputs the following text:
ca e na ŋa va o sa;
þa ša ra la ła.
ma a pa fa ga ta ča;
în ja i da ða.
ar ħo ên ôn ân uħo;
carþ taŋ neŋ es nem.
elo cenvos.
But there's a catch: for each letter (any character whose general category in Unicode starts with L) in your source, you get a penalty of 20 characters! (For reference, the text to be printed has 81 letters.)
The Perl 6 code below has 145 bytes and 84 letters, so it gets a score of 1,845:
say "ca e na ŋa va o sa;
þa ša ra la ła.
ma a pa fa ga ta ča;
în ja i da ða.
ar ħo ên ôn ân uħo;
carþ taŋ neŋ es nem.
elo cenvos."
The code below has 152 bytes and 70 letters, so it gets a score of 1,552:
$_="C e N ŋa V o S;
Þ Š R L Ł.
M a P F G T Č;
în J i D Ð.
ar ħo ên ôn ân uħo;
Crþ Tŋ neŋ es nem.
elo cenvos.";s:g/<:Lu>/$/.lc~'a'/;.say
Standard loopholes are forbidden.
Originally, I thought of forbidding letters altogether, but I don't think there are many languages that make this possible. You're more than welcome to try.
(ŋarâþ crîþ [ˈŋaɹa̰θ kɹḭθ] is one of my conlangs. I wanted to capitalise its name here, but I get the ugly big eng here. Oh well, the language doesn't use capital letters in its romanisation anyway.)
Edit: realised that one of the lines is wrong, but I'll keep it since there are already answers. The correct version of the third line is ma a fa ga pa ta ča; at your choice, you may choose to produce the corrected text instead.
code-challenge kolmogorov-complexity restricted-source
asked Apr 15 at 19:53
bb94bb94
1,7267 silver badges16 bronze badges
$endgroup$
add a comment
|
$begingroup$
Your goal is to write a program that takes no input and outputs the following text:
ca e na ŋa va o sa;
þa ša ra la ła.
ma a pa fa ga ta ča;
în ja i da ða.
ar ħo ên ôn ân uħo;
carþ taŋ neŋ es nem.
elo cenvos.
But there's a catch: for each letter (any character whose general category in Unicode starts with L) in your source, you get a penalty of 20 characters! (For reference, the text to be printed has 81 letters.)
The Perl 6 code below has 145 bytes and 84 letters, so it gets a score of 1,845:
say "ca e na ŋa va o sa;
þa ša ra la ła.
ma a pa fa ga ta ča;
în ja i da ða.
ar ħo ên ôn ân uħo;
carþ taŋ neŋ es nem.
elo cenvos."
The code below has 152 bytes and 70 letters, so it gets a score of 1,552:
$_="C e N ŋa V o S;
Þ Š R L Ł.
M a P F G T Č;
în J i D Ð.
ar ħo ên ôn ân uħo;
Crþ Tŋ neŋ es nem.
elo cenvos.";s:g/<:Lu>/$/.lc~'a'/;.say
Standard loopholes are forbidden.
Originally, I thought of forbidding letters altogether, but I don't think there are many languages that make this possible. You're more than welcome to try.
(ŋarâþ crîþ [ˈŋaɹa̰θ kɹḭθ] is one of my conlangs. I wanted to capitalise its name here, but I get the ugly big eng here. Oh well, the language doesn't use capital letters in its romanisation anyway.)
Edit: realised that one of the lines is wrong, but I'll keep it since there are already answers. The correct version of the third line is ma a fa ga pa ta ča; at your choice, you may choose to produce the corrected text instead.
code-challenge kolmogorov-complexity restricted-source
asked Apr 15 at 19:53
bb94bb94
1,7267 silver badges16 bronze badges
$endgroup$
11
$begingroup$
kolmogorov-complexity, restricted-source, and special scoring are all sorts of things that benefit greatly from careful consideration in the sandbox. Currently, it seems like the best approach to this challenge would be to just write out all of the codepoints in decimal then turn them into text with a builtin, with some shortcut to encode all of theas--or not, depending on how many letters it would take, because 20 characters is a really big penalty (although when everything else is scored by bytes, it's not quite well defined...)!
$endgroup$
– Unrelated String
Apr 15 at 20:09
4
$begingroup$
And considering the invocation of Unicode, some explicit rules governing special codepages as used by most golflangs are probably called for (alongside maybe a link to a script to validate scoring).
$endgroup$
– Unrelated String
Apr 15 at 21:00
add a comment
|
$begingroup$
Your goal is to write a program that takes no input and outputs the following text:
ca e na ŋa va o sa;
þa ša ra la ła.
ma a pa fa ga ta ča;
în ja i da ða.
ar ħo ên ôn ân uħo;
carþ taŋ neŋ es nem.
elo cenvos.
But there's a catch: for each letter (any character whose general category in Unicode starts with L) in your source, you get a penalty of 20 characters! (For reference, the text to be printed has 81 letters.)
The Perl 6 code below has 145 bytes and 84 letters, so it gets a score of 1,845:
say "ca e na ŋa va o sa;
þa ša ra la ła.
ma a pa fa ga ta ča;
în ja i da ða.
ar ħo ên ôn ân uħo;
carþ taŋ neŋ es nem.
elo cenvos."
The code below has 152 bytes and 70 letters, so it gets a score of 1,552:
$_="C e N ŋa V o S;
Þ Š R L Ł.
M a P F G T Č;
în J i D Ð.
ar ħo ên ôn ân uħo;
Crþ Tŋ neŋ es nem.
elo cenvos.";s:g/<:Lu>/$/.lc~'a'/;.say
Standard loopholes are forbidden.
Originally, I thought of forbidding letters altogether, but I don't think there are many languages that make this possible. You're more than welcome to try.
(ŋarâþ crîþ [ˈŋaɹa̰θ kɹḭθ] is one of my conlangs. I wanted to capitalise its name here, but I get the ugly big eng here. Oh well, the language doesn't use capital letters in its romanisation anyway.)
Edit: realised that one of the lines is wrong, but I'll keep it since there are already answers. The correct version of the third line is ma a fa ga pa ta ča; at your choice, you may choose to produce the corrected text instead.
code-challenge kolmogorov-complexity restricted-source
asked Apr 15 at 19:53
bb94bb94
1,7267 silver badges16 bronze badges
$endgroup$
Your goal is to write a program that takes no input and outputs the following text:
ca e na ŋa va o sa;
þa ša ra la ła.
ma a pa fa ga ta ča;
în ja i da ða.
ar ħo ên ôn ân uħo;
carþ taŋ neŋ es nem.
elo cenvos.
But there's a catch: for each letter (any character whose general category in Unicode starts with L) in your source, you get a penalty of 20 characters! (For reference, the text to be printed has 81 letters.)
The Perl 6 code below has 145 bytes and 84 letters, so it gets a score of 1,845:
say "ca e na ŋa va o sa;
þa ša ra la ła.
ma a pa fa ga ta ča;
în ja i da ða.
ar ħo ên ôn ân uħo;
carþ taŋ neŋ es nem.
elo cenvos."
The code below has 152 bytes and 70 letters, so it gets a score of 1,552:
$_="C e N ŋa V o S;
Þ Š R L Ł.
M a P F G T Č;
în J i D Ð.
ar ħo ên ôn ân uħo;
Crþ Tŋ neŋ es nem.
elo cenvos.";s:g/<:Lu>/$/.lc~'a'/;.say
Standard loopholes are forbidden.
Originally, I thought of forbidding letters altogether, but I don't think there are many languages that make this possible. You're more than welcome to try.
(ŋarâþ crîþ [ˈŋaɹa̰θ kɹḭθ] is one of my conlangs. I wanted to capitalise its name here, but I get the ugly big eng here. Oh well, the language doesn't use capital letters in its romanisation anyway.)
Edit: realised that one of the lines is wrong, but I'll keep it since there are already answers. The correct version of the third line is ma a fa ga pa ta ča; at your choice, you may choose to produce the corrected text instead.
code-challenge kolmogorov-complexity restricted-source
code-challenge kolmogorov-complexity restricted-source
asked Apr 15 at 19:53
bb94bb94
1,7267 silver badges16 bronze badges
asked Apr 15 at 19:53
bb94bb94
1,7267 silver badges16 bronze badges
edited Apr 22 at 19:15
bb94
asked Apr 15 at 19:53
bb94bb94
1,7267 silver badges16 bronze badges
asked Apr 15 at 19:53
bb94bb94
1,7267 silver badges16 bronze badges
asked Apr 15 at 19:53
bb94bb94
1,7267 silver badges16 bronze badges
1,7267 silver badges16 bronze badges
11
$begingroup$
kolmogorov-complexity, restricted-source, and special scoring are all sorts of things that benefit greatly from careful consideration in the sandbox. Currently, it seems like the best approach to this challenge would be to just write out all of the codepoints in decimal then turn them into text with a builtin, with some shortcut to encode all of theas--or not, depending on how many letters it would take, because 20 characters is a really big penalty (although when everything else is scored by bytes, it's not quite well defined...)!
$endgroup$
– Unrelated String
Apr 15 at 20:09
4
$begingroup$
And considering the invocation of Unicode, some explicit rules governing special codepages as used by most golflangs are probably called for (alongside maybe a link to a script to validate scoring).
$endgroup$
– Unrelated String
Apr 15 at 21:00
add a comment
|
11
$begingroup$
kolmogorov-complexity, restricted-source, and special scoring are all sorts of things that benefit greatly from careful consideration in the sandbox. Currently, it seems like the best approach to this challenge would be to just write out all of the codepoints in decimal then turn them into text with a builtin, with some shortcut to encode all of theas--or not, depending on how many letters it would take, because 20 characters is a really big penalty (although when everything else is scored by bytes, it's not quite well defined...)!
$endgroup$
– Unrelated String
Apr 15 at 20:09
4
$begingroup$
And considering the invocation of Unicode, some explicit rules governing special codepages as used by most golflangs are probably called for (alongside maybe a link to a script to validate scoring).
$endgroup$
– Unrelated String
Apr 15 at 21:00
11
11
$begingroup$
kolmogorov-complexity, restricted-source, and special scoring are all sorts of things that benefit greatly from careful consideration in the sandbox. Currently, it seems like the best approach to this challenge would be to just write out all of the codepoints in decimal then turn them into text with a builtin, with some shortcut to encode all of the
as--or not, depending on how many letters it would take, because 20 characters is a really big penalty (although when everything else is scored by bytes, it's not quite well defined...)!$endgroup$
– Unrelated String
Apr 15 at 20:09
$begingroup$
kolmogorov-complexity, restricted-source, and special scoring are all sorts of things that benefit greatly from careful consideration in the sandbox. Currently, it seems like the best approach to this challenge would be to just write out all of the codepoints in decimal then turn them into text with a builtin, with some shortcut to encode all of the
as--or not, depending on how many letters it would take, because 20 characters is a really big penalty (although when everything else is scored by bytes, it's not quite well defined...)!$endgroup$
– Unrelated String
Apr 15 at 20:09
4
4
$begingroup$
And considering the invocation of Unicode, some explicit rules governing special codepages as used by most golflangs are probably called for (alongside maybe a link to a script to validate scoring).
$endgroup$
– Unrelated String
Apr 15 at 21:00
$begingroup$
And considering the invocation of Unicode, some explicit rules governing special codepages as used by most golflangs are probably called for (alongside maybe a link to a script to validate scoring).
$endgroup$
– Unrelated String
Apr 15 at 21:00
add a comment
|
13 Answers
13
active
oldest
votes
$begingroup$
7, 410 characters, 154 bytes in 7's encoding, 0 letters = score 154
55104010504200144434451510201304004220120504005434473340353241135014335450302052254241052253052244241052335452241114014241310052340435303052335442302052335500302052335430302052313340435303135014243241310335514052312241341351052302245341351525755102440304030434030421030442030424030455733413512410523142410523030523112411350143355142410523414252410523102410523002410523413342411145257551220304010420030455741403
Try it online!
In a challenge that dislikes using letters, what better language to use than one consisting only of digits?
This is a full program that exits via crashing, so there's extraneous output to stderr, but stdout is correct.
Explanation
A 7 program, on its first iteration, simply pushes a number of elements to the stack (because out of the 12 commands that exist in 7, only 8 of them can be represented in a source program, and those 8 are specialised for writing code to push particular data structures to the stack). This program does not use the 6 command (which is the simplest way to create nested structures, but otherwise tends not to appear literally in a source program), so it's only the 7 commands that determine the structure; 7 pushes a new empty element to the top of stack (whereas the 0…5 commands just append to the top of stack). We can thus add whitespace to the program to show its structure:
551040105042001444344515102013040042201205040054344 7
33403532411350143354503020522542410522530522442410523354522411140142413100523
40435303052335442302052335500302052335430302052313340435303135014243241310335
514052312241341351052302245341351525 7
55102440304030434030421030442030424030455 7
33413512410523142410523030523112411350143355142410523414252410523102410523002
41052341334241114525 7
551220304010420030455 7
41403
The elements near the end of the program are pushed last, so are on top of the stack at the start of the second iteration. On this iteration, and all future iterations, the 7 interpreter automatically makes a copy of the top of the stack and interprets it as a program. The literal 41403 pushes the (non-literal, live code) 47463 (7 has 12 commands but only 8 of them have names; as such, I use bold to show the code, and non-bold to show the literal that generates that code, meaning that, e.g. 4 is the command that appends 4 to the top stack element). So the program that runs on the second iteration is 47463. Here's what that does:
47463
4 Swap top two stack elements, add an empty element in between
7 Add an empty stack element to the top of stack
4 Swap top two stack elements, add an empty element in between
6 Work out which commands would generate the top stack element;
append that to the element below (and pop the old top of stack)
3 Output the top stack element, pop the element below
This is easier to understand if we look at what happens to the stack:
- … d c b a
47463(code to run:47463) - … d c b
47463empty a (code to run:7463) - … d c b
47463empty a empty (code to run:463) - … d c b
47463empty empty empty a (code to run:63) - … d c b
47463empty empty "a" (code to run:3) - … d c b
47463empty (code to run: empty)
In other words, we take the top of stack a, work out what code is most likely to have produced it, and output that code. The 7 interpreter automatically pops empty elements from the top of stack at the end of an iteration, so we end up with the 47463 back on top of the stack, just as in the original program. It should be easy to see what happens next: we end up churning through every stack element one after another, outputting them all, until the stack underflows and the program crashes. So we've basically created a simple output loop that looks at the program's source code to determine what to output (we're not outputting the data structures that were pushes to the stack by our 0…5 commands, we're instead recreating what commands were used by looking at what structures were created, and outputting those). Thus, the first piece of data output is 551220304010420030455 (the source code that generates the second-from-top stack element), the second is 3341351…114525 (the source code that generates the third-from-top stack element), and so on.
Obviously, though, these pieces of source code aren't being output unencoded. 7 contains several different domain-specific languages for encoding output; once a domain-specific language is chosen, it remains in use until explicitly cleared, but if none of the languages have been chosen yet, the first digit of the code being output determines which of the languages to use. In this program, only two languages are used: 551 and 3.
551 is pretty simple: it's basically the old Baudot/teletype code used to transmit letters over teletypes, as a 5-bit character set, but modified to make all the letters lowercase. So the first chunk of code to be output decodes like this:
551 22 03 04 01 04 20 03 04 55
c a SP e SP n a SP reset output format
As can be seen, we're fitting each character into two octal digits, which is a pretty decent compression ratio. Pairs of digits in the 0-5 range give us 36 possibilities, as opposed to the 32 possibilities that Baudot needs, so the remaining four are used for special commands; in this case, the 55 at the end clears the remembered output format, letting us use a different format for the next piece of output we produce.
3 is conceptually even simpler, but with a twist. The basic idea is to take groups of three digits (again, in the 0-5 range, as those are the digits for which we can guarantee that we can recreate the original source code from its output), interpret them as a three-digit number in base 6, and just output it as a byte in binary (thus letting us output the multibyte characters in the desired output simply by outputting multiple bytes). The twist, though, comes from the fact that there are only 216 three-digit numbers (with possible leading zeroes) in base 6, but 256 possible bytes. 7 gets round this by linking numbers from 332₆ = 128₁₀ upwards to two different bytes; 332 can output either byte 128 or 192, 333 either byte 129 or 193, and so on, up to 515 which outputs either byte 191 or 255.
How does the program know which of the two possibilities to output? It's possible to use triplets of digits from 520 upwards to control this explicitly, but in this program we don't have to: 7's default is to pick all the ambiguous bytes in such a way that the output is valid UTF-8! It turns out that there's always at most one way to do this, so as long as it's UTF-8 we want (and we do in this case), we can just leave it ambiguous and the program works anyway.
The end of each of the 3… sections is 525, which resets the output format, letting us go back to 551 for the next section.
$endgroup$
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This is either 410 bytes + 0 letters in the unpacked representation, or 154 bytes + lots of letters in the packed representation. Counting the bytes in one and the letters in the other one seems cheaty.
$endgroup$
– Grimy
Sep 5 at 16:18
1
$begingroup$
@Grimy: You're confusing bytes with characters. The packed representation consists of 154 bytes in 7's encoding that encode 410 octal digits, each of which is a digit not a letter. Your reasoning implies that, say,ɓin Jelly is not a letter (because its encoding in Jelly's encoding corresponds to the control code "CSI" if interpreted in a typical 8-bit character set, rather than a letter). Just like Jelly, 7 also uses a custom encoding; but because 7 uses no letters, the encoding has no need to encode letters and thus can't.
$endgroup$
– ais523
Sep 5 at 20:16
add a comment
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$begingroup$
Haskell, 0 letters, 423 bytes = score 423
(['10'..]!!)<$>[89,87,22,91,22,100,87,22,321,87,22,108,87,22,101,22,105,87,49,0,244,87,22,343,87,22,104,87,22,98,87,22,312,87,36,0,99,87,22,87,22,102,87,22,92,87,22,93,87,22,106,87,22,259,87,49,0,228,100,22,96,87,22,95,22,90,87,22,230,87,36,0,87,104,22,285,101,22,224,100,22,234,100,22,216,100,22,107,285,101,49,0,89,87,104,244,22,106,87,321,22,100,91,321,22,91,105,22,100,91,99,36,0,91,98,101,22,89,91,100,108,101,105,36]
Try it online!
answered Apr 15 at 21:40
niminimi
34k3 gold badges27 silver badges91 bronze badges
$endgroup$
add a comment
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$begingroup$
Jelly, 274 260 212 bytes + 2 letters = 314 300 252
-48 bytes thanks to Nick Kennedy
“19ב+49;7883,8220,8216,7884Ọ“19937801,1169680277365253“38“68112“;107¤+1+“@“&%"("/%"@%"6%"0"3%$!<%" %"2%"-%"?%#!.%"%"1%")%"*%"4%"=%$!9/",%"+"'%":%#!%2">0"8/";/"7/"5>0$!&%2<"4%@"/(@"(3"/(.#!(-0"&(/603#“_32¤”;";/V
(Uses !"#$%&'()*+,-./0123456789:;<=>?@V_¤×Ọ‘“” of which V and Ọ are Unicode letters and are used once each)
Try it online!
answered Apr 15 at 22:55
Jonathan AllanJonathan Allan
60.5k5 gold badges44 silver badges188 bronze badges
$endgroup$
$begingroup$
212 bytes plus two letters
$endgroup$
– Nick Kennedy
Apr 17 at 22:58
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Verification
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– Nick Kennedy
Apr 17 at 23:23
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@NickKennedy I'd played around with golfing the number, but didn't step back and look to just offset the ordinals, good stuff - thanks!
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– Jonathan Allan
Apr 18 at 7:17
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$begingroup$
PowerShell, scores 601 546
-join(67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,78,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14|%[char]($_+32))
Try it online!
Naive approach; I just took the code points and converted them to decimal, subtracted 32, then this code treats them as a char before -joining it back together into a single string.
answered Apr 15 at 20:21
AdmBorkBorkAdmBorkBork
29.2k4 gold badges74 silver badges252 bronze badges
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901, ouch
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– ASCII-only
Apr 16 at 0:22
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686 :/
$endgroup$
– ASCII-only
Apr 16 at 0:28
add a comment
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$begingroup$
Jelly, 321 bytes + 2 letters = score 361
3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304b354Ọ
Try it online!
This is hideous and someone can definitely do better.
Verify score.
answered Apr 15 at 20:35
HyperNeutrinoHyperNeutrino
20.6k4 gold badges41 silver badges155 bronze badges
$endgroup$
1
$begingroup$
actually less bad than it seems
$endgroup$
– ASCII-only
Apr 16 at 0:06
add a comment
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$begingroup$
Python 3, 380 bytes + 5 letters = 480
print("""143141 145 156141 513141 166141 157 163141;
376141 541141 162141 154141 502141.
155141 141 160141 146141 147141 164141 415141;
356156 152141 151 144141 360141.
141162 447157 352156 364156 342156 165447157;
143141162376 164141513 156145513 145163 156145155.
145154157 143145156166157163.""")
Try it online!
answered Apr 16 at 15:03
LynnLynn
52.2k9 gold badges103 silver badges238 bronze badges
$endgroup$
add a comment
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$begingroup$
Retina, 140 characters, 159 bytes, 14 letters = score 439
%# ' 1# !# 9# 2 6#;¶þ# š# 5# /# ł#.¶0# # 3# (# )# 7# č#;¶î1 ,# + &# ð#.¶#5 ħ2 ê1 ô1 â1 8ħ2;¶%#5þ 7#! 1'! '6 1'0.¶'/2 %'1926.
T`!--/-9`ŋ`-
Try it online! Edit: Saved 1 letter by switching from K` to a newline. Now also works in Retina 0.8.2 (but the title would be too long).
answered Apr 16 at 16:56
ShaggyShaggy
21.8k3 gold badges21 silver badges73 bronze badges
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$begingroup$
PHP -a, 402 bytes + 200 penalty = 602 score
foreach([67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,8,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14] as $i)echo ''.mb_chr($i+32);
Port of Artermis Fowl's answer, and my first codegolf entry!
Leaves me wishing that chr() could just support UTF-8; those extra 3 bytes + 40 characters hurts!
answered Apr 17 at 0:10
videovideo
111 bronze badge
$endgroup$
$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
Apr 17 at 12:15
add a comment
|
$begingroup$
05AB1E, score 209 (189 bytes + 20 penalty for 1 letter)
•£?;:'%¢;.'¡£/':¢?'¢°':¢°#@¢«>#%¡¤;®[¢:¥¢:©¢:¦¢;®¢>#¡£#¨¢#&¢+¢#,¢:§¡¤#¬¢#@¢#)¢#(¢#<¢#¢#/¡£#¯¢#.¢#>¢#±¢#«¡¤#?¢;¢#¢#°¢#:¢'¢#%•[₅‰`©®_#∞158+902201401301670804020409010150250102709022¾¡.¥>:ç?
Try it online!
The only letter is ç. The currency symbols €£¢ are not considered letters in Unicode.
answered Sep 5 at 16:14
GrimyGrimy
8,22818 silver badges37 bronze badges
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$begingroup$
Python 3, 397 bytes + 19 letters = 777 score
print(''.join(chr(i+32)for i in[67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,78,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14]))
Try it online!
Port of AdmBorkBork's answer.
answered Apr 15 at 20:41
Artemis FowlArtemis Fowl
3402 silver badges13 bronze badges
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$begingroup$
1 less letter?
$endgroup$
– ASCII-only
Apr 16 at 12:28
1
$begingroup$
score 732
$endgroup$
– ASCII-only
Apr 16 at 12:29
$begingroup$
562, -2 if using python 2
$endgroup$
– ASCII-only
Apr 16 at 12:33
$begingroup$
TIO doesn't work at my organization, so I'll have to wait to get home to add those.
$endgroup$
– Artemis Fowl
Apr 16 at 14:17
add a comment
|
$begingroup$
R, 384 bytes + 12 letters * 20 points = 684 score
Not terribly original.
cat(intToUtf8(c(67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,78,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14)+32))
Try it online!
answered Apr 17 at 5:25
CT HallCT Hall
54111 bronze badges
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$begingroup$
05AB1E, score 383 365 (325 bytes + 2 letters * 20 penalty)
3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304 354вç.««
Port of @HyperNeutrino's Jelly answer.
Will try to improve here on. The number is divisible by a bunch of numbers, but none of them would save any bytes unfortunately, and the larger divisors compressed contain at least 1 letter..
-18 (+2 bytes and -20 penalty) thanks to @Grimy, replacing the letter J (join) with .«« (reduce by concatenating).
Try it online.
answered Apr 16 at 8:11
Kevin CruijssenKevin Cruijssen
51.4k7 gold badges85 silver badges248 bronze badges
$endgroup$
1
$begingroup$
Jcan be.««for -18. Or for a completely different approach, see my answer.
$endgroup$
– Grimy
Sep 5 at 16:15
$begingroup$
@Grimy Thanks! :) And nice answer!
$endgroup$
– Kevin Cruijssen
Sep 5 at 16:52
add a comment
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$begingroup$
7, 410 characters, 154 bytes in 7's encoding, 0 letters = score 154
55104010504200144434451510201304004220120504005434473340353241135014335450302052254241052253052244241052335452241114014241310052340435303052335442302052335500302052335430302052313340435303135014243241310335514052312241341351052302245341351525755102440304030434030421030442030424030455733413512410523142410523030523112411350143355142410523414252410523102410523002410523413342411145257551220304010420030455741403
Try it online!
In a challenge that dislikes using letters, what better language to use than one consisting only of digits?
This is a full program that exits via crashing, so there's extraneous output to stderr, but stdout is correct.
Explanation
A 7 program, on its first iteration, simply pushes a number of elements to the stack (because out of the 12 commands that exist in 7, only 8 of them can be represented in a source program, and those 8 are specialised for writing code to push particular data structures to the stack). This program does not use the 6 command (which is the simplest way to create nested structures, but otherwise tends not to appear literally in a source program), so it's only the 7 commands that determine the structure; 7 pushes a new empty element to the top of stack (whereas the 0…5 commands just append to the top of stack). We can thus add whitespace to the program to show its structure:
551040105042001444344515102013040042201205040054344 7
33403532411350143354503020522542410522530522442410523354522411140142413100523
40435303052335442302052335500302052335430302052313340435303135014243241310335
514052312241341351052302245341351525 7
55102440304030434030421030442030424030455 7
33413512410523142410523030523112411350143355142410523414252410523102410523002
41052341334241114525 7
551220304010420030455 7
41403
The elements near the end of the program are pushed last, so are on top of the stack at the start of the second iteration. On this iteration, and all future iterations, the 7 interpreter automatically makes a copy of the top of the stack and interprets it as a program. The literal 41403 pushes the (non-literal, live code) 47463 (7 has 12 commands but only 8 of them have names; as such, I use bold to show the code, and non-bold to show the literal that generates that code, meaning that, e.g. 4 is the command that appends 4 to the top stack element). So the program that runs on the second iteration is 47463. Here's what that does:
47463
4 Swap top two stack elements, add an empty element in between
7 Add an empty stack element to the top of stack
4 Swap top two stack elements, add an empty element in between
6 Work out which commands would generate the top stack element;
append that to the element below (and pop the old top of stack)
3 Output the top stack element, pop the element below
This is easier to understand if we look at what happens to the stack:
- … d c b a
47463(code to run:47463) - … d c b
47463empty a (code to run:7463) - … d c b
47463empty a empty (code to run:463) - … d c b
47463empty empty empty a (code to run:63) - … d c b
47463empty empty "a" (code to run:3) - … d c b
47463empty (code to run: empty)
In other words, we take the top of stack a, work out what code is most likely to have produced it, and output that code. The 7 interpreter automatically pops empty elements from the top of stack at the end of an iteration, so we end up with the 47463 back on top of the stack, just as in the original program. It should be easy to see what happens next: we end up churning through every stack element one after another, outputting them all, until the stack underflows and the program crashes. So we've basically created a simple output loop that looks at the program's source code to determine what to output (we're not outputting the data structures that were pushes to the stack by our 0…5 commands, we're instead recreating what commands were used by looking at what structures were created, and outputting those). Thus, the first piece of data output is 551220304010420030455 (the source code that generates the second-from-top stack element), the second is 3341351…114525 (the source code that generates the third-from-top stack element), and so on.
Obviously, though, these pieces of source code aren't being output unencoded. 7 contains several different domain-specific languages for encoding output; once a domain-specific language is chosen, it remains in use until explicitly cleared, but if none of the languages have been chosen yet, the first digit of the code being output determines which of the languages to use. In this program, only two languages are used: 551 and 3.
551 is pretty simple: it's basically the old Baudot/teletype code used to transmit letters over teletypes, as a 5-bit character set, but modified to make all the letters lowercase. So the first chunk of code to be output decodes like this:
551 22 03 04 01 04 20 03 04 55
c a SP e SP n a SP reset output format
As can be seen, we're fitting each character into two octal digits, which is a pretty decent compression ratio. Pairs of digits in the 0-5 range give us 36 possibilities, as opposed to the 32 possibilities that Baudot needs, so the remaining four are used for special commands; in this case, the 55 at the end clears the remembered output format, letting us use a different format for the next piece of output we produce.
3 is conceptually even simpler, but with a twist. The basic idea is to take groups of three digits (again, in the 0-5 range, as those are the digits for which we can guarantee that we can recreate the original source code from its output), interpret them as a three-digit number in base 6, and just output it as a byte in binary (thus letting us output the multibyte characters in the desired output simply by outputting multiple bytes). The twist, though, comes from the fact that there are only 216 three-digit numbers (with possible leading zeroes) in base 6, but 256 possible bytes. 7 gets round this by linking numbers from 332₆ = 128₁₀ upwards to two different bytes; 332 can output either byte 128 or 192, 333 either byte 129 or 193, and so on, up to 515 which outputs either byte 191 or 255.
How does the program know which of the two possibilities to output? It's possible to use triplets of digits from 520 upwards to control this explicitly, but in this program we don't have to: 7's default is to pick all the ambiguous bytes in such a way that the output is valid UTF-8! It turns out that there's always at most one way to do this, so as long as it's UTF-8 we want (and we do in this case), we can just leave it ambiguous and the program works anyway.
The end of each of the 3… sections is 525, which resets the output format, letting us go back to 551 for the next section.
$endgroup$
$begingroup$
This is either 410 bytes + 0 letters in the unpacked representation, or 154 bytes + lots of letters in the packed representation. Counting the bytes in one and the letters in the other one seems cheaty.
$endgroup$
– Grimy
Sep 5 at 16:18
1
$begingroup$
@Grimy: You're confusing bytes with characters. The packed representation consists of 154 bytes in 7's encoding that encode 410 octal digits, each of which is a digit not a letter. Your reasoning implies that, say,ɓin Jelly is not a letter (because its encoding in Jelly's encoding corresponds to the control code "CSI" if interpreted in a typical 8-bit character set, rather than a letter). Just like Jelly, 7 also uses a custom encoding; but because 7 uses no letters, the encoding has no need to encode letters and thus can't.
$endgroup$
– ais523
Sep 5 at 20:16
add a comment
|
$begingroup$
7, 410 characters, 154 bytes in 7's encoding, 0 letters = score 154
55104010504200144434451510201304004220120504005434473340353241135014335450302052254241052253052244241052335452241114014241310052340435303052335442302052335500302052335430302052313340435303135014243241310335514052312241341351052302245341351525755102440304030434030421030442030424030455733413512410523142410523030523112411350143355142410523414252410523102410523002410523413342411145257551220304010420030455741403
Try it online!
In a challenge that dislikes using letters, what better language to use than one consisting only of digits?
This is a full program that exits via crashing, so there's extraneous output to stderr, but stdout is correct.
Explanation
A 7 program, on its first iteration, simply pushes a number of elements to the stack (because out of the 12 commands that exist in 7, only 8 of them can be represented in a source program, and those 8 are specialised for writing code to push particular data structures to the stack). This program does not use the 6 command (which is the simplest way to create nested structures, but otherwise tends not to appear literally in a source program), so it's only the 7 commands that determine the structure; 7 pushes a new empty element to the top of stack (whereas the 0…5 commands just append to the top of stack). We can thus add whitespace to the program to show its structure:
551040105042001444344515102013040042201205040054344 7
33403532411350143354503020522542410522530522442410523354522411140142413100523
40435303052335442302052335500302052335430302052313340435303135014243241310335
514052312241341351052302245341351525 7
55102440304030434030421030442030424030455 7
33413512410523142410523030523112411350143355142410523414252410523102410523002
41052341334241114525 7
551220304010420030455 7
41403
The elements near the end of the program are pushed last, so are on top of the stack at the start of the second iteration. On this iteration, and all future iterations, the 7 interpreter automatically makes a copy of the top of the stack and interprets it as a program. The literal 41403 pushes the (non-literal, live code) 47463 (7 has 12 commands but only 8 of them have names; as such, I use bold to show the code, and non-bold to show the literal that generates that code, meaning that, e.g. 4 is the command that appends 4 to the top stack element). So the program that runs on the second iteration is 47463. Here's what that does:
47463
4 Swap top two stack elements, add an empty element in between
7 Add an empty stack element to the top of stack
4 Swap top two stack elements, add an empty element in between
6 Work out which commands would generate the top stack element;
append that to the element below (and pop the old top of stack)
3 Output the top stack element, pop the element below
This is easier to understand if we look at what happens to the stack:
- … d c b a
47463(code to run:47463) - … d c b
47463empty a (code to run:7463) - … d c b
47463empty a empty (code to run:463) - … d c b
47463empty empty empty a (code to run:63) - … d c b
47463empty empty "a" (code to run:3) - … d c b
47463empty (code to run: empty)
In other words, we take the top of stack a, work out what code is most likely to have produced it, and output that code. The 7 interpreter automatically pops empty elements from the top of stack at the end of an iteration, so we end up with the 47463 back on top of the stack, just as in the original program. It should be easy to see what happens next: we end up churning through every stack element one after another, outputting them all, until the stack underflows and the program crashes. So we've basically created a simple output loop that looks at the program's source code to determine what to output (we're not outputting the data structures that were pushes to the stack by our 0…5 commands, we're instead recreating what commands were used by looking at what structures were created, and outputting those). Thus, the first piece of data output is 551220304010420030455 (the source code that generates the second-from-top stack element), the second is 3341351…114525 (the source code that generates the third-from-top stack element), and so on.
Obviously, though, these pieces of source code aren't being output unencoded. 7 contains several different domain-specific languages for encoding output; once a domain-specific language is chosen, it remains in use until explicitly cleared, but if none of the languages have been chosen yet, the first digit of the code being output determines which of the languages to use. In this program, only two languages are used: 551 and 3.
551 is pretty simple: it's basically the old Baudot/teletype code used to transmit letters over teletypes, as a 5-bit character set, but modified to make all the letters lowercase. So the first chunk of code to be output decodes like this:
551 22 03 04 01 04 20 03 04 55
c a SP e SP n a SP reset output format
As can be seen, we're fitting each character into two octal digits, which is a pretty decent compression ratio. Pairs of digits in the 0-5 range give us 36 possibilities, as opposed to the 32 possibilities that Baudot needs, so the remaining four are used for special commands; in this case, the 55 at the end clears the remembered output format, letting us use a different format for the next piece of output we produce.
3 is conceptually even simpler, but with a twist. The basic idea is to take groups of three digits (again, in the 0-5 range, as those are the digits for which we can guarantee that we can recreate the original source code from its output), interpret them as a three-digit number in base 6, and just output it as a byte in binary (thus letting us output the multibyte characters in the desired output simply by outputting multiple bytes). The twist, though, comes from the fact that there are only 216 three-digit numbers (with possible leading zeroes) in base 6, but 256 possible bytes. 7 gets round this by linking numbers from 332₆ = 128₁₀ upwards to two different bytes; 332 can output either byte 128 or 192, 333 either byte 129 or 193, and so on, up to 515 which outputs either byte 191 or 255.
How does the program know which of the two possibilities to output? It's possible to use triplets of digits from 520 upwards to control this explicitly, but in this program we don't have to: 7's default is to pick all the ambiguous bytes in such a way that the output is valid UTF-8! It turns out that there's always at most one way to do this, so as long as it's UTF-8 we want (and we do in this case), we can just leave it ambiguous and the program works anyway.
The end of each of the 3… sections is 525, which resets the output format, letting us go back to 551 for the next section.
$endgroup$
$begingroup$
This is either 410 bytes + 0 letters in the unpacked representation, or 154 bytes + lots of letters in the packed representation. Counting the bytes in one and the letters in the other one seems cheaty.
$endgroup$
– Grimy
Sep 5 at 16:18
1
$begingroup$
@Grimy: You're confusing bytes with characters. The packed representation consists of 154 bytes in 7's encoding that encode 410 octal digits, each of which is a digit not a letter. Your reasoning implies that, say,ɓin Jelly is not a letter (because its encoding in Jelly's encoding corresponds to the control code "CSI" if interpreted in a typical 8-bit character set, rather than a letter). Just like Jelly, 7 also uses a custom encoding; but because 7 uses no letters, the encoding has no need to encode letters and thus can't.
$endgroup$
– ais523
Sep 5 at 20:16
add a comment
|
$begingroup$
7, 410 characters, 154 bytes in 7's encoding, 0 letters = score 154
55104010504200144434451510201304004220120504005434473340353241135014335450302052254241052253052244241052335452241114014241310052340435303052335442302052335500302052335430302052313340435303135014243241310335514052312241341351052302245341351525755102440304030434030421030442030424030455733413512410523142410523030523112411350143355142410523414252410523102410523002410523413342411145257551220304010420030455741403
Try it online!
In a challenge that dislikes using letters, what better language to use than one consisting only of digits?
This is a full program that exits via crashing, so there's extraneous output to stderr, but stdout is correct.
Explanation
A 7 program, on its first iteration, simply pushes a number of elements to the stack (because out of the 12 commands that exist in 7, only 8 of them can be represented in a source program, and those 8 are specialised for writing code to push particular data structures to the stack). This program does not use the 6 command (which is the simplest way to create nested structures, but otherwise tends not to appear literally in a source program), so it's only the 7 commands that determine the structure; 7 pushes a new empty element to the top of stack (whereas the 0…5 commands just append to the top of stack). We can thus add whitespace to the program to show its structure:
551040105042001444344515102013040042201205040054344 7
33403532411350143354503020522542410522530522442410523354522411140142413100523
40435303052335442302052335500302052335430302052313340435303135014243241310335
514052312241341351052302245341351525 7
55102440304030434030421030442030424030455 7
33413512410523142410523030523112411350143355142410523414252410523102410523002
41052341334241114525 7
551220304010420030455 7
41403
The elements near the end of the program are pushed last, so are on top of the stack at the start of the second iteration. On this iteration, and all future iterations, the 7 interpreter automatically makes a copy of the top of the stack and interprets it as a program. The literal 41403 pushes the (non-literal, live code) 47463 (7 has 12 commands but only 8 of them have names; as such, I use bold to show the code, and non-bold to show the literal that generates that code, meaning that, e.g. 4 is the command that appends 4 to the top stack element). So the program that runs on the second iteration is 47463. Here's what that does:
47463
4 Swap top two stack elements, add an empty element in between
7 Add an empty stack element to the top of stack
4 Swap top two stack elements, add an empty element in between
6 Work out which commands would generate the top stack element;
append that to the element below (and pop the old top of stack)
3 Output the top stack element, pop the element below
This is easier to understand if we look at what happens to the stack:
- … d c b a
47463(code to run:47463) - … d c b
47463empty a (code to run:7463) - … d c b
47463empty a empty (code to run:463) - … d c b
47463empty empty empty a (code to run:63) - … d c b
47463empty empty "a" (code to run:3) - … d c b
47463empty (code to run: empty)
In other words, we take the top of stack a, work out what code is most likely to have produced it, and output that code. The 7 interpreter automatically pops empty elements from the top of stack at the end of an iteration, so we end up with the 47463 back on top of the stack, just as in the original program. It should be easy to see what happens next: we end up churning through every stack element one after another, outputting them all, until the stack underflows and the program crashes. So we've basically created a simple output loop that looks at the program's source code to determine what to output (we're not outputting the data structures that were pushes to the stack by our 0…5 commands, we're instead recreating what commands were used by looking at what structures were created, and outputting those). Thus, the first piece of data output is 551220304010420030455 (the source code that generates the second-from-top stack element), the second is 3341351…114525 (the source code that generates the third-from-top stack element), and so on.
Obviously, though, these pieces of source code aren't being output unencoded. 7 contains several different domain-specific languages for encoding output; once a domain-specific language is chosen, it remains in use until explicitly cleared, but if none of the languages have been chosen yet, the first digit of the code being output determines which of the languages to use. In this program, only two languages are used: 551 and 3.
551 is pretty simple: it's basically the old Baudot/teletype code used to transmit letters over teletypes, as a 5-bit character set, but modified to make all the letters lowercase. So the first chunk of code to be output decodes like this:
551 22 03 04 01 04 20 03 04 55
c a SP e SP n a SP reset output format
As can be seen, we're fitting each character into two octal digits, which is a pretty decent compression ratio. Pairs of digits in the 0-5 range give us 36 possibilities, as opposed to the 32 possibilities that Baudot needs, so the remaining four are used for special commands; in this case, the 55 at the end clears the remembered output format, letting us use a different format for the next piece of output we produce.
3 is conceptually even simpler, but with a twist. The basic idea is to take groups of three digits (again, in the 0-5 range, as those are the digits for which we can guarantee that we can recreate the original source code from its output), interpret them as a three-digit number in base 6, and just output it as a byte in binary (thus letting us output the multibyte characters in the desired output simply by outputting multiple bytes). The twist, though, comes from the fact that there are only 216 three-digit numbers (with possible leading zeroes) in base 6, but 256 possible bytes. 7 gets round this by linking numbers from 332₆ = 128₁₀ upwards to two different bytes; 332 can output either byte 128 or 192, 333 either byte 129 or 193, and so on, up to 515 which outputs either byte 191 or 255.
How does the program know which of the two possibilities to output? It's possible to use triplets of digits from 520 upwards to control this explicitly, but in this program we don't have to: 7's default is to pick all the ambiguous bytes in such a way that the output is valid UTF-8! It turns out that there's always at most one way to do this, so as long as it's UTF-8 we want (and we do in this case), we can just leave it ambiguous and the program works anyway.
The end of each of the 3… sections is 525, which resets the output format, letting us go back to 551 for the next section.
$endgroup$
7, 410 characters, 154 bytes in 7's encoding, 0 letters = score 154
55104010504200144434451510201304004220120504005434473340353241135014335450302052254241052253052244241052335452241114014241310052340435303052335442302052335500302052335430302052313340435303135014243241310335514052312241341351052302245341351525755102440304030434030421030442030424030455733413512410523142410523030523112411350143355142410523414252410523102410523002410523413342411145257551220304010420030455741403
Try it online!
In a challenge that dislikes using letters, what better language to use than one consisting only of digits?
This is a full program that exits via crashing, so there's extraneous output to stderr, but stdout is correct.
Explanation
A 7 program, on its first iteration, simply pushes a number of elements to the stack (because out of the 12 commands that exist in 7, only 8 of them can be represented in a source program, and those 8 are specialised for writing code to push particular data structures to the stack). This program does not use the 6 command (which is the simplest way to create nested structures, but otherwise tends not to appear literally in a source program), so it's only the 7 commands that determine the structure; 7 pushes a new empty element to the top of stack (whereas the 0…5 commands just append to the top of stack). We can thus add whitespace to the program to show its structure:
551040105042001444344515102013040042201205040054344 7
33403532411350143354503020522542410522530522442410523354522411140142413100523
40435303052335442302052335500302052335430302052313340435303135014243241310335
514052312241341351052302245341351525 7
55102440304030434030421030442030424030455 7
33413512410523142410523030523112411350143355142410523414252410523102410523002
41052341334241114525 7
551220304010420030455 7
41403
The elements near the end of the program are pushed last, so are on top of the stack at the start of the second iteration. On this iteration, and all future iterations, the 7 interpreter automatically makes a copy of the top of the stack and interprets it as a program. The literal 41403 pushes the (non-literal, live code) 47463 (7 has 12 commands but only 8 of them have names; as such, I use bold to show the code, and non-bold to show the literal that generates that code, meaning that, e.g. 4 is the command that appends 4 to the top stack element). So the program that runs on the second iteration is 47463. Here's what that does:
47463
4 Swap top two stack elements, add an empty element in between
7 Add an empty stack element to the top of stack
4 Swap top two stack elements, add an empty element in between
6 Work out which commands would generate the top stack element;
append that to the element below (and pop the old top of stack)
3 Output the top stack element, pop the element below
This is easier to understand if we look at what happens to the stack:
- … d c b a
47463(code to run:47463) - … d c b
47463empty a (code to run:7463) - … d c b
47463empty a empty (code to run:463) - … d c b
47463empty empty empty a (code to run:63) - … d c b
47463empty empty "a" (code to run:3) - … d c b
47463empty (code to run: empty)
In other words, we take the top of stack a, work out what code is most likely to have produced it, and output that code. The 7 interpreter automatically pops empty elements from the top of stack at the end of an iteration, so we end up with the 47463 back on top of the stack, just as in the original program. It should be easy to see what happens next: we end up churning through every stack element one after another, outputting them all, until the stack underflows and the program crashes. So we've basically created a simple output loop that looks at the program's source code to determine what to output (we're not outputting the data structures that were pushes to the stack by our 0…5 commands, we're instead recreating what commands were used by looking at what structures were created, and outputting those). Thus, the first piece of data output is 551220304010420030455 (the source code that generates the second-from-top stack element), the second is 3341351…114525 (the source code that generates the third-from-top stack element), and so on.
Obviously, though, these pieces of source code aren't being output unencoded. 7 contains several different domain-specific languages for encoding output; once a domain-specific language is chosen, it remains in use until explicitly cleared, but if none of the languages have been chosen yet, the first digit of the code being output determines which of the languages to use. In this program, only two languages are used: 551 and 3.
551 is pretty simple: it's basically the old Baudot/teletype code used to transmit letters over teletypes, as a 5-bit character set, but modified to make all the letters lowercase. So the first chunk of code to be output decodes like this:
551 22 03 04 01 04 20 03 04 55
c a SP e SP n a SP reset output format
As can be seen, we're fitting each character into two octal digits, which is a pretty decent compression ratio. Pairs of digits in the 0-5 range give us 36 possibilities, as opposed to the 32 possibilities that Baudot needs, so the remaining four are used for special commands; in this case, the 55 at the end clears the remembered output format, letting us use a different format for the next piece of output we produce.
3 is conceptually even simpler, but with a twist. The basic idea is to take groups of three digits (again, in the 0-5 range, as those are the digits for which we can guarantee that we can recreate the original source code from its output), interpret them as a three-digit number in base 6, and just output it as a byte in binary (thus letting us output the multibyte characters in the desired output simply by outputting multiple bytes). The twist, though, comes from the fact that there are only 216 three-digit numbers (with possible leading zeroes) in base 6, but 256 possible bytes. 7 gets round this by linking numbers from 332₆ = 128₁₀ upwards to two different bytes; 332 can output either byte 128 or 192, 333 either byte 129 or 193, and so on, up to 515 which outputs either byte 191 or 255.
How does the program know which of the two possibilities to output? It's possible to use triplets of digits from 520 upwards to control this explicitly, but in this program we don't have to: 7's default is to pick all the ambiguous bytes in such a way that the output is valid UTF-8! It turns out that there's always at most one way to do this, so as long as it's UTF-8 we want (and we do in this case), we can just leave it ambiguous and the program works anyway.
The end of each of the 3… sections is 525, which resets the output format, letting us go back to 551 for the next section.
answered Apr 16 at 2:02
community wiki
ais523
$begingroup$
This is either 410 bytes + 0 letters in the unpacked representation, or 154 bytes + lots of letters in the packed representation. Counting the bytes in one and the letters in the other one seems cheaty.
$endgroup$
– Grimy
Sep 5 at 16:18
1
$begingroup$
@Grimy: You're confusing bytes with characters. The packed representation consists of 154 bytes in 7's encoding that encode 410 octal digits, each of which is a digit not a letter. Your reasoning implies that, say,ɓin Jelly is not a letter (because its encoding in Jelly's encoding corresponds to the control code "CSI" if interpreted in a typical 8-bit character set, rather than a letter). Just like Jelly, 7 also uses a custom encoding; but because 7 uses no letters, the encoding has no need to encode letters and thus can't.
$endgroup$
– ais523
Sep 5 at 20:16
add a comment
|
$begingroup$
This is either 410 bytes + 0 letters in the unpacked representation, or 154 bytes + lots of letters in the packed representation. Counting the bytes in one and the letters in the other one seems cheaty.
$endgroup$
– Grimy
Sep 5 at 16:18
1
$begingroup$
@Grimy: You're confusing bytes with characters. The packed representation consists of 154 bytes in 7's encoding that encode 410 octal digits, each of which is a digit not a letter. Your reasoning implies that, say,ɓin Jelly is not a letter (because its encoding in Jelly's encoding corresponds to the control code "CSI" if interpreted in a typical 8-bit character set, rather than a letter). Just like Jelly, 7 also uses a custom encoding; but because 7 uses no letters, the encoding has no need to encode letters and thus can't.
$endgroup$
– ais523
Sep 5 at 20:16
$begingroup$
This is either 410 bytes + 0 letters in the unpacked representation, or 154 bytes + lots of letters in the packed representation. Counting the bytes in one and the letters in the other one seems cheaty.
$endgroup$
– Grimy
Sep 5 at 16:18
$begingroup$
This is either 410 bytes + 0 letters in the unpacked representation, or 154 bytes + lots of letters in the packed representation. Counting the bytes in one and the letters in the other one seems cheaty.
$endgroup$
– Grimy
Sep 5 at 16:18
1
1
$begingroup$
@Grimy: You're confusing bytes with characters. The packed representation consists of 154 bytes in 7's encoding that encode 410 octal digits, each of which is a digit not a letter. Your reasoning implies that, say,
ɓ in Jelly is not a letter (because its encoding in Jelly's encoding corresponds to the control code "CSI" if interpreted in a typical 8-bit character set, rather than a letter). Just like Jelly, 7 also uses a custom encoding; but because 7 uses no letters, the encoding has no need to encode letters and thus can't.$endgroup$
– ais523
Sep 5 at 20:16
$begingroup$
@Grimy: You're confusing bytes with characters. The packed representation consists of 154 bytes in 7's encoding that encode 410 octal digits, each of which is a digit not a letter. Your reasoning implies that, say,
ɓ in Jelly is not a letter (because its encoding in Jelly's encoding corresponds to the control code "CSI" if interpreted in a typical 8-bit character set, rather than a letter). Just like Jelly, 7 also uses a custom encoding; but because 7 uses no letters, the encoding has no need to encode letters and thus can't.$endgroup$
– ais523
Sep 5 at 20:16
add a comment
|
$begingroup$
Haskell, 0 letters, 423 bytes = score 423
(['10'..]!!)<$>[89,87,22,91,22,100,87,22,321,87,22,108,87,22,101,22,105,87,49,0,244,87,22,343,87,22,104,87,22,98,87,22,312,87,36,0,99,87,22,87,22,102,87,22,92,87,22,93,87,22,106,87,22,259,87,49,0,228,100,22,96,87,22,95,22,90,87,22,230,87,36,0,87,104,22,285,101,22,224,100,22,234,100,22,216,100,22,107,285,101,49,0,89,87,104,244,22,106,87,321,22,100,91,321,22,91,105,22,100,91,99,36,0,91,98,101,22,89,91,100,108,101,105,36]
Try it online!
answered Apr 15 at 21:40
niminimi
34k3 gold badges27 silver badges91 bronze badges
$endgroup$
add a comment
|
$begingroup$
Haskell, 0 letters, 423 bytes = score 423
(['10'..]!!)<$>[89,87,22,91,22,100,87,22,321,87,22,108,87,22,101,22,105,87,49,0,244,87,22,343,87,22,104,87,22,98,87,22,312,87,36,0,99,87,22,87,22,102,87,22,92,87,22,93,87,22,106,87,22,259,87,49,0,228,100,22,96,87,22,95,22,90,87,22,230,87,36,0,87,104,22,285,101,22,224,100,22,234,100,22,216,100,22,107,285,101,49,0,89,87,104,244,22,106,87,321,22,100,91,321,22,91,105,22,100,91,99,36,0,91,98,101,22,89,91,100,108,101,105,36]
Try it online!
answered Apr 15 at 21:40
niminimi
34k3 gold badges27 silver badges91 bronze badges
$endgroup$
add a comment
|
$begingroup$
Haskell, 0 letters, 423 bytes = score 423
(['10'..]!!)<$>[89,87,22,91,22,100,87,22,321,87,22,108,87,22,101,22,105,87,49,0,244,87,22,343,87,22,104,87,22,98,87,22,312,87,36,0,99,87,22,87,22,102,87,22,92,87,22,93,87,22,106,87,22,259,87,49,0,228,100,22,96,87,22,95,22,90,87,22,230,87,36,0,87,104,22,285,101,22,224,100,22,234,100,22,216,100,22,107,285,101,49,0,89,87,104,244,22,106,87,321,22,100,91,321,22,91,105,22,100,91,99,36,0,91,98,101,22,89,91,100,108,101,105,36]
Try it online!
answered Apr 15 at 21:40
niminimi
34k3 gold badges27 silver badges91 bronze badges
$endgroup$
Haskell, 0 letters, 423 bytes = score 423
(['10'..]!!)<$>[89,87,22,91,22,100,87,22,321,87,22,108,87,22,101,22,105,87,49,0,244,87,22,343,87,22,104,87,22,98,87,22,312,87,36,0,99,87,22,87,22,102,87,22,92,87,22,93,87,22,106,87,22,259,87,49,0,228,100,22,96,87,22,95,22,90,87,22,230,87,36,0,87,104,22,285,101,22,224,100,22,234,100,22,216,100,22,107,285,101,49,0,89,87,104,244,22,106,87,321,22,100,91,321,22,91,105,22,100,91,99,36,0,91,98,101,22,89,91,100,108,101,105,36]
Try it online!
answered Apr 15 at 21:40
niminimi
34k3 gold badges27 silver badges91 bronze badges
answered Apr 15 at 21:40
niminimi
34k3 gold badges27 silver badges91 bronze badges
answered Apr 15 at 21:40
niminimi
34k3 gold badges27 silver badges91 bronze badges
answered Apr 15 at 21:40
niminimi
34k3 gold badges27 silver badges91 bronze badges
34k3 gold badges27 silver badges91 bronze badges
add a comment
|
add a comment
|
$begingroup$
Jelly, 274 260 212 bytes + 2 letters = 314 300 252
-48 bytes thanks to Nick Kennedy
“19ב+49;7883,8220,8216,7884Ọ“19937801,1169680277365253“38“68112“;107¤+1+“@“&%"("/%"@%"6%"0"3%$!<%" %"2%"-%"?%#!.%"%"1%")%"*%"4%"=%$!9/",%"+"'%":%#!%2">0"8/";/"7/"5>0$!&%2<"4%@"/(@"(3"/(.#!(-0"&(/603#“_32¤”;";/V
(Uses !"#$%&'()*+,-./0123456789:;<=>?@V_¤×Ọ‘“” of which V and Ọ are Unicode letters and are used once each)
Try it online!
answered Apr 15 at 22:55
Jonathan AllanJonathan Allan
60.5k5 gold badges44 silver badges188 bronze badges
$endgroup$
$begingroup$
212 bytes plus two letters
$endgroup$
– Nick Kennedy
Apr 17 at 22:58
$begingroup$
Verification
$endgroup$
– Nick Kennedy
Apr 17 at 23:23
$begingroup$
@NickKennedy I'd played around with golfing the number, but didn't step back and look to just offset the ordinals, good stuff - thanks!
$endgroup$
– Jonathan Allan
Apr 18 at 7:17
add a comment
|
$begingroup$
Jelly, 274 260 212 bytes + 2 letters = 314 300 252
-48 bytes thanks to Nick Kennedy
“19ב+49;7883,8220,8216,7884Ọ“19937801,1169680277365253“38“68112“;107¤+1+“@“&%"("/%"@%"6%"0"3%$!<%" %"2%"-%"?%#!.%"%"1%")%"*%"4%"=%$!9/",%"+"'%":%#!%2">0"8/";/"7/"5>0$!&%2<"4%@"/(@"(3"/(.#!(-0"&(/603#“_32¤”;";/V
(Uses !"#$%&'()*+,-./0123456789:;<=>?@V_¤×Ọ‘“” of which V and Ọ are Unicode letters and are used once each)
Try it online!
answered Apr 15 at 22:55
Jonathan AllanJonathan Allan
60.5k5 gold badges44 silver badges188 bronze badges
$endgroup$
$begingroup$
212 bytes plus two letters
$endgroup$
– Nick Kennedy
Apr 17 at 22:58
$begingroup$
Verification
$endgroup$
– Nick Kennedy
Apr 17 at 23:23
$begingroup$
@NickKennedy I'd played around with golfing the number, but didn't step back and look to just offset the ordinals, good stuff - thanks!
$endgroup$
– Jonathan Allan
Apr 18 at 7:17
add a comment
|
$begingroup$
Jelly, 274 260 212 bytes + 2 letters = 314 300 252
-48 bytes thanks to Nick Kennedy
“19ב+49;7883,8220,8216,7884Ọ“19937801,1169680277365253“38“68112“;107¤+1+“@“&%"("/%"@%"6%"0"3%$!<%" %"2%"-%"?%#!.%"%"1%")%"*%"4%"=%$!9/",%"+"'%":%#!%2">0"8/";/"7/"5>0$!&%2<"4%@"/(@"(3"/(.#!(-0"&(/603#“_32¤”;";/V
(Uses !"#$%&'()*+,-./0123456789:;<=>?@V_¤×Ọ‘“” of which V and Ọ are Unicode letters and are used once each)
Try it online!
answered Apr 15 at 22:55
Jonathan AllanJonathan Allan
60.5k5 gold badges44 silver badges188 bronze badges
$endgroup$
Jelly, 274 260 212 bytes + 2 letters = 314 300 252
-48 bytes thanks to Nick Kennedy
“19ב+49;7883,8220,8216,7884Ọ“19937801,1169680277365253“38“68112“;107¤+1+“@“&%"("/%"@%"6%"0"3%$!<%" %"2%"-%"?%#!.%"%"1%")%"*%"4%"=%$!9/",%"+"'%":%#!%2">0"8/";/"7/"5>0$!&%2<"4%@"/(@"(3"/(.#!(-0"&(/603#“_32¤”;";/V
(Uses !"#$%&'()*+,-./0123456789:;<=>?@V_¤×Ọ‘“” of which V and Ọ are Unicode letters and are used once each)
Try it online!
answered Apr 15 at 22:55
Jonathan AllanJonathan Allan
60.5k5 gold badges44 silver badges188 bronze badges
edited Apr 18 at 8:38
answered Apr 15 at 22:55
Jonathan AllanJonathan Allan
60.5k5 gold badges44 silver badges188 bronze badges
answered Apr 15 at 22:55
Jonathan AllanJonathan Allan
60.5k5 gold badges44 silver badges188 bronze badges
answered Apr 15 at 22:55
Jonathan AllanJonathan Allan
60.5k5 gold badges44 silver badges188 bronze badges
60.5k5 gold badges44 silver badges188 bronze badges
$begingroup$
212 bytes plus two letters
$endgroup$
– Nick Kennedy
Apr 17 at 22:58
$begingroup$
Verification
$endgroup$
– Nick Kennedy
Apr 17 at 23:23
$begingroup$
@NickKennedy I'd played around with golfing the number, but didn't step back and look to just offset the ordinals, good stuff - thanks!
$endgroup$
– Jonathan Allan
Apr 18 at 7:17
add a comment
|
$begingroup$
212 bytes plus two letters
$endgroup$
– Nick Kennedy
Apr 17 at 22:58
$begingroup$
Verification
$endgroup$
– Nick Kennedy
Apr 17 at 23:23
$begingroup$
@NickKennedy I'd played around with golfing the number, but didn't step back and look to just offset the ordinals, good stuff - thanks!
$endgroup$
– Jonathan Allan
Apr 18 at 7:17
$begingroup$
212 bytes plus two letters
$endgroup$
– Nick Kennedy
Apr 17 at 22:58
$begingroup$
212 bytes plus two letters
$endgroup$
– Nick Kennedy
Apr 17 at 22:58
$begingroup$
Verification
$endgroup$
– Nick Kennedy
Apr 17 at 23:23
$begingroup$
Verification
$endgroup$
– Nick Kennedy
Apr 17 at 23:23
$begingroup$
@NickKennedy I'd played around with golfing the number, but didn't step back and look to just offset the ordinals, good stuff - thanks!
$endgroup$
– Jonathan Allan
Apr 18 at 7:17
$begingroup$
@NickKennedy I'd played around with golfing the number, but didn't step back and look to just offset the ordinals, good stuff - thanks!
$endgroup$
– Jonathan Allan
Apr 18 at 7:17
add a comment
|
$begingroup$
PowerShell, scores 601 546
-join(67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,78,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14|%[char]($_+32))
Try it online!
Naive approach; I just took the code points and converted them to decimal, subtracted 32, then this code treats them as a char before -joining it back together into a single string.
answered Apr 15 at 20:21
AdmBorkBorkAdmBorkBork
29.2k4 gold badges74 silver badges252 bronze badges
$endgroup$
$begingroup$
901, ouch
$endgroup$
– ASCII-only
Apr 16 at 0:22
$begingroup$
686 :/
$endgroup$
– ASCII-only
Apr 16 at 0:28
add a comment
|
$begingroup$
PowerShell, scores 601 546
-join(67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,78,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14|%[char]($_+32))
Try it online!
Naive approach; I just took the code points and converted them to decimal, subtracted 32, then this code treats them as a char before -joining it back together into a single string.
answered Apr 15 at 20:21
AdmBorkBorkAdmBorkBork
29.2k4 gold badges74 silver badges252 bronze badges
$endgroup$
$begingroup$
901, ouch
$endgroup$
– ASCII-only
Apr 16 at 0:22
$begingroup$
686 :/
$endgroup$
– ASCII-only
Apr 16 at 0:28
add a comment
|
$begingroup$
PowerShell, scores 601 546
-join(67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,78,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14|%[char]($_+32))
Try it online!
Naive approach; I just took the code points and converted them to decimal, subtracted 32, then this code treats them as a char before -joining it back together into a single string.
answered Apr 15 at 20:21
AdmBorkBorkAdmBorkBork
29.2k4 gold badges74 silver badges252 bronze badges
$endgroup$
PowerShell, scores 601 546
-join(67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,78,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14|%[char]($_+32))
Try it online!
Naive approach; I just took the code points and converted them to decimal, subtracted 32, then this code treats them as a char before -joining it back together into a single string.
answered Apr 15 at 20:21
AdmBorkBorkAdmBorkBork
29.2k4 gold badges74 silver badges252 bronze badges
answered Apr 15 at 20:21
AdmBorkBorkAdmBorkBork
29.2k4 gold badges74 silver badges252 bronze badges
answered Apr 15 at 20:21
AdmBorkBorkAdmBorkBork
29.2k4 gold badges74 silver badges252 bronze badges
answered Apr 15 at 20:21
AdmBorkBorkAdmBorkBork
29.2k4 gold badges74 silver badges252 bronze badges
29.2k4 gold badges74 silver badges252 bronze badges
$begingroup$
901, ouch
$endgroup$
– ASCII-only
Apr 16 at 0:22
$begingroup$
686 :/
$endgroup$
– ASCII-only
Apr 16 at 0:28
add a comment
|
$begingroup$
901, ouch
$endgroup$
– ASCII-only
Apr 16 at 0:22
$begingroup$
686 :/
$endgroup$
– ASCII-only
Apr 16 at 0:28
$begingroup$
901, ouch
$endgroup$
– ASCII-only
Apr 16 at 0:22
$begingroup$
901, ouch
$endgroup$
– ASCII-only
Apr 16 at 0:22
$begingroup$
686 :/
$endgroup$
– ASCII-only
Apr 16 at 0:28
$begingroup$
686 :/
$endgroup$
– ASCII-only
Apr 16 at 0:28
add a comment
|
$begingroup$
Jelly, 321 bytes + 2 letters = score 361
3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304b354Ọ
Try it online!
This is hideous and someone can definitely do better.
Verify score.
answered Apr 15 at 20:35
HyperNeutrinoHyperNeutrino
20.6k4 gold badges41 silver badges155 bronze badges
$endgroup$
1
$begingroup$
actually less bad than it seems
$endgroup$
– ASCII-only
Apr 16 at 0:06
add a comment
|
$begingroup$
Jelly, 321 bytes + 2 letters = score 361
3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304b354Ọ
Try it online!
This is hideous and someone can definitely do better.
Verify score.
answered Apr 15 at 20:35
HyperNeutrinoHyperNeutrino
20.6k4 gold badges41 silver badges155 bronze badges
$endgroup$
1
$begingroup$
actually less bad than it seems
$endgroup$
– ASCII-only
Apr 16 at 0:06
add a comment
|
$begingroup$
Jelly, 321 bytes + 2 letters = score 361
3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304b354Ọ
Try it online!
This is hideous and someone can definitely do better.
Verify score.
answered Apr 15 at 20:35
HyperNeutrinoHyperNeutrino
20.6k4 gold badges41 silver badges155 bronze badges
$endgroup$
Jelly, 321 bytes + 2 letters = score 361
3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304b354Ọ
Try it online!
This is hideous and someone can definitely do better.
Verify score.
answered Apr 15 at 20:35
HyperNeutrinoHyperNeutrino
20.6k4 gold badges41 silver badges155 bronze badges
answered Apr 15 at 20:35
HyperNeutrinoHyperNeutrino
20.6k4 gold badges41 silver badges155 bronze badges
answered Apr 15 at 20:35
HyperNeutrinoHyperNeutrino
20.6k4 gold badges41 silver badges155 bronze badges
answered Apr 15 at 20:35
HyperNeutrinoHyperNeutrino
20.6k4 gold badges41 silver badges155 bronze badges
20.6k4 gold badges41 silver badges155 bronze badges
1
$begingroup$
actually less bad than it seems
$endgroup$
– ASCII-only
Apr 16 at 0:06
add a comment
|
1
$begingroup$
actually less bad than it seems
$endgroup$
– ASCII-only
Apr 16 at 0:06
1
1
$begingroup$
actually less bad than it seems
$endgroup$
– ASCII-only
Apr 16 at 0:06
$begingroup$
actually less bad than it seems
$endgroup$
– ASCII-only
Apr 16 at 0:06
add a comment
|
$begingroup$
Python 3, 380 bytes + 5 letters = 480
print("""143141 145 156141 513141 166141 157 163141;
376141 541141 162141 154141 502141.
155141 141 160141 146141 147141 164141 415141;
356156 152141 151 144141 360141.
141162 447157 352156 364156 342156 165447157;
143141162376 164141513 156145513 145163 156145155.
145154157 143145156166157163.""")
Try it online!
answered Apr 16 at 15:03
LynnLynn
52.2k9 gold badges103 silver badges238 bronze badges
$endgroup$
add a comment
|
$begingroup$
Python 3, 380 bytes + 5 letters = 480
print("""143141 145 156141 513141 166141 157 163141;
376141 541141 162141 154141 502141.
155141 141 160141 146141 147141 164141 415141;
356156 152141 151 144141 360141.
141162 447157 352156 364156 342156 165447157;
143141162376 164141513 156145513 145163 156145155.
145154157 143145156166157163.""")
Try it online!
answered Apr 16 at 15:03
LynnLynn
52.2k9 gold badges103 silver badges238 bronze badges
$endgroup$
add a comment
|
$begingroup$
Python 3, 380 bytes + 5 letters = 480
print("""143141 145 156141 513141 166141 157 163141;
376141 541141 162141 154141 502141.
155141 141 160141 146141 147141 164141 415141;
356156 152141 151 144141 360141.
141162 447157 352156 364156 342156 165447157;
143141162376 164141513 156145513 145163 156145155.
145154157 143145156166157163.""")
Try it online!
answered Apr 16 at 15:03
LynnLynn
52.2k9 gold badges103 silver badges238 bronze badges
$endgroup$
Python 3, 380 bytes + 5 letters = 480
print("""143141 145 156141 513141 166141 157 163141;
376141 541141 162141 154141 502141.
155141 141 160141 146141 147141 164141 415141;
356156 152141 151 144141 360141.
141162 447157 352156 364156 342156 165447157;
143141162376 164141513 156145513 145163 156145155.
145154157 143145156166157163.""")
Try it online!
answered Apr 16 at 15:03
LynnLynn
52.2k9 gold badges103 silver badges238 bronze badges
edited Apr 18 at 13:58
answered Apr 16 at 15:03
LynnLynn
52.2k9 gold badges103 silver badges238 bronze badges
answered Apr 16 at 15:03
LynnLynn
52.2k9 gold badges103 silver badges238 bronze badges
answered Apr 16 at 15:03
LynnLynn
52.2k9 gold badges103 silver badges238 bronze badges
52.2k9 gold badges103 silver badges238 bronze badges
add a comment
|
add a comment
|
$begingroup$
Retina, 140 characters, 159 bytes, 14 letters = score 439
%# ' 1# !# 9# 2 6#;¶þ# š# 5# /# ł#.¶0# # 3# (# )# 7# č#;¶î1 ,# + &# ð#.¶#5 ħ2 ê1 ô1 â1 8ħ2;¶%#5þ 7#! 1'! '6 1'0.¶'/2 %'1926.
T`!--/-9`ŋ`-improve this answer
answered Apr 15 at 22:06
NeilNeil
89.1k8 gold badges46 silver badges188 bronze badges
$endgroup$
add a comment
¬
Try it
answered Apr 16 at 16:56
ShaggyShaggy
21.8k3 gold badges21 silver badges73 bronze badges
edited Apr 16 at 17:27
answered Apr 16 at 16:56
ShaggyShaggy
21.8k3 gold badges21 silver badges73 bronze badges
answered Apr 16 at 16:56
ShaggyShaggy
21.8k3 gold badges21 silver badges73 bronze badges
answered Apr 16 at 16:56
ShaggyShaggy
21.8k3 gold badges21 silver badges73 bronze badges
21.8k3 gold badges21 silver badges73 bronze badges
add a comment
|
add a comment
|
$begingroup$
PHP -a, 402 bytes + 200 penalty = 602 score
foreach([67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,8,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14] as $i)echo ''.mb_chr($i+32);
Port of Artermis Fowl's answer, and my first codegolf entry!
Leaves me wishing that chr() could just support UTF-8; those extra 3 bytes + 40 characters hurts!
answered Apr 17 at 0:10
videovideo
111 bronze badge
$endgroup$
$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
Apr 17 at 12:15
add a comment
|
$begingroup$
PHP -a, 402 bytes + 200 penalty = 602 score
foreach([67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,8,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14] as $i)echo ''.mb_chr($i+32);
Port of Artermis Fowl's answer, and my first codegolf entry!
Leaves me wishing that chr() could just support UTF-8; those extra 3 bytes + 40 characters hurts!
answered Apr 17 at 0:10
videovideo
111 bronze badge
$endgroup$
$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
Apr 17 at 12:15
add a comment
|
$begingroup$
PHP -a, 402 bytes + 200 penalty = 602 score
foreach([67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,8,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14] as $i)echo ''.mb_chr($i+32);
Port of Artermis Fowl's answer, and my first codegolf entry!
Leaves me wishing that chr() could just support UTF-8; those extra 3 bytes + 40 characters hurts!
answered Apr 17 at 0:10
videovideo
111 bronze badge
$endgroup$
PHP -a, 402 bytes + 200 penalty = 602 score
foreach([67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,8,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14] as $i)echo ''.mb_chr($i+32);
Port of Artermis Fowl's answer, and my first codegolf entry!
Leaves me wishing that chr() could just support UTF-8; those extra 3 bytes + 40 characters hurts!
answered Apr 17 at 0:10
videovideo
111 bronze badge
answered Apr 17 at 0:10
videovideo
111 bronze badge
answered Apr 17 at 0:10
videovideo
111 bronze badge
answered Apr 17 at 0:10
videovideo
111 bronze badge
111 bronze badge
$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
Apr 17 at 12:15
add a comment
|
$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
Apr 17 at 12:15
$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
Apr 17 at 12:15
$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
Apr 17 at 12:15
add a comment
|
$begingroup$
05AB1E, score 209 (189 bytes + 20 penalty for 1 letter)
•£?;:'%¢;.'¡£/':¢?'¢°':¢°#@¢«>#%¡¤;®[¢:¥¢:©¢:¦¢;®¢>#¡£#¨¢#&¢+¢#,¢:§¡¤#¬¢#@¢#)¢#(¢#<¢#¢#/¡£#¯¢#.¢#>¢#±¢#«¡¤#?¢;¢#¢#°¢#:¢'¢#%•[₅‰`©®_#∞158+902201401301670804020409010150250102709022¾¡.¥>:ç?
Try it online!
The only letter is ç. The currency symbols €£¢ are not considered letters in Unicode.
answered Sep 5 at 16:14
GrimyGrimy
8,22818 silver badges37 bronze badges
$endgroup$
add a comment
|
$begingroup$
05AB1E, score 209 (189 bytes + 20 penalty for 1 letter)
•£?;:'%¢;.'¡£/':¢?'¢°':¢°#@¢«>#%¡¤;®[¢:¥¢:©¢:¦¢;®¢>#¡£#¨¢#&¢+¢#,¢:§¡¤#¬¢#@¢#)¢#(¢#<¢#¢#/¡£#¯¢#.¢#>¢#±¢#«¡¤#?¢;¢#¢#°¢#:¢'¢#%•[₅‰`©®_#∞158+902201401301670804020409010150250102709022¾¡.¥>:ç?
Try it online!
The only letter is ç. The currency symbols €£¢ are not considered letters in Unicode.
answered Sep 5 at 16:14
GrimyGrimy
8,22818 silver badges37 bronze badges
$endgroup$
add a comment
|
$begingroup$
05AB1E, score 209 (189 bytes + 20 penalty for 1 letter)
•£?;:'%¢;.'¡£/':¢?'¢°':¢°#@¢«>#%¡¤;®[¢:¥¢:©¢:¦¢;®¢>#¡£#¨¢#&¢+¢#,¢:§¡¤#¬¢#@¢#)¢#(¢#<¢#¢#/¡£#¯¢#.¢#>¢#±¢#«¡¤#?¢;¢#¢#°¢#:¢'¢#%•[₅‰`©®_#∞158+902201401301670804020409010150250102709022¾¡.¥>:ç?
Try it online!
The only letter is ç. The currency symbols €£¢ are not considered letters in Unicode.
answered Sep 5 at 16:14
GrimyGrimy
8,22818 silver badges37 bronze badges
$endgroup$
05AB1E, score 209 (189 bytes + 20 penalty for 1 letter)
•£?;:'%¢;.'¡£/':¢?'¢°':¢°#@¢«>#%¡¤;®[¢:¥¢:©¢:¦¢;®¢>#¡£#¨¢#&¢+¢#,¢:§¡¤#¬¢#@¢#)¢#(¢#<¢#¢#/¡£#¯¢#.¢#>¢#±¢#«¡¤#?¢;¢#¢#°¢#:¢'¢#%•[₅‰`©®_#∞158+902201401301670804020409010150250102709022¾¡.¥>:ç?
Try it online!
The only letter is ç. The currency symbols €£¢ are not considered letters in Unicode.
answered Sep 5 at 16:14
GrimyGrimy
8,22818 silver badges37 bronze badges
answered Sep 5 at 16:14
GrimyGrimy
8,22818 silver badges37 bronze badges
answered Sep 5 at 16:14
GrimyGrimy
8,22818 silver badges37 bronze badges
answered Sep 5 at 16:14
GrimyGrimy
8,22818 silver badges37 bronze badges
8,22818 silver badges37 bronze badges
add a comment
|
add a comment
|
$begingroup$
Python 3, 397 bytes + 19 letters = 777 score
print(''.join(chr(i+32)for i in[67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,78,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14]))
Try it online!
Port of AdmBorkBork's answer.
answered Apr 15 at 20:41
Artemis FowlArtemis Fowl
3402 silver badges13 bronze badges
$endgroup$
$begingroup$
1 less letter?
$endgroup$
– ASCII-only
Apr 16 at 12:28
1
$begingroup$
score 732
$endgroup$
– ASCII-only
Apr 16 at 12:29
$begingroup$
562, -2 if using python 2
$endgroup$
– ASCII-only
Apr 16 at 12:33
$begingroup$
TIO doesn't work at my organization, so I'll have to wait to get home to add those.
$endgroup$
– Artemis Fowl
Apr 16 at 14:17
add a comment
|
$begingroup$
Python 3, 397 bytes + 19 letters = 777 score
print(''.join(chr(i+32)for i in[67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,78,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14]))
Try it online!
Port of AdmBorkBork's answer.
answered Apr 15 at 20:41
Artemis FowlArtemis Fowl
3402 silver badges13 bronze badges
$endgroup$
$begingroup$
1 less letter?
$endgroup$
– ASCII-only
Apr 16 at 12:28
1
$begingroup$
score 732
$endgroup$
– ASCII-only
Apr 16 at 12:29
$begingroup$
562, -2 if using python 2
$endgroup$
– ASCII-only
Apr 16 at 12:33
$begingroup$
TIO doesn't work at my organization, so I'll have to wait to get home to add those.
$endgroup$
– Artemis Fowl
Apr 16 at 14:17
add a comment
|
$begingroup$
Python 3, 397 bytes + 19 letters = 777 score
print(''.join(chr(i+32)for i in[67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,78,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14]))
Try it online!
Port of AdmBorkBork's answer.
answered Apr 15 at 20:41
Artemis FowlArtemis Fowl
3402 silver badges13 bronze badges
$endgroup$
Python 3, 397 bytes + 19 letters = 777 score
print(''.join(chr(i+32)for i in[67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,78,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14]))
Try it online!
Port of AdmBorkBork's answer.
answered Apr 15 at 20:41
Artemis FowlArtemis Fowl
3402 silver badges13 bronze badges
edited Apr 15 at 20:53
answered Apr 15 at 20:41
Artemis FowlArtemis Fowl
3402 silver badges13 bronze badges
answered Apr 15 at 20:41
Artemis FowlArtemis Fowl
3402 silver badges13 bronze badges
answered Apr 15 at 20:41
Artemis FowlArtemis Fowl
3402 silver badges13 bronze badges
3402 silver badges13 bronze badges
$begingroup$
1 less letter?
$endgroup$
– ASCII-only
Apr 16 at 12:28
1
$begingroup$
score 732
$endgroup$
– ASCII-only
Apr 16 at 12:29
$begingroup$
562, -2 if using python 2
$endgroup$
– ASCII-only
Apr 16 at 12:33
$begingroup$
TIO doesn't work at my organization, so I'll have to wait to get home to add those.
$endgroup$
– Artemis Fowl
Apr 16 at 14:17
add a comment
|
$begingroup$
1 less letter?
$endgroup$
– ASCII-only
Apr 16 at 12:28
1
$begingroup$
score 732
$endgroup$
– ASCII-only
Apr 16 at 12:29
$begingroup$
562, -2 if using python 2
$endgroup$
– ASCII-only
Apr 16 at 12:33
$begingroup$
TIO doesn't work at my organization, so I'll have to wait to get home to add those.
$endgroup$
– Artemis Fowl
Apr 16 at 14:17
$begingroup$
1 less letter?
$endgroup$
– ASCII-only
Apr 16 at 12:28
$begingroup$
1 less letter?
$endgroup$
– ASCII-only
Apr 16 at 12:28
1
1
$begingroup$
score 732
$endgroup$
– ASCII-only
Apr 16 at 12:29
$begingroup$
score 732
$endgroup$
– ASCII-only
Apr 16 at 12:29
$begingroup$
562, -2 if using python 2
$endgroup$
– ASCII-only
Apr 16 at 12:33
$begingroup$
562, -2 if using python 2
$endgroup$
– ASCII-only
Apr 16 at 12:33
$begingroup$
TIO doesn't work at my organization, so I'll have to wait to get home to add those.
$endgroup$
– Artemis Fowl
Apr 16 at 14:17
$begingroup$
TIO doesn't work at my organization, so I'll have to wait to get home to add those.
$endgroup$
– Artemis Fowl
Apr 16 at 14:17
add a comment
|
$begingroup$
R, 384 bytes + 12 letters * 20 points = 684 score
Not terribly original.
cat(intToUtf8(c(67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,78,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14)+32))
Try it online!
answered Apr 17 at 5:25
CT HallCT Hall
54111 bronze badges
$endgroup$
add a comment
|
$begingroup$
R, 384 bytes + 12 letters * 20 points = 684 score
Not terribly original.
cat(intToUtf8(c(67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,78,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14)+32))
Try it online!
answered Apr 17 at 5:25
CT HallCT Hall
54111 bronze badges
$endgroup$
add a comment
|
$begingroup$
R, 384 bytes + 12 letters * 20 points = 684 score
Not terribly original.
cat(intToUtf8(c(67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,78,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14)+32))
Try it online!
answered Apr 17 at 5:25
CT HallCT Hall
54111 bronze badges
$endgroup$
R, 384 bytes + 12 letters * 20 points = 684 score
Not terribly original.
cat(intToUtf8(c(67,65,0,69,0,78,65,0,299,65,0,86,65,0,79,0,83,65,27,-22,222,65,0,321,65,0,82,65,0,76,65,0,290,65,14,-22,77,65,0,65,0,80,65,0,70,65,0,71,65,0,84,65,0,237,65,27,-22,206,78,0,74,65,0,73,0,68,65,0,208,65,14,-22,65,82,0,263,79,0,202,78,0,212,78,0,194,78,0,85,263,79,27,-22,67,65,82,222,0,84,65,299,0,78,69,299,0,69,83,0,78,69,77,14,-22,69,76,79,0,67,69,78,86,79,83,14)+32))
Try it online!
answered Apr 17 at 5:25
CT HallCT Hall
54111 bronze badges
answered Apr 17 at 5:25
CT HallCT Hall
54111 bronze badges
answered Apr 17 at 5:25
CT HallCT Hall
54111 bronze badges
answered Apr 17 at 5:25
CT HallCT Hall
54111 bronze badges
54111 bronze badges
add a comment
|
add a comment
|
$begingroup$
05AB1E, score 383 365 (325 bytes + 2 letters * 20 penalty)
3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304 354вç.««
Port of @HyperNeutrino's Jelly answer.
Will try to improve here on. The number is divisible by a bunch of numbers, but none of them would save any bytes unfortunately, and the larger divisors compressed contain at least 1 letter..
-18 (+2 bytes and -20 penalty) thanks to @Grimy, replacing the letter J (join) with .«« (reduce by concatenating).
Try it online.
answered Apr 16 at 8:11
Kevin CruijssenKevin Cruijssen
51.4k7 gold badges85 silver badges248 bronze badges
$endgroup$
1
$begingroup$
Jcan be.««for -18. Or for a completely different approach, see my answer.
$endgroup$
– Grimy
Sep 5 at 16:15
$begingroup$
@Grimy Thanks! :) And nice answer!
$endgroup$
– Kevin Cruijssen
Sep 5 at 16:52
add a comment
|
$begingroup$
05AB1E, score 383 365 (325 bytes + 2 letters * 20 penalty)
3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304 354вç.««
Port of @HyperNeutrino's Jelly answer.
Will try to improve here on. The number is divisible by a bunch of numbers, but none of them would save any bytes unfortunately, and the larger divisors compressed contain at least 1 letter..
-18 (+2 bytes and -20 penalty) thanks to @Grimy, replacing the letter J (join) with .«« (reduce by concatenating).
Try it online.
answered Apr 16 at 8:11
Kevin CruijssenKevin Cruijssen
51.4k7 gold badges85 silver badges248 bronze badges
$endgroup$
1
$begingroup$
Jcan be.««for -18. Or for a completely different approach, see my answer.
$endgroup$
– Grimy
Sep 5 at 16:15
$begingroup$
@Grimy Thanks! :) And nice answer!
$endgroup$
– Kevin Cruijssen
Sep 5 at 16:52
add a comment
|
$begingroup$
05AB1E, score 383 365 (325 bytes + 2 letters * 20 penalty)
3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304 354вç.««
Port of @HyperNeutrino's Jelly answer.
Will try to improve here on. The number is divisible by a bunch of numbers, but none of them would save any bytes unfortunately, and the larger divisors compressed contain at least 1 letter..
-18 (+2 bytes and -20 penalty) thanks to @Grimy, replacing the letter J (join) with .«« (reduce by concatenating).
Try it online.
answered Apr 16 at 8:11
Kevin CruijssenKevin Cruijssen
51.4k7 gold badges85 silver badges248 bronze badges
$endgroup$
05AB1E, score 383 365 (325 bytes + 2 letters * 20 penalty)
3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304 354вç.««
Port of @HyperNeutrino's Jelly answer.
Will try to improve here on. The number is divisible by a bunch of numbers, but none of them would save any bytes unfortunately, and the larger divisors compressed contain at least 1 letter..
-18 (+2 bytes and -20 penalty) thanks to @Grimy, replacing the letter J (join) with .«« (reduce by concatenating).
Try it online.
answered Apr 16 at 8:11
Kevin CruijssenKevin Cruijssen
51.4k7 gold badges85 silver badges248 bronze badges
edited Sep 5 at 16:51
answered Apr 16 at 8:11
Kevin CruijssenKevin Cruijssen
51.4k7 gold badges85 silver badges248 bronze badges
answered Apr 16 at 8:11
Kevin CruijssenKevin Cruijssen
51.4k7 gold badges85 silver badges248 bronze badges
answered Apr 16 at 8:11
Kevin CruijssenKevin Cruijssen
51.4k7 gold badges85 silver badges248 bronze badges
51.4k7 gold badges85 silver badges248 bronze badges
1
$begingroup$
Jcan be.««for -18. Or for a completely different approach, see my answer.
$endgroup$
– Grimy
Sep 5 at 16:15
$begingroup$
@Grimy Thanks! :) And nice answer!
$endgroup$
– Kevin Cruijssen
Sep 5 at 16:52
add a comment
|
1
$begingroup$
Jcan be.««for -18. Or for a completely different approach, see my answer.
$endgroup$
– Grimy
Sep 5 at 16:15
$begingroup$
@Grimy Thanks! :) And nice answer!
$endgroup$
– Kevin Cruijssen
Sep 5 at 16:52
1
1
$begingroup$
J can be .«« for -18. Or for a completely different approach, see my answer.$endgroup$
– Grimy
Sep 5 at 16:15
$begingroup$
J can be .«« for -18. Or for a completely different approach, see my answer.$endgroup$
– Grimy
Sep 5 at 16:15
$begingroup$
@Grimy Thanks! :) And nice answer!
$endgroup$
– Kevin Cruijssen
Sep 5 at 16:52
$begingroup$
@Grimy Thanks! :) And nice answer!
$endgroup$
– Kevin Cruijssen
Sep 5 at 16:52
add a comment
|
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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$begingroup$
kolmogorov-complexity, restricted-source, and special scoring are all sorts of things that benefit greatly from careful consideration in the sandbox. Currently, it seems like the best approach to this challenge would be to just write out all of the codepoints in decimal then turn them into text with a builtin, with some shortcut to encode all of the
as--or not, depending on how many letters it would take, because 20 characters is a really big penalty (although when everything else is scored by bytes, it's not quite well defined...)!$endgroup$
– Unrelated String
Apr 15 at 20:09
4
$begingroup$
And considering the invocation of Unicode, some explicit rules governing special codepages as used by most golflangs are probably called for (alongside maybe a link to a script to validate scoring).
$endgroup$
– Unrelated String
Apr 15 at 21:00