Otdfbt

Question

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

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...)! — Apr 15 at 20:09
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). — Apr 15 at 21:00

score 19 · Accepted Answer · 2019-04-16 02:02:18Z

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 47463 empty a (code to run: 7463)

… d c b 47463 empty a empty (code to run: 463)

… d c b 47463 empty empty empty a (code to run: 63)

… d c b 47463 empty empty "a" (code to run: 3)

… d c b 47463 empty (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.

niminimi 32.9k32489 · Accepted Answer · 2019-04-15 21:40:17Z

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!

score 6 · Accepted Answer · 2019-04-18 08:38:57Z

Jelly, _{^{274 260 212 bytes + 2 letters}} = 314 300 252

-48 bytes thanks to Nick Kennedy

“19×‘+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!

@NickKennedy I'd played around with golfing the number, but didn't step back and look to just offset the ordinals, good stuff - thanks! — Apr 18 at 7:17

AdmBorkBorkAdmBorkBork 28k468241 · Accepted Answer · 2019-04-15 20:21:38Z

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.

$begingroup$
901, ouch
$endgroup$
– ASCII-only
Apr 16 at 0:22 — Apr 16 at 0:22
$begingroup$
686 :/
$endgroup$
– ASCII-only
Apr 16 at 0:28 — Apr 16 at 0:28

HyperNeutrinoHyperNeutrino 19k437148 · Accepted Answer · 2019-04-15 20:35:51Z

Jelly, 321 bytes + 2 letters = score 361

3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304b354Ọ

Try it online!

This is hideous and someone can definitely do better.

Verify score.

$begingroup$
actually less bad than it seems
$endgroup$
– ASCII-only
Apr 16 at 0:06 — Apr 16 at 0:06

score 2 · Accepted Answer · 2019-04-18 13:58:41Z

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!

score 1 · Accepted Answer · 2019-04-16 17:27:52Z

1

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).

edited Apr 16 at 17:27

answered Apr 16 at 16:56

Shaggy

19.1k21768

add a comment |

1

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

video

111

New contributor

$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
Apr 17 at 12:15

add a comment |

0

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.

edited Apr 15 at 20:53

answered Apr 15 at 20:41

Artemis Fowl

27112

$begingroup$
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 |

0

05AB1E, score 383 (323 bytes + 3 letters * 20 penalty)

3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304 354вçJ

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..

Try it online.

edited Apr 16 at 9:23

answered Apr 16 at 8:11

Kevin Cruijssen

43.4k573222

add a comment |

0

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 Hall

51111

add a comment |

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19

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 47463 empty a (code to run: 7463)

… d c b 47463 empty a empty (code to run: 463)

… d c b 47463 empty empty empty a (code to run: 63)

… d c b 47463 empty empty "a" (code to run: 3)

… d c b 47463 empty (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

add a comment |

19

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 47463 empty a (code to run: 7463)

… d c b 47463 empty a empty (code to run: 463)

… d c b 47463 empty empty empty a (code to run: 63)

… d c b 47463 empty empty "a" (code to run: 3)

… d c b 47463 empty (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

add a comment |

19

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 47463 empty a (code to run: 7463)

… d c b 47463 empty a empty (code to run: 463)

… d c b 47463 empty empty empty a (code to run: 63)

… d c b 47463 empty empty "a" (code to run: 3)

… d c b 47463 empty (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

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 47463 empty a (code to run: 7463)

… d c b 47463 empty a empty (code to run: 463)

… d c b 47463 empty empty empty a (code to run: 63)

… d c b 47463 empty empty "a" (code to run: 3)

… d c b 47463 empty (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

answered Apr 16 at 2:02

community wiki

ais523

community wiki

ais523

community wiki

ais523

add a comment |

9

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

nimi

32.9k32489

add a comment |

9

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

nimi

32.9k32489

add a comment |

9

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

nimi

32.9k32489

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

nimi

32.9k32489

answered Apr 15 at 21:40

nimi

32.9k32489

answered Apr 15 at 21:40

nimi

32.9k32489

answered Apr 15 at 21:40

nimi

32.9k32489

add a comment |

6

Jelly, _{^{274 260 212 bytes + 2 letters}} = 314 300 252

-48 bytes thanks to Nick Kennedy

“19×‘+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!

edited Apr 18 at 8:38

answered Apr 15 at 22:55

Jonathan Allan

54.8k537175

$begingroup$
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 |

6

Jelly, _{^{274 260 212 bytes + 2 letters}} = 314 300 252

-48 bytes thanks to Nick Kennedy

“19×‘+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!

edited Apr 18 at 8:38

answered Apr 15 at 22:55

Jonathan Allan

54.8k537175

$begingroup$
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 |

6

Jelly, _{^{274 260 212 bytes + 2 letters}} = 314 300 252

-48 bytes thanks to Nick Kennedy

“19×‘+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!

edited Apr 18 at 8:38

answered Apr 15 at 22:55

Jonathan Allan

54.8k537175

Jelly, _{^{274 260 212 bytes + 2 letters}} = 314 300 252

-48 bytes thanks to Nick Kennedy

“19×‘+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!

edited Apr 18 at 8:38

answered Apr 15 at 22:55

Jonathan Allan

54.8k537175

edited Apr 18 at 8:38

answered Apr 15 at 22:55

Jonathan Allan

54.8k537175

answered Apr 15 at 22:55

Jonathan Allan

54.8k537175

answered Apr 15 at 22:55

Jonathan Allan

54.8k537175

$begingroup$
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

212 bytes plus two letters

– Nick Kennedy
Apr 17 at 22:58

Verification

– Nick Kennedy
Apr 17 at 23:23

@NickKennedy I'd played around with golfing the number, but didn't step back and look to just offset the ordinals, good stuff - thanks!

– Jonathan Allan
Apr 18 at 7:17

add a comment |

3

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

AdmBorkBork

28k468241

$begingroup$
901, ouch
$endgroup$
– ASCII-only
Apr 16 at 0:22

$begingroup$
686 :/
$endgroup$
– ASCII-only
Apr 16 at 0:28

add a comment |

3

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

AdmBorkBork

28k468241

$begingroup$
901, ouch
$endgroup$
– ASCII-only
Apr 16 at 0:22

$begingroup$
686 :/
$endgroup$
– ASCII-only
Apr 16 at 0:28

add a comment |

3

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

AdmBorkBork

28k468241

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

AdmBorkBork

28k468241

answered Apr 15 at 20:21

AdmBorkBork

28k468241

answered Apr 15 at 20:21

AdmBorkBork

28k468241

answered Apr 15 at 20:21

AdmBorkBork

28k468241

$begingroup$
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

901, ouch

– ASCII-only
Apr 16 at 0:22

686 :/

– ASCII-only
Apr 16 at 0:28

add a comment |

3

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

HyperNeutrino

19k437148

1

$begingroup$
actually less bad than it seems
$endgroup$
– ASCII-only
Apr 16 at 0:06

add a comment |

3

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

HyperNeutrino

19k437148

1

$begingroup$
actually less bad than it seems
$endgroup$
– ASCII-only
Apr 16 at 0:06

add a comment |

3

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

HyperNeutrino

19k437148

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

HyperNeutrino

19k437148

answered Apr 15 at 20:35

HyperNeutrino

19k437148

answered Apr 15 at 20:35

HyperNeutrino

19k437148

answered Apr 15 at 20:35

HyperNeutrino

19k437148

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

actually less bad than it seems

– ASCII-only
Apr 16 at 0:06

add a comment |

2

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!

edited Apr 18 at 13:58

answered Apr 16 at 15:03

Lynn

51.1k899234

add a comment |

2

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!

edited Apr 18 at 13:58

answered Apr 16 at 15:03

Lynn

51.1k899234

add a comment |

2

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!

edited Apr 18 at 13:58

answered Apr 16 at 15:03

Lynn

51.1k899234

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!

edited Apr 18 at 13:58

answered Apr 16 at 15:03

Lynn

51.1k899234

edited Apr 18 at 13:58

answered Apr 16 at 15:03

Lynn

51.1k899234

answered Apr 16 at 15:03

Lynn

51.1k899234

answered Apr 16 at 15:03

Lynn

51.1k899234

add a comment |

1

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

edited Apr 16 at 17:27

answered Apr 16 at 16:56

Shaggy

19.1k21768

Japt `-S`, 304 286 bytes + 2 1 letters = 344 306

Well, this is just god-awful!

"3 1
5
14 1
235 1
22 1
15
19 1 -37 -86 158 1
257 1
18 1
12 1
226 1 -50 -86 13 1
1
16 1
6 1
7 1
20 1
173 1 -37 -86 142 14
10 1
9
4 1
144 1 -50 -86 1 18
199 15
138 14
148 14
130 14
21 199 15 -37 -86 3 1 18 158
20 1 235
14 5 235
5 19
14 5 13 -50 -86 5 12 15
3 5 14 22 15 19 -50"·®¸®°d96 ¬

Try it

edited Apr 16 at 17:27

answered Apr 16 at 16:56

Shaggy

19.1k21768

edited Apr 16 at 17:27

answered Apr 16 at 16:56

Shaggy

19.1k21768

answered Apr 16 at 16:56

Shaggy

19.1k21768

answered Apr 16 at 16:56

Shaggy

19.1k21768

add a comment |

1

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

video

111

New contributor

$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
Apr 17 at 12:15

add a comment |

1

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

video

111

New contributor

$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
Apr 17 at 12:15

add a comment |

1

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

video

111

New contributor

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

video

111

New contributor

answered Apr 17 at 0:10

video

111

New contributor

answered Apr 17 at 0:10

video

111

answered Apr 17 at 0:10

video

111

New contributor

video is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
Apr 17 at 12:15

add a comment |

$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
Apr 17 at 12:15

Welcome to PPCG :)

– Shaggy
Apr 17 at 12:15

add a comment |

0

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.

edited Apr 15 at 20:53

answered Apr 15 at 20:41

Artemis Fowl

27112

$begingroup$
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 |

0

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.

edited Apr 15 at 20:53

answered Apr 15 at 20:41

Artemis Fowl

27112

$begingroup$
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 |

0

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.

edited Apr 15 at 20:53

answered Apr 15 at 20:41

Artemis Fowl

27112

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.

edited Apr 15 at 20:53

answered Apr 15 at 20:41

Artemis Fowl

27112

edited Apr 15 at 20:53

answered Apr 15 at 20:41

Artemis Fowl

27112

answered Apr 15 at 20:41

Artemis Fowl

27112

answered Apr 15 at 20:41

Artemis Fowl

27112

$begingroup$
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

1 less letter?

– ASCII-only
Apr 16 at 12:28

1

score 732

– ASCII-only
Apr 16 at 12:29

562, -2 if using python 2

– ASCII-only
Apr 16 at 12:33

TIO doesn't work at my organization, so I'll have to wait to get home to add those.

– Artemis Fowl
Apr 16 at 14:17

add a comment |

0

05AB1E, score 383 (323 bytes + 3 letters * 20 penalty)

3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304 354вçJ

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..

Try it online.

edited Apr 16 at 9:23

answered Apr 16 at 8:11

Kevin Cruijssen

43.4k573222

add a comment |

0

05AB1E, score 383 (323 bytes + 3 letters * 20 penalty)

3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304 354вçJ

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..

Try it online.

edited Apr 16 at 9:23

answered Apr 16 at 8:11

Kevin Cruijssen

43.4k573222

add a comment |

0

05AB1E, score 383 (323 bytes + 3 letters * 20 penalty)

3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304 354вçJ

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..

Try it online.

edited Apr 16 at 9:23

answered Apr 16 at 8:11

Kevin Cruijssen

43.4k573222

05AB1E, score 383 (323 bytes + 3 letters * 20 penalty)

3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304 354вçJ

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..

Try it online.

edited Apr 16 at 9:23

answered Apr 16 at 8:11

Kevin Cruijssen

43.4k573222

edited Apr 16 at 9:23

answered Apr 16 at 8:11

Kevin Cruijssen

43.4k573222

answered Apr 16 at 8:11

Kevin Cruijssen

43.4k573222

answered Apr 16 at 8:11

Kevin Cruijssen

43.4k573222

add a comment |

0

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 Hall

51111

add a comment |

0

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 Hall

51111

add a comment |

0

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 Hall

51111

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 Hall

51111

answered Apr 17 at 5:25

CT Hall

51111

answered Apr 17 at 5:25

CT Hall

51111

answered Apr 17 at 5:25

CT Hall

51111

add a comment |

If this is an answer to a challenge…

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…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.

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More generally…

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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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videovideo 111 · Accepted Answer · 2019-04-17 00:10:25Z

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!

$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
Apr 17 at 12:15 — Apr 17 at 12:15

score 0 · Accepted Answer · 2019-04-15 20:53:19Z

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.

TIO doesn't work at my organization, so I'll have to wait to get home to add those. — Apr 16 at 14:17

score 0 · Accepted Answer · 2019-04-16 09:23:36Z

05AB1E, score 383 (323 bytes + 3 letters * 20 penalty)

3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304 354вçJ

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..

Try it online.

CT HallCT Hall 51111 · Accepted Answer · 2019-04-17 05:25:55Z

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!

score 19 · Accepted Answer · 2019-04-16 02:02:18Z

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 47463 empty a (code to run: 7463)

… d c b 47463 empty a empty (code to run: 463)

… d c b 47463 empty empty empty a (code to run: 63)

… d c b 47463 empty empty "a" (code to run: 3)

… d c b 47463 empty (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.

niminimi 32.9k32489 · Accepted Answer · 2019-04-15 21:40:17Z

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!

score 6 · Accepted Answer · 2019-04-18 08:38:57Z

Jelly, _{^{274 260 212 bytes + 2 letters}} = 314 300 252

-48 bytes thanks to Nick Kennedy

“19×‘+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!

@NickKennedy I'd played around with golfing the number, but didn't step back and look to just offset the ordinals, good stuff - thanks! — Apr 18 at 7:17

AdmBorkBorkAdmBorkBork 28k468241 · Accepted Answer · 2019-04-15 20:21:38Z

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.

$begingroup$
901, ouch
$endgroup$
– ASCII-only
Apr 16 at 0:22 — Apr 16 at 0:22
$begingroup$
686 :/
$endgroup$
– ASCII-only
Apr 16 at 0:28 — Apr 16 at 0:28

HyperNeutrinoHyperNeutrino 19k437148 · Accepted Answer · 2019-04-15 20:35:51Z

Jelly, 321 bytes + 2 letters = score 361

3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304b354Ọ

Try it online!

This is hideous and someone can definitely do better.

Verify score.

$begingroup$
actually less bad than it seems
$endgroup$
– ASCII-only
Apr 16 at 0:06 — Apr 16 at 0:06

score 2 · Accepted Answer · 2019-04-18 13:58:41Z

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!

score 1 · Accepted Answer · 2019-04-16 17:27:52Z

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

videovideo 111 · Accepted Answer · 2019-04-17 00:10:25Z

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!

$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
Apr 17 at 12:15 — Apr 17 at 12:15

score 0 · Accepted Answer · 2019-04-15 20:53:19Z

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.

TIO doesn't work at my organization, so I'll have to wait to get home to add those. — Apr 16 at 14:17

score 0 · Accepted Answer · 2019-04-16 09:23:36Z

05AB1E, score 383 (323 bytes + 3 letters * 20 penalty)

3343781777797791350694255572961968519437585132057650209974147122192542459108221624793330048943528237823681411832154316740173721249435700067706302064570847610741421342406380917446310820012503592770000532190167243585300911078873144513786923305473352724133578818457026824110152529235136461572588027747840738399150398304 354вçJ

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..

Try it online.

CT HallCT Hall 51111 · Accepted Answer · 2019-04-17 05:25:55Z

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!

12 Answers 12

7, 410 characters, 154 bytes in 7's encoding, 0 letters = score 154

Explanation

Haskell, 0 letters, 423 bytes = score 423

Jelly, 274 260 212 bytes + 2 letters = 314 300 252

PowerShell, scores 601 546

Jelly, 321 bytes + 2 letters = score 361

Python 3, 380 bytes + 5 letters = 480

Retina, 140 characters, 159 bytes, 14 letters = score 439

PHP -a, 402 bytes + 200 penalty = 602 score

Python 3, 397 bytes + 19 letters = 777 score

05AB1E, score 383 (323 bytes + 3 letters * 20 penalty)

R, 384 bytes + 12 letters * 20 points = 684 score

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12 Answers 12

12 Answers 12

7, 410 characters, 154 bytes in 7's encoding, 0 letters = score 154

Explanation

7, 410 characters, 154 bytes in 7's encoding, 0 letters = score 154

Explanation

7, 410 characters, 154 bytes in 7's encoding, 0 letters = score 154

Explanation

7, 410 characters, 154 bytes in 7's encoding, 0 letters = score 154

Explanation

Haskell, 0 letters, 423 bytes = score 423

Haskell, 0 letters, 423 bytes = score 423

Haskell, 0 letters, 423 bytes = score 423

Haskell, 0 letters, 423 bytes = score 423

Jelly, 274 260 212 bytes + 2 letters = 314 300 252

Jelly, 274 260 212 bytes + 2 letters = 314 300 252

Jelly, 274 260 212 bytes + 2 letters = 314 300 252

Jelly, 274 260 212 bytes + 2 letters = 314 300 252

PowerShell, scores 601 546

PowerShell, scores 601 546

PowerShell, scores 601 546

PowerShell, scores 601 546

Jelly, 321 bytes + 2 letters = score 361

Jelly, 321 bytes + 2 letters = score 361

Jelly, 321 bytes + 2 letters = score 361

Jelly, 321 bytes + 2 letters = score 361

Python 3, 380 bytes + 5 letters = 480

Python 3, 380 bytes + 5 letters = 480

Python 3, 380 bytes + 5 letters = 480

Python 3, 380 bytes + 5 letters = 480

Retina, 140 characters, 159 bytes, 14 letters = score 439

Japt -S, 304 286 bytes + 2 1 letters = 344 306

PHP -a, 402 bytes + 200 penalty = 602 score

PHP -a, 402 bytes + 200 penalty = 602 score

PHP -a, 402 bytes + 200 penalty = 602 score

PHP -a, 402 bytes + 200 penalty = 602 score

Python 3, 397 bytes + 19 letters = 777 score

Python 3, 397 bytes + 19 letters = 777 score

Python 3, 397 bytes + 19 letters = 777 score

Python 3, 397 bytes + 19 letters = 777 score

05AB1E, score 383 (323 bytes + 3 letters * 20 penalty)

05AB1E, score 383 (323 bytes + 3 letters * 20 penalty)

05AB1E, score 383 (323 bytes + 3 letters * 20 penalty)

05AB1E, score 383 (323 bytes + 3 letters * 20 penalty)

R, 384 bytes + 12 letters * 20 points = 684 score

R, 384 bytes + 12 letters * 20 points = 684 score

R, 384 bytes + 12 letters * 20 points = 684 score

R, 384 bytes + 12 letters * 20 points = 684 score

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12 Answers
12

Jelly, _{^{274 260 212 bytes + 2 letters}} = 314 300 252

12 Answers
12

12 Answers
12

Jelly, _{^{274 260 212 bytes + 2 letters}} = 314 300 252

Jelly, _{^{274 260 212 bytes + 2 letters}} = 314 300 252

Jelly, _{^{274 260 212 bytes + 2 letters}} = 314 300 252

Jelly, _{^{274 260 212 bytes + 2 letters}} = 314 300 252

Japt `-S`, 304 286 bytes + 2 1 letters = 344 306