43
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Create a program that outputs Hello, World!

The winner is the source code that is the biggest, subject to the constraint that it is irreducible

For this challenge, irreducible means there does not exist a program that outputs Hello, World! that can be constructed by removing characters from the original program.

For example

console.log(`Hello, ` + `World!`);

can have the the bracketed characters removed

console.log(`Hello, [` + `]World!`)[;] -> console.log(`Hello, World!`)

And it will still output Hello, World, so it is not irreducible.

Extra Rules:

  • Whitespace is ignored with regards to both irreducibility and size.
  • Unary or anything that operates in a similar fashion (I'll make this more clear if I have to) is banned.
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  • 21
    \$\begingroup\$ Would code-bowling be an appropriate tag? \$\endgroup\$ – Quintec Dec 10 '19 at 3:30
  • 7
    \$\begingroup\$ Why Unary is banned? This question would be code-golf in Unary like languages. And it is almost impossible to post an answer in such languages: Since it is hard to prove the answer fit the requirement. \$\endgroup\$ – tsh Dec 10 '19 at 6:28
  • 7
    \$\begingroup\$ I feel like the post should at least ask for a detailed explanation of why the code can have no sub-sequence which would output the string. \$\endgroup\$ – Jonathan Allan Dec 10 '19 at 9:05
  • 4
    \$\begingroup\$ I'm impressed. A code-bowling that actually works and doesn't need to impose restrictions like "number of different ascii bytes" (that inherently have a maximum score). \$\endgroup\$ – Draco18s no longer trusts SE Dec 10 '19 at 22:56
  • 2
    \$\begingroup\$ And now this is the first code-bowling where the winning score involves double-arrow notation. \$\endgroup\$ – Bubbler Dec 12 '19 at 3:51

15 Answers 15

34
\$\begingroup\$

Keg, \$3.3 * 10^{8912898}\$ bytes

...÷⅍+ℤ,(􏿿|⅍+)ℤ,(􏿿|⅍+)ℤ,(􏿿|⅍+)ℤ,(􏿿|⅍+)ℤ,⅍+ℤ,⅍+ℤ,⅍+ℤ,(􏿿|⅍+)ℤ,(􏿿|⅍+)ℤ,(􏿿|⅍+)ℤ,(􏿿|⅍+)ℤ,⅍+ℤ,

Program as seen in the raw editor enter image description here

So where that ... is, replace it with \$3.3 * 10^{8912898}\$ 0s. Then, add a ! after those 0s. This splits the number into chunks, divides these chunks by 􏿿 and outputs them.

I owe a special thanks to @JoKing for helping me generate a truly irreducible score... it took me a while to understand it, but it got there in the end.

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  • 7
    \$\begingroup\$ Congratulations, you're the first one to beat Unary (which, based on the best golfed Brainfxxk answer, would score around 1e65). \$\endgroup\$ – Bubbler Dec 10 '19 at 7:58
  • 1
    \$\begingroup\$ This program can be reduced in length by over 100 orders of magnitude. \$\endgroup\$ – A. Rex Dec 10 '19 at 16:52
  • 3
    \$\begingroup\$ @A.Rex That number at the start cannot be formed just by removing characters in the program \$\endgroup\$ – Embodiment of Ignorance Dec 10 '19 at 17:56
  • 14
    \$\begingroup\$ An explanation of how this is irreducible would be appreciated \$\endgroup\$ – Jo King Dec 10 '19 at 22:37
  • 3
    \$\begingroup\$ What programming language is the generator even written in? And why can this not be reduced? \$\endgroup\$ – Fabian Röling Dec 11 '19 at 14:25
28
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APL (Dyalog Unicode), \$ \approx (10 \uparrow)^{625} 10.024549 \gg 10 \uparrow \uparrow 626 \$ bytes

⎕UCS(13⌿*⌿2+|⊃⎕RL⊣⎕RL←⍪⍨1)⊤≢⍬⍬⍬...

The exact byte count is

$$ \text{byte count before zilde} + \text{byte count of single zilde} \times \text{length of zilde chain} \\ = 52 + 3 \times (72x^{12} + 101x^{11} + 108x^{10} + 108x^9 + \\ 111x^8 + 44x^7 + 32x^6 + 87x^5 + 111x^4 + 114x^3 + 108x^2 + 100x + 33) $$ where $$ x = \text{power tower of (} 2 + \text{abs(internal state of Mersenne twister with seed 1))} $$

At the end of the power tower is $$ 1338723255 \uparrow (1025549142 \uparrow (1187976951 \uparrow (683638953 \uparrow 1197707996))) \\ \approx 10^{10^{10^{10^{10^{10.02454898892586}}}}} = (10 \uparrow)^5 10.02454898892586 $$

How it works

⎕UCS(13⌿*⌿2+|⊃⎕RL⊣⎕RL←⍪⍨1)⊤≢⍬⍬⍬...
                      ⍪⍨1  ⍝ 1 1
                  ⎕RL←     ⍝ Set RNG to Mersenne twister with seed 1
             ⊃⎕RL⊣         ⍝ Query the internal state,
                           ⍝   a vector of 625 integers starting with 0 0 1
          2+|              ⍝ 2+abs(above)
        *⌿                 ⍝ RTL reduce with exponentiation
     13⌿                   ⍝ 13 copies of it

⎕UCS(13⌿ .. )⊤≢⍬⍬⍬... part is classic. Without going through ⎕RL, the highest single number you can get is *⌿2+⍪⍨1, which is 3*3 or 27. The only other system function that can be made from ⎕RL is ⎕R, which doesn't work with numbers (it's PCRE replace function).


Previous answers

APL (Dyalog Unicode), ≈ \$ 10^{10^{10^{10^{10^{10^{10^{10^{8.56784}}}}}}}} \$ bytes

⎕UCS(13⌿*⌿9⌿9)⊤≢⍬⍬⍬...

The exact byte count is

$$ \text{byte count before zilde} + \text{byte count of single zilde} \times \text{length of zilde chain} \\ = 28 + 3 \times (72x^{12} + 101x^{11} + 108x^{10} + 108x^9 + \\ 111x^8 + 44x^7 + 32x^6 + 87x^5 + 111x^4 + 114x^3 + 108x^2 + 100x + 33) $$ where $$ x = 9^{9^{9^{9^{9^{9^{9^{9^{9}}}}}}}} $$

13⌿*⌿9⌿9 does this: make 9 copies of 9, reduce it by power (RTL), and again make 13 copies of it. Since we need 13 digits (to convert to 13-character string), we can't touch 13⌿. The only operator that contributes to making a number higher than 114 is * (power), so we can't remove it either. If any char is removed from 9⌿9, we simply get a 9 or 99. If of *⌿ is removed, the * becomes "power of e" instead, so ⎕UCS will error out.


APL (Dyalog Unicode), ≈ 1.6198e1175 bytes

⎕UCS≢¨⊆⍨(1129⍴11)⊤≢⍬⍬⍬...

No TIO this time. The trailing section contains exactly 53993507850758146386142744333468571550524632618191311756757485604204265568989694819322874649587028378175296725206816716551117407353231278417177118257863382844631479973422921647480947817133933420316923525318614797726155835511430816083805415403840314027172595512518082931013938879825791513465541135540380972893540597585182649169086261680393824120050960955555779401997457521829600791475089390923918230965201076266924917726191938715916934581792969501864553539221144133166874835664827114498421480824828796840502396309900816750358227125625773629167173751259375816861525751727161646360849365621299142542955252895691416097697138666219891143191120795244088895226859387434190840571233858421367397012295189969102957711871947102112735851782580022907564984257273029144801079437982463987025764736538779486298171952179565363242464045693851301934849205024515654449551352138803030923347653267563857979066215774822906142655781986485757538363542188567283809498107727208952982783066735317822771494040957372930876073109323855643561468444882121936718611178126143503679181305866975597866998548105699612159571516074201535860599366981549403015140492393925783660483338575307748583670119709554797435941 zilde symbols. Still no whitespace.

How it works

⎕UCS≢¨⊆⍨(1129⍴11)⊤≢⍬⍬⍬...
                  ≢  ⍝ Length
        (1129⍴11)⊤   ⍝ Convert to 1129 base-11 digits
                     ⍝ which gives 1 1 1 1 .. 2 1 1 1 .. 2 1 1 1 ..
      ⊆⍨             ⍝ Chunkify at 2's as starting position,
                     ⍝ including the leading chunk of ones
    ≢¨               ⍝ Length of each chunk
⎕UCS                 ⍝ Convert codepoints to string

The "chunked unary" encoding requires at least 1129 digits, and can't contain zero digits in order to be exactly 1129 digits. The base 11 was carefully chosen to avoid cracks using lower bases.


APL (Dyalog Unicode), 1169882994796837103436970021 ≈ 1.17e27 bytes

⎕UCS(13⍴115)⊤≢⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬⍬...

Try it online!

The trailing section contains exactly 389960998265612367812323333 zilde symbols. There's no whitespace.

How it works

⎕UCS converts from Unicode codepoints to characters. You don't get any string without it.

does the only numerical calculation in this program. What it does is, take a single number on its right and convert into the base on its left. Since we want an array of length 13, length-13 array is required on its left side. Since the largest digit value we will need is 114, I chose base 115 to tightly fit the requirements of left side of .

The right side of is ≢⍬⍬⍬⍬⍬⍬..., which computes the length of the array containing exactly 389960998265612367812323333 s. is an empty numeric vector, so you can't get any meaningful number from it, other than the length of the vector.

Finally, a comment about the byte count. In challenges, we use SBCS to count the APL symbols as one byte each. But the default source encoding is actually UTF-8, making APL symbols take 2 or 3 bytes.

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  • 1
    \$\begingroup\$ This is the first answer I've seen on here with a score measured with up-arrow notation. \$\endgroup\$ – Skyler Dec 12 '19 at 20:06
  • 1
    \$\begingroup\$ @Skyler We once had this challenge to generate big numbers in limited code size. \$\endgroup\$ – Bubbler Dec 12 '19 at 23:21
22
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Python 3, 44619 8.13*1011515 ~\$10^{10^{5749}}\$ bytes

The program itself is rather too large to fit in the observable universe on Stack Overflow, so this is a link to a program that can generate it... if you change the rather large number generator to a ~9 instead.

Try it online!

This can certainly be longer, but I'm going to post it now since I'm not 100% sure this is irreducible. It is basically made up of several irreducible components stacked together.

This relies mostly on a very large number n, generated by the state of the Mersenne twister with the seed 8 reduced by multiplication exponentation, i.e ~functools.reduce(pow,b), where b is taken from:

import functools
from random import *
seed(8)
d=getstate()
a,b,c=d

First, it execs the reverse of a very long string, separated by n waste bytes in the form:

exec("a...b...c...d"[::~n])

You can't remove the .s, unless you also remove the [::~n] reversing the string. If you remove the .s and the [::~n] then the resulting program won't really work since it is backwards, and the only print part is at the end, so no output without that.

The next portion looks like:

print(''.join([chr(len(...)), n ''s, chr(len(...)))][::~n]))

Similarly, you can't remove the ''s without also taking out the [::~n]. You also can't reduce chr(len([0]+[0]+...+[0])) and still get the character you want, nor can I find anyway of modifying it with the ''s to still get the character as well as another list item.

Please point any mistakes I have made or any ways you can think of to reduce this.

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  • 2
    \$\begingroup\$ What about -9 -> ~9? Or mabye -0xf? \$\endgroup\$ – tsh Dec 10 '19 at 6:17
  • \$\begingroup\$ @tsh Huh, good ideas. Couldn't it be made pretty much arbitrary with -ord('.')? Up to the max size of strings \$\endgroup\$ – Jo King Dec 10 '19 at 6:34
  • \$\begingroup\$ Good idea... So, using ~ord('􏿿') instead would made this program 111411x large? \$\endgroup\$ – tsh Dec 10 '19 at 7:35
  • 2
    \$\begingroup\$ This looks easy to considerably reduce, unless I missed something. \$\endgroup\$ – Grimmy Dec 10 '19 at 14:42
  • 1
    \$\begingroup\$ @Grimy Oops, how about functools.reduce(int.__mul__,b)? \$\endgroup\$ – Jo King Dec 10 '19 at 16:32
17
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Malbolge, 265 bytes

Malbolge is too concise to win this competition.

('BA@?>=<;:9876543210/.-,+*)('&%$#"!~}|{zyxwvutsrqponmlkjihgfedcba`_^A\[ZYXWVUTSRQPONMLKJIHGFEDCBA@?>=<;:9876543210/.-,+*)('&%$#"!~}|{zyxwvutsrqponmlkjihgfedcba`_^]\[ZYXWVUTSRQPONMLKJIHGFEDC<$#9]=<;:9y1U/432+ONo'&+$Hi!~}$#z@xw|uzs9Zpunsrk1RQgfe+iba'Hdcb[!_XW{[TSRvc

Try it online!

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  • 11
    \$\begingroup\$ +1 for simply using Malbolge in an actual programming competition. \$\endgroup\$ – COTO Dec 10 '19 at 22:00
13
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JavaScript, 9007199254741154 bytes

n=0
a=Number.MAX_SAFE_INTEGER-1
b="Hlo, world![...]e"
console.log(`${b[n]}${b[a]}${b[++n]}${b[n]}${b[++n]}${b[++n]}${b[++n]}${b[++n]}${b[++n]}${b[++n]}${b[++n]}${b[++n]}${b[++n]}`)

Where [...] is \$2^{52}-14\$ ? characters.

The -1 is needed since strings cannot be longer than \$2^{52}-1\$ characters.

a cannot be reduced to a smaller number other than 1, which would not work since the e in the string cannot be put that early in the string than the !.

The console.log cannot be reduced, since none of the characters are needed in the string "Hello, world!".

Overall, there's no way to get the character e without going past the ! in the string, and there's no way to get a number bigger than 11 other than a.

This could probably be improved by using charAt, but I'm not confident that couldn't be reduced in some way and it would not change the order of magnitude.

| improve this answer | |
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  • \$\begingroup\$ How about removing n, and instead decrementing a until it reaches 0, and then back up again to a-1 and down etc... The string would basically look like "eo l![...]drW,lH". \$\endgroup\$ – Jo King Dec 12 '19 at 23:28
9
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Python 3, more than \$X\$ with \$X=I \uparrow^{I\uparrow^I I} I\$ with \$I=5735816763073854918203775149089\$

The code is obtained by replacing the ... in the following code template with \$X\$ 1s.

Q=...
exec((762032390336888825205597926747277417487750831511159354629396897028305422383809626318366885737257014369722764755200848545489986725226445234793373668671810533881138176176448303759321782505578156778007890400705523398552371012795706771315225018503248709400507137379430210053109689578295520735219645943760052105586918960055263333607993034635218854740823639508170672552847248741914750040981085688131560558606928438081257071207856256012302413393203594650299912375359565857111459962764512316315869934115819878468802797551498049081777877631940391679271936869727271355722250026194813663937402797387251113215482921743182358788283759577057493220552247028564950916113993279648626087865361138353729963114793860621704499349445249742987180001965672667483449648323146579902833012079301328244612376274244962633770509185016854868366255206566065732320472204953606291770458667922914208019769351151138241745701181139210).to_bytes(ord("ŵ"),"big"))

Shortened version if you don't want to/can't scroll that far

Q=...
exec((762...digits...210).to_bytes(ord("ŵ"),"big"))

Why it prints Hello, World!

The expression passed to exec evaluates to the string

def arrows(a, n, b):
    if n == 1:
        return a**b
    if b == 0:
        return 1
    return arrows(a, n - 1, arrows(a, n, b - 1))

def fast_function(n):
    return arrows(n, arrows(n, n, n), n)

def inverse(n):
    i = 1
    while True:
        if fast_function(i) == n:
            return i
        i += 1

print(inverse(len(str(Q))).to_bytes(13, 'big').decode())

How? The big pile of digits in the original code is just this string but we interpret the bit sequence given by it's ascii encoding as a binary representation of a number (this is obtained using int.from_bytes(...,'big'), 'big' for endianness). Then we get the string by the inverse method .to_bytes(length,'big'). We use a trick to meet the requirements and instead of having the length in decimal we use ord("ŵ") which evaluates to 373 because that's the unicode codepoint of ŵ.

arrows computes \$A \uparrow^N B\$ using a formula from Wikipedia.

The number \$I\$ is given by int.from_bytes(b'Hello, World!', 'big'). This way, Hello, World! is encoded in that very long string of ones. Then we decode this string by inverting the function with which it was defined.

Why it's irreducible

It's not very a formal proof, but I think it works. The fact that my code structure is very simple helps.

So to print Hello, World! we need to call some sort of a printing function. Clearly, the only hope for that is by the exec function. The doc says that exec must take a string or a code object. Well, actually we are passing a bytes object which also works. If it didn't work, we could just use .decode('ascii'), but because bytes work this would make a reduction possible. So to obtain such an object (byte, code or string) we will need to use either the quotation marks or the to_bytes method. There's only a few options with quotation marks and none of them work. So we must use to_bytes. Because we encoded the string length as ord("ŵ"), any change to the parameters of to_bytes will break the program. So we must change the big integer that encodes our code. It's statistically obvious, but if you want you can also computationally check that if we delete any single character or a pair or a triplet of characters from it, we will get garbage, and for sure not a working code that prints Hello, World!. If we delete 4 or more characters, we will get an integer more than 256 times smaller. Because we interpret that with the same length (ord("ŵ")), we will get a null byte, and exec will throw ValueError: source code string cannot contain null bytes.

Extra

Code for generating answers:

code=b'''
<insert code to be executed>
'''
print('Q=...')
print('exec(('+str(int.from_bytes(code,"big"))+').to_bytes(ord("'+chr(len(code))+'"),"big"))')

Note that chr(len(code)) won't always exist (especially for very long codes), so this answer doesn't easily generalize to any very fast function. In my opinion if someone comes up with a way to do it for any code this question will be ultimately "solved" (edit: actually, Higman's lemma implies that the score on this question is bounded, so this is impossible).

| improve this answer | |
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  • \$\begingroup\$ @Bubbler yeah, that edit was stupid, I should've thought about it more (both times). Thanks for pointing the flaw out. Do you think this approach is salvageable? Should I delete the post and make a new one when I come up with the save or is it ok to keep it wrong and edit? \$\endgroup\$ – acupoftea Dec 13 '19 at 2:25
  • 1
    \$\begingroup\$ The rule of thumb here is to delete the answer, (hopefully) edit the post with your corrected answer, and then undelete it. \$\endgroup\$ – Bubbler Dec 13 '19 at 2:38
  • 1
    \$\begingroup\$ Well, this does make things harder to crack. I did find by random experimentation that decode('utf-8') can be reduced to just decode(), though, for what that's worth. Try it online! \$\endgroup\$ – Value Ink Dec 13 '19 at 5:04
  • \$\begingroup\$ Code point may go up to 65535. So, in theory, v you can have up to 65535 byte program, although at that point we might have greater chance that there are some program that works. I think using functions such as Loader's Derive function(one of the fastest growing computable "function") could work, although to make sure we might need to check somewhere around 142444900 programs (assuming 950 byte Loader's Derive+inverse function+other stuff), although >70% would be eliminated immediately because of invalid characters very early, let alone defining and keeping the inverse function. \$\endgroup\$ – Naruyoko Dec 14 '19 at 3:58
  • 1
    \$\begingroup\$ I am not thinking about using template, just saying. It is possible. \$\endgroup\$ – Naruyoko Dec 15 '19 at 17:58
6
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05AB1E, ~\$ 2.5 * 10^{1777} \$ bytes

The program starts with

0>>>

and ends with

>>>₃!B

The middle part contains only >s. The total number of > characters is:

25098162498309588141154755713074390728103069770760074052238917714173876569868691148634176042549175869030448613081100685727537868105068700724415925612269913808499593749918259101545058124654299723726788764334516211684517256948150101180347289022916451558063426031977347451784256897939400652192236258404382348291844743098281742132582049622009384267074810165806373625784728505734586204138892008754810475549403699039426186131241703961770535892728517146019084030050442530148258880465488890083243702565168936069732829928523360052189688531653618029376719759781260232103800093423138357627776562507740948193791067875062805362373001178316850530206611655698994809536157618843298702262299360918288261564223252149018550707768612220426942031389689583329954564583964489539401022510402195332590853218889942437696056695694081344105843161388328303942196356740993490866574068963912863542108279785861191372068053541704843746894440836367259040196995776642457675255580282220480198375934046309922417713428449142775746485165498609097611793368868901625760889393125891944629947649779133088521446313250650588664709735304384247130277354954932511616598273575895960608533708122096438984165676161316003824004917770330565834832795106416101932812692556113632137070814564345428247405370236327568550073970242283021250795797947677972536187886102436517154778645480506007742084298028704070794564223287823038090829494405820111266559120056874535142759744222946196301541177968199011163222223586545099389340200154682693649402549469667375585074560462157985842615966304441485063393013230239088187979990342162755440071831240592907396920570963455610552369098113994760814140693782610141274535146253381742786984559554283939414217235549169221491345428712601180598448294078655886898659207237704837686666199040000000000000000000095

Try it online! (using an integer literal instead of all the >s, since the full version obviously isn't gonna fit on TIO). Or try the code that generated this number.

Proof of validity:

  • B is the only operator in the source that can return a string, thus it cannot be removed.
  • B on an empty stack returns an empty string, and on a single-element stack it returns 10. So we absolutely need two elements on the stack. Since 0 and are the only operators in the source that increase the size of the stack, neither can be removed.
  • If ! was removed, the resulting program would look like 0 (some number of '>'s) ₃B. This converts the number of >s to base 95. Since ! is not a base-95 digit in 05AB1E, this cannot produce the string Hello, World!. Thus ! cannot be removed.
  • Since neither 0, , !, nor B can be removed, the program is guaranteed to look like 0 (some number of '>'s) ₃!B. This converts the number of >s to base 95! (where ! means factorial). This is an injective operation: only one number will yield Hello, World! when converted to base 95!. Thus, the number of >s cannot be altered.
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6
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Ruby, 1862 ... 1.3*1056 bytes

eval ("🆒"*33425716343835650854570646632771294442636521632690315021).size.digits(999).map(&:chr).join

Don't try it online!

Packed the whole program into a single base-999 number (1 digit for each character).

Decode, convert to ascii characters, join and evaluate.

| improve this answer | |
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  • 6
    \$\begingroup\$ Obligatory "replace a with some multi-byte unicode character" -- 10138 bytes \$\endgroup\$ – Value Ink Dec 11 '19 at 1:46
  • \$\begingroup\$ On second thought this can probably be reduced by changing the puts section of the eval to $><<, since each of those characters has a lower byte code. \$\endgroup\$ – Value Ink Dec 13 '19 at 2:21
6
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Runic Enchantments, 12186 15596231 bytes

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^+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"
"n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+n+
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n"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+
\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\k@

Try it online!

Above code generates only an H and operates as a platform by which modifications are made in order to achieve the listed score. Previous explanation in edit history.

Further?

Further.

"̌ * " +᷾ "̌ + . ᷾.
. . . . * . . .
n . . . " . . .
. . . . 1 . . .
. . . . ? . . .
͐. . + . " * " n̂
" * " n̂ n͐ + . .

This pattern is not reducible (ignore the spaces, they are used to align things so that the modifier characters are visible, as they don't combine with non-alpha characters; . are NOP code points that can be subject to alteration). See this tio for a version without and this one for full accurate placement with spaces.

The pattern uses 74 bytes to represent -4, or 18.5 bytes per -1 (up from the baseline 7 bytes). And we need a lot of -1s. 117704 of them to be precise.

Proof of irreducibility:

  • No entire rows can be removed, the 1? lines appear to be, but due to the duplicated " (one above, one below), this will result in the string *n͐ being multiplied against the next value on the stack, which (as the string is first) is not a valid action. the n..." row can't be removed along with those two, as the n is needed in the leftmost column
  • There is no 8-way contiguous vertical collection of . that can be removed (reducing pattern width), as the lower right corner is inaccessible and no other . exist on the bottom row.
  • Directional change characters can't be removed without bypassing the needed number of -1 operations. The closest we can get is changing the bottom row to "*"n+.., but this would only be viable if the initial value "rb"n (9141) minus the desired character was both:
    1. Divisible by 4
    2. Constructable from the available bytes (e: 2260/"inch", !: 2277/"failure", the rest are not divisible by 4)
  • Or if "rb" can be reduced to either "r" or "b" without adversely affecting the output (note that "r " and " b" are not valid and will produce a -1 instead of a positive value). Given a vertical arrangement of the setup and output sequences and the ability to place these bytes such that this is prevented, this will not be possible.
    • Six rows need to be prevented from wholesale deletion (partial deletion results in "r." or ".b" where . is a non-empty byte)
    • Putting rb and k$ on paired rows
    • entry and terminator each on their own rows
    • A reflector on the top row above a ^ entry point protects the top row
    • Any row with the " around rb is automatically protected, as these bytes need to remain

Thus, the entry column is _^"rb"n͐, the output/reset between letters is two columns, ̌.."rb"!͐ / ᷾..$k..̂., and the last column (aligning k$ with the initial rb) is ..;$k.̂. (executed bottom to top).

The two columns allow the code to bloat slightly, adding seven more . positions over keeping input and reset on a single column (k$"rb"ǹ) as well as an additional (more expensive) .

This represents code entry, one copy of the pattern, one output, and final output (a newline) and termination. Not the extra ! needed on the last iteration of the pattern so that flow is correctly handled (this costs a measly 13*3 bytes; along with 7*4 bytes at the end of rows that we can't enforce not being removed).

Finally, replace all . and * with 􀀀 (or other four-byte codepoint).

I've done my best to accurately account for the overhead increases and costs after computing the raw pattern's byte sum, but I recognize I may be off by a small amount.

Ohgodwhy; it gets bigger

I believe this is irreducible.

DL.....DL..
R."p"...U..
..RD....D.L
.U.L"̌a"L.RU
RDU.\..../.
U.L...DL\.D
..ULUL...DL
...ULUL.D.L
....U...LRU
>.0....U../@
."R"h"D.R+U
.t...DL.UL.
."..D/..≠.D
.RU/LU..L.?
R.D....RU.n
U"z"᷾.R5URU.
..R:l͍U..U.L

Try it online!
Sequence >.0 simply to set the stack to contain an initial value (and should be considered ...)

All the UDLR/\ would be replaced by their 2-byte modifier counterparts and every cell not otherwise filled with a command is given a 4-byte unicode character. This works out to (I think) 647 bytes to generate a single -1. On top of that, our initial value rises from "rb"n (9141) to 914,100,000 by making each initial letter get constructed from "rb"nYC. As YC occupy space, they would irreparably damage the redirection pattern.

Roughly speaking (accurate overhead between letters not verified, but we're talking about a few dozen bytes at most), this works out to a score of 7,688,494,370,086 (7.6 trillion).

Breaking this would require finding a way to bypass the maze of pointer redirections without (a) breaking the pattern--as a large number of the paths redirect away an infinite loop or an invalid command--and still (b) generating a 5 character long string prior to hitting the n, due to a :l͍5≠? jump check. It costs a few bytes to make this check (11 bytes lost), but reinforces the durability of the pattern.

The string pathz was chosen as a 5-character long sequence where every right-truncated string ("p", "pa", "pat", "path") was present in the word dictionary (meaning that when calling n on it, a positive value will be generated), forcing the program to trace out a meandering route to pick up each letter before hitting the "z" to get a string not in the dictionary. This can still be improved as paths is also present in the dictionary (leading to a 6-length string pathsz), but at this point I'm starting to look for something that will increase the byte count exponentially and searching the word dictionary for longer strings with the desired characteristics is tedious.

| improve this answer | |
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JavaScript, 24855 bytes

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Try it online!

This was made using jsfuck.com with the input console.log("Hello, World!"). This is the smallest program I could make. Smallest because, due to the fact that there's only 6 characters, I'm afraid that a bigger one could easily remove characters to come up with a smaller one.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ alert("Hello, World!") is shorter, and has less of each type of character: (format is # of [], # of (), # of +, # of !): (6696, 1248, 5770, 3197) and (6317, 1180, 5438, 3015) respectively for console.log and for alert \$\endgroup\$ – Embodiment of Ignorance Dec 10 '19 at 5:31
  • \$\begingroup\$ I believe convert (function d(){return function l() {return function r(){return function o() {return function W(){return function e() {return function H(){console.log(H.name+e.name+l.name+l.name+o.name+', '+W.name+o.name+r.name+l.name+d.name+'!')}}}}}}}()()()()()()()) to JSFuck would be longer. But I cannot prove it fit the requirement. \$\endgroup\$ – tsh Dec 10 '19 at 5:45
  • \$\begingroup\$ @EmbodimentofIgnorance would alert typically be counted as an output? It doesn't work on TIO. \$\endgroup\$ – Mason Dec 10 '19 at 5:52
  • 1
    \$\begingroup\$ I've seen other answers use alert before, so I believe it is a valid output format \$\endgroup\$ – Embodiment of Ignorance Dec 10 '19 at 6:00
  • 3
    \$\begingroup\$ Excellent find! I am using this generator for all of my production code. \$\endgroup\$ – wha7ever Dec 11 '19 at 19:20
4
\$\begingroup\$

Wren, 1220 bytes

Finally found a use of Wren's verbose syntax!

I'm pretty surprised that none of those answers use the perfect byte-waster log. The eclipses are things that are too long for this program.

["...","...","...",...].each{|i|System.write(String.fromCodePoint(i.count))}

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Explanation

["...","...","...",...] // Those constants too long for this program
.each{|i|} // Each, the perfect byte-waster!
System.write() // Verbose output to STDOUT...
String.fromCodePoint() // Convert to code points
i.count // Length of i
| improve this answer | |
\$\endgroup\$
4
\$\begingroup\$

C# (Visual C# Interactive Compiler), 10658 bytes

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C# allows unicode escapes in identifiers, and I decided to take full advantage of the fact here.

Non-escaped code:

Write(String.Join(String.Empty,(String.Empty+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'¥'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'¥'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'¥'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'¥'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'¥'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'¥'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'¥'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'¥'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'¥'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'¥'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'¥'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'¥'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö'+'Ö').Split('¥').Select(l=>(Char)l.Length)))

The program creates a string of ÖÖÖÖ...¥ÖÖÖÖ...... (... are placeholders) and splits it along ¥. For each resulting string in the array, it takes the length of the section and casts it into a character.

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ The trailing semi-colon can be removed. Apart from that I can't think of anything to remove, so +1 from me. :) \$\endgroup\$ – Kevin Cruijssen Dec 10 '19 at 8:25
  • 1
    \$\begingroup\$ You have JSFuck as input to your code in the TIO link \$\endgroup\$ – LiefdeWen Dec 10 '19 at 9:39
  • 1
    \$\begingroup\$ For the top one, simply deleting entire if(...) does work, so it's invalid. \$\endgroup\$ – Bubbler Dec 11 '19 at 6:51
  • \$\begingroup\$ Easily reducible. \$\endgroup\$ – Grimmy Dec 11 '19 at 8:39
  • \$\begingroup\$ Can't you reduce System.Numerics.BigInteger to int? \$\endgroup\$ – Fabian Röling Dec 11 '19 at 14:32
2
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Golfscript, 144,202,216,216,222,088,064,174,222,228,216,200,081 bytes

"...",`3/{~2/}%''+

where ... is replaced by 144202216216222088064174222228216200066 0's

Explanation:

"...",`                  Count 0's and convert to string
       3/{   }%          Map over each group of 3 digits:
          ~2/                Divide by 2
               ''+           Convert ascii codepoints to string

The ascii codepoints for "Hello, World!" are as follows:

072 101 108 108 111 044 032 087 111 114 108 100 033

Concatenating them together isn't enough to be able to access them by splitting them into threes because the integer would be interpreted as 72101... rather than 072101..., meaning that splitting it into threes would yield 721 as the first codepoint instead of 072. To remedy this, I doubled it and then divided each individual codepoint by 2 after they're separated out. This has the added bonus of nearly doubling the length of the code.

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1
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Charcoal, 4022948789620328394911726 144202216216222088064174222228216200073 bytes

↑E↨L´´...´´φ℅ι

Explanation: ´´...´´ represents a string literal of length 72101108108111044032087111114108100033. However, ´ is the quote character in Charcoal, so it needs to be doubled in order to quote itself, thus requiring 144202216216222088064174222228216200066 ´s. The length is then converted to base 1000 using the predefined variable that contains 1000, and the values are then converted to their Unicode character values. However, the result is still an array, so the is used to output the characters correctly.

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1
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naz, 2308 4548 bytes

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Explanation

This commit adds the n command to naz, which negates the value of a specific variable.

For each character in the string Hello, World!, the above program follows the following steps:

  • Subtract 1 from the register 127 times.
  • Store the resulting value of -127 in variable 1.
  • Negate variable 1, and set the register to the new value.
  • Subtract 1 enough times to reach the ASCII value of the character, then output it.
  • Multiply the register by 0 and store the resulting value of 0 in variable 1.

I believe this is irreducible because of the following (but I'd love to be proven wrong):

  • Removing any one number or any one letter will create a broken instruction and throw an error.
  • Removing any instance of 1s will result in an incorrect character being output.
  • Removing any instance of 1o will result in a character being omitted from the output.
  • Removing any instance of 0m will eventually result in an incorrect or completely invalid output value.
  • Removing any instance of 2x will either result in an invalid output value or allow the value in the register to go out of bounds, unless you remove the first instance, in which case an error will be thrown because the variable will be used before its value is first declared.
  • Removing any instance of 1v will throw the error improper use of opcode 2.
  • Removing any instance of 1n will allow the value in the register to go out of bounds.

The previous 2308-byte solution:

1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1o0m1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1o0m1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1o0m1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1o0m1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1o0m1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1o0m1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1o0m1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1o0m1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1o0m1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1o0m1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1o0m1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1o0m1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1a1o
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\$\endgroup\$

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