Biggest Irreducible Hello World

Question

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.

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. — tsh, Dec 10, 2019 at 6:28
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. — Jonathan Allan, Dec 10, 2019 at 9:05
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). — Draco18s no longer trusts SE, Dec 10, 2019 at 22:56
A bit late but maybe it's worth a comment - it can actually be proven that for a fixed language, this question has bounded score - if not, that means we can indefinitely make larger and larger strings (programs) none of which is a subsequence of another, and that contradicts Higman's lemma. You can find a nice proof in my Math SE question (and answer). — acupoftea, Apr 25, 2020 at 8:26

lyxal · Accepted Answer · 2021-06-09 12:30:07Z

50

+50

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

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

Program as seen in the raw editor

So where that ... is, replace it with \$3.3 * 10^{8912898}\$ 0s. Then, add a ! after those 0s. This splits the number into chunks of length 􏿿 through concatenation and then outputs the corresponding ascii character.

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.

Why can't this be reduced?

Remove any of the leading 0s, and then it has the wrong number at the start, meaning that the wrong letters will be printed
Remove any of the ÷⅍+ℤ,(􏿿|⅍+)ℤ,(􏿿|⅍+)ℤ,(􏿿|⅍+)ℤ,(􏿿|⅍+)ℤ,⅍+ℤ,⅍+ℤ,⅍+ℤ,(􏿿|⅍+)ℤ,(􏿿|⅍+)ℤ,(􏿿|⅍+)ℤ,(􏿿|⅍+)ℤ,⅍+ℤ, part, and the chunking will go wrong.

Score calculation: https://chat.stackexchange.com/transcript/message/52895760#52895760

edited Jun 9, 2021 at 12:30

answered Dec 10, 2019 at 3:37

lyxal

32.5k2 gold badges56 silver badges138 bronze badges

9

\$\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, 2019 at 7:58
\$\begingroup\$ Could you perhaps make this bigger by forming the numbers you are using in some creative ways? \$\endgroup\$
– Jonathan Allan
Dec 10, 2019 at 9:03
21

\$\begingroup\$ An explanation of how this is irreducible would be appreciated \$\endgroup\$
– Jo King ♦
Dec 10, 2019 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, 2019 at 14:25
1

\$\begingroup\$ @FabianRöling I believe the generator was written in Python 3 \$\endgroup\$
– Anthony Pham
Dec 11, 2019 at 15:48

| Show 10 more comments

Bubbler · Accepted Answer · 2019-12-13 07:51:13Z

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 code-golf 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.

This is the first answer I've seen on here with a score measured with up-arrow notation. — Skyler, Dec 12, 2019 at 20:06
@Skyler We once had this challenge to generate big numbers in limited code size. — Bubbler, Dec 12, 2019 at 23:21

score 31 · Accepted Answer · 2019-12-10 23:14:18Z

31

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!

edited Dec 10, 2019 at 23:14

answered Dec 10, 2019 at 4:29

user85052

19

\$\begingroup\$ +1 for simply using Malbolge in an actual programming competition. \$\endgroup\$
– COTO
Dec 10, 2019 at 22:00

Add a comment |

Jo King · Accepted Answer · 2019-12-11 00:35:29Z

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.

@tsh Huh, good ideas. Couldn't it be made pretty much arbitrary with -ord('.')? Up to the max size of strings — Jo King, Dec 10, 2019 at 6:34
This looks easy to considerably reduce, unless I missed something. — Grimmy, Dec 10, 2019 at 14:42
@Grimmy Ah, I didn't realise pow was it's own function unlike int.__mul__ — Jo King, Dec 11, 2019 at 0:36

Community · Accepted Answer · 2020-06-17 09:04:33Z

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.

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". — Jo King, Dec 12, 2019 at 23:28

Toby Speight · Accepted Answer · 2021-10-28 13:22:32Z

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 its 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.

To print Hello, World! we need to call some sort of 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).

@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? — acupoftea, Dec 13, 2019 at 2:25
The rule of thumb here is to delete the answer, (hopefully) edit the post with your corrected answer, and then undelete it. — Bubbler, Dec 13, 2019 at 2:38
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! — Value Ink, Dec 13, 2019 at 5:04
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. — Naruyoko, Dec 14, 2019 at 3:58
I am not thinking about using template, just saying. It is possible. — Naruyoko, Dec 15, 2019 at 17:58

G B · Accepted Answer · 2019-12-13 08:06:33Z

9

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.

edited Dec 13, 2019 at 8:06

answered Dec 10, 2019 at 9:10

G B

21.3k1 gold badge19 silver badges51 bronze badges

8

\$\begingroup\$ Obligatory "replace a with some multi-byte unicode character" -- 10138 bytes \$\endgroup\$
– Value Ink
Dec 11, 2019 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, 2019 at 2:21
1

\$\begingroup\$ I'm getting a compile error. What version did you test this on? bignum too big to convert into long (RangeError) \$\endgroup\$
– user974006
Mar 11, 2022 at 8:11
1

\$\begingroup\$ It might have worked when I first posted it, December 2019, but I'm not even sure. \$\endgroup\$
– G B
Mar 11, 2022 at 8:56

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

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.

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 — Gymhgy, Dec 10, 2019 at 5:31
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. — tsh, Dec 10, 2019 at 5:45
I've seen other answers use alert before, so I believe it is a valid output format — Gymhgy, Dec 10, 2019 at 6:00
Excellent find! I am using this generator for all of my production code. — Vanity Slug - codidact.com, Dec 11, 2019 at 19:20
@tsh I believe that code are reduced to (function l(){return function o(){return function e(){return function H(){console.log(H.name+e.name+l.name+l.name+o.name+', World!')}}}}}}})()()()() — Naruyoko, Dec 13, 2019 at 19:14

Grimmy · Accepted Answer · 2019-12-11 01:08:06Z

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 ₃ 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, ₃, !, 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.

Draco18s no longer trusts SE · Accepted Answer · 2019-12-14 04:34:53Z

Runic Enchantments, 12186 15596231 bytes

_/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\
^+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"
"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+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"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+"+
\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\k@

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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

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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.

score 5 · Accepted Answer · 2019-12-10 09:48:53Z

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

Gymhgy · Accepted Answer · 2019-12-13 03:27:15Z

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.

The trailing semi-colon can be removed. Apart from that I can't think of anything to remove, so +1 from me. :) — Kevin Cruijssen, Dec 10, 2019 at 8:25
For the top one, simply deleting entire if(...) does work, so it's invalid. — Bubbler, Dec 11, 2019 at 6:51

Pseudo Nym · Accepted Answer · 2019-12-13 17:08:56Z

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.

FarSeenNomic · Accepted Answer · 2022-03-11 07:20:53Z

Mathematica, 10^{10¹²⁸⁴³³} at least 10↑↑11.29 bytes

FromCharacterCode[StringLength["aaaa...aaaa"]~IntegerDigits~Nest[Hyperfactorial, 10, 9]]

old:

 FromCharacterCode[StringLength["aaaa...aaaa"]~IntegerDigits~Hyperfactorial[Hyperfactorial[3]]]

Where "aaaa...aaaa" is FromDigits[ToCharacterCode["Hello, World!"], Hyperfactorial[Hyperfactorial[3]]] a's long. That's about 10¹²⁸⁴³³ digits long. The key here is that the "W" is character code 114, which means the base of the number can't even drop below 114, or the code won't output properly. I believe Hyperfactorial to be the fastest growing function, with Hyperfactorial[3] = 108 and Hyperfactorial[4] = 27648, meaning that nesting 2 Hyperfactorial is the maximum number, and any attempt to reduce it will drop the base below 114.

Edit: This is not the biggest answer, Nest[Hyperfactorial, 10, 9] as a base would be bigger, as it cannot be reduced to a number greater than 114. Sadly, I'm not at the level to even start getting close bounds, so lets go with a very poor lower bound, which would still be a valid score.

Ω(Log10[Hyperfactorial[x]]) is x, where Ω is lower bound notation. It's really closer to x², but compared to the power of logerithms, they may as well be the same. This means that Hyperfactorial nested 9 times is at minimum (10)↑↑(10). The number of characters is 10 to this power, so (10)↑↑(11).

Can this be reduced by removing the 0 from 10 and decreasing the number of a? — emanresu A, Mar 11, 2022 at 7:59
@emanresuA It cannot - Hyperfactorial[0] is 1, and Hyperfactorial[1] is 1. This means the by removing the 0 from 10, the base 'collapses' to 1. If the decoding base is less than 114, it cannot decode the "W" for "World". So removing the 0, with any number big or small of a's, would never output Hello, World! — FarSeenNomic, Mar 11, 2022 at 15:28

PkmnQ · Accepted Answer · 2020-09-27 03:49:29Z

3

05AB1E, 5,697,476,345,285,646,822,141,672,875,833 bytes

….g257B

The … is actually 1,424,369,086,321,411,705,535,418,218,957 …aaa's.

Explanation

…       # Push 1,424,369,086,321,411,705,535,418,218,957 things
 .g     # and count how many things there are on the stack
   257B # then convert it to base 257

I think …aaa is the longest irreducible literal possible, so I use it here.

edited Sep 27, 2020 at 3:49

answered Sep 26, 2020 at 11:55

PkmnQ

2,6938 silver badges20 bronze badges

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Neil · Accepted Answer · 2019-12-11 12:01:09Z

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.

Seggan · Accepted Answer · 2022-03-11 16:47:08Z

2

Jelly, 17,502,248,526,741,541,454,928,712,794 \$3.14 * 10^{1896}\$ bytes

³!¤“...”LbỌ

Try it online!

No, I did not force TIO to execute that many bytes of code :) I hardcoded the value in. See the TIO link for the exact number of characters.

³!¤“...”LbỌ - Main niladic link
   “...”    - The real meat of the program. a very long string of characters. See the TIO link for the exact number of characters
        L   - Takes the length of the string
³!¤      b  - Convert from base 10 to base 100!
          Ọ - Turn the array into a string

Irreducible because:

If you omit a character in the literal, the resultant number will be wrong
If you change ³!¤, the base conversion will be incorrect
If you omit any other byte, the program will either not print "Hello, World!" or do strange things

Funny that Jelly is a golfing language, and yet you can make such a long irreducible program for the simplest of things.

EDIT: Thanks @Bubbler for fixing my output format and adding some bytes EDIT: Fixed the score

edited Mar 11, 2022 at 16:47

answered Oct 28, 2021 at 1:31

Seggan

5,27121 silver badges47 bronze badges

\$\begingroup\$ Yours is missing a comma in the output, and is not irreducible because I can remove _1 and a bunch of chars in the literal at once to get the same string. This should work instead. (This is how to get the literal correctly) \$\endgroup\$
– Bubbler
Oct 28, 2021 at 2:16
\$\begingroup\$ I think ³!¤ (100!) works as an irreducible base. \$\endgroup\$
– Bubbler
Oct 28, 2021 at 2:26
\$\begingroup\$ Hmm earlier I tried without the _1 and it have gibberish... I got the number by doing “Hello World!’ \$\endgroup\$
– Seggan
Oct 28, 2021 at 13:14

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sporeball · Accepted Answer · 2020-03-23 07:54:54Z

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

FarSeenNomic · Accepted Answer · 2022-03-25 04:46:35Z

BrainF***, 2992 bytes

-{184}.-{227}.-{249}>+{254}[+<.>]<-{253}.+{189}.+{244}.-{201}.-{232}.-{253}.+{250}.+{248}.+{189}.

Where w{n} means w repeated n times. BF is cyclical mod 256, so instead of going "up" 72 to get to "H", you can go "down" 184 to get there. -{227} just adds 29 to get to "e", and so on with these basic building blocks, except for the double "ll". I made a counter set to -2 (254), then add 1 both times to run the output (.) twice.

Full program:

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------.-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------.--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------->++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++[+<.>]<-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------.+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++.++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++.---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------.----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------.-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------.++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++.++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++.+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++.

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Biggest Irreducible Hello World

19 Answers 19

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

Why can't this be reduced?

APL (Dyalog Unicode), \$ \approx (10 \uparrow)^{625} 10.024549 \gg 10 \uparrow \uparrow 626 \$ bytes

How it works

Previous answers

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

APL (Dyalog Unicode), ≈ 1.6198e1175 bytes

How it works

APL (Dyalog Unicode), 1169882994796837103436970021 ≈ 1.17e27 bytes

How it works

Malbolge, 265 bytes

Python 3, 44619 8.13*1011515 ~\$10^{10^{5749}}\$ bytes

JavaScript, 9007199254741154 bytes

Python 3, more than \$X\$ with \$X=I \uparrow^{I\uparrow^I I} I\$ with \$I=5735816763073854918203775149089\$

Why it prints Hello, World!

Why it's irreducible

Extra

Ruby, 1862 ... 1.3*1056 bytes

JavaScript, 24855 bytes

05AB1E, ~\$ 2.5 * 10^{1777} \$ bytes

Runic Enchantments, 12186 15596231 bytes

Ohgodwhy; it gets bigger

Wren, 1220 bytes

Explanation

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

Golfscript, 144,202,216,216,222,088,064,174,222,228,216,200,081 bytes

Explanation:

Mathematica, 1010128433 at least 10↑↑11.29 bytes

05AB1E, 5,697,476,345,285,646,822,141,672,875,833 bytes

Explanation

Charcoal, 4022948789620328394911726 144202216216222088064174222228216200073 bytes

Jelly, 17,502,248,526,741,541,454,928,712,794 \$3.14 * 10^{1896}\$ bytes

naz, 2308 4548 bytes

BrainF***, 2992 bytes

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Python 3, 44619 8.13*10¹¹⁵¹⁵ ~\$10^{10^{5749}}\$ bytes

Why it prints `Hello, World!`

Ruby, 1862 ... 1.3*10⁵⁶ bytes

Mathematica, 10^{10¹²⁸⁴³³} at least 10↑↑11.29 bytes

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