Largest Number Printable

Question

Your goal is to write a program that prints a number. The bigger the number, the more points you'll get. But be careful! Code length is both limited and heavily weighted in the scoring function. Your printed number will be divided by the cube of the number of bytes you used for your solution.

So, let's say you printed 10000000 and your code is 100 bytes long. Your final score will be \$\frac {10000000} {100^3} = 10\$.

There are other rules to follow, in order to make this challenge a bit harder.

• You cannot use digits in your code (0123456789);
• You can use mathematical/physical/etc. constants, but only if they are less than 10. (e.g. You can use \$\pi \approx 3.14\$ but you can't use the Avogadro constant \$= 6\times 10^{23}\$)
• Recursion is allowed but the generated number needs to be finite (so infinite is not accepted as a solution. Your program needs to terminate correctly, assuming unbounded time and memory, and generate the requested output);
• You cannot use the operations * (multiply), / (divide), ^ (power) nor any other way to indicate them (e.g. 2 div 2 is not allowed);
• Your program can output more than one number, if you need it to do that. Only the highest one will count for scoring;
• However, you can concatenate strings: this means that any sequence of adjacent digits will be considered as a single number;
• Your code will be run as-is. This means that the end-user cannot edit any line of code, nor he can input a number or anything else;
• Maximum code length is 100 bytes.

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Leaderboard

1. Steven H., Pyth \$\approx f_{\Gamma_0+7}(256^{26})\$
2. Simply Beautiful Art, Ruby \$\approx f_{\Gamma_0}(3)\$
3. Peter Taylor, GolfScript \$\approx f_{\varepsilon_0+\omega+1}(17)\$
4. r.e.s., GolfScript \$\approx f_{\varepsilon_0}^9(126)\approx f_{\varepsilon_0+1}(9)\$ ^[1]
5. Simply Beautiful Art, Ruby \$\approx f_{\omega^{\omega2}+1}(126^22^{126})\$
6. eaglgenes101, Julia \$\approx f_{\omega^3}(127)\$
7. col6y, Python 3, \$\approx 127\to126\to\dots\to2\to1\approx f_{\omega^2}(127)\$ ^[1][3]
8. Toeofdoom, Haskell, \$\approx a_{20}(1)\approx f_{\omega+1}(18)\$ ^[1]
9. Fraxtil, dc, \$\approx 15\uparrow^{166665}15\$ ^[3]
10. Magenta, Python, \$\approx\mathrm{ack}(126,126)\approx10\uparrow^{124}129\$
11. Kendall Frey, ECMAScript 6, \$\approx1000\uparrow^43\$ ^[1]
12. Ilmari Karonen, GolfScript, \$\approx10\uparrow^310^{377}\$ ^[1]
13. Aiden4, Rust, \$\approx10\uparrow^3127\$
14. BlackCap, Haskell, \$\approx10\uparrow\uparrow65503\$
15. recursive, Python, \$\approx2\uparrow\uparrow11\approx10\uparrow\uparrow8.63297\$ ^[1][3]
16. n.m., Haskell, \$\approx2\uparrow\uparrow7\approx10\uparrow\uparrow4.63297\$ ^[1]
17. David Yaw, C, \$\approx10^{10^{4\times10^{22}}}\approx10\uparrow\uparrow4.11821\$ ^[2]
18. primo, Perl, \$\approx10^{(12750684161!)^{5\times2^{27}}}\approx10\uparrow\uparrow4.11369\$
19. Art, C, \$\approx10^{10^{2\times10^6}}\approx10\uparrow\uparrow3.80587\$
20. Robert Sørlie, x86, \$\approx10^{2^{2^{19}+32}}\approx10\uparrow\uparrow3.71585\$
21. Tobia, APL, \$\approx10^{10^{353}}\approx10\uparrow\uparrow3.40616\$
22. Darren Stone, C, \$\approx10^{10^{97.61735}}\approx10\uparrow\uparrow3.29875\$
23. ecksemmess, C, \$\approx10^{2^{320}}\approx10\uparrow\uparrow3.29749\$
24. Adam Speight, vb.net, \$\approx10^{5000\times2^{256}}\approx10\uparrow\uparrow3.28039\$
25. Joshua, bash, \$\approx10^{10^{15}}\approx10\uparrow\uparrow3.07282\$

_Footnotes

1. ^{If every electron in the universe were a qubit, and every superposition thereof could be gainfully used to store information (which, as long as you don't actually need to know what's being stored is theoretically possible), this program requires more memory than could possibly exist, and therefore cannot be run - now, or at any conceiveable point in the future. If the author intended to print a value larger than ≈10↑↑3.26 all at once, this condition applies.}
2. ^{This program requires more memory than currently exists, but not so much that it couldn't theoretically be stored on a meager number of qubits, and therefore a computer may one day exist which could run this program.}
3. ^{All interpreters currently available issue a runtime error, or the program otherwise fails to execute as the author intended.}
4. ^{Running this program will cause irreparable damage to your system.}

---

Edit @primo: I've updated a portion of the scoreboard using a hopefully easier to compare notation, with decimals to denote the logarithmic distance to the next higher power. For example \$10↑↑2.5 = 10^{10^{\sqrt {10}}}\$. I've also changed some scores if I believed to user's analysis to be faulty, feel free to dispute any of these.

Explanation of this notation:

If \$0 \le b \lt 1\$, then \$a \uparrow\uparrow b = a^b\$.

If \$b \ge 1\$, then \$a \uparrow\uparrow b = a^{a \uparrow\uparrow (b-1)}\$.

If \$b \lt 0\$, then \$a \uparrow\uparrow b = \log_a(a \uparrow\uparrow (b+1))\$

An implementation of this notation is provided in Python that let's you test reasonably sized values.

Answer 1

Windows 2000 - Windows 8 (3907172 / 23³ = 321)

NOTE: DON'T F'ING RUN THIS!

Save the following to a batch file and run it as Administrator.

CD|Format D:/FS:FAT/V/Q

Output when run on a 4TB drive with the first printed number in bold.

Insert new disk for drive D:
and press ENTER when ready... The type of the file system is NTFS.
The new file system is FAT.
QuickFormatting 3907172M
The volume is too big for FAT16/12.

Answer 2

GolfScript, score: way too much

OK, how big a number can we print in a few chars of GolfScript?

Let's start with the following code (thanks, Ben!), which prints 126:

'~'(

Next, let's repeat it 126 times, giving us a number equal to about 1.26126 × 10³⁷⁷:

'~'(.`*

(That's string repetition, not multiplication, so it should be OK under the rules.)

Now, let's repeat that 378-digit number a little over 10³⁷⁷ times:

'~'(.`*.~*

You'll never actually see this program finish, since it tries to compute a number with about 10³⁸⁰ ≈ 2¹¹⁴⁰ digits. No computer ever built could store a number that big, nor could such a computer ever be built using known physics; the number of atoms in the observable universe is estimated to be about 10⁸⁰, so even if we could somehow use all the matter in the universe to store this huge number, we'd still somehow have to cram about 10³⁸⁰ / 10⁸⁰ = 10³⁰⁰ digits into each atom!

But let's assume that we have God's own GolfScript interpreter, capable of running such a calculation, and that we're still not satisfied. OK, let's do that again!

'~'(.`*.~*.~*

The output of this program, if it could complete, would have about 10¹⁰³⁸³ digits, and so would equal approximately 10^1010383.

But wait! That program is getting kind of repetitive... why don't we turn it into a loop?

'~'(.`*.{.~*}*

Here, the loop body gets run about 10³⁷⁷ times, giving us a theoretical output consisting of about 10^10⋰10377 digits or so, where the tower of iterated powers of 10 is about 10³⁷⁷ steps long. (Actually, that's a gross underestimate, since I'm neglecting the fact that the number being repeated also gets longer every time, but relatively speaking that's a minor issue.)

But we're not done yet. Let's add another loop!

'~'(.`*.{.{.~*}*}*

To even properly write down an approximation of such numbers requires esoteric mathematical notation. For example, in Knuth up-arrow notation, the number (theoretically) output by the program above should be about 10 ↑³ 10³⁷⁷, give or take a few (or 10³⁷⁷) powers of ten, assuming I did the math right.

Numbers like this get way beyond just "incredibly huge", and into the realm of "inconceivable". As in, not only is it impossible to count up to or to write down such numbers (we crossed beyond that point already at the third example above), but they literally have no conceivable use or existence outside abstract mathematics. We can prove, from the axioms of mathematics, that such numbers exist, just like we can prove from the GolfScript specification that program above would compute them, if the limits of reality and available storage space did not intervene), but there's literally nothing in the physical universe that we could use them to count or measure in any sense.

Still, mathematicians do sometimes make use of even larger numbers. (Theoretically) computing numbers that large takes a little bit more work — instead of just nesting more loops one by one, we need to use recursion to telescope the depth of the nested loops. Still, in principle, it should be possible to write a short GolfScript program (well under 100 bytes, I would expect) to (theoretically) compute any number expressible in, say, Conway chained arrow notation; the details are left as an exercise. ;-)

Answer 3

JavaScript 44 chars

This may seem a little cheaty:

alert((Math.PI+''+Math.E).replace(/\./g,""))

Score = 31415926535897932718281828459045 / 44^3 ≈ 3.688007904758867e+26 ≈ 10↑↑2.1536134004

Answer 4

C, score = 10^1097.61735/98³ ≈ 10↑↑2.29874984

unsigned long a,b,c,d,e;main(){while(++a)while(++b)while(++c)while(++d)while(++e)printf("%lu",a);}

I appreciate the help in scoring. Any insights or corrections are appreciated. Here is my method:

n = the concatenation of every number from 1 to 2⁶⁴-1, repeated (2⁶⁴-1)⁴ times. First, here's how I'm estimating (low) the cumulative number of digits from 1 to 2⁶⁴-1 (the "subsequence"): The final number in the subsequence sequence is 2⁶⁴-1 = 18446744073709551615 with 20 digits. Thus, more than 90% of the numbers in the subsequence (those starting with 1..9) have 19 digits. Let's assume the remaining 10% average 10 digits. It will be much more than that, but this is a low estimate for easy math and no cheating. That subsequence gets repeated (2⁶⁴-1)⁴ times, so the length of n will be at least (0.9×(2⁶⁴-1)×19 + 0.1×(2⁶⁴-1)×10) × (2⁶⁴-1)⁴ = 3.86613 × 10⁹⁷ digits. In the comments below, @primo confirms the length of n to be 4.1433x10⁹⁷. So n itself will be 10 to that power, or 10^1097.61735.

l = 98 chars of code

score = n/l³ = 10^1097.61735/98³

Requirement: Must run on a 64-bit computer where sizeof(long) == 8. Mac and Linux will do it.

Answer 5

GolfScript; score at least f_{ε_0+ω+1}(17) / 1000

Following r.e.s.'s suggestion to use the Lifetime of a worm answer for this question, I present two programs which vastly improve on his derivation of Howard's solution.

They share a common prefix, modulo the function name:

,:z){.[]+{\):i\.z={.z+.({<}+??\((\+.@<i*\+}{(;}if.}do;}:g~g

computes g(g(1)) = g(5) where g(x) = worm_lifetime(x, [x]) grows roughly as f_ε0 (which r.e.s. notes is "the function in the fast-growing hierarchy that grows at roughly the same rate as the Goodstein function").

The slightly easier (!) to analyse is

,:z){.[]+{\):i\.z={.z+.({<}+??\((\+.@<i*\+}{(;}if.}do;}:g~g.{.{.{.{.{.{.{.{.{.{g}*}*}*}*}*}*}*}*}*}*

.{foo}* maps x to foo^x x.

,:z){[]+z\{\):i\.z={.z+.({<}+??\((\+.@<i*\+}{(;}if.}do;}:g~g.{g}*

thus gives g^(g(5)) ( g(5) ); the further 8 levels of iteration are similar to arrow chaining. To express in simple terms: if h_0 = g and h_{i+1} (x) = h_i^x (x) then we calculate h_10 (g(5)).

I think this second program almost certainly scores far better. This time the label assigned to function g is a newline (sic).

,:z){.[]+{\):i\.z={.z+.({<}+??\((\+.@<i*\+}{(;}if.}do;}:
~
{.['.{
}*'n/]*zip n*~}:^~^^^^^^^^^^^^^^^^

This time I make better use of ^ as a different function.

.['.{
}*'n/]*zip n*~

takes x on the stack, and leaves x followed by a string containing x copies of .{ followed by g followed by x copies of }*; it then evaluates the string. Since I had a better place to burn spare characters, we start with j_0 = g; if j_{i+1} (x) = j_i^x (x) then the first evaluation of ^ computes j_{g(5)} (g(5)) (which I'm pretty sure already beats the previous program). I then execute ^ 16 more times; so if k_0 = g(5) and k_{i+1} = j_{k_i} (k_i) then it calculates k_17. I'm grateful (again) to r.e.s. for estimating that k_i >> f_{ε_0+ω+1}(i).

Answer 6

Python 3 - 99 chars - (most likely) significantly larger than Graham's number

I've come up with a more quickly increasing function based on an extension of the Ackermann function.

A=lambda a,b,*c:A(~-a,A(a,~-b,*c)if b else a,*c)if a else(A(b,*c)if c else-~b);A(*range(ord('~')))

http://fora.xkcd.com/viewtopic.php?f=17&t=31598 inspired me, but you don't need to look there to understand my number.

Here is the modified version of the ackermann function that I'll be using in my analysis:

A(b)=b+1
A(0,b,...)=A(b,...)
A(a,0,...)=A(a-1,1,...)
A(a,b,...)=A(a-1,A(a,b-1,...),...)

My function A in the code above is technically not the same, but it is actually stronger, with the following statement to replace the third line of the above definition:

A(a,0,...)=A(a-1,a,...)

(a has to be at least 1, so it has to be stronger)

But for my purposes I will assume that it is the same as the simpler one, because the analysis is already partially done for Ackermann's function, and therefore for this function when it has two arguments.

My function is guaranteed to eventually stop recursing because it always either: removes an argument, decrements the first argument, or keeps the same first argument and decrements the second argument.

Analysis of size

Graham's number, AFAIK, can be represented as G(64) using:

G(n) = g^n(4)
g(n) = 3 ↑^(n) 3

Where a ↑^(n) b is knuth's up-arrow notation.

As well:

A(a,b) = 2 ↑^(a-2) (b+3) - 3
A(a,0) ≈ 2 ↑^(a-2) 3
g(n) ≈ A(n+2,0) // although it will be somewhat smaller due to using 2 instead of 3. Using a number larger than 0 should resolve this.
g(n) ≈ A(n+2,100) // this should be good enough for my purposes.

g(g(n)) ≈ A(A(n+2,100),100)

A(1,a+1,100) ≈ A(0,A(1,a,100),100) = A(A(1,a,100),100)

g^k(n) ≈ A(A(A(A(...(A(n+2,100)+2)...,100)+2,100)+2,100)+2,100) // where there are k instances of A(_,100)
A(1,a,100) ≈ A(A(A(A(...(A(100+2),100)...,100),100),100),100)

g^k(100) ≈ A(1,k,100)
g^k(4) < A(1,k,100) // in general
g^64(4) < A(1,64,100)

The number expressed in the program above is A(0,1,2,3,4,...,123,124,125).

Since g^64(4) is Graham's number, and assuming my math is correct then it is less than A(1,64,100), my number is significantly larger than Graham's number.

Please point out any mistakes in my math - although if there aren't any, this should be the largest number computed so far to answer this question.

Answer 7

Perl - score ≈ 10↑↑4.1

$_=$^Fx($]<<-$]),/(?<R>(((((((((((((((((((.(?&R))*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*(??{print})/

Once again abusing perl's regex engine to grind through an unimaginable amount of combinations, this time using a recursive descent.

In the inner most of the expression, we have a bare . to prevent infinite recursion, and thus limiting the levels of recursion to the length of the string.

What we'll end up with is this:

/((((((((((((((((((((.[ ])*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*/
   ___________________/ \_____________________________________
  /                                                           \
  (((((((((((((((((((.[ ])*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*
   ___________________/ \_____________________________________
  /                                                           \
  (((((((((((((((((((.[ ])*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*
   ___________________/ \_____________________________________
  /                    .                                      \
                       .
                       .

... repeated 671088640 times, for a total of 12750684161 nestings - which quite thoroughly puts my previous attempt of 23 nestings to shame. Remarkably, perl doesn't even choke on this (once again, memory usage holds steady at about 1.3GB), although it will take quite a while before the first print statement is even issued.

From my previous analysis below, it can be concluded that the number of digits output will be on the order of (!12750684161)^671088640, where !k is the Left Factorial of k (see A003422). We can approximate this as (k-1)!, which is strictly smaller, but on the same order of magnitude.

And if we ask wolframalpha:

...which barely changes my score at all. I thought for sure that'd be at least 10↑↑5. I guess the difference between 10↑↑4 and 10↑↑4.1 is a lot bigger than you'd think.

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Perl - score ≈ 10↑↑4

$_=$^Fx($]<<-$]),/((((((((((((((((((((((.*.*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*)*(??{print})/

Abusing the perl regex engine to do some combinatorics for us. The embedded codeblock
(??{print}) will insert its result directly into the regex. Since $_ is composed entirely of 2s (and the result of print is always 1), this can never match, and sends perl spinning through all possible combinations, of which there's quite a few.

Constants used

• $^F - the maximum system file handle, typically 2.
• $] - the perl version number, similar to 5.016002.

$_ is then a string containing the digit 2 repeated 671088640 times. Memory usage is constant at about 1.3GB, output begins immediately.

Analysis

Let's define P_k(n) to be the number of times the print statement is executed, where k is the number of nestings, and n is the length of the string plus one (just because I don't feel like writing n+1 everywhere).

(.*.*)*
P₂(n) = [2, 8, 28, 96, 328, 1120, 3824, 13056, ...]

((.*.*)*)*
P₃(n) = [3, 18, 123, 900, 6693, 49926, 372615, 2781192, ...]

(((.*.*)*)*)*
P₄(n) = [4, 56, 1044, 20272, 394940, 7696008, 149970676, 2922453344, ...]

((((.*.*)*)*)*)*
P₅(n) = [5, 250, 16695, 1126580, 76039585, 5132387790, 346417023515, 23381856413800, ...]

(((((.*.*)*)*)*)*)*
P₆(n) = [6, 1452, 445698, 137050584, 42142941390, 12958920156996, ...]

((((((.*.*)*)*)*)*)*)*
P₇(n) = [7, 10094, 17634981, 30817120348, 53852913389555, ...]

etc. In general, the formula can be generalized as the following:

where

That is, the Left Factorial of k, i.e. the sum of all factorials less than k (see A003422).

I've been unable to determine closed forms for D_k and E_k, but this doesn't matter too much, if we observe that

and

With 23 nestings, this gives us an approximate score of:

This should be nearly exact, actually.

But to put this into a notation that's a bit easier to visualize, we can approximate the base of the inner exponent:

and then the exponent itself:

and then ask wolframalpha:

which you may as well just call 10↑↑4 and be done with it.

Answer 8

Javascript, 10↑↑↑↑210

100 chars:

z=~~Math.E+'';o={get f(){for(i=z;i--;)z+=i}};o.f;for(i=z;i--;)for(j=z;j--;)for(k=z;k--;)o.f;alert(z)

Based on the observation that maximally iterating f is the optimal way to go, I replaced the 13 calls to f with 3 levels of nested loops calling f, z times each (while f keeps increasing z).

I estimated the score analytically on a piece of paper—I'll type it up if anyone is interested in seeing it.

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Improved Score: 10↑↑13

Javascript, in exactly 100 characters, again:

z=~~Math.E+'';__defineGetter__('f',function(){for(i=z;i--;)z+=i});f;f;f;f;f;f;f;f;f;f;f;f;f;alert(z)

This improves my original answer in three ways—

1. Defining z on the global scope saves us from having to type o.z each time.

2. It's possible to define a getter on the global scope (window) and type f instead of o.f.

3. Having more iterations of f is worth more than starting with a larger number, so instead of (Math.E+'').replace('.','') (=2718281828459045, 27 chars), it's better to use ~~Math.E+'' (=2, 11 chars), and use the salvaged characters to call f many more times.

Since, as analyzed further below, each iteration produces, from a number in the order of magnitude M, a larger number in the order of magnitude 10^M, this code produces after each iteration

1. 210 ∼ O(10²)
2. O(10¹⁰²) ∼ O(10↑↑2)
3. O(10^10↑↑2) = O(10↑↑3)
4. O(10^10↑↑3) = O(10↑↑4)
5. O(10^10↑↑4) = O(10↑↑5)
6. O(10^10↑↑5) = O(10↑↑6)
7. O(10^10↑↑6) = O(10↑↑7)
8. O(10^10↑↑7) = O(10↑↑8)
9. O(10^10↑↑8) = O(10↑↑9)
10. O(10^10↑↑9) = O(10↑↑10)
11. O(10^10↑↑10) = O(10↑↑11)
12. O(10^10↑↑11) = O(10↑↑12)
13. O(10^10↑↑12) = O(10↑↑13)

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Score: ∼10^1010101016 ≈ 10↑↑6.080669764

Javascript, in exactly 100 characters:

o={'z':(Math.E+'').replace('.',''),get f(){i=o.z;while(i--){o.z+=i}}};o.f;o.f;o.f;o.f;o.f;alert(o.z)

Each o.f invokes the while loop, for a total of 5 loops. After only the first iteration, the score is already over 10^{42381398144233621}. By the second iteration, Mathematica was unable to compute even the number of digits in the result.

Here's a walkthrough of the code:

Init

Start with 2718281828459045 by removing the decimal point from Math.E.

Iteration 1

Concatenate the decreasing sequence of numbers,

• 2718281828459045
• 2718281828459044
• 2718281828459043
• ...
• 3
• 2
• 1
• 0

to form a new (gigantic) number,

• 271828182845904527182818284590442718281828459043...9876543210.

How many digits are in this number? Well, it's the concatenation of

• 1718281828459046 16-digit numbers
• 900000000000000 15-digit numbers
• 90000000000000 14-digit numbers,
• 9000000000000 13-digit numbers
• ...
• 900 3-digit numbers
• 90 2-digit numbers
• 10 1-digit numbers

In Mathematica,

In[1]:= 1718281828459046*16+Sum[9*10^i*(i+1),{i,-1,14}]+1
Out[1]= 42381398144233626

In other words, it's 2.72⋅10^{42381398144233625}.

Making my score, after only the first iteration, 2.72⋅10^{42381398144233619}.

Iteration 2

But that's only the beginning. Now, repeat the steps, starting with the gigantic number! That is, concatenate the decreasing sequence of numbers,

• 271828182845904527182818284590442718281828459043...9876543210
• 271828182845904527182818284590442718281828459043...9876543209
• 271828182845904527182818284590442718281828459043...9876543208
• ...
• 3
• 2
• 1
• 0

So, what's my new score, Mathematica?

In[2]:= 1.718281828459046*10^42381398144233624*42381398144233625 + Sum[9*10^i*(i + 1), {i, -1, 42381398144233623}] + 1

During evaluation of In[2]:= General::ovfl: Overflow occurred in computation. >>

During evaluation of In[2]:= General::ovfl: Overflow occurred in computation. >>

Out[2]= Overflow[]

Iteration 3

Repeat.

Iteration 4

Repeat.

Iteration 5

Repeat.

---

Analytical Score

In the first iteration, we calculated the number of digits in the concatenation of the decreasing sequence starting at 2718281828459045, by counting the number of digits in

• 1718281828459046 16-digit numbers
• 900000000000000 15-digit numbers
• 90000000000000 14-digit numbers,
• 9000000000000 13-digit numbers
• ...
• 900 3-digit numbers
• 90 2-digit numbers
• 10 1-digit numbers

This sum can be represented by the formula,

        

where Z denotes the starting number (e.g. 2718281828459045) and O_Z denotes its order of magnitude (e.g. 15, since Z ∼ 10¹⁵). Using equivalences for finite sums, the above can be expressed explicitly as

        

which, if we take 9 ≈ 10, reduces even further to

        

and, finally, expanding terms and ordering them by decreasing order of magnitude, we get

        

Now, since we're only interested in the order of magnitude of the result, let's substitute Z with "a number in the order of magnitude of O_Z," i.e. 10^O_Z—

        

Finally, the 2nd and 3rd terms cancel out, and the last two terms can be dropped (their size is trivial), leaving us with

        

from which the first term wins out.

Restated, f takes a number in the order of magnitude of M and produces a number approximately in the order of magnitude of M(10^M).

The first iteration can easily be checked by hand. 2718281828459045 is a number in the order of magnitude of 15—therefore f should produce a number in the order of magnitude of 15(10¹⁵) ∼ 10¹⁶. Indeed, the number produced is, from before, 2.72⋅10^{42381398144233625}—that is, 10^{42381398144233625} ∼ 10¹⁰¹⁶.

Noting that M is not a significant factor in M(10^M), the order of magnitude of the result of each iteration, then, follows a simple pattern of tetration:

1. 10¹⁶
2. 10¹⁰¹⁶
3. 10¹⁰¹⁰¹⁶
4. 10^10101016
5. 10^1010101016

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LaTeX sources

(Z-10^{\mathcal{O}_Z}+1)(\mathcal{O}_Z+1)+\sum_{k=0}^{\mathcal{O}_Z-1}{(9\cdot10^k(k+1))}+1

(Z-10^{\mathcal{O}_Z}+1)(\mathcal{O}_Z+1)+\frac{10-\mathcal{O}_Z10^{\mathcal{O}_Z}+(\mathcal{O}_Z-1)10^{\mathcal{O}_Z+1}}{9}+10^{\mathcal{O}_Z}

(Z-10^{\mathcal{O}_Z}+1)(\mathcal{O}_Z+1)+\mathcal{O}_Z10^{\mathcal{O}_Z}-\mathcal{O}_Z10^{\mathcal{O}_Z-1}+1

Z\mathcal{O}_Z+Z-10^{\mathcal{O}_Z}-\mathcal{O}_Z10^{\mathcal{O}_Z-1}+\mathcal{O}_Z+2

\mathcal{O}_Z10^{\mathcal{O}_Z}+10^{\mathcal{O}_Z}-10^{\mathcal{O}_Z}-\mathcal{O}_Z10^{\mathcal{O}_Z-1}+\mathcal{O}_Z+2

\mathcal{O}_Z10^{\mathcal{O}_Z}-\mathcal{O}_Z10^{\mathcal{O}_Z-1}

Answer 9

APL, 10↑↑3.4

Here's my revised attempt:

{⍞←⎕D}⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⍣n⊢n←⍎⎕D

100 char/byte* program, running on current hardware (uses a negligible amount of memory and regular 32-bit int variables) although it will take a very long time to complete.

You can actually run it on an APL interpreter and it will start printing digits. If allowed to complete, it will have printed a number with 10 × 123456789⁴⁴ digits.

Therefore the score is 10^{10 × 12345678944} / 100³ ≈ 10¹⁰³⁵³ ≈ 10↑↑3.406161

Explanation

• ⎕D is a predefined constant string equal to '0123456789'
• n←⍎⎕D defines n to be the number represented by that string: 123456789 (which is < 2³¹ and therefore can be used as a loop control variable)
• {⍞←⎕D} will print the 10 digits to standard output, without a newline
• {⍞←⎕D}⍣n will do it n times (⍣ is the "power operator": it's neither *, /, nor ^, because it's not a math operation, it's a kind of loop)
• {⍞←n}⍣n⍣n will repeat the previous operation n times, therefore printing the 10 digits n² times
• {⍞←n}⍣n⍣n⍣n will do it n³ times
• I could fit 44 ⍣n in there, so it prints n⁴⁴ times the string '0123456789'.

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
^{*: APL can be written in its own (legacy) single-byte charset that maps APL symbols to the upper 128 byte values. Therefore, for the purpose of scoring, a program of N chars that only uses ASCII characters and APL symbols can be considered to be N bytes long.}

Answer 10

Haskell, score: (2^2265536-3)/1000000 ≈ 2↑↑7 ≈ 10↑↑4.6329710779

o=round$sin pi
i=succ o
q=i+i+i+i
m!n|m==o=n+i
 |n==o=(m-i)!i
 |True=(m-i)!(m!(n-i))
main=print$q!q

This program is exactly 100 bytes of pure Haskell code. It will print the fourth Ackermann number, eventually consuming all available energy, matter and time of the Universe and beyond in the process (thus slightly exceeding the soft limit of 5 seconds).

Answer 11

Ruby, \$f_{\varphi(1,0,0)}(5)\$ (non-deterministic)

More precisely: \$f_{\varphi(\varphi(\varphi(\varphi(\varphi(16381,\omega^2),\omega^2),\omega^2),\omega^2),\omega^2)}(1048577)\$.

*h=[[[[n=$$]]]]
f=->a{b,c,d=a;b ?a==b ?~-a:a==a-[$.]?[f[b],d,[b,f[c],d]]:d:n+=n}
h=f[h]while h!=p(n)

Try it online!

_{*$. == 0 is a global variable.}

_{*Relies on $$, the process ID. I suppose the rules never said random behavior isn't allowed. I'm assuming the maximum PID to be 32768.}

Ruby, \$f_{\varphi(1,0,0)}(3)\$ (deterministic)

More precisely: \$f_{\varphi(\varphi(\varphi(\omega+\omega,\omega^2),\omega^2),\omega)}(16)\$.

*h=[[[n=-~$.]]]
f=->a{b,c,d=a;b ?a==b ?~-a:a==a-[$.]?[f[b],d,[b,f[c],d]]:d:n+=n}
h=f[h]while h!=p(n)

Try it online!

_{*Previously removed the prints, not sure why but they're back since removing them doesn't really let us improve much. Probably just +1 (or +2 at best) inside the \$f_{\varphi(1,0,0)}(n+\dots)\$.}

Ungolfed:

_{*See math explanation below.}

n = 32768         # n = "count"
h = [[[[[n]]]]]   # h0 = 32768
                  # h1 = "count op(h0) count"
                  # ...
                  # h = h5 = "count op(h4) count"

f = -> a {        # f[a] = reduce(a)
  b, c, d = a                     # truncate to first 3 elements
  if a == [] || a.nil?            # if a == "count"
    n += n                          # increment counter
    return n                        # and replace with counter
  elsif a.is_a? Integer           # if a == int
    return a - 1                    # decrement int
  elsif a != a-[0]                # if a == "++d"
    return d                        # return just d
  else                            # if a == "d op c" where b = op  (*note out of order)
    return [f[b], d, [b, f[c], d]]  # (d op reduce(c)) reduce(op) d
  end
}

while h != n  # reduce until h == n
  h = f[h]
  p(n)
end

_{*We write "x op(y) z" = [y, z, x] so that reducing it reduces y before z and so that [t] = "count op(t) count".}

Math explanation:

The code may be approximately understood as a combination of symbolic manipulation and recursive reduction of algebraic expressions.

Let ++a denote a+1, the successor of a.

We then define reduce("x") as x-1, meaning:

• reduce("++a") = a.

We then try to work out reduce("x") for algebraic expressions:

• reduce("a+b") = a + reduce("b") = reduce("++(a + reduce("b"))")
• reduce("a*b") = reduce("a*reduce("b") + a")
• reduce("a^b") = reduce("a^reduce("b") * a")

Generalizing this, we get:

• reduce("a op b") = reduce("(a op reduce(b)) reduce("op") a")

where:

• reduce("*") = "+"
• reduce("^") = "*"

and more generally:

• reduce("op(n)") = op(reduce("n"))
• x op(0) y = ++x
• x op(1) y = x+y
• x op(2) y = x*y
• x op(3) y = x^y

for simplicity we also define:

• x op 0 = ++x

We then introduce symbols. The only symbol is "count", where reduce("count") is the amount of times reduce has been called.

We then apply x = reduce(x) until x = 0 and print the count.

Example:

"count^3"

1:  reduce("count^3")
2:  reduce("count^reduce(3) reduce("^") count")
      Note: we evaluate right-to-left
3:  reduce("count^reduce(3) * count")
4:  reduce("count^2 * count")
5:  reduce("count^2 * reduce("count") reduce("*") count^2")
6:  reduce("count^2 * reduce("count") + count^2")
7:  reduce("count^2 * 6 + count^2")
8:  reduce("count^2 * 6 + reduce("count^2") reduce("+") ...")
9:  reduce("++(count^2 * 6 + reduce("count^2"))")
10: reduce("++(count^2 * 6 + count^reduce(2) reduce("^") count)")
11: reduce("++(count^2 * 6 + count^reduce(2) * count)")
12: reduce("++(count^2 * 5 + count^1 * count)")

"count^2 * 6 + count^1 * count"
(continue applying reduce(...) until you get 0)

Note that the growth relies on the 'lazy' evaluation of "count". Despite how simple it is to normally cube an integer, fully reducing "count^3" actually results in count > 10^10^10^10^10^10^5.

Let g(x, n) be the final count after fully reducing x with an initial count = n.

So from the previous example, g("count^3", 0) > 10^10^10^10^10^10^5.

For some intuition on the very fast growth, notice that g("a+b", n) > g(a, g(b, n)). This is because we have to reduce b before a. This means addition allows recursion, and we also know that multiplication expands into lots of addition.

Then note that

• g("count", n) ~ 2 * n
• g("count + count", n) ~ 2 * 2 * n = 4 * n
• g("count + count + ...", n) ~ 2 * 2 * ... * n = 2^k * n
• g("count * k", n) ~ 2^k * n
• g("count * count", n) ~ 2^n
• g("count ^ 2", n) ~ 2^n
• g("count ^ 2 + count ^ 2", n) ~ 2^2^n
• g("count ^ 2 * k", n) ~ 2^2^...^n
• g("count ^ 2 * count", n) ~ n↑↑n
• g("count ^ 3", n) ~ n↑↑n
• g("count ^ 4", n) ~ n↑↑↑n

Some other values of interest:

For the last 3, we define:

• h(0, y) = y
• reduce(h(x, y)) = "count op(h(reduce(x), y)) count"

g("count^k", n) ~ n↑↑...k arrows...↑↑n
g("count^count", n) ~ n↑↑...n arrows...↑↑n ~ Ack(n,n)
g("count^(++count)", 20) ~ Toeofdoom's a_20(1)
g("count^(++count)", 64) ~ Graham's number
g("count^(count+1)", 15) ~ eaglgenes101's number  (note count+1 turns into ++(++count))
g("count op(4) 2", 127) ~ col6y's number
g("(count op(4) 3) * count", 3) ~ my older/other number
g("count op(6) 1", 9) ~ r.e.s.'s number
g("(count op(6) 1) * (++count op(4) 1)", 17) ~ Peter Taylor's number
g(h(3, "count + count"), 126) ~ this deterministic number
g(h(5, 16381), 1048577) ~ this non-deterministic number
g("h("count", "count") * count^7", 256^26) ~ Steve H.'s number

Answer 12

Python, 2↑↑11 / 830584 ≈ 10↑↑8.632971 (Knuth up arrow notation)

print True<<(True<<(True<<(True<<(True<<(True<<(True<<(True<<(True<<(True<<True<<True)))))))))

Probably no computer has enough memory to successfully run this, but that's not really the program's fault. With the minimum system requirements satisfied, it does work.

Yes, this is doing bit shifting on boolean values. True gets coerced to 1 in this context. Python has arbitrary length integers.

Answer 13

GolfScript,   ≈ f_ε0(f_ε0(f_ε0(f_ε0(f_ε0(f_ε0(f_ε0(f_ε0(f_ε0(126)))))))))

This is shamelessly adapted from another answer by @Howard, and incorporates suggestions by @Peter Taylor.

[[[[[[[[[,:o;'~'(]{o:?~%{(.{[(]{:^o='oo',$o+o=<}{\(@\+}/}{,:^}if;^?):?)*\+.}do;?}:f~]f]f]f]f]f]f]f]f

My understanding of GolfScript is limited, but I believe the * and ^ operators above are not the arithmetic operators forbidden by the OP.

(I will happily delete this if @Howard wants to submit his own version, which would doubtless be superior to this one anyway.)

This program computes a number that's approximately f_ε0(f_ε0(f_ε0(f_ε0(f_ε0(f_ε0(f_ε0(f_ε0(f_ε0(126))))))))) -- a nine-fold iteration of f_ε0 -- where f_ε0 is the function in the fast-growing hierarchy that grows at roughly the same rate as the Goodstein function. (f_ε0 grows so fast that the growth rates of Friedman's n(k) function and of k-fold Conway chained arrows are virtually insignificant even in comparison to just a single non-iterated f_ε0.)

Answer 14

dc, 100 characters

[lnA A-=ilbA A-=jlaSalbB A--SbLndB A--SnSnlhxlhx]sh[LaLnLb1+sbq]si[LbLnLasbq]sjFsaFsbFFFFFFsnlhxclbp

Given enough time and memory, this will calculate a number around 15 ↑¹⁶⁶⁶⁶⁶⁵ 15. I had originally implemented the hyperoperation function, but it required too many characters for this challenge, so I removed the n = 2, b = 0 and n >= 3, b = 0 conditions, turning the n = 1, b = 0 condition into n >= 1, b = 0.

The only arithmetic operators used here are addition and subtraction.

EDIT: as promised in comments, here is a breakdown of what this code does:

[            # start "hyperoperation" macro
lnA A-=i     # if n == 0 call macro i
lbA A-=j     # if b == 0 call macro j
laSa         # push a onto a's stack
lbB A--Sb    # push b-1 onto b's stack
LndB A--SnSn # replace the top value on n with n-1, then push n onto n's stack
lhxlhx       # call macro h twice
]sh          # store this macro in h

[            # start "increment" macro (called when n=0, the operation beneath addition)
LaLnLb       # pop a, b, and n
F+sb         # replace the top value on b with b+15
q            # return
]si          # store this macro in i

[            # start "base case" macro (called when n>0 and b=0)
LbLnLa       # pop b, n, and a
sb           # replace the top value on b with a
q            # return
]sj          # store this macro in j

Fsa          # store F (15) in a
Fsb          # store F (15) in b
FFFFFFsn     # store FFFFFF "base 10" (150000+15000+1500+150+15=1666665) in n
lhx          # load and call macro h
lbp          # load and print b

As noted, this deviates from the hyperoperation function in that the base cases for multiplication and higher are replaced with the base case for addition. This code behaves as though a*0 = a^0 = a↑0 = a↑↑0 ... = a, instead of the mathematically correct a*0 = 0 and a^0 = a↑0 = a↑↑0 ... = 1. As a result, it computes values that are a bit higher than they should be, but that's not a big deal since we are aiming for bigger numbers. :)

EDIT: I just noticed that a digit slipped into the code by accident, in the macro that performs increments for n=0. I've removed it by replacing it with 'F' (15), which has the side effect of scaling each increment operation by 15. I'm not sure how much this affects the final result, but it's probably a lot bigger now.

Answer 15

GolfScript \$ \approx 3.673 \times 10^{374} = 10 \uparrow\uparrow 2.70760 \$

'~'(.`*

I think the * is allowed since it indicates string repetition, not multiplication.

Explanation: '~'( will leave 126 (the ASCII value of "~") on the stack. Then copy the number, convert it to a string, and do string repetition 126 times. This gives 126126126126... which is approximately 1.26 e+377. The solution is 7 characters, so divide by 7^3, for a score of approximately 3.673e+374

Answer 16

Japt, 47 bytes, BMS[26] ~ f_PTO(Z2)(26)

Code snippet

NUCLEAR BOMB ALERT

This is probably the largest number ever posted on this site that is less than Loader's Number. We utilize the Bashicu Matrix System, which is a ridiculously powerful notation that is proven to terminate, but whose strength is unknown. We know BMS is at least as strong as collapsing inaccessible cardinals with respect to standard fundamental sequences, though that was an understatement. BMS is conjectured to reach the limit of second order arithmetic.

In all of its full glory, here is the code. I could probably optimize this more (idk what to do with the other 50 bytes) or port this to a more familiar language. If you want to copy me, you are free to do so, but credit me:

;T°?UÊ?ßUcUsX=Uo)m_<X-X%BÊ?Z:Z+UÊ-X:T:ßBÊo cBÊo

Try it here!

The rest of this post will first show what BMS is, and how it is so strong. Then, we introduce Flattened Address Matric System (FAMS) and how it is another representation of BMS. Finally, we analyze the program part and show how this program calculates BMS.

Resources:

• Intro to BMS
• Visual BMS
• Expansion Playground
• Googology Discord (they have a section entirely dedicated to BMS)

What is BMS anyways?

Now although I could start listing expansion rules, I figured the resources above do a better job showing the power of BMS. Thus, I would focus more on the structuring on BMS.

The Bashicu Matrix System is a system of matrices with an expansion rule, causing these matrices to behave similar to ordinals. They are lexicographically ordered, and there are many subsystems: PrSS which is 1-row BMS, PSS which is 2-row BMS, TSS which is 3-row BMS, and so on.

We represent a BMS matrix like this: (a,b,c)(d,e,f)(g,h,j)(k,l,m),... which corresponds to 3-row BMS, though extra rows are possible. Each 3-tuple corresponds to a BMS column. Moreover, each element/number has exactly one parent corresponding to its row level within its column. Thus, try to think of BMS as multitrees or a direct acyclic graph; that is trees that are layered on top of each other, and each column having a level 1 parent, a level 2 parent, and so on.

It could be possible that a parent does not exist for a particular level within a column. This is usually represented in BMS by a zero, or a self-address in FAMS. Let me show an example.

Here, you can see Level 1 parents in black connections, the Level 2 parents in red connections, and the Level 3 parents in blue connections.

Now here is the neat part: when expanding a BMS matrix, all of the parent-child relationships are preserved except possibly the rightmost node. Read the resources above more to understand BMS more.

Introducing FAMS

We can make BMS easier to understand by replacing each element with an address of its parent (i.e. what column its parent refers to). Parent addresses are always smaller than their children. Some elements may not have parents; in that case we let those elements be their own parents (so we have a self-referential address). This notation is called AMS, or Address Matrix System.

To go from AMS to FAMS, we flatten the entire matrix, so (a,b,c)(d,e,f)(g,h,i),... becomes {3}[a,b,c,d,e,f,g,h,i,...], where the {3} indicates that there are three rows.

FAMS has a few more simplifications - mainly that trailing zeroes in BMS, or trailing self-references in FAMS, are omitted.

Some examples of BMS matrices and their correspondence to FAMS:

• (0,0)(1,1) = {2}[0,1,0,1]
• (0,0)(1,1)(2,0) = {2}[0,1,0,1,2]
• (0,0)(1,1)(2,0)(1,1) = {2}[0,1,0,1,2,5,0,1]
• (0,0)(1,1)(2,1)(0,0)(1,0) = {2}[0,1,0,1,2,3,6,7,6]
• (0,0,0)(1,1,0)(2,2,0) = {3}[0,1,2,0,1,5,3,4]
• (0,0,0)(1,1,1) = {3}[0,1,2,0,1,2]

Here is code converting between FAMS and BMS

function FAMS_to_BMS(arr,rows) {
    
    for (let i=0;i<arr.length;i++) { //loop over all
        arr[i]=arr[i]==i?0:1+arr[arr[i]]       
    }
    
    // Initialize the resulting array of blocks
    let result = [];
    
    // Loop through the array in steps of 'size'
    for (let i = 0; i < arr.length; i += rows) {
        // Extract a block of 'size' elements
        let block = arr.slice(i, i + rows);
        
        // If the block is smaller than the desired size, pad it with zeroes
        while (block.length < rows) {
            block.push(0);
        }
        
        // Add the block to the result array
        result.push(block);
    }
    
    return result;
}

function parent(matrix,row,col) { //returns column that is a parent
    if (row==-1) {
        return col-1
    }
    let c = col
    let element = matrix[col][row]
    while(c >= 0 && matrix[c][row]>=element) {
        c = parent(matrix,row-1,c)
    }
    return c >= 0 ? c : -1 // cannot find
}

function BMS_to_FAMS(matrix,rows) {
    arr = []
    for (let i=0;i<matrix.length;i++) {
        for(j=0;j<rows;j++) {
            let p = parent(matrix,j,i)
            arr.push(p==-1?i*rows+j:p*rows+j)
        }
    }
    while (arr[arr.length-1]==arr.length-1) {
        arr.pop()
    }
    return arr
}

More importantly, expansion in FAMS is ridiculously easy:

function Expand(matrix,rows,amount) {
    child = matrix.pop()
    ascension = matrix.length - child 
    pre_expand = Array(amount).map((x,y)=>y+1) //array 1,2...amount
    append = pre_expand.flatMap(q=>
        matrix.slice(child).map(x=>x>=child-child%rows?x+ascension*q:x))
    return matrix.concat(append)
}

In the code below, you would find out that we expand with q=1. This actually corresponds to the Slow Growing Hierarchy, with base equal to 2. It turns out that this expansion is nondegenerate starting from [0,1,2,...,n,0,1,2,...,n], since at each non-successor step we remove 1 element and add a (multiple of number of rows) number of elements. Another possible concern is that SGH is much weaker than the FGH, though there are "catching points" where SGH has similar strength to the FGH. The first catching point starts at ψ(Ω_ω), but BMS goes waaaaaaay beyond that.

Analysis

This is best done by converting to ungolfed JavaScript.

ALPHABET = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"                   // ; (built-in, variable B)
rows = 26                                                 // BÊ (our "constant", we just take length of alphabet)
T = 0                                                     // (implicit start)
matrix = Array(rows).fill("").map((x,y)=>y)               // BÊo (creates 0,1,2,...,25 array)
matrix = matrix.concat(matrix)                            // cBÊo (duplicates array)
while (matrix.length) {                                   // ß (run program again loop)
    child = matrix.pop()                                  // X=Uo  (sets child)
    ascension = matrix.length - child                     // UÊ-X (sets ascension)
    matrix = matrix.concat(matrix.slice(child)            // UcUsX=Uo)
        .map(x=>x>=child-child%rows?x+ascension:x))       // m_<X-X%BÊ?Z:Z+UÊ-X
    T++                                                   // T°
}
alert(T)               //implicit Japt output, using Japt st:out

Immediately, one sees that this program does not implement Bashicu Matrix System, but rather Flattened Address Matrix System (FAMS), which is equivalent to BMS. In FAMS, there is a constant which denotes the number of rows (rows=26 here), which means that every 26 entries form a single column. Each entry is an address that refers to the 0-index of its parent, though some entries have no parents, in which their addresses refers to themselves.

The large number in this post is {26}(0,1,2,3,...,25;0,1,2,3,...,25), which corresponds to (0,0,...,0,0)(1,1,...,1,1) in BMS. This number is massive. To really understand the size of this number, we need to compare BMS with known ordinal correspondences.

Size Comparison

• (0) corresponds to ordinal 1
• (0)(0) corresponds to ordinal 2
• (0)(1) corresponds to ordinal ω
• (0)(1)(0)(1) corresponds to ordinal ω2, beating Graham's Number
• (0)(1)(1) corresponds to ordinal ω²
• (0)(1)(1)(1) corresponds to ordinal ω³
• (0)(1)(2) corresponds to ordinal ω^ω
• (0)(1)(2)(3) corresponds to ordinal ω^ωω
• (0,0)(1,1) corresponds to ordinal φ(1,0)
• (0,0)(1,1)(1,0)(2,1) corresponds to ordinal φ(1,0)^2
• (0,0)(1,1)(1,1) corresponds to ordinal φ(1,1)
• (0,0)(1,1)(2,0) corresponds to ordinal φ(1,ω)
• (0,0)(1,1)(2,1) corresponds to ordinal φ(2,0)
• (0,0)(1,1)(2,1)(2,1) corresponds to ordinal φ(3,0)
• (0,0)(1,1)(2,1)(3,0) corresponds to ordinal φ(ω,0)
• (0,0)(1,1)(2,1)(3,1) corresponds to ordinal φ(1,0,0)
• (0,0)(1,1)(2,1)(3,1)(4,0) corresponds to the Small Veblen Ordinal, beating all other entries on this list
• (0,0)(1,1)(2,1)(3,1)(4,1) corresponds to the Large Veblen Ordinal, beating TREE(3)
• (0,0)(1,1)(2,2) corresponds to the Bachmann-Howard Ordinal
• (0,0)(1,1)(2,2)(3,3) corresponds to ψ(Ω₃), collapsing Extended Buchholz OCF
• (0,0,0)(1,1,1) corresponds to ψ(Ω_ω); the limit of the Π₁-CA₀ subsystem of Z₂ and also the first Catching ordinal
• (0,0,0)(1,1,1)(2,1,0) corresponds to ψ(Ω_ω*Ω)
• (0,0,0)(1,1,1)(2,1,0)(1,1,1) corresponds to ψ(Ω_ω^2), which is the first time BMS really takes off
• (0,0,0)(1,1,1)(2,1,1)(3,1,0)(2,0,0) corresponds to ψ(ψ_I(0))
• (0,0,0)(1,1,1)(2,1,1)(3,1,1) corresponds to ψ(ψ_Iω(0)) ...
• (0,0,0)(1,1,1)(2,2,1)(3,0,0) corresponds to the limit of Strong Array Notation ...
• (0,0,0)(1,1,1)(2,2,2) is expected to correspond to the limit of Π₂-CA₀ subsystem of Z₂
• (0,0,0,0,0,...)(1,1,1,1,1,...) is expected to correspond to the limit of Z₂

Conclusion

The Bashicu Matrix System is known to be very powerful. Analysis of BMS is still a very active field in Googology, and there are a few extensions (such as the Y-sequence). This BMS number is smaller than Loader's Number, since BMS is bounded above by the limit of second order arithmetic.

Answer 17

Ruby, probabilistically infinite, 54 characters

x='a'.ord
x+=x while x.times.map(&:rand).uniq[x/x]
p x

x is initialized to 97. We then iterate the following procedure: Generate x random numbers between 0 and 1. If they are all the same, then terminate and print x. Otherwise, double x and repeat. Since Ruby's random numbers have 17 digits of precision, the odds of terminating at any step are 1 in (10e17)^x. The probability of terminating within n steps is therefore the sum for x=1 to n of (1/10e17)^(2^n), which converges to 1/10e34. This means that for any number, no matter how large, it is overwhelmingly unlikely that this program outputs a lesser number.

Now, of course, the philosophical question is whether a program that has less than a 1 in 10^34 chance of terminating by step n for any n can be said to ever terminate. If we assume not only infinite time and power, but that the program is given the ability to run at increasing speed at a rate that exceeds the rate at which the probability of terminating decreases, we can, I believe, in fact make the probability of terminating by time t arbitrarily close to 1.

Answer 18

No more limit on runtime? OK then.

Does the program need to be runnable on modern computers?

Both solutions using a 64-bit compile, so that long is a 64-bit integer.

C: greater than 10^(264-1)264, which is itself greater than 10^{10355393490465494856447} ≈ 10↑↑4.11820744

long z;void f(long n){long i=z;while(--i){if(n)f(n+~z);printf("%lu",~z);}}main(){f(~z);}

88 characters.

To make these formulas easier, I'll use t = 2^64-1 = 18446744073709551615.

main will call f with a parameter of t, which will loop t times, each time printing the value t, and calling f with a parameter of t-1.

Total digits printed: 20 * t.

Each of those calls to f with a parameter of t-1 will iterate t times, printing the value t, and calling f with a parameter of t-2.

Total digits printed: 20 * (t + t*t)

I tried this program using the equivalent of 3-bit integers (I set i = 8 and had main call f(7)). It hit the print statement 6725600 times. That works out to 7^8 + 7^7 + 7^6 + 7^5 + 7^4 + 7^3 + 7^2 + 7 Therefore, I believe that this is the final count for the full program:

Total digits printed: 20 * (t + t*t + t^3 + ... + t^(t-1) + t^t + t^(2^64))

I'm not sure how to calculate (2⁶⁴-1)²⁶⁴. That summation is smaller than (2⁶⁴)²⁶⁴, and I need a power of two to do this calculation. Therefore, I'll calculate (2⁶⁴)^264-1. It's smaller than the real result, but since it's a power of two, I can convert it to a power of 10 for comparison with other results.

Does anyone know how to perform that summation, or how to convert (2⁶⁴-1)²⁶⁴ to 10ⁿ?

20 * 2^64^(2^64-1)
20 * 2^64^18446744073709551615
20 * 2^(64*18446744073709551615)
20 * 2^1180591620717411303360
10 * 2^1180591620717411303361
divide that exponent by log base 2 of 10 to switch the base of the exponent to powers of 10.
1180591620717411303361 / 3.321928094887362347870319429489390175864831393024580612054756 = 
355393490465494856446
10 * 10 ^ 355393490465494856446
10 ^ 355393490465494856447

But remember, that's the number of digits printed. The value of the integer is 10 raised to that power, so 10 ^ 10 ^ 355393490465494856447

This program will have a stack depth of 2^64. That's 2^72 bytes of memory just to store the loop counters. That's 4 Billion Terabytes of loop counters. Not to mention the other things that would go on the stack for 2^64 levels of recursion.

Edit: Corrected a pair of typos, and used a more precise value for log2(10).

Edit 2: Wait a second, I've got a loop that the printf is outside of. Let's fix that. Added initializing i.

Edit 3: Dang it, I screwed up the math on the previous edit. Fixed.

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This one will run on modern computers, though it won't finish any time soon.

C: 10^10^136 ≈ 10↑↑3.329100567

#define w(q) while(++q)
long a,b,c,d,e,f,g,x;main(){w(a)w(b)w(c)w(d)w(e)w(f)w(g)printf("%lu",~x);}

98 Characters.

This will print the bitwise-inverse of zero, 2^64-1, once for each iteration. 2^64-1 is a 20 digit number.

Number of digits = 20 * (2^64-1)^7 = 14536774485912137805470195290264863598250876154813037507443495139872713780096227571027903270680672445638775618778303705182042800542187500

Rounding the program length to 100 characters, Score = printed number / 1,000,000

Score = 10 ^ 14536774485912137805470195290264863598250876154813037507443495139872713780096227571027903270680672445638775618778303705182042800542187494

Answer 19

C

(With apologies to Darren Stone)

long n,o,p,q,r;main(){while(--n){while(--o){while(--p){while(--q){while(--r){putchar('z'-'A');}}}}}}

n = 2^64 digit number (9...)

l = 100 chars of code

score ≈ 1e+2135987035920910082395021706169552114602704522356652769947041607822219725780640550022962086936570 ≈ 10↑↑3.2974890744

[ Score = n^5/l^3 = (10^(2^320)-1)/(100^3) = (10^2135987035920910082395021706169552114602704522356652769947041607822219725780640550022962086936576-1)/(10^6) ]

Note that I deserve to be flogged mercilessly for this answer, but couldn't resist. I don't recommend acting like me on stackexchange, for obvious reasons. :-P

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EDIT: It would be even harder to resist the temptation to go with something like

long n;main(){putchar('z'-'A');putchar('e');putchar('+');while(--n){putchar('z'-'A');}

...but I suppose that an intended but unspecified rule was that the entire run of digits making up the number must be printed.

Answer 20

Pyth, f_ψ(Ω^Ω)+7(256²⁶)/1000000

=CGL&=.<GG?+Ibt]Z?htb?eb[XbhhZyeby@bhZhb)hbXbhhZyeb@,tb&bG<bhZ=Y[tZGG)VGVGVGVGVGVGVGVG=[YYY)uyFYHpG)

SimplyBeautifulArt has a fantastic explanation of a function that both of our solutions share, namely that I define a function y[b,G] with a global =g[h,n]. The major differences are as follows:

• I begin my value at the base 256 representation of the ASCII codes for the string "abcdefghijklmnopqrstuvwxyz".

• I use bit shifting .< to increase the value of G instead of addition.

• G gets incremented each step that y gets run including the recursive cases, while SBA's n gets incremented outside of his g and thus is updated less frequently.

• Instead of being satisfied with simply doing [[],n,n] (which my code represents as Y=[-1,G,G], I nest Y into itself as Y=[Y,Y,Y] G times, increasing G by calling Y=y[Y,G] 2(x+1) times, where x is as many times as it takes the new version of Y to reach 0 by these repeated applications. Y's value doesn't actually get reset to 0, because we calculate x by counting upwards using the reduce-until-seen-before builtin u.

• I then wrap the entirety of the above into 7 for loops (VG), which will repeat the key inner loop that nests Y and increases G until repeat(y[Y,G],Y=0 G times.

• Unwrapping all of these nested Ys results in a number that, to the best of my understanding, blows all the other solutions out of the water. I'll hold off on adding my solution to the leaderboard until someone else confirms my math.

Answer 21

New Ruby: score ~ f_ω^ω2+1(126²2¹²⁶)

where f_α(n) is the fast growing hierarchy.

n=?~.ord;H=->a{b,*c=a;eval"b ?H[b==$.?c:[b==~$.?n:b-(b<=>$.)]*n+c]:p(n+=n);"*n};eval"H[~n];".*n*n<<n

Try it online!

The *n are just string and array multiplication, so they should be fine.

Ungolfed code:

n = 126
H =-> a {
    b, *c = a
    n.times{
        case b
        when nil
            puts(n += n)
        when 0
            H[c]
        when -1
            H[[n]*n+c]
        else
            H[[b.-b<=>0]*n+c]
        end
    }
}
(n*n<<n).times{H[~n]}

where b.-b<=>0 returns an integer that is 1 closer to 0 than b.

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Explanation:

It prints n at the start of every call of H.

H[[]] doubles n (n times), i.e. n = n<<n.

H[[0,a,b,c,...,z]] calls H[[a,b,c,...,z]] (n times).

H[[k+1,a,b,c,...,z]] calls H[[k]*n+[a,b,c,...,z]] (n times), where [k]*n = [k,k,...,k].

H[[-1,a,b,c,...,z]] calls H[[n]*n+[a,b,c,...,z]] (n times).

H[[-(k+1),a,b,c,...,z]] calls H[[-k]*n+[a,b,c,...,z]] (n times).

H[k] = H[[k]].

My program initializes n = 126, then calls H[-n-1] 126²2¹²⁶ times.

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Examples:

H[[0]] will call H[[]] which applies n = n<<n (n times).

H[[0,0]] will call H[[0]] (n times).

H[[1]] will call H[[0]*n] (n times).

H[[-1]] will call H[[n]*n] (n times).

H[[-1,-1]] will call H[[n]*n+[-1]] (n times).

H[[-3]] will call H[[-2]*n] (n times).

Try it online!

---

See revisions for other cool things.

Answer 22

Haskell - Ackermann function applied to its result 20 times - 99 characters

This is the best haskell solution I can come up with based on the ackermann function - you may notice some similarities to n.m.'s solution, the i=round$log pi was inspired from there and the rest is coincidence :D

i=round$log pi
n?m|m<i=n+i|n<i=i?(m-i)|True=(n-i)?m?(m-i)
a n=n?n
b=a.a.a.a
main=print$b$b$b$b$b$i

It runs the ackermann function on itself 20 times, starting at one, the sequence being

• 1,
• 3,
• 61,
• a(61,61),
• a(a(61,61),a(61,61)) --- we will call this a₂(61), or a₄(1) ---
• a₃(61)
• ...
• a₁₈(61), or a₂₀(1). I think this is approximately g₁₈ (see below).

As for the estimation, wikipedia says:

a(m,n) = 2↑^m-2(n+3) - 3

From this we can see a3(1) = a(61,61) = 2↑⁵⁹64 + 3, which is clearly greater than g1 = 3↑⁴3, unless the 3 at the start is far more important than I think. After that, each level does the following (discarding the insignificant constants in a_n):

• g_n = 3↑^g_n-13
• a_n ~= 2↑^a_n-1(a_n-1)

If these are approximately equivalent, then a₂₀(1) ~= g₁₈. The final term in a_n, (a_n-1) is far greater than 3, so it is potentially higher than g₁₈. I'll see if I can figure out if that would boost it even a single iteration and report back.

Answer 23

x86 machine code - 100 bytes (Assembled as MSDOS .com file)

Note: may bend the rules a little

This program will output 2^(65536*8+32) nines which would put the score at (10^2524320-1) / 1000000

As a counter this program uses the entire stack (64kiB) plus two 16bit registers

Assembled code:

8A3E61018CD289166101892663018ED331E4BB3A01438A2627
018827A0300130E4FEC4FEC4FEC410E4FEC400E431C95139CC
75FB31D231C931DBCD3F4175F94275F45941750839D4740D59
4174F85131C939D475F9EBDD8B266301A161018ED0C3535858

Assembly:

ORG 0x100

SECTION .TEXT
            mov bh, [b_ss]
            mov dx, ss
            mov [b_ss], dx
            mov [b_sp], sp
            mov ss, bx
            xor sp, sp
            mov bx, inthackdst
            inc bx
            mov ah, [inthacksrc]
            mov [bx], ah
            mov al, [nine]
            xor ah, ah
            inc ah
            inc ah
            inc ah
inthacksrc: adc ah, ah
            inc ah
            add ah, ah
            xor cx, cx
fillstack:  push cx
nine:       cmp sp, cx
            jnz fillstack
regloop:    xor dx, dx
dxloop:     xor cx, cx
cxloop:     xor bx, bx
inthackdst: int '?'
            inc cx
            jnz cxloop
            inc dx
            jnz dxloop
            pop cx
            inc cx
            jnz restack
popmore:    cmp sp, dx
            jz end
            pop cx
            inc cx
            jz popmore
restack:    push cx
            xor cx, cx
            cmp sp, dx
            jnz restack
            jmp regloop
end:        mov sp, [b_sp]
            mov ax, [b_ss]
            mov ss, ax
            ret

b_ss:       dw 'SX'
b_sp:       db 'X'

Answer 24

R - 49 41 characters of code, 4.03624169270483442*10^5928 ≈ 10↑↑2.576681348

set.seed(T)
cat(abs(.Random.seed),sep="")

will print out [reproducing here just the start]:

403624169270483442010614603558397222347416148937479386587122217348........

Answer 25

ECMAScript 6 - 10^3↑↑↑↑3 / 884736

(3↑↑↑↑3 is G(1) where G(64) is Graham's number)

u=-~[v=+[]+[]]+[];v+=e=v+v+v;D=x=>x.substr(u);K=(n,b)=>b[u]?n?K(D(n),K(n,D(b))):b+b+b:e;u+K(v,e)

Output: 10^3↑↑↑↑3

Hints:

G is the function where G(64) is Graham's number. Input is an integer. Output is a unary string written with 0. Removed for brevity.

K is the Knuth up-arrow function a ↑ⁿ b where a is implicitly 3. Input is n, a unary string, and b, a unary string. Output is a unary string.

u is "1".

v is "0000", or G(0)

e is "000".

Answer 26

C

The file size is 45 bytes.

The program is:

main(){long d='~~~~~~~~';while(--d)printf("%ld",d);}

And the number produced is larger than 10^(10^(10^1.305451600608433)).

The file I redirected std out to is currently over 16 Gb, and still growing.

The program would terminate in a reasonable amount of time if I had a better computer.

My score is uncomputable with double precision floating point.

Answer 27

GNU Bash, 10^40964096² / 80^3 ≈ 10↑↑2.072820169

C=$(stat -c %s /) sh -c 'dd if=/dev/zero bs=$C$C count=$C$C|tr \\$((C-C)) $SHLVL'

C = 4096 on any reasonable system. SHLVL is a small positive integer (usually either 1 or 2 depending on whether /bin/sh is bash or not).

64 bit UNIX only:

Score: ~ 10^(40964096409640964096*40964096409640964096) / 88^3

C=$(stat -c %s /) sh -c 'dd if=/dev/zero bs=$C$C$C$C$C count=$C$C$C$C$C|tr \\$((C-C)) $SHLVL'

Answer 28

C, 10^10^2485766 ≈ 10↑↑3.805871804

unsigned a['~'<<'\v'],l='~'<<'\v',i,z;main(){while(*a<~z)for(i=l;printf("%u",~z),i--&&!++a[i];);}

We create an array of 258048 unsigned integers. It couldn't be unsigned longs because that made the program too long. They are unsigned because I don't want to use undefined behavior, this code is proper C (other than the lack of return from main()) and will compile and run on any normal machine, it will keep running for a long time though. This size is the biggest we can legally express without using non-ascii characters.

We loop through the array starting from the last element. We print the digits of 2^32-1, increment the element and drop the loop if the element hasn't wrapped to 0. This way we'll loop (2^32 - 1)^254048 = 2^8257536 times, printing 10 digits each time.

Here's example code that shows the principle in a more limited data range:

#include <stdio.h>
unsigned int a[3],l=3,i,f;

int
main(int argc, char *argc){
        while (*a<4) {
        for (i = l; i-- && (a[i] = (a[i] + 1) % 5) == 0;);
            for (f = 0; f < l; f++)
                printf("%lu ", a[f]);
            printf("\n");
        }
}

The result is roughly 10^10^2485766 divided by a million which is still roughly 10^10^2485766.

Answer 29

Powershell (2.53e107976 / 72³ = 6.78e107970 ≈ 10↑↑1.701853371)

This takes far more than 5 seconds to run.

-join(-split(gci \ -r -EA:SilentlyContinue|select Length))-replace"[^\d]"

It retrieves and concatenates the byte length of every file on your current drive. Regex strips out any non-digit characters.

Answer 30

Python 3, score = ack(126,126)/100^3

g=len('"');i=ord('~');f=lambda m,n:(f(m-g,f(m,n-g))if n else f(m-g,g))if m else n+g
print(f(i,i))

The f function is the ackermann function, which i have just enough space to invoke.

Edit: previously "else n+1", which was in violation of challenge rules- kudos to Simply Beautiful Art.