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.


  1. Steven H., Pyth \$\approx f_{\varphi(1,0,0)+7}(256^{26})\$
  2. Simply Beautiful Art, Ruby \$\approx f_{\varphi(1,0,0)}(3)\$
  3. Peter Taylor, GolfScript \$\approx f_{\varepsilon_0+\omega+1}(17)\$
  4. r.e.s., GolfScript \$\approx f_{\epsilon_0}^9(126)\approx f_{\epsilon_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\$


  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.

  • 23
    \$\begingroup\$ Has someone explicitly said "base 10" yet? \$\endgroup\$
    – keshlam
    Commented Jan 9, 2014 at 14:42
  • 2
    \$\begingroup\$ Does the large number count if it's say 12e10 (12*10^10) as 12*10^10? \$\endgroup\$
    – hichris123
    Commented Jan 9, 2014 at 19:36
  • 6
    \$\begingroup\$ I think a better constraint instead of forbidding *, /, and ^, would've been to allow only linear operations, e.g. +, -, ++, --, +=, -=, etc. Otherwise, coders can take advantage of Knuth's up-arrow/Ackermann library functions if made available in their language of choice, which seems like cheating. \$\endgroup\$ Commented Jan 10, 2014 at 0:19
  • 25
    \$\begingroup\$ I'm still waiting to see someone earn footnote [4]. \$\endgroup\$ Commented May 18, 2017 at 15:41
  • 8
    \$\begingroup\$ Is a zero-byte program valid? \$\endgroup\$
    – pxeger
    Commented Oct 21, 2021 at 15:43

95 Answers 95


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


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.

  • 26
    \$\begingroup\$ Sheer unadulterated genius! \$\endgroup\$ Commented Jan 9, 2014 at 0:34
  • 9
    \$\begingroup\$ I think you're supposed to cube the solution length in which I get about 321 as score Your printed number will be divided for the number of bytes you used for your solution^3. \$\endgroup\$
    – Cruncher
    Commented Jan 9, 2014 at 14:16
  • 1
    \$\begingroup\$ 77 upvotes, and yet... I note the score is 321... \$\endgroup\$ Commented May 19, 2017 at 0:54
  • 9
    \$\begingroup\$ @SimplyBeautifulArt, it's not the score, but the journey. :-D \$\endgroup\$ Commented May 19, 2017 at 2:04
  • 7
    \$\begingroup\$ Apparently so, one that gave many a good laugh. Now if only we could get this up to the leaderboard...someone needs to earn the "irreparable damage" tag ;) \$\endgroup\$ Commented May 19, 2017 at 11:10

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 101010383.

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 1010⋰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. ;-)

  • 10
    \$\begingroup\$ "...No computer ever built could store a number that big... Correct me if I'm wrong, but I don't think that applies here. Isn't it just repeatedly "storing" and printing 3 digits at a time (?) so no need to store the final result. \$\endgroup\$ Commented Jan 9, 2014 at 19:16
  • 14
    \$\begingroup\$ @KevinFegan: That is true — the number is incredibly repetitive, so it would be easy to compress. But then we're no longer really storing the number itself, but rather some abstract formula from which the number may, theoretically, be computed; indeed, one of the most compact such formulas is probably the GolfScript program above that generates it. Also, once we go a step further to the next program, even "printing" the digits one at a time before discarding them becomes impractical — there's simply no known way to carry out that many steps of classical computation in the universe. \$\endgroup\$ Commented Jan 9, 2014 at 22:46
  • 6
    \$\begingroup\$ How about actually pushing this to the limit, see how big precisely you can really make it in GolfScript within 100 chars? As it stands, your result is less than Graham's number (which my Haskell solution "approximates"), but as you say GolfScript can probably go even further. \$\endgroup\$ Commented Jan 10, 2014 at 0:55
  • 3
    \$\begingroup\$ @leftaroundabout: I managed to write a Conway arrow notation evaluator in 80 chars of GolfScript, although it doesn't pass all the requirements of this challenge (it uses numeric constants and arithmetic operators). It could probably be improved, but I thought I might pose that as a new challenge. \$\endgroup\$ Commented Jan 10, 2014 at 0:59
  • 1
    \$\begingroup\$ @Tobia: GolfScript integers are bignums, so in principle they have no size limit. That said, the reference interpreter, which is written in Ruby, does limit string length to less than 2^31 or 2^63 bytes (depending on platform), so you're right, that step is going to fail on the reference interpreter. That said, the string length limit is not part of the GolfScript spec (which, admittedly, is pretty vague), so it's arguably a bug in the interpreter rather than a fundamental limitation. \$\endgroup\$ Commented Jan 19, 2014 at 19:39

JavaScript 44 chars

This may seem a little cheaty:


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

  • 11
    \$\begingroup\$ No rules bent at all: ;) * Can't use 0123456789 [check] * Use any language in which digits are valid characters; [check] * You can use mathematic/physic/etc. constants <10. [check, used 2] * Recursion is allowed but the generated number needs to be finite; [check, no recursion] Can't use *, /, ^; [check] Your program can output more than one number; [check] You can concatenate strings; [check] Your code will be run as-is; [check] Max code length: 100 bytes; [check] Needs to terminate w/i 5 sec [check] \$\endgroup\$ Commented Jan 9, 2014 at 23:45
  • \$\begingroup\$ Shave off 2 characters by passing "." to replace instead of /\./g \$\endgroup\$
    – gengkev
    Commented Jan 12, 2014 at 21:10
  • 1
    \$\begingroup\$ @gengkev Sadly, only using .replace(".","") only removes the first . character; I have to use the global replace to replace ALL . characters from the string... \$\endgroup\$ Commented Jan 12, 2014 at 23:55
  • \$\begingroup\$ You can do m=Math,p=m.PI,e=m.E,s="",alert((p*p*p+s+e*e*e).replace(/\./g,s)) instead, your score is then 3100627668029981620085536923187664 / 63^3 = 1.240017943838551e+28 \$\endgroup\$
    – A.M.K
    Commented Jan 22, 2014 at 2:58
  • 1
    \$\begingroup\$ @Cory For one, I'm not going to repeat a constant, otherwise everyone would be using it... Second, I really don't have a second argument... \$\endgroup\$ Commented Jun 19, 2014 at 23:28

C, score = 101097.61735/983 ≈ 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 101097.61735.

l = 98 chars of code

score = n/l3 = 101097.61735/983

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

  • 2
    \$\begingroup\$ In C, 'z' is the constant value 122. Right? \$\endgroup\$
    – primo
    Commented Jan 9, 2014 at 5:31
  • 1
    \$\begingroup\$ @klingt.net, it should not matter. The while loop will see 2^64-1 iterations, regardless of sign. With a signed value, n gets progressively smaller (more negative), then wraps to maximum positive, then continues to decrease until it hits 0 and exits. \$\endgroup\$ Commented Jan 9, 2014 at 9:53
  • 1
    \$\begingroup\$ i think printf("%d",n) will make the number much larger. Also, 64-bit computer doesn't mean 64-bit longs, for example Windows use the LLP64 model so long is still 32 bits \$\endgroup\$
    – phuclv
    Commented Jan 9, 2014 at 10:59
  • 3
    \$\begingroup\$ it should not matter It does. Signed integer overflow is undefined behavior in C, so it's impossible to predict what will happen when your code is executed. It might violate the finitude requirement. \$\endgroup\$
    – Dennis
    Commented Jan 9, 2014 at 14:55
  • 2
    \$\begingroup\$ Hi @primo. Thanks for putting in the time to check this. Appreciate it. I see what you're saying. My final exponent is wrong. It should be 2.0 instead of 97. 10^10^10^2.00 = 10^10^97.6. I will reflect that in my score now. \$\endgroup\$ Commented Jan 15, 2014 at 21:49

GolfScript; score at least fε_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:


computes g(g(1)) = g(5) where g(x) = worm_lifetime(x, [x]) grows roughly as fε0 (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


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


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).

}*'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 >> fε_0+ω+1(i).

  • \$\begingroup\$ If I'm not mistaken, the number your program computes (call it n) can be written n = f^9 (g(3)), where f(x) = g^(4x) (x), and g(x) is the lifetime of worm [x]. If we treat g as being approximately the same as f_eps_0 in the fast-growing hierarchy, then my "back-of-envelope" calculations show that f_(eps_0 + 2)(9) < n < f_(eps_0 + 2)(10). Of course it's the current winner -- by far. \$\endgroup\$
    – r.e.s.
    Commented Jan 20, 2014 at 18:58
  • \$\begingroup\$ @r.e.s., I think that's underestimating it quite a lot. .{foo}* maps x to foo^x (x). If we take h_0 (x) = g^4 (x) and h_{i+1} (x) = h_i^x (x) then the value calculated is h_9 (g(3)). Your f(x) = g^(4x) (x) = h_0^x (x) = h_1 (x). \$\endgroup\$ Commented Jan 20, 2014 at 19:12
  • \$\begingroup\$ (This pertains to your original program -- I just saw that you've made some edits.) Ohhh... I misunderstood how the * works. It is safe to say that h_0(x) = g^4(x) >> f_eps_0(x); consequently, the relation h_{i+1} (x) = h_i^x (x) effectively defines an "accelerated" fast-growing hierarchy such that h_i(x) >> f_(eps_0 + i)(x). I.e., the computed number h_9 (g(3)) is certainly much greater than f_(eps_0 + 9)(g(3)). As for g(3), I think I can show that it's greater than g_4, the fourth number in the g_i sequence used to define Graham's number (which is g_64). \$\endgroup\$
    – r.e.s.
    Commented Jan 20, 2014 at 21:46
  • \$\begingroup\$ @r.e.s., so j_i ~ f_{eps_0 + i}; does that make k_i ~ f_{eps_0 + i omega + i^2}? \$\endgroup\$ Commented Jan 20, 2014 at 23:50
  • \$\begingroup\$ Given what you wrote, I get k_i ~ f_{ε_0 + ω}^i (k_0). Here's the reasoning: k_{i+1} = j_{k_i} (k_i) = j_ω (k_i) ~ f_{ε_0 + ω} (k_i) ~ f_{ε_0 + ω}^2 (k_{i-1}) ... ~ f_{ε_0 + ω}^{i+1} (k_0), so k_i ~ f_{ε_0 + ω}^i (k_0). A very conservative lower bound on k_i, entirely in terms of the fast-growing hierarchy, is then k_i >> f_{ε_0 + ω}^i (i) = f_{ε_0 + ω + 1} (i). \$\endgroup\$
    – r.e.s.
    Commented Jan 21, 2014 at 0:39

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:


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 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.

  • 5
    \$\begingroup\$ Looks great; apparently your "modified Ackermann" is exactly a Conway-chain evaluator. \$\endgroup\$ Commented Jan 13, 2014 at 11:27
  • 1
    \$\begingroup\$ @leftaroundabout Not quite, but I think that it has about the same recursive strength. Also - zeroes aren't valid in chains, so you'll want to drop the zero from your Conway chain in the scores list. \$\endgroup\$
    – Cel Skeggs
    Commented Jan 14, 2014 at 2:25
  • 1
    \$\begingroup\$ Why did you do range(ord('~'))? Couldn't you have done range(125) for fewer bytes, which would allow you to squeeze in a higher number like range(A(9,9,9))? \$\endgroup\$ Commented Dec 29, 2016 at 17:20
  • 2
    \$\begingroup\$ @Challenger5: rule 1 says "You cannot use digits in your code (0123456789)" \$\endgroup\$
    – Cel Skeggs
    Commented Dec 29, 2016 at 21:49
  • \$\begingroup\$ @CelSkeggs: Oh, I forgot about that. \$\endgroup\$ Commented Dec 30, 2016 at 4:22

Perl - score ≈ 10↑↑4.1


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.

Perl - score ≈ 10↑↑4


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.


Let's define Pk(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:


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 Dk and Ek, but this doesn't matter too much, if we observe that


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.

  • 1
    \$\begingroup\$ So, this will only be a valid solution so long as the version number remains lower than 10? \$\endgroup\$
    – Mr Lister
    Commented Jan 9, 2014 at 14:38
  • 3
    \$\begingroup\$ @MrLister Yes. Fortunately, no major version higher than 6 exists, and even that's not considered to be fully 'ready', despite having been originally announced in 2000. \$\endgroup\$
    – primo
    Commented Jan 9, 2014 at 14:43
  • \$\begingroup\$ @primo You do realize that you will have to revise this answer once Perl goes into a version number > 10, right? ;) \$\endgroup\$ Commented Jan 10, 2014 at 3:10
  • 3
    \$\begingroup\$ @Eliseod'Annunzio If I'm still alive when that day arrives - if ever - I promise to come back and fix it. \$\endgroup\$
    – primo
    Commented Jan 10, 2014 at 8:19
  • 2
    \$\begingroup\$ A running solution that surpasses 10↑↑4. That's impressive. Bravo! \$\endgroup\$
    – Tobia
    Commented Jan 19, 2014 at 20:16

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.

Improved Score: 10↑↑13

Javascript, in exactly 100 characters, again:


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 10M, this code produces after each iteration

  1. 210 ∼ O(102)
  2. O(10102) ∼ O(10↑↑2)
  3. O(1010↑↑2) = O(10↑↑3)
  4. O(1010↑↑3) = O(10↑↑4)
  5. O(1010↑↑4) = O(10↑↑5)
  6. O(1010↑↑5) = O(10↑↑6)
  7. O(1010↑↑6) = O(10↑↑7)
  8. O(1010↑↑7) = O(10↑↑8)
  9. O(1010↑↑8) = O(10↑↑9)
  10. O(1010↑↑9) = O(10↑↑10)
  11. O(1010↑↑10) = O(10↑↑11)
  12. O(1010↑↑11) = O(10↑↑12)
  13. O(1010↑↑12) = O(10↑↑13)

Score: ∼101010101016 ≈ 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 1042381398144233621. By the second iteration, Mathematica was unable to compute even the number of digits in the result.

Here's a walkthrough of the code:


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⋅1042381398144233625.

Making my score, after only the first iteration, 2.72⋅1042381398144233619.

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


Iteration 4


Iteration 5


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,

        enter image description here

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

        enter image description here

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

        enter image description here

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

        enter image description here

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 OZ," i.e. 10OZ

        enter image description here

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

        enter image description here

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(10M).

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⋅1042381398144233625—that is, 1042381398144233625 ∼ 101016.

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

  1. 1016
  2. 101016
  3. 10101016
  4. 1010101016
  5. 101010101016

LaTeX sources






  • \$\begingroup\$ My reckoning about your score is based on the observation that f does something like take the number z to its own power. So that's something like ↑↑↑. Of course the score is not 2↑↑↑2, sorry... more like, 2↑↑↑5+1 it seems. Would you agree, should I put that in the leaderboard? \$\endgroup\$ Commented Jan 10, 2014 at 2:33
  • \$\begingroup\$ @leftaroundabout - Thanks for looking into it again. I don't feel comfortable enough with up-arrow notation to say whether your suggestion sounds right or not, but I calculated the order of magnitude of my score (see edit) if you'd like to update the leaderboard with that. \$\endgroup\$ Commented Jan 10, 2014 at 8:07
  • \$\begingroup\$ Excellent! I'm not at all firm with up-arrows either. So actually you have "only" a tower of power; I'm afraid that places you two spots lower in the ranking. Kudos for properly analysing the result; my estimations have probably yet more flaws in them, but I felt someone should at least try to get some order in the answers. \$\endgroup\$ Commented Jan 10, 2014 at 9:35
  • 1
    \$\begingroup\$ Your score is wrong. Whenever you start a loop with i=o.z;while(i--)... you are not executing the loop o.z times, because a loop is based on an integer variable and o.z contains a string larger than the largest representable integer, depending on your interpreter's word size. Supposing for your benefit that your interpreter won't barf on converting such a string to int, i will start each time with its largest representable integer value, let's say 2^63, and not with the current value of o.z. \$\endgroup\$
    – Tobia
    Commented Jan 10, 2014 at 21:34
  • 2
    \$\begingroup\$ @acheong87 Don't remove yourself, you just have to recompute your score, capping the loop variables to 2^63 or such. PS: leave your analytical score posted here, it's very instructive! \$\endgroup\$
    – Tobia
    Commented Jan 11, 2014 at 16:06

APL, 10↑↑3.4

Here's my revised attempt:


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 1010 × 12345678944 / 1003 ≈ 1010353 ≈ 10↑↑3.406161


  • ⎕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.

  • \$\begingroup\$ Your printed number will be divided for the number of bytes you used for your solution^3., you're dividing by 100 right now. \$\endgroup\$ Commented Jan 9, 2014 at 18:30
  • 2
    \$\begingroup\$ @ToastyMallows - looks like 100 cubed (100^3) to me. \$\endgroup\$ Commented Jan 9, 2014 at 19:36
  • 1
    \$\begingroup\$ I know but it's bytes, not characters. \$\endgroup\$ Commented Jan 9, 2014 at 20:03
  • 1
    \$\begingroup\$ @ToastyMallows Read the end notes on the answer. \$\endgroup\$ Commented May 17, 2017 at 13:25
  • \$\begingroup\$ Change {⍞←⎕D} to ⍞← which saves you three bytes that you can use to add one more ⍣n and make ⊢n←⍎⎕D into ⌽⍕n←⍎⎕D for an 80-fold increase. If you allow running with ⎕PP←17 then use ×⍨ instead of ⌽⍕ which almost doubles the number of digits printed. \$\endgroup\$
    – Adám
    Commented Jan 15, 2019 at 12:51

Haskell, score: (22265536-3)/1000000 ≈ 2↑↑7 ≈ 10↑↑4.6329710779

o=round$sin pi
i=succ o

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).

  • 1
    \$\begingroup\$ o=length[] gets you an extra !q at the end and saves you a byte on top of that. \$\endgroup\$ Commented Jul 1, 2019 at 21:56

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)\$.

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)\$.

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)\$.


*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

while h != n  # reduce until h == n
  h = f[h]

*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")


  • 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.



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

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.

  • \$\begingroup\$ Your code doesn't run. Only print True<<(True<<(True<<(True<<True<<True))) does, and that outputs a 19k string. \$\endgroup\$
    – Gabe
    Commented Jan 9, 2014 at 22:56
  • \$\begingroup\$ What are that minimum system requirements? \$\endgroup\$
    – user14373
    Commented Jan 10, 2014 at 10:15
  • 10
    \$\begingroup\$ Could you not make it shorter by defining t=True and then using t after? \$\endgroup\$
    – Bob
    Commented Jan 16, 2014 at 12:51
  • 1
    \$\begingroup\$ Better yet, just make a loop that does these nestings for you. \$\endgroup\$ Commented May 14, 2017 at 0:24
  • \$\begingroup\$ This fails for me: $python -c 'print True<<(True<<(True<<(True<<(True<<(True<<(True<<(True<<(True<<(True<<True<<True)))))))))' Traceback (most recent call last): File "<string>", line 1, in <module> OverflowError: long int too large to convert to int \$\endgroup\$ Commented May 19, 2017 at 13:17

GolfScript,   ≈ fε0(fε0(fε0(fε0(fε0(fε0(fε0(fε0(fε0(126)))))))))

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


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 fε0(fε0(fε0(fε0(fε0(fε0(fε0(fε0(fε0(126))))))))) -- a nine-fold iteration of fε0 -- where fε0 is the function in the fast-growing hierarchy that grows at roughly the same rate as the Goodstein function. (fε0 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 fε0.)

  • \$\begingroup\$ '',:o;'oo',:t; just assigns the values 0 to o and 2 to t; if that's just to work around lack of digits then it can be abbreviated heavily to ,:o)):t;, except that there's no reason to delete t in the first place because you can write expr:t;{...}:f;[[[t]f]f]f as [[[expr:t]{...}:f~]f]f saving a further 3 chars. \$\endgroup\$ Commented Jan 20, 2014 at 0:26
  • \$\begingroup\$ Still no need to pop o: I'm pretty sure that [0 126]f will be one larger than [126]f so you save a char and bump the output. Although you're leaving an empty string in there, which probably breaks things: it might be better to start [[,:o'~'=] \$\endgroup\$ Commented Jan 20, 2014 at 8:44
  • \$\begingroup\$ Oh, and the [ are unnecessary since you don't have anything else on the stack. \$\endgroup\$ Commented Jan 20, 2014 at 10:39
  • \$\begingroup\$ Ha... scrolling these answers, and then I see this... and then I notice the accepted answer... hm...... \$\endgroup\$ Commented May 13, 2017 at 14:13
  • \$\begingroup\$ @SimplyBeautifulArt I'm not sure what you mean, but the accepted answer does compute a very much larger number than this one (assuming both are as claimed). \$\endgroup\$
    – r.e.s.
    Commented May 14, 2017 at 3:30

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.

  • \$\begingroup\$ I have no idea what this code does... can only assume it's correct. Perhaps you could explain a little? \$\endgroup\$ Commented Jan 10, 2014 at 12:27
  • \$\begingroup\$ I'll explain the code piece by piece when I have time later tonight. \$\endgroup\$
    – ashastral
    Commented Jan 10, 2014 at 18:14
  • \$\begingroup\$ Well, I spaced on that explanation, but I've added it now. Hope it clears things up. \$\endgroup\$
    – ashastral
    Commented Jan 13, 2014 at 4:28
  • \$\begingroup\$ dc-1.06.95-2 terminates immediately, having printed nothing. \$\endgroup\$
    – primo
    Commented Jan 16, 2014 at 17:05
  • 1
    \$\begingroup\$ I wouldn't expect it to work on any existing machine, given the magnitude of the value it will try to generate. I have the same version of dc and it segfaults after a few seconds. I'm assuming "theoretically correct" answers are permitted here, since there's no criteria for resource consumption. \$\endgroup\$
    – ashastral
    Commented Jan 16, 2014 at 19:16

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

  • \$\begingroup\$ I converted your score into tetration notation described at the end of the question post. It was done using print(0.126126... * TetrationDecimal(10) ** (126 * 1000)). \$\endgroup\$ Commented May 29, 2022 at 20:38

Ruby, probabilistically infinite, 54 characters

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.

  • 3
    \$\begingroup\$ this depends on the number generator which in most languages is unlikely be able to generate 97 times the same number \$\endgroup\$ Commented Jan 10, 2014 at 15:00
  • 1
    \$\begingroup\$ Good point, so in addition to assuming continually rapidly increasing computation power, I also need to assume a perfect source of randomness and a Ruby implementation that uses it. \$\endgroup\$
    – histocrat
    Commented Jan 10, 2014 at 19:09

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 1010355393490465494856447 ≈ 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 = 
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.

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

  • \$\begingroup\$ Maybe. %u was printing 32-bit numbers even with a 64-bit compile, so I just did the ll out of habit from writing in a 32-bit compiler. \$\endgroup\$
    – David Yaw
    Commented Jan 9, 2014 at 19:28
  • \$\begingroup\$ I think %llu would be for long long, and %lu would be correct for long. \$\endgroup\$
    – tomlogic
    Commented Jan 9, 2014 at 20:12
  • \$\begingroup\$ Fixed. Force of habit: %u is always 32-bit, %llu is always 64-bit, whether compiling as 32 or 64 bit. However, the solution here requires that long be 64-bit, so you're right, %lu is sufficient. \$\endgroup\$
    – David Yaw
    Commented Jan 9, 2014 at 20:42
  • \$\begingroup\$ Your variables on the stack are not guaranteed to be initialized to 0. In the second program, just put them outside of any function. In the first one, you'll have to initialize i. \$\endgroup\$
    – Art
    Commented Jan 10, 2014 at 12:39
  • \$\begingroup\$ Also, long overflow is undefined behavior and many modern compilers will just optimize it away if they detect it, you probably want to use unsigned long. \$\endgroup\$
    – Art
    Commented Jan 10, 2014 at 12:41


(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

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.

  • 1
    \$\begingroup\$ #DEFINE C while(-- long n,o,p,q,r,s,t;main(){Cn){Co){Cp){Cq){Cr{Cs{Ct){putchar('z'-'A');}}}}}}}} \$\endgroup\$ Commented Jan 9, 2014 at 12:36
  • \$\begingroup\$ @RobAu You're a genius! Make it an answer. I'm sure it'd be the winner. I think you forgot a couple ), but that's okay, because you're only at 96 characters right now. \$\endgroup\$ Commented Jan 9, 2014 at 23:14
  • \$\begingroup\$ For everyone that didn't get the sarcasm: see codegolf.stackexchange.com/a/18060/7021 for an even better solution ;) \$\endgroup\$ Commented Jan 10, 2014 at 13:45

Pyth, fψ(ΩΩ)+7(25626)/1000000


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.

  • \$\begingroup\$ I'd say the most noticeable difference in strength between our programs is that in [b,c,d], I'm limited to natural values of c, whereas you are not. \$\endgroup\$ Commented Dec 3, 2017 at 18:29
  • \$\begingroup\$ I didn't even realize that you didn't allow for [[],[],n] in your program. \$\endgroup\$
    – Steven H.
    Commented Dec 3, 2017 at 21:22
  • \$\begingroup\$ Doing so would cost me 1 more byte. \$\endgroup\$ Commented Dec 3, 2017 at 21:33
  • \$\begingroup\$ Woo... I finally managed to do f[[n,[],n],n]! x'D \$\endgroup\$ Commented May 18, 2019 at 19:55
  • \$\begingroup\$ I rewrote my explanation (linked in your answer) to what I hope is more understandable than before (lol). If you care, feel free to let me know what you think. \$\endgroup\$ Commented Jun 2, 2021 at 4:57

New Ruby: score ~ fωω2+1(12622126)

where fα(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
        case b
        when nil
            puts(n += n)
        when 0
        when -1

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


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.


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.

  • \$\begingroup\$ Let us continue this discussion in chat. \$\endgroup\$
    – hyper-neutrino
    Commented May 12, 2017 at 14:35
  • \$\begingroup\$ It's actually 103 bytes, you had a trailing newline I think. \$\endgroup\$
    – Riker
    Commented May 13, 2017 at 14:37
  • \$\begingroup\$ @Riker I believe you copied and pasted from here. Note there should be an unprintable character on the second line, hence 104 bytes. \$\endgroup\$ Commented May 13, 2017 at 14:42
  • \$\begingroup\$ @SimplyBeautifulArt ah, okay. I thought I copied the character. Sorry. \$\endgroup\$
    – Riker
    Commented May 13, 2017 at 14:43
  • \$\begingroup\$ @Riker Nah, its not even there due to Stackexchange not letting me hide invisible characters everywhere. \$\endgroup\$ Commented May 13, 2017 at 14:47

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
a n=n?n

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 an):

  • gn = 3↑gn-13
  • an ~= 2↑an-1(an-1)

If these are approximately equivalent, then a20(1) ~= g18. The final term in an, (an-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.

  • \$\begingroup\$ Your analysis is correct and g<sub>18</sub> is a good approximation. \$\endgroup\$ Commented Jun 20, 2017 at 17:37
  • \$\begingroup\$ length"a" saves a couple bytes and allows you another .a \$\endgroup\$ Commented Jul 1, 2019 at 21:54

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 (102524320-1) / 1000000

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

Assembled code:



ORG 0x100

            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

b_ss:       dw 'SX'
b_sp:       db 'X'
  • \$\begingroup\$ You obviously never ran this. It overwrites its code and crashes. \$\endgroup\$
    – Joshua
    Commented Jan 3, 2016 at 20:52

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


will print out [reproducing here just the start]:

  • 2
    \$\begingroup\$ I don't think you need to include the number in the post. It takes up a lot of space on mobile as well. \$\endgroup\$ Commented May 12, 2017 at 14:02
  • \$\begingroup\$ @totallyhuman I agree, maybe the first 100 digits, max \$\endgroup\$
    – tuskiomi
    Commented May 12, 2017 at 15:07
  • \$\begingroup\$ @totallyhuman ok thanks done :) \$\endgroup\$
    – lebatsnok
    Commented May 19, 2017 at 6:05
  • \$\begingroup\$ cat is a weird function in that the first argument is .... So everything before the first named argument goes to ... (and will be cat'ed), which is why sep must be named -- otherwise one could shorten it as cat(abs(.Random.seed),,"") \$\endgroup\$
    – lebatsnok
    Commented May 17, 2018 at 7:48

ECMAScript 6 - 10^3↑↑↑↑3 / 884736

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


Output: 10^3↑↑↑↑3


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".

  • \$\begingroup\$ Maximum code length is 100 bytes; Otherwise this is near unbeatable \$\endgroup\$
    – Cruncher
    Commented Jan 9, 2014 at 14:16
  • \$\begingroup\$ @Cruncher Aaah, I missed that \$\endgroup\$ Commented Jan 9, 2014 at 14:24
  • \$\begingroup\$ Ahh, I hate you now. Everytime I try to fathom the size of Graham's number my head hurts. \$\endgroup\$
    – Cruncher
    Commented Jan 9, 2014 at 14:34
  • \$\begingroup\$ also, doesn't Graham's number count as a constant > 10? \$\endgroup\$ Commented Jan 9, 2014 at 14:35
  • 1
    \$\begingroup\$ Now to determine if mine beats Ilmari's. \$\endgroup\$ Commented Jan 9, 2014 at 15:00


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.


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'
  • \$\begingroup\$ SHLVL is the level of bash as subbash: bash -c 'bash -c "echo \$SHLVL"' \$\endgroup\$ Commented Jan 3, 2016 at 7:41
  • \$\begingroup\$ stat --printf don't work. Try stat -c %s \$\endgroup\$ Commented Jan 3, 2016 at 7:41
  • \$\begingroup\$ @F.Hauri: --printf works for me but so does -c so that shaved a few bytes. Thanks. \$\endgroup\$
    – Joshua
    Commented Jan 3, 2016 at 20:42

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;

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]);

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

  • \$\begingroup\$ Best C implementation, by far. Why use 5 variables, when you can use an array of 258048? \$\endgroup\$
    – primo
    Commented Jan 16, 2014 at 17:31

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.

  • \$\begingroup\$ Rule 1 says no digits allowed, you have a 0 in there. \$\endgroup\$
    – Kyle Kanos
    Commented Jan 9, 2014 at 3:05
  • \$\begingroup\$ Damn, I do too. There goes my character count. \$\endgroup\$ Commented Jan 9, 2014 at 3:54
  • \$\begingroup\$ You can use -ea(+'') to reduce the size ('' converted to a number is 0, which the enum value of SilentlyContinue). You can use \D for the replacement regex which is the same as [^\d]. And you can just use %{$_.Length} instead of select Length which gets rid of the column headers. And then you can get rid of the -split and -replace as well, leaving you with -join(gci \ -ea(+'')-r|%{$_.Length}) which is 37 characters shorter (I also reordered the parameters because the parentheses are needed anyway because of +''). \$\endgroup\$
    – Joey
    Commented Jan 11, 2014 at 12:01

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

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.

  • \$\begingroup\$ You can increase your number by changing f(m-g,g) to f(m-g,m). \$\endgroup\$ Commented May 13, 2017 at 22:36
  • \$\begingroup\$ or f(m-g,i). Also, at the end of the first line, you use a number. I believe you meant to use n+g, whereupon I will point out n+n will be larger. \$\endgroup\$ Commented May 13, 2017 at 22:56
  • \$\begingroup\$ You can save a few bytes by changing len('"') for True \$\endgroup\$ Commented May 18, 2017 at 20:16
  • \$\begingroup\$ And use ord('^?') (where ^? is the DEL character, ASCII 127) for a bigger number. EDIT never mind, that's not "Printable". \$\endgroup\$ Commented May 18, 2017 at 20:29
  • \$\begingroup\$ @BrianMinton Who says it has to be printable? \$\endgroup\$ Commented May 18, 2017 at 23:51

JavaScript 98 chars


generates 2.718e+239622337 ≈ 10↑↑2.9232195202

For score of just slightly more than 2.718e+239622331 ≈ 10↑↑2.9232195197

which is the largest I can make it without the browser crashing.

(console.log(a) will show you the full output)

Don't run these:


would output 2.718+e121333054704 ≈ 10↑↑3.0189898069 (aka 2.718*10^(1.213*10^12) to compare to the longer answer:

more extreme version, if it didn't crash your browser: (80 char)


which would create a number around the same size as e * 10^(10^19) ≈ 10↑↑3.106786869689

Edit: updated code original solution only generated 2.718e+464


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