Making any integer with four 2s
177 comments
·February 23, 2025horsawlarway
Someone
> I feel like the second you allow functions you've thrown the spirit of the game.
+, - (both binary and unary), ×, ÷ are functions, as is raising to a power. Why would you allow them?
As always in this kind of things, one can disagree about what constitutes an elementary function, but I don’t think taking square roots should be disqualified in this puzzle.
> Ex, the gamma function is (n-1)!
And 2 is just S(S(0)) (https://en.wikipedia.org/wiki/Peano_axioms)
> If I can hide numbers in a function call... It's trivially easy to always succeed.
I wouldn’t call the construction given by Paul Dirac trivial. Do you think it is, or do you know of a simpler one?
timerol
> And 2 is just S(S(0))
This is a good example of why you need rules on which functions are allowed. Repeated application of the successor function makes the entire exercise trivial
Karellen
But also, if the criteria for allowed functions is that they are "reasonable, elemental" (as per the fine article), then I think it would be quite hard to come up with a set of rules to encode that in a way that includes log() and sqrt(), but not S(). It's hard to imagine a function that is less elemental than S() (except maybe the identity function), or why its inclusion would be unreasonable when the other two aren't.
jonahx
This is a great point. I think what you're really responding to is that it's a game without clear rules, and so part of the "game" is thinking about creative interpretations of the rules themselves and pushing the boundary of what others originally assumed the rules to be.
Granted there is creativity in this sort of game -- indeed, most "games" in life are like this -- but it's quite a different thing from winning a game with clearly defined rules like chess, or this game with the set of allowed operations specified up front.
quietbritishjim
I think a key distinction is that those are functions of two parameters. You can't just use them as many times as you like "for free" like the square root trick at the end of the article, because you need to "spend" at least one extra 2 on the second parameter each time.
That's not the whole story of course, you still need to agree on the set of allowed operations, but I think it makes a big difference even though it seems incidental at first.
nwallin
Surely we accept unary minus as one of our functions? Once we accept one unary function we're just quibbling over details.
I agree that you need to define and agree upon a finite set of allowed operations before playing the game. IMHO, square root, logarithm/exp, floor/round/ceiling, sin/cos/tan ought to be included in the list. But that's just like, my opinion, man.
thaumasiotes
> I wouldn’t call the construction given by Paul Dirac trivial. Do you think it is
Yes? It's doing exactly the thing that your parent comment complains about in the gamma function, introducing additional constants (in this case, mostly 2s) that, for no particular reason, don't count.
Why would you interpret squaring as consuming a 2, but square rooting as not consuming a 2?
zeroonetwothree
Because of our standard notation for those
the__alchemist
That's exactly the point. What exactly, is the set of allowable functions used for the problem? I think you, and the post you reply to, are stating the same thing.
Where do you draw a line between "Functions available on a 4-function calculator" and "Functions I can make up specifically to generate a target integer"? I think you have to rigidly define this, or the game loses meaning.
jrockway
Right, I invented the jrockway function which is defined as f(2, 2, 2, 2) = 7 so I made 7. Maybe the rule is "someone else has to invent the function", but I invented that so you can use it. (Please make me a the__alchemist function.)
Maybe the rule should be that the function has to be invented before the inventor has knowledge of this game. But now I'm just going through /usr/bin looking for binaries where the 2222th byte is 0x7.
null
dooglius
It sounds like you've just found one for >=2: 2, S(2), S(S(2)), ...
Someone
That, somewhat ironically, typically isn’t included in the set of elementary functions. ‘plus’ is, but ‘plus one’ isn’t.
omoikane
Maybe they meant symbolic operators feel alright but named functions feel like cheating, so 2+2+2+⌊√2⌋ is fine but 2+2+2+floor(sqrt(2)) is not.
null
the__alchemist
This was my initial thought once we got to the Gamma function.
My reasoning is (I'm pretty sure it's the same as yours), why is the gamma function allowed, but not others? I could insert arbitrary functions to make the game arbitrarily solvable.
While this hit me at the Gamma introduction, I think it leads back to the beginning: It's a poorly defined problem from the rules at the start of the article. It should instead define the set of allowable functions (or operations) explicitly. I think you could modify this to retain the intent of showing how the problem scales with knowledge level.
Boldened15
It’s just about having fun at the end of the day, the gamma function and square root are considered fundamental enough. But if one wants they could try to limit to different subsets of functions and prove which numbers are possible or impossible to achieve just with those.
They also say “mathematical tools” not arbitrary functions.
vesinisa
I think you have a point, but as others have commented "allowing functions" is not the problem, as fundamental math operations are functions. But if we limit ourselves to only functions that map (tuples of) integers to integers ((Z, Z, ...) -> Z), the spirit of the original game is retained. This disallows sqrt and log, but retains addition, subtraction and multiplication (but not division). Factorial (n!) is allowed, as is exponentiation to a non-negative power.
Wonder if someone could come up with general solution within these constraints.
amluto
I think that, if you are restricted to a finite list of n-ary functions, n>=2, each returning a single value, then you can’t do it, as you will only have finitely many valid expressions.
This may be easier to see in a stack machine / RPN model. An expression is a list of operations, drawn from a finite set, each of is either “push the number 2” or something that decreases the stack size by at least 1. And you need exactly 4 pushes. So a valid expression has four pushes and at most 3 other operations, because otherwise the stack would underflow. This gives a finite number of possible expressions, but there are an infinite number of integers, so it can’t work.
throwway120385
Are you sure the number of possible expressions is finite? Or is it countably infinite? I could in theory make an arbitrarily long expression out of the four fundamental operators, and because I can derive -1 from a finite set of operations it is possible to derive any integer from any even integer. You can also derive zero from the rules.
vesinisa
I didn't mean to say that n would need to be 2 or more, unary functions (I even mentioned factorial) should be fine. 1-tuple is valid tuple :)
And yes I think your analysis that only allowing n-ary funcs with n>=2 would make general solution very much impossible since you can only have a limited number of inputs.
energy123
f_n(a,b,c,d) = n is a mapping from Z^4 to Z
vlovich123
The Dirac solution doesn’t involve gamma, just N square roots and 2 logarithms.
Biganon
But the square root has a hidden operand, 2. We don't write it because by convention the default root is 2, but that still feel like cheating to me.
necovek
This is why people have brought up the successor function too: A + B, is, by definition B applications of Succ() on A:
A + B = Succ(Succ(Succ(...Succ(A))))
The mathematical root probably first appeared as a square root and was later extended to support other exponents.
But is there any fun in this? As noted elsewhere, the game is in finding the rules, and a solution within those rules.
[]Since all the natural numbers other than 1 are defined using a Succ() functions, there's a trivial solution. But if we only limit ourselves to this most elementary operation, we can't get a 1 because that's an axiom in itself ("There exists 1" or "There is a set of cardinality 1").
lblume
I would argue that the Gamma function is more fundamental than the factorial operation. But you are still correct that if arbitrary functions were allowed, the game would degenerate to triviality.
TOGoS
Inorite. If we're allowing any old function, then I can just define 12345 as
Onetwothreefourfive()-2+2-2+2MisterKent
You don't even need gamma for 7:
2 + 2 + 2 + floor(sqrt(2))
Which feels at least more in the spirit of the challenge than gamma.
account42
The article already has a solution for 7 (and any other number) that doesn't use the gamma function.
tantalor
> use any mathematical operations
Okay, then this is easy, just use the successor function.
S(n) = n+1
6 = 2*2*2-2
7 = S(2*2*2-2)
8 = S(S(2*2*2-2))
Hamuko
I feel like having a 1 on the first line is cheating given the constraint "using no other digits".
tantalor
By that argment you shouldn't be able to use factorial function either, because that has a "hidden 1" too:
fac(n) = n * fac(n-1) n>0
= 1 n=0Hamuko
The one here isn't hidden in any way – it's on the very first line. You might as well just define X = 7 and do 2-2+2-2+X.
tessierashpool9
log is also using another digit.
venusenvy47
I took a lot of math in my schooling, and have continued using math on a daily basis in my engineering career. I even subscribe to many math channels on YouTube, but this is the first time in my life that I've even heard of this function. I know there aren't real rules to this puzzle, but this function doesn't seem well-known at all.
tantalor
It has a rich history going back to the formalization of arithmetic in the 1800s.
https://en.wikipedia.org/wiki/Peano_axioms
It's probably something that only folks who study the foundations of mathematics would know.
You may have also heard about it if you learned about Gödel's incompleteness theorems, or read Gödel, Escher, Bach
empath75
The successor function is how natural numbers are defined in most axiomatic arithmetic systems.
A natural number is either zero or a successor of a natural number.
Addition is defined as a recursive application of successor functions.
m + 0 = m,
m + S(n) = S(m + n).
S(S(S(0))) + S(S(0)) = S( S(S(S(0))) + S(0) ) = S( S( S(S(S(0))) + 0 ) ) = S(S(S(S(S(0)))))
There's nothing really in the definition of the successor function that necessarily requires that it's interpreted as n+1, though. It's just an interpretation from the context in which it's used. It could represent any operation as long as it is isomorphic to adding one -- but there's nothing special about "adding one". You could have it represent multiplying by a constant, and interpret "zero" as the number one.
So a number in this interpretation is either 1; or 2 times a natural number.
1, 2, 4, 8, 16, 32 -- you're working only with powers of 2 now.
The above "addition" rule above still works, but now it represents "multiplication" instead of addition. I'll replace all the S's in the above example with x2
2x2x2x1 "+" 2x2x1 = 2x(2x2x2x1 + 2x1) = 2x(2x(2x2x2x1 + 1) = 2x2x2x2x2x1 = 32
So now, instead of addition, we've recursively defined multiplication where the successor function is interpreted as multiplying by 2. There's an infinite number of ways that you can interpret the successor function.
So basically, I do think it's cheating, and if you do want to define it as n+1, it would be even simpler to just define a function that takes any number to the desired output.
kccqzy
I'm not sure what kind of math schooling you did, but did you learn to construct the natural numbers from scratch?
This may have been the fault of math education. In my college people learn real analysis (constructing the real numbers) before they learn to construct the natural numbers, which is backwards to me. I recommend learning it: constructing the natural numbers from just sets in the tradition of Zermelo–Fraenkel is mind blowing the first time you see it. Of course you could just use Peano axioms without touching set theory too.
Aardwolf
Finding the shortest expression with 4 2's for a given integer would be a more interesting challenge
0xfffafaCrash
I had the same thought. Also with square roots hiding 2s behind notation, etc. The whole project isn’t really very coherent without specifying what specific operators you can use (and how many times).
svat
See also: "Representing numbers using only one 4" written by a 26-year-old Donald Knuth in 1964 (https://www.jstor.org/stable/2689238 reprinted as Chapter 10 of his Selected Papers on Fun and Games) — it uses the single digit 4, and the three operations √x (square root), ⌊x⌋ (floor, i.e. greater integer not greater than), and x! (factorial), and ends with a (still unsolved) conjecture about whether every integer can be represented in this way.
The appendix (written for the book in 2011) points out an earlier (1962) 1.5-page paper π in Four 4's by J. H. Conway and M. J. T. Guy, written when they were students at Cambridge, that has a similar idea: https://archive.org/details/eureka-25/page/18/mode/1up?view=...
For example,
5 = ⌊√√√√√(4!)!⌋
tasn
Maybe it's just me, but writing sqrt(2+2) instead of sqrt(2*2) or sqrt(2^2) was such an odd choice. It obfuscates the reason why 2=sqrt(2+2), and unnecessarily so.
mmooss
Good point and feedback, but an odd choice by the author?
It could be the phenomenon of the author's cognitive bandwidth being consumed by everything in the article, including each argument, the overall argument, the writing, the formatting, etc. etc., and with time pressures. The critic can focus at their leisure on one point, with bandwidth to spare - and so it's obvious! :)
tasn
I agree it potentially wasn't a conscious choice, but it's still interesting nonetheless.
I wasn't criticizing him for this, but rather fascinated that this is the variant that was chosen.
hinkley
Speaking of odd choices:
12 = 2 * (2+2+2)
Is a hell or a lot simpler than using complex numbers. Might be a different example for that would have been better.
eliben
Sorry, the intention was just to show a cool use of complex numbers, not claim this is the simplest method to generate 12 which is pretty simple, as you demonstrate.
axus
Maybe there's a "golf score" somewhere that rewards less expensive operations. The "Dirac hack" would run up a lot of points.
eliben
Really? But why? All of 2+2, 2*2 2^2 are trivially 4, and sqrt(4)=2 so why is the + more odd than others?
tasn
Because sqrt is the reverse of 2^2 and 2*2 (which is 2^2 unwrapped). Though there's no direct relationship between sqrt and 2+2 other than that it happens to be equal to 2*2.
Or put differently: N = sqrt(N^2) or sqrt(N * N) for every positive N, but x = sqrt(x + x) or sqrt(x + 2) is only true for x = 2 for both or x = 0 for the first representation.
gpm
If 2*2 is 2^2 unwrapped, then surely 2+2 is 2*2 unwrapped, thus 2+2 is 2^2 unwrapped^2 and is also natural via transitivity? :P
Lerc
I think my preference is more towards conciseness.
I made a stack machine with single character instructions and needed to solve a variation of this problem. I had just the digits 0 through 9. The characters '23' would be push 2 followed by push 3. To actually represent the number 23 you would use
45*3+
or something similar.
Tools at hand.
The digits 0 through 9
'P': Pi
'*': (a * b),
'/': (a / b),
'-': (a - b),
'+': (a + b),
's': sin(a),
'c': cos(a),
'q': sqrt(a),
'l': log(a),
'~': abs(a),
'#': round(a),
'$': Math.floor(a),
'C': clamp(a),
'<': min(a, b),
'>': max(a, b),
'^': pow(a, b),
'a': atan2(a, b),
'%': positiveMod(a, b),
'!': (1 - a),
'?': (a <= 0 ? 0 : 1)
'o': a xor b scaled by c; ((a*c) xor (b*c))/c
'd': duplicate the top stack entry
':': swap the top two stack entries
';': swap the top and third stack entries
(Next time I post something like this I am not going to use my phone)
Y_Y
The answer in general may be uncomputable.
nwallin
Kolmogorov complexity is uncomputable because it admits Turing complete languages, and reduces to the halting problem. If the language you admit isn't Turing complete, then the Kolmogorov complexity of the thing is computable.
It looks like OP's language is not Turing complete. It always terminates. You can just do a breadth first search on the program space. The first program you get that outputs the number you want is the shortest program.
If it were Turing complete, you can't do this, because eventually you'll find a program that just keeps running for like a really long time. Is it running because the program never halts? Or is will it halt eventually and output the number you want? You can't know for sure.
null
Lerc
AHH but in practice you are limited to 50 characters.
Which gives you a finite problem. The VM cannot loop or define functions (yet, anyway) so it doesn't go all busy beaver on you.
Y_Y
I thought it would be possible to use the (undocumented?) T instruction and use the time variable as persistent memory, but it seems to reset each step. It would be nice if there were a way to store things across time steps!
saulpw
What about making each digit be the instruction "times 10 plus digit", with a different instruction to push a 0, like a space? Then you can represent 23 with " 23".
rixed
Still, some numbers would admit a shorter sequence of instructions
Nuzzerino
Reminds me of https://www.hacker.org/hvm/ (2008)
Dylan16807
I suspect a whole lot of numbers are going to get encoded in base 9 or 10 as mostly repetitions of digit + * digit + * or equivalent.
kazinator
> There's just one small wrinkle: it uses three instances of the digit 2, not four.
One small wrinkle, if you ignore the fact that the root notation conceals exponentiation by 1/2, by making that common value a default.
That's a lot of hidden 2's!
xelxebar
Root notation isn't really _concealing_ anything. The fact that it's mostly equivalent to exponentiation by half is a theorem. Do we need to admit that 2 is concealing 1+1, thus making the game impossible?
Given the prevalence of quadratic polynomials over higher-order ones, sqrt does feel somewhat more fundamental than arbitrary exponentiation.
Lots of people have pointed out the farcity of the game after allowing fancy functions, but IMHO, there is a lot of fun in just finding satisfying solutions, without the need for specific rule limitations.
kazinator
The root notation is a syntax with a place for a number, to specify the degree. For instance ∛ -> cube root(s). When the degree is 2, it can be omitted.
So yes, the notation conceals something: its default value.
xelxebar
I have a function f: R -> R such that f(0)=1 and f'(x)=f(x). Is this concealing a parameter? It turns out we can deduce that f(x)=e^x, so f(x)=pow(e,x), so maybe? But wait, is pow concealing another parameter? Indeed, exponentiation is just (a completion over) repeated multiplication; multiplication is repeated addition, and addition repeated succesorship. It's straightforward to show that pow(x,y)=f(2,x,y) where f(n,x,y) encodes said hierarchy. But wait! f is a function, which is a set of 2-tuples, so we have another "hidden" parameter. It's also essentially part of Fast growing hierarchy[0] functions. Or what if we extend to complex numbers? Is the imaginary axis degree of freedom another hidden parameter?
Et cetera ad infinitum...
So, if you want to say that sqrt(x) has a hidden parameter, you're just making an arbitrary claim unless you accept it is concealing an unbounded number of hidden parameters.
Or we can just do the obvious thing and say that sqrt(x) has one parameter.
[0]:https://googology.fandom.com/wiki/Fast-growing_hierarchy
SilasX
Related: there as a reverse engineering/CTF challenge (which shall remain nameless to prevent you from cheating) where my solution involved injecting shellcode that adds specific number to the stack pointer. But your shellcode -- including the number(s) you add -- can only involve bytes from the ascii alphanumeric set.
So I used a SAT solver to find a combination of numbers, not using prohibited bytes, that add up to the number I really want.
https://docs.google.com/presentation/d/19K7SK1L49reoFgjEPKCF...
pil0u
This reminds me of this mobile game Tchisla[0] where you have to build all numbers up to 1000 (10000?) using only a given digit and a couple of operators (including sqrt and !)
It was a lot of fun, you tend to develop strategies and the game has a simple, efficient UX. Fair warning, it is very time consuming.
[0] https://apps.apple.com/fr/app/tchisla-number-puzzle/id110062...
virgulino
There's the classic “four fours”, which I learned as a child in the book “The Man Who Counted”.
dwheeler
That's version I learned as a kid. You might enjoy this page of mine:
The Definitive Four Fours Answer Key https://dwheeler.com/fourfours/
floodle
I love this:
> Note that these are large files; both are over 1.6 Megabytes, so don’t load these if you have a slow Internet connection
Just shows how far internet speeds have come...
TZubiri
That's the one!
That's how I learned about false induction. I also liked the one about the men who were lined up and had something on their backs and they had to guess what it was.
null
ziofill
This is amazing, but there are a lot of 2's hiding in those sqrt symbols
madflame991
Here are some values that are (understandably) not listed on the blog. They happen only due to the limited precision of floating point formats.
128 = √(2 / √√(√2 - (2 / √2)))
8192 = √√(2 / ((√2 * √2) - 2))
16384 = (2 / √√(√2 - (2 / √2)))
67108864 = √(2 / ((√2 * √2) - 2))
134217728 = (2 / √(√2 - (2 / √2)))
4503599627370496 = (2 / ((√2 * √2) - 2))
9007199254740992 = (2 / (√2 - (2 / √2)))
6369051672525773 = (√2 / (√2 - (2 / √2)))
My old solvers from what feels like a previous life: https://madflame991.blogspot.com/2013/02/four-fours.html https://madflame991.blogspot.com/2013/02/return-of-four-four...
That was fun
lifthrasiir
When everything is an IEEE 754 floating point number, a mathematically "linear" function can indeed be coerced into anything: http://tom7.org/grad/
svat
Nice! Looking into them a bit deeper, they all rely on two facts involving quantities that are off by 1 ulp:
√2 - 2/√2 as a float64 is exactly equal to 2^{-52}
√2 * √2 as a float64 is exactly equal to 2 + 2^{-51}
√(2 / √√(√2 - (2 / √2))) := √(2 / √√(2^{-52})) = √(2 / 2^{-13}) = 2^7 = 128
√√(2 / ((√2 * √2) - 2)) := √√(2 / 2^{-51}) = √√(2^{52}) = 2^{13} = 8192
6369051672525773 = 2^{52} * 1.4142135623730951454746218587388284504413604736328125
BobbyTables2
Very clever, but using an arbitrary number of square roots seems almost cheating since it’s practically another symbol for a “2” (exponent of 1/2)
mandarax8
I would allow it, given they physically write down all N square root on a piece of paper.
I feel like the second you allow functions you've thrown the spirit of the game.
Ex, the gamma function is (n-1)! So now you're making 7 with four twos and a one. You've broken the spirit.
If I can hide numbers in a function call... It's trivially easy to always succeed.