Making any integer with four 2s (2025)

223 pointsposted 4 days ago
by germanrabbit
(eli.thegreenplace.net)

57 Comments

hahajk

4 days ago

i know everyone plays with their own rules, but from my perspective, a square root spends a 2 - it's not an operator like + or ! . there's an implicit "2" since you could also take the cube root or fourth root

jjmarr

4 days ago

You can use `sin(arctan())` as the repeatable operator instead of sqrt(). Then "unwrap" by applying ^-2 - (2^-2)

3=(sin(arctan(sin(arctan(sin(arctan(2))))))^−2 − (2)^−2

shiandow

4 days ago

There's a 2 in the definition of sin :-)

Though it really depends which definitions you find more fundamental.

Still, exp and log seem like good choices.

JadeNB

3 days ago

> There's a 2 in the definition of sin :-)

How?

shiandow

3 days ago

sin(x) = (e^ix - e^-ix) / 2i

anthk

3 days ago

Keep it simple.

2^2 - (2 / 2).

olmo23

3 days ago

2 + 2 - (2/2).

schiffern

4 days ago

Perfect answer, I hate it :)

JumpCrisscross

4 days ago

> it's not an operator like + or !

Squaring is to multiplication as is the latter to addition. If roots are invalid then so should be multiplications.

penteract

4 days ago

Exponentiation is to multiplication as multiplication is to addition. Squaring is to multiplication as doubling is to addition.

jolmg

4 days ago

> Squaring is to multiplication as doubling is to addition.

as halving is to addition.

penteract

4 days ago

Square rooting is to multiplication as halving is to addition.

SamBam

4 days ago

GP didn't say roots are invalid, just that they burn a two.

If we agree that 2^2 uses two twos (and, indeed, that 2^3 isn't valid, because it uses a 3), then what's the equivalent of a square root?

Of course to my simple math brain, a square root is ^1/2, which includes the digit 1. But I agree there is someone 2-ish about the square root.

direwolf20

4 days ago

It can be written as ²√2 and there's also ³√2 and so on

schiffern

4 days ago

Yep.

It would be like if we had a stylistic convention that you can drop the second number in an addition and it becomes 1. So "6 +" means "6 + 1."

Now I can trivially make any number, but I'm just 'exploiting' a convenience feature in the notation:

0 = 2 + 2 - 2 - 2

1 = 2 + 2 - 2 - 2 +

2 = 2 + 2 - 2 - 2 + +

Etc

user

4 days ago

[deleted]

danparsonson

4 days ago

Would you elaborate on that? That doesn't make any sense to me

drfloyd51

4 days ago

Multiplication is repeated addition.

2x3= 2+2+2 AxB= A+A B times.

Raising to a power is repeated multiplication.

2^3= 2x2x2 A^B = AxA B times

Roots are the inverse of powers

Division is the inverse of multiplications.

Given this… the OC’s comment about giving exceptions to powers in the rules is nonsense because they are still operations.

mojomark

4 days ago

Multiplication is not the same as repeated addition, which you can observe as soon as you try the two different operations on fractions or negative numbers. For example, you can't add a number to itself 1/3 times, but you can certainly scale a number by 1/3.

Multiplication is a scaling operation.

jibalt

4 days ago

Because it's just wrong. hahajk said that taking the 2nd root spends a 2, not that "roots are invalid". (FWIW, I'm not necessarily endorsing the view that the square root symbol implicitly spends a 2, I'm just saying that the counterarguments given are logically wrong ... actually more like semantically wrong.)

The comment below by drfloyd51 is similarly wrong ... hahajk didn't say anything about "giving exceptions to powers in the rules".

And the comment by speglt is also wrong:

> If taking a square root is an inherent use of a two, then dividing a number by another number should also be a use of a two, because there are two operands.

The number of operands isn't relevant; the degree of the root is.

And from SideQuark we get

> One could even claim division is raising to the -1 power, thereby disqualifying it in exactly the same manner you disqualify sqrt.

But hahajk said nothing about anything being disqualified. (Not to mention that division is not raising to the -1 power ... raising what? n/d = n * d^-1, but this is completely irrelevant to hahajk's argument.)

spieglt

4 days ago

Multiplication is adding numbers to themselves. Exponentiation is multiplying numbers by themselves. Ditto division/taking roots. Squaring a number is just the instance of exponentiation that happens to have two operands. If taking a square root is an inherent use of a two, then dividing a number by another number should also be a use of a two, because there are two operands.

Biganon

4 days ago

Yes, taking the nth root is an operator. Taking the SQUARE root is this operator with the operand 2, which is usually graphically omited. I totally agree with the top comment: it's burning one of the 2's.

jibalt

3 days ago

The obvious counterargument is that the problem statement is

> Given exactly four instances of the digit 2 and some target natural number, use any mathematical operations to generate the target number with these 2s, using no other digits.

and the square root symbol is a mathematical operation and is not a digit. That it is equivalent to taking the 2nd root isn't relevant ... the whole puzzle is about finding and using equivalents.

SideQuark

4 days ago

Alternatively, one can claim sqrt is a single symbol, and there are extensions requiring a digit. Log, ln, also have hidden bases, but they’re also valid symbols. Cos/sin may be in radians or degrees, both in common use, and no real place to denote it.

One could even claim division is raising to the -1 power, thereby disqualifying it in exactly the same manner you disqualify sqrt. I’ve never seen sqrt with a 2 index; power -1 is seen quite often for division.

Go figure….

Someone

4 days ago

It’s true cheating, but if you assume the Peano axioms, you can write any integer with one zero without having to resort to complex functions such as √. For example (if I didn’t miscount the S-es or parentheses),

  S(S(S(S(S(S(S(0)))))))) = 7

rav

4 days ago

You don't even need zero in that case, you can use the fact that 2 - 2 = 0 to use the four 2s:

  S(S(S(S(S(S(S(2+2-2+2)))))))) = 7
...which is essentially the same as Dirac's general solution (which also derives a "successor operator" akin to S that doesn't spend any numeric literals).

EDIT: I realize that this was already posted in a comment by shoo several hours ago.

vessenes

4 days ago

You’ve got S(4) in the center there

jasomill

3 days ago

From the text, the inner term was clearly intended to be "2-2+2-2".

user

4 days ago

[deleted]

adas0693

4 days ago

there is a related problem (inspired by an interview questions from ITA software, a part of Google now) that asks to find the smallest positive integer that cannot be expressed using a digit K (in [2,9]) N times in an expression that involves the four basic arithmetic operations and the parentheses: +, -, *, /, (, and ).

to make the problem more interesting, the list of operations can be extended to include exponentiation and concatenation.

for example, using the four basic operations for N=4, and K in [2,9], we get:

7 cannot be computed in 4 2s; 11 cannot be computed in 4 3s; 10 cannot be computed in 4 4s; 8 cannot be computed in 4 5s; 9 cannot be computed in 4 6s; 4 cannot be computed in 4 7s; 5 cannot be computed in 4 8s; 4 cannot be computed in 4 9s;

catlifeonmars

4 days ago

This feels very suspiciously like a finite field problem.

genter

4 days ago

Are you allowed to brute force it?

argee

4 days ago

That's how you do it right? It's a BFS with 4 being the max depth, and you print whatever numbers in the problem set which you didn't find.

twoodfin

4 days ago

Given the problems ITA solved, thinking about the question that way makes perfect sense.

dooglius

4 days ago

Figuring that out is presumably part of the interview (and would depend on N)

omoikane

3 days ago

A recent IOCCC entry solves a related game of making integers out of variable number of constants (e.g. make expression using only "2" and basic operations, but any number of "2" can be used as opposed to always four):

https://www.ioccc.org/2025/kurdyukov/index.html

igor47

3 days ago

I had a teacher in middle school who would assign a version of this over winter break with the digits of the new year. I had a lot of fun with it in 1996. Sad to think she must've stopped in 2000...

bajard

4 days ago

22 is cheating because 22 = 2*10+2 is arbitrary in base 10. If you write 22 in some base, you have to write the base, and that's an extra number.

teo_zero

4 days ago

Never understood why 7 is considered difficult...

7 = ⌈ 2.2² + 2 ⌉

anthk

4 days ago

(2 << 2) - ( 2 / 2)

blueblock

4 days ago

Rounding spoils the fun of the puzzle.

nostrademons

4 days ago

It seems very much in the spirit of allowing concatenation and square roots...

jtari3333

4 days ago

Rounding is a completely valid mathematical operation.

user

4 days ago

[deleted]

AnimalMuppet

4 days ago

It is difficult when the rules do not allow exponentiation.

Jtsummers

4 days ago

  7 = ⌈ 2.2 + 2 + 2⌉
Would still work.

user

4 days ago

[deleted]

shoo

4 days ago

There's a direct trivial solution for any natural number using the successor function S(a) = a + 1 from Peano arithmetic.

Encode 0 as 2 + 2 - 2 - 2 then apply S a times to that 0.

Only depends upon natural numbers & Peano stuff, no need for construction of rationals, irrationals, logarithms, etc.

edflsafoiewq

3 days ago

This and Dirac's trick are both writing the number in unary. Is there a way to write it with o(n) symbols?

shoo

3 days ago

That's a fun question.

If we're allowed to introduce our own operators, we could concoct something like decimal notation

a . b := 10a + b

So we could write 90210 as 9 . 0 . 2 . 1 . 0 = S(S(S(S(S(S(S(S(S(0))))))))) . 0 . S(S(0)) . S(0) . 0

That doesn't quite work as it still depends upon a bunch of literal zeros. We could replace 0 with an operator Z() := 0, but that seems to be very much against the spirit of the challenge.

We could rewrite our ersatz decimal expression to only require a single zero & then evaluate a vector of functions f : N -> N across that zero, resulting in a vector of natural numbers, then reduce that vector to a single number with our `.` decimal notation operator.

Let M(fs, x) := (fs_1(x), ..., fs_n(x))

Let R(x) := x_1 . x_2 . ... . x_n

90210 = R(M((SSSSSSSSS, Id, SS, S, Id), 0))

the number of symbols to write a d-digit number this way is bounded by 10 + 10d = 10 + 10log_10(n)

Is this cheating / against the spirit of the challenge? Perhaps!