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;
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
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.
Can you not? 9+(9(1/3)) [adding nine to itself a third of a time] is nine plus one third of nine {or nine divided by three} which is one of the three in three times three that equals nine which is three therefore 9+(9(1/3))=11 because 9+3=11
By dividing by 3 you are introducing a scaler into the operation.
Multiplication also flips signs (direction). That's why a negative * negative = positive, whereas a negative + negative = negative. The addition operator does not flip direction.
Multiplication is not generally the same as multiple addition operations. However, it has the same result for positive whole numbers.
Multiplication is the generalization of repeated addition. Repeated addition can be generalized to fractional values.
In the integers, it is the same thing, with a rule for negative times.
In the rational numbers, you can generalize to adding a number to itself a rational number of times. Adding a/b to itself p/q times is defined as adding a to itself p times, with a divisor that is adding b to itself q times.
So adding the rational 5 (5/1) with itself 2_and-1/3 (4/3) times just means (5+5+5+5) / (1 + 1 + 1) = 20/3, a rational that we can also write as 6
-and-2/3. Only addition of integers is used.
Which is the same as rational multiplication (a/b)(p/q) = (ap)/(b*q).
But it's true that you can't say that rational multiplication is the same as _integer_ repeated addition.
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.
Because it's just wrong. hahajk said that taking the 2nd root spends a 2, not that "roots are invalid".
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.
I think you misunderstand. They’re saying (maybe not very well) that the square root is not really a unary operator, it’s a binary operator of which the “square root” symbol just elides the 2 in the left.
But the “rules” of this game are necessarily arbitrary.
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.
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.
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.
I think the series \sum_{n = 0}^\infty 1/((2n + 1)!)x^{2n + 1}, to which the definition in terms of exponentials reduces, is about equivalently simple, but, sure, it has a 2 in it (several 2's in it!) too.
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!
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.
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):
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...
62 comments
[ 0.17 ms ] story [ 15.3 ms ] threadHowever, you can solve this by using the alternative construction of nesting logarithms, factorials and the floor function as much as you please.
And JFYI since December is near: https://rant.gulbrandsen.priv.no/hack/advent-calendar-for-ne...
7 = ⌈ 2.2² + 2 ⌉
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;
Squaring is to multiplication as is the latter to addition. If roots are invalid then so should be multiplications.
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.
Multiplication is a scaling operation.
I like it when people who are clearly smart do things like this! Makes me feel better.
Reminds me of this story: https://news.ycombinator.com/item?id=5663709
Multiplication also flips signs (direction). That's why a negative * negative = positive, whereas a negative + negative = negative. The addition operator does not flip direction.
Multiplication is not generally the same as multiple addition operations. However, it has the same result for positive whole numbers.
In the integers, it is the same thing, with a rule for negative times.
In the rational numbers, you can generalize to adding a number to itself a rational number of times. Adding a/b to itself p/q times is defined as adding a to itself p times, with a divisor that is adding b to itself q times.
So adding the rational 5 (5/1) with itself 2_and-1/3 (4/3) times just means (5+5+5+5) / (1 + 1 + 1) = 20/3, a rational that we can also write as 6 -and-2/3. Only addition of integers is used.
Which is the same as rational multiplication (a/b)(p/q) = (ap)/(b*q).
But it's true that you can't say that rational multiplication is the same as _integer_ repeated addition.
Learn logarithms.
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.
But the “rules” of this game are necessarily arbitrary.
as halving is to addition.
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.
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
> 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.
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….
3=(sin(arctan(sin(arctan(sin(arctan(2))))))^−2 − (2)^−2
Though it really depends which definitions you find more fundamental.
Still, exp and log seem like good choices.
How?
2^2 - (2 / 2).
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!
EDIT: I realize that this was already posted in a comment by shoo several hours ago.
https://www.ioccc.org/2025/kurdyukov/index.html