The emoji problem (2022)
24 comments
·May 20, 2025OisinMoran
I love this genre! I've started calling it “Dantzig Sniping” and here’s my own one: https://x.com/TheOisinMoran/status/1298305686082744320
Some more context and related ones here: https://x.com/TheOisinMoran/status/1299124512240398336
przemub
I thought you named it for Gdańsk (Danzig) and was wondering what’s got sniped there :)
jsheard
> A guy by the name of Sridhar Ramesh
Sridhar is a pro follow, you don't see many people with PhDs in both math and shitposting.
adzm
On a related tangent, when I was teaching my younger kids about math and helping with homework, I would often rewrite things as a formula, or rewrite the formula itself once they got to that point. However instead of things like x I would use things like fluffy cloud, star, etc. They thought it was annoying but it still kept them interested and have said they've done the same when helping their peers. It's easy to forget what it was like learning these abstractions, and it was important to show that x was nothing special, it could be a sun, it could be the phrase "total number of kittens"
oytis
It's year 2025, why doesn't the author use actual fruit emoji for variable names?
voidUpdate
While current year is 2025, the year the language was made in probably isnt
socalgal2
Good or not, many languages choose not to support emoji, even newer ones.
The reasons are complicated but, many people prefer unicode normalization so that different forms of what appear to be the same word are considered the same word. People argue whether or not this is important but it can certainly be argued that it would be frustrating to get an error like
let café = 1;
café += 1; // error, unknown identifier 'café'
But, choosing a normalization affects emoji as well. Worse, when new ones are added the normalization rules can change.
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ykonstant
I remember the problem when it first appeared, we had a good laugh at the number theory seminar :)
gcanyon
Once upon a time I was good at math -- like, win state-wide competitions, get a scholarship good. It's sad that I can easily follow the path laid out here, but I am nowhere close to being able to lay out this course on my own.
dmurray
Nobody can do this on their own. You don't get to do this by having a great aptitude for high school maths and then being exposed to one or two more concepts that just intuitively click - which is enough to be competitive at the level you describe, though not enough to make the US Olympiad team.
You get the level of understanding necessary to create this blog post by already having worked through several textbooks and papers on number theory, elliptic curves, etc. So you don't do it on your own, you do it with help from Poincaré and Euler and Diophantes and whoever else.
Based on your description, you absolutely have the aptitude to do this if you pursued maths further academically. And also by having great teaching and explaining skills - that part might be harder to learn.
patapong
The picture says 95% are unable to solve this - I wonder what the correct percentage is. Closer to 99.999% I would guess, based on your explanation.
amenghra
Originally, more like 99.999995%
ysofunny
you seem to be missing this idea that mathematics must be doable alone, individually
this appears to be what sets mathematics appart from science? I mean, it's not like Euler, or Poincaré, and so on, are really there with you when "doing math"
point being, mathematics is not a team game. but everything else is??
gus_massa
When I'm teaching congruence or the Gauss method to solve equation systems, I tell my students: "You can solve the midterm alone or with the ghost of Gauss sit on the side chair."
mcphage
> point being, mathematics is not a team game. but everything else is??
Sorry, it’s early, and I can’t tell if you’re being sarcastic. I hope you are? Mathematics is very much a team game.
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less_less
The theory of elliptic curves goes amazingly deep. Scratching the surface slightly more, to fill in the article's "ignore the labels like 2P":
Intersecting the curve with lines the way the author does is, perhaps shockingly, a commutative group operation, known as point addition. You define this operation by saying that the three points A,B,C on a line sum to zero: that is A+B+C=0 or in other words, A+B = -C. Reflecting across the curve's line of symmetry is negation (there's an alternative definition that extends to curves without reflection symmetry). Combining the two defines an operation A+B which adds two points on the curve and gives a third one: draw the line from A to B, intersect it with the curve to find a point -A-B (using the same type of formula given in this article) and then reflect it to get A+B. This addition operation obviously commutes (meaning, the line between A and B is the same as the line between B and A), but surprisingly it also associates and you get a group operation.
(For math olympiad nerds out there: so the union of a conic and a line is also a bivariate cubic equation. You can carry out the same "addition" operation there. Again the operation clearly commutes. But it also associates! This is basically Pascal's theorem.)
The theory of elliptic curves is also the basis of elliptic curve cryptography. In that case, instead of the curve being over the reals, all the calculations are done mod some prime p, which destroys structure based on continuity and prevents the numbers from becoming too large. There are a bunch of subtleties here but the key is that you can still straightforwardly compute addition in this context, with basically the same formulas. Then from addition you can get n*A, both for small integers n (eg 5*A = A+A+A+A+A) but also for large n (e.g. 2*A = A+A; 66*A = 2*2*2*2*2*2*A + 2*A). This "scalar multiplication" operation is a one-way operation for appropriately chosen curves: it's easy to calculate n*A from (n,A), but as far as we know it is hard to calculate n from (A,n*A) ... at least if you can't build a large quantum computer.
This gives you mix of easy operations (eg, addition and scalar multiplication) plus problems believed to be hard, which is a great starting point to build cryptography. There are a lot of important technical details, though fewer than with the new lattice schemes.
(Further comment on the olympiad thing: so you can do this with conic+line too, extending from addition to multiplication and using the same formulas mod p. But it's not as good: it's not hard to find a projection that sends the line to infinity and the conic to the unit circle or similar, and then the group operation becomes equivalent to multiplying the points' coordinates as complex numbers or similar. If the group operation is equivalent to multiplication, then scalarmul becomes equivalent to exponentiation, which ends up being in Fp or Fp^2 depending on some Legendre symbol or other. Exponentiation is still potentially secure: it's basically classical Diffie-Hellman. However, more attacks are known on exponentiation in Fp or Fp^2: the attacks don't outright break it but you need p to be much bigger.)
Edit: unescaped stars make italics.
raverbashing
So the meme was a crap-post and never "supposed" to be solved anyway (in human terms?)
yorwba
At least it was a problem with a known solution obtainable by a standard procedure in not too many steps. Could've also picked a deceptively simple Diophantine equation where it is unknown whether it even has any solutions or not: https://thehighergeometer.wordpress.com/2021/07/27/diophanti...
Xplan
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chuckadams
The original "95% of people can't solve this" was always just rage-inducing tracking clickbait anyway, painfully easy problems to make you feel superior if you were stupid enough to believe the claim in the first place. 95% didn't bother because the tone of them was about as annoying as another forwarded email from grandma listing "fun facts" about how Coca-Cola will dissolve a tooth.
Past tense because you don't really see those anymore, so it's actually kind of funny now to see it come up "ironically" with a puzzle that really means it.
There is a great quora answer for this one: https://www.quora.com/How-do-you-find-the-positive-integer-s...