Readit News logoReadit News
Posted by u/MCSP 2 years ago
Show HN: I made a puzzle game that gently introduces my favorite math mysteriesrahulilango.com/coloring/...
This is the first iteration of a short game I’m making that tries to interactively explain some of my favorite math questions / ideas. My goal is mostly to get the player curious and not necessarily to explain absolutely everything.

There were a lot of fun technical parts to building this:

- For implementation reasons, it’s much easier if the lines all have integer intersection points with each other. To do this, when a new line is added I “cheat” by rounding intersections to integers and then splitting the old lines at the intersection into new linds (with potentially different slopes) going through the rounded point

- I had to draw semi accurate maps of actual places (UK, South America, US west coast) in the HTML canvas using just line segments. I tried a few different solutions, including using SVG data. I ended up using the topojson library to give nice line approximations to GeoJSON maps

- I use a simple backtracking algorithm to handle the live coloring of graphs

- I use turf.js’s polygonize function to handle finding polygons from line segments (very happy I didn’t have to implement this myself!)

- I wanted to make the game as mobile friendly as possible (don’t think I’ve nailed this quite yet)

There were also a few tradeoffs I made:

- I wanted give links earlier in the game for players to learn more, but I decided to wait until the end to maintain the flow of the game

- In order to make the game more mobile-friendly, I generally stuck to maps with a small number of regions (at least for maps people have to interact with them). So for the most part all of the instances in the game are “easy”

bgoated01 · 2 years ago
I showed this to my two kids, and we all three enjoyed it. The zero knowledge proof portion didnt really click for me, but we liked the four color map theorem stuff. This led me to download some maps for my kids to attempt coloring on paper, and also got me wondering about how this holds or doesn't on non-euclidian spaces. Turns out the maximum is four colors on a sphere, but 7 colors on a torus! More details here: https://mathworld.wolfram.com/TorusColoring.html

Thanks for leading us down this mathematical rabbit hole today.

ianseyer · 2 years ago
The best analogy of zk-proofs I've heard is to suppose you have found Waldo in "Where's Waldo," and want to prove that you have done this without revealing the location.

You could take a piece of paper (much larger than the picture/book), and cut out a waldo-shaped hole it and position the paper such that he is shown in the hole. Then, when you show it to the challenger, they know that you have found him without you revealing where he is.

delecti · 2 years ago
It took me a minute to fully get this, so I'm adding this so it's a bit more obvious for anyone else: the piece of paper is much larger than the picture/book so that it can hide the book's relative position underneath it.
nroets · 2 years ago
But it's a simplification: One iteration is enough to detect lying.

In a real ZK proof the probability of the prover lying reduces after each iteration but never reached 0.

isaacfung · 2 years ago
How do I know if it is the original "Where's Waldo" under the paper?
IshKebab · 2 years ago
I think that conveys what a zero knowledge proof achieves but it doesn't really correspond to any real zero knowledge proof algorithms. You can't do that over the phone, which is kind of the whole point.
dmurray · 2 years ago
Four colours also isn't enough for some real-world country maps. Countries with enclaves (like Alaska, Kaliningrad or Sint-Maarten, though only the last actually makes 4- colouring impossible) change the topology: you can think of what shape the earth would have if there was also a tunnel connecting Alaska and New York.
Blammar · 2 years ago
Torus, of course, but how does that relate to enclaves, which are planar ?
pwmtr · 2 years ago
This video helped me a lot to understand the basics of zero knowledge proof: https://www.youtube.com/watch?v=fOGdb1CTu5c
wonger_ · 2 years ago
Great video, thanks. My takeaways:

- The physical demonstrations at 1:16 and 3:48 were helpful

- thinking of proofs as between a prover and a verifier, not as classical deterministic proofs

- insight into the name "zero-knowledge", as in "you can already predict the answer, so you're not gaining any knowledge from that interaction"

guyomes · 2 years ago
And it turns out that on the Klein bottle, the maximum is 6 colors. More generally, this number depends directly on the number of holes of the surface [1].

[1]: https://en.wikipedia.org/wiki/Heawood_conjecture

enlyth · 2 years ago
Regarding the zero knowledge proof, there'd be a chance that the two random ones you reveal are the same color, if three weren't enough.

So by doing this over and over again, if the two you choose are always different colors, you approach a 100% certainty that it's legit.

You never really get a 100% proof but the more times you repeat the closer you are to being sure. At 99.999999% after repeating this enough times, you'd most likely be satisfied.

Aardwolf · 2 years ago
When I read the zero knowledge proof part the first time, my thought was:

The person can just always give two different colors (and never two same colors) when you reveal two post it notes, so you could never proof them false.

Only after re-reading I realized _all_ the colors are to be hidden under the post it notes _beforehand_, not at the time you choose two post it notes.

Aardwolf · 2 years ago
Does the torus one apply to a wrap-around 2D map then? So if the edges don't wrap, the max amount of colors needed is 4, but if the edges do wrap, you could make a map requiring 7 colors?

EDIT: yep looks like it, the leftmost picture under 'snapshots' here is a 2D map of what's on the torus and looking at e.g. the blue band, it touches all 6 others (just barely the cyan and yellow ones with a few pixels of its tips when wrapping between top/bottom): https://demonstrations.wolfram.com/SevenColoringOfATorus/

armchairhacker · 2 years ago
Here's my explanation of the ZKP:

First, both you and the oracle know what the uncolored map looks like, and what the 3 colors are.

Step 1: The oracle sends the entire 3-colored map, but asymmetrically encrypted. So you can't see the map, but the oracle can't "un-send" it. If no 3-coloration of the map exists, the oracle has to send you something, and because you know what the map and colors look like, the only thing they can send is a map where some two adjacent regions have the same color.

Steps 2&3: You randomly pick two adjacent regions and ask the oracle for their decryption keys, so you can see their colors. When you initially received the encrypted map, you could not know whether it had two same-colored adjacent regions. But, if you happen to randomly choose the same-colored regions, the oracle has no way of not telling you. If the oracle refuses to send the decryption keys for those regions, or sends ones that don't successfully decrypt, you'll know something is up, and can assume the regions are the same color. And the only keys that successfully decrypt the regions' colors, decrypt into their original colors; so the regions will only decrypt into different colors, if they were different colors in the encrypted map the oracle originally sent.

To further illuminate: the "random pick after the map is received" ensures that the oracle must send you a map where all adjacent regions have different colors, even though you can't see all the colors yourself. Otherwise, the oracle can't guarantee that you won't ask for the two regions with different colors (because they sent the map before you randomly pick), and if you do ask for those regions, they can't respond in a way that re-assures you the colors are different (because the only way to do so is send keys that decrypt the regions into separate colors, and since they decrypt into the same color, such keys don't exist).

Step 4: Repeat steps 1-3 an unbounded number of times. This is necessary because in a single iteration, there's a chance that the oracle sends a map with two adjacent same-colored regions, but you pick two different regions; so the map is un-3-colorable, but you don't find out. In fact, this is a very high chance if it's a large map and only a single pair of adjacent regions have the same color. But more iterations increase the probability that you do find out indefinitely; with enough iterations, the probability that one of them you get lucky and select the same-colored region is 99.999...%.

Also, each time you repeat steps 1-3, the oracle sends the map with a different coloration. Otherwise, you'd slowly reveal the colors to get the fully-colored map, so it wouldn't be a zero-knowledge proof (two colors in each of two maps with different colors doesn't give you any more information than two colors in one map, so even with unbounded iterations the original coloration isn't revealed). If the map isn't 3-colorable, the randomization doesn't affect the probability: when the oracle randomizes the map, they could choose an entirely different coloration and give two different adjacent regions the same color, but the probability of you randomly choosing those two regions stays the same, so the probability of "getting lucky" in one of enough iterations also stays the same, at 99.999...%.

helboi4 · 2 years ago
same that was extremely fun but im not sure i'm going to grasp the zk proof thing
franciscop · 2 years ago
[spoiler alert]

I knew that 4 colours sufficed for any arbitrary map from back in the day when I learned this, but still I found it VERY rewarding by attempting to draw a map that needed 5 colors, and how intuitive this demo was for getting a "feel" for a thing that I knew only theoretically! Like I needed an impossible geometry to fit, either an area that stretched to a zero-width path (which would becomes a point, and thus 2 areas, so doesn't fit) or some other "impossible" geometry. Loved it, congrats on a really well executed idea!

8organicbits · 2 years ago
It was really enjoyable to try. I got an optimal 4-color map on the first try, so I was over confident. My approach was something like: put a bunch of the 4-color maps next to a long Chile-like thing. When that didn't work I added borders until my device couldn't render it any longer. Very very fun.
genewitch · 2 years ago
there has to be a layman's explanation. I knew the easiest way to get four colors was to put a split square inside another split square "donut", but the reason you can't force 5 colors is that word "inside". There has to be a nice, tidy "verbal proof" that no matter what, one or more colors will be "trapped" inside at most 3 other colors.
waterproof · 2 years ago
You would think. The easier, five-color theorem proof fits in a few paragraphs but the four-color theorem really resists a simple explanation. Even the simplified 1990s version of the proof (which came ~20 years after the original proof and 100 years after the 5CT proof) required enumeration of hundreds of individual cases.

https://en.m.wikipedia.org/wiki/Four_color_theorem

zamadatix · 2 years ago
Loved the interactions and flow overall but I'm a bit lost on the zero knowledge proof example. I'm familiar with the concept but I don't follow how the example is one. E.g. "By repeating the process enough times, the probability that you never catch me becomes smaller than, say, getting struck by lightning" doesn't seem to show it's a proof? If I pick a hundred numbers it'll look like I just proved some black box function which happens to be Sin[n] + 0.999999999999 is always positive even though I'd be able to clearly show it negative with the knowledge of the function.

It feels like something that got detached from the things that make it work during simplification. Or it could be that I just have a misunderstanding/oversight in the zero knowledge proof :).

In an unrelated note: I colored the larger graph and it didn't even play along!

MCSP · 2 years ago
Very glad you enjoyed it!

For the ZK example, the math behind it is this: if there are m bordering regions and I am lying, you have a 1/m chance of catching me each time. Thus after k repetitions the chance you haven't caught me is (1-1/m)^k \approx e^{-k/m} which is extremely small for k sufficiently larger than m.

Now, you may rightfully say: hey that's still not a "proof," you could still be lying! There are two responses to this:

1. The probability can be made incredibly small, like smaller than the the chance, say, your computer got hit by a gamma ray burst that would flip bits from 0 to 1 (I really have no idea if this actually happens but people have said it to me).

2. It turns out it is mathematically impossible to get the zero knowledge property if you want true proofs (i.e., no probability of being wrong). So, there's a trade off: if you want zero knowledge, you have to accept some (small) failure probability

P.S. Adding an easter egg for coloring the larger graph is on the todo list :)

xg15 · 2 years ago
Yeah, I got tripped up by that formulation as well and it's actually something that annoys me with a lot of algorithms that have some properties proven in a limit: It's "easy" (or at least possible) to mathematically prove that in the limit of some variable, the property will hold: If you repeat the challenge increasingly often, the probability of being lied to will get arbitrarily close to zero; for sufficiently large input sizes, some algorithm runs in linear time; with sufficiently large amounts of training data and iterations, some prediction error will become arbitrarily small, etc etc.

But none of that is telling you how much is "sufficient", or even which order of magnitude we're talking about. If the quantity has a real life cost, this would result in enormous practical differences.

(With the formula you have given for the ZK proof, we're at least one step further: You can start with the desired probability, e.g. the gamma ray burst und calculate the required minimum k from that - also, it's easy to see that the color problem lends itself well to such proofs because the probability of failure drops exponentially quickly with growing k, so the actual k you choose can be relatively small. But if all you have is a proof in the limit, that's not possible)

ncruces · 2 years ago
The problem with doing this on a computer is getting us to believe you didn't just make up the colors as we tell you to reveal them (after being “dishonest” before).
zamadatix · 2 years ago
Thanks for the explanation, it seems the definition was slightly different than I assumed it to be previously and that was my missing link to it all making sense. Thanks also for the demonstration to share this info!

Looking forward to the easter egg :)

quuxplusone · 2 years ago
OP: Although I'm not really in the target audience for this demo (I already knew all the punchlines), it does occur to me that it might be helpful to readers like the parent-commenter — and even perhaps thought-provoking to us know-it-alls — if you provided a mode at the end of the demo where the graph was in fact not three-colorable, and the computer actually would lie about its being three-colorable. So it would generate a "three-coloring" with a flaw somewhere, and display its representation as products of primes, and you'd get to choose two adjacent products and receive their factors... and so you could see for yourself exactly how long it took for you to luck into catching the computer in one of its lies.

And the demo could also tell you the expected number of iterations to catch the lie with (50/90/99%) probability. It'd be a pretty large number even for such a small graph, I'd bet.

(Of course the computer could also lie about the factorizations, since it's unlikely a human would bother to catch it in that kind of lie; but let's assume it doesn't ever do that.)

Readers might also be interested in the https://mathworld.wolfram.com/McGregorMap.html (reported, on 1 April 1975, to require five colors!)

zamadatix · 2 years ago
For me it was less about the idea of how likely you'd quickly it converges to you almost certainly outing them vs misunderstanding the idea that a zero knowledge proof is about, more or less, the "limit" of the validation behavior to an arbitrary point choosable by the tester not necessarily an actual guarantee you can finitely reach the conclusion.

Prior to this I'd only seen "proof" in math where it has meant you can absolutely guarantee there to bo no counterexample not just that it seems impossibly unlikely there could be a counterexample. E.g. the Tarry-Escott problem where we have proof there is no sets exists with n=4 and m=5 even though we haven't ever found numerical values of sets matching that description or Merten's conjecture where the smallest counterexample is estimated to be so large (~10 billion digits) we've not even been able to find the first counterexample value despite knowing it exists due to a proof. On the other side of things we have things like the Goldbach conjecture or Riemann Hypothesis where we've poured our hearts, brains, and souls into trying to find a counterexample or proof and don't claim to have either yet.

Adjusting to that definition of "proof" for the context it all makes a lot more sense now.

raincole · 2 years ago
I've got another problem about this zero knowledge proof. The digital version doesn't make a lot of sense to me. It depends on the fact we don't have a fast integer factorization algorithm. But integer factorization is not proven to be NP-complete, and 3-coloring is NP-complete.

So isn't it possible that there is a polynomial time algorithm for integer factorization, but no polynominal time algorithm for 3-coloring, and therefore the "zero knowledge proof" actually reveals the answer?

dmurray · 2 years ago
I think you're right, and integer-factorization is often used in these examples as a process that is hard to do but easy to verify. There are plenty of other processes that could be substituted in, e.g. reversing SHA256 hashes, that would likely be even less tractable to the target audience.

However, if P = NP, there is no process that works here - there's nothing that is hard to do but easy to demonstrate, and therefore no zero knowledge proofs exist.

Actually, that's not true either. It requires the definition that all polynomial-time algorithms run quickly and all superpolynomial ones run slowly. This is not an accurate definition for all practical problem sizes and this is where the analogies all break down. Polynomial vs nonpolynomial is more interesting to complexity theorists than "how many years would this actually take with a fast computer".

DoktorL · 2 years ago
> E.g. "By repeating the process enough times, the probability that you never catch me becomes smaller than, say, getting struck by lightning" doesn't seem to show it's a proof?

That's fine though, because the point isn't really to publish math papers without disclosing proofs. For example, presenting a valid digital signature is sometimes colloquially called a proof that you had the private key, even though there is 1 in gazillion chance that you didn't. For such practical tasks, very high chance tends to be good enough.

esquivalience · 2 years ago
I think the answer is that each time you reveal the colours, you observe that they are within the set of three colours illustrated at the beginning of the proof. Whichever you reveal, you never find a fourth colour.

This confused me at first.

zamadatix · 2 years ago
For anyone confused by this response I had edited my comment after reading https://news.ycombinator.com/item?id=40740557 but before equivalence had hit reply and now their reply is left hanging. Sorry esquivalience! To summarize the linked answer on trusting the second dot isn't just randomly assigned: keep the context as physical post-its. Barring something like a matter bending psychic you'd be able to tell the dot under the second post-it was swapped as you made your pick.

That still leaves how to rely on chance of picks for a proof though.

gavindean90 · 2 years ago
Reminds me of different colored swans
chromy · 2 years ago
The Republic of Ireland (the west most region on the first page) isn't part of the United Kingdom. The term for the group of regions shown is 'the British Isles'. See https://qntm.org/uk While it seems like a trivial distinction the whole thing is somewhat fraught (https://en.wikipedia.org/wiki/The_Troubles).
romwell · 2 years ago
It's about as "trivial" a distinction as considering Crimea (or the entirety of Ukraine, for that matter) a part of Russia.

Many, many people have died for this triviality.

"Somewhat fraught" is a very interesting choice of words, but then again, so is "The Troubles" (when the subject matter is decades of bombings and killings).

djeastm · 2 years ago
Naming things is hard
MCSP · 2 years ago
Whoops! Switched to "British Isles" -- update should be percolating now
darajava · 2 years ago
I prefer the term “Atlantic Archipelago”. The “British Isles” encompassing a non-british sovereign state is contentious. Other good terms are “Britain and Ireland” or the “British-Irish isles”
nirolo · 2 years ago
You might want to get into touch with some museums on science topics to see if they are up to show it. I live in Germany at the moment and know of at least two MINT focused museums that let visitors engage a lot with their (sometimes digital) exhibits and this here checks all the boxes to make a great exhibit.

Very well done, I'll try to see if my children will enjoy that already too.

wrsh07 · 2 years ago
Absolutely!! Talk to MoMath in NYC too

I'm happy to find a contact there if you want

tzs · 2 years ago
> Step 1. I put a <purple circle>, <blue circle>, or <red circle> color dot on each region, but hide it under a post-it

> Step 2. You click any two bordering regions, and I pull off their post-its

> Step 3. You check that the revealed colors are different

I would add explicitly in step 1 that you tell me what 3 colors you used, and in step 3 that I check that the revealed colors are different from each other and are both from that set of 3. Similar in the explanation of why this should convince me.

gkoberger · 2 years ago
This was one of the cooler teaching examples I've ever played with... awesome job! Appreciate the warning that the 5 color map is "very difficult". It felt easy enough, and I would have spent an hour on it!

This was so much cooler than just being told that 4 colors is enough for every map – this one will stick with me.

It would be wonderful if schools taught a bit more like this – I almost felt like I discovered it myself!

baruchel · 2 years ago
Hi, my two cents; you claim "Although mathematicians believe their proof is correct, it is too complex to verify without computer assistance", but I'm not sure "believe" is the correct verb since the proof has been formally verified (see for instance https://github.com/coq-community/fourcolor for a formal verification in Coq).

I understand that you want to emphasize the fact that no human can understand the proof with a full overview, but I wonder whether the current sentence will not make people think mathematicians are not perfectly sure of the proof.

TheNewAndy · 2 years ago
I would consider a proof to be a "repeatable argument". One you could show to someone and expect them to be convinced by it. I think it is a defensible viewpoint that a proof in coq is not 100% convincing. If you think otherwise, then how can you reconcile the existence of falso (a coq verified proof of "false")?

https://github.com/clarus/falso

Proof and belief I think are pretty strongly intertwined, but I'm not going to pretend to have a particularly rigorous philosophy on the matter. Similarly, when the proof of Fermat's last theorem was published, I don't know if I should consider that to be a proof because it is well beyond my comprehension. I have no reason to question it, but should I consider it a proof? I know that people smarter than me (e.g. Wiles) thought the original version of it was a proof, but it had a subtle error in it which required a fix. While I haven't looked at the proof and revision, I would be surprised if I could look at the two versions as labelled and tell which one is the correct version.

MCSP · 2 years ago
Good point! I'll try to think about a better way to phrase this (happy to hear suggestions)
lcnPylGDnU4H9OF · 2 years ago
"Although mathematicians have been able to prove this, the proof is too complex to verify without computer assistance."

Or some such.

ifdefdebug · 2 years ago
s/believe/know
valenterry · 2 years ago
Nah, actually I agree with you. What counts as believe and what as fact is rather abitrary. Is 2+2=4 a fact? Is global warming a fact? What about man-made global warming? Ask 100 people whether something is a fact or a believe.

To top that up, it's fact that there have been "proves" that were wrong (or maybe that's just my believe? :^]) even for a long time.

Hence, I think we can say that there are 4 options for a theorem:

1) Some mathematician believes the theorem is correct (but can't prove it)

2) Some mathematician believes the theorem is incorrect (but can't prove it)

3) Some mathematician believes the proof of a theorem is correct

4) Some mathematician believes the proof of a theorem is incorrect

Proving that a proof is correct is kind of meaningless. At that point it's all believe anyways.

gergo_barany · 2 years ago
That Coq proof is not "without computer assistance". No Coq proof is, as Coq is literally an "assistant" that runs on a computer.

And those many jobXtoY.v and taskXtoY.v files sure look like they also do the same as the Appel and Haken proof, namely enumerate lots and lots of cases that are then machine-checked. So I don't think the computerized Coq proof is really qualitatively different from other computerized proofs that enumerate so many cases that a manual check would be impractical.