How to keep an open secret with mathematics.

Shamir's Secret Sharing on wikipedia.
en.wikipedia.org/wiki/Shamir%...
Teaching resource! This was made by a high school teacher in the US to use with Algebra 1, Algebra 2 and Pre-calc. It's a work in progress but you can get it here: www.dropbox.com/s/snee4of46ig...
I made the animated parabola in GeoGebra with help from my mate Ben Sparks. Here is the GeoGebra5 file if you'd like to have a play:
www.dropbox.com/s/zydcy38eeda...
Read about finite fields on MathWorld:
mathworld.wolfram.com/FiniteFi...
Shamir Secret Sharing used with Bitcoin:
blog.trezor.io/shamir-backup-...
And for balance, an argument against using Shamir Sharing with Bitcoin, but rather Multisignature.
blog.keys.casa/shamirs-secret...
CORRECTIONS
- None yet, let me know if you spot any mistakes!
Thanks again, as always, to Jane Street for supporting this channel.
www.janestreet.com/
Thanks to my Patreon supporters who help make these videos possible. Here is a random subset:
Michael Lehenbauer
Peter Gerrard
Matthew Roberts
Ben White
Jan Strohbeck
Lucas Werkmeister
Support my channel and I can make more maths videos:
/ standupmaths
Filming and editing by Alex Genn-Bash
Music by Howard Carter
Design by Simon Wright
MATT PARKER: Stand-up Mathematician
Website: standupmaths.com/
Maths book: wwwh.umble-pi.com
Nerdy maths toys: mathsgear.co.uk/

[@spewkkake]

SUMS: S: stand U: up M: math S: s

[@BaldAndroid]

Acronym

[@KiroOsexXIII]

@@Ymparipyorailija That's a Parker spelling of Parker.

[@thetntsheep4075]

So for Americans it's just "SUM"

[@Ymparipyorailija]

Oof. In the true Parker manner, my comment is incorrect in multiple ways.

[@scottdebrestian9875]

A Parker password.

[@jaxielrivera4640]

Matt: There is no meaning behind 41,366,362 Me: No, there has to be. I googled the number and one of the things that came up was an adobe file image (with the number as its file ID) and it’s a 3D animated couple kissing and hugging, so the secret meaning is that he loves math. There it is. I got it guys, you’re welcome.

[@U014B]

Sauce?

[@yerwol]

@@U014B he..... _literally_ told you how he found it. Go to google, put the number in, and click 'images'.....

[@iAmTheSquidThing]

There has to be some meaning here. Let's work it out together. What's the 41st letter of the alphabet?

[@lexingtonbrython1897]

@@iAmTheSquidThing Hurricanes start using Greek letters if there's more than 26 named storms in a season 41 - 26 = 15 15th Greek letter is Omicron (just 'o') so that doesnt help; 15th Latin letter is also O by sheer luck

[@I25mI25]

@@yerwol It is possible that you wont find that image with google, since google factors a lot more than the search term itself into the results it is showing you. While it is likely, that everybody can find that image (especially when adding adobe), different configurations could theoretically cause you to not get that image as a result.

[@rq4740]

Teachers would have a much easier time getting their students to care about quadratics if they told them that they can solve polynomials to access someone's financial accounts

[@dekunut6416]

So true lol

[@JoostMehrtens]

My teacher told us we need math to maintain our own financial accounts, how boring...

[@nickmartin7418]

Matt: This number is random and has no meaning. Everyone: That number definitely has meaning.

[@Leyrann]

I'm lazy, I just go through the comments to see if it has a meaning or not.

[@numero7mojeangering]

@@Leyrann same

[@Yoshiiro]

The clever-one : My password are the last 6 digits of Pi. The mathematician : It's impossible to guess. The engineer : I'll iterate from 000000 to 999999, should take a couple of seconds.

[@calansmith655]

Yoshiiro Took me a second to get it, but that is genius

[@nibblrrr7124]

lol (well, technically the mathematician would say the clever-one's password was ill-defined, and probably conclude that they don't know what they're talking about :P)

[@zeikjt]

@@calansmith655 Haha, same!

[@delve_]

Yo, would some kind soul be willing to explain? I don't think I get it.

[@zeikjt]

@@delve_ since it's only 6 digits you can write a program to try out all possible combinations of digits really easily (this is known as brute forcing) instead of trying anything more complicated like messing around with pi

[@GammaFn.]

That's your password? "sums"? That doesn't add up.

[@roberttomsiii3728]

Lol

[@MisterHunterWolf]

Lol

[@karlneff]

Lol

[@laurihei]

Lol

[@-YELDAH]

Lol

[@faastex]

By assuring that number had no hidden meaning, I'm now even more convinced it has a hidden meaning.

[@roberttomsiii3728]

Have you figured put that answer yet? Gonna have to find the Reddit for this now ;)

[@vampyricon7026]

BUT WHAT IS THE HIDDEN MEANING?

[@jayd2279]

Well dividing it up like the other one doesn't quite work, the only meaningful ways would be 4|1|3|6|6|3|6|2 = DACFFCFB or 4|13|6|6|3|6|2 = DMFFCFB

[@jayd2279]

You have any other ideas?

[@faastex]

I'm too lazy to work it out, but I trust in you guys

[@lexingtonbrython1897]

Matt's valuables are: a business card Level 2 Menger sponge, a ream of brown paper, a copy of Humble Pi, and a framed picture of the Parker Square

[@thefountainpendesk]

What about things to make and do in the 4th dimension

[@OrcinusDrake]

@@thefountainpendesk that's encoded in micro engravings on the business card

[@amydebuitleir]

@@thefountainpendesk Wow, someone else likes both fountain pens, maths, and Matt Parker! At least. I assume you like fountain pens. If not, strange choice of usernames. ;)

[@thefountainpendesk]

@@amydebuitleir yeah I use em all the time

[@isidore551]

Horcruxes

[@user-hi4sm3ig5j]

That's complicated. I think I'll stick with my trusty '1234'. Nobody will guess that.

[@LeoStaley]

That's the code to my luggage!

[@lexingtonbrython1897]

I'm sticking with hunter2

[@LeoStaley]

@@lexingtonbrython1897 I'm pretty sure most places won't accept "*******" as a password.

[@AA-100]

'password' is an even better password

[@hebl47]

Amazing! I have the same password for my planetary shield!

[@yuvalne]

Referring to Tom's video without risking your brand deal. Clever.

[@Imthefake]

what video?

[@typo691]

@@Imthefake Tom Scott's video on VPN sponsorships.

[@matthewellisor5835]

What are you Nording about?

[@green0563]

Yup, loving that subtle reference.

[@rareroe305]

I already watched it, but Tom's video is in my sidebar for this one.

[@arnet95]

Shamir secret sharing is such a brilliant and beautiful idea.

[@michaellin4553]

It has many hidden uses as well, such as cryptographic commitments

[@terner1234]

שמיר

[@peppybocan]

Yep, it is beautiful.

[@lolledopke]

I logged in to Matt's bank account, but because he's a mathematician I couldn't steal anything :(

[@crispoman]

Well, he's a complex guy, so all his money is imaginary.

[@Bruno_Noobador]

69 likes. Nice

[@liquidkey8204]

@@crispoman Oooh good one

[@second-handyt3958]

@@crispoman this comment is better than the one you replied to

[@cosmicdeven5976]

Toms really out here risking all of his friends sponsorship deals😂😂

[@MephieStopheles]

41.36, 63.62 are Long/Lat's to a location in the desert of Uzbekistan I think he's telling us where he hid the bodies.

[@gbear1005]

I am there... really.. no bodies

[@aldobernaltvbernal8745]

That's very close to Lake Aral, which used to be the 4th biggest lake in the world until the USSR drained it

[@astropgn]

9:30 "And they said no." Hahaha matt's timing is really funny!

[@djones02]

Thank you for clarifying the VPN claims. I'm sick of all the false advertising in VPN ads.

[@robnorris4770]

Now Matt has to change his password to something else we could never guess, like “muss”.

[@vampyricon7026]

Leon Muss: Leader of the movement to make sure the human race stays off Mars... At least until 2026.

[@konstantinkh]

It gets better. You aren't restricted to prime-sized sets. Any finite field will do, so you can use splitting fields. These can be p^n, where p is prime and n is any integer. And yes, 2 is a valid prime for this purpose. So if you need to encode, say, 256 bit key, there's a field over which you can define a polynomial that will admit any 256 bit number, and you'll need exactly the right number of people to come together with their parts of information to know it. The algebra of such fields gets a little intense, though.

[@lierdakil]

9:23 "The only people who will see it [data] are you [...] and ExpressVPN" Not strictly true. Once the tunnel terminates -- and it eventually has to -- "the data" travels through the Internet unencrypted (unless it would've been encrypted without VPN also). Which means that once "the data" left the relative safety of the encrypted tunnel, it can (in theory) be seen by any number of third parties. In essence, using a VPN is trading your ISP for your VPN service. I mean, if you don't trust your ISP, or if you want to circumvent geographical restrictions, sure, VPN services are great for that. But those otherwise offer no more security than there is already there. In fact, those offer arguably _less_ security, since (a) VPN services are kinda shady with their stated no-logs policies often being in direct conflict with local laws, and (b) VPN services can be a prime target for malicious agents and/or government agencies -- if a VPN is compromised, it's statistically speaking less "secure" (in terms of possible data leaks) than not using a VPN at all.

[@nibblrrr7124]

Yep. The selling points of VPNs are mostly overblown, esp. in the age of widespread HTTPS. If you trust their intentions & capabilities, they can provide some anonymity for your _metadata_ to third parties, at the cost of them knowing your entire browsing history. But a vague sense of "security" apparently is easier to market than an all-eggs-in-one-basket mass surveillance self-defence strategy. Or illicit filesharing. :^) Tor e.g. is better for anonymity - it doesn't require you to trust _any_ single party. (And no, "Three Letter Agency runs exit relays" is not even in the same ballpark as "Three Letter Agency has a backdoor at your VPN provider")

[@rene0]

You're missing the point of a VPN. The primary goal more often than not is not to encrypt the message, but to conceal the clients identity to the remote party. The VPN merely serves as a proxy and the encryption is a bonus. It offers little protection against state-level targeted attacks but is usually sufficient for small civil offenses like copying music. Informed users will not have the illusion of high-level protection, only journalists seem to.

[@wolvenmoonstone8138]

video aside i really appreciated the honesty of that ad spot. VPN companies tend to be really manipulative and oversell themselves but not this time it fits the rest of the video better this way

[@zacozacoify]

For some reason I thought it was going to generalise into planes/hyper-planes or higher dimensional lines instead of polynomials.

[@robertwiesner6825]

Thank you so much for this video. We talked about this in one of our university courses and I didn't understand it very much. Having it explained like this with pictures makes it a TON more understandable.

[@boudmaths]

At 7:48 you said: "preferably a prime" Well it's not an option, I mean for the integers mod n be a field n has to be a prime

[@ManuelBTC21]

A power of a prime also works, but then you need to not just do modulo operations, you also have to divide by a reducing polynomial.

[@DeGuerre]

In most cryptographic applications, 2^k for some k is more usual because anything in binary is trivial to represent as an element of the field. There's another gotcha here, in that working in this field is a little more complicated. In field GF(p), the elements are just the integers mod p. In GF(p^k), the elements are polynomials of degree k-1 with coefficients in GF(p). (GF means Galois Field, by the way, but just think "finite field".)

[@francomiranda706]

3:10 "oh hes obviously talking about prime numbers, awesome" 3:24 "what..."

[@SSardonic]

Did the same thing. 3:10 "factorize. Factorize! FACTORIZE! :D" 3:24 "... oh :/" Unsurprisingly, his version is better

[@SillyMakesVids]

I finally understand the use of polynomials in cryptography!

[@defenestrated23]

Equation-over-finite-field-modulo-huge-prime is the basis of a TON of cryptography. RSA, Diffie-Hellman and elliptic curve crypto of course come to mind. Oh and I think Reed-Solomon error correction codes are related as well. Computer security: Brought to you by Galois

[@MK-13337]

To get a finite field you *need* the number to be prime if you take your field as the congruence classes. If you take modulo of a composite number it's going to be a ring

[@christianbarnay2499]

Not only primes. Powers of primes also generate finite fields. They are not used with RSA because RSA is weaker on those fields. But elliptic curve cryptography doesn't have this issue.

[@MK-13337]

@@christianbarnay2499 If I recall correctly (I took algebra years ago), you get a finite field from p^n if you consider some polynomials or roots of degree p^n..? I might very well be wrong and I'm too lazy to check but what I meant was if you just have [x] mod p^n (the congruence classes modulo p^n) then p*p^(n-1) = p^n =0 mod p^n and we no longer have a field.

[@christianbarnay2499]

@@MK-13337 Oh, you're right I was not fully awake when I read your comment. p^n fields are not modular.

[@JasonSmith0]

I always imagined this is how horcruxes are implemented.

[@danieln7777]

That's incredible! Thanks for showing me this awesome math! I wrote my own program that does this and it was a lot of fun. Thanks for inspiring me today!

[@thekmotr135]

Or maybe use: Random x,y,z (similar digit count as “sums”) Use formula 19,211,319 + x + y + z = Result 1. Friend gets x 2. Friend gets y 3. Friend gets z 4. Friend gets Result Everyone is needed

[@wingracer1614]

You don't want everyone to be needed. What if you and friend 1 die in a car crash? The remaining friends can't access your secret.

[@sublivion5024]

Having 1 or 2 values would make checking all values easier

[@NStripleseven]

That’d be an interesting social experiment. If you took some really high-value polynomial, handed out points to, say, 1 thousand people around the world, and also gave each of those people a name of another person who was participating, how long would it take for those people to find each other and figure out what the polynomial was?

[@Shaquille69355]

Guys I just signed into his KZbin channel.

[@benjaminsonntag7262]

For us linux user, the package named "ssss" allows you to do a Shamir Secret Sharing easily (using the commnd line ;) )

[@nibblrrr7124]

Neat, didn't know that one! (Though it seems to be some guy's hobby project from 2006 & not actively maintained, so maybe don't rely on it for super important stuff. ^^)

[@JNCressey]

Bob: What does the fourth S stand for? Alice: It's a secret.

[@Slikx666]

I've got dyslexia, it's funny what smelling mistakes are in my passwords.

[@mmburgess11]

I hide all my passwords in the comments section of "Watch Later" maths videos under a pseudonym....No self-respecting crook would ever watch maths videos, let alone all the comments.

[@willemvandebeek]

Happy new year and best wishes for 12020, Matt and company! :)

[@isaiahlopez51]

This is so amazing!! Love this video!

[@ethanjensen661]

So fun seeing you in MathVengers, Euler's game this year!

[@midik13]

excited to see this topic covered. My master's thesis was on this subject!:)

[@MrChernobyl22]

Awesome! So simple and yet very clever

[@Archeious]

I use SSS at work. We use it to unencrypt servers on boot. The servers requires a very long manual password or 3 authentication servers have to agree to unlock the server. This is nice as it mean if a server reboots in the middle of the night it will automatically unlock (as long as there are authentication servers). Check out Tang and Clevis if you are interested in running the service locally.

[@nibblrrr7124]

Ooooh that is cool! Thanks for the tip! (Maybe add _Network-Bound Disk Encryption_ to your search query; turns out clevises & tangs are also mechanical parts. ^^)

[@Archeious]

@@nibblrrr7124 LOL Good idea. I forgot about that. Man it was a pain, but I now know what a mechanical tang/clevis is.

[@marcoantonio5662]

What a nice explanation! Thanks a lot!

[@AlexanderRafferty]

I was expecting him to simply split the password into several binary strings that need to be XORd to recover the original password.

[@radadadadee]

That's clever but not as clever as Shamir's idea because you can't set a min threshold to unlock the secret. With XOR'd bits you need all of the pieces to unlock.

[@K0nomi]

1:24 when the impostor

[@bagelmeister2295]

Is su_s

[@glowstonelovepad9294]

my feet is sus

[@PsychoMuffinSDM]

If someone told me this when I was in middle school learning about y=mx+b, I think I would have been so much more intrigued. And I liked math already too!

[@DiegoMathemagician]

These ideas are so neat wow

[@__mk_km__]

You can use a simpler system if all of the keys are required to get the secret. Generate N-1 random numbers. Then take the reciprocal of a multiple of those numbers, multiplied by the secret(everything in a finite field of course). Those N numbers are the keys. To get the secret, you just multiply all of the numbers together. The last key contains most of the information, provided you can recover it. But factorising a large number is very hard(and AFAIK can only be done with bruteforce if in a field) Shamir Secret Sharing System is more flexible though, since you can generate however many keys you like.

[@NoMoreUsersAvailible]

Some HSM (hardware security modules) vendors use Shamir Secret Sharing to implement their operator threshold systems. For example, the nCipher line (which was bought by Thales a while back and just recently was sold to Entrust) uses it to export the security world master keys to the backup/admin card sets in an M of N threshold.

[@jameshouseweart]

Finally a real vpn add. Toms video about them is really informative.

[@phasm42]

Brilliantly explained.

[@madacol]

Awesome!. Loved this topic!

[@infundere]

nicely explained, good job.

[@compugab]

Simple yet effective!!! I like that!!!

[@patrickboner]

Thanks for the password

[@samuelalphabet5360]

im only 30 seconds in but imma have to call "Stand Up MathS" a Parker Password AND a Parker acronym EDIT: Aight I watched the rest of the video this is pretty cool

[@thecoolkid440]

You could do a focus video on all of RSA from a math perspective. You could do an entire video on each attack on RSA and how those attacks are solved. The math behind cryptography is beautiful. I would encourage everyone to look up Ron Rivest, Adi Shamir, Leonard Adleman. The developers and the namesake of RSA which is widley used today with little competition. The only other system that comec close (off the top of my head) is ECC.

Great video! Thanks a lot!

[@slunce12]

This was probably the firsts VNP ad spot that didn't make me feel all icky :D Thanks Matt!

[@MintyBlaziken]

Somebody graph the Bee movie script

[@juanpods]

boi... HE SUS

[@ThomasGiles]

Interesting stuff! So if you've cut the poly into tons of dots... how would those 2 people (or whatever) get back to their original value to then work out the secret value where it crosses Y?

[@I_Echion]

Late congratulations on 500k

[@danielb7006]

It's okay, he said he would celebrate at 2^19

[@TintagelEmrys]

I would have thought you also could have just given everyone a factor of the secret, so only once all are together could they know what the full number is.

[@ethanshaw8256]

Love the video! Expected prime factorization and got something much cooler. Where can I buy your shirt? It’s beautiful

[@RJSnyper]

I had something similar that I used for a long time where I would write two or three words nearby to whatever needed the password. The password wasn't the words, but rather the result of some order of operations done on those letters (converted into numbers) that was then translated back into letters to form a random string. All you have to do is remember a fairly basic order of operations, but the odds of somebody finding several words, converting them to numbers, running them through 3 or 4 specific math operations, and then converting them back into a string of letters that don't actually spell anything, seems quite low.

[@clandestin011]

"The solution of course, is to not use real numbers" the way he said cracked me up so bad

[@smallbar2012]

Big fan of that Parker Transition at 8:18.

[@user-oc3sl4to9m]

Surely it would’ve been easier to get his number then do some operation to it ( for example divide by 3)have the answer as one of the secrets then the reverse operation as the other (x3). Neither secret would make it easier to find the original answer You could split it up to any number of people by doing more operations to get back to the first secret

[@MechazoidApocalypse]

Can someone explain how that finite field works? How do you get the answer from 3 points for example?

[@CraigNull]

Take Z_5, the field with elements 0, 1, 2, 3, 4. Don't know how familiar you are with how you get the answer with real numbers so here's an over-abundance of details. There's an quadratic polynomial p(x) = a0 + a1*x + a2*x^2 with unknown coefficients a0, a1, a2 (where a0 is the password). You have the three points (xi, yi) for i = 1, 2, 3. Construct 3x3 matrix M where the (i, j) element is xi^(j-1) and make column vector y from the yi. Have a be the unknown column vector with entries a0, a1, a2. If you keep the order of things proper and consistent and more careful than I can be in a youtube comment then the equation M*a = y, if you multiply out the left side, is just saying p(xi) = yi, one row each for i = 1, 2, 3. To solve for vector a you invert matrix M and multiply it through on the left. The inverse of M will always exist so long as the xi are distinct. Z_5 hasn't entered the picture yet, because everything I said so far is true no matter what the field is. For Z_5 the elements in M are just one of 0, 1, 2, 3, 4. The inverse can be computed entirely in Z_5. It's not hard to do by hand, just keep modding by 5. The weird part is multiplicative inverses. Just keep in mind you don't have things like "1/2", you have to use the quirky multiplication table of Z_5, where the inverse of 2 is 3 (since they multiply to 1), etc. So once you have M inverse, M^(-1), you compute vector a with a = M^(-1)*y. In this cryptographic application you only need one of the rows, since a0 is all you're after

[@michaljanecek1103]

This was awesome!

[@IsYitzach]

I just got done watching a video on Reed-Solomon. Its a related algorithm. This algorithm is how break up a Reed-Solomon transmission so that N people can cooperate to get enough information together to work out the Reed-Solomon algorithm and reconstruct the missing data, ie error correct.

[@mahousenshi00]

I'm a hard of hearing person and have a math degree and like yours stand ups videos but I can't follow your channel videos because there's no subtitles. I know the auto subtitles are horrible but still better than nothing.

[@nitfumble]

Leaving a nice comment, because I like Matt's videos!

[@vylbird8014]

If you calculate more points on your polynomial than you need, and pass them all to one person, you get something that looks a lot like Reed-Solomon forward error correction coding.

[@nibblrrr7124]

oooh, fascinating! * zoom in on claude shannon in the background, amusedly eating popcorn *

[@therabbit3307]

This is the first KZbin video I watch in 2020.

[@Seth_M-T]

I'm so sorry.

[@therabbit3307]

@@Seth_M-T Wait. This is actually really funny

[@Richardincancale]

Nice and interesting - thanks. But did you think the quite loud background music would help? Your voice is fine ‘a Capella’!

[@gordonrichardson2972]

The combination of low audio volume and background music makes the narration barely intelligible. Good way to hide information in the video...

[@olgierdvoneverec4135]

I use hashing, meaning i also split my passwords in 2 except one of the parts is the way in witch you have to transform the other part for it to work.

[@latt.qcd9221]

Could you, instead, come up with a polynomial that's actually the polynomial involved in finding the solution to a system of equations using Krylov methods? In Krylov methods, you have a system Ax = b and you attempt at coming up with an approximate solution y. You define a "residual" as r = b - Ay and the algorithm seeks to make the norm of r as small as possible. Using a Krylov subspace (i.e. span{b, Ab, A^2b, ..., A^(m-1)b}) you can take the approximate solution y to be an element of the subspace and express the residual as a polynomial, r = p(A)b. You can work out what the coefficients of the polynomial are supposed to be, but this gives you a way of "encoding" a polynomial which can contain information about the password in much the same way as the Shamir method, but you can give people rows of a matrix and also give people single elements of the right-hand side vector 'b.' That way, you've changed the problem from not only finding the right polynomial, but also having to solve a large system of linear equations and you need to know how to reconstruct the matrix and right-hand side to begin with. Not to mention, you'd also need to know to what precision you need to find the solution in order to find the residual the approximates the polynomial to a degree sufficient for finding the polynomial that encodes the data.

[@PlainPlaneOfficial]

I may have misunderstood the explanation why dividing polynomials by a big primes makes this more secure but I don't really get it. Polynomials are always continuous so if you have two of the three points, how can you limit the numbers you need to search to find the intersect? Wouldn't you still need to search all numbers to find that third coefficient of the polynomial? How can a function f(x) = a_0 + a_1*x + a_2*x^2 + ... not be continuous? And if you divide everything by a big prime, wouldn't you only have to search the numbers up to that big prime? Is the prime itself known to everyone?

[@zolv]

I think a better solution would be to move to higher dimensions and not polynomials. Surface needs 3 points to be defined in 3D. Hyper-surface in 4D needs 4 points and so on. There is no need to use a prime division reminder to mess up to remove giving a clue.

[@PaulMansfield]

Check out the Vault application from Hashicorp.

[@musikSkool]

Additive OTPC with wraparound for when you run out of letters. "test" = 20 5 19 20 Add these two together to get the secret, "test" "pine" = 16 9 14 5 "dveo" = 4 22 5 15 Notice how "i" and "v" are 9 and 22, that makes 31, if it is larger than 26 you subtract 26, making 5, or "e" in the output. Any number can be added to your number to get any output. Unhackable with only one number.

[@musikSkool]

In the case where you don't want anyone to know the length of the secret, simply add a bunch of null characters and use 27 as the wraparound. Example: "test" = 20 5 19 20 "pinepine" = 16 9 14 5 16 9 14 5 "dweokrmv" = 4 23 5 15 11 18 13 22 If the numbers add up to 27 that is a null character and not a letter, it isn't added to the output. Notice how the "v" in the previous example became "w" in this one. It is because the wraparound number is 27 now instead of 26, allowing for the null character. This secret word can be any length of letters up to 8, disguising the length of the secret. We call this "padding" in cryptology.

[@jan_harald]

that's a neat trick in case you die or get in an accident or such, sure the trick is to always keep one of the required parts with you, so that you will always know when somebody's got your password, and use this sharing for your password manager's password (if you're not big brain and don't use a stateless manager)

[@drewdurant3835]

Happy new year!!

[@MatthewStinar]

Just last week I was trying to remember the name of this. I knew it was something secret algorithm, but that wasn't nearly enough to find it by it's name. Just like his example, I was looking for some sort of break glass mechanism for multiple people to cooperate to retrieve sensitive information they otherwise should have access to individually.

[@kultek]

Now another great way to answer my students’ question of “When will I ever use this?!”

[@darksentinel082]

perfect for my upcoming puzzle~

[@matthewstuckenbruck5834]

I wish he'd shown us an example of how one could reduce the number of possibilities by seeing it as a continuous function.

[@CraigNull]

I think, strictly speaking, what happens is the probability density of what the unknown number could be sharpens considerably but no possibility is completely ruled out with incomplete information. For instance, maybe the search space "reduces" in the sense that what was previously 10^200 equally like possibilities is now 95% certainty the solution is among just 10^10 possibilities.

[@gyroninjamodder]

@@CraigNull Isn't it infinite in both cases?

[@jetison333]

@@gyroninjamodder technically yes, but we are only sharing finite data so the number will be an integer or only have so many decimals.

[@DeGuerre]

Remember, the secret is an integer, and the coefficients of the polynomial are too.

[@gm90rt24]

I missunderstand you because I do not know alot at english but you are great. happy new year for you from iraq😍

[@eduardosuela7291]

What about instead of growing the polinomial order, increase X dimension? You would calculate the intercept of a hiperplane defined by the points you had distributed

[@radadadadee]

The intercept of a plane with a 3D surface is a 2D curve, not a single point. It wouldn't work.

[@eduardosuela7291]

@@radadadadee a hiperplane intersecting one linear axis gives a point on that line. If your line is the z axis it works. It gives an "intercept" coordinate on z axis

[@IMYTnNERDEE2]

like the action music behind

[@mina86]

There are hardware encryption devices which use the scheme as well.

[@chraman169]

0:00 - 0:10 would be my hair loss

[@CraigNull]

At first I thought you were heading for a system where you write your password as a sum of N integers (chosen in some suitably random way) then give each person some k < N of those numbers That's pretty simple but I assume it has major problems if it doesn't even merit passing mention?

[@piperboy98]

Hm. I think there might be an equivalence here through the method of finite differences. At least if the x positions given out are evenly spaced, the points allow you to construct the difference table with just sums and differences, so when you get to x=0 (the secret) it is just a weighted sum of the originally given y values. If the weight is exactly known just from your own x position then it is equivalent to just give the already weighted numbers and it is the same as you suggested. I wouldn't be surprised if there were an equivalence.

[@gyroninjamodder]

The share distribution part of your idea does not work. Imagine k = 2 and n = 3 there is no way to distribute it. This is assuming you have no way to check if you do indeed have the secret, else for this case you can reduce the search space to 2 possible secrets by doing your scheme for every possible 2 people that can recover your secret.

[@piperboy98]

Oh yeah... It can't be 100% equivalent since there is no way to give out to an arbitrary number of people such that ANY subset of N people can recover it, while that is possible with the polynomial construction. You'd have to give some people the same number and then some groups would still be missing numbers. So I think your method is equivalent to the special case of the polynomial method where there are exactly N x-values in circulation, but the polynomial method is more general.

[@aakashmaniar9494]

The fact that the number has no meaning makes that number special & meaningful

[@fuckoffgoogle12]

That is incredibly clever.

[@aaronr.9644]

pretty cool!

[@CaptainMagnus]

This sounds oddly similar to Reed-Solomon but without the error correction properties.

[@MrNacknime]

In the use case of this for multi party computation against actively cheating adversaries there is also error correction similar to Reed-Solomon

[@DaviddeKloet]

As far as I know, the only difference with Reed-Solomon is that with RS, you want it to take up less space so many of the points on your polynomial represent part of the data, while with SSS, all other points that are not the original data are chosen completely at random.

[@IceMetalPunk]

Wait, but... once you switch to using modulos, doesn't that break the ability to reconstruct the polynomial even with all the pieces?

[@christianbarnay2499]

Polynomial interpolation on a finite modular field works almost the same way as on real numbers.

[@skyemegakitty]

Is this related to elliptic-curve cryptography, or are they both just based on intractability?

[@raj.svc.google911]

The background "music" gave me a headache (listening thru' the headphones).