Kill the Mathematical Hydra | Infinite Series

[@requiembeeblebroxx]

Alternatively, try to make it regrow so many heads that its heart can no longer pump blood to all of them and it dies of hypoxia, OOTS-style.

[@lorekeeper685]

Anya Thurmes or atack the body

[@psdnmstr4609]

A man of culture, I see

[@bookworm3696]

it didn't die, it lives eternally supplying food for a food chain.

[@What-thaW]

Or if it just collapses

[@darknessblades]

haha true, or that moving all the heads is giving it a large energy deficient, forcing it to shut down one of its own heads

[@tf2hark362]

In case anyone gets confused wanting to learn more - the fact that all well ordered descending sequences terminate is not usually called the "Well Ordering Theorem", it's just a property of well ordered sets. The "Well Ordering Theorem" usually denotes Choice/Zorn's Lemma, which is a different statement.

[@jwylde2]

"The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?" Jerry Bona

[@martixy2]

Damn, this is not a video you can watch casually at 4AM. Gotta remember that about this channel.

[@whatisthis2809]

Hey my sleep time

[@StefanReich]

Lol... Literally just did that

[@whatisthis2809]

It's 4:04am and my phone gave me an error notification, send help

[@abdalrahman6618]

What about burning the hydra????? 😃😃😃😃

[@cazymike87]

Yeah, lets just make it a DvD Hidra instead.

[@abdalrahman6618]

Mike D 😣😣😣🙍🙍🙍

[@KilosWorld]

Jeesh how did you guys get on a video for smart guys when you're talking about DVDing a Hydra

[@funnygeeks8126]

let's define the question more clearly. How many seconds would it take to burn the hydra? (assuming it takes 1 second for a head to burn). Knowing only the outermost heads burn, the number of seconds would be the equal to the length of the longest chain of heads. But that's too easy, let's assume the fire doesn't immediately spread to all the outer heads, but is restricted, so that if there is a fire set to a node, the fire will only go up the chain to the heads above it, and spreads to a node if its direct leaf node has been burned up. now that's a question. (but with a similar answer)

[@jyotishkaraychoudhury4762]

It's a mathematical hydra Not a real one. (Well....I know they aren't real)

[@coatduck]

I think personally I subscribe to the theory that the best way to defeat a hydra is to cut off it's heads until it's just a big ball of heads. It's only a tactical advantage for so long . . .

[@valerfox2155]

Any advantage can be exsploted

[@ocean_0602]

I just wanted to appreciate the awesome eye contact with the camera. It's a subtle detail but so often overlooked and underappreciated! So many channels have hosts blatantly looking back and forth at the script and it's a huge distraction. But here it's almost as if you're speaking directly to us which makes the videos subconsciously so much more captivating. Amazing channel, keep up with good work Kelsey and PBS!

[@rhodesclosed9563]

Mont Blanc Don't they have the teleprompter positioned so it looks like the host is looking at the camera?

[@ocean_0602]

Yeah probably, which is fine because it still creates the illusion that she's looking at the audience. Many other channels can't pull this off as well as PBS hosts.

[@nicolararesfranco9772]

I completely agree!

[@euphrosyne7386]

She does that because she knows what she is talking about

[@pixelfairy]

how do you survive N hydra head attacks while your busy chopping heads off?

[@eac-ox2ly]

Plot armor.

[@pixelfairy]

some also suggested getting on the hydras back. another way to frame this, how much ammo would you need, assuming you were skilled enough to keep out of striking range and could reliably pull off head shots.

[@michaelpotvin5406]

attack in the good order.

[@brownmold]

Roll Natural 20's. Lots of them.

[@spalshie7662]

Dodge-rolling

[@pedrocrb123]

I think thats killing a mosquito with an atomic bomb

[@quantuminfidel8364]

Except that this isn't a mosquito, it's a freagin hydra and you are not killing it with an atomic bomb but with a sword

[@pedrocrb123]

I was talking abou using too complicated maths to solve a simple problem, just like using too overpowered weapons to kill a simple mosquito

[@quantuminfidel8364]

I know you meant it's overkill but Do you have a simpler answer to this problem?

[@fgvcosmic6752]

@@quantuminfidel8364 fire

[@vitorschroederdosanjos6539]

@@quantuminfidel8364 I think you could solve it by considering every hydra as a body with a 1 neck-head tower and disconsider everything else (as if you can decrease to 1 the largest of necks, and all necks can be made the eventually become the largest (or a copie) than all necks will eventually become lenth 1 than you just have a hydra with a finite amount of trivial heads I don't think this is "THE PROOF" because it's easier, I think the hydra's number tells us something more about the fundamental structure of the animal and that's really the point of using it, idk

[@MikeyG_f-of-x]

This is my favorite channel.

[@GargaGaming]

y?

[@alexsere3061]

ƒ(x) I am not liking This channel, it goes over the concept in minutes, not providing any insight, I want to learn but I cant if all I hear is extremely Surface content or a theorem with no insight, I honestly want to learn hyperreals btw if someone knows where I can please tell me

[@pokestep]

Alex Sere You're looking for something else than this channel provides so it's no wonder you're dissatisfied

[@MikeyG_f-of-x]

Alex Sere​ I like that this channel gives an overview to complex theorems, and provides some sense of it's application. Going into great detail would keep it from being fun and bite-sized. The way it is, it hopefully introduces concepts that people would want to seek out more in depth.

[@justinward3679]

GargaGaming because x squared

[@talkingcowthatwasthereallalong]

What about the Hydra's feelings? I can't do those maths, I'm a potato basically... Care about your hydracidal tendencies, not my potatoness...

[@whatthefunction9140]

I've learned that mathematicians have way too much time.

[@dave5194]

Dylan T There are some problems in mathematics have taken centuries to solve, through the work of many generations, some still haven't been solved for longer. It's pretty amazing.

[@mikeo759]

Everyone has the same quantity of time per minute. Precisely one minute per minute. Some people may, in the end, have more time total simply by outliving others. SO I conclude that you have inferred from the video that mathematicians outlive non-mathematicians? Or, perhaps you consider the solving of complex mathematical problems that advance humanity's understanding of the nature of the universe a waste of the limited amount of time we're awarded? When that time could be better spent commenting on a youtube video...

[@whatthefunction9140]

I was just joking. I'm a math major myself. What I really learned is that math geeks have no sense of humor!

[@car-keys]

Dylan T It really wasn't that funny though

[@dave5194]

Dylan T​ humor? Never heard of it. Sounds like a very dangerous illness, you should get that checked

[@DavidAnderson-cw7oq]

2:40 my strategy: keep chopping off heads until it’s heart can no longer pump enough blood to all of them

[@MultivectorAnalysis]

Another great video! I see that some viewers already pointed out the issue with omega*2 vs 2*omega, but I noticed one more small typo. The graphic @9:48 has one too many nodes. The head immediately above the body should not be there. I really like this hydra example using transfinite ordinals. Keep up the good work!

[@bentupper4614]

Small correction at 6:55: The next limit ordinal after w should be w*2, not 2*w (omega = w). Since ordinal arithmetic is not commutative, these two are not equal. 2*w is actually equal to w.

[@Farzriyaz]

yeah infinite ordinals are ω+1, ω+2, ω+3... ω*2 or ω+ω, ω*3, ω*4... ω^2 or ω*ω, ω^3, ω^4... and so on... and sadly aleph null is the farthest this list goes, [aleph][subscript 0]+1=[aleph][subscript 0]

[@Farzriyaz]

plus in ordinal math if S is smaller than B, S+B=B and B+S=S+B in regular math

[@pasarebird02]

@@Farzriyaz >In ordinal maths if S

[@julesharris6383]

@harrybotnton7800 wow I didn’t know that

[@ViewtifulSam]

as someone whose research is mostly in combinatorial optimization (so I might say I'm conditioned to thinking really finitely), when you first mentioned ordinals to tackle this problem I was like "what, why??" but geez this solution is so pretty. I really like the way it elegantly encodes a very natural form of complexity on this tree. I wonder if it can be translated into finite things in an equally meaningful way, but anyway I'm more than fine with this proof. thanks for the great episode!

[@oooBASTIooo]

It actually cannot be done with purely finitistic means. That is one of the most interesting things about this hydra. You cannot formalize the proof in weak theories.

[@ViewtifulSam]

oooBASTIooo oh, that's really neat, now I'm curious as to why that is... I'll try to look it up. thank you

[@oooBASTIooo]

kurser.math.su.se/moodle19/file.php/855/Kirby-Paris_1982.pdf

[@dliessmgg]

By my calculations, it takes 3 812 798 742 507 chops to get from a hydra with four heads on top of each other to a hydra with 7 625 597 484 987 heads next to each other. So to get to a hydra without any heads, you'd need around 11 trillion chops. (Or 11 billion if you're from mainland Europe like me.)

[@empireempire3545]

plus one for 11 billion :D

[@DrGerbils]

I got 2,541,865,828,329 stalks of 1 head each. Total number of chops to kill it was 3,812,798,742,507

[@nockieboy]

Dliess Mgg Get to da choppaaaaah!

[@DrGerbils]

+Dliess Mgg I looked again at the last part of my calculation. You're right. I had 2,541,865,828,329 more heads to chop to get to the spikey ball.

[@enzy9864]

Does this have anything to do with the number TREE(3)? That number is very big and equally interesting. I would love it if you did a series on unimaginably huge numbers like TREE(3) and Graham's Number and SCG(13) and the fast-growing hierarchy.

[@benjaminprzybocki7391]

Ilan Goldman TREE(n) and the hydra are both about trees in graph theory, so in that sense they're somewhat related. Hydras can also generate extremely large numbers, like the Buchholz hydra gives the function BH(n), which is comparable to SCG(n). Googology certainly is a really interesting niche field of math.

[@ardorpraxis9661]

Yeah Graham's Number has been tackled really well on several other channels but I have yet to see a clear description of TREE(3) that I could actually follow. I would love to see this!

[@benjaminprzybocki7391]

Ardor Praxis This explanation is pretty good: cp4space.wordpress.com/2012/12/19/fast-growing-2/ I should make a video on it (and SCG(n)). It seems like a lot of people want to understand it, and I believe I have a grasp on the definition of the number (but not the proof of why the number is finite).

[@lock_ray]

Benjamin Przybocki I finally understood what that sequence means .-. thank you for that link. Still baffled about the size of the number itself though

[@benjaminprzybocki7391]

Lock Ray It is unimaginably large. TREE(3) makes Graham's number seem tiny. What's even more mind blowing is that SCG(3) makes TREE(3) seem tiny.

[@morris1818]

Hercules is the Roman name. The Greek name is Heracles.

[@-_Nuke_-]

Or to be even more precise, its "Ηρακλής"

[@davidwuhrer6704]

Correct to within ε ☺

[@GoldenKingStudio]

Man, I'm normally a pedant about that and I wasn't even paying attention. Thanks for making me annoyed out of nowhere, I appreciate that. We must strike when people are being wrong on the internet! ;D

[@janeza382]

All about Greece is myth. Greece is modern state since 1831 while term Hellene describe ancient people whit many tribes and nations who have formed Hellenism as culture and religion in Greko-Roman era. So Greeks are Romans or Romans are Greeks?! Hellenes believe that they are descend as Hercullides as describing the human race from ancient times and that today Hellene are exclusive Greeks. It is more confusing as Latin , Hellenic and Cyrillic alphabets have same origin but same race was somehow in modern era divided in Latins, Greeks and Slavs. In top of that modern neo-Greek history claim Heracles as hero but he was living mathematician,then for matter of truth Hercules was ancient mythical hero. Problem whit modern Greeks is that they have lost ancient pronunciation and as they read outcome is in contradiction.At the and ancient city states had cultural wars among them from as arguing what myth was truth or false for their dominance in Hellenism.

[@davidwuhrer6704]

Η, ιτ'ς ολ Γρίκ του μι.

[@jledragon]

I think this is all the starting point for explaining TREE(3), which by the way is a number far bigger than even Graham's number

[@douglaspantz]

i mean, according to this link (googology.wikia.org/wiki/Kirby-Paris_hydra), the maximum amount needed to kill a hydra of length 4 is over grahams number Edit: This is actually a different variant where the hydra spawns 1 copy, then 2 copies, then 3 copies, etc.

[@CDolph296]

FINALLY someone does a video on this. I first saw this in a seminar years ago and it was definitely the most mind-blowing lecture I'd ever seen (I hadn't seen Godel's incompleteness theorem or the Halting Problem yet, so I was still in the nieve mindset that math could solve itself). Great video!

[@anon8109]

This feels like using a hammer to push in a thumbtack. There are never infinitely many heads so why resort to infinite ordinals for a proof? I bet induction over natural numbers suffices. Since we're dealing with trees the induction probably is nested. One induction for the depth, and one for the height.

[@pbsinfiniteseries]

I completely agree. It seems like overkill! But just wait until next week when we explore exactly this. :)

[@orbital1337]

That's the truly fascinating thing about this problem: it's a very simple problem about completely concrete and finite objects which cannot be proven in Peano arithmetic. That means that it in some sense *requires* infinitary mathematics.

[@justunderreality]

I would imagine it is to keep a classification system. Since level 2 is always higher than level 1 then the infinite keeps at a higher value - regardless of how many heads are (or will be) at level 1.

[@Cyber1zed]

So that we learn about the topic. Isn't it better to learn a complex topic with a simple problem first?

[@eofirdavid]

I think that as long as you keep the number of copies that grow after each cut fixed, then probably taking some big enough integer instead of omega should still work (bigger than the number of copies). Trying to generalize for larger number of copies should suggest to look for a number "bigger" than all the integers. I still don't know if you can find such a potential function when you assume that the number of copies increase in time, and here is where the strength of omega really comes into play.

[@Karan-wz3yg]

I have calculated the answer of four lined hyra challenge .It will have 7,625,597,484,987 heads and the total no. of steps to defeat this hyra are 11,438,395,749,194. I loved infinite series .Interesting mathematics. Pbs space time is also my favorite.

[@ShioPK]

Why advanced math on YT is so simple and logical, yet basic math on an exam makes no sense and looks like some form of black magic?

[@bored_person]

Bored teachers.

[@aCaptAmerica]

Wentyliasz Szprychownik it's easy to think you understand something when someone else is doing all the work.

[@TheRolemodel1337]

because you need practice for algebra here its a very simple concept and simple transformations also in many geometric exercises you need to find procedure for solving the task yourself which is the hard part not following someone elses way :p

[@ecarpioxp]

Yeap, the illusion of competence.

[@wiktordykas5513]

gitare siema

[@chrstfer2452]

This is without a doubt my favorite math channel already, and you're only just starting out. I love it, keep up the good work.

[@joshuafife6792]

"You'll eventually chop off all the Hydras head" Me-"Not if you die"

[@vitorschroederdosanjos6539]

"You don't need to be a Greek hero, just a mathematician" Pitagoras: am I a joke to you?

[@seanpeery7780]

This would be a very boring Hydra to fight.

[@marian519]

Would you rather fight one that increases its depth at every chop?

[@seanpeery7780]

There's so many heads at certain points that the heads individually can't be any kind of real threat or do any real damage or they'd insta-kill at a certain point while you fight them the correct way. So it would be less tedious to just ignore the Hydra and walk around it as it gets AoO on you then spawning massively more heads just to defeat it.

[@KaelynWillingham]

How about a Hydra that grows N heads for every Nth cut, and one of those new heads spits fire?

[@schrodingerscat3912]

Shh.. did you hear that Hercules?.. What was that sound??... AAAAAAAAAAAAAHHHHHHHHH!!!!! (gets eaten by Hydra) Hydra: RAAAAAAAAAAAAAA!!!! (pulls sword out) Hydra: Johnny is traveling at a speed of 25mph while Sue is lagging by a factor 5. Given only a 2 hour duration of time to reach the same destination, how much money would each have spent on gas for every mile traveled if gasoline cost $1.25 per mile? Round to the nearest tenth and bubble one of my heads in clearly, no calculators allowed.

[@papa515]

I've seen this shown true before but this particular demonstration is the best I've see. Very well done. The 'ordinal-proof' is really cool.

[@Kram1032]

Nice episode once more :) - Having played this game before, strategies vary a LOT in how many heads they produce. Is the fastest strategy something simple? Either "always pick the highest head" or "always pick the lowest head", perhaps? And unrelatedly, did you really censor "damn"? lol

[@Horinius]

@4:30 The hydra will look like a spiky ball. In 2020, we would call it coronavirus-like 😂

[@Nethershaw]

Wait a second. Is that... the Ackermann function? en.wikipedia.org/wiki/Ackermann_function The question at 4:48 sounds an awful lot like A(4,2)... you can write a tiny program to evaluate the function in a handful of lines, but with those parameters, good luck getting to the end of it before you have used all your working memory. *Edit:* here it is in Scala. gist.github.com/nethershaw/26c3d8089e0acb35af4d0e46022c5f7c

[@fractuz]

No, but it is another fast growing function. Goodstein's function (the function that tells you for every n how many moves you need at the minimum to defeat a hydra of height n) actually grows considerably faster than the Ackermann function. In the "fast-growing hierarchy", the Ackermann function sits roughly at index ω, whereas Goodstein's function is at ε_0, a much larger infinite ordinal. ε_0 is so large that, when you raise ω to the power of ε_0, you end up with ε_0 again. In a sense, it is the first ordinal that is so large that exponentiation stops having an effect.

[@scipio6142]

wanja .hentze i

[@UMosNyu]

Regarding the hydra that grows more heads pre chop: status 0: 3 heads attached to the body. first chop: 2 head + 1 new head. 2nd chop: 2 heads + 2 new heads. 3rd chop: 3 heads + 3 new heads 4th: 5 + 4 new. I cannot see this getting to zero... even not "eventually"

[@Idriel007]

But the height eventually decreases until the heads are directly attached to the body, after that chopping those heads will decrease the total ammount of heads until the total goes to 0.

[@martingutlbauer9071]

When a head directly attached to the body is chopped off the rule says that no new head is growing out of the body.

[@empireempire3545]

Watch again. Notice that heads attached to the body directly cannot regrow. So the hydra with all the heads attached to the body is one step from being defeated - it cannot regrow anything. And lo and behold - You can ALWAYS reduce the hydra to the state when the heads are attached to the body - so they cannot regrow.

[@UMosNyu]

Wow... yeah you are right. I had forgotten about this fact. I guess I was too baffled about her statement.

[@bobbyharper8710]

Every regrowth after the first chop is shorter than the original height. Call it diluted or weaker regrowth till it's all gone.

[@minhkhangtran6948]

So...what happen with infinihydra? Like if you chopped up 1, infinite will replace it?

[@pXnTilde]

I was thinking the same thing

[@minhkhangtran6948]

Well, I did the initial math (really basic, like only-next-step), and if the pattern is right...no change in complexity. Basically you would take infinite time to defeat infinihydra

[@ScubaDaveGSXR]

Well in math, not all infinities are equal.. some are bigger, some are smaller. You just have to use a larger infinity cut. Crazy right? Math!

[@adolfodef]

You can defeat the "infinity hydra" (1 chop replicates infinite heads one level below), in less than infinite time IF you are allowed to chop all "valid heads" on the current step at the same time [with a literal "infinity+1 sword"]. tvtropes.org/pmwiki/pmwiki.php/Main/InfinityPlusOneSword

[@minhkhangtran6948]

Well true, but that would defeating the point isn't it?

[@max007manga]

I think the problem is easier to solve "in reverse" so to speak. Imagine you have n heads attached to 1 head that is attached to the main body : if you cut it until you only have heads directly attached to the body, you will get 3^n heads (you cut once you get 3 groups with n-1 heads then 3^2 groups with n-2 all the way down to 0) or in "reverse" you need 3^n heads to get one head attached to the body with n heads attached to it. Now comes the interesting part : we know that 3 heads attached to the same head/body can "fuse" to become a line of 2 heads, so 3 heads attached to 1 head on the body cane fuse to become a line of 2 on 1 head or a line of 3. And how many heads do we need to get this group of 3 on one head ? 3^3=27 exactly the answer we got by cutting a line of 3. If we want a line of 4 we can apply the same reasoning to get a group of 27 with 3^27 (3^3^3) heads attached to body who will then fuse to become a line of 3 on 1 head or a line of 4. If we keep going we find that to get a line of n heads we need 3^3^3...^3 with exactly n-1 3's.

[@nictimus24]

Hail Hydra!

[@blblbl2750]

Looks like for a line hydra of lenght n, there will be 3^3^3^3...^0 (with n 3 exponents and one 0 exponent) heads on the final ball. So we get (3^(3^(3^(3^0)))) = 3^27 for a line of lenght 4.

[@geodraws]

I got 3^27 too but didn't figure out the general form, nice solution!

[@spacemario]

6:54 it's actually ω*2, not 2*ω. Remember that order matters.

[@thatguyq4783]

PBS came out with a new channel.... and there goes my social life. I am seriously pumped up for this! Use to love math so much and I am trying to get my life back together and go back to school, so this came at the perfect time for me. Thanks PBS for making free great quality educational videos!

[@thisaccountisdead9060]

This is like the Bruce Lee fight in the hall of mirrors at the end of Enter The Dragon - there is only one evil Han that is reflected in all the mirrors, and each mirror reflects another mirror image... so to get to the real Han, Bruce must break every Han he sees (even if it is a reflection) until he gets to the real one... Though it differs in that in this case (if it was an Enter The Dragon Hydra) then some of the heads attached to the body would be mirror images and not the real Hydra... I think? A Hydra in a hall of mirrors - Yeh, I'd love that :P

[@kamoroso94]

Poly Urethane That was actually a really cool analogy, thanks!

[@thisaccountisdead9060]

Kyle Amoroso ha ha... be like water my friend :)

[@eval_is_evil]

Poly Urethane yeah but honestly that last fight was pretty lame...

[@eval_is_evil]

Poly Urethane the water analogy is great...I heard it in the movie that was the life story of Bruce Lee with a bit of mythical stories in it. Dragon : The Bruce Lee Story ? or something? It was superbly acted by Jason (?) Lee (no,not a relative of Bruce ) . Fighting sequences were awesome ....go watch it

[@bangboom123]

"...And so on" Suddenly, I am put in mind of Kelsey doing a Slavoj Zizek impersonation.

[@kingpet]

The hydra looks like cute plushie

[@VincentGonzalezVeg]

thatd be a great rage plushie

[@OrbitalNebula]

Maan this is amazing! It's super rare to find a video on KZbin about infinite ordinals..

[@OrbitalNebula]

Heyy

[@jojo989GD]

we

[@kristiankember8973]

7:48 I'm not too sure on what you said...correct me if I'm making a mistake. you said ω^3ω however in ordinals you'd need the 3 after the ω, otherwise the order is equivalent to ω^ω

[@phucminhnguyenle250]

I think I can make a function of all ordinals that returns rational numbers such that f(a)

[@zairaner1489]

That will fail..

[@MirorR3fl3ction]

wow, that was an incredibly good explaination of Omega and ordinals. So is Omega basically is an algebraic version of infinity?

[@KohuGaly]

omega is simply infinity in ordinal numbers. Which is different than infinity in cardinal numbers (which is called alef 0). They are both the same thing in slightly different contexts. They are both algebraic as far as I can tell, they just don't behave intuitively.

[@fatihnri2484]

Where do you know about alef 0?

[@columbus8myhw]

Kind of, but that thinking won't get you all of the ordinals. For example, after ω and ω^ω and ω^ω^ω and ω^ω^ω^ω and so on, we get a new ordinal called ε_0. And there's a higher ordinal called ε_1, and a higher ordinal called ε_2, and even ordinals called ε_ω or ε_ε_0 or ε_ε_ε_0 or… and things even higher than that. I don't know if thinking of ω algebraically helps you get to ε_0 or beyond.

[@DarkestValar]

your way works, but i prefer ε_(n+1) = ω^((ε_n)+1), ω^ω^((ε_n)+1), ω^ω^ω^((ε_n)+1) ...

[@TehAarex]

And there's further ordinals, like ζ_0 = lim(ε_0,ε_ε_0,ε_ε_ε_0,...) and ζ_(n+1) = lim(ε_(ζ_n+1),ε_ε_(ζ_n+1),ε_ε_ε_(ζ_n+1),...) Oh hello Nish, remember me?

[@RaidChampion]

Technically speaking, something should be said about the order in which the ordinals are added, as ordinal addition is non-commutative. Consider the hydra which first has three heads: two heads which are stacked upon one another and one head which stands on itself. Depending on the order in which we add the ordinals, this hydra is assigned the ordinal omega + 1 or the ordinal 1 + omega, the latter being equal to omega. But in this last case, chopping off the single head would not decrease the ordinal number associated to the hydra!

[@DrGerbils]

The graphic at 4:31 is wrong.Each of the 3 necks on the right should have 26 heads, not 27, attached to a central head.

[@pbsinfiniteseries]

They should. And wow, I'm impressed with your eyesight. :)

[@thesentientneuron6550]

Yeah I thought that too

[@kevincsellak296]

yeah, I saw that too... though I counted two heads with 27 heads attached, and one with 28. But I don't know how I came to that result.

[@TalysAlankil]

3:15 "You can't lose!" you know, unless the hydra eats you.

[@yeetrpg]

what did i've done wrong to you? i now must kill hydra... its 3 o'clock at night here!

[@Mrnothing1777]

this reminds me of Zorn's Lemma and i did imagine that the Hydra is sticking to the ceiling , then i did think of a giant with lot of feet , so chopping feet with the help of transfinite recursion until reaching the maximal element ; and that is exactly what is the demonstration of the Lemma is about. (constructing maximal chains )

[@add852]

Hydra was so easy to defeat.. until they introduced math

[@ahoj7720]

Excellent! This is a fun variation on the theorem of Goodstein, which deals with integers only, but is not provable in the context of Peano arithmetics (and this non-provability has been proved!), but only by using enumerable ordinals.

[@wwguee3338]

I can do it easily, ever heard of a little thing called explosives

[@dawidaleksanderwitkowski7455]

Great video, but I think there is an error at minute 9:49 (when we have omega cubed) the drawing is wrong, there should be 4 heads attached to the body such that the closest head to the body is an ordinal 3. Anyway thank you for creating this amazing channel :)

[@davidwuhrer6704]

Assuming that the hydra only grows copies of the remaining head trees, how did the head trees get so high in the first place?

[@frankschneider6156]

An axiom defined it to be that way

[@oooBASTIooo]

When a hydra eats chicken, it grows heads on top of heads. Unfortunately this hydra ate all the chicken, so it cannot grow anymore...

[@sinithparanga2481]

David Wührer actually a very good question. It is impossible to have any of the starting position. Could every hydra have the same origin? Could we do the math backwards and then define the above?

[@davidwuhrer6704]

Sinithparanga I don't think we could. The inverse operation of chopping off heads is not a function. We would need additional operators. Building a hydra seems to be a lot more fun than defeating one.

[@ravernot8889]

this channel is getting better- remember, don't be one of the other pbs channel hydra heads.

[@0xDACA]

notification squad, engage

[@thanasisgiannakopoulos2315]

דניאל כהן הלל i'm here

[@0xDACA]

Agent Six איך עלית עליי? זה בגלל השם בעברית? , דרך אגב עכשיו חצי שעה אחרי חצות, לך לישון.

[@favorite89103]

דניאל כהן הלל גם אתה

[@0xDACA]

Moshe Sherman גם אתה

[@kriiistofel]

there is 10 minutes to midnight at my place

[@t3st1221]

Why must a decreasing serie using infinite ordinal reach zero in a finite number of move ? Using w for omega, w, w-1, w-2, w-3... is a decreasing ordinal serie but it would take an infinite amount of step to reach zero (w steps to be exact) EDIT: just got it, subtraction is way more tricky with ordinal and w-1 'doesn't exist

[@m.a.malcolmR828]

i am commenting on behalf of my five-year-old son, KM. He says: I, KM, said that mathematically, the spiky ball that is made from a line of height 4 should have 64 heads.

[@marvinwatson3096]

This is the most exhilarating KZbin channel in a while. Thank you.

[@DavidRoberts]

Ah, kids.... assuming PA is consistent and ε_0 is well-ordered ;-) You know the Fields Medallist Vladimir Voevodsky said in a public forum that arithmetic is probably inconsistent? (but of course we can't prove this!)

[@columbus8myhw]

But, come on, ε_0 is _definitely_ well-ordered!

[@AlcyonEldara]

video.ias.edu/voevodsky-80th (Always funny to hear Deligne's english) I'd be seriously happy if we had to rebuild our logic from scratch. First-order is so boring ....

[@DavidRoberts]

We have to assume a basic formal system in which to work, in order to prove that, and if we can prove that, then the formal system PA is consistent (a theorem of Gentzen showed that if we can prove ε_0 well-ordered, then we can prove PA is consistent), which is, while likely, unprovable. Sometimes I like to be contrarian about such things (and yes, I am a professional mathematician :-)

[@morgengabe1]

David Roberts, Kurt Gödel did. That is, if arithmetic as we know it is a complete system, then it must be inconsistent.

[@DavidRoberts]

I'm not sure what you mean. Gödel did what? Phrased more straightforwardly, what I mean was things: while PA is likely consistent, this fact is unprovable. Regarding my experience talking about things that are probably true but are independent of axiom systems: dx.doi.org/10.1007/s11225-015-9603-6 :-)

[@liopowers]

Mathematical Hydra was really supportive by being inspiration for life problems that someone can face. I ll use it to calm my self and others.

[@SupLuiKir]

That's not how hydras work, thus we're not really talking about hydras. Stop calling it that.

[@buschtoens]

· 0xFFF1 are you the Hydra police? of course that's not exactly how a Hydra works. But a Hydra is the closest matching concrete metaphor for this mathematical abstract object. Stop whining about and learn to appreciate the great work that has been put into this video.

[@MavicityHerself]

· 0xFFF1 huuu i thought i was the only one! T_T chop of a head, two more grow back. I thought that was what they were going to solve.

[@JaceSomers]

I wonder if there is any finite rate that the hydra can regenerate such that it can't be slain I'm a finite amount of steps. The amount that it regenerates would have to increase as more heads are cut off (the video goes over that) and it can't be linear since that can be done in a finite amount of steps. I'm thinking that a growth rate of a^n may work, where a is a natural number over 1 and n is the amount of steps taken so far.

[@NicholasRoge]

just for the record, because I spent way too much time, a single headed hydra with _n_ neck segments that spawns _s_ tree copies every time a head is cut off will have a maximum of _s_^(_n_-1) heads with one neck segment and would take _s_^(0)+_s_^(1)+...+ _s_^(_n_-2)+_s_^(_n_-1) cuts to kill such a beast.

[@DanGRV]

For the stack of height 4 we need 3^3*(3^25-1)/2 total cuts to kill the hydra. Also, at the last step (when only heads of height 1 remain) there will be 3^27 heads. How did I arrive to this conclusion? Solving for the case of a head of height k > 0 with exactly n heads attached to it (i.e. those n heads are of height k+1) With 1 cut, we get 3 heads of height k, each with n-1 heads of height k+1. With another 3 cuts, we get 9 heads of height k, each with n-2 heads of height k+1. With another 9 cuts, we get 27 heads of height k, each with n-3 heads of height k+1. And so on... Until we have 3^n heads of height k, each with 0 heads of height k+1. In a nutshell: from 1 head of height k with n heads of height k+1 attached to it, we go to 3^n heads of height k with no heads of height k+1, after 1+3+9+...+3^(n-1) = (3^n-1)/2 cuts.

[@caseyhale1726]

There will be 7,625,597,484,987 or 3^27 heads, and it will take 11,438,396,227,495 or 3^0+3^1+3^2...+3^27+15 steps to kill the hydra.

[@robertroy1435]

I love these videos. As an engineer I have always loved mathematics and these videos are like math snacks. Tasty but not too filling.

[@noobexception2175]

(Yes, 2 years later, ikr) Well, I guess I have figured it out. Let A(H) be a function that returns number of heads attached to a body for a hydra described by H. Also, to speak in general, let N be numbers of segments that regrow. Now, if there are some heads growing from the one attached to the body, the hydra can be described by w^H. We can notice, that basically, we want to reduce H to some number, and we'll get a ball of heads sticked to one. We can tell from definition, that there are A(H) of this heads. So we got a hydra that is described by w^A(H). Now we shall set a formula for A(w^k), where k is a natural number. Notice, that if we cut off a head of w^k, we get (N + 1) segments described by w^(k-1). Then A(w^k) = A( (N + 1) * w^(k - 1) ). Since these segments are independent we can tell that : A(w^k) = (N + 1)A(w^(k-1)). By solving this recurrence we get : A(w^k) = (N + 1)^k Since in our case N = 2, A(w^k) = 3^k. Now, we have this formula, let's use it. Hydra of height 4 is described by w^w^w. A(w^w^w) = A(w^A(w^w)) = A( w^A( w^A(w) ) ) = A(w^A(w^3)) = A(w^(3^3)) = A(w^27) = 3^27. And that shall be the answer :D

[@aesome12345]

It's interesting that in computer science a two dimensional array could simulate the same thing as the omega ordinal, and throughout your explanation, I kept drawing parallels between the two.

[@tannerjacobi3497]

Ordinal numbers are a neat way to write the final boss's health bar.

[@cr1216]

I am having hard time understanding the decreasing order reaching zero part. What comes before omega, for instance?

[@cr1216]

I mean if we say the element before omega is omega-1, then it would take infinite time to reach zero, wouldn't it?

[@Benson_Bear]

@@cr1216 Watching this video late, but, the answer to your question is that there is no element in the ordering of ordinals that is "just before" ω. ω is a "limit ordinal", and is not generated from an earlier ordinal by the successor operation. Simple example, suppose you dislike big numbers, but you dislike odd numbers even more: any even number is to be preferred to any odd number. So, your preference ordering is 0,2,4,6, ..... 1,3,5,7... Here 1 is position ω in the ordering. There is nothing that comes immediately before it. So when someone moves from 1 downward in this order, they are forced to jump over an infinite number of even numbers and pick a specific even number. That's roughly the idea of how you must get down to zero in a finite time.

[@powerflame]

There are extremely many googological functions who work like that. it decreases a little bit but expands a lot. however, they will 100% terminate, even if it takes over the length of the remaining time for the universe

[@tianchenzheng7464]

It looks weird the first second, but become really intuitive immediately. If the hydra doesn't grow in height, each chop will make it shorter and eventually makes all head height 1 which means he is dead.

[@TheJulioToboso]

I love your new show! I can't wait to see what's next. Maybe some fractals or Fourier analysis? Would be nice

[@amaarquadri]

It's crazy how the proof techniques leans heavily on infinite math to prove something about a finite hydra. It somehow shows how thinking about infinity is important to our world even though no physical quantity can ever truly be infinite.

[@ReadyPlayerYou]

What would happen if the structure of the hydra was a fractal? For example, if each head had 2 heads. It obviously could not be defeated in a finite amount of steps, but if each beheading took half the time that the last one took, could you do it in a finite amount of time? This is only one example of a fractal and so could you adjust the time difference to the kind of fractal? For example, if it is possible to do it in a finite amount of time, would a fractal where each head has three attached require the time to be divided by three, and so on?

[@8BitRip]

I did it after you told me too. Turns out it was super simple after 2 minutes. They don't think it be like it is, but it do

[@Kelly_Jane]

"Meself," Hagrid continued, "I think we might 'ave a Parisian hydra on our 'ands. They're no threat to a wizard, yeh've just got to keep holdin' 'em off long enough, and there's no way yeh can lose. I mean literally no way yeh can lose so long's yeh keep fightin'. Trouble is, against a Parisian hydra, most creatures give up long before. Takes a while to cut down all the heads, yeh see." "Bah," said the foreign boy. "In Durmstrang we learn to fight Buchholz hydra. Unimaginably more tedious to fight! I mean literally, cannot imagine. First-years not believe us when we tell them winning is possible! Instructor must give second order, iterate until they comprehend." -HPMOR

[@pluspiping]

You're right, they absolutely did not teach me the difference between ordinals and cardinals in high school. Or college (I stopped after Calc 3). But this is a really fascinating tool and I feel like I understand how useful it can be now! Thank you! I'll be excited to watch your other videos

[@Benson_Bear]

Although they do indicate, grammatically, the difference between the way natural numbers can be used: 1st and 1, 2nd and 2, etc. So they kind of point out that each natural number is both and ordinal and a cardinal.

[@MrRyanroberson1]

By 2:13 the visualization got me thinking to always cut the head farthest from the body, it doesn't matter in which order, so long as there is no farther head from the body than your choice, the heads would essentially degrade as a sum, for the example at 2:13 one should chop all the left heads of the pairs, then you would have six head-on heads, chop those and the result is one head, with 13 heads on it, chopping of one you have three with 12, 9 with 11, so on unwil a body with 3^12 heads on it, finishing the beast is easy from there. I'm going to watch the video now to see how I did.

[@pabloagsutinnavavieyra2308]

Hmmm I just have a little, kind of obnoxious, question on your solution. So, I think, we use omega to represent a more generalized case of our hydra diagrams, but then why the last head avaliable for us to cut isn't also labeled with omega to the first power and going on from there? If it was the case, each of the avaliable heads for us to chop could be a mere representation of an infinite set of heads. Having a finite number for them, but going with infinite cardinals to the heads "below" adds a little bit of (finite) oddness to the explanation, for me. Still pretty awesome video. Just two days before it came out I was talking about subreal numbers with a friend, and this problem was a nice way to go around some of the concepts behind it. Great job!

[@rainetheplanet]

Level 1: A branch of height 2 Level 2: A head with 2 heads on top Level 3: A branch of height 3 Level 4: A branch of height 4 Level 5: A branch of height 10

[@columbus8myhw]

By the _well-ordering theorem_, you probably mean well-ordering principle or something else. (The well-ordering theorem is the one that says that any set can be given an order that well-orders it; it's equivalent to the axiom of choice.)

[@iIO_OIi]

The sun rising at 3 pm here does sound odd, though since measuring time by 24 segments per planetary rotation is already somewhat abstract and simple only seems reasonable because we grew up with this idea, it's not any less logical, debatably more logical, while I like this abstract measurement and how it helps us schedule things, I still can't deny that currently the most logical way to tell time is day or night, and within that how far along the moon is, or how long shadows are & what direction they're stretching to (unless you're in a forest... particularly the Giant Redwood forest...)

[@xBieux]

Ik this has nothing to do with the video, but can you explain the TREE(3) and other infinitely huge numbers? I tried learning about it but never found an easy explanation to it, and all those lines and heads made me remember of what it is supposedly based on.

[@TheGamingG810]

Original ordinal is ω^ω^ω. After you cut the first head, there will be 3 top heads (ω^ω^3) . After you cut all the forking branches, there will be 9 top heads (ω^9ω) , and after you cut the branches, there will be 27 top heads growing out of a single head (ω^27). After you cut these heads, there will be 7,625,597,484,987 heads, and about 11 trillion steps.

[@sickone3060]

So from watching your poisoned wine, hydra, and hierarchy of infinites made me think of a game called Nim and this question. If you are playing a game of Nim with infinite heaps can you still ensure your victory mathematically? I ask in pure curiosity and fascination. Btw awesome show!

[@zioscozio]

While the order of cuts doesn't matter in terms of overall convergence, it does in terms of number of cuts to kill the hydra.

[@chrisdock8804]

That was simply amazing. This channel should be world famous.

[@funnyvalentine175]

I got 243 heads in the ball and 374 chops total for the bonus question. A length four will produce 9 length 3s in 5 moves and each length 3 takes 41 chops to kill. Similarly there are 9 3 length which will each turn into 27 heads, so 9*27 is 243 heads in ball.

[@lilig.8161]

NOOoo ! You can't do n*ω (it would still be equal to ω), sounds weird but you should have done a thing like ω*n here

[@sdmartens22]

At 9:48 the diagram is incorrect(it's too high), i sincerely appreciate your content. Thank you.

[@ardorpraxis9661]

I noticed this as well.

[@HarkunwarSinghKochar]

Loved this video, KZbin notification woke me up at 4 am but was worth anyway.

[@TheLolle97]

Awesome video! You again hit just the right topic to expand my mathematical understanding.

[@alexandrugheorghe5610]

Nice work! I really enjoyed this episode from start to finish. Thank you.

[@flymypg]

This channel is maturing wonderfully. I really like needing to hit pause: It makes the video interactive for me, instead of my being a passive vidiot. And here I was thinking Cardinal vs. Ordinal was a Catholic thing about Priests and Congregants and the rise of Protestantism. I was way off on that one. ;^)

[@meganheffel6085]

So greatful for this channel and the spacetime channel! Interesting material and amazing teachers!!!! THANK YOU!!!!!

[@alazrabed]

In response to the the two questions around 4:30. I'd guess the hydra will have, at the final stage, three to the twenty-seven heads. And, in total, Heracles would need forty one plus four third of three to the twenty-seven chops in order to complete the second of his twelve labors. Sounds like a quick one.

[@rickzegooene]

let me give the challenge a try... so if you have n heads that grow on another "base" head, if you chop one off, there will be 3 branches, each having n-1 heads. if you chop one head off from each branch, there will be 3*3=9 branches, each having n-2 heads. repeat the process, by the time you get to each branch having 0 heads (which means that it will just be the base head), and by this time you will have 3^n of those branches. if it's a straight line of x heads, once you reduce the maximum length to x-1 heads, you will have 3^1=3 heads at the tip. to reduce the maximum branch length to x-2 heads, since you already have 3 heads at the tip, you will get 3^3=27 heads (which is what you get when cutting a line of 3 heads). By this logic, if you want to chop 4 heads, you will end up getting 3^27 heads. as for how many steps it takes, if you have a branch of n heads, you chop once to get 3 branches of n-1 heads, and then take 3 chops to cut off one head from each branch to get 9 brahches of n-2 heads, and so on. by the time you get to branches of 0 heads, you will have taken (1+3+3^2+3^3+...+3^(n-1)) steps. to get from a line of 4 heads to spike ball, it takes 1+(1+3+9)+(1+3+9+...+3^26) steps, and then another 3^27 steps to clean up the spike ball.