Splitting Rent with Triangles | Infinite Series

[@pbsinfiniteseries]

Hey guys! We know Mathologer just released an awesome video on the same topic. (There's an appendix addressing this at the end of the video.) We recorded this beforehand and, after speaking with Mathologer, decided to go ahead and release it. Hope you enjoy having two perspectives on this fun topic! Also, 3Blue1Brown just released a video on a related topic, Borsuk-Ulam, which is totally worth checking out. :)

[@aviralrastogi]

PBS Infinite Series You should pin this comment so everyone can see this at top of comments (for who didn't watch the complete video). Btw, love your videos.

[@Antediluvian137]

It's great to see these 3 channels supporting each other :)

[@TheJaredtheJaredlong]

You should do an episode about zero-sum games so that people can learn how youtube viewership is _not_ zero sum.

[@unvergebeneid]

What are the odds, though :)

[@chrstfer2452]

Antediluvian it really is awesome that these three channels are doing similar stuff in different ways. I absolutely love all three.

[@ansambel3170]

I tried to use this while eating pizza with my 4 friends, but they ate it all, while i was drawing 4 dimensional simplexes

[@ashboon1625]

Much Wow.

[@alexwang982]

You mean 5 dimensional simplexes? 5 people

[@seanmurphy7727]

@@alexwang982nah you need n-1 dimensional simplexes

[@InToTheNetherMc]

I'm really tired of reading comments about "copying" other channels. These channels are about mathematics, and this is just another topic that can be explained in a video. Following this logic means Mathologer "copied" this from New York Times, and they "copied" it from the original paper etc. Makes absolutely no sense.

[@bonbonpony]

Unless the presentation has pretty much the same structure, wording etc. One can talk about the same subject in different ways, can't one?

[@bonbonpony]

You wouldn't, because mathematical formulae are not copyrightable. Ideas cannot be copyrighted. Forms of expression of those ideas can. If you chiseled your "2+2=4" in stone as a fancy sculpture, or painted a beautiful painting with colourful 2+2=4, then yes, you could copyright that painting (actually, it would be copyrighted without you needing to do anything special), but the copyright would apply to the painting or sculpture, not the idea of adding numbers. As for the above video, there is absolutely no problem in the fact that PBS made a video about the same idea suspiciously shortly after Mathologer. Everyone could. But the _form of expression_ in both videos is quite similar, which raises all those controversies.

[@codycast]

Lol. I was going to say the same. "Copied". Give me a break. I have no idea who/what "mathloger" is but the fact they were linked by this channel might bring him/them some new viewers that never heard of him before.

[@ipadair7345]

Those of you, who have seen Mathologer's video: _11:25_ . There is an explanation given.

[@judgeg2906]

People who comment on a video without viewing it to the end. You find dozens of them in here.

[@GoldenKingStudio]

*Insert Vi Hart Here:* Trianglestriangelstrianglestrianglestrianglestrianglestriangles.... Anyway, is there a formula to determine how many N roommates can have secret preferences for this to still work?

[@ReaperUnreal]

I think it's about time for a triangle party.

[@moritzkockritz5710]

now I'm just sad she doesn't make (math) videos any more D:

[@Noah-fn5jq]

Theoretically, yes. You can use a n-1 simplex and perform the same analysis. Although building the analysis would take some work. Note: I misread the question. I think the above still addresses it, but it may not if I misunderstood what "secret preferences" means.

[@fredvand.6183]

GoldenKingStudio I think only one (with that method anyway). Here's my intuition on the subject; feel free to TL;DR if you feel the need. The idea is that with n people your simplex has n vertices, but you only know the preferences of n-1 people. You need to label the vertices with information in such a way that there are multiple possibilities, so no matter what Moe (or the missing person) chooses, there is an arrangement in which the others are still happy. You need two (or n-1) pieces of information for each of three (or n) vertices, for a total of six (or (n-1)*n) pieces of information. By way of conceptual analogy, you can imagine looking at room A and labeling it with sticky notes. If Moe picks that room, then he gets it, so there's no question there. There are two (n-1) other options, so you put two (n-1) sticky notes: one saying which person gets it if Moe picks room B, and one saying who gets it if he picks C (and so on for all n rooms). Then you look at B and label it the same you did for A, etc. When you've covered all three (n) rooms, you have two (n-1) sticky notes on each. So, you need at least two (n-1) pieces of information for each to place Joe, Curly, etc. no matter what Moe picks. With only one person absent (i.e., Moe on vacation), you do have enough information. You can get the preference of each of n-1 people for each of n vertices, for a total of, yes, (n-1)*n. However, with two people absent and higher-dimensional (simplexes? simplices? simplici?) triangles, you only get (n-2)*n, which doesn't give you enough. P.S. I know this is a bit hand-wavy and certainly isn't rigorous by any stretch of the imagination, but coming from someone familiar with computer science, it's intuitive. Also, since this is KZbin and no one is actually going to read this far, long live the Pacific Northwest Tree Octopus.

[@NoActuallyGo-KCUF-Yourself]

Great Vi Hart impression. ;-)

[@liesdamnlies3372]

What I learned, is that I probably don't want mathematicians as roommates.

[@Tangerinetaco]

😂

[@yujiokitani4492]

what's the red circle on her shoulder?

[@RiRiDingetjes]

It's illuminati in a completely differentiable form

[@ruinaS2]

Yuji Okitani Clearly a werewolf bite. Duh

[@robertvralph]

A tattoo, you can see it poking out in any video where she's wearing a sleeveless garment.

[@fossilfighters101]

+

[@pfeffer1729]

Funnily enough, me and a couple of friends were just moving in together, and I looked up algorithms to solve our room distribution, stumbling into Su's paper. I was very surprised to find both mathologer and infinite series doing a video on it so soon after it happened.

[@daseinbot]

That's great actually, because I understood almost nothing from Mathologer's video.

[@bloodyadaku]

The issue I have with this analogy is that just because one person chooses their optimal room out of the three choices at each given price point, doesn't mean that that is the optimal solution out of all possible solutions. That is, since there can be more than one inscribed triangle with a solution, we cannot determine WHICH solution is the best one using this method.

[@jetison333]

well how would you define which solution is the 'best one'. Any of the appropriate triangles are envy-free, which means that no one person would want a different room under that price distribution. so none would be better than the others.

[@f1r3fox235]

Here is my (not formally) proof of why there is always an odd number of labeled rooms: If you get stuck at every path you can go starting at an outside door, than you obviously have an odd number of labeled rooms you get stuck in. If there are n outside doors (with n = 2*x + 1, x € N), than everytime you will exit the triangle through an other door, this door and the one you started, can't end in a labeled room. So you have n-2 doors left (which is an odd number), that could lead to a labeled room. So there only could be an odd number of path's you will get stuck. If there is a labeled room you can't enter from outside, it must have (only) one door. If you start inside this room an got through the door, than the next will either have no other door, so it is a labeled room too, or you can go through all the following rooms, until you will get stuck. So for every room you can't enter from outside, there must be another room with only one door, where you finaly get stuck, which is also a labeled room. Since you have always an odd number of rooms you can enter from outside and always an even number of rooms you can't, there is always an odd number of labeled rooms inside the big triangle. I appologize for my bad english, it is not my native language. But i hope you guys can understand it though :)

[@bjornmu]

I see one possible problem here which means this method may not always work. You make the assumption that if any one room is free, that room will always be preferred. That may well not hold. Say if the least desirable room is free but the best room costs only 10%, then you may well choose the best room because it's worth the difference. But maybe that doesn't matter since that will never be the optimal split anyway?

[@JaapVersteegh]

Same thought here. The assumption that a free room is always preferred seems not necessarily rational. Maybe you can avoid it by extending the triangle to negative rent prices? At some point getting money for a room will convince someone to take it..

[@pbsinfiniteseries]

That's a good point. It might be that in real life you'd choose a cheap room over a free room. But, in order to make all the math work out here, we assume everyone would choose a free room over a non-free room. It's an assumption of the model, but not necessarily always accurate.

[@benjaminv3748]

PBS Infinite Series Though, if looking at it realistically, why would any 2 persons hire 3 rooms and then give one away for free? Since these "points" would never be selected it should be fine to just select, whatever suits the math, those sides' numbers. I mean, it won't have an impact on a RL situation. So the math works in any case :) What I'm trying ot say is: Any of the inner vertecies being predetermined could affect and represent RL occasions but the outmost ones with free rooms would simply never occur. Hope that makes sense?

[@benjaminv3748]

Timaeus Bouma But, realistically speaking, does this ever happen? I've never heard of a, say 3, group of people where one person lives for free and has no housing costs and where the other 2 just split tje rent. IF this would be the case though, couldn't you just lower the dimenson of the simplex and completely exclude the 3rd room since it's free anyway and one person is ready to take it for that. I like your getting paid idea though, I think it's quite unrealistic but what do I know. And nevertheless it's an intrieguing idea for this particular splitting method! What would happen if you set negative numbers?

[@benjaminv3748]

Timaeus Bouma Fair enough :) Good thinking, thanks for that!

[@oker59]

The filling of 4-dimensional space in between the three space can be explained by the way a sphere goes through a plain . . . first a point appears, then the point grows into ever larger circles, and then finally ever smaller circles till a point is left, and then the point disappers. The tetrahedrons ever smaller inside the tetrahedron . . . to infinity corresponds, and visually proves this explanation of higher dimensions.

[@dayliss413]

I think it's unfortunate that this came out so soon after mathologer did a video on the same topic. I'm worried people will think one channel stole from the other even though I'm pretty sure the choice of topics and use of the same proofs is just a coincidence. Both this and mathologer's are great videos though!

[@Ermude10]

Well, for us viewers it's the perfect timing. This means we can solidify these concepts through two different videos.

[@dayliss413]

Very true! It's always good to hear multiple different voices to reinforce understanding.

[@TheDaniel2718]

I don't think it's a coincidence, I think the channels thought right around now would be a good time to release them so all the college students who watch these videos can try it out when they go apartment hunting for next year.

[@NickCombs]

I think of it more as a collaboration. Ideas are meant to be shared, not owned.

[@kuijpers5870]

The weakness of the method is that it is unlikely that you would prefer a free room half the size of a room with a rent of one dollar a month.

[@donaldhobson8873]

Easier way of dividing up rooms, for non mathematician room mates. Pick rooms A,B. Let mates X,Y,Z choose prices Px, Py, Pz (which can be negative) for the amount they prefer room A to room B. So if Px=£30 then person X will not mind between rooms A,B if room A is £30 more expensive. Choose Pm the median of Px, Py, Pz. Bid for room C. Whoever is willing to pay most for room C gets it at that price. The person who chose a higher price at the Px, Py, Pz. stage gets room A. The remaining rent is divided between the people in rooms A,B with the person in room A paying Pm more.

[@equicomplex]

5:12 "And if you break an N-simplex into smaller N-simplices" - But the tetrahedron is not a space-filling polyhedron. When the outer triangles have four small triangles each, the tetrahedron has four tetrahedra and an octahedron in the middle.

[@stevethecatcouch6532]

The regular tetrahedron is not space filling, but simplices don't have to be regular.

[@laksharora2559]

Because the total doors on the edges are an odd number, the total one door rooms we find through this method will be odd; either one will be found through a single walk through the rooms until a dead one is reached or none will be found if one just comes out again thus consuming two and a total even number of exterior doors in such cases. Thus through this method we will only find an odd number of rooms. But the total rooms we find are not exhaustive. We will always find at least one and in total, an odd number of rooms but besides what we are able to find, there may be a number of pairs of rooms present which are connected through just one door each. Therefore an even number of more rooms are possible which allow fair division. Thus, the total number of triangles are odd + even = ODD number of triangles.

[@yaeldillies]

About the proof of the odd number of 1 - 2 - 3 labeled triangles, we use the same method of 1 - 2 "doors" as showed in the video : We already know that there's an odd number of 1 - 2 - 3 triangles accessibles by the 1 - 2 side (as there's an odd number of 1 - 2 doors on it). Then, if a 1 - 2 - 3 triangle isn't connected by 1 - 2 doors to the 1 - 2 side of the big triangle, we can walk through 1 - 2 doors until we reach another 1 - 2 - 3 triangle, which keeps odd the number of these triangles. It works as well in higher dimensions by recurrency : Suppose that Sperner's lemma is true for an N-simplex (with an odd number of 1 - 2 - ... - N labeled N-simplices) Then on the 1 - 2 - ... - N side of a N+1-simplex we have an odd number of 1 - 2 - ... - N N-simplices and we can walk through 1 - 2 - ... - N doors until we find an 1 - 2 - ... - N+1 N+1-simplice * and with the same arguments than before, there's well an odd number of 1 - 2 - ... - N+1 labeled N+1-simplices in our N+1-simplex *Proof that an N+1-simplice has either zero, one or two 1 - 2 - ... - N doors : Suppose we have a N+1-simplice that has at least one 1 - 2 - ... - N door. The point that isn't in this door is either N+1, in that case the simplice has only one door 1 - 2 - ... - N and is labeled 1 - 2 - ... - N+1, either 1, 2, ... or N, in which case the simplice has a second door and not more.

[@yaeldillies]

I may haven't been that understandable...

[@TheLoneGnu]

Mini challenge: Let t0, t1 and t2 be the number of small triangles with 0, 1 or 2 colored sides. If we count the number of small triangles with one colored side (t1) once, the number of small triangles with two colored sides (t2) twice, and the number of colored sides on the outside of the big triangle (s) once, we will have counted the each colored side twice (once for each side of the side), that is: 2*c = t1 + 2*t2 + s. This rewrites to t1 = 2*(c-t2) - s. No since the number of colored sides on the outside of the big triangle (s) is odd then so is t1. And t1 is exactly the number of small triangles with tree different values on the corners.

[@sinecurve9999]

Given a necklace with n types of gems, prove that the gems can be divided evenly among two people with exactly n cuts (clasp doesn't count as a cut).

[@typhoonf6]

So much effort... I find the sacred rule of dibs is a much simpler way to settle these things.

[@stevethecatcouch6532]

Here's the streamlined version of my proof that there are an odd number of one door rooms. The path from each outside door leads back outside or to a one door room. There are an odd number of outside doors. An even number are involved in each path from outside to outside, leaving an odd number of paths leading to a one door room. That means there are an odd number of accessible one door rooms. If a one door room cannot be reached from the outside, any path originating in it must end in another one door room, so they come in pairs. So there are an even number of inaccessible one door rooms. Added to the odd number of accessible one door rooms gives us an odd number for the total one door rooms.

[@svenkummetz2434]

After looking at the times version of this, I assume that they use some adaptive version of this, since their error decreases over time. I assume that, upon reaching a sub simplex with all vertices colored differently(thus after asking the person assigned to the empty vertex) they subdivide it again. When leaving the simplex again they will need to recolor the previously empty vertex. This will create a new exit on one of the incident edges, this is the exit we take.

[@ArjunSAriyil]

Mini Challenge Answer: Since there are odd number of edge doors, there has to be an odd number of fully labeled rooms in which you get stuck if you start walking through an edge door. Mark these rooms as Set1 If you start walking through doors from any other fully labeled room, you will again get stuck in another fully labeled room. So all these fully labeled rooms exists in pairs. Mark these rooms as Set2 Since number of rooms in Set1 is odd and number of rooms in Set2 is even, total number of fully labeled rooms will be an odd number

[@TheTechBoostChannel]

3 of my classmates and I are currently stuck on the division of work on an assignment that is due tomorrow. Only the assignment is a synthesis of Francis Su's "Rental Harmony"

[@khalidabduljaleel]

Oh, I have fallen in love with Kelsey.

[@glenplonk]

Here is an answer to the first question you asked (why is the number of fully labelled triangle always odd) : imagine there is a fully labelled triangle in the middle of the big triangle, it has one door which leads you to an other room. Or it has one door (the one you just went trough) and you are in a fully labelled room, or it has two doors and you can continue your walk until you reach the first case (You can't go outside, I can't give a short explanation but it is ... trivial)

[@KyleZager]

What I got out of this video: don't become roommates with a mathematician.

[@jacobdawson2109]

After watching this, I can't help but wonder if it can't be used in a kind of iterative bisection algorithm. Where you continuously subdivide each triangle till you get whatever arbitrary precision you desire, focusing only on the best sub-triangle from the previous step.

[@jetison333]

yep! in mathologers video theres a link in the description to a site that does just that. not sure how it works, but im fairly sure its what it does.

[@NonTwinBrothers]

My man Simmons just wanted to split up a cake

[@kevinocta9716]

What a nifty and useful slice of math.

[@Valdagast]

So if there are two people, you use a one-dimensional triangle? What does that look like?

[@JaapVersteegh]

A line?

[@yaeldillies]

An edge. Yeah, it blow your mind

[@stevethecatcouch6532]

And in the video she proved that if you label one end 1 and the other 2, with the points in the middle labeled either 1 or 2, there will be an odd number of segments labeled 1 on one end and 2 on the other. The label one end A and the other B and alternate between the two for the rest of the points. (There ought to be an even number of points.) Any segment with a 1 on one end and 2 on the other is a fair distribution of the rent.

[@adityakhanna113]

I love how almost every comment reply points towards 3Blue1Brown. His is literally the best math channel on KZbin (sorry Kelsey and Brady!)

[@vladimirivanov1562]

Good luck dividing the rent between 4 or 5 people on a sheet of paper with this method :D

[@thisaccountisdead9060]

LOL - I'm going to have to watch this again another time.... I could only make it half way through before I could feel a panic attack coming on. Seriously.

[@arjunabetta4572]

Infinite Series is more mindblowing than PBS Spacetime.

[@knife_wizard]

So... I found this channel from the crossover you did with PBS Spacetime. Are there any other PBS channels that cover interesting material like this that I should be aware of? I know of Crash Course, which is also fantastic, but I'm curious if there's other stuff out there as well, that I don't know of.

[@holothurieabysse]

Wondering if we can solve this problem with a formulation in terms of Nash's equilibrium (because Sperners' lemma (or equivalent) is essential in the proof of existence of such equilibriums)

[@oscarsmith3942]

Is there a unique fair division? What about a "best" one? Does the hidden preference technique always result in the same fair division as the original?

[@Maroof_64]

Splitting a triangle into tri-forces... 2:57 "Switch"... Hope I'm not the only one hyped for Zelda: Breath of the Wild :D!

[@Thee_Sinner]

I have no clue what this was about, but I watched it.

[@diribigal]

Spiddit is another good fair dovision website implementing a similar algorithm.

[@Kram1032]

Quantum Mechanics really is largely Linear Algebra. However, typically with the (for Linear Algebra) non-standard notation of BraKets. - These are absolutely awesome at being super short hands for Linear Algebra but initially it's a little hard to wrap your head around them. Once you get them, you'll love them. I recommend looking into courses that specifically use them _together_ with courses that focus on Linear Algebra in "Standard" notation. To be honest, for lots of Math subjects, specifically well-established ones like Arithmetic, (planar) Trigonometry, Linear Algebra or (basic) Calculus, the hardest part is really learning how any given notation corresponds to concepts in a given theory. Once you get that, most of the rest will be pretty easy. (This does not apply as much to stuff that's currently on a research- or specialist level. That stuff will be extra hard, among other things because, for instance, key theorems might not yet be established or, at the very least, there may only be a handful of people who actually know them, and it'll take time for them to convey these results to broader groups)

[@Sejiko]

I have seen a lot of cubes hidden in this big triangle picture.

[@GelidGanef]

Already saw Mathologer AND 3Brown1Blue do this. But this was totally worth it for the epilogue! Thanks for adding your take on this!

[@cocoa1996]

For the guy who wanted to learn Linear Algebra, there's a KZbin lecture series from MIT's opencourseware if you want to go in-depth. MIT also provides homework assignments and their answers and everything. It's pretty awesome Link to lecture series: kzbin.info/www/bejne/kHyWgGdmZ9yZZ8U

[@chuvzzz]

I have a few lingering questions about the stability and optimality of these results... i.e., is it possible to cheat or game the system? That is, can one of the roommates, by lying about their actual preferences, make it so that the apparent even split actually favours him/her? If not, do they always make things worse for themselves by lying, or can they just end up screwing someone else instead? And a last question, which I suspect must true: Can they simply just make things worse for everybody (including themselves), period? Or can the other roommates shield themselves from "doom" by simply answering honestly?

[@rmsgrey]

Yes, it's possible to game the system if you know your roommates preferences. Look at all fair outcomes if you're all honest about your preferences. Those can be divided into 3 groups based on which room you would get. Within each group, there will be some variation in the price you pay. For each outcome where you'd be paying more than the minimum for that room, you can avoid getting that outcome by lying about your preference. Of course, while you're telling an entirely justified tactical falsehood in order to try to improve your position, your treacherous friends may be telling outrageous lies to try to improve theirs and end up with two or more of you in the wrong rooms. Of course, if that happens, you can always quietly swap rooms afterwards - since you're all three only lying about rooms when there's a fair division available, you'll always be able to revert to the true fair division for whichever outcome you end up with.

[@chuvzzz]

Hmmm I don't know. I'm not really getting a clear picture here, I think there is more to it? Also, if you lie about a single preference, aren't you potentially opening yourself to other outcomes that didn't exist before, or no?

[@rmsgrey]

Potentially, yes. When you switch one preference, you change three triangles, potentially creating new "fair" outcomes, but that would mean the others' preferences being weird - if you switch from 1 to 2 or 3 to avoid a fair outcome, that means there are fair outcomes where 1 is cheaper and you still prefer it, but both the others don't; for that switch to create a new "fair" outcome, one of the others has to prefer 1 nearby despite not wanting it cheaper for another division. If everyone's willing to lie about their preferences to avoid unwanted fair outcomes, then, yeah, there's a thin shell of potential "fair" outcomes that could accidentally be triggered. In general, I feel it's better to use the proof to show that there are fair outcomes rather than to use it to pick one. Knowing that there must be fair outcomes if the assumptions of the proof are met, you can then determine something about the range of available fair outcomes and pick one by some other method.

[@ArjunSAriyil]

To get a clear picture consider the simple case of two persons A and B trying to share a big room and a small room. A prefers big room and is ready to pay upto 70% of rent for It. B is ready to take small room for 40% of rent. One fair division can be A taking big room with 65% rent and B taking small room with 35% rent. A can shift the game in his favour by lying he is ready to pay only upto 61%. Or else B can shift the game in his favour by lying that he is ready to take small room with 31% rent or less. Of course it gets screwed up if both lies as they would end up with their opposite choices.

[@chuvzzz]

Hmmm I see although that is not the kind of "gaming" I was thinking about. I think what you are taking about is due to the grid not being discretized finely enough, isn't it? I was thinking of gaming the system by changing the locations of where the fair splits will happen, given that the other two have already chosen. To me this sounds a lot like the case where one of the roommates' preferences is unknown that she mentioned in the video, although I cant quite put my head around how those are related precisely. I think they don't amount to quite the same thing...

[@IlTrojo]

I'd say the idea to prove that the number of rooms with just one door is odd is to count door *sides*: this is clearly twice the number of doors, hence even. Now place one person in each room plus one outside, and ask each of them how many doors their room has; add them up and you get the same number as before, hence an even number. Now, whatever the number of two-doored rooms, the number of doors accounted for by those people in two-doored rooms must be even (because each room has two doors). Hence, also the difference between these two numbers must be even, and this is the number of doors accounted for by the people in one-doored rooms plus the person outside. This last person sees an odd number of doors, hence the number of doors seen by everyone in one-doored rooms must be odd; that is, there must be an odd number of one-doored rooms.

[@stevethecatcouch6532]

+IlTrojo Brilliant. It was a bit hard to follow, because I originally understood "plus one outside" to mean "plus one outside each room", instead of "one outside the big triangle.

[@IlTrojo]

It is probably easier this way. Just apply the method shown in the video over and over until all the doors to the outside are used up. This will give an odd number of one-doored rooms. We are left to count the number of one-doored rooms that are not connected with the outside. Pick one and start walking: you must end in another. This shows that the set of one-doored rooms can be split into pairs, hence it has an even number of elements, and we are done.

[@IlTrojo]

+Steve's Mathy Stuff Sorry, I only just read your answer, I had not seen it when I wrote the last comment.

[@sprites75]

i'm seeing this two weekd after , but in the second example there are 5 triangles which are 3-labeled. which one do you pick then? the explanation proved at least one existed though multiple can , with different results each.

[@omhekde9033]

this went above my head!!

[@MorRobots]

So this is a negotiating model where you distribute positions in n-space in a zero sum system?

[@aviel04]

It took me a day to realize I didn't see this video after I already watched the mathloger one on the same topic hahaha

[@TonyCorin]

Thanks again, Infinite Series.

[@andriypredmyrskyy7791]

this sounds nice and all, but wouldn’t I have to ask each roomate for their opinion on a room for each triangle vertex in my model? It seems like that gets really tedious really fast.

[@SlimThrull]

The mini-challenge is the same as Seven Bridges of Königsberg problem, isn't it?

[@DrJTPhysio]

The most accurate way to split rent!

[@NarsisAusian]

It looks like its possible to have more than one fair division so how do we choose which one to use? Also what happens if some people prefer one fair division over another? Is that even possible?

[@kennethgewerth7303]

How three math grad students got evicted because they were late with the rent.

[@brunoschultz4346]

So could this be applied to an election? Basically that triangle could be the election ballet meaning the result would make 100% of the people satisfied with the outcome.

[@wilsoo3659]

I was thinking the same thing! But it seems a bit different as Sperner's splits n things among n people, whereas an election splits n candidates among the entire population? So you want to determine the real popularity weights of each candidate. I think preferential voting just about covers it...

[@quarkyquasar893]

Where is Vi Hart? She must be very happy today 2 videos about triangles :D

[@PhrontDoor]

It's like a geometric stable-marriage algorithm .

[@veggiet2009]

Oh I got excited when you started talking simplexes...

[@박수진-i5j]

What if I pick up another fully labeled triangle which is diffrent from your choice? Does the result change? I need your answers :)

[@josephgroves3176]

Woah that hairdo change

[@LuxinNocte]

I have a question: Will this method compute a "best" scenario, or are there some "least suck-y" scenarios, where everyone is just happy they didn't have to take the other rooms? Or is that the same?

[@zairaner1489]

Theoretically possible

[@magne14527]

At 3:50, how do you get to the other two triangles with one door? They're not connected to the outside

[@ElectricIguana]

"What's your point, Vanessa?"

[@tondopx5616]

It is kind of a letdown to arrive at an approximate solution "within a dollar" after going through all that. Wouldn't it be better to pick the center of the small triangle as the point according to whose x, y, z coordinates the rent would be split? Or better still - a point within the small triangle weighted by the position of the small triangle in the big picture, as the center of small triangle would not really be fair if the small triangle was close to an edge or corner of the big one.

[@JWentu]

I am another one who thinks that the statement at 8:00 is wrong. If I am allergic to fitted carpet and the room with the fitted carpet is free, I will NOT chose it

[@akiranishiki137]

This method is so beautiful and elegant, but I think it would be difficult to do this with actual room mates that are frightened by math.. like just too many people.

[@rhamph]

Indeed, and the definition of "fair" isn't necessarily what would make people happiest. If you have a hard limit of 30% of the total rent then you would never compete for a much larger room when it's over that, allowing a higher income roommate to dominate that section of the graph. Mind you this method also doesn't promise unique solutions; it's a set of solutions to which each roommate says "I would not change places with you". Real people have an instinctual concept of fairness. If the payments are vastly different from their perceived value they'd be unhappy regardless of how you derived those payments. Still interesting though.

[@rmsgrey]

Actually, if someone has a hard limit of 30% of the rent, then there are regions of the triangle where they're unable to chose a room because they each cost more than that, and the system breaks down...

[@pritishpatil9351]

You can make the triangle bigger and make the price of the rooms negative. So that you get paid to live in the room. Doing this will allow you to at some point say, yes that one particular room for $10 is worth less that this other room for $-10000. So, at some point, you will prefer the correct room at each edge/vertex. (Another way to think about it is to color an infinite grid, starting from the center until you reach a triangle which satisfies the sperner lemma properties. Some cheap room cannot be worth more than all the money in the world.) So what if the optimal solution happens to be in one of the negative regions? It means that at least one the people in your group would pay you for them to live with you in a particular apartment than live by themselves. (I suspect that will not happen, so although there seems to be this weird assumption required that you will choose the free room, we can make do without it by using a bigger triangle as the feasible solution would not be around that point. )

[@captainpuffinpuffinson4769]

could you make a video series about the KdV equation, or non linear waves in general? I am working on this subject and find a disturbing lack of easy to explain methods to my friends and family

[@fredvand.6183]

The definition of "fair" used here sounds rather like the definition of "stable" in the context of the stable marriage problem. Actually, on second thought, this does seem almost like a stable marriage problem: if two people simultaneously prefer the other's room/price, the situation is unstable, and they swap... although that doesn't take into account distributing the price around, so I guess it's not a perfect analogy.

[@MuffinsAPlenty]

Interesting connection! But it's not exactly the same thing. In the stable marriage problem, you can have a stable situation which would not be considered "fair" under the definition of fair used in this video. The point of fair here is that no one would prefer anyone else's room to their own room at the current price distribution. It's actually a bit stronger than the stability in the stable marriage problem. If you only cared about stability, Larry could prefer Moe's room to Larry's own room at the current prices, but as long as Moe prefers his own room too, then the situation is stable (since Moe will be unwilling to switch).

[@Jadinandrews]

Besides splitting rent or cutting cake, has Sperner's lemma been used for any other more interesting problems?

[@KohuGaly]

YEEYY I got mentioned!! :-D OK, but what happens when one of the rooms is much better than the other one, that you'd rather pay for it instead of getting the other one for free? This would mean one of the sides of the triangle has all 3 numbers on it instead of one. In that situation does the algorithm break?

[@stevethecatcouch6532]

Given the symmetry of the situation, we can assume without loss of generality that Moe will always prefer room 3 to room 1 even if 1 is free. So all the M's on the right hand side are labeled 3 instead of 1. There is still a small triangle with all three numbers. I haven't worked out a proof that it will always work. I'll get back to you. Using the app I simulated 3 roommates, 1 of whom wanted room 3 regardless of price, one of whom wanted room 2 regardless of price and one who didn't care. Three iterations took me to a split of $41.67, $479.17, $479.17. More iterations slowly converged on $0, $500, $500 as expected. If you are hypothesizing that there is a room no one will take even with negative rent, then it's not a 3 bedroom apartment; it's a 2 bedroom apartment and a cardboard box with drainage problems.

[@mansertwo]

theres been three heckin proofs of this in the last two weeks

[@levi12howell]

I feel like the modified version with the unknown isn't really a proof that "it can be done" but that it can be done under certain rules and not in all cases

[@levi12howell]

Mainly the rule that the two ppl can't have the same favorite room

[@khoavo5758]

Splitting rent with triangles mini-challenge: 1. Kesley found an odd number of fully numbered triangles (FLTs) that lead to the outside 2. To each of the other FLTs you can wander around until you find another one (but there's only one possible path). Therefore, the number of landlocked FLTs are even 3. odd + even = odd: tada! If you find something wrong, feel free to discuss!

[@aman-qj5sx]

Isn't there any "competition" between the different math youtube channels? How come you recommend each other's videos?

[@kordellcurl7559]

What if only one person showed up to tell what room he wants. Is it possible to divide the rent among the 3 people equally using the method. Also if no one showed.

[@HoD999x]

so what happens if they all want the same room no matter the price, or if they want someone else to be in a specific room for a specific price?

[@stevethecatcouch6532]

In the first case, there is no way to make everyone happy. In the second case, if your roommate is that controlling, division of the rent is the least of your concerns.

[@MKD1101]

is it possible to have 1.2, 2.3, -4 dimensions? why do dimensions exist in whole numbers only?

[@CuulX]

Watch the 3blue1brown video on fractal dimensions "Fractals are not self-similar"! watch?v=gB9n2gHsHN4

[@yonatanbeer3475]

thank you. that was very interesting

[@Zerotan]

I found it odd that you were encouraging further subdivisions of triangles like you were approaching a limit. what happens to Sperner's Lemma when you move from a discrete realm to a continuous one?

[@addemfrench]

I wonder if this has some application in search result ranking.

[@pankajnegi9795]

nice explanation..gr8 work

[@mohammedhubail1607]

What's the reb line in your right hand ?!

[@ebigunso]

What if there was a room that was so bad that nobody wanted it even if it was free? Would that break this method?

[@Roshkin]

Why is this the easiest proof that proves this lemma? It seems there are many abstractions, are they all necessary?

[@R.Instro]

Just saying, it may not be "fair," but college students are college students: nobody wants to pay 50% of the rent/utilities on 1 room in a 3-room apartment where the rooms are all even vaguely similar in size/quality. I'd have been laughed at if I'd tried to get people to do anything other than just divide the total cost by 3 in such a case.

[@bakhtiarmeraj7789]

Wait a minute!!!!....Spenner's Lemma give fair distribution according to the opinion of the person for room...Why not distribute it according to the facilities available in those rooms?..

[@b43xoit]

Can this be applied to invent a voting system for legislatures?

[@badlydrawnturtle8484]

You end by saying that you link to 3Blue1Brown's introduction to linear algebra videos in the description. I think you forgot to actually do that.

[@pbsinfiniteseries]

Included now. Thanks!

[@U014B]

Yeah, this works in the Mathematical realm, but what about the Physical?

[@manaspratimbiswas7004]

Do SPERNER"S LEMMA is kind of a proof of the cosmological principle?

[@adrian_zombturtle148]

I a keept on being distracted cos i wanted more information about that red circle on her shoulder

[@iamstickfigure]

Well, regarding the question of splitting evenly when one of the roommates isn't there, just pick the two best rooms and leave the shittiest room for the person who hasn't shown up. Lol.

[@gabrielz6047]

lemma puts on glasses ligma

[@deldarel]

I guess you made this before the Mathologer released their video about the exact same subject 2 weeks ago.