This video presents a comprehensive analysis of which squares can be tiled by various polyominoes (connected shapes made of unit squares). The speaker systematically examines monominoes (1x1 squares), dominoes (2x1), trominoes (3-cell shapes), tetrominoes (4-cell shapes), and pentominoes (5-cell shapes). Key findings include: monominoes tile all squares; dominoes tile only even-sized squares; L-trominoes tile all multiples of 3 except 3x3; I-tetrominoes, T-tetrominoes, and L-tetrominoes tile only multiples of 4; S/Z tetrominoes cannot tile any square. For pentominoes, the speaker demonstrates that most can tile all multiples of 5 except 5x5, while X, F, T, Z, U, and W pentominoes cannot tile any square. The speaker also explores computational approaches to determine tiling possibilities, particularly for the challenging case of tiling 15x15 squares with P-pentominoes without reflections.
Is This Pentomino Tileable?本站添加:
Um hi everybody. Um, apparently this is third take, but I'm going to be doing this um relatively testing kind of thing where I'm going to talk about the problem of tiling squares with polyominos. Um uh specifically I'm thinking about small polyominos um polyominos which are small enough um that are interesting to consider and I'm considering like for a specific polyomino which squares can you tile um so that's the problem that I'm going to try to consider here right so for instance I mean there are some results that uh that are pretty um famous um even though they're they're not technically scores so here I have an 8 by8 grid right And then I can take away these two corners and I it's it's relatively famous that if you take polyominos um specifically dominoes like these ones then you cannot tile the entire board with these dominoes. You can try but you find that after you tile um after you tile the majority of them um there isn't enough space. specifically. Um, if you tell them, you'll get one here, one here, one here, one here, one here, one here, one here, and then down.
And then you'll find that here you have a missing space over here. And whatever you do over here doesn't matter, and you're not going to be able to cover it in this pattern. Uh, and there is a proof for why you can't cover it.
And uh, specifically, it's has to do with coloring. Um, and it's if you color every even one in a specific color. So, let's say if I color these black, then I can do this um kind of checkerboard pattern in which each domino I'm not I'm not going to do it for the right half, but in which each domino covers one white and one black square.
Um, and the entire board here has 32 black squares and 30 white squares. And therefore, since each domino covers one black square and one white square, we're going to take 31 dominoes to cover six uh 62 squares. Um, but that will cover 31 black squares and 31 white squares instead of the 30 um 32 black squares and 30 white squares.
So that is the rationale.
Okay.
Now, um I want to talk about um specifically we're going to limit our scope to squares because um once you get to bigger polyominos um then it's harder to analyze um it's harder to analyze non-square shapes, right? Especially weird shapes. And I did consider doing this for rectangles, but it didn't quite work. So, we're going to be doing squares.
So um continuing back we're going to start from the simplest polyomino um arguably the monomino um anyone has a guess for what what squares a monomino can tile >> I guess that >> when you say when you say squares >> are we only considering like n byn squares is that the question here >> yeah n by n square okay all of them positive integer Okay, >> cool.
Um, lattice can find the monotile. Um, I'm not sure how how how that would work, but okay, I'm I'm reading from the chat. Um, as Talia is saying, um, what about tiling squares with um, uh, basically uh, tiling really big tiles and trying to encode I'm assuming trying to encode some satisfactory problem inside there. I think that's interesting, but I don't really see how it can be done. But yes, um one one by one squares, right? Um and one by one squares can tile all squares.
Uh one by one polyominos can tile all squares. Um easy. Yay. Now let's consider dominoes. Um, and because it's squares, uh, the the result is kind of still, uh, very easy. Um, um, any guesses, I guess. Only even n.
Yep. Only even n and um, right. So, I'm going to outline the proof. This similar proof is probably going to come up like if you want to do it rigorously, it's obviously going to come up like a bazillion times. and by business I'm thinking like 10 um throughout this talk but basically if n is odd then you get an odd total number of squares that's obviously not tileable with an even number with an integer number of dominoes because um you're going to be tiling an even number of squares and so it has to be an even by even and for an even by even you just make a construction and the construction is trivial because you can just tell it in um 2x two um I can call monomers, but that's like a biology term. Um, I'll call these units, right? Um, I can tell these with these 2x2 units, and that will allow me to tell any, um, 2x2 square.
Yep.
Um, next up, um, 1x3. Um, right. For 1x3, um, I I got I guess I'll Yeah. 3N by 3N as in the chat. Um again uh the proof is easy. If it's not three that's obviously not possible otherwise we built this structure. Um the more interesting trooino um is well the other tromino and that's going to be if this decides to work um the lroino right and I'm going to I was going to attempt to draw this here but it looks like this. Um, any guesses for what the elmanino can tile? And this is, I think, significantly harder.
>> Yeah, I'll need a minute to think about that.
>> This is what this entire talk is probably going to be.
I have no idea why this is just not open.
Can try 6N by 6N. Um, the construction is obvious and because Excalibur isn't allowing me to create um isn't allowing me to create a new shape.
All I have to do do with this um you have one here, one here, and then this is a 2x3 unit. So if you repeat this 2x3 unit six times then you get a 6x6 unit like this and so it can tall a 6x6 which means it can tile a 6Nx 6N. Yeah this yeah it can obviously tile a square.
Then the question remaining is is there some some K such that like K times 3N is is possible where K is not even like is it possible you could do a 9 by9? I'm not immediately seeing the proof for that but >> Yep. Um I mean it's it's obvious that if n is not a multiple of three then that's you obviously can't tile it. If it's a multiple of six, uh, we've shown that you can tile it, but what if it's a 9 by9? And that is a great question, and I'll I'll leave that as a further exercise.
And we're I'm going to wait Um I might have a hunch.
Um I I maybe you can run with me with this. Um but >> I will >> I think um if you draw my my hypothesis here is that it's not going to be possible for any for any odd um like KN like like for any KN with K being odd. Um >> odd multiple.
>> Yeah. Yeah. Yeah. Um so do me a favor. Draw like an L shape in the corner. Like there's there's two possibilities for what how how a piece can go in the corner. Um, you want Okay, I'll draw.
>> Yeah, let's do both of them.
>> Okay.
>> Um, >> I'm gonna give you this and I'm give you this.
>> Yeah, exactly. So, for the top one, there's only one possible way that a that a like a a different colored L shape can be like snuck into that there.
Yeah, exactly. And it's like that. And so, you're forced to construct that same shape from earlier. Um now for the bottom one uh it's almost the same thing. Um the only issue is there's one different possibility where we can construct you know where I was trying to go with this is like you know if we're if we're forced to keep constructing those 2x3 blocks on the top then um like we can't use those to tile in uh what's it called? A three times an odd dimension block, right?
Uhhuh. So, because because six is even, so and that that has an odd number of cells. Um the only problem with this argument is, you know, on the bottom one, there's like one way that you can wedge another L shape into that. No, sorry, there's multiple ways that you can wedge an L-shape into it.
>> Yeah, there are, I believe, two ways.
Uh, yep, two ways. Cuz essentially if you create a new corner, >> you can exactly >> or >> Yeah. Or the other orientation.
>> Yes. Yes. Yes. Yes. Yes.
>> Um shucks. That might not go anywhere as easily as I thought it would.
Uh well, okay. on the left side you have to now make like like the bottom the bottom sort of like the the cranny that is left is now forced.
Um >> yeah. Okay.
>> Yeah. So that's also forced. Let's just make those all green.
>> Um yeah. I don't know. I don't know if if I can take this anywhere.
I'm I'm going to basically spoil the puzzle unless uh anyone wants to give it more time.
Um okay, I'm going to spoil the puzzle.
Um depending on Yeah. Okay. So, this is a construction of a 9x9 tile. Whoa. um here.
>> So, it's possible >> I >> Yeah, I put this um in the in the cover page.
>> I should have should have paid more attention.
>> I amum I don't know. I'm not completely surprised that nobody uh picked up on that. But, but anyhow, um I noticed this and I found this um very amusing. I was um okay backstory cuz this is I believe this is new for um the person who's not uh people who are not previously in the discord call or like people who are viewing the recording. I was on I was participating in HMMT and we had like several teams.
So we were like scheduled together. So I was with a bunch of math nerds and we were looking at this problem specifically um of you know what things can you tie with what and this is the puzzle which intrigued me cuz I I thought for a very long time that um 9x9 wasn't possible cuz okay cuz obviously 3x3 isn't possible and uh the rationale is basically exactly this um like you can't tell it at all Um but yeah um for 9 by9 it was possible and you do it like this and I'll admit that it kind of looks ugly. Um but I did manage to construct this um while we were on a bus um which which meant I I I I just um imagined this but I did end up constructing this just without pen and paper and I'm really proud of it. But yes, um you can see this is an essentially a 5 by9 and with with like a bunch of rectangles which I've colored in red which is like the ordinary bits and some blue leftovers like which is just weird.
>> Uh so you're the you're the first whoever did this.
>> I am definitely first.
>> Um I am okay. So two reasons. Um there is a lot of buzzing. Um it it's probably not just my problem, but like there's a lot of pausing in here. Anyways, so I am definitely not the first person.
Um um first, this is a relatively easy to find construction. Second, I later found out that this is um apparently a duplicate of a problem in ISL.
Um I can attempt to search it up. Um, it's an ISOL problem about um what shapes can you tile with trinino? I'm not sure if I can find it directly, but yes. Um, there was a uh puzzle uh there was like a problem in like uh something um math olympiad or something.
I'm not really a heavy olympiad person.
um about how um about which rectangles you can fully tile with um Latroinos that sorry Lromino um and the result is that you can tile because you can tile a 5x9 that allows to tile like um odd stuff that is big enough and then um for even stuff you use the 2x3 module and so that's how it works.
Okay, so this is the construction for 9 by9. Um, any updated hypothesis on what a Lu Lromino can tile uh specifically like for squares?
>> Multiube in the chat says every odd multiple of three from nine upwards >> which okay um uh any explanation in mind? I mean that is correct.
Um, well, I guess I mean you can just uh you can't just naively tile it with 6x6 blocks, but you almost can. I feel like um you can tile it with 3x6 blocks, I believe, if I'm imagining that right.
>> Yeah. Yeah, you can tile it by 3x6 blocks because you have 2x3 blocks um from which you can make a 6x3 block. Um which means that for any like for any 3x3 end you can always just extend this by six um because you can go out um here and then repeat this pattern over. So assuming that we the same construction here um we can always just >> um have this repeat and then similarly for the other half. So I'm just not going to draw that.
And that is the construction. So it's um multiples of three except three cuz um it just says nah.
Okay. And that is all the trinos. And at at this pace, we're going to be sailing past.
Okay. Um, we've gone monominos. I I'm not actually sure what these are called. Um, we've gone dominoes.
We've goneos.
Monominino seems like maybe it's anyways. We've got we've done, dominoes, dominoes. Our next up obviously is Tetrooinino. And I'm also a huge Tetris fan. Um so I'm going to look at this problem. So um what do you all think the easiest tetrome would be? Um to start with 2x two. Yeah. Um any any guesses? Uh yeah. Uh a 2x two would be able to cover um any even by even square uh because we literally just have a 2x two unit over here that we can repeat across multiple places and that gives us the construction uh which is really really easy.
Okay.
Um any oh I guess upwards uh there upwards there is one that is uh really obvious um but there are uh some other path that we can take so I don't think there's an obvious easiest one obviously all forn by 4n for the l to trino interesting do you mean that 4nx by 4n is sufficient or do you mean that it's both sufficient and necessary so um just to be clear alter trino is this and we're saying that um elteinos can tile 4x4 8x8 etc but it can't tile um let's say 6x6 or 10 x 10 uh I'm assuming that's not necessarily just sufficient yep okay okay um which tetromea okay I'm going to pick a tetromea then and I'm going to pick the one which I think is the easiest to work with I'm going to talk about the S or a Z to Tino. Um it it's funny. It's it's interesting if you try to analyze like what happens if you allow or don't allow um flipping. Um in the end I believe that none um like I know for a fact that none of the results change for troops. Um even if you consider like you don't allow flipping.
But let's take the S/Z piece and let's look at um which squares can the S sh uh S piece tile.
>> Well, you got to tuck one under the like the overhang there. Um >> yeah, >> well I mean even if if we don't consider um if we don't consider flips, then the other overhang is already unreachable.
Um >> yeah, >> but uh Yeah. So, you're you're screwed from the start. So, I mean, I guess we have to assume that we're dealing with with flips being legal. But in both of these cases, you'll just kind of run into the other corner.
>> Yeah. Like a ladder.
>> Yep. So, the result for SNSD, which is why why I said it's um one of the easiest ones, is um you'll eventually run into a corner because every piece you fill will end up just creating another um uh cornered point in which you have to dedicate another S/Z to fill. And um you'll eventually run into the corner or if you really want to, you can probably induction this, but uh I think it's obvious enough. So, uh, S/Z pieces unfortunately cannot tile any square.
Okay.
Now, let's look at the other pieces. I mean, let's track our progress here, right? I mean, okay. So, Monominos can tile all squares. So, um, multiples of one, dominoes multiples of two.
Right? So specifically I'm going to say the O monomino is multiples of one. The I doino is multiples of two. The iroo is multiples of three.
Elroino is multiples of three except three. Um oino is multiples of two.
S slash Z is um none or I mean I was going to write multiples of zero just to um keep it like in the same pattern, but I'll I guess I'll go with this.
What else do we have left? Well, we have I pieces, we have T pieces, and we have L pieces.
>> Is an I just four in a row?
>> Yeah, I is just Okay, >> I guess that's going to be pretty This is a T. This is an I. This is an L.
>> I think I should just naively be four by four by uh >> yeah, multiples of four.
>> Yep. Okay.
We can start with um any of T and L. Um we already know that four a multiple of four is going to be a sufficient condition.
Um, specifically for the Tpiece, you can always tell it like this.
>> Oh, I forgot you could like twist it.
Yeah, you could possibly have >> Yeah. Um, with the eyepieces, you can always do it like this. So, that's uh really nice. With the L pieces, you can do it like this. Um, and this is even like without flipping.
Um so with t i and l uh four by um like multiple of four is a sufficient condition because we have um these units.
Yeah. Um as I very accurately draws it with ascart.
Yes. Um the sufficient condition is the easy part. Um the harder part is the necessary. Um, at this point I feel like you can kind of just choose your favorite um tetromeino um and see see if you can um get a result.
Yeah, you you can you Yeah, I didn't mention this. I I felt like this implied but yeah, I didn't mention this explicitly.
You also roll out odd by odd because you don't have an even number of like or like divisible by four number of squares. But um that means that you have to consider the cases for 4 4k + 2 which is the only uh which is the only leftover cases.
Yeah. Um so just to explicitly mention it um we know we're we know that multiples of four work but maybe multiples of two work. Like we don't know. I mean I do but me too. I wrote a proof about it a long time ago.
Should I just tell you the answer for the T or >> H? Okay. Um I okay you can you can go ahead with the explanation.
>> Uh for the t it's only 4x4 like multiples of two like odd multiples of two doesn't work and there's some kind of proof based on induction that it's a little bit complicated. So it's too long ago for me I don't remember the proof but yeah it has to be a multiple of four for the T and for the L I don't I don't remember. Okay. Um I'm going to say it is indeed true that okay you continue never mind. Um but yes it is true that TPS requires um multiples of four but um the proof is actually quite simple and doesn't involve induction at all. So remember how um uh in our original um motivation, right? The motivation was if I had like a checkerboard pattern like this.
Um I'm going to copy that over.
If I have a checkerboard pattern um remove the two corners, then this can't be telled with dominoes because the number of black and number of whites don't match here. It's a kind of similar case with the um with with the tetromeo except it's uh it's kind of like a flipped condition.
I'll leave the checkerboard here.
There's um answers in the comments like answers in the chat um which I've already seen.
But you can think about why this means that um I guess I'll take it to 6x6. Why 6x6 cannot be tiled with T or T.
Troos.
Okay. So you have two types of T tetromeos on the checker board. uh some which are darker and some which are brighter.
Basically, some of them have three dark uh fields and the other one have three white ones. So, and yeah, for the corners, we already know I mean the the the color of the corner dictates what the color of the the terminal has to be that fits in the corner.
Um, >> right. Kind of. It's not that complicated. Um, okay. I'll I'll >> Oh, so it's like Okay, >> I'll explain. So, >> uh I know I think I got it. Uh the number of black squares right now, if you count them, it is uh 3 * 6. So, 18.
that is not a multiple of four because we the original size of the square is like uh not a multiple of four. Yep. And that is a problem because uh we need the same amount of black T's and white T's, right? So we know that a black tea has three black squares and a white tea has one black square. So a pair of a black T and a white T should have four black squares. And if we like pair all of them, then the entire number of black squares must be a multiple of four. And that's impossible.
>> Yep. I had a basically similar basically identical proof in mind, but yes. Um so basically that idea um because there's an equal number of black and white um squares and each T is either so if you color uh if you take this T like this then it covers more black squares than white and if you color it like this it takes more white squares than black but critically it's either three black and one white or three white and one black and to have the blacks and whites cancel out you need to have an equal number of black um T's and white T's basically or like um so you'd have to have so basically each black T has to pair up with a white T such that their colors can cancel out but each pair has four black squares and four white squares um whereas there's not a multiple of four squares um total or there is not a sorry there's not a multiple of four total black squares or multiple of total of white squares. So for each pair you have four black scores and that will be impossible to cover the 18 black squares.
Um my initial proof would be that um since you're going to be um using up three black squares uh or one black square every time and you're going to have like um an odd number of total uh or like okay let me just write this out.
Um my my idea is that since you have 36 total squares and then you have what what what was I going to say? Um you have 36 total squares of which 18 are black, 18 are white. So So if I take the black minus white count, it's going to be zero. But each t um ah I I see. Okay. Uh each t either contributes two more blacks or two more whites. So no matter what you do, the black minus white counter must always shift like in either direction by two.
And so after a total of nine t's, which is an odd number, it can't return back to zero because it started at zero. And I I realized that if you know that there's a total of nine T's, then you can end it right here. Um because you you you straight up cannot pair it together.
um because you have nine total T's and so there must be more black T's than white T's or there must be more white T's than black T's and that will mean that the colors won't be able to cancel out there so that's uh I guess that's a more straightforward proof than either okay so the T piece is done the Tpiece is only multiples of four now we're going to continue to the other cases of I and L >> think I have a proof for the I um I claim the same the same fact that it is only doable for multiples of four.
>> Um and I I I think this works. So the idea is you're going to do another tiling pattern. It's not going to be checkerboard this time. Um what I have is like a a larger checkerboard of um size like each each checker is like 2 by two.
Yeah, like that. Um, and then yeah, exactly. So that would be like a 6x6 grid. Now, you'll notice anywhere you place an eyepiece on this grid, it will always still take two dark squares and two light squares. So, um, however, if you count it up, this one has four more dark squares than light squares. So, there is not going to be a way that you can possibly tile the whole the whole grid.
>> Yep. Um and that is the same proof that I have in mind. So this is actually >> Yeah. So um Yep. Uh you can only tile multiples of four. Um because of actually the very s a very similar argument to the one with um the initial domino one because with the initial domino what you have is more black squares and white squares because you remove two white squares. But here you have more black squares than white squares because of the way that you tiled them or sorry because of the way that you color them. And remarkably uh no matter where you put this um eyepiece um it will always span two black squares and two white squares.
And therefore it will be impossible to tile this square with um eyepieces because you can't get the black and white difference.
And that is the result for I.
That leaves us with L or J. Pick your favorite letter.
So what do we think about the result for L or J.
So sawdust in um chat says uh I suspect there are outlier cases for the L that allows all multiples of two the possibility for wonky cases etc. Oh I also wanted to point out that the eye tiling was like uh I also got it from the video that was mentioned earlier that I I do remember that video and having watched it. So that that was a nice video.
>> Is it the polymath video?
>> Yeah, the polyth um >> he's in this he's in this discord.
>> Yeah, I I did uh remember that being mentioned.
I'm busy trying to tile one of these, but I'm not getting anywhere.
I will give a hint in like a couple of minutes.
I don't have a a checkerboard that I can work this out on. So, I'm going to be thinking about it as I'm talking about it, but it seems like actually for the L tetromeino, there's something with the way that it has to like land in the corners that that leaves on the on the edge a certain number of squares that must be covered.
And maybe, no, maybe maybe there's still wonky cases that are allowed no matter what, but Okay, I will give as a hint that um the result is indeed that um so with the space argument it's very iffy and I I don't see a universe where it works out really nicely and I'm going to follow up with why I think that like I think soon and by soon I mean a lot later um ah probably not a lot later but anyways um the result for L pieces is is actually you can't tile all multiples of four uh sorry all multiples of two um and the argument is a coloring argument so I'll give you all more time to think about it um are reflections allowed um it doesn't matter um specifically um what I'm what I'm saying is because if you look at this construction uh even if you don't allow reflections you can still tell multiples of four and I'll tell you that even if you do uh if you yeah even if you do allow reflections you can'tile um non multiple of four multiple of twos so it's like for tetromedas it doesn't really matter considering four colors on the board. I think if you tile it with four colors, depending on how how how you are thinking of tiling it, cuz I see two obvious ways of tiling it with like four colors, but depending on how how you do it, it it could work. it. You don't have to though.
Do I know if it's true in general or just for L to Trumino? I will be uh yeah, I will be pleasantly I will be very surprised if it's general. Um I I don't immediately have a counter example, but I feel like if you give me 10 minutes, I can cook up a counter example.
way. If any of you all want any hints or just for me to reveal the solution, I'll be happy to. I just want to um give you all time to think about this cuz I feel like this is findable in a way.
>> This is like definitely solvable.
I was trying to think of like coloring things based on knights moves, but I'm not sure. Yeah.
Interesting. Um I I'll get back to that idea later. Um uh just to work, what is your idea?
Color rows/c columns the same.
Interesting.
Anyways, um uh Z, do you have a specific idea for what you mean by uh tiling by knights or just like a general idea for uh I mean if I'm only thinking about one like not flipping then the thing one like L should always cover some kind of knight's move between the two ends.
And I had an idea, but then I didn't have an idea.
So, I was trying to color. Yeah. What you're Yeah, basically what you're doing right now. Something like that.
>> Okay.
I don't know if this works. Um I mean, uh this obviously doesn't work if we're considering um reflections, or at least I I feel like this shouldn't work. But let me think about what happens if we don't consider that.
Uh I don't know.
How many of these are there?
Yeah, I think um if you color like odd rows and even rows differently or columns, I think you can do it very similar to the T argument.
Yeah, I mean I know that that works, but I don't want to explore something that I know works.
Um, let me So, let me look at this.
Um, did I did I forget the color of this? I did. Okay, so I have five colors, which is interesting. Um, each L will tile two of the same color and ah, this is weird. So, I can tile two of the same color and two other colors, but they can't be like blue and green and they can't be like white and red. That's weird.
I feel like there there's a possibility in which you can like weirdly group these um to have an argument that works.
But anyways, returning to at least um the other idea, you can indeed color the rows alternatingly and you get this. Um this is also the solution that I had in mind whereby wherever you put an L piece it will either be three white one black or it must be three black one white. So for instance if I do it like this you get that. If I do it like this right even if I consider flipping you get three black one white and this thing works. Um I am curious about uh pitchback's um other thing which works or if it's just >> I have one I think I've got one that works using a four coloring um >> um >> where where each one each one each each L shape has none of one of them one of two of them and two of the last one. Um, and then because you got an even number of exactly two and you need to use exactly nine L shapes, you're left with two out of the four with an odd number covered at the end, which doesn't work cuz you got an even number of all of them, I think.
>> Okay. So, yeah, uh, this indeed also works because, um, if you really think about it, it's just coloring like it's just, uh, increasing more separation within the um, which will end up working as well. Oh yeah, I do the same thing.
Yeah, >> cool.
>> So if you do the coloring like as multitude says color according to n mod m what the fish m2 m2 then it does also work um specifically because even if you just take mu of one of the rows/c columns it works. So L's are also multiples of four um exclusively.
This is where we tread into penttoinos, also known as I want to die.
Um because I Okay, last week when I was trying to do this, I started looking I was like, "Ah, yeah, I'll I'll go into this blind." And then I I did a little bit of a run through with a friend. I was like, "Okay, I I feel like pentomos should be easier." I realized they weren't. Um did some research. uh they weren't but we'll start with the simple ones and uh specifically the ones which we know how to solve.
Um um I I see typing in chat but I'll I guess I'll address it when it's out. Um okay so first how do we how do we list all of the the how how do we list all of the pentas? Um we start with the tetromeos. So, I guess that's an O, that's an S. Um, that's a T, and that's an I, and that's an L. And we try to add stuff onto them. And so, with an O, wherever we add, we get a P. Um, with an S, wherever we add, we uh we either get a P or we get an N or we get an F. I still find it very amusing that they had to have names for all of these.
Um, here you can potentially get a T, potentially get a Y, um, potentially get an F, potentially get an X, right? Um, this you can potentially get an get a you can potentially get a Y, potentially get an L, potentially get an I.
Uh, I'll put it like this.
the L, you can potentially get uh you can potentially get a T, potentially get an F, potentially get a Z, potentially get an L, potentially get a U, um potentially get a P, um potentially get a V as well, um and potentially get an N, potentially get a Y. Okay, so these are all of the 12 penttoas, which again I still um find it very fascinating that they they have letter names for all of them.
>> Please elongate I by one.
>> Oh. Oh gosh.
Ah.
Well, okay. We're going to start with the easiest one, which this time is I.
Cuz >> is that the easiest?
Um I don't think that's the easiest.
>> Uh it's one of the easiest.
>> Yeah, it's it's one of the easiest.
>> Let's let's >> I guess it's Yeah, I guess it is just 5N. I wanted to just discard like there's like four of them here that I think we can just discard easily.
>> Yeah, I was going to I was going to get that. Um let me uh because I really want a way to organize these. So let's organize these by alphabetic. Uh let's not Okay. So here what do we have? We have starting with the F pentomino M pentomino T pentomino X pentomino.
Um where is my W?
Uh oh gosh. Where? Okay. Yeah, I just forgot about W. Oh, here we go.
Um, P N P N Y Z. You can see that they really like the later letters. Um, U, L, um, F, T, X, uh, V, W, and I.
I was gonna I'm I'm just gonna Yeah. I is multiples of five because if it is a multiple of five, then you don't have a multiple of five amount of squares.
That's clearly impossible. But if you do have a multiple of five, then here's a construction. So that's a unit.
Okay.
And we're going to go with the discarding. So >> all right, I choose to discard the X.
>> Yep. Uh we're going to >> you can never fit it into a corner.
>> Yeah. Okay. So um the easiest most methodical way is like to fit this into a corner you can consider the bounding box and the bounding box has to like one of the corners has to be uh on the corner of the bounding box and x has none of these. Um so uh x is discarded.
So not um what else can we discard?
you discard F. Okay. Uh F. Um specifically, the only thing that's capable of being a corner is this. And if we attempt to chuck it into a corner, you'll find Oh my god, I forgot that that doesn't work. If you attempt to chuck it inside a corner, you'll find that you can't tiling uh you can't tile these remaining spaces.
So F does not work. Um fails and yeah um a lot of these just fail because of corners for you.
>> Yeah.
>> Sorry.
>> Okay. You can >> I was going to say the T I think the T fails cuz uh once again there's only two that you can fit into the corner. And in both of those cases there's like you know length two block.
So I'm just going to speedrun through these cuz this is not uh this is not really fun. Um, for the end piece, um, if you consider putting it on a corner, let me take a smaller stroke with. If you consider putting it on the corner, you'll have stuff that's left that you can only tile with another piece of the same thing, and you'll just get into repetition and it eventually fails. Same thing if the corner is like the other side. So, N also fails by N.
N is none. Um, same thing for the U. If you Okay, let's start with T. Same thing for the T. The only thing that can be the only thing that can go on a corner are these, which if they go on a corner, that means you have these two and that fails. So, that's impossible. So, I'm just going to mark here T is none before I forget about it.
Um, we're also going to look at we're also going to look at Z. same reason. Um corner um two remaining that you can't tile at all.
Uh another another one is going to be uh we can also look at U. Um if you put U on a corner then you have this space which you can only tile with this and then you have this corner which is impossible to tile. So U is impossible to tile. So that's T, Z, and U, right? W is impossible to tile. Um because uh one of the corners must be this.
And so you have this, which is only possible to tile with this, which is only possible to tile with another repetition. And you have to go on and on and up.
Okay, so these are all impossible.
So I'm going to write here T is none, U is none, Z is none, W is none. Um, great.
Which leaves us with these, which leaves us with these three. Uh, these, sorry, these four.
>> Um, >> what else do >> I I think I'm assuming that one's the L.
The one that Well, wait, there's two things that look like an L here. Um the the that one. Yeah, that one. I think I'm convinced that that one's not possible. Um because there's so there's two different ways that you can uh fill in the sort of inner crevice in inside of the like the like the center piece in the bounding box.
>> Yeah. There's like two there's two ways that you can reach that spot >> if it's in the corner. Um right. And it has to be, you know, it has to be in the corner with the with the corner piece in the corner because if it's not, then you'll have Yeah. Yeah. Exactly. Um >> so that's one that's one option. There's a second option which is to like >> Yeah, exactly. Um and I >> there are more scenarios though. Um this can go down the corner, but >> Right. That's what I was That's what I was saying with like you know if it's like that then there's no way to reach that that inner corner there. So like that's that's not possible. Um >> Yep.
>> So this is just >> case work. Um basically if it's like this then this has to be reached with this and this is impossible to reach.
>> Yep. Um, if you go it like this and you take this path cuz you have to reach this, >> you either reach like this in which case I believe you can only reach that with that and then you can only reach uh >> Yeah, exactly. You can only reach that >> with either this but that's impossible.
So you go down and then this one also goes here and this one is also forced to go here and that's just hill and on the uh on the final case to the right um if it looks like this with this being the corner then um where do we go?
Um, this has to be filled with either this going to the left, which is impossible, or going to the right, but then this is impossible. So, this is impossible. And basically V is impossible.
Basically V sucks.
Um, now what?
Okay, hold on a moment. I'm going to deal with some um You all can hear me, right? Uh my internet disconnected and I'm going to have to deal with some stuff.
>> Yep, we can hear you.
Hey, I am back.
I am I'm back from fixing some internet issues.
Well, one fact that I've discovered at least is that the top two that you have there can definitely tile um odd multi or even multiples of five like so multiples of 10.
Yep. Okay. So, uh we have here some um easy results. Um, if you have this, then you can take that. And if you have this, then you can take that. And this allows you Oh, and this allows you to have a 2x5, which you can make uh which you can eventually make into a 10 x 10 by just having 10 copies. And then you can make that into a 10k by 10k.
So, we're going to write um multiples of 10 multiples of five maybe and then same thing with what do we think about y okay I am going to basically I am going to end the thinking here. Um I mean I I'm going to take any conjectures but I won't give um too much more time to think because there isn't much easy progress here.
Like I don't believe there's any progress you can make within like a reasonable amount of time. So I I'll be taking any hypothesis that anyone may may have regarding these and then uh afterwards I'm going to show you the results which I've discovered online via searching.
But yes, I'm suspecting that they might be tileable for sufficiently large even odd multiples of five. Um, I'm not sure which ones though, but maybe all of them.
>> Uh, I'm I'm I'm going to be taking I I'll I'll take any hypothesis even though I don't see any per se. Or uh do I hold? Okay. Uh my VPN disconnected.
Untilable. I guess Y has a corner problem. PNL are multiples of 10 because of normal checkerboard coloring. um PNL are multiples of 10 because of normal checkerboard coloring is not true. Um because originally what we had is um uh what what did we have?
Um yeah. Okay. So originally what we had was if there are multiples of two but not four. Let me demonstrate. If there if it's a multiple of two but not four and then you decide to do a checkerboard pattern.
So like this, you have the same number of black and white squares. And so if one of them covers like more black squares and one of them covers more white squares, it fails. Um the thing which doesn't hold on true is that if you try to extend this to 5x5, you begin with a different number of white and black cells which throws you off.
um which means that the result does not uh cannot hold.
This is where um it gets interesting. So I have actually cached these results because I knew I was going to get to them. They are in fact all tileable um ridiculously. Um they can tile all five multiples of five except five.
Holy Wait, what?
Wait, that was >> I think I think I might be a genius. I think I might be a genius.
>> Wait, that was so quick.
>> Holy.
>> Where's this?
>> Well, I mean, I I guess it's a 10 x 10.
So, it's not as hard as a 15 by 15, but but still like like my first thought was that that was a 15 by 10, but still that was >> my hypothesis was right.
>> Yeah. Um, wow.
>> Oh, that's cool.
>> Yeah. Um, 15. That's a 15 by 16 tiling, is it not?
Um, I'm looking at the pentomino by multi- tube. Um, >> oh, it is.
>> This looks too that looks too too too simple.
Is okay. Those tilings are crazy looking.
>> These tilings are crazy. Okay. Okay.
Let's look at this. Is arguably the most sane one. This is the tiling for Okay.
So, because we have a 2x5, that means that we'll be able to get a 5x10. Uh because 10's a multiple of two. So, that means we'll be able to extend these by 10 at a time. So, that that's why we know that this can extend. And here it's similar except uh this just a straight up construction for 5x10. We'll get to that later.
Um, for for for P um um it's it's uh it's straightforward to check that you can't have a 5x5 tiled for each of these pieces. Um and um what you do is you create this this insane shape um which works. And then you pat it around with a bunch of 2x5s to create a 15x5 and you could extend it. I feel like this is like maybe you can discover it within the time frame.
Um, and this is L. Um, same thing but with a bunch more pieces like with a bunch more patternless pieces.
I also have to say I I'm not necessarily certain that these are the most elegant ones. I'm pretty sure they are, but I don't know. Um, these are re uh results of searching online. Um this is uh 15 by 15 um with construction by L. I remember I did ask. Okay. So because I I was typing like um which squares can you tile with? Okay. Um can you tile a 15x 15 square with L um penttoinos? And you know how Google has this like automatically automatic thing? I I could have sworn it told me on something about three color.
Yeah. And it's just like something um about um just random colorings.
So >> Well, there it is again. It just lied to you, didn't it? If you scroll up.
>> Yeah. Well, yeah, but like exactly this.
It was like But it was like specifically in like it specifically gave me a reason why. So, I was trying to understand it, but I realized that that was >> Um, but yeah, I I I would be really surprised if I can find it because I remember that it took me some decent amount of um yeah, leomen tilings. I believe this is this would be correct.
But yes, um yeah, that's that's the side which I will show. But yes, now on to the crazy stuff. There's a 5x10 tiling when using white pieces. Um, and oh, by the way, all of these use flips. So, and I'm pretty sure the rotating wouldn't be possible. I haven't explored it. I'm probably going to explore it like right after I finish talking. Um, >> well, actually, well, I don't know if you have time. I mean, I I think I have a way that you could disprove the P's working. Um, >> no, I mean, right after I finish this section of >> Okay. Yeah, >> go ahead. I have I have not looked into it at all actually which I feel ashamed.
Um anyways this is a 5x10 tiling with Y pieces and so you can use that to make a 10x10 Y. That tiling is gorgeous. Um yeah I feel like this is also uh I actually just read this is also like a symmetric kind of construction which makes it extra nice. Um, and you have this monstrosity. Um, I don't I don't I don't even want to what?
Like I was I was like doing this. I was like, "Okay, I can maybe like color this together." Uh, but then after a while, I just give up. I can certainly still try.
Let's see. These pair up together nicely.
Um, these pair up together nicely.
These pair up together nicely. But it's it's it's just that this is so much like just more um non-standard than like all of the other ones.
It's like for for all of these I could get um these relatively saint tileings.
This is just a 15x5 monstrosity. Um and again by the same argument because I have this 5x10 I can extend this to make like um basically bigger and better monstrosities. Um, okay. And, um, all of these are thanks to polyomin tilings. Um, as you can see, I looked into a lot of, um, this actually shows you what you can tile with what pieces. So, if you look at like if you're just interested in a random, I don't know, uh, monomino plus hypomino cuz cuz why not? Um, then you can show all of the information that other people has. Um so like like this um it's like um reptiles etc. Uh and so for instance if you wanted to look at the result with um that LPS piece then you can show you can get and you can get this and a a lot of stuff. Okay. So I I will be sending this in case anyone wants to um look at this.
So wait, when you select two things there, what is it doing? Are you talking about like it's asking about what's possible >> if you have either of them available?
>> I think it's both like uh so you have this.
>> I see. Gotcha.
>> Um so so yeah, that's smallest rectangle, small square. Um, I think it also assumes that you have to use both of them because otherwise that'd be 4x4.
>> What the heck is that 70o that they have listed?
>> Oh. Um, >> okay. Um, this is a specific tile set with the 70o, 66, and 20. I think apparently >> that's just um >> why >> it's just apparently exists. I don't I don't exactly know how it works. Um but yes um they can uh they can apparently tile something and I believe that it's probably there because it's used in the construction of something but I'm not really sure but yeah um uh this is just a fantastic resource. Anyways, getting back to the original um because I remember somebody said something about um disproving um 4k plus two when you have like L's um like L tetromeas. I feel like this makes me lean into thinking that that proof approach doesn't quite work. So that that's that's specifically why.
Okay. Um we're okay. So I'll first um everything that I've prepared is done.
So I'll I'll I'll start exploring random that I don't know. Uh and I'll start with um what if you don't allow reflections.
Uh in which case I think you get into corner problems.
Okay. Yeah, you do get into corner problems because um if you look at one of the corners which is like this uh does it >> I don't see a corner problem there.
>> I don't see I don't see one either. I I Okay. Yeah. Uh what else?
What if I think about the ways it tiles?
Okay. You know what? Before I Okay, I do I want to even try this problem cuz I feel like there just exists a super stupid tiling.
>> Whether you do or don't, I'm going to write a solver after this to try and find them.
>> I mean, I'm I am aware that other people have wrote solvers, but >> that's not the same cuz it's not like result matter. Solvers are fun because you can optimize them and I don't No, I'm kidding, but I I want to try anyway.
>> Yeah, I mean, sure. Go ahead.
>> Yeah, like that. That's not immediately looking unfeasible.
>> Just going to try random stuff.
That seems like you're in a Wait, that looks like you're kind of in a bind, though.
You can never fill that bottom left corner now.
>> Yeah.
O that's >> Wait, but you can still you can still make the 5 by 5 by two block >> in case that wasn't obvious. I mean, that was probably obvious.
>> You mean like here?
>> All right. Yeah, I was >> um Yeah. Anyway, uh that what we're trying to do is um do this but without flipping the pieces or like without reflection like only rotation.
>> Uh what? Oh. Oh, the the thing above.
Right. Right. Right. Sorry.
>> Yeah. So, I was trying to see if cuz you can still get the 2x5. So, you can still get 10k 10 by like you can still get multiples of 10 without uh with uh without reflection. But can you get 15 by 15? Um, and I don't know if I want to try.
Okay. How feasible is this? If I go into like what's the probability of me frauding myself?
Um, all right. I'll try more manual and see if I discover anything, but I strongly think I won't.
It's just like it's so absurd. Why would this even work? But also like why would this not I mean because I would be really surprised if there was just like an actual legit proof. I'm just more so like cuz like there might be some tiling which can prove that it's impossible but at the same time that feels really weird like it's just not something that I feel like should exist.
Oh, is this just uh that's just unfillable.
Uh, well, I guess I can What can I do about this?
>> I mean, if we just do a checkerboard, then a P should always cover like three blacks and two whites or three whites and two blacks. Right.
>> Right. Yeah. But you start with more blacks than whites. Like, supposing you have a 15 by 15. I mean, keep in mind whatever proof you do here has to make use of the fact that it already worked on the on the case where like reflections are legal.
>> So like it has to differentiate between reflection and non-reflection.
>> Yeah.
Ah that's actually this is really hard in one direction.
This like much harder in one direction than it is in like in the other.
>> How so? Like I like why not just fill in like uh I'm not sure what what the problem is. Like try and you see here just put your mouse somewhere and I'll like direct you to where I think you might want to put something. Okay.
>> Actually go right. Yeah. Go right a little bit. Maybe like four four cells.
Yeah. that up to the the thing. Yeah. Up one and two to the right. Like I think you can fill in a nice box there. Yeah, that one. And then the cell one to the right of it. No, no, no, no, no. Not not up there.
>> I think if you do that then >> I But like I'm concerned with this part because right now this part is impossible.
>> Trust me. Trust me, trust me. Just uh just fill in the next one that's now forced and then that one I'm pretty sure will fall into place if I'm doing this right in my head. So now flip the Yeah.
Put one more in there. But it's like a P facing rightward. Oh, wait. Shoot. We don't have reflection. Scrap. Never mind.
>> We don't have reflection. And that's like I think >> Yep.
>> Just straight up screwed.
>> I've been doing this in my head as though we did.
>> I feel like the night checkerboard might help this time.
>> Yeah, I'm trying it right now because the night trick, if you remember last time, it gave us exactly five colors, right? I feel like there's something there.
also feel like that's possible.
Interesting cuz these five by like these um that pattern repeats every like five.
The kind of annoying thing here though is that like Oh yeah. Yeah. Cuz you do have two pieces that are separated by a knight's move in this case. However, because like the knight's grid is like is like chai roll and so is the piece.
You might just have to choose the right direction of knight tiling and then it works.
>> Yeah. Huh.
>> That seems >> um I don't know. I don't I don't like I don't see an obvious way of it working, but I don't know.
I'm going to try it.
>> It's weird. I If we uh like can I share this in chat?
>> Okay.
>> If we do this uh >> um with with my current orientation of P, I'm assuming.
>> Yes. With your current orientation like the Q shape, then the Q covers every color exactly once.
>> Yeah. I Yeah, I see.
And I feel like there's something there.
Just an idea.
>> Yeah.
>> But immediately. Yeah. Just >> look. It's okay. Do I want to Do I want to potentially f myself? What would be the best way to cut this?
I would I've started the corner.
Did do I want to detect?
Uh yeah, I don't know what the easiest way to do it here is cuz also I've never coded something remotely like that.
Surely I have to try though.
I mean, uh, in case anyone doesn't know, you you can you're you're free to leave at any time. I'm going to attempt to code this and I will I will regret this.
>> Nah, I shall >> I am having flashbacks.
>> Um, I will call this unreing.
Um, in case you haven't figured out, my naming conventions are out of the window. bad.
Okay.
N equals 15.
So, we're um n equals 15. So, we're dealing with a Oh my gosh, I just wanted to type this cuz I've been using Discord way too much.
Um, m M byn.
Surely I can just hard code this.
Uh, yeah, I'll hard code this. Um, you know, you can do 7 * 15, right?
>> What do you mean?
>> With this shape, you can do like a rectangle of 7 * 15.
Wait, how >> uh looking at it in some uh Wait, I'll try to take a screenshot some.
>> Okay, do do try that. I'm going to continue with this uh programming approach because I that's surely going to be beneficial at some point.
That's >> just looking at some old uh paper that I remembered. Some high school students wrote this to complete high school in Switzerland in German. Some Yeah.
>> Uh interesting new countries to write fascinating um high school papers on orientations. We're going to have how do I want to encode this? Do I really want to use a bit map? Can I not use a bit map? I I feel like I have to use a bit map to get this to run like even sufficiently reasonably fast.
>> Uh you're trying to have a 15 by 15, right?
>> Yeah.
Uh yeah, >> I don't >> Yeah, I don't I don't know how to find >> I don't know how crazy you want to get about this, but um there's a way that you can do um like you can use SIMD to like do uh like a pack like a four uh 64-bit integers um and like make a 16x6 box.
Um >> but that's kind of crazy. I'm well I mean I'm in Python so I'm not worrying about like over um oh 7 by 15. Yeah but that has flipping.
>> Oh okay >> wait no that doesn't Oh does it have >> the two the two bottom >> Oh yeah you're right.
>> So that's that's not what I want.
37. That's not what I want.
Great. Got all of this perfectly readable readable code.
>> What are you writing?
>> Um, these are the orientations for the piece.
Uh, I guess here >> what's your notation?
>> Uh, these are just the bit maps by rows.
>> What do you mean by that?
Um specifically these are the one one and then uh one one zero one one um one >> and that that would be um zero 011 >> I I think I see problem >> well I guess if you're in Python you could just do a 16 by6 bit board um it's not going to be fast but it would >> at least work.
>> No, I'm like trying to think about how to how do I want to encode this? Ah, I think I'll just hardcode this somewhere else.
>> Oh, I know. I'm telling you. I'm telling you how to encode it.
>> I mean, it would work. I It wouldn't be It wouldn't be >> If two swap is just saying that you you don't like pre-calculate those numbers here. You just do a bit board and then every iteration you just like do put one there, go right by one, put one, go right by one, put one that to make three blocks and then like just >> Oh, so below that.
>> So if I have like something like this, um I'll I'll be like pasting this and duplicating it on multiple times to see like which places remain empty.
Uh >> it's like a recursive function. In each step, the function is going to put one block in the uh in the leftmost corner of the I mean if you look at all the unfilled blocks, the recursive function is going to look at the topmost a top leftmost unfilled block >> and try four different iterations of the Q block. And for each iteration, it's going to put one block there. And then it's going to recursive call back on itself and then put the next block. And then if it everything ends, then it's going to try the next rotation, the next rotation, and so on four times.
>> Okay. I Right. Okay. So I want to put it somewhere in my recursive function.
Good. Uh so not here. And yeah, so the representation that I'm suggesting is instead of doing that, you just have like if you just have each like bit like this is a 16x6 grid that you're working in or a 15 x 15 because it's Python, we don't care about the integer bounds. Um, you can just call the like you can call a Pshape um like in binary 1 one then 12 zeros then one one and then zeros until the rest of the 15x5 board is filled.
>> Yeah. Yeah. I'm just worrying about Yeah. because I was like doing this orientation because I know that if I do that orientation, I can calculate to get that bit map and hopefully get it to be betterish.
>> Wait, what do you mean?
>> But I was I was uh I was wondering about how I would u make sure that PS don't overflow in which case I think I'll just add a border around. Well, I mean you can but I mean if you do the well if you do the bitboard approach then you can just check that that the number is smaller than than uh 2 to the 15 by 15 in order to check if it's in bounds >> and that >> I'm just worried that if I put something here it's going to allow me to put it and then that's >> No no no it wouldn't it wouldn't it wouldn't cuz you like if if if the thing Uh, oh yeah, I do see what you mean. Um, but what you can Well, you can just put it in an envelope of 16 by 16 and then um >> Yeah. Yeah. So, like, yeah, that does work. So, I I do see that now. So, I'm going to uh code that, which >> and then if you really want to get fancy, you can transfer it over to like some C and SIMD approach.
>> Yeah, not doing that. Well, I mean, but but at least with this approach, it it like easily extends is what I mean.
>> Yeah. Okay.
>> So, um I'll need my grid. My grid will be ah whatever. I'll just use a for loop. I Yeah, I think for loop is probably better best for this.
Is this a thing? Surely. Um, >> it is in in C.
>> I don't know about Python.
>> I mean, it's not giving me errors, so I'm assuming it is. I I have like seen this like I think once, so I like this might be a thing. Um, I want I don't care about efficiency. It's probably going to get optimized anyways.
Is it 2 to the 15 minus one? Right. So, that's going to give me 15 once and I bit shift um 16 every time.
Should that just um deal with all of my issues?
It seems so.
I'm still really worried about messed up things that can happen, but I think this works. Wait, what? What's What's What What are you What's your question? Grid shift.
>> I'm >> What are you doing there?
>> I am um making a uh this a grid.
>> Yeah.
>> Um out of a 16x6 area. So >> Oh, you're just clearing out all of the You're just clearing out the um the 15 by 15 top left corner. Is that what you're doing?
>> Yeah, I guess it's the bottom right. Uh depending on how you you want to represent it.
>> Sure. Okay.
But wait, 2 to the 15 minus one. I'm not sure what you're that that doesn't work, right?
>> Unless I'm misunderstanding. I'm pretty sure you're you're like that's >> I'm I'm uh I'm I'm creating a row of like 15 cells. Um and then just >> Okay. Yeah. Yeah. Yeah. Yeah. Yeah. Uh, grid plus equals does is that right?
>> Uh, one way to test out. Um, >> yeah, >> cuz I mean I'm going to need this anyways unless I somehow manage to not write a bug, which >> h yep grid. Um, I * I + J mod 2. Um, okay.
And I'll print that. No, I'll print that.
Um, and hopefully this works. Okay.
Okay. I cannot fit int into an in index sized integer. Oh, is it is it that?
Whoa. Nice. Is this This is what I want.
This is what I want. Um, obviously that's bad mistake. I also wanted it to go from left to right reading them. Uh uh I guess least significant digit being top left is fine to work with. Yeah, I I'll just Okay.
Right. So I have this. Now what I want to do is I probably want to start with the least significant bit.
Um I'll start with the least significant bit. Um, what's uh I've forgotten cuz I don't write bit bit maps a lot. Uh, what's the best way to get like least significant bit?
>> Um, >> the least significant bit bids. Uh, >> end with one and then check if it's one or not zero.
something that gives you the Yeah, that gives you it. But yeah, not the position.
>> Yeah. No, not you want the position of it, I think.
>> Yeah, >> I don't need position, but um uh is this what works?
>> Well, you could just you could just shift it left by one and then check where there's like >> uh also it's not tilda. It's supposed to minus I think x and minus x >> probably. I I was I was going to check it. Um, you know, probably just make >> Wait, what is that doing?
>> Uh, least significant digit of the like um uh the last one that appears >> is that is that but how does that work?
X and minus X like there's some two tw's complement stuff going on.
>> Yeah.
>> Yep. Minus X is tilda X + one. So, >> normally if you add and tilda x, everything's going to be canceled out, but you added one.
>> Yep.
>> I was going to say like left shift by one and then like look at the intersection between them. But yeah, that's that's even that's even cleaner.
>> Like I remember something like this existed. So, so just yeah, I should check for that. um SD. Now what I need to do, I need to check if all of these are within um pattern.
Um it's just um whether the pattern is within the grid. That work? That works.
Okay.
Um, oh, uh, I I may as well check what's the behavior with this. Good. Um, possible grid.
Um, okay. I'm going to have to make a cache anyways, but let's deal with uh I may as well deal with that now.
Uh, this is going to be a boolean value.
Do I want this to be a boolean value?
>> Do I really want this to be a boolean value?
>> Sorry, why are you doing the cache?
>> Um uh doing the cache just so it might get um slightly faster with duplicate grids. Uh >> no, but you you'll never hit duplicate grids.
>> What?
>> You'll necessarily not hit a duplicate grid. each like search is unique because you'll >> uh yeah if you're going for the method I said earlier like four rotations trying every placement then you'll never get a duplicate you need to go a dynamic programming approach completely different approach to ever get a cash something >> really I was thinking like maybe like there's some setup in which after you place a certain amount of um pieces the like the space you've covered becomes the same.
>> Well, I mean, no, that couldn't happen.
I mean, you do still have to, if you really wanted to, you could concern yourself with like like symmetry according to like D4, like the group D4 like uh >> I mean, I'm going to start left. So, I I don't think D4 is occurring like anytime, >> right? Exactly. Yeah. Exactly. So, >> I feel like surely there must be like some sort of case where like >> No, like you're you're you're filling you're filling each corner uniquely like >> Well, I mean, but like each corner has multiple choices and sometimes I might run into duplicates. Like, I'm not sure if it will happen. I mean, >> I don't have a proof for you, but that doesn't seem possible. Like, um, >> okay, I will then I'll then I'll do this later. like like you could imagine checking equality by tracing back along the the um uh um from fun tools and yeah probably a good idea. I always just write a cache to my myself.
>> Okay, here's a here's a here's a quick proof that that wouldn't be an issue. So suppose you have two grids that are that are the same and like they're you know not necessarily full but like they're the same grid. Um which means you reach them both via different paths through recursion but they look the same. So you can sort of imagine tracing the recursive algorithm through both of them. They both started in the top left corner. Then they would have had to choose the next possible thing. Um >> Mhm.
Uh I'm not sure exactly how to phrase.
>> I mean I I have a construction for uh how an an instance of how it might occur. It's just not very probable. So it's like okay.
>> No no no no. But it's not possible. It's not possible cuz like >> surely if I if I have like if I have like this then I can get the same thing just like this rotated 90°.
Like surely there's nothing preventing me from getting both of these.
What do you mean by get?
>> Like I like my recursive algorithm can decide to place this here.
>> Oh, you mean the entire envelope as opposed to the the actual tiling pattern. Yes. Sorry. I thought you were saying the Yeah. Okay. In that in that case, yes, I agree with you.
>> Yeah. Yeah. So, I was like, I'm obviously like going to be caching the >> Yeah. Sorry. Yes, that that that totally makes sense. Uh never mind. But I was wondering about what the whether but I still think it's it's an interesting thing. Does it uh is it even worth it to consider?
>> Yeah, I mean I mean it's definitely possible that you just run this and it just immediately finds something. So I would maybe you know save it for later.
>> Okay. Well, we'll do that later then.
L equals grid. Um L, sorry, L equals L of grid.
Um that's probably going to be LSB, but whatever. Um, L is zero. Return true. Um, placement given some X. I'll probably rename this to X. Um, what are the possible placements? Um, >> wait. Sorry if I'm being slow here, but what does your LSD do again? It's the It gives you >> won't this always return one?
>> Um no. So like um if you give it like two and negative two then >> Oh, it doesn't give you the index. It gives you the It just gives you the mask highlighted. Okay. Sorry. Never mind.
I'm I'm I'm back with you. Never mind.
>> Placement return. Um right. I guess this could just be x times um one one.
It's just going to be a bunch of zeros.
Oh my god.
Um Oh my gosh.
Now that that's 15.
So, if I were to look at it um in my current orientation, I mean, I don't even care about the orientation because it's like once it goes like the it's not going to matter anyways. One, one, that's a one. That's a one. One, right?
One, one. It's going to be one one. And then down like a full row, but that's >> missing a zero.
>> Like missing a zero where? like >> um before the in the middle like with all the other zeros >> like you have >> 15 >> 15 things in the same row and there's like three ones that >> Oh, >> right cuz I have six, >> right? Well, no, but but Well, yeah, you would need another zero there and then another zero after the I think you need two more zeros, right?
I am okay. How would I I'm trying to like So this would be I'm taking one and then one holy.
>> Yeah.
>> And then I'm I'm I'm gapping I'm uh I'm completing this 16, right?
>> So this would >> So that would all be zeros >> and then you have three ones.
>> Oh, you're doing it in the top right. I see.
Okay. And um similarly I'll be having one one complete the entire 16 and then one one.
>> Okay. Yeah. Yeah. I think you're I think that works.
And then I'll be having um that in which case I'll be having um one one complete the 16 and then one one and complete the 16 and then gap one and then I'll have one one gap and then finish the finish the row kind of and then one one and then one one. So these are the placements. Um and multiply by X.
Um yeah, whatever.
This is so not going to be the best way to do it, but whatever.
Which rectangles are tolerable with Pentto with? Oh okay. Um to answer Rstone gamers um question in the chat were yeah which yeah which technical rectangles are tellable with peep pentominals without reflection. Um the result for peep dominals with reflection is um all multiples of five except 50 uh all multiples of five except five itself. And for um without reflection, we know that um multiples of 10 are possible. Um obviously non multiples of five are impossible. So we're just looking at the other cases. Um five is obviously impossible. So we're looking at 15 and we're trying to make a program.
If x is equal to zero, return true.
Otherwise for P in placement I guess I'll call this placements if contains what did I call it? Oh within grid um pattern being P if possible. Um, grid minus P return true.
Um, return false.
Okay, what am I missing here?
Potentially um for a P within placements X if within group P if possible.
If it's within and I can take that away and it's possible then I can have it. Um otherwise return false.
Also realize that's such a stupid.
We love constants that never appear using I I'm seeing a block of text by multi 2, but I'm assuming it doesn't go anywhere.
I'm I mean I'm unsure. Um this is okay.
Grid. Um print possible grid. If this runs within 5 seconds, I'll be pleasantly surprised.
What? Okay. Um, uh, print grid.
Oh, that's a lot of cases. Holy Um, so, oh, okay.
I'm going to cut it off after some Oh.
Ah.
I'm I'm going to do this cuz I don't care.
>> I think I got a code working.
I think I could help.
Um, you can, yeah, you can just do yours.
I mean, I my code returned false, but I'm just not sure if it's just um bad.
>> Yeah, I would be surprised if it finished that entire like um look at look at my absolutely befuddling code. I just don't want to deal with What do you mean? Oh, of course.
I just don't want to deal with global variables by subtly introducing them.
>> Okay, so it's trying to make these. I mean, obviously, preferably I would want to take care of these right away, but that feels like hard to construct, like hard to prove.
Okay. Um, you said you had you had code or like you have what?
>> Need to debug.
>> Okay.
>> Okay. I'm Look, I'm going to I'm going to throw away this spaghetti.
I'm going to see if what's a better way to see if this code is working.
>> By working, do you mean like exploring the entire recursive tree?
>> Yeah, >> I don't believe for a second that that's completing the entire >> Yeah, but like how do I how do I even begin to debunk this? Um >> because well I would I mean maybe just get some statistics on like >> um I've reforgotten how how how the library works. is not that.
>> What's that?
>> Um um thing which visualizes how many time how many times each line runs basically.
>> Oh, that's cool.
>> I'm confirming that.
>> Are there any squares? Are there any squares we do know you can do with a with this five uh what's it called?
Pental without rotations.
>> Multiples of 10. Um, >> right.
>> I'm also >> gonna be going to chat between very soon.
>> Um, for just visualization stuff.
>> Are there any odd are there any odd sizes we know you can do or is that does it seem none?
>> What >> are there any odd size squares that we know?
>> That's what we're trying to figure out.
>> No.
>> Yeah, we don't know any. And it looks like probably not 15. Five is obviously impossible.
This is >> this Nothing about this seems extremely suspicious.
>> Wow, that's Wait, is it saying that it got called 2 million times? It's faster than I thought.
>> Yeah, that's the number of times it got called.
You said five by five is not possible.
Um how what what why was that?
>> Um u because I um checked possibilities.
It's just not possible.
>> Well, it's already known to not be possible even when you include symmetries. So >> yeah.
>> Right. But that's just is that just by there's not like a nice proof of that.
It's just people have checked and it's not.
>> Yeah. Seems like it Yeah.
Okay, I'm going to tell GBD to allow me to >> Did you print out your pieces themselves?
>> Yeah, I mean I I printed out the resulting grid, which is basically that I can do that again. Oh my god.
>> What is What do you mean by resulting grid?
>> Uh like because I'm taking away pieces from the grid. So I'm just like the the grid like the new grid without the pieces. But that's basically the basically the same thing.
>> Wait, sorry. Is that what you're doing right now? What?
>> Uh, no. But what I meant is like I want to see the I want to see the actual definitions of the pieces and see that it matches the shape that we expect.
>> Oh. Um, these are there. You can >> Yeah, but I mean like can we print them and look at the shape?
>> Um, let's see. Okay, I guess.
>> Um, that seems like an easy place for a bug to hide. So I mean one one way that we could easily debug this is you know increase the size of that by four by just adding the the flips and we should see true.
>> Oh okay that makes sense.
>> Yeah but I also I also am very dubious of my own ability at judging whether these are correctly written or not. So, you know, I would just be flipping all of these that I I'm doing rotation by 180°. Um, >> yeah, that's why that's why I think you should print them.
>> No, but no, I was like um when I was generating the first uh you know what um let's let's do that first.
Um placements. Yeah, placements.
Um that that that Okay. I mean clearly I mean that's not a bug necessarily, but I missed it. Did it work? I mean, I guess so.
>> I'm assuming this is working because in reality, this will be filtered as not able to be placed.
>> Yeah. Maybe just make that like eight times or I don't know, it's like two to the >> two to the 17th or something. I don't know.
>> That that that which works. Yep.
>> Yeah. Yes.
And um right um actually yeah good suggestion about testing that how do I want to how do I want to do this one one one a bunch of stuff and then one one so I can instead make this one one one a bunch of stuff and then 011 a bunch of stuff and then 011 okay that's flipped so what I'm doing is then um flip from top to bottom. Okay, so that would be one one and then zero.
Um and then a bunch of ones and then a one. Okay, then that would be um flip the top from top to bottom and that would be 1 zero and then one one. Okay.
And then that would be one one and then one zero.
So that would be one.
Um I mean obviously going to be testing that.
Um um and it should um yeah um that that that um that that that and that should be Oh no that that that does not work. That's one of these is a duplicate.
Um, >> I have bad news.
>> Okay, >> I have I have debugged the code. I think it's working fine now.
And my code cannot handle the case of size 12 which means you know I wanted it to say I cannot I cannot fill 12 x 12 grid with the five penttomino uh and the 12 thing I I put in 12 and it's not giving me an answer for a minute. So there's no way this works for 15.
>> Okay. Well, yeah, but it's easy to find if there is an answer like like if there's or well, it might be easy to find if there is an answer.
>> I mean, but I'm also finding no answer.
>> I'm also finding that there's no answer.
So, that's not an explanation, >> but I would case is not Yeah. giving me an answer right away either.
>> I would like to see if you can send that.
>> Right. I mean, >> okay.
>> How is that running that fast?
>> I have no idea.
>> Can you print it? Can we Can we print the >> what's it called? The output.
>> Not kind of. Yeah, I guess kind of.
>> Um like if we can like color things along the way somehow.
>> Um I guess um you can look at it by um what?
Oh, it's that's just Okay. Yeah, that's going to be How would I do that?
>> I mean, you could just like I don't know like at the end like when you traverse back up the recursive stack, you could just print things. Um >> Oh, yeah, true. Um, if Okay, that's probably a good time to introduce a rapper.
That's kind of insane, though. I can't believe it that that that seems to imply that it's actually functioning. I don't know how they how I mean I I imagine there must be some kind of um I forget the technical term for it but like you know where you where Python identifies that something here can be like rewritten in some kind of lower level language or something.
>> Yeah, >> cuz I can't imagine that this is just actually executing in plain Python that that fast.
Wait, I thought it was bugged, but now I believe it's not bugged. And I think I just maybe this definitely tells me that it's impossible to do 15 by 15.
>> I still there's still a chance that I have made a mistake, but I think it works. So, I'm going to share it. Uh, >> how would I share this? Where should I do I just copy it and paste it on this?
>> We've already code. I would I would I'm just mainly interested in the code.
>> Yeah.
>> When I said that about the recursive stack thing, is that is that what you're showing here?
>> Yeah. Um >> maybe we could just maybe we could just um like uh so I did I didn't implement the part where you wait I it says my message is too long. I've hit the 2,00 character count limit. Get intro.
>> Um, you can send it as a txt or you can like you can just drag the follower over or if >> Right. Right. Right.
>> This >> um but yes, I I I think this works.
>> Um like Okay. The the way you can view this is you can start with this and then um you can see pieces subtracted uh if you scroll up. And I'm pretty sure this works.
>> Cool. I'm going to be sending this code.
Um, this is like with a nitro. Holy Um, I wonder if I mean we should be able to rather easily extend this to the other two uh shapes, right?
>> Yeah, >> that is good.
There is no shot this works for 25, right?
>> Well, try it. I mean, maybe maybe it'll just find one right away.
>> Um 26 25.
Um I don't think it works for 25. Yep.
It doesn't work for 25. At least according to me.
I mean, if it works, I'll be so happy, but surely not.
Like, that's that's way too insane. Um Um underscore here, underscore here, underscore here. I'm just adding 10.
Adding 10 bits to everything I see. Um there's a non-zero chance. I just forget to change something obvious, but whatever. Um, impossible grid. Yep. No, it's not running.
>> What?
>> Oh, I found what?
>> Hell yeah.
>> Oh, no. No, no, no. This is wood for a >> Oh. Oh, okay.
>> Um, but still that's like super fast.
That's like way faster than it should be.
>> I was expecting it would be instant for like if there is one.
Yeah, probably true. Uh I'm just going to leave this running. There's no shot this finishes since I'm surely. Um do you all uh preferably put it in spoilers?
I mean I mean I'm I'm liking um doing code. So and also like to to what extent your solution is.
Okay. So we look at okay I believe we don't need any more of that.
Still waiting on Zo by the way. um commenting on it.
>> Oh uh I guess you don't um well I guess you can send an uncommented version first but >> Oh yeah.
>> Um P that's P L which okay I'll set this as my primary standard for L.
um this exact problem being with P specifically or or like with the other pentas as well because if it is with P specifically I'd be interested to just know a solution because that allows me to like maybe tackle the other problems with no flips. Okay. Um, share but spoiler.
Um, it's going to be a that's going to one one 0 0 0.
I guess I can just um do a left to right. Okay. one one one um and then the full full shenanigans the full 16 and then one rotating it the it's just going to be this backwards.
Yeah. Yeah. I I assume that was just a link. Um um 1 0 0 0 um one. Wait. Okay, that's um that's one line. Holy We have three lines.
One, two, three, and then four.
And that's just going to be reverse. So one one. And then >> I'm going to send the file in the chat.
>> Okay. One, two, three.
Um. Oh, that's just a debug. Okay.
Boom. Boom. Boom. Boom. Yep. Okay.
False. I mean, okay. I don't know if I want to trust it though.
Undian fraud. Okay.
Holy For kilobytes. Um, I'm just going to name it fraud.
Not with the same.
Oh, it's a size 25.
Yep. So, what you're seeing is it's it succeeded in placing 24. That's the first line before it found some impossible square to fill. Uh, starting from the top left.
>> Okay.
>> And the numbers below.
So I I just wrote it like that to reconstruct this. If you look at it yeah the comment the number denotes which rotation has been placed at that point.
So I haven't wrote the code to reconstruct it yet but we can reconstruct it like manually because the code decided to put the one case the rotation one case at the top left.
So >> yeah, if we place that then the next empty space is going to be like uh I don't know. I I I yeah I didn't make the reconstructing part.
Well, I I can do that later. But I think uh if you put in like size equals 10 then it correctly says that you can place all how much is it all 20 penttoinos in the grid. So it works for size 10 and 20.
>> Okay. Check empty shapes all shapes.
Oh, are these intervals?
>> So, um, wait, could you like increase the width of your window a little bit?
Is it or is it possible to like shrink the size of the fonts? Because I try to >> Yeah.
>> Um, font size. I usually deal with super um large font sizes, so that's why.
>> Yep. So on the top I just define the bitboard which is B and then I have a simple get and set function uh final you can just uh ignore that part. Get and set function uh and then >> get and set >> uh I see. Okay.
Uh-huh. And the check is going to check the first empty cell.
So that's like if not get B, that's how you check the first empty cell. Then you're going to try all four shapes.
And the results are going to be putting in yeah the result from putting in the four rotations.
You're just gonna find out which rotation resulted in the maximum number of uh penttoinos and then I just do something with that. Yeah.
>> Okay. So, this is trying to maximize the which I guess makes sense. So, because mine would just not uh mine would just um not care about the number maximizing the number. I would just end if there isn't one. Um, and all the things below are just like I didn't have time so I just copy and pasted. Uh, I I made one and then copy and pasteed. So each of the functions check for one kind of penttoino.
>> I denoted it with a little yeah circles on the top. Just >> check if we can place that pentomino in the current position. Yeah. In the J position. and place it and do that.
>> Check. And so check would be the number of stuff you can place. Um, changing 25 would work, I'm assuming.
>> Yes. Yes.
>> And so this would be possible with 10.
Okay. So 20 means possible.
>> Yep.
>> Okay. And Okay. And how would I interpret this?
>> Two would mean >> I think I think I didn't make it correctly. So you can just ignore that part.
>> I kind of see how it worksish.
>> I I I could like fix it so that it shows how it how to place things.
>> But yeah, >> I see how how to read this. I if it's a two, I place a two down and it overrides any numbers that it has just because the code hasn't handled >> and Y.
>> Okay. And um and that just repeats.
Okay. And >> do I dare look at the 15 case?
>> It puts 14.
>> Yeah, 14. And then it found something impossible.
>> Okay.
Um, oh gosh. Okay. Yeah, I can't read this.
Um, yeah, I cannot read this.
Well, um, okay. I'm going to check out the math link.
Hoping that it contains something. Um, and also I'm going to going to be in Tomino be proven that if five by n but only by considering um okay I'm the for 10 by n trying it this way then it must be okay.
Yeah. Yeah. Yeah.
does not need to be made up of 5x two rectangles. Consider this.
One side must clearly be a multiple of five. The other side does not need to be even. And you could have 10 by 7 using rectangles.
You would have 10 by 7. You have 10 by 17. That doesn't make that doesn't then m is even. Okay.
This is the important thing. Okay.
Because if you could call an MN, right? Okay.
What?
Let G be the whatever the boundary what what I mean obviously I can't read this if anyone if if anyone can read just um feel free to tell me, but I can't.
>> Oh, by the way, I found a bug, >> right? I found a bug.
>> Where is it?
>> I'm going to fix it and I'm going to send it again. But yeah, you can ignore my code for now.
It actually does I does the site um take care of um rectifiable.
Oh, whoa. This just there's just a this page.
Okay. I mean, that would have been vastly more helpful.
Um, so basically this like a multi-dimensional, not quite a coloring argument that's kind of a coloring argument.
Okay.
I mean, wait, no, this is a more general paper, is it not?
Great. We're note to self. I was also considering trying to explore into auto tilings, but I don't think we're getting to it.
I can't believe I'm using SIHO because this web page wouldn't load.
Okay, I'm assuming I solved the bug and uh it looks like it stops at 44 for 15 by 15. I'm going to send it again. So anyway, it's impossible. 15 is impossible.
>> It stops at what?
>> 44.
>> Okay. So, >> 44 block phentos placed.
>> All right. So, all but one.
>> Yep.
>> It should be available. Um, I'm going to send this link. Can anyone access this link or is it just me?
Ah, dandon froth.
Amazing.
What is the change?
>> Uh, the bug was there's one pentomino which is blank. O then o that's check_3.
I forgot that the top left square on that case would not be empty. the the top left square of that doesn't I don't have to check for that at all. Yeah.
>> So I change that one and now it takes quite a while but it does give you the answer to 15 within a minute.
>> Okay, I see. I'll leave that running in the background. Meanwhile, y'all can check that thing that uh that link and I that I sent and um if anyone can find anything like uh check to see if you can open the link, I guess. Oh, thank you.
Yeah, that's just me then.
Did I just completely miss this?
We need one very young person to even out.
Oh. Um, I guess I'll have to degrade myself by two years.
>> Wait, aren't we still recording?
Uh, >> we are still recording. Um, >> do you want me to clip that out?
>> Admittedly, I haven't really been paying attention for the last 10 minutes, so um I I I don't >> I do >> Huh?
>> I don't have a preference for whether it is clipped out.
>> All right. I didn't see any like names on the poll answers, so I think we're fine.
>> Yeah.
Oh no. Is this a non-English paper? Oh great, good. Thank god. Uh classical problem is the following.
Um can 25 um 10 label ones and fives. Okay. So that's where we're at. Okay. Introduce a variable.
Okay. System of linear equations.
Oh, that's obvious, right? I'm stupid.
Um, linear algebra.
I am.
visible by 16 xy xy x -2 y x1 Okay.
Is this ah so there are two methods homology which and the path method.
Okay. Where are the results?
Oh, it's just a it's just a random result in the middle of the paper.
The hell?
>> Yep. The next page has something. Yeah, the next page is the end.
>> Wait.
Oh, there are multiple pages. A 1x5.
Okay.
So is the proof is just the proof just ends at 147 I think.
>> Oh 147. Yeah. Yeah. But I was like looking for Yeah. I I know that therefore like that's a proof but I was like uh are there other results for the other things that we're considering?
What?
Um, okay. Yuri ask um what what is the other what is the final other thing that we're considering?
L pieces. Okay. Oh, watch out for the L piece.
And it's just references, email communication.
Please have this be like some Okay.
Okay. Hopefully I can read this quote unquote um clever geometric proof.
You can proof this.
>> What is this?
What set of penttos?
that adjacent to a border with I4 part they can't be placed interruptedly so they can't they can't be separated by more than one cell.
So it's like a bounding kind of thing.
Wait. Okay.
Four, right? because that doesn't work in place xxx but without continuation.
So boundary wipes can be separated only by one cell.
Okay. So it is just a fat number two is even simpler.
This is so true.
Finally, I proved impossibility of tyling.
Oh, this is like a computational proof, but this is somewhat somewhat faster.
Dissection of rectangle into crosses.
All partial crosses are covered by fixed pentoinos.
The number of crosses that there is a possible.
All pentomino except border ones can be ped. So each pair covers a pair of adjacent crosses.
It's an odd number of crosses. Um, okay. So, it's a bunch of it's a bunch of casework, right? So, it's just a bunch of case.
I'll send this out, I guess. But yeah, I can also absolutely not understand like the general proof.
Anyways, um how are we feeling? Should we look into auto tilings like like tilings of themselves but larger or should we end it right here?
I will probably have to get going soon.
I am going to be assuming that. Okay, I will.
Okay, I don't know if there's a lot to explore about all the tailings. Um, I'm going to be honest and like so like maybe and maybe not.
So, I'll I'll I'll look into that a bit and if there's anything interesting, I'll schedule something. But for the meantime, I have something scheduled which is um I I will probably schedule something because I think this has worked out like non-disastrous enough. I will be scheduling and um can you fold every um polyomino into rectangles um thing which will come up eventually in which I will talk about um folding um folding shapes into rectangles. More on that later. uh more on that probably like in the actual presentation. But yes, um any parting words or additional commentary?
>> Thank you very much. I'm going to cut the recording right now.
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