Sophie's previous polyominoes video: ruclips.net/video/ONdgXYEBihA/видео.html More Sophie on Numberphile: bit.ly/Sophie_Numberphile Tic-Tac-Toe with Xs only - ruclips.net/video/ktPvjr1tiKk/видео.html Three board Tic-Tac-Toe - ruclips.net/video/h09XU8t8eUM/видео.html
Fun problem! It's not too hard to show that no domino blocking strategy can stop snakey. Tile the plane with dominoes. Either there are three squares in a row somewhere which are covered by parallel dominoes, which can only happen in three distinct ways, or there aren't, which can only happen in two. In all five cases, you can fit a snakey somewhere that doesn't cover a domino. Unfortunately, that doesn't actually give us a winning strategy, and it's not enough to show there isn't any *other* blocking strategy.
Well yes but no. This proof is not complete since dominos need not be connected. You could equally tile the board (or infinite plane) like this: X Y Y X That still yields a valid strategy for player O (albeit not a winning one in this case). I suspect that even those "disconnected dominos" are not sufficient to find a tiling, but proving this seems like a lot more effort.
@@patrickwienhoft7987 True. I actually wrote up some terrible python code to test exactly that. I can show that even with disconnected dominoes (or partial matchings of squares), there's no way to block the snakey on a 6x6 grid or larger. We'll need another clever approach to make that work.
This is a pretty interesting problem! c: The personification & animating of the different polyminoes was cute and the domino strategy for preventing certain polyminoes from being made was cool to witness!
At the age of 10 to the age of 16 I was playing this variant of tic tac toe that in Sweden was called Luffarschack. It was a 15*15 grid and your aim was to get five in a row horizontally, vertically or diaganolly. My fascination in math made me to take the 4th place in the swedish championships back in 2005. The maths I used to make advantage came from graph theory. I feel like the problem presented in this video could be solved by the theories defined in that kind of mathematics.
@@columbus8myhw It's the same, yeah. Grid size can vary (usually you'd play just on grid paper) while in Gomoku you just use the Go board, and Gomoku also apparently has a ton of extra advanced rules to compensate for first player advantage. I tried to figure out whether the two were related (ie. whether western five-in-a-row games were derived from Gomoku) but couldn't find definite answers.
I applied and got into King’s College London because of Ms. MacLean. As a lifetime American, I didn’t know my education would take such a veering trajectory, but I’m happy. Hopefully I end up going there, or at least to the UK for my tertiary and quaternary education. I’m very excited! Ms. MacLean is the best!
6:05 “you’re not saying you’re going to win as player 2” In fact you can definitely say you can’t win as player 2 (unless player 1 makes a blunder) using a strategy stealing argument.
As a Tetris player, it hurts me physically to see the pieces being called by something else than "S-piece", "T-Piece", "L-piece" and so on, but at least we can agree in that the O-piece is a loser.
I think the true proof that someone plays Tetris rather than call themselves a "tetris player" simply because they like the names of the pieces is the fact that they agree that S and Z are much better pieces than O. all my homies hate O
in maths class me and my friend used toy play this "infinite" tic tac toe variant where we had to get 5 in a row (or diagonally) using the graph paper we had as our notebooks, it seemed like it was always possible to defend but when the tic tac toe structure got very big it became very hard to have oversight of everything so you'd usually end up with a win or loss eventually. But drawing from that experience I find it very unlikely that it is possible to build a snakey without the other player stopping it.
This is actually similar to a well-known game, gomoku. The only difference is that it's played on a finite board, but it's large enough that it's essentially infinite.
By the proof shown here, you can see that it's always possible to defend. Five in line is an endless game if played perfectly. Of course, nobody plays perfectly, and that's what makes it fun. But you'll never beat a computer.
It should. Tic on a regular 3•3 board is solved, you can force a draw. Sophie says that in the intro, too. @4:09 on the brown paper there's the correct ≥4.
Great video as always. Love Dr. MacLean's enthusiasm and style of presentation. One question: At 4:04 the graphic says that TIC will win with n >= 3. But if n = 3 then the game will end in a draw, I think.
Given that the English version of this game is called "Noughts and Crosses" I've always been against the idea of the player going first playing 'X'. For me it should be 'O'.
@@Matthew-bu7fg In Finnish it's called ristinolla, literally "cross-zero". Of course the presenters are British but the game is played around the world.
this feels like the outline of an interesting expnsion to the game of tic-tc-toe, maybe each player draws the polyomino they have to create on a board (maybe a large one so there's enough space) and idk if keep them hidden would make it too difficult to compete, but there's enough different strategies that it could be quite interesting.
Have them both revealed, but the first player has a polyomino n+1 larger. That's a level of complication I can't begin to imagine, because whichever player is at a disadvantage can focus on blocking moves, but also vary their strategy to start making their own piece. Parrying!
Not sure if it was mentioned at some point, but I guess that mirrored versions of the shapes are also sufficient, right? Otherwise, the example at 3:37 is problematic as the O could be placed in the bottom right corner such that the original shape cannot be completed anymore.
I think you're right about them not mentioning it, but, since the mirrored versions weren't included in their lists of shapes, it can be assumed that any rotation or reflection is allowed to successfully make the shape.
The name Snakey makes it very inviting to attempt Some Terrible Python Code™. Requires quite some thinking though, as the space of possibilities is sprawling uncontrollably.
I believe you can show that by preventing 1P from forming a 4 in a row using the optimal blocking moves, when 1P finally manages to get one, it is impossible for them to create a position with two winning plays at once
I have a proof for snakey being a winner on the condition that the response of player 2 is always next to where player 1 just played. (Forming a domino grid like the ones shown in the video for the other hexominos). I'll post it at some point (or correct my comment if I've made a mistake).
Under the "dominoes" strategies you mentioned there is no way for player 2 to ever ein either. This is because 2 always plays in a "domino" that has already been played in by 1, therefore 2 can never fill a domino either. It's a minor inaccuracy as you say that the strategy stops 1 from winning and 2 will either draw or win where in fact it's a straight draw strategy. Deviation from the strategy could lead to a win for 2, but then again it could lead to a win for 1.
Has it already been proved that there exists no partition of a (let's say infinite) square grid into pairs (not necessarily dominoes) such that every Snakey covers at least one pair?
Just wrote some terrible python code to brute force this, and it seems it's not possible to use any partition (or even partial partition) to block anything larger than 5x5. If there is a blocking strategy, it doesn't blindly use pairs.
@@aziidio We need to block both the snakey and its mirror image. A diagonal can only block one of the two. We would need a mix of diagonal directions to make it work, and there's not quite enough space to do that.
Mass= V and D. Volume is what is used. Density is what is not. E=mc2. C= (frequency, point, amplitude) 2 = parallelogram. (Not all squared?). So you are blocking something that has fluidity on the circumstances or not. Its just change in axis. How planes shortes route is not a straight line.
For french speakers (or using subtitles), I strongly encourage checking this video out: Ce que vous ne savez pas sur le morpion - Aline Parreau by Le Myriogon, a great video talking about tictactoe on an infinite plane with very similar ideas. There are even cooler tilings here to stop certain strategies and more unsolved problems to tackle! I spent hours trying to figure out the two unsolved ones but its so tricky!
Wait, doesn’t a basic diagonal blocking strategy make snakey a loser? I.e. tile the plane by 2x2 squares and always play diagonally from where player 1 played.
3:32 Actually, Snooty should be able to win in 4 moves by skipping the (useless) move bottom left and playing top middle directly or am I missing something substantial?
don't think about numberblocks don't think about numberblocks don't think about numberblocks DON'T THINK ABOUT NUMBERBLOCKS DON'T THINK ABOUT NUMBERBLOCKS DON'T THINK ABOUT NUMBERBLOCKS AAAAAAAAA YOU ARE NOW ON FOUR'S HITLIST I'M SORRYYYYYYYYYY
Given that going first usually confers an advantage in tic-tac-toe games, how would things change with the following tweak: the second player can make two non-connecting moves, then alternate?
Maybe the basic strategy with dominos (2x1) is not enough? Is there a known algorithm to win with more complex strategy, or is this unsolved problem only with dominos?
For the pentamino winners, does the first player get to make the mirror image shape to win, or do they still win even if they have to choose the handedness of the pentomino beforehand?
There can't be for a game like this. Imagine that there is a strategy that lets the second player win. That strategy would take a certain number of moves. The play is symmetric, so nothing would stop the first player from playing that strategy. But since they went first, they will reach the winning number of moves before the second player can.
Because a move _anywhere_ can only have a positive effect for the person who played it, you can use the "strategy stealing" argument to show that the second player cannot ever have a forcing win: Assume that a strategy exists so that the second player forcingly wins. Then the first player can start by playing "anywhere", and then completely ignore that move, thus in effect becoming the second player. Now, by the power of the assumption above, the first player will have a winning strategy! This is, of course, a contradiction, so a strategy for a forcing victory for the second player cannot exist in this game.
No, because player 1 is always 1 move ahead. If a strategy existed that player 2 could use to win, then player 1 could use that strategy (plus have an extra square claimed)
Not in this puzzle, but in other related games like connect 4, there are some board sizes like 8x8 where the second player can force a win. On the standard 7x6 board player one can force a win.
I think at 3:33 you didn't show the optimal strategy for O. The first O move needs to be in a corner to force 5 moves, with the first O on the edge X can win in 4 moves
I'm not sure I see how X can win in 4 with the O on the edge. What is the sequence of the moves? I assume the one shown in the video (opposite of the O) is not optimal?
Does adding a second shape winner change BL into winners or do they need not to have the same shared domino pattern to change from winner to loser? Like if you paired Boxy and F does the first player always win? If you pair Boxy and S you still have a never ending game?
One thing I didn't notice in the video, but might make a difference in solving this, is: do we care about chirality? I.e., can player one achieve snakey with reflections or not? It's entirely possible that reflections being allowed is a winner but chirality mattering is a loser.
The 'El' pattern wouldn't work on an n=5 grid 100% of the time but I think if you had it so when ever X plays in an el tile, O plays in the center or the closest leg of the tile. This makes it so whenever X tries to build a the 'el' part of snakey, O starts placing in a checker pattern and when X is trying to build the worm part, O tries to block it. I'm not sure I could formalize this into a proof, but I played a couple of games and it seems to work, so take this with a grain of salt.
@ , still doesn't matter. The only thing an odd number of columns would do is give an additional space in each row that can basically be ignored. 'O' will still play in a pairwise way with 'X', which will effectively stop any attempt to create "Boxy", while 'X' would be wasting a turn filling in that last odd square, allowing 'O' to play anywhere.
I kinda think the basic losers that don’t need to be proven because they contain a lower count basic loser could be called “inherent (basic) losers” because, well, that trait makes it obvious.
Sophie's previous polyominoes video: ruclips.net/video/ONdgXYEBihA/видео.html
More Sophie on Numberphile: bit.ly/Sophie_Numberphile
Tic-Tac-Toe with Xs only - ruclips.net/video/ktPvjr1tiKk/видео.html
Three board Tic-Tac-Toe - ruclips.net/video/h09XU8t8eUM/видео.html
Instead of playing games, Mathematicians prefer to think about what would happen if they hypothetically played a game.
Yes, exactly.
Commoners PLAY games, mathematicians SOLVE games
Very soon extension to 3D and n-D😅
And programmers make games 😎
I was thinking what I would respond to this comment when I noticed something that did not fit in the margin...
I haven't heard the phrase "basic loser" so often in ten minutes since that conversation I had yesterday!
That's a sign mate. RUclips has chosen you
@@Zwebbbel but is the sign + or -?
Ya basic!
@@Bluhbear It's ±
I love how excited the guest is each time she explains something
I didn't have "watching a mathematician needlessly and relentlessy bully Tetris pieces" on my RUclips bingo card today.
Looking forward to the casual comment that proves snakey's category
Herringbone
Spiral
Fun problem!
It's not too hard to show that no domino blocking strategy can stop snakey. Tile the plane with dominoes. Either there are three squares in a row somewhere which are covered by parallel dominoes, which can only happen in three distinct ways, or there aren't, which can only happen in two. In all five cases, you can fit a snakey somewhere that doesn't cover a domino.
Unfortunately, that doesn't actually give us a winning strategy, and it's not enough to show there isn't any *other* blocking strategy.
I wondered if you could alternate strategies
I'm not sure I follow. What does it mean to have three squares covered by 'parallel dominoes'? A domino is only two squares.
@@scottdebrestian9875 I mean three vertical (or three horizontal) dominoes in a row, like |||. They might also be offset like the ones at 5:47.
Well yes but no. This proof is not complete since dominos need not be connected. You could equally tile the board (or infinite plane) like this:
X Y
Y X
That still yields a valid strategy for player O (albeit not a winning one in this case).
I suspect that even those "disconnected dominos" are not sufficient to find a tiling, but proving this seems like a lot more effort.
@@patrickwienhoft7987 True. I actually wrote up some terrible python code to test exactly that. I can show that even with disconnected dominoes (or partial matchings of squares), there's no way to block the snakey on a 6x6 grid or larger. We'll need another clever approach to make that work.
This is a pretty interesting problem! c: The personification & animating of the different polyminoes was cute and the domino strategy for preventing certain polyminoes from being made was cool to witness!
This is why I love this channel. You just know someone's going to have solved this in a few months time
Maybe Matt Parker can write some terrible Python code while giving it a go?
@@ideallyyours In no time we'll have a Parker strategy that *almost* blocks Snakey
At the age of 10 to the age of 16 I was playing this variant of tic tac toe that in Sweden was called Luffarschack. It was a 15*15 grid and your aim was to get five in a row horizontally, vertically or diaganolly. My fascination in math made me to take the 4th place in the swedish championships back in 2005.
The maths I used to make advantage came from graph theory. I feel like the problem presented in this video could be solved by the theories defined in that kind of mathematics.
Is this the same as Gomoku?
@@columbus8myhw Not quite but pretty much the same, yes
@@columbus8myhw It's the same, yeah. Grid size can vary (usually you'd play just on grid paper) while in Gomoku you just use the Go board, and Gomoku also apparently has a ton of extra advanced rules to compensate for first player advantage. I tried to figure out whether the two were related (ie. whether western five-in-a-row games were derived from Gomoku) but couldn't find definite answers.
In America, we had a version that I played in the 80s called "Pegity"
I applied and got into King’s College London because of Ms. MacLean. As a lifetime American, I didn’t know my education would take such a veering trajectory, but I’m happy. Hopefully I end up going there, or at least to the UK for my tertiary and quaternary education. I’m very excited!
Ms. MacLean is the best!
The animations are superb. Great video. Thanks.
wait I need a follow up video! what in particular makes the snakey problem hard? i have so many questions
I love Sophie's videos
6:05 “you’re not saying you’re going to win as player 2”
In fact you can definitely say you can’t win as player 2 (unless player 1 makes a blunder) using a strategy stealing argument.
As a Tetris player, it hurts me physically to see the pieces being called by something else than "S-piece", "T-Piece", "L-piece" and so on, but at least we can agree in that the O-piece is a loser.
S-piece is peak anti-whimsy polynomino newspeak
Smashboy!
I think the true proof that someone plays Tetris rather than call themselves a "tetris player" simply because they like the names of the pieces is the fact that they agree that S and Z are much better pieces than O. all my homies hate O
Dont you mean "Cleveland Z" and "Rhode Island Z"; "Orange Ricky" and "Blue Ricky"; "Teewee", "Smashboy", and "Hero"?
7:56
“Beautiful friend here~
**Basic loser”**
I love her energy
xD
It's always wonderful to see Sophie Maclean!
RJ
man i love sophie, she's such an energetic host
4:47 adds a new meaning to "Be there or be square"
in maths class me and my friend used toy play this "infinite" tic tac toe variant where we had to get 5 in a row (or diagonally) using the graph paper we had as our notebooks, it seemed like it was always possible to defend but when the tic tac toe structure got very big it became very hard to have oversight of everything so you'd usually end up with a win or loss eventually. But drawing from that experience I find it very unlikely that it is possible to build a snakey without the other player stopping it.
This is actually similar to a well-known game, gomoku. The only difference is that it's played on a finite board, but it's large enough that it's essentially infinite.
By the proof shown here, you can see that it's always possible to defend. Five in line is an endless game if played perfectly. Of course, nobody plays perfectly, and that's what makes it fun. But you'll never beat a computer.
The table at 4:00 should show n>=4 for the first row I guess?
It should. Tic on a regular 3•3 board is solved, you can force a draw. Sophie says that in the intro, too. @4:09 on the brown paper there's the correct ≥4.
That is my question too
is there a 3D version of this?
This can easily be extended to 3D
~ 3:35 the snooty/tippy could've won in 1 guess less if the 3rd X wasn't placed in the bottom left corner, but instead in the top middle
Nice catch!
Was looking for this comment, I cant actually figure out a way cross needs 5 moves to win, I think it can always be done in 4
Great video as always. Love Dr. MacLean's enthusiasm and style of presentation. One question: At 4:04 the graphic says that TIC will win with n >= 3. But if n = 3 then the game will end in a draw, I think.
this feels like something a SAT solver would be good for
FYI: It has been shown that Snakey is a winner if you give X an extra move at the start of the game.
Given that the English version of this game is called "Noughts and Crosses" I've always been against the idea of the player going first playing 'X'. For me it should be 'O'.
@@Matthew-bu7fg In Finnish it's called ristinolla, literally "cross-zero". Of course the presenters are British but the game is played around the world.
As a Tetris player, I find all these tetramino names hilarious😅
this feels like the outline of an interesting expnsion to the game of tic-tc-toe, maybe each player draws the polyomino they have to create on a board (maybe a large one so there's enough space) and idk if keep them hidden would make it too difficult to compete, but there's enough different strategies that it could be quite interesting.
Have them both revealed, but the first player has a polyomino n+1 larger.
That's a level of complication I can't begin to imagine, because whichever player is at a disadvantage can focus on blocking moves, but also vary their strategy to start making their own piece. Parrying!
Not sure if it was mentioned at some point, but I guess that mirrored versions of the shapes are also sufficient, right?
Otherwise, the example at 3:37 is problematic as the O could be placed in the bottom right corner such that the original shape cannot be completed anymore.
I think you're right about them not mentioning it, but, since the mirrored versions weren't included in their lists of shapes, it can be assumed that any rotation or reflection is allowed to successfully make the shape.
The name Snakey makes it very inviting to attempt Some Terrible Python Code™. Requires quite some thinking though, as the space of possibilities is sprawling uncontrollably.
I believe you can show that by preventing 1P from forming a 4 in a row using the optimal blocking moves, when 1P finally manages to get one, it is impossible for them to create a position with two winning plays at once
Can Someone check that? He may be onto something.
Awesome episode
we got el and elly, now what did they do to ell. she's a vital part of this operation
Boxy is my spirit tetromino
A basic loser?
I have a proof for snakey being a winner on the condition that the response of player 2 is always next to where player 1 just played. (Forming a domino grid like the ones shown in the video for the other hexominos). I'll post it at some point (or correct my comment if I've made a mistake).
Reminds me of dimer tilings 🤔 amazing video, definitely need a follow up
Under the "dominoes" strategies you mentioned there is no way for player 2 to ever ein either. This is because 2 always plays in a "domino" that has already been played in by 1, therefore 2 can never fill a domino either.
It's a minor inaccuracy as you say that the strategy stops 1 from winning and 2 will either draw or win where in fact it's a straight draw strategy. Deviation from the strategy could lead to a win for 2, but then again it could lead to a win for 1.
Has it already been proved that there exists no partition of a (let's say infinite) square grid into pairs (not necessarily dominoes) such that every Snakey covers at least one pair?
Just wrote some terrible python code to brute force this, and it seems it's not possible to use any partition (or even partial partition) to block anything larger than 5x5. If there is a blocking strategy, it doesn't blindly use pairs.
@@mostly_mental How does a diagonal blocking strategy fail?
@@aziidio We need to block both the snakey and its mirror image. A diagonal can only block one of the two. We would need a mix of diagonal directions to make it work, and there's not quite enough space to do that.
I love me a good game rigg- I mean combinatorial analysis of games. Thanks Sophie!
Poor Boxy 😥
Mass= V and D. Volume is what is used. Density is what is not. E=mc2. C= (frequency, point, amplitude) 2 = parallelogram. (Not all squared?). So you are blocking something that has fluidity on the circumstances or not. Its just change in axis. How planes shortes route is not a straight line.
Saw the thumbnail and just hoped it was gonna be a Sophie video
Am I the only one who read the text on the thumbnail first as “Snacky Tic-Tac-Toe”? 😂
Please explain Euclid book 5 definition 5 🙌💯💯
For french speakers (or using subtitles), I strongly encourage checking this video out: Ce que vous ne savez pas sur le morpion - Aline Parreau by Le Myriogon, a great video talking about tictactoe on an infinite plane with very similar ideas. There are even cooler tilings here to stop certain strategies and more unsolved problems to tackle! I spent hours trying to figure out the two unsolved ones but its so tricky!
Wait, doesn’t a basic diagonal blocking strategy make snakey a loser? I.e. tile the plane by 2x2 squares and always play diagonally from where player 1 played.
@@aziidio That would not stop a Snakey that is facing left and with its head on the bottom right of a square.
@@supermarc Ah, gotcha, thanks!
Interestring. Thank you.
“Don’t be a square!” still rings true in Tic-Tac-Toe. Poor Boxy.
in fact, we should give boxy some love
"Prevent the inner loser" sounds like a great New Year's Resolution ... right up there with "Stop Procrastinating" ... is 5! Day too late to start?
Whelp better luck next year
_Prevent the Inner Loser_ sounds like a self-help course hosted by Troy McClure
sorry to break it to you, but currently you are 4! days late for that. 5! days too late would place you sometime in may
Could someone provide a link to any academic papers on this topic? I'd like to see how close the problem is to being solved.
The animations are the winner here
Great video, but calling 3:22 Snooty is a crime! His name is Squiggly :(
3:32 Actually, Snooty should be able to win in 4 moves by skipping the (useless) move bottom left and playing top middle directly or am I missing something substantial?
At 4:00 you have an error on screen with Tic's board size.
Never thought I'd get emotionally attached to a hexomino
don't think about numberblocks don't think about numberblocks don't think about numberblocks DON'T THINK ABOUT NUMBERBLOCKS
DON'T THINK ABOUT NUMBERBLOCKS
DON'T THINK ABOUT NUMBERBLOCKS
AAAAAAAAA YOU ARE NOW ON FOUR'S HITLIST I'M SORRYYYYYYYYYY
I love snakey! ❤
"This is a different kind of loser to the others" - I felt that
Given that going first usually confers an advantage in tic-tac-toe games, how would things change with the following tweak: the second player can make two non-connecting moves, then alternate?
4:20 LINE PIECE 🗣️🗣️🗣️
"EL" is too cute and I want to save him from the brown paper.
Maybe the basic strategy with dominos (2x1) is not enough? Is there a known algorithm to win with more complex strategy, or is this unsolved problem only with dominos?
For the pentamino winners, does the first player get to make the mirror image shape to win, or do they still win even if they have to choose the handedness of the pentomino beforehand?
is there a case of guaranteed loser? like, the second player always has a winning strategy?
There can't be for a game like this. Imagine that there is a strategy that lets the second player win. That strategy would take a certain number of moves. The play is symmetric, so nothing would stop the first player from playing that strategy. But since they went first, they will reach the winning number of moves before the second player can.
Because a move _anywhere_ can only have a positive effect for the person who played it, you can use the "strategy stealing" argument to show that the second player cannot ever have a forcing win: Assume that a strategy exists so that the second player forcingly wins. Then the first player can start by playing "anywhere", and then completely ignore that move, thus in effect becoming the second player. Now, by the power of the assumption above, the first player will have a winning strategy! This is, of course, a contradiction, so a strategy for a forcing victory for the second player cannot exist in this game.
No, because player 1 is always 1 move ahead. If a strategy existed that player 2 could use to win, then player 1 could use that strategy (plus have an extra square claimed)
Not in this puzzle, but in other related games like connect 4, there are some board sizes like 8x8 where the second player can force a win. On the standard 7x6 board player one can force a win.
@@binaryagenda That's why it's crucial to check that the extra move cannot have a negative consequence before applying the strategy stealing argument.
If you do manage to win, does EL open the secret elevator to level D?
Note that the second player can definitely not force a win.
3:36 Couldn't snooty win in four turns if you skip the third move shown?
Fun fact: Tetri is short for tetrimino, which is how the creator of a certain game spelled tetromino. That game involves multiple Tetris.
4:43 When you write an algebra "x" as your normal "x", you are definitely a mathematician.
there's only one x.
)(
Can I win if I match the reflected shape (i.e. flipped over)? e.g. if I'm trying to create an L then can I also win with a J?
then the blocking strategy could be a diagonal or something funny like a knight move
I think at 3:33 you didn't show the optimal strategy for O. The first O move needs to be in a corner to force 5 moves, with the first O on the edge X can win in 4 moves
I'm not sure I see how X can win in 4 with the O on the edge. What is the sequence of the moves? I assume the one shown in the video (opposite of the O) is not optimal?
@@DukeBG opposite of the O is the correct move, it's the third move that is not optimal in the video, it should have been top center
Good catch!
Please make videos on colour prediction number trick 🙏🙏🙏 please 🙏❣️❣️
Does adding a second shape winner change BL into winners or do they need not to have the same shared domino pattern to change from winner to loser? Like if you paired Boxy and F does the first player always win? If you pair Boxy and S you still have a never ending game?
Great video
Hey! What about filling the grey boxes on extreme left? 5:25 They are single...
I'm going to solve this.
I do think that snakey looses : would be very astonished otherwise.
I’ll never say no to giving snakey some love.
I wonder whether I'm a basic loser, or is there somewhere in me a smaller subset, the original basic loser.
-Tiling the plane with a1b1, b2b3, b4b5, b6c6 proves Snakey is a basic loser?- nvm. it doesn't.
"ok new rules. We just allow a row of length 3.............. see! the first player win with this column of length 3"
One thing I didn't notice in the video, but might make a difference in solving this, is: do we care about chirality? I.e., can player one achieve snakey with reflections or not? It's entirely possible that reflections being allowed is a winner but chirality mattering is a loser.
Since all of the domino tilings shown in the video are symmetrical, I think that all reflections would also have to be failures
You didn't mention dominos though!
The solution is that the domino is always a winner, after 2 turns
The plus pentomino contains boxy on a rotated grid, so to speak.
Now do it on a hexagonal grid with triangle -omino
After playing around with this for a few hours I believe Snakey is a loser, but he cannot be stopped by a repeating domino pattern.
Wouldn't you be able to prevent snakey using an 'El' pattern?
I think you could also do it with an imperfect domino tiling that includes a single tile in the center and groups of three domino's around it
The 'El' pattern wouldn't work on an n=5 grid 100% of the time but I think if you had it so when ever X plays in an el tile, O plays in the center or the closest leg of the tile. This makes it so whenever X tries to build a the 'el' part of snakey, O starts placing in a checker pattern and when X is trying to build the worm part, O tries to block it. I'm not sure I could formalize this into a proof, but I played a couple of games and it seems to work, so take this with a grain of salt.
2:42 Real Civil Engineer would be proud.
Sneaky Snakey could be his own category of nobody knows.
I don't remember anyone before having prepared a brown paper in advance 😅 6:15
There is an ambiguity : Can you flip the snakey or not ?
dont mind the haters boxy, youre still great
Sophie holds the pen on a strange fashion.
But how much does things change if you're allowed to wrap around? Could Boxy win that way?
Wraparound wouldn't matter, because the defensive strategy would wraparound as well.
@@SgtSupamanunless you have odd dimensions
@ , still doesn't matter. The only thing an odd number of columns would do is give an additional space in each row that can basically be ignored. 'O' will still play in a pairwise way with 'X', which will effectively stop any attempt to create "Boxy", while 'X' would be wasting a turn filling in that last odd square, allowing 'O' to play anywhere.
What's the title and description of this video? The RUclips auto-translation makes no sense.
"The Snakey Hexomino (unsolved Tic-Tac-Toe problem) - Numberphile"
@ thank you. I don’t know why anyone thought translated titles would make sense.
very cool
8:12 the we dont know is a numberphile trademark
I kinda think the basic losers that don’t need to be proven because they contain a lower count basic loser could be called “inherent (basic) losers” because, well, that trait makes it obvious.