Symmetry Puzzles

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[@tomasstana5423]

The one with ants is my favourite. So counter intuitive from one point of view, yet so intuitive from a different point of view ...

[@nullvideoyoutuber2967]

Tbh i dont get it if they would be points ok but ants have a lenght and if they bounce its like they wallk and then suddenly jump forward on the place where the other ant is as they meet (sry for english)

[@ViratKohli-jj3wj]

@@nullvideoyoutuber2967 What are trying to say. I don't get it

[@nullvideoyoutuber2967]

@@ViratKohli-jj3wj idk

[@tomasstana5423]

@@nullvideoyoutuber2967 That only means in configuration where some collisions happen, the collisions only speed things up compared to the analog with them just passing. But that does not change the answer, the maximum is still one minute, because there is configuration with no collisions that takes one minute.

[@maxdixonman]

​@@tomasstana5423let's say that there are 2 ants facing each other that have to move an arbitrary distance 4 and 5, they each move 1 towards each other before colliding so and 1 is 3 away from the end and ant 2 is 4 away when they collide ant 1 is now 4 away and ant 2 is now 3 away from the end. They can keep hitting each other and keep swapping distances but the time they move keeps them going towards the end

[@RC32Smiths01]

Always enjoy Symmetry puzzles, they just feel so right because of the equality and cleverness. Quite awesome as always!

[@WillToWinvlog]

The universe is a symmetry puzzle

[@pendragon7600]

clearly, the ants must be assumed to have zero length. Otherwise, a collision is not equivalent to the ants just passing through one another.. they both teleport forward by the ant length.

[@maswinkels]

Yes. I was surprised how he forgot to mention that.

[@zachstar]

Yeah I guess we'll say negligible compared to the length of the rope.

[@samuelthecamel]

@@zachstar In true engineer fashion.

[@digitig]

Or the time they take to turn around is the same as the time they take to walk one ant's length.

[@Lance0]

@@zachstar is it 1/750,000m?

[@rupen42]

Every time one of these videos comes out, I regret switching out of math as a major. This is so good. I love math.

[@bingbong2179]

Spent all day on a 3rd year stats course assignment. Trust me it is far from sunshine and roses

[@rupen42]

@@bingbong2179 I'm aware. I was a math major. And the workload was still easier/less time consuming than my current one (econ).

[@bingbong2179]

@@rupen42 Maybe it's a uni difference, cause the one econ subject I did was the easiest thing in the world

[@roygalaasen]

The patreon calling himself AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA though... lol

[@__mk_km__]

Don't pay attention to the guy named *S*

[@Nulono]

That's AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA, thankyouverymuch.

[@roygalaasen]

Nulono I was waiting for someone to count the exact number of As. It is now correct, but I still don’t know how many.

[@soula1568]

@@roygalaasen it's 128

[@isaakvandaalen3899]

The thing with puzzles that abuse symmetry like this is that they're deceptively simple. At first glance it may seem like you need to do a bunch of calculating and you couldn't possibly account for every variable, but by abusing the symmetries of the puzzles we find a lot of the complexities cancel out and we're left with a very simple solution. The table one, for example, I solved with the following thoughts: -The table can be any finite size, and so can the coin, so really the actual sizes don't matter, but rather the ratio between the size of the table and the size of the coin. -Let's say the table is the same size as the coin. Obviously whoever goes first will win, as they will occupy the entire table. -Now let's say the table is large enough to barely fit two coins (TableDiameter=2coins). This game is possible to lose if you go first, but only if you place your coin perfectly on the edge, allowing your opponent to fit theirs next to it. If you go first and place your coin in the center, you will always win this game. -What this means is that we can treat tables with a diameter that is a multiple of 2 times the coin's size, then we can treat that game as though it were 1 coin size smaller. -Now let's say the table diameter of 3 coin lengths. Geometrically speaking, you can only fit a maximum of seven coins on such a table. This means that if you go first and place your in the center, there are a maximum of six coin positions left to fill. You can guarantee that this drops down to four by mirroring your opponent's placement, and then based on your opponent's next move you can either win or leave them with a choice to fill one of two remaining positions and allow you to win the turn after. -Now we've basically found our winning strategy, and all larger games are essentially just longer versions of this. Placing a coin in the center will win every time, so long as we mimic our opponent's placements afterwards. EDIT: I realise that to some this may still look confusing, but basically I just solved the games where the table was really really small and so you couldn't fit many coins on it. I just brute force solved them, found a winning strategy, and then proved that all larger tables are essentially just bigger versions of the smaller game - thus the same winning strategy still applies.

[@louf7178]

"Abuse"

[@isaakvandaalen3899]

@@louf7178 I don't mean it like it's a bad thing. I just mean that it's taking advantage of the properties of symmetry to make the puzzle seem harder than it really is.

[@PapaFlammy69]

Maybe :v

[@UnknownRager96]

Zach Star: uploads Brilliant: brilliant

[@adityachk2002]

It’s funny how i cant be relaxed until I watch new video as soon as it uploads....you binge entertainment.,..i binge education \

[@oleyespop]

Adityachk2002 yup :)

[@sirk603]

I love this channel. I’m only in 7th grade so a lot of things are confusing, but thinking about it and trying to understand it is really enjoyable.

[@jimmysyar889]

WhaT 8 dAyS eArLY! somEtHINg iS wRONG WITh tHe TiME! HOW dID YOU dO tHat!

[@obibellowme]

Andrew Salinas hE pRoBaBlY hAs A pAtReOn lol

[@Joseph-gq9nd]

He broke time symmetry

[@randomname5580]

Yo wtf

[@roygalaasen]

Noah Borch he does indeed have a patreon. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA. Just check the credits!

[@jimmysyar889]

roygalaasen wow how’d you know that was me?

[@tales9476]

I would note that the labeling of each tile on the solitare checker-board as 'H','V','D' was somewhat more confusing that sticking with the more abstract notion of 'elements'. The relationship between the tiles is not that of 'Horizontal move' 'Vertical move' and 'Diagonal move', but rather the distinct elements would be 'Tile which contains the piece you are moving', 'Tile being jumped', 'Tile to which the moved piece will be relocated', and the null tile (representing a tile that cannot be reached by a move).

[@pantheracheshire]

I have always loved group theory, but I didn't know it's application to the peg puzzles! Thank you for another great video!!

[@CoachEzFit]

There is undoubtedly a lot of math in Chess, and I wonder if some of these concepts can apply there, or is Chess just too complex for that?

[@trelligan42]

Yes and yes. Except in extremely distorted versions, the 'state space' of possible chess positions is far too large for any conceivable computer memory to hold. Many of the successful strategies are statistical, where games are evaluated and general patterns are noticed. Okay, some of the simpler endgames have been exhaustively researched, but that's minor.

[@kindanooby2988]

I believe any position with 7 of any piece (combining both sides) have actually been “solved” and are available for download (assuming you have enough space in your computer)

[@RespecterAlexander]

@@kindanooby2988If you have 18.4 TB of storage

[@паша-н3у5н]

8:51, i'm super confused about this part. With this configuration, we have HHHHHVVVVDD, which, when we reduce yields H, which suddenly means the problem is solvable.

[@BinJWu]

We’ll, this only means that we didn’t prove it was unsolvable. Notice that once you label one diagonal, there are two ways to label the other diagonal. Only one of the two set ups gives you an N as the final result

[@arshsverma]

5:37, that reminded me of quaternions!

[@clementdato6328]

Similar but not completely the same. Multiplication is commutative in Klein-4 but not in quaternion space.

[@abramosia]

They are related. If you take quaternion units and ignore the signs, you get the Klein group. (In more technical terms, if you take the quotient of quaternions by its center, you get a group isomorphic to Klein group)

[@EpicMathTime]

We can also recognize the Klein-four group as the set {(1,1), (1,-1), (-1,1), (-1,-1)} under coordinate-wise multiplication.

[@DanteKG.]

For those less mathematically inclined reading this: in the context of quaternions, if we multiply "i" and "j" we get "k". But if we multiply "j" and "i" (reverse order) we get "-k". So multiplying 2 vectors we will always get the third but it's sign will depend on the order of the factors

[@hassanakhtar7874]

@@EpicMathTime you are on like every video 😯

[@VaradMahashabde]

That square peg solitaire was a great example. Mind applying it to MPMP#6? (Matt Parker's Math Puzzles)?

[@boium.]

That second game is called nim, and it's a game with a lot of beautiful underlying mathematics. It is connected to a tower of fields of characteristic two, namely the fields of order 2,4,16,256,... You can read John Conway's book On Numbers and Games to know more about this. (I've discussed this game for my bachelor's thesis)

[@theproofessayist8441]

Excellent video - these puzzles are divine. I think I had an example of the pin board with one empty space one in my house as a toy.

[@BrainDamageNeko]

According to my knowledge, this is the MOST under rated channel in KZbin. You deserve really more subscribers.

[@A-RonGG]

I knew you for funny skits and now i know you for fun maths

[@diabbluiszegpidelgado1674]

The solution of the last puzzle it's brilliant!

[@Adomas_B]

That last puzzle would work great as a street scam

[@vinitsancheti3408]

That ant question is just insane.

[@hganti]

How is the checkerboard peg solitaire not equivalent to the regular peg solitaire, but rotated by 45°? It seems the problem is that only the boundary is different, which is why all the initially occupied tiles would have an identity product. If you added a single additional space, it becomes solvable, so there's nothing about the checkerboard, fundamentally, that makes it unsolvable (aside from the typical shape of a checkerboard).

[@thatpaxyton]

3 seconds at most, I’m waving the string around my head the entire time

[@brainfreeze7979]

I feel like I want to animate the ant one myself. Intuitively I would have said 30 seconds (-ish). Since no any would completely travel from one side to the other. Even with the oscillations of hitting other ants and negating the length of the ant (50 ants over a metre would still combine to a fair length), I feel the more correct answer would need to take the length of ant into consideration but more so the number of ants. Because one hop is negated by the fact we need the ant to be off the rope, but then every other hop advances that ant. I'd like to see a better breakdown of that question.

[@ion9084]

That's what I'm confused about, the ants would be spread across the 1 metre rope,so how is the answer 1min?

[@momom6197]

@@ion9084 1 minute is the minimum time that guarantees all ants are off the rope. Worst case scenario, there's an ant at the end of the rope walking towards the other end: it takes 1 minute to that ant to walk off the rope if it passes through the others, and by that time every other ant will have walked off as well. Since it cannot take longer than 1 minute, and there are scenarii in which you need 1 minute, then 1 minute is the minimum amount of time you have to wait to be certain all ants are off the rope, independent of the ants' starting positions.

[@brendenkasprzyk]

Thabk you for your videos man! I really love your stuff so keep it up!

[@ChrisSutherlandPhys]

The physicists dream

[@PositronQ]

Yeah all is = with all and that but the efficient is the magic of recursions

[@WontTrout]

Interestingly, if you extend the board twice but represent the extension differently, it is solvable one way and not the other! 1. Extend the board to the right side, adding another column of three Ds. Multiply it up and you get DDD = D as the result. Your video would make me think it is solvable in this case. 2. Extend the top of the board, adding a row of HDV = N. This time the resulting multiplication would still yield N and it is not solvable! First way the symmetry is destroyed, second way it is conserved.

[@ysquaredyobozo]

Question about the peg solitaire, can you be certain that a solution does exist if the product of the start state is not N, or is it only a tool for proving the non-existence of a solution, and not the inverse

[@momom6197]

It allows you to bound the set of possible solutions. For example, in the rotated example, the end product is H, therefore the only possible solution is H (though it does not tell you if such a solution exists, with a bit more reflection you can conclude that there is no solution).

[@hithere-cz4oc]

The coins game can be solved by taking the bitwise xor of the stack sizes, if it’s zero, player one loses, if it’s nonzero player one can win. The trick is to give your opponent an xor of zero by making a smart play (so in this case, give them 10 xor 10).

[@morkovija]

daym! blasted by the explainer 2000 once again

[@HexCopper]

I was first introduced to group theory via advanced inorganic chemistry where we’re using group theory to find how molecules bond to each other and predicting there spectrum. Now I just want to learn more about it as a pure math 😅

[@DragonPanda4]

The image made me think of just like, putting the rope in water. That'll get all the ants off

[@FernTheRobot]

Wait I have a question: At 8:10 Zach concluded that if we removed the upper D spot, the product of the rest of the board will be N, thus this problem has no solution. But the example at 8:40 has the V tile removed, resulting the final product to be H instead of N, so shouldn't there exist an H solution? I know those two problems are fundamentally the same, so why do we prefer the D tile solution over the V tile solution? Can we say that this "group theory" way of problem solving is kinda "flawed" if we can somehow solve the same problem in different ways and get different results?

[@anishkulkarni8297]

The group theory approach cannot tell you about the existence of a solution .So even if the product is H you can't say that a H solution exists.

[@momom6197]

There is no flaw here. The first example shows there is no solution, and the rotational argument is valid. The calculation you do shows that the only possible solution is H. You have not shown nor disproved the existence of a solution, though you have bounded the set of possible solutions. As it turns out, this set is empty, but you need more information/reasoning to prove it. For example, using symmetry, you could show similarly that the only possible solution is V. Therefore, since any solution must be V and H simultaneously, there is no solution.

[@Syrahl696]

This is my crack at the ant puzzle before watching the solution. Obviously, we start at "one minute". If an ant starts at one end of the rope, facing the other, and doesn't collide with any ants, it will take 1 minute for that ant to get off the rope. To get this hypothetical ant to spend longer than one minute on the rope, we would need to engineer a collision that would happen after this ant reaches the midpoint. Any single ant placed anywhere on the rope will not do it - by the 30 second mark, any such ant will have already reached or crossed the midpoint, either moving towards our first ant, in which case or collision doesn't extend the time on the rope, or away, in which case there will not be a collision. Can we make this collision happen with 2 ants? No. The father apart these 2 new ants are, the closer the collision must be to the midpoint. The closer these two ants are to start with, the more time one of them will spend moving towards the midpoint after they collide. I don't know how to write an equation for this, but it seems intuitable at this point that the ant that would collide with the very first ant will always end up at or past the midpoint by the time the first 30 seconds are up. So we couldn't make it take longer than 1inute with 1 ant (obviously). And we couldn't do it with 2 ants. And we couldn't do it with 3 ants. I can't think of any way to add another ant that would mess that up, so... No matter how many ants there are, after 30 seconds are up they should all be moving away from the midpoint, at which point it would only take 30 seconds for the rope to clear off.

[@momom6197]

This is interesting! I like the approach of trying to make a specific ant spend more than 1 minute on the rope. You have shown that with 2 ants, you cannot have the collision happen after the midsection. With three ants, in order to have the collision happen after the midsection, you'd need a collision between the 2 farther ants before the first quarter, which is equivalent to the problem with 2 ants on a shorter rope, so it's impossible. Likewise, you can go on to show by recursion that it's impossible to make your ant starting from the edge collide after the midsection. Unfortunately, that is not enough to conclude. You also need to show that an ant that would not start from the edge would not spend more than 1 minute, for example by going back and forth repeatedly. To do that, you can consider the product of a collision between two ants, and how it matters for the ants facing the right (arbitrary, but what matters is always considering the same direction). Unfortunately, this is generalized by the passing-through argument.

[@yanivshemtov9430]

Easier to win the first game, place the coin so it’s not centered but covers the center spot, turn one win

[@hhill5489]

Ok that last one was freakin cool

[@manswind3417]

Whoa, great video Zach! Btw, if I'm not wrong, the earlier examples covered in this video can fairly well be solved using the concepts of invariants and algorithms. Yes, I understand that they're perhaps simplified versions of group theory and transformations, meant for high school students but I believe that the core ideas involved in both the cases are the same.

[@Nikioko]

1. I will go first in the game and place my first coin in the centre of the table. And every subsequent opposite to my opponent. 2. I will go first, take 5 coins from the stack of 15 coins and then do whatever my opponent does with the other stack.

[@Nikioko]

4:07: You are supposed to end in the middle.

[@nikolakosanovic9931]

For the second one you can remove 9 from 10 and follow oponent with a parity of a number they removed from other pile

[@arunsahoo3145]

intresting video

[@zenithparsec]

Ants have length, and you need to assuming it takes the same time to turn around for an ant as it does for them to keep walking. You never said they were spherical cows/points.

[@bostangpalaguna228]

Always love the content of this channel. ❤❤ I wonder where you get the Idea/inspiration for the video

[@bostangpalaguna228]

Brilliant.org?

[@NickDolgy]

First! Thank you! Very interesting!

[@jimmysyar889]

:(

[@arshsverma]

How is this comment of 1 week ago?

[@obibellowme]

Arsh Shankar Verma patreon

[@jimmysyar889]

Noah Borch lol

[@arshsverma]

Noah Borch Oh, ok

[@Expert15696]

MAJOR PREP

[@Expert15696]

MAJOR PREP

[@hamiltonianpathondodecahed5236]

Zach King --{walks away}--

[@nestam6844]

If its about balance the oppent also just looses if your first move is a coin in the middle. Table will always be out of balance after that.

[@isabeldel-ponte1313]

How we know all ants are off the rope, when we shake it off!

[@cmyk8964]

So if you have another open spot to start with, it’s possible?

[@TheShoo]

The first puzzle is strange in that the first player can't place their coin anywhere but the center. So whoever goes first is handed the winning strategy.

[@samuelpak669]

I think some of those games were combinatorial reasonings as well.

[@avivmatas3340]

this was really informative

[@donutman4020]

The unit quaternions (1, -1, i, -i, j, -j, k, & -k) form a Klein-8 group

[@violet_broregarde]

How did you know the Klein product of the solitaire game wouldn't change?

[@danielyuan9862]

On every move, you remove two different Klein values and add the third one into the product. This doesn't affect the total Klein product.

[@dimitrisvain]

Not a minute passed and... like! like! like!

[@smhemant9111]

I really liked that coin stack game

[@BleachWizz]

is there an intuitive way to see what kind of group you're looking at when you give it a problem. like you showed that this last group can be configured on the board thus both are the same thing, but if I have the board how do I find the group it belongs?

[@zuriellanderos5788]

So could you still solve the checkers problem if you start the game with the missing peg being somewhere on a black tile that is at the edge of board?

[@skya6863]

No then you can't even start playing

[@pinkraven4402]

Ok but like, how can the table be balance when red guy places a coin?

[@TheParkourPenguin]

You're affirming the consequent in your explanation of the "checkered peg solitaire" problem. By representing the board squares as elements of the Klein four-group, you've lost information about the problem as it was originally presented. It isn't a reexamination of the problem- it's a simplification. There are constraints on valid manipulations of the state of the board that are not imposed on elements of the Kelin four-group. For example, viewing the board at 7:45, the H and V elements in the bottom left corner cannot produce a D element anywhere else on the board. The problem is more constrained than the simplification. More so, every constraint in the simplification is also a constraint in the original problem (it's a strict subset). Board states that can be derived from manipulations of some initial board under the game's constraints can also be equivalently evaluated under the constraints imposed by the simplification; however, the converse is not necessarily true- i.e. there exist constraints that would prevent the board from doing something that could be done under the simplification. For example, your evaluation of the board at 8:10 - 8:15 leaving two pegs on squares labeled V means there exists an evaluation of "HHHHHVVVVVD" that would result in "VV"; however, "HHHHHVVVVVD -> N" does not imply you could (validly) manipulate the board from the initial state such that there are no tiles left. Your inference at 8:02 is making the same logical fallacy: affirming the consequent. An obvious counterexample to your logic is at 8:40. With the initial blank tile as the centre-left "V" tile and the same mapping of tiles to elements in the Klein four-group, the expression "HHHHHVVVVDD" results in "H". This does not mean you will be left with one peg- that would likewise be affirming the consequent.

[@marvinliraDE]

Year comment is a year old, but I will still answer it for anyone interested. You made a good point and I agree that this part of the explanation was a bit too short and can easily be misunderstood. "HHHHHVVVVVD = N" does not imply that the board displaying "HHHHHVVVVVD" can be turned into a board displaying "N" through valid turns. However, the argument in the video is still valid. Doing a valid move on the board creates a new board which represents the same element of the Klein Group. Thus, if it was possible to make valid moves such that only one stone is left, "HHHHHVVVVVD" would have to be equal to the element "H", "V" or "D". As it is none of them but "N", this is not possible.

[@hollygraham102]

1:47 A better strategy is to go first and take all but one coin from one pile. If the other player takes that coin then you take the whole other pile for the win. If they take the whole other pile, well then you are left with just one coin, take it and win. If they only take a portion of the other pile (their only move to not lose immediately) Then you take all the coins but one from that pile. Now there are two piles of one each. Your opponent chooses one and then you take the other.

[@ThePianistOS]

This fails if the opponent uses the balancing strategy: if you take all the coins but one in a pile, the opponent will take all but one in the second pile. Then you are forced to take one of the coins, and there will be only one pile left.

[@simplesimon534]

AWESOME

[@manassivakumar1349]

this video is amazing make a video on asymmetry

[@decodedbunny101]

You know, I love problems like this, I make a game that’s exploits these

[@KabeerGoraKEC]

Who was the author of the book you showed at the end of the video at 9:02

[@berketozlu]

Oh nice

[@wyboo2019]

guessing the thumbnail problem and trying to solve it: obviously if the ants can just pass through each other, the solution is obvious so i'm assuming they cannot pass through each other. also assuming the ants are zero-dimensional because otherwise the problem becomes way harder if the ants bounce off each other, it seems more difficult, but when two ants come head-to-head and then bounce off one another, you just end up with two ants going out either side, so mathematically its the same as the case when the ants just pass through each other, so the solution is the same: (distance of stick)/(speed of an ant) i dont remember the specific numbers

[@rudygarcia3451]

How interesting!

[@evanobrien1407]

Hi Zack, can you do another book recommendations video, appreciate it thanks!

[@entertainment_hub5012]

Hey Zach why don't you do a video about avionics engineering

[@brainfreeze7979]

with the nim coins (I'm use to playing this where the player who is forced to take the last one loses so this is harder for me to articulate), if the first player takes all but the last two coins of either pile, the other player can take all of their pile, leaving player one to take all of the first and winning. If the other player takes all but the last, the first player takes only one from their pile. One each left, player one wins. If player two leaves more than one, player one takes only one from their pile, still leaving two piles. Player one should always win. Player two can never take all of one pile because that's an automatic win for player one. But me now searches for the video on the solitaire video. I have NEVER solved that one. Lol

[@the_cheese_cultist]

start with two piles of 3 player 1 makes it 3 2 player 2 makes it 2 2 to not lose, player 1 will have to make it 2 1 player 2 makes it 1 1 and no matter what player 1 does, player 2 wins also, "last player loses" is called Misére rules

[@inara6350]

Never played these puzzles and I probably never will, but I watched the whole video and I loved it. Great content, yet so underrated.

[@yarakharam5343]

why did you change your name from major prep to zach star and whats up with that picture?

[@noisyash4234]

With puzzle 2 wouldn't you also win if you took all but one from either pile first

[@calledahlin4197]

No then the other player take all but one from the other pile forcing you to finish one pile (as both now only has 1 coin), then he can finish the other pile and win

[@paperboy1790]

3:05 not very long if i shake the rope

[@tetraedri_1834]

Does group theory really need that many prereqs? Granted, it is quite abstract and not easy if you have not been exposed to proof-heavy math before, but as a topic it is mostly self-contained, at least in bachelor level.

[@zachstar]

At my school the undergrad abstract algebra course required two linear algebra courses, the calculus series, and a proofs class to be taken before hand. So most math majors could take it after two years probably but definitely not something for a first year.

[@wewladstbh]

I did introductonary group theory in the first year of my degree so it is possible at least

[@schwenke069]

Fact: Best I can tell. Two fire ants that exist on a string (or absolutely anywhere in Texas) will become 1000 fire ants in 2 days. Find a way to eliminate fire ants ... you'll be the richest person in the world.

[@swaglidangadu8127]

couldn't you take 14 coins of the first one and if the next person takes for example 3 of the other you take 6 of the other and you win automatically?

[@EtienneQuarez]

You have to be the one doing the symmetric action to win. If you take 14 coins the other player will "copy" you and take 9 coins from the first pile

[@sadeepweerasinghe]

great

[@hdt1948]

suggestion : problems with lcm and gcd

[@samuelthecamel]

2:58 you mean like Goomba physics?

[@mohammadaman175]

I’m not a student. I’m 33 years old. I watch your videos as entertainment. I wish I had this content available for me when I was in school.

[@amadeuszsitnicki3433]

Can't understand from 6:56 ...

[@jiajianhou426]

After watching this: why did I pay $40k a year to go to college lmao.

[@diygarygaming]

You should make a fan discord.

[@crazy4hitman755]

Ahaha I feel so stupid, I calculated the ant problem and it took my like 10 minutes to come up with 1 min 3:30

[@yanivshemtov9430]

Ok but, first turn place a slightly off centered coin, there’s no way to mirror it and they lose.

[@oliverdowning1543]

Actually if two ant heads bump then the switch direction then it’s equivalent to them teleporting past each other, not walking

[@karsonio3543]

In the context of this problem and for what the question is asking, it doesn't matter

[@trelligan42]

What _Karson Io_ said. The 'ants' are points with a (binary) direction, and they 'change direction' instantaneously. Points can approach arbitrarily closely, effectively 'ghosting' through each other.

[@oliverdowning1543]

Karson Io yes it doesn’t matter and in fact shortens the time taken but it’s interesting to think about

[@mohamedsamsudeen7694]

Just leave 1 coin on the side you take you will win

[@the_cheese_cultist]

opponent takes all coins but 1 from the other pile

[@WillToWinvlog]

1:39 you ask the question, but don't leave the problem on the screen for us to look at... Next time put the problem on the screen after asking the question, and give us a chance to pause. We don't want to skip around the video, risking spoiling the problem.

[@talshafir10]

Was I the only one who saw an avocado at the beginning?

[@Diamond19X96]

What I just don't get is why the game is only impossible to win if one of the 4 black spots in the middle is blank. When we write down the letters for the spots, the order doesn't matter and the end result is conserved, hence if f.e the coin on the black spot labelled V is removed I could as well just remove any of the black spots labelled V instead and end up with the same result. Same applies for H labelled spots. For me that'd lead to the conclusion that the game is impossible to win if you remove any of the coins.

[@shreddedtwopack6625]

I know I’m late, but I think he was just saying how this result extended to the other four center squares, not that it ONLY applied to those center four squares. Your analysis is right, and so is your conclusion about the problem.

[@Nikkikkikkiz]

usaco.org/index.php?page=viewproblem2&cpid=967 this is the ants puzzle

[@glitchmc9764]

For the Coin Stack game, couldn't you just take all but a single coin from either stack to start? Then your opponent would be forced to either take the last coin of that stack (which would win you the game), or to take some coins from the other untouched stack. At which point you could once again take all but the last coin of the second stack, which would just leave two stacks of one coin each. So they would be forced to pull one of those coins, and then you win by taking the other. Win in three turns guaranteed.

[@patprell251]

Came here looking for this! (y)

[@zachstar]

If you did that, the opponent could take all but one from the other stack, now you are forced to remove the last coin from either stack and your opponent will win.

[@glitchmc9764]

You are absolutely right, I don't know how I didn't consider that

[@patprell251]

Haha yeah I didn't think of the case where my opponent is clever too

[@ayden8885]

Hello

[@katakana1]

Currently there are NO dislikes. Don't ruin it...

[@shambosaha9727]

7 people ruined it already

[@pabloariza2295]

pai sho?