The great Simon Singh tweeted this video! Here's the tweet: twitter.com/SLSingh/status/1021646697607426053 Simon Singh has written some of the best popular books on mathematics, including "Fermat's Enigma" and more recently "The Simpsons and Their Mathematical Secrets" (which I received as a birthday present, since my friends know I enjoy his books). Do check out his books!

@ecpgieicg

5 жыл бұрын

Proof is incomplete. The statement “Bob will always have a move” (whenever a 4x2 region is revisited) is nontrivial.

@matthewstuckenbruck5834

5 жыл бұрын

If the chessboard was labeled so that the rows, from top to bottom, A-H and the columns were labeled, from left to right, 1-8, suppose Alice puts her knight on square C1. Bob would put his knight in square A2. Suppose Alice moves to square E2. Where would Bob move? Am I missing something?

@philiplauren7024

5 жыл бұрын

The destroyer 94 If you didn’t already know, they’re both playing with the same night. Besides, even now when they are both playing with the same night Alice can’t move from either A2 or C1 to E2. Is there a possibility I missed something in your questions/statement (/Opposition)?

@darknessblade7480

5 жыл бұрын

you can play the game here also another theory what is the highest posible score you can get by not moving over the already moved spaces but still complete the entire map

@Heisenberg612

5 жыл бұрын

It’s just a theory a GAME THEORY

ofcourse bob will win his last name is fischer

@raghav4866

5 жыл бұрын

Only chess players will understand this!

@axemenace6637

5 жыл бұрын

@@raghav4866 I understood it and I don't play chess. Hence, your premise is falsw

@voidblessing

5 жыл бұрын

@@raghav4866 ok gatekeeper

@Carricklie

5 жыл бұрын

Booby fisher

@speedyx3493

5 жыл бұрын

No, Bob will win cause he is man...

Math Olympiad: *Sweats* Chess Grand master: *Laughs*

@belligerentapj3685

3 жыл бұрын

It might be tough for a chess grandmaster as well

@NateRobinson633

3 жыл бұрын

@Pulzanicus pulzanic 5 minutes? Give Hikaru 30 seconds before he starts calling it easy

@InternetMadnez

3 жыл бұрын

@Pulzanicus pulzanic take and take and take - free juicer - take. I blundered. Why am i so bad at Chess? Somehow he missed it too. Ok let's premove now.

@user-ii9ig7vq1e

3 жыл бұрын

@@InternetMadnez chat chat i dont get it chat why are you laughing chat

@leonardokazinga1536

3 жыл бұрын

Yeah lol. I am a chess player and I solved this.

"Did you figure it out?" yea i just whipped up my handy coloring graphs and chessboard png i always have those for random youtube videos

@jantemili

3 жыл бұрын

@@aftabkhan-EAEB imagine having such an attitude towards math

@parthibhayat

3 жыл бұрын

@@aftabkhan-EAEB I am in 9th grade and the book is still colorful, but yeah they aren't as big as 3rd grade

@nomatyx5789

3 жыл бұрын

@@aftabkhan-EAEB well then clearly they don't know game theory. It's not like that's a bad thing

@pyro1694

3 жыл бұрын

@@aftabkhan-EAEB You be calling everyone a handicap child lmao

@diomyiasis4164

3 жыл бұрын

@@pyro1694 he wants to feel special by calling everyone else special lol

So basically, if Bob loses he has no excuse

@aryankhullar7101

5 жыл бұрын

He cannot technically

@RandomStuff-xn9wy

5 жыл бұрын

@@aryankhullar7101 Of course he 'can', the point of the video is that he can always win when using the right strategy. But there is no guarantee that Bob will always use the right strategy, or even know it. If you didn't know the answer to this video, just as I didn't, then you wouldn't even be knowing what strategy to play in the first place and you would go back to square 1 and could as well just lose.

@magic_cfw

5 жыл бұрын

But Alice said she'll play Smash Ultimate with him if he loses.

@Nuggetsupreme

5 жыл бұрын

Alsp if alice starts she will win and if bon start he will wil since the amount of moves is odd the first to start is always the first to win if used the strategy

@itsahmd295

4 жыл бұрын

No body wins, No kings are in the game lol

"I couldn't figure it out, CAN YOU FIGURE IT OUT"

@icycloud6823

5 жыл бұрын

It's not like he is some kind of genius above us mortals, where if he couldn't figure it out, It means we can't figure out. I think you underestimate some people out there. Not everybody who watches these videos come to learn or see some problem that they wouldn't be able to figure out themselves. I'm sure there are people who watch these videos who can solve the problems he gives.

@ctsilver_1642

5 жыл бұрын

Diamond Bolt r/iamverysmart

@kapilbusawah7169

5 жыл бұрын

CTsilver _ no. This is just a true statement and also quite a useful piece of life information. There are some people who are familiar with these problems that know the solution but are curious if Presh has come across a new strategy. Presh is a very smart intellectual which cannot be denied but as shown, Sebastian helped him with the solution. A viewer similar to anyone of us provided the solution.

@pllurple

5 жыл бұрын

AlexD r/nobodygivesafuck

@pt6810

5 жыл бұрын

jerckï72 Dora the explorer reference?

What if Alice knows this as well, and insists that you go first?

@mudasirjamil9942

5 жыл бұрын

😂, good one

@lukesdewhurst

5 жыл бұрын

Second player will always win. 8x8=64. First player places piece which =63 squares left. Playing optimally will result in the second player taking the last available space. Dont know the reason for needing all the colours

@ColmarrPlays

5 жыл бұрын

SinfulMonk That approach assumes you can move from any square on the board to any other square on the board, but you can't. You can only use the knight's normal movement. That movement needs to be taken into account.

@lukesdewhurst

4 жыл бұрын

@족발러버 if there are 2 squares ledt lent of course player 2 wins.... player 1 plays now there is 1 square left. Not too hard to get that is it?

@dozzayeet7412

4 жыл бұрын

@Luke Dewhurst Due to how the knight moves, you need to come up with a strategy that works 100% of the time and explains how. You can't just say "theoretically he should win" you need to give the method of how he should win

Very clever solution. I had another way in my head about how Alice can only move to dark squares ever and Bob only to white and since it starts on a white tile then Alice would automatically run out first.

@yellowhero5413

6 жыл бұрын

Julian Lapenna that's a correct way and will yield the same ans but you have to prove that Alice can't trap Bob before that and thus the videos strategy comes in

@richkillertsm6664

6 жыл бұрын

I had the same thing in my mind.It's not a proof, but it's a good intuition to decide who will win.

@bsman321

6 жыл бұрын

same here it took me 2 sec to do it... too lazy to really solve it

@philhutchinson5885

6 жыл бұрын

That's assuming that there are enough legal ways to move up until you clear every single tile, and that there's no strategy you could devise which would generate a win up until then.

@julianlapenna7707

6 жыл бұрын

Phil Hutchinson that's a very good point thank you I didn't think of that!

OK, at first I didn't understand how this proves anything, but I think that is because you left out a step. Yes, Alice can return to one of the previous sections, but will always have to choose a different color square than when they were previously there. Since it is a new color, Bob will always have the matching one. So, the important part of the strategy is that Bob is always making sure that each pair is eliminated together in consecutive moves. This prevents Alice from ever returning to a block where the matching color was already used.

@XenophonSoulis

6 жыл бұрын

This formation also gives Bob a square to move each time.

@hassankassem2858

5 жыл бұрын

If u start on a yellow, It’s impossible to end up on another color if u put that 4*2 color pattern throughout the board

@TwisterHunt

5 жыл бұрын

@@hassankassem2858 nope, it's not

@za_arto

5 жыл бұрын

Oh ok so that's why i was suspecious of this strategy, i knew something wasn't right but he just didn't/forgot to mention that the horse (Alice's turn) can return to a visited region, but of course i didn't think all the way deep

@detectivl5811

5 жыл бұрын

I don't think he had to mention something that obvious...

When he say "game theory", i felt it.

@geckogeico2212

3 жыл бұрын

But that's just a theory... A GAME THEORY

@skiggywiggy8386

3 жыл бұрын

That’s just what they call it

@DavidBrit101

2 жыл бұрын

Game theory first term

A beautifully simple solution to what I see as a very hard puzzle

@darkghoul4049

3 жыл бұрын

Yes

@guilhermecaiado5384

2 жыл бұрын

The answer is 2x4. He literally gave the answer at the start of the explanation, but didnt explained how he got to that number.

@tomdekler9280

Жыл бұрын

@@guilhermecaiado5384 That's not an answer, that's a small part of an example of a winning strategy. You could play the same game and flip it into horizontal regions of 4x2 and it'll have the same result, naturally. I'm willing to bet there's other winning strategies, or other ways of framing the same winning strategy.

@guilhermecaiado5384

Жыл бұрын

@@tomdekler9280 No, im talking about maths to prove it.

Before I see the complicated mathematical reasoning, I’m going to guess that player 2 always wins. My logic behind this guess: the knight switches color tiles each move, if we assume all tiles can be touched, player 1 will lose one pace before player 2.

@gremlinn7

2 жыл бұрын

Yep, that's a good intuitive place to start, looking for the possibility of a Bob-favoring strategy rather than an Alice-favoring one.

@gremlinn7

2 жыл бұрын

I think it also might help to be familiar with other games (such as the "21 game" described on en.wikipedia.org/wiki/Nim ) in which the second player always has the chance to counter the first player's previous move in some way so as to "balance" it some way and keep a certain quantity within a certain state. The maintenance of the ability to balance it then implies the simpler condition of even being able to make a move in the first place, and thus not immediately lose the game. In the "21 game" player 2 is always able to keep the number modulo 4 equal to 0 by countering a 1 with a 3, a 2 with a 2, and a 3 with a 1, and thus force the number 20 to come up on player 1's turn. In this problem, a person who successfully came up with it might do so by looking for a condition player 2 could force to be balanced on every turn. Turns out that with the strategy found, this condition is "keep the number of visited squares of any knight-linked pair in any 4x2 region either 0 or 2". If Alice changes that quantity from 0 to 1, then Bob makes the moves that changes it to 2, from which it must remain the rest of the game.

You need to clarify that Alice can choose any square, and then Bob makes the first move.

@BigDBrian

6 жыл бұрын

Adam Pugh It was made clear that in the first move, alice places the Knight, then it'll be bob's turn and the moves will follow Knight movement patterns. Not really ambiguous

@9adam4

6 жыл бұрын

mrBorkD It was not clear, either that Alice can place the knight on any square, or that Bob moves it first. Neither was stated explicitly in the problem description.

@rmsgrey

6 жыл бұрын

Alice getting to choose the initial square or not is a moot point - if she can't win when she chooses the square, then she still can't win when she doesn't. That Bob takes the first move once the knight is on the board is clearly implicit from the explanation, but I agree it should have been explicit in the initial problem statement.

@BigDBrian

6 жыл бұрын

"Alice starts by placing a knight on the chessboard. Then they take turns moving the knight to a new square" while this doesn't explicitly make clear that after placing the knight, bob gets the next move, it does make clear that Alice gets to "place the knight" i.e. she can choose any square.

@denzelbacarra1082

6 жыл бұрын

So basically, Alice puts a knight on a random tile. Thinking that, 4x2 regions will eventually pile up. Bob's only goal is to fill the only legal move for the knight in that 4x2 regions, and since Alice can't move it on the same tile where the knight has been, she is forced to move it on another 4x2 region. Bob then takes the legal move again and that leaves Alice to move into another region. Even if you encounter the same region, Bob always has the legal move since Alice will only move it to another. The whole chessboard, marking the 4x2 region system, leaves 4 moves for Bob and therefore, he can occupy the 32 squares since 4x2 regions are 8 in total. Removing the first position will only leave Alice in an odd possibility, specifically 31 since (64-1=63) then (63-32=31). Bob takes the first move and Alice will make another move so we can say that every turn leaves them at the same number of moves. (Every turn = 1 Bob, 1 Alice) and since Bob took the first turn, the 32nd tile he placed will leave the 32nd turn into 32 Bob, 31 Alice. This time, all squares are occupied, so the first position won't matter anyway

fantastic! its such an easy proof for a problem that seems mindbogglingly complex. This problem really shows the importance of breaking down problems into smaller pieces and then putting them back together. Kind of like a divide and conquer in algorithm design.

@codyruchian

5 жыл бұрын

In this essay I will...

@xTORREfiolx

4 жыл бұрын

I didn't think the strategy out, but simply counting the possible moves gives the outcome away. In each square of the board, there's an even number of possible moves; the only squares where it's not true are the ones adjacent to the corners, because starting from there you have only 3 moves possible. However, there are 8 of such squares, and 8*3 = 24 possible moves. This method didn't take into account the fact that you can't return on the same square, but it doesn't matter: as long as the total of moves is even, Bob can always assure his win if he plays optimally, and if the total number of moves is odd then Alice can always assure her win. Think of it as trying to draw a sumple home without lifting the pencil: each vertex has either an even number of lines that depart from it, or there is an even number of vertexes (dunno if that's the right plural) that have an odd number of lines that depart from them. If this condition is satified, you can finish to draw the house, and the same principle applies to this problem, where each vertex (square) only has 2 lines (because no player can return on that square, so there's one line to get into the square and one line to move out from that square).

@cl9408

3 жыл бұрын

you just described "analysis"

@hybmnzz2658

3 жыл бұрын

More like the insanity of game theory

I wonder if this game will end in less than 64 moves, each time Bob will block a square Alice will find a way to escape until all 64 squares are occupied.

@jiogcyihsugyiocjfdoivhphvw6821

2 жыл бұрын

if both players play optimally, no

@tomdekler9280

Жыл бұрын

@@jiogcyihsugyiocjfdoivhphvw6821 Proof?

Wow, the solution is so simple once it’s explained, even though the question itself seems so difficult to prove

@brianofphobos8862

2 жыл бұрын

"Brother and sisters I have non, but that man's father, is my fathers son." Solution: That man is my son. Simple once you know the solution.

*Knights tour was a java coding project we did in class...its pretty fun!* You go around the edges then slowly come back to the middle.

Presh That was brilliant. I'm reasonably good at maths, and love your videos, but some go over my head.But you explained this one beautifully. :)

That end art where the letters rearrange themselves is so cool

I didn't figure out this one. So wonderful and clever solution! Thanks for sharing it.

I immediately recognized that optimal play by both players would end with a full knight tour. Since there are 64 squares to visit, and Alice is filling the first square, then Bob will visit all even squares, including the last unvisited square.

@Daniel-fv1ff

6 жыл бұрын

Why does optimal play result in a knight tour? Maybe one player has a strategy that will cause them to win earlier.

@flamewingsonic

6 жыл бұрын

Daniel Macculloch Any movement which is not part of a knight's tour will lead to a player losing in fewer moves than a knight's tour. But you are right, this is not enough; I still have to prove that if player 1 plays anything that will truncate the knight's tour, player 2 will still have the last move and win. And this is not obvious.

@BigDBrian

6 жыл бұрын

flamewing ⁠ I thought that too at first, but quickly rejected the idea due to the logical flaws listed above

@alandouglas2789

6 жыл бұрын

flamewing ⁠ wrong, you did not

@jaxxinator5999

6 жыл бұрын

After working on it, I think that optimal play should result in a full knight tour save three corner squares. This comes from the fact that for 3 out of 4 corners, all squares within reach of corner squares are an even number of steps away from start but one of the three corners they are an odd number of steps away. This means that Alice will always be able to ovoid going to three of the corners reducing the visited squares to 61 and thus Alice can always win. Edit: 2 out of 4 squares within corner square are an even number of steps away so Bob will win (I confess this is after looking at the solution.)

Actually there's an easier solution that ends in the same answer. A very interesting property with chess knights is that they always move to a square with the other color. So if the knight is in a white square it can only move to a black square and viceversa. This means they won't get in each other's way. Also, when Alice places the knight she's taking up one square. When bob moves the knight, there will always be an even number of squares available. This means the 64th square will be taken up by bob. The only thing he has to make sure is to not get himself trapped without squares to move. With this information, you can deduct that bob can win, without the 4x2 rule.

@giovannilunardi1261

2 жыл бұрын

I think this explanation is clearer biut less coloured..😂

@add-me_for_1v195

2 жыл бұрын

are you kidding me you said my comment 10 months before mine XDDDD very nice thinking man i grant you 200 iq XD

@Gnaaal

2 жыл бұрын

Even if the properties of the board and game means that Bob always wins in the end, unless someone screws up before the last square is moved to, it is still possible for Bob to lose since he can also screw up. The solution in the video guarantees the win though, i.e. it guards against getting trapped. No matter what square Alice moves to, there will always be the other side of the "square pair" inside the region open for Bob. The terminology of saying "Bob can always win" is a bit misleading though.

@kylen6430

2 жыл бұрын

“The only thing he has to make sure is to not get himself trapped without squares to move”. Yes, that is what the 4x2 method is for

@mattc3581

2 жыл бұрын

That's not really a solution I'm afraid. It would be possible for Bob to have already blocked off the only squares available in some case, so though there are available spaces somewhere on the board he can't necessarily move to them if his strategy hasn't been correct. The solution given here is possibly slightly more complex than it needs to be however. The 4*2 blocks are helpful to demonstrate a system where Bob has a unique move he should choose from any position Alice leaves him. Importantly every spot Alice can finish has a unique followup move from Bob and every spot Bob can move to has a unique position on the board that he moves there from. What Alice does and which sector she is in is at that point irrelevant, he has a defined move available for any square she leaves him on and that move is always available since she can't have left him on the trigger square for that move previously, thus if she has a legal move he always has his follow up move pre-defined and always wins.

It is difficult but you have made it easy to follow. Such an amazing work. Thank you!

Wow. Such a simple answer from what seemed to be such a complex question. That's what I like to see, Presh!

I have played chess for a long time and the knight is my favorite. I also love math and I have been watching your videos for a long time. I did not manage to solve the puzzle but I had a thing through it and found some interesting ideas. Keep up the great work on these videos!

@strideman1680

Жыл бұрын

I've always thought standard chess theory is wrong in assigning approximately equal value to knights and bishops. It seems to me knights are much more valuable than bishops, as no other piece, including the queen, can jump over pieces, plus the fact bishops are permanently restricted from half the board. I think the advantage of being the only piece on the board that can jump over pieces is undervalued by most players. I always trade bishops for knights whenever possible.

@olegpronin982

Жыл бұрын

@@strideman1680 knights are often better when there are a lot of pieces on the board. However in endgame bishops are usually better.

I didn’t work out the 2x4 I just figured it would always be the second person winning because it would ultimately be an odd number of moves.

That is truly clever. I definitely wouldn’t have thought of that.

Seems I missed this one when it was posted. Very good.

Duck. I thought Alice always wins, but her putting the knight on the board is her first move. If the chest piece was already there (randomly) and THAN Alice's first move was to move it, than Alice would always win.

@rayhanq8116

3 жыл бұрын

yeah whoever goes first wins that’s it

1:40 But it’s just a theory. A GAME THEORY!! Thank you for watching

How can you move 65 times when theres only 64 squares. It should be 63 moves, as the knight starts at a square and is not considered a move

I love you math problem videos, but I love your game theory videos a ton more! Are there other channels that have game theory puzzles?

This was interesting

But that's just a theory, *A GAME THEORY*

@socialexperimentable

6 жыл бұрын

it's not though. if both players are playing optimally the first person to move would win. if bob placed the piece and alice went first, who was also playing optimally she would win. the problem assumes the winning player knows this solution.

@noodlesinabowl3682

6 жыл бұрын

chazm00 he made a joke. Don’t over analyse it. (Which I guess is kinda ironic)

@m4rvin927

6 жыл бұрын

chazm00 1:41 cmon man it's a joke

@socialexperimentable

6 жыл бұрын

dont tell me what and what not to do sir/madam!

@daguido742

6 жыл бұрын

chazm00 whooosh

Two things that do need to be kept in mind here is that it is relevant who actually makes the first knight move, as well as the fact that a knight's tour is possible. The knight's tour probably becomes less relevant if not irrelevant if the 4x2 theory is applied by both players, since I am unsure if knight's tours can be completed taking into consideration this condition.

@yurenchu

2 жыл бұрын

Yes, there exists at least one tiling for which there exists at least one (open) Knight's Tour for which each even-numbered square lies in the same 4x2 region (or 2x4 region) as the previous odd-numbered square. The tiling for which I found a solution is not the same tiling as in the video, though. Consider the following tiling (each tile given by two diagonally opposed "corner" squares): [ A1/B4 ], [ A5/B8 ], [ C1/F2 ], [ C3/D6 ], [ E3/F6 ], [ C7/F8 ] [ G1/H4 ], [ G5/H8 ] For this tiling, here is a Knight's Tour that also satisfies the algorithm that each pair of squares { #(2k-1), #(2k) } are in the same 4x2 tile (or 2x4 tile): A8, B6, A5, B2, D1, F2, H1, G3, H5, G7, E8, C7, D5, C3, E4, F6, H7, G5, H3, G1, E2, C1, A2, B4, A6, B8, D7, F8, E6, F4, D3, C5, B7, A5, B3, A1, C2, E1, G2, H4, G6, H8, F7, D8, C6, D4, F3, E5, G4, H2, F1, D2, B1, A3, B5, A7, C8, E7, G8, H6, F5, E3, C4, D6. Here are the steps given in an 8-by-8 diagram (square #1 at A8, square #64 at D6): +----+-----+-----+----+-----+-----+----+-----+ | 01 | 26 | 57 | 44 | 11 | 28 | 59 | 42 | | 56 | 33 | 12 | 27 | 58 | 43 | 10 | 17 | | 25 | 02 | 45 | 64 | 29 | 16 | 41 | 60 | | 34 | 55 | 32 | 13 | 48 | 61 | 18 | 09 | | 03 | 24 | 63 | 46 | 15 | 30 | 49 | 40 | | 54 | 35 | 14 | 31 | 62 | 47 | 08 | 19 | | 23 | 04 | 37 | 52 | 21 | 06 | 39 | 50 | | 36 | 53 | 22 | 05 | 38 | 51 | 20 | 07 | +----+-----+-----+----+-----+-----+----+-----+ I hope that helps.

Presh forgot to expand out on what happens if Alice moves to a different 4x2 rectangle on her move, after moving out of the initial 4x2 rectangle. Alice could move in a few different directions on her second move. Bob can still win if he sees what Alice is up to, though.

@jiogcyihsugyiocjfdoivhphvw6821

2 жыл бұрын

he did explain that

As presented, the proof that Bob can force a win is missing a step for completeness. The proof shows that Bob can force Alice to exit a 4x2 region, but does not examine what happens when an "Alice" move returns to a region in which Alice and Bob have already played. Since untouched regions and regions that have already seen paired Alice/Bob moves are not identical, the form of the proof requires an argument at the sub-region (graph color pair) level. Note first that: (1) Alice will always play on the same board color (black or white); (2) Bob will always play on the opposite board color (white or black); (3) each of the 4x2 regions contains four color pairs (green, teal, yellow, and purple); and (4) a color pair within each region includes one square of each board color (green in a 4x2 region is one black square and one white square, teal in a 4x2 region is one black square and one white square, etc.). Assuming Bob follows the strategy described, then for each 4x2 region, Alice might return up to three times. Each time Alice returns to a 4x2 region, she must play to a graph coloring pair (green, teal, purple, or yellow) that was not previously exhausted by an Alice/Bob move pair. When Alice does, Bob will always be able to play to the opposite board color (black/white) square of the same graph coloring color. That will both exhaust one of the graph coloring color pairs for the region and (returning us to the proof as presented) force Alice to move outside the 4x2 region on her next move.

@elwinwinter

6 жыл бұрын

Shawn Winnie going to a region they've already been doesn't change anything. Bob can do the exact same, going to the only possible spot in that 4x2, resulting in alice having the do the exact same thing, moving out of that 4x2?

@uroscolovic1357

6 жыл бұрын

Yoy are right someone with little knowlege in math like me might not understand the video but with your explanation everything is crystal clear thans!

@stevenattanasso2003

5 жыл бұрын

No missing step , You just said it in a much more verbose fashion .......

@amilemon

5 жыл бұрын

I didn't reed

Great solution tricky question

there is a neat little trick/puzzle where you have to visit each and every square once and not more with a knight and if you know the moves it does not matter where the knight starts

Your videos have been so helpful to me and these are much manifest to resist in any way so easy peasy.

I missed this guy.. idk y KZread stopped recommending him

@TKVisme

5 жыл бұрын

Itachi best brother

@arnavpandey3823

4 жыл бұрын

Subscribe

Once you know what they are doing, u can actually trap them as 2nd player. You need to work your way back into a prior 4x2, while leaving the available 4x2 escapes unable to be jumpped into. hard to explain in text though.

@preschoolPT

5 жыл бұрын

Exactly

@Gnaaal

2 жыл бұрын

If by second player you mean Alice (the one to place it then make the second actual movement on the board), she can never win over Bob if he follows this strategy. Never.

Another thing that I wanna point out (this is unrelated to the puzzle) is that the colors match perfectly despite one seemingly lighter than the other. It is but another illusion.

an incredible proof! Thank you man!!!!:)

Btw I think that bob losses if he choses to go to a différent 4x2 region

But if you take a 4x4 there can be executed up to 4 moves. Ex. (R4,C1)→ (R2,C2) (R2,C2)→(R3,C4) (R3,C4)→(R1,C3) (R1,C3)→(R2,C1) R=Row, C=Column So I dont know if this is the way to prove it. Also there is a thing called "Knights tour" where there a knight can go from any position and fill the entire board. Which means that there are 64 moves in including the act of putting the knight on the board. So without that there are 63 moves to be done so the one to go first always wins indeed.

@yurenchu

2 жыл бұрын

Your post presents the following four moves: 1. (R4,C1) → (R2,C2) 2. (R2,C2) → (R3,C4) 3. (R3,C4) → (R1,C3) 4. (R1,C3) → (R2,C1) After those four moves, we can certainly do ten more moves: 5. (R2,C1) → (R4,C2) 6. (R4,C2) → (R2,C3) 7. (R2,C3) → (R1,C1) 8. (R1,C1) → (R3,C2) 9. (R3,C2) → (R2,C4) 10. (R2,C4) → (R4,C3) 11. (R4,C3) → (R3,C1) 12. (R3,C1) → (R1,C2) 13. (R1,C2) → (R3,C3) 14. (R3,C3) → (R1,C4) And now it's Bob's move, and since he can't move, he loses.

Damn...that's smart. Never would have thought it in a million years

I just figured that there were an equal number of black and white tiles. That's 2*32. The knight can only move from one color to the other. That creates either a black-white or a white-black sequence that can be repeated up to 32 times. Even if, theoretically, the knight could move to every tile, the person to finish the 32nd sequence would be the winner, as there is no 65th tile to move the knight to. This would only prove that Bob wins though, didn't exactly provide a strategy for him.

@OndrejHolecek

5 жыл бұрын

but you didn't consider Alice could win in less than 32 moves

in the first move, 2 squares are covered so this means there can't be more than 63 moves, not 65 like you said at the end of the video

@abcd123906

6 жыл бұрын

iCEW0LF Was thinking the same thing and started looking for others in the comments who had noticed that. He just got n+1 mixed up with n-1 :-) Thanks for confirming my suspicion.

@Nomnomlick

6 жыл бұрын

The first move is placing the knight, so 1 square. Then each subsequent move adds 1 more square. You can do 64 legal moves. The 65th ends the game.

@yellowhero5413

6 жыл бұрын

iCEW0LF both are moving the same knight

@abcd123906

6 жыл бұрын

Alex Burns Oh, ok, gotcha; yeah, makes sense now

@Lovuschka

6 жыл бұрын

Well, it depends on what we call a move. It should be 65, as Alice places the piece (move 1). Bob moves (move 2). At move 64 all squares have been visited once. Move 65 then results in Alice losing, as no free squares are left.

Oh, this is a variation of the Knight's Tour, isn't it?

@ounobaga1829

4 жыл бұрын

It's really not

I have been working on a similar problem for several decades , and have actually been able to solve my puzzle with about 20 different patterns . Start with the knight in either of the standard starting spaces and move the knight around the board to new spaces , until you have placed it on every square once , and end up back on the starting point . I was Kreskin do it with his back to the board and calling the squares out to another person who was moving the piece for him .

Very interesting and well explained, but good luck keeping track of which squares have already been used and which haven't.

Doesn’t this only take into account a game where bob goes first?

@rise7308

5 жыл бұрын

Bradley Hudson i think if alice go 1st and stayed on the 4x2 box. bob can win if he makes it to vertical box (2x4). haven't fully confirmed this. can anyone confirm?

Presh, you give us really good math questions. But when you say that you didn't figure it out by yourself, it really breaks my heart for some reason... 😳

seems like a more accessible illustration of how the second person always gets to take the last square on the board since there is an even number of them

I’m impressed that someone managed to get that into an email, that alone would stump me.

MY G I JUST GOT A 105 MINUTE LONG LEGO TRAILER ON THIS VIDEO WHAT THE ACTUAL HECK

@harjitsingh7308

5 жыл бұрын

Lego is love, lego is life

There's a similar problem with Queens instead of the Knights.

@dipborah7978

5 жыл бұрын

The 8-queen problem, did my final year project on a related problem.

Probably said already but Alice can easily break this tactic by making repeated sideways L moves instead of vertical L moves. In the video example, If Alice were to move into the next section with a horizontal L, and then repeat that to loop BACK into that previous section after leaving it on a horizontal L, Alice effectively can counter Bob's vertical L strategy and route him wherever she wants.

The graph has a perfect matching. Just consider any matching, and what Bob does is that he always moves the knight to the square to which the square the knight currently is on is matched to. So Bob wins.

what if when the knight is placed allice moves it in the second square of that region

@abhinavjain2985

5 жыл бұрын

Dude Alice's first move is placing the knight See it is simple 1)If alice moves the knight after placement means her squares get used later than Bob which according to theory means that alice wins 2)If you think that the placement sqaure is not booked then Bob can move back thus you contradict yourself(which i hope you don't (just to make sure u understand in the first go)) In your sense the first to move always wins The same thing

Thanks for your videos, Fresh Towel Locker.

I think there needs to be more specification that "new square" means a square that you haven't landed on before. At first I assumed that they meant the first person who couldn't move the knight at all looses and I was like "umm... that's literally an impossible circumstance on an empty chess board..." lol

You should clarify that the first square (the one alice places the knight on) can be revisited, as that changes the result. I for example, thought the winner would be alice because the starting square is "one it has been on before".

Hey Presh, you inspire me to do a lot of things and I really enjoy every single video you upload. I’m a junior in high school and have never gotten less than an A in any subject on report cards and I really appreciate math (algebra and trigonometry so far, taking calculus next year) and I wanted to ask if you had any suggestions for becoming a mathematician or something like that. I’m not sure if you can contact me or anything but with your input I think I can get into an excellent college and be want I want to be.

@Scobson

6 жыл бұрын

Tyler Cenko look up khan academy on the we website/KZread. They teach a tonne of math topics

@JuditeTormentor

6 жыл бұрын

I echo what Scott said. Khan Academy is a great resource if you are a visual/auditory learner. As a student and Teacher, I've utilized Khan Academy to much benefit. You should also check out the free coursework offered by MIT. They literally have entire courses (lectures and exams) for you to use at your disposal.

@tcenko

6 жыл бұрын

JuditeTormentor any suggestions on “getting my name out there?” What should I do while I’m taking my last year in high school to try and make it to the college that I want? Is it strictly gpa and sat scores that let me go to a top college?

@aldin7158

6 жыл бұрын

Tyler Cenko Send him an email, he will reply. I know because he replied to me within 48 hours. And I wish you good luck in math 😊

@RitobanRoyChowdhury

6 жыл бұрын

Along with Khan Academy, I recommend the 3blue1brown's and Eddie Woo's youtube channels.

*Very good*

@stevenattanasso2003

5 жыл бұрын

Very pretty .....

There is even more intuitive solution. Each move of knight is changing color of tile on which it is standing (it moves from black tile on white and vice versa), therefore if it stand on white color there are 32 moves (32 black tiles) for the first one who moves, and only 31 moves for second player (because 1 tile of this color is already used at the beginning), therefore second player has 1 less available spots and thus in ideal game where players use every tile - second player always loses

This is a special case of a harder problem, where you have a graph G and players move to vertices and the player who traps the other wins. Turns out the second player has a winning strategy if and only if the graph has a perfect matching, and it’s easy to check the knight’s graph has one.

5:24 You mean the game has to end in 63 moves or fewer, since there are only 64 squares and Alice already occupied 1 of the at the start so 64-1= 63 possible squares left?

@waffleuser

3 жыл бұрын

The starting square counts as a move

@danielschneider9358

3 жыл бұрын

@@waffleuser so we're at 64 Moves ...what's the 65th?

Bruuuuuh... The add I just got... *BRUUUUUUHHHH!!!* *[COMMOOOON MY PEEEERIIOOOOOOOD!!!!]*

I had solved a variant of this problem. It asked if a knight on the chessboard can cover the entire chessboard in a series of moves. It was a programming question and the answer is yes. I have been since wondering if one can analytically prove that a solution exists and it is unique. The video helps orients me to think in graph theory terms. Thanks.

We had this challenge at high school except it wasn't a game and it makes more sense as a game. The objective was only to fill in the entire board using a knight . Just spiral it around the board

After learning this, I am the new Bob in the town 😎

@abhinavjain2985

5 жыл бұрын

😂👍

@pianoguy9300

4 жыл бұрын

*KING BOB!!!*

The principal: if Alice is able to move the knight to ANY of the 8 regions, Bob will always have a square in THAT region to move the knight to. This is because of the strategy Bob uses. Alice places the knight = first move. Bob does the even moves and wins because there are 64 moves/squares.

Nice logic!! My logic was that as there are only 64 squares and assuming both of them are playing optimally they will eventually run out of squares thus the one who is starting first will always lose.

my first thought on the problem was that it was a flawed game, given that your dealing with even numbers it to me meant that the winner was predetermined by who went first or second, very cool video to see other peoples thought processes.

Now that's what we call a math riddle, not things such as 2 + 2 * 2 = 8, nice video :)

@hexagonist23

6 жыл бұрын

2 + 2 * 2 = 6

@GourangaPL

6 жыл бұрын

no, it's 8, because 2+2 = 4 and times 2 = 8 :) just trolling, after 5 years of math studies let's assume i know how it works :)

@rmsgrey

6 жыл бұрын

Which is why infix notation sucks and it should be written *,+,2,2,2 (or 2,2,2,+,*)

@AA-100

6 жыл бұрын

Pedmas rules. Multiplication comes before addition therefore making 6.

@rmsgrey

6 жыл бұрын

+Super Destroyer If you apply the multiplication first, then you get 2+(2*2) -> 2+4 which is 6, not 8.

1:45 And that’s just a theory, a game theory.

@aerosniper620

6 жыл бұрын

Matthew Morcos lol

I actually had to solve the Knight's Tour problem for my computer science class. The winning strategy is to use Warnsdorff's Heuristic: always move the knight to the square where the fewest number of possible moves could result.

Optimaly, the first one to move the horse wins, if you run out of tiles then the last place will be the second colour that horse steped on (it doesn t move to same colour it stands on)

Me? I'm just here searching answers

How do you come to solving the problem in that way? :O Who came up with that? ^^ And especially how?

@jaspermooren5883

6 жыл бұрын

Languste Well it's basically an easy to remember way of turning 64 squares into 32 pairs. And that ia what matters, the pair. As long as you have 32 pairs you can constantly pick the other of the pair and it will be imposible for A to find a square that does not have its pair left, because when it is A's turn, there are always only two option, either both of the pair are free, or neighter are free. This way B always has a move. It's a very elegant solution to the problem. The entire box thing is just an easy way for humans to remember, but any solution in which there are 32 pairs it would work.

@brandonkellner2920

6 жыл бұрын

It's probably easier to come up with the solution if you consider the 4x4 version of the chess board. If Bob doesn't stay on the same side, Bob can lose.

@andymcl92

6 жыл бұрын

There's another cool puzzle that has a proof that similarly made me think "Wow, that's really neat but how did someone think of it?" in a numberphile video about pebbling a chessboard and freeing clones: kzread.info/dash/bejne/nnqFqbWAo7vIiKg.html

@THEMithrandir09

6 жыл бұрын

If you know about gametheory and nim then you will automatically try to reduce it to that.

I just had the idea of moving the knight into the corner after the 2 squares were already taken at some point

@Alegost1

5 жыл бұрын

that can happen but it is situational at best

Amazing!! Thanks so much

I felt the no game no life vibes

I don't think anything needs to be changed for the audio part. The explanation is crystal clear and surprisingly simple. However I am not a fan of slides showing every spoken word. I'd rather see interesting visual designs, such as chess pieces, boards or anything relevant to the topic.

@alfredkjeller8420

6 жыл бұрын

For us that try to solve the problems it is alot easier to have the the rules showing in the slides.

@MrPictor

6 жыл бұрын

Alfred Kjeller Sure, I definitely agree with you about that. The rules usually take a few words if you look at earlier videos.

@Seth_Hezekiah

6 жыл бұрын

Not everyone learns the same way you do sir.

@dlevi67

6 жыл бұрын

Not all of us have audio turned on... I often listen to music while "reading videos".

@myc763

6 жыл бұрын

Well Mr Pictor, some of us are Wordors.

You should have shown a full game played out I'd like to see that

I solved it in a simpler way: the knight can only move to opposite coloured squares (if it were on white one, it can only move to a black one) and for the players to move across the whole board, without leaving any square, they would have to end up on a black square (given the knight started on a white square). Which means that the person that makes the first move wins, because he will keep moving to dark squares and eventually (after 63 moves) the other player will run out of moves.

@user-zj9rr6yc4u

3 жыл бұрын

a4 b6 d5 c7 a8 and Alice has won. With imperfect play obviously what you say doesn't hold. It is a pretty big assumption that perfect play makes it impossible for there to be a way to end the game before all squares have been visited without that also being a Bob win. It sounds plausible and as the video proofs Bobs victory is actually guaranteed but without such a proof it is just an educated guess.

You're overanalyzing. Each move goes from either white to black, or black to white. If the Knights initial placement counts as a square that can't be landed on again, and if both players are playing optimally, the best strategy is to go first.

@abhinavjain2985

5 жыл бұрын

@@arthurmp3243 🤣🤣🤣👍

*_BUT HEY, THATS JUST A THEORY, A GAME THEORY!_*

A lot of these older MYD videos are more like "check out this hard puzzle and watch the clever solution," rather than "solve this cool puzzle."

Took me 20 seconds to figure out, feels good man. in short: It is possible to divide the chessboard into pairs of squares, which are "L" apart. (trick that I've used in some other combinatorics problem, such as "how many knights can stand on the chessboard and not attack each other") Bob wins, because he always moves the knight to the second square from the pair, Alice just entered into, thus he always has a move and after his move, pair becomes unavailable.

Simple. The solution is to finish the video and re watch it with the solution in mind. Boom. Case closed.

Not clear enough; bob cant trap alice, only visit all squares (optimaly), bob always wins

@supasf

6 жыл бұрын

IU ER I guess the person who makes the first move always loses?

@XenophonSoulis

6 жыл бұрын

I played this game by moving both players myself and I managed to trap Alice in just 13 moves by each player.

@stevenattanasso2003

5 жыл бұрын

Stop playing with Yourself and then touch the chess pieces .... You are gross .....

@XenophonSoulis

5 жыл бұрын

No, I played with a chessboard designed in the Paint of the Windows. I didn't touch any pieces. And why am I gross???

@stevenattanasso2003

5 жыл бұрын

OK , it was an attempt at humor .... " playing with myself " is a metaphor for masturbation in English , I don't know if English is a second language for You ( if so , You speak it well ) or if You are "playing" with Me now but either way have a great life ..... ( and get a Girlfriend )

That's a fun game and challenging problem. I didn't solve it even though I knew about the "knight's tour". What this video lacked was a discussion of how one might come to the idea of breaking the board into 2x4 regions.

@poro3246

Жыл бұрын

it's the only way to split the board in equal portions to which this rule of unique pairs applies. (of course you can swap 4x2 and 2x4). The rule also applies to 3x2 but you can't split the board into only 3x2s.

@tomdekler9280

Жыл бұрын

@@poro3246 there's no matching pair for the middle two squares of a 3x2 region.

The problem with this solution is that a knight doesn't have to move 2 squares first then one square. The knight can move 1 square then 2 squares. So using the grid from above the knight can move one square up to purple for example and then over 2 squares landed in another grid also putting the knight in a color square it didn't start in. I don't know if the outcome would be different as you can still use all the squares in the grid this way but after 2 turns Bob would be locked out of using the same color squares method cause he couldn't move back to yellow.

Mega FacePalm to this trick.... 😫😫😫😫😫😫😫😫

Hello intellectuals how be ye

@filmishit

6 жыл бұрын

יוחנן דוד pretty good matey how bout t'you?

@user-eo5bh2zg2f

6 жыл бұрын

I'm doin swell!

@matanrubin8726

6 жыл бұрын

מאה אחוז

I came up with the right answer, but with much less formal reasoning (but did not come up with an optimal strategy.) There are an even number of squares that can be occupied. The first player takes odd moves, the second player takes even moves. It's possible to occupy every space on the board. Bob wins by forcing an even move victory. Alice wins by forcing an odd move victory. There are more even move solutions than odd move solutions. Since, there is no backtracking, Bob has more control over the sequence the knight proceeds through the board. All Bob needs to do is prevent leaving the Knight in a location that leads to an odd turn victory. The caveat is that I'm not 100% sure I've proven the board starts in an even turn victory solution. It makes intuitive sense to me, and I find it difficult to imagine it could be wrong. But I'm unsatisfied with just "the longest game is this way, so it must be the default." But ultimately, as long as that assumption is correct, the concept of Bob being able to maintain advantage should hold true.

Thats nutty and a really clever solution.