# Can You Solve The Knight On A Chessboard Riddle? Math Olympiad Problem

## 100 thoughts on “Can You Solve The Knight On A Chessboard Riddle? Math Olympiad Problem”

1. MindYourDecisions says:

The great Simon Singh tweeted this video! Here's the tweet: https://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!

2. Revibe DF says:

Or bob went firstš¤£

3. Talking Mango says:

Or you can write a program to brute Force this in your programming class

4. k says:

A GAME THEORY

5. TurboJ says:

I beat this riddle when I was 14.
Professor Layton was a thing. Had the exact puzzle.

6. JOSH_Went_Kapow says:

"But That's Just A Theory! A Game Theory!"

7. SierraRain says:

Did Anyone Else Notice He Messed Up A 4×2 He Did 4 Vertically When Its Supposed To Be 4 Horizontally

8. The Red Hood says:

I figured it out that the 3nd player will always win but i didnt really think too complex lol

9. 1,000 subs with no videos says:

I don't even understand the question rigjt

10. Neville Lamberti says:

Not very interesting.

11. Itachi's Will says:

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

12. xDennNiex says:

so the rule is "the knight can not go were it has been on before" but, in the explaination, at the 4th move it moves were it has been at the start. so either the rule is insufficient or the explanaition is wrong.

13. One in a billion says:

Took me a while to think why Alice is forced to new 4×2 Square lol. Very nice

14. Jake Rakazak says:

The optimal strategy would be
1. Alice places the night in any 4×2 region,
2. Bob moves the knight to the only legal position in that same 4×2 region,
3. Alice is forced to move to a new region
4. Bob moves back to the first 4×2 region,
5. Alice does step 2.
Continue.

15. Deusdekis says:

You explain the problem as if the targeted audience was a bunch of 5yo kids

16. Ian Wu says:

Nope bob doesnt always win

17. CarlJo :3 says:

Who here actually tried to solve the problem before continuing the video

18. LynxDDragon says:

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

19. letter h says:

nice thing to program an ai to do this

20. Caleb Cali says:

Mind = Blown

This only implies when Alice starts the game

22. Ashish Sharma says:

Why wont alice win? You sexiest piece of mathematician

23. valerian o o o says:

8.000th like. Nice video

24. LOL _cryz says:

But hey!!! THATS JUST A THEORY, A GAME THEORY

25. Nandakishor K says:

This is trivial

26. Albert Kelvin says:

My question is:
1. is there a 50% chance that Alice will ever win? because if not this is not a game. You can call it a game when everyone have a same winning chance.
2. If Bob does not move his knight in the 4×2 region, but seeking a new 4×2 region, did Alice can win?

27. wyeth sutton says:

You could have just said bob ends in an even number my dude you also should have made the vid 10 mins

28. Raul QuindosMorales says:

There are 64 squares. If Alice starts, she will start with 1, then Bob with the 2 move, and so on and so forth. Alice will represent the odd numbers, and since the last number is an even number, Bob will have the last turn, making Alice loose. My math might not be completely applicable, but the answer is a 100%

29. Kinny says:

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

30. MZ107 says:

Of course Bob wins, he is a man.

31. The Neon says:

No one can beat bob

32. RQ5-Hunter says:

This was awesome solution. It would be fun to challenge friends to this knowing that I will always win if I can start tho to make it fair we would need to make turns who starts so I need to also come up with alternative strategy of swaying the tide of the game and blocking some squares. The hardest part tho is to remember which squares we used already, there would need to be some kind of marking every time you visit one..

The solution I was thinking of is that because there is 64 squares, when one places the knight there are 63 possible squares left and if they would both play optimally, they would use ALL the tiles so that the one who puts the knight to the board would always lose. The problem in this is tho that of course the other player would try block this from happening, but I don't know if it's possible or not to be able to do so if they both play optimally. This would need more thinking and/or trial-and-error method to think about more and I'm too lazy to do that but there's a problem for someone else to solve if they want to. š

33. Bob Babas says:

Can someone prove this mathematically?

34. Jugvir Sidhu says:

Canāt she just go back to the same square?

35. Aaron Turner says:

Bob can lose if he deviates to a new region. Alice can then take on the legal square method and win.

36. Fluffy Fwaff says:

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.

37. Justin Didderen says:

Actually this is completely wrong. If Bob made the first move, alice would of won. It's like tic-tac-toe. The person who made the first move has a strategical advantage.

38. Frankie Kimbrell says:

Thanks for your videos, Fresh Towel Locker.

39. Augustine J. Contreras says:

Are you sure its not a you tube channel

40. Chris Dalton says:

Orrrr you can just not make the first move and then youāll be the person to make the last possible move. Just donāt go first!

41. Tim Barnett says:

I thought of a math visual teaching aid while sitting at my desk with a headache. I know it works but NOT WHY, think you could help me understand? Put both hands in front of you, palms facing you an fingers facing each other but not touching. Make little fingers=6, ring finger=7, middle=8, first/pointing finger=9 on both hands. Example: just touch two 9 (first/pointing) together. (2 fingers touching an all fingers below are valued at 10 each). So two 9 fingers plus all below (6 fingers) equals 8 finger times 10 each for value of 80. Now you have two thumbs (one thumb on each hand) above two fingers touching. MULTIPLY what's above to other hand. ( this case we have one thumb on one hand MULTIPLY one thumb above fingers touching on other hand) 1 thumb times (X) one thumb equals ONE plus 80 is 81 (correct, seems difficult here but very easy). Second example: touch two middle (8s) fingers together. You have two fingers touching plus four below finger touching for tens value of 60. Now MULTIPLY two fingers above to two finger touching on other hand for 4 total= 64 (8X8=64). Third example: 6X9 touch little finger (6) to first finger (9). Two fingers touching an everything below (valued 10s) three below for 50. Now MULTIPLY fingers above to other hand 1X4=4 total (6X9=54) this sounds hard but it's just a visual identity that multiplies all the six through nines. (Note: this even works for tens, try it: two thumbs together for two touching and 8 below for 10X10=100 with no fingers above so no one's! Try this is saves elementary school students much effect by just putting hands in front of them for identity! BUT WHY DOES IS WORK FOR SIX THROUGH NINE AN what other operations like this could function? [email protected]

Doesnāt this only take into account a game where bob goes first?

You talk too slow

44. Sanctum Sanctorum says:

I donāt know how I came up with the solution.
But I did it the exact same way!

I suck at math!!

I have always loved chess and have played since I was 5!!

I divided the board into 8 4×2 squares just as they did here and realized that the second player will always win as there are an even number of moves to be made in the game before it ends!

I feel like a flipping genius right now!!!!!!

š„³š„³š„³š„³š„³š„³š„³š„³š„³š„³š„³š„³

EDIT: I think the fact that I solved it and didnāt need to pause the video was why I am so ecstatic right now!!

š¤

45. Slumpy says:

Pee pee poo poo man

46. Bhavy Doshi says:

"Bob can always win, no matter how Alice plays"
That's sexist broš¤£

47. Darren Bonner says:

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.

48. joops110 says:

Bob wins because he's a man. No contest.

49. oldschooldiy says:

My solution was much simpler! If the object of the game is to always find a new square, then Bob will always win because Alice has fewer squares to play with! As she starts the game by placing the knight on a square, she will always have fewer squares to move to!

50. Ardaffa Randra says:

What if Bob mistakenly move into a different 4×2 region and Alice knows this strategy?

51. Ballist Exx says:

I tried for 3 months to solve this puzzle… 3 MONTHS!!! šŖš£

52. Matthew Hernand says:

iām doing my IB Math SL IA on this topic

53. Cobbler says:

If the startkng square counts this doesnt work

54. Harold Kid says:

Anyone else think he sounds kind of like Kermit the Frog šø

55. irfan naufal says:

bob always playing against noob

56. Swapnil Katakwar says:

So Alice has to go first in order for bob to counter move? Or else it wonāt work

57. Finchbird 1 says:

I found a way for alicia to win.
She starts the game on the bottom purple square then Bob uses the corresponding color for his move then she moves to a new section but what if Bob then moves to the blue square in the previous section and then alice goes to the other blue square from this point Bob is now forced to always move to a new section

58. FaZe Finna Nut says:

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

59. GeneralMerten says:

i fiddle it out this riddle is sexist. The girl always loses Lol XD

60. Sniper Thief says:

"Bob will always win no matter how Alice plays"
T r i g g e r e d f e m i n i s t s i n c o m i n g

61. Ophis says:

Maybe i'm wrong (or i missed something) but you left out some info, that the original square the knight starts on can be visited again as you do that in the example.
For me "you can't visite a square again" implies the first position before people do a move is also one (which would implie the oposite result)

62. K H says:

This story is so confusing… who is Bob again??

63. Sener S says:

64. Clariphi says:

Just play your turn once, then don't play again. Since she can no longer move, you win.

65. Swinde says:

So you are saying that whoever takes the first move and their opponent using this strategy wins? Can the first mover win?

66. aroze123 says:

was ur 666th dislike. thank me later

67. Autarch says:

I'd like to see the solution where you move 65 times on a 64 square chess board without being on a square twice.

68. RiceSalt says:

if bob would be the 1st one to put the horse and alice accidentally did 1 legal move (inside 4×2). can bob counter it if he change it to 2×4 box strat? can anyone confirm?

69. DRY says:

But thatās just a theory… a GAME THEORY

70. Rikard Nilsson says:

I could…but it's too much work so I'm just going to wait for the answer.

71. Kasumi Ninja says:

Me? I'm just here searching answers

72. NCERT TV says:

Kuch smajh me nhi aaya

73. lieutenant smackahoe says:

Who else went straight to the solution

74. name2 says:

But that's not a theory.
IT'S A GAME THEORY!
Thanks for watching.

75. arthur brosens says:

This is a really cool trick

76. Magic Domo says:

Hey I hate to break it to you but bob can't do that last move you showed. That's were the piece started. If it happened like this then alic would get the upper hand and Bob would loose.

77. MrTR909 says:

Why is the same logic not applying to Alice. She can use the same tactics by 90 degree rotation the 4×2 box. Shouldn't we see a draw, or is the advantage on the beginner side?

78. Querterus says:

There is just 1 problem to your solution. The knight start at 1 spot and for example in your showcase Bob could not usw that place because that was the starting place for the knight. (at 5.45 you moved it to the place were it should start befor Alice make her first move)

79. Josh Boyd says:

The solution is actually so simple I thought it was too easy to be right… literally choose one of the 8 squares. Bob goes first, so one of the squares will be taken. From there, just count the squares off naming them Alice and Bob, then whoever has the last square wins.

80. Alex Kramer says:

But why would that work? After those 8 regions are gone only 16 tiles were used and thus there are still plenty of opportunities to place it? What is beneficient for bob to stay inside it?

81. BuddyBaac says:

If you follow the example moves the video shows, the moves end when Bob actually moved the chess piece back to where it started. So technically bob lost because the piece went into a space it was in before.

82. Daren Gauthier says:

In 65 or less moves?, I think I solved it in 4 moves. Six tops

83. Joseph Williams says:

What if the game starts with the piece already the board? Like, yāknow, a game of chess does.

84. Alex Mocuta says:

What if Alice starts by moving to the corect colorcoded square, then Bob must enter another zone. What will it happen?

Surprisingly got it in 2 minutes

86. PrinzEugen says:

Well, this riddle is miles ahead from the entire Bright side riddles and 7 sec riddle channel

*Edit: i wrote that comment before watching entirely. In the end, the solution is really good

87. Mathijs Hertgers says:

But how many turns does it take for Bob to win?

88. eles horea says:

Math say that Bob can win, But don't forget that math say that if you flip a coin it will be 50% chance for eighter side to be up at landing, But we all know that is not true. So, can YOU really win playing like Bob did?

89. Angela Hsiao says:

Wow, the solution is so simple once itās explained, even though the question itself seems so difficult to prove

90. Nat Man says:

If both sides play optimally the game will end in 33 moves as the knight always moves to the opposite colour and there are nly 32 squares of each colour

91. Frank Erdman says:

This is a good model for a pyramid sceme – in any given metropolitan area, the schemer sells her products to a set of people (the first knight square visit) and then that set of people sell the product to a set of their friends in the same metro area (the second knight square visit). After this, the metro area is inundated and the schemer has to move on to another metro area. Unlike a chess board, though, with a big enough board (metro areas) she can eventually come back (say in 10 years) to her original metro area because by then there will be a new set of suckers to be had. This is how many pyramid schemes keep on trucking decade after decade. We prove (mathematically) that provided there are enough metro areas, and a certain "decay rate" of how "used up" a certain metro area is, that you can actually do a pyarmid scheme indefinitely. Goodness, I like math!

92. Dragic Magic says:

But that's just a theory, a game theory.

93. Jorge NuĆ±ez says:

But what if Alice goes into a 4×2 region with all the square used? Is that possible?

94. Simeon BorovanskĆ½ says:

I am good at chess and math so easy

95. Der Moos says:

Brilliant!

96. Andreas Foerster says:

Great problem and solution!

97. Ruige Verfaillie says:

The move Bob makes is not a legal move…

98. Ryan Ye says:

change this knight into a queen and what will happen?

99. Swamyveerandra Gopi says:

Why can't Alice move knight to old 4×2 region finding new colour in old 4×2 region? If this happens then proof is incomplete

100. éŗ„åå¤ says:

This isn't a complete proof.
Let's name the grids by using coordinates. The left bottom is (0,0) and the right top is (7,7).That means the range of x and y is 0 to 7. Which means the strategy mentioned would only work in this range. The example in the video shows that Alice chose (1,5) and so Bob chose the corresponding place which is (0,7). The strategy is whenever Alice moves as (x1+a,y1+b), Bob moves as (x2+a,y2+b) onto Alice's corresponding place. But how about Alice moves from (1,5) to (3,6)? Alice moves as x+2,y+1, so Bob will move to (0+2,7+1) which is (2,8). But this exceeds the edge of the game board. In fact (2,8) is not existent. Therefore this strategy is unusable in this situation.

Actually there must be a strategy which can make one side always win. And every game which doesn't affected by fortune must have a strategy that can lead to inevitable winning for one side. I think although the proof is not completed, it is already very close. The last step is to find out another strategy to deal with the previous problem and make the method in video work again.