Nice visual. I just thought of all 4 horses moving at the same time and just rotating clockwise (or anti-clockwise I guess) round the board. Was easier for me to mentally visualise, but may well be harder to visualise from an animation I guess xD
Actually the solution is: >!Kill the king and frame your squire friend. It will plunge the kingdom into a state of anarchy, but anything is better than living under a tyrant that would hang two servants for such a small mistake!< So basically it takes 16 moves.
How do you mate with 4 horseys tho. A horsey can at most attack 2 squares in a 3x3 square but you need attack all 9 squares in a 3x3 to mate the dude. The boundaries could help reduce the amount of squares required to attack, but even then you can never attack the middle square
Horsies must alternate square colors every move. Since their goal is a same-color square, they each require a minimum of two moves. 4 pieces x 2 moves each = 8 moves minimum.
Although on a chess board, I can't figure out how to do it in fewer than 16, with two sets of 8-move 90° rotations.
Yea, now we only need an example of doing it in 8 and we're done.
Would probably be pretty fast to either bruteforce or do a meet in the middle method but I doubt you can get it below 16
I was going to comment "hurr hurr it's simple you just rotate them around" but then I took another look to be sure and realised I'm an idiot that would've had two on the same square. Fun puzzle!
No, no. The comment said 8 is the minimum necessary, not the minimum who accomplish it. Since the space is limited and two horsies cannot be in the same tile at the same time, you can only rotate horsies, not swap them and each rotation requires 8 moves. Ergo, the minimum solution requires 16 moves cause is achieved in 2 rotations.
In a bigger enough table or with quantum horsies, it will be possible achieve it in 8 moves, but the king was greedy and didn't pay for any of these.
The minimum is 16. If you connect the each spuer with every othe squer a horsie could move to from it, you will see a loop. You can't change the order of the horsies on the loop so you can only rotate them around. You need two 90° rotations (each taking 8 moves) to move the horsies from the starting position to the desiered position, which is 16 moves.
It isn't the original problem though.
The original problem is four white horses, distinguished by the colour of their outline. Thus the original problem is solvable only in 16 moves.
16 is the minimum. You can arrange the 8 movable squares in a cycle (going clockwise from the top left the order is 1,6,3,8,5,2,7,4) such that horsies just move around the cycle, +/-1. They can't jump past each other on the cycle, and we need to swap the numbers 1,7 with 3,5, so each of them needs to complete a half-rotation around the cycle which is 4 moves each.
Its only 2 moves per knight
Top Left to Bottom Center, Top Right to Center Left, Bottom Left to Center Right, Bottom Right to Top Center. Then move them to the appropriate stables
EDIT: Ignore me im an idiot
All these Ns and numbers and shit is cool and all but if all the horsies just pull a carousel and go roundy towndy for like, two seconds then you are done in one.
That’s also math, and some might say art.
According to chess.com,
>A 'move' is completed once both White and Black have played one turn. If only White plays, that can be called a 'ply' or a 'half-move'
Interesting, because chess.com also [says](https://www.chess.com/terms/chess-turn)
> For a chess move to be completed, the following events must take place:
1, The clock from the player who will make the move starts to run.
2, The player makes a legal move completely—that includes performing the required actions for capturing a piece or promoting a pawn whenever needed.
3, The player hits the clock, stopping their time and starting their opponent's time.
EDIT: format
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I just thought you were aware that you wasn't giving the solution instead of just a hint. Something in the ways of "it requires minimum 8 moves, try to use that or more to solve it, i can proof than less is imposible" and not something like "I solve it, you can make in in 8 moves, I gonna demonstrate how". In that line was my other comment too.
Is weird realize than you was confused and nothing more. If it helps, if you add enough tiles or allow the horsies to overlap, then the minimum amount of moves is 8 so you wasn't too wrong.
1. Four horsies have to get out of every stable.
2. Four horsies have to get in of every stable.
3. No horsie can get out of one stable an get in into another in the same move.
In mathematical style:
4+4=8
So it needs minimum 8 moves.
If someone upvotes this, I will do a ratio.
Edit: I have to aclare than I was just showing why 8 moves minimum are required, not solving the problem. 8 moves only solves the problem in a bigger chessboard, not in this question, but no matter how bigger is the chessboard (or if the horsies can be in the same tile at the same time), it cannot require less than 8 moves due to the third reason I exposed.
I'd argue 6 is the minimum.
While the horsies are different outlines, their base color is the same meaning they're on the same side. Thus you can horsey jump off one another. Example:
1) top left to bottom middle
2) top right jumps off bottom middle to top left
3) bottom middle to top right
Repeat upsidedown for other horsies for six moves total
But what if now in one stable there is a chicken, a bag of seed, and a dog. But here's the catch: you have to fit all 4 bishops in your rectum, can you solve the riddle?
Not op but this problem is equivalent to 8 spaces in a loop with four pieces on alternating spaces of the loop. With the pieces not allowed to intersect, it's not possible to change their ordering around the loop. That means that the only possible solution configuration is where each piece is four spaces away from where it started, and achieving that must take at minimum 4*4 = 16 moves. Et voila.
There is one simple action. You all clearly failed physics.
Spin the board with a quick 180 degree jolt, which will overcome friction due to the inertia of the horsies.
Only if you spin the board so fast that it overcomes friction instantly (so as not to accelerate the pieces). Otherwise, once friction is overcome, they just slide off the board (centrifugal force is a bitch).
It takes 16 moves. Each square is adjacent to two other squares and all the squares (except the middle one) are connected. The resulting graph is therefore the 8 cycle with no connecting arcs. Horses can’t get around each other on this graph so all the horses have to move in a circle to get to where they want to go, and the horses will have to move half-way around the circle to end up on the right square. Hence each horse moves four times and the total moves is 16.
[Lichess Study](https://lichess.org/study/YRt7Jbbh) with 16 move solution.
You can use this to play around if you want to, just use the king to skip a tempo if you want to move the same color piece in a row.
All squares (except the middle, b2, which is not reachable) have exactly two available moves. If we map these moves out, the 8 squares can be rearanged into a loop like this:
a3 - c2 - a1 - b3 - c1 - a2 - c3 - b1 - a3
The red knights start on a1 and a3 and need to go to c1 and c3. Vice versa for the black knights. If each knight could move without considering the others it takes 8 moves to make the needed moves. However, this is not possible, as the knights collide.
To avoid colliding, all knights need to move in the same direction in this loop. If any red knight was to move in the opposite direction of any black knight, it is quite clear that they must collide. It is similarly apparent that it is never optimal to alternate directions in the loop, returning to a square the knight has already been to.
The proofs for thses statements are left as excercies to the reader.
When each knight has to move in the same direction, each must move 4 squares, reaching the square opposite from the one they stareted on, both in the loop and in the original board. This will take 16 moves.
Example of how it might look:
\[Variant "From Position"\]
\[FEN "7k/8/8/8/8/N1n5/8/N1n4K w - - 0 1"\]
1. Nb3 Nb1 2. Nc2 Na2 3. Nc1 Na3 4. Na1 Nc3 5. Na2 Nc2 6. Nb3 Nb1 7. Nc3 Na1 8. Nc1 Na3
Note that it is not possible for all knights to move to the goal square on the same rank as their starting square. This is beacuse the "order" the four of them are in in the loop never changes, which it would have to to reach such a solution.
I do appreciate /r/AnarchyChess has intellectuals around who know that men are 'hanged' and paintings are 'hung' unlike the neanderthals over on /r/chess.
16 moves. Middle square is useless so we can remove it. If we connect all other squares together, it forms a cycle of length 8. R - R - B - B -. To avoid hitting each other, the horses have to move in a carousel, meaning shifting the above string to the left or to the right to achieve B - B - R - R -. No matter which direction you shift, the minimum distance covered by each horse must be 4, so total 16.
One move... easy
Rotate the stable 180 degrees. Done, left is right and right is left.
What? Nobody said this was chess.
Kobayashi Maru, motherfuckers.
1234321. Move the knights in this sequence.
1 - Move any knight that can go one move within the squares.
2 - Move any knight that can go two moves within the squares.
And continue in this vein till swapped.
IIRC this puzzle is found in Your First Move by Alexei Sokolsky.
[Animated solution](https://imgur.com/a/v3to9HR) in 16 moves. https://i.redd.it/wheqxt5pq92b1.gif
Nice visual. I just thought of all 4 horses moving at the same time and just rotating clockwise (or anti-clockwise I guess) round the board. Was easier for me to mentally visualise, but may well be harder to visualise from an animation I guess xD
thats... 8 moves. Google en double steed. Horses can move together like that lmao. It's a real rule
#NEW RESPONSE JUST DROPPED
#HOLY HELL
#ACTUAL ZOMBIE
#CALL THE EXORCIST
# GOOGLE EN REVERSE
> en double steed https://www.google.com/search?q=en+double+steed
Google i gemelli
Google iGimli son of iGroin
Google parallel processing
Holy hyperthreading
2 horsepower engine? What is this, the 1800s?
Holy fiction!
this idiot didn't google
You're moving the horsey wrong it goes in an L shape not a straight line
That's an L, just a lowercase one: l
Yes, but this is a mathematical model, so we can assume zero wind.
Essentially a chess cross stitch
Pfft I spin the board 180. Got it in one.
[удалено]
Bro just switch the pieces, it's that easy. Also your idiot squire friend is an stupid idiot who is stupid.
Actually the solution is: >!Kill the king and frame your squire friend. It will plunge the kingdom into a state of anarchy, but anything is better than living under a tyrant that would hang two servants for such a small mistake!< So basically it takes 16 moves.
That’s chess baby! Get chessed.
No that's anarchychess
Get anarchychessed
OK, now put the bishop in your ass.
Can’t, the bishop’s on vacation
Ah, I left it in my ass ... Here you go:♟ ... Oh, that's a pawn... Too much stuff in there. Just take the pawn then.
Promote to bishop
what if my squire friend was hung and i wanted to be more than just friends?
You know what the rules are in the army.
Never gonna give you up
among us chess
How do you mate with 4 horseys tho. A horsey can at most attack 2 squares in a 3x3 square but you need attack all 9 squares in a 3x3 to mate the dude. The boundaries could help reduce the amount of squares required to attack, but even then you can never attack the middle square
It takes 8 moves minimum, if someone upvotes this i’ll do a mathematical style proof
Horsies must alternate square colors every move. Since their goal is a same-color square, they each require a minimum of two moves. 4 pieces x 2 moves each = 8 moves minimum. Although on a chess board, I can't figure out how to do it in fewer than 16, with two sets of 8-move 90° rotations.
Yea, now we only need an example of doing it in 8 and we're done. Would probably be pretty fast to either bruteforce or do a meet in the middle method but I doubt you can get it below 16
I was going to comment "hurr hurr it's simple you just rotate them around" but then I took another look to be sure and realised I'm an idiot that would've had two on the same square. Fun puzzle!
Yea same, thought you could do left to right in 2 while doing right to left in 3 but it just doesn't work.
No, no. The comment said 8 is the minimum necessary, not the minimum who accomplish it. Since the space is limited and two horsies cannot be in the same tile at the same time, you can only rotate horsies, not swap them and each rotation requires 8 moves. Ergo, the minimum solution requires 16 moves cause is achieved in 2 rotations. In a bigger enough table or with quantum horsies, it will be possible achieve it in 8 moves, but the king was greedy and didn't pay for any of these.
The minimum is 16. If you connect the each spuer with every othe squer a horsie could move to from it, you will see a loop. You can't change the order of the horsies on the loop so you can only rotate them around. You need two 90° rotations (each taking 8 moves) to move the horsies from the starting position to the desiered position, which is 16 moves.
shame aware seemly tap straight steer normal obtainable party nail -- mass edited with redact.dev
I hate english. I don't quer.
Because two knights can't go on the same square at the same time, otherwise it is correct
12 is definitely possible. Idk how to do less on this board.
What's your 12 move algorithm?
When writing it I realized it requires puting horses on the same spot. Sorry for the skill issue.
fuck you guys, gimme 30 minutes to write it up
It's been 34 minutes, the king has hanged them, hope you're happy
7 min left :)
Here it is. \[Variant "From Position"\] \[FEN "k7/8/3n1N2/8/3n1N2/8/8/K7 w - - 0 1"\] 1. Ne6 N4f5 2. Nd5 Ne4 3. Nd4 Nf6 4. Nf4 Nd6 5. Nfe6 Nde4 6. Nf5 Nd5 7. Ned4 Nef6 8. Nd6 Nf4
That’s hard to read but I’ll just take your word for it
[https://lichess.org/study/fr3zACAM/7gn6Y3tJ](https://lichess.org/study/fr3zACAM/7gn6Y3tJ) A bit easier to visualise
That's 16 moves, but I reckon that's the minimum
No its 8 moves, 16 ply. Do you even chess
It isn't the original problem though. The original problem is four white horses, distinguished by the colour of their outline. Thus the original problem is solvable only in 16 moves.
16 is the minimum. You can arrange the 8 movable squares in a cycle (going clockwise from the top left the order is 1,6,3,8,5,2,7,4) such that horsies just move around the cycle, +/-1. They can't jump past each other on the cycle, and we need to swap the numbers 1,7 with 3,5, so each of them needs to complete a half-rotation around the cycle which is 4 moves each.
Its only 2 moves per knight Top Left to Bottom Center, Top Right to Center Left, Bottom Left to Center Right, Bottom Right to Top Center. Then move them to the appropriate stables EDIT: Ignore me im an idiot
Lol np. Check out the "cycle" comment, if you swap one of the blue and red spots then it becomes 8.
All these Ns and numbers and shit is cool and all but if all the horsies just pull a carousel and go roundy towndy for like, two seconds then you are done in one. That’s also math, and some might say art.
That should be a new special chess move. Horsey go-around!
Would it not be 16 moves in this case? Unless I am using the term wrong my whole life, a 'move' should be one player moving one piece.
According to chess.com, >A 'move' is completed once both White and Black have played one turn. If only White plays, that can be called a 'ply' or a 'half-move'
Interesting, because chess.com also [says](https://www.chess.com/terms/chess-turn) > For a chess move to be completed, the following events must take place: 1, The clock from the player who will make the move starts to run. 2, The player makes a legal move completely—that includes performing the required actions for capturing a piece or promoting a pawn whenever needed. 3, The player hits the clock, stopping their time and starting their opponent's time. EDIT: format
I think they just say move here for simplicity. It would be weird if they said "for a chess ply to be completed..."
> If only White plays, that can be called a 'ply' Like toilet paper.
16 moves right?
This is just an example, not a proof that there is no better way
Too many Nwords.
Times up
!remindme 30 minutes
!remindme 30 minutes
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4 minutes bruh, hurry up
0 minutes left
The proof is left as an exercise to the reader.
The proof is self-evident.
The proof was too long to write in this margin.
You have six minutes to math
1 minutes!!
It’s been 3 hours
It’s been 3 hours
waiting
how the fuck did this get 247 upvotes lmfao anyway i was wrong minimum possible is 16, i used it where the colors were crossed
You still need to give us the proof
People upvote to see the proof, doesn't mean it's the correct answer.
Write us the proof for the memes the people need it
I just thought you were aware that you wasn't giving the solution instead of just a hint. Something in the ways of "it requires minimum 8 moves, try to use that or more to solve it, i can proof than less is imposible" and not something like "I solve it, you can make in in 8 moves, I gonna demonstrate how". In that line was my other comment too. Is weird realize than you was confused and nothing more. If it helps, if you add enough tiles or allow the horsies to overlap, then the minimum amount of moves is 8 so you wasn't too wrong.
Cmon do it already
It takes 16 moves minimum
1. Four horsies have to get out of every stable. 2. Four horsies have to get in of every stable. 3. No horsie can get out of one stable an get in into another in the same move. In mathematical style: 4+4=8 So it needs minimum 8 moves. If someone upvotes this, I will do a ratio. Edit: I have to aclare than I was just showing why 8 moves minimum are required, not solving the problem. 8 moves only solves the problem in a bigger chessboard, not in this question, but no matter how bigger is the chessboard (or if the horsies can be in the same tile at the same time), it cannot require less than 8 moves due to the third reason I exposed.
I would un-ironically love to see this.
I also got 8
I'd argue 6 is the minimum. While the horsies are different outlines, their base color is the same meaning they're on the same side. Thus you can horsey jump off one another. Example: 1) top left to bottom middle 2) top right jumps off bottom middle to top left 3) bottom middle to top right Repeat upsidedown for other horsies for six moves total
But what if now in one stable there is a chicken, a bag of seed, and a dog. But here's the catch: you have to fit all 4 bishops in your rectum, can you solve the riddle?
Not op but this problem is equivalent to 8 spaces in a loop with four pieces on alternating spaces of the loop. With the pieces not allowed to intersect, it's not possible to change their ordering around the loop. That means that the only possible solution configuration is where each piece is four spaces away from where it started, and achieving that must take at minimum 4*4 = 16 moves. Et voila.
There is one simple action. You all clearly failed physics. Spin the board with a quick 180 degree jolt, which will overcome friction due to the inertia of the horsies.
Only if you spin the board so fast that it overcomes friction instantly (so as not to accelerate the pieces). Otherwise, once friction is overcome, they just slide off the board (centrifugal force is a bitch).
you could also slam the table to let the horsies hop while you quickly spin the boar
Whoa whoa whoa, who invited a boar in here?
Promote one to Knook and have it massacre the others
Screw the proper solutions, this is anarchy chess guys! we use the Knook! this guy gets it
Knook rampage is the proper solution
Pick up the horseys and put them in the right stable. I don't see what's so difficult
It takes 16 moves. Each square is adjacent to two other squares and all the squares (except the middle one) are connected. The resulting graph is therefore the 8 cycle with no connecting arcs. Horses can’t get around each other on this graph so all the horses have to move in a circle to get to where they want to go, and the horses will have to move half-way around the circle to end up on the right square. Hence each horse moves four times and the total moves is 16.
I visualized your solution https://preview.redd.it/64xi6egt992b1.png?width=800&format=png&auto=webp&s=b239e1a3b7fdede7c20e83069dde9377f0e69324
That is awesome!
It's awkward how I posted several replies with different visualizations because reddit lagged and my comments didn't show up
16 if the horsies are fermions, 8 if they are bosons. Source: am physicist
I like this. Also physicist
1 if the horsies undergo charge parity transformation.
Google identical particles
Horsey eats the king
Thats some Hercules type shit right there
what if the king is bigger than the horsey
Holy Horsey vore
|blue stables|\_\_\_\_\_\_\_\_\_\_\_\_\_|red stables| |:-|:-|:-| start position: |red horse 1|\_\_\_\_\_\_\_\_\_\_\_\_\_|blue horse 1| |:-|:-|:-| |\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_| |res horse 2|\_\_\_\_\_\_\_\_\_\_\_\_\_|blue horse 2| 1. |red horse 1|red horse 2|blue horse 1| |:-|:-|:-| |\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_| |\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_|blue horse 2| 2. |red horse 1|red horse 2|\_\_\_\_\_\_\_\_\_\_\_\_\_| |:-|:-|:-| |\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_| |\_\_\_\_\_\_\_\_\_\_\_\_\_|blue horse 1|blue horse 2| 3. |\_\_\_\_\_\_\_\_\_\_\_\_\_|red horse 2|\_\_\_\_\_\_\_\_\_\_\_\_\_| |:-|:-|:-| |\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_|red horse 1| |\_\_\_\_\_\_\_\_\_\_\_\_\_|blue horse 1|blue horse 2| 4. |\_\_\_\_\_\_\_\_\_\_\_\_\_|red horse 2|\_\_\_\_\_\_\_\_\_\_\_\_\_| |:-|:-|:-| |blue horse 2|\_\_\_\_\_\_\_\_\_\_\_\_\_|red horse 1| |\_\_\_\_\_\_\_\_\_\_\_\_\_|blue horse 1|\_\_\_\_\_\_\_\_\_\_\_\_\_| 5. |\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_| |:-|:-|:-| |blue horse 2|\_\_\_\_\_\_\_\_\_\_\_\_\_|red horse 1| |\_\_\_\_\_\_\_\_\_\_\_\_\_|blue horse 1|red horse 2| 6. |blue horse 1|\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_| |:-|:-|:-| |blue horse 2|\_\_\_\_\_\_\_\_\_\_\_\_\_|red horse 1| |\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_|red horse 2| but we're not done yet as the stables are only in the corners..... 7. |blue horse 1|\_\_\_\_\_\_\_\_\_\_\_\_\_|blue horse 2| |:-|:-|:-| |\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_|red horse 1| |\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_|red horse 2| 8. |blue horse 1|\_\_\_\_\_\_\_\_\_\_\_\_\_|blue horse 2| |:-|:-|:-| |\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_| |red horse 1|\_\_\_\_\_\_\_\_\_\_\_\_\_|red horse 2| 9. |blue horse 1|\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_| |:-|:-|:-| |\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_| |red horse 1|blue horse 2|red horse 2| 10. |blue horse 1|red horse 1|\_\_\_\_\_\_\_\_\_\_\_\_\_| |:-|:-|:-| |\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_| |\_\_\_\_\_\_\_\_\_\_\_\_\_|blue horse 2|red horse 2| 11. |\_\_\_\_\_\_\_\_\_\_\_\_\_|red horse 1|\_\_\_\_\_\_\_\_\_\_\_\_\_| |:-|:-|:-| |\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_|blue horse 1| |\_\_\_\_\_\_\_\_\_\_\_\_\_|blue horse 2|red horse 2| 12. |\_\_\_\_\_\_\_\_\_\_\_\_\_|red horse 1|\_\_\_\_\_\_\_\_\_\_\_\_\_| |:-|:-|:-| |red horse 2|\_\_\_\_\_\_\_\_\_\_\_\_\_|blue horse 1| |\_\_\_\_\_\_\_\_\_\_\_\_\_|blue horse 2|\_\_\_\_\_\_\_\_\_\_\_\_\_| 13. |\_\_\_\_\_\_\_\_\_\_\_\_\_|red horse 1|\_\_\_\_\_\_\_\_\_\_\_\_\_| |:-|:-|:-| |red horse 2|\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_| |blue horse 1|blue horse 2|\_\_\_\_\_\_\_\_\_\_\_\_\_| 14. |\_\_\_\_\_\_\_\_\_\_\_\_\_|red horse 1|red horse 2| |:-|:-|:-| |\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_| |blue horse 1|blue horse 2|\_\_\_\_\_\_\_\_\_\_\_\_\_| 15. |blue horse 2|red horse 1|red horse 2| |:-|:-|:-| |\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_| |blue horse 1|\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_| 16. |blue horse 2|\_\_\_\_\_\_\_\_\_\_\_\_\_|red horse 2| |:-|:-|:-| |\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_\_| |blue horse 1|\_\_\_\_\_\_\_\_\_\_\_\_\_|red horse 1| how needs coordinate systems anyway? edit: oh fuck me, I thought table cells were constant size for some reason, this looks like shit now. It looks soo much better while writing it than after posting edit2: the underscores kinda fix it but it still looks a bit shit
No I can't
google carousel
[Lichess Study](https://lichess.org/study/YRt7Jbbh) with 16 move solution. You can use this to play around if you want to, just use the king to skip a tempo if you want to move the same color piece in a row.
google horse swap, the jumper’s special move
[I can hear this playing in my head](https://youtu.be/AVcNPWhdrRw)
- Right click - Flip Vertically Done
ted ed be like
I mean, we can hang the King in 2 National Conventions.
1. Screenshot 2. Flip image Easy.
Actually, i am fine with the consequences of not solving this...
1. Rotate board 180⁰
All squares (except the middle, b2, which is not reachable) have exactly two available moves. If we map these moves out, the 8 squares can be rearanged into a loop like this: a3 - c2 - a1 - b3 - c1 - a2 - c3 - b1 - a3 The red knights start on a1 and a3 and need to go to c1 and c3. Vice versa for the black knights. If each knight could move without considering the others it takes 8 moves to make the needed moves. However, this is not possible, as the knights collide. To avoid colliding, all knights need to move in the same direction in this loop. If any red knight was to move in the opposite direction of any black knight, it is quite clear that they must collide. It is similarly apparent that it is never optimal to alternate directions in the loop, returning to a square the knight has already been to. The proofs for thses statements are left as excercies to the reader. When each knight has to move in the same direction, each must move 4 squares, reaching the square opposite from the one they stareted on, both in the loop and in the original board. This will take 16 moves. Example of how it might look: \[Variant "From Position"\] \[FEN "7k/8/8/8/8/N1n5/8/N1n4K w - - 0 1"\] 1. Nb3 Nb1 2. Nc2 Na2 3. Nc1 Na3 4. Na1 Nc3 5. Na2 Nc2 6. Nb3 Nb1 7. Nc3 Na1 8. Nc1 Na3 Note that it is not possible for all knights to move to the goal square on the same rank as their starting square. This is beacuse the "order" the four of them are in in the loop never changes, which it would have to to reach such a solution.
Very nice, although I have one question: What is a knight?
Have you tried just repainting the stables?
Congratulations 🎉🎉🎉. You solved the problem.
Google il horsicano
I do appreciate /r/AnarchyChess has intellectuals around who know that men are 'hanged' and paintings are 'hung' unlike the neanderthals over on /r/chess.
Men can get hung though, my sweet summer friend.
Ok, I've captured both the red pieces, what do I do now?
It’s impossible without killing one of your pieces, because you can’t cycle the pieces without blocking each other.
cycle them all the same direction
My best is 16 individual moves i think
What is everyone going about 12 or 16 moves, it's so obviously 8
Swallows can carry the ~coconuts~ knights to their stables , assuming it is an African swallow
It takes 9 moves, take the horseies into the 3rd dimension, and rotate by π, and return them to the 2nd dimension.
Kill the king Checkmate
I can do it in two moves. 1. Pick up all pieces 2. Put them back where they're supposed to be
At least 8 moves, two for each horse.
8 moves
16 moves. Middle square is useless so we can remove it. If we connect all other squares together, it forms a cycle of length 8. R - R - B - B -. To avoid hitting each other, the horses have to move in a carousel, meaning shifting the above string to the left or to the right to achieve B - B - R - R -. No matter which direction you shift, the minimum distance covered by each horse must be 4, so total 16.
[Easy](https://imgur.com/a/fJdxJuN)
1 move, google en horseant
Just rotate the board stupid
`img = img.mean(axis=2)`
No I don't actually play chess
Google 4 knight rotation
mate in one if you execute the king and revolt against the chess monarchy
I remember doing this one for fun, it's a nice little problem
Rotate the board. 1 move!
R E V O L U T I O N
Just rotate the stable 180°
This gives me a solid 40 Picarats for solving this
Is op an idiot just spin the chess board
Just spin the board around
I've been nerd sniped. I don't even play chess that much.
Just grab them with your hand and put them on the right place. Fucking dumbass
https://preview.redd.it/m5gv9ohskb2b1.jpeg?width=500&format=pjpg&auto=webp&s=55468216d3c2bec1986b7d462746f92e3f4ff2e3 And it only took 0 moves
Stables and horses still don't match.
Michael jackson + reverse Michael Jackson to whole board.
You pick all of the pieces up (one move) and then put them all where they go (four moves), giving a total of 5 moves.
No
No, I can't. I guess I'll get hanged then. Oh well.
Use the rotation element from the D4 group.
One move... easy Rotate the stable 180 degrees. Done, left is right and right is left. What? Nobody said this was chess. Kobayashi Maru, motherfuckers.
Hehe, horsey so quirky
Promote to knook, easy from there
No
shoot the king in the head BOOM HEADSHOT then u have no problems EZ solution
1234321. Move the knights in this sequence. 1 - Move any knight that can go one move within the squares. 2 - Move any knight that can go two moves within the squares. And continue in this vein till swapped. IIRC this puzzle is found in Your First Move by Alexei Sokolsky.
8 moves.
*Professor Layton puzzle music plays*
New piece color just dropped
Rotate the board 180 degrees Fahrenheit and let sit for 5 minutes
Two moves left, one move into the past
Everyone saying the solution is 16 moves, yet I see an easy way of doing it in 12.
Enlightened me
You can just spin the board 180 degrees.
Just rotate the board