[soop]

Another chess question.

On a regular chessboard, what is the minimum number of knights you can place so that each knight (using its regular movement pattern) is in check by another knight.

[Guest]

2--if A is in check with B, then B is in check with A...2 knights, both in check

Now if you meant they start with 4 in their original positions and we add more along the way, then that's a larger solution

[Guest]

Another chess question.

On a regular chessboard, what is the minimum number of knights you can place so that each knight (using its regular movement pattern) is in check by another knight.

Two?

Just 2 knights in a position to attach each other.

Or did I misunderstand the problem?

[soop]

Or did I misunderstand the problem?

Two?

Just 2 knights in a position to attach each other.

Oooops... I was going to make it:

What is the minimum number of knights you can place knights on a chessboard (using their regular movement pattern) so they fullfill the following critera:

1 - Each knight is in check from another night

2 - No knight cannot escape from check

But for some reason I considered them to render the same result. Does that one make more sense?

Sorry for being dumb :#

Edited August 28, 2008 by soop

[Guest]

Oooops... I was going to make it:

What is the minimum number of knights you can place knights on a chessboard (using their regular movement pattern) so they fullfill the following critera:

1 - Each knight is in check from another night

2 - No knight cannot escape from check

But for some reason I considered them to render the same result. Does that one make more sense?

Sorry for being dumb :#

Do you mean, perhaps, the MAXIMUM number that meet (1) and (2)? (2) seems to be a restriction that decreases the maximum, but doesn't affect the minimum. The minimum is still the same as the answers given above.

[Guest]

Well the fact that they are essentially 'checkmated' changes the minimum. You can choose where they are placed on the board, but you have to have it so every knight placed on the board can only jump to squares that can be attacked by another knight

[soop]

Do you mean, perhaps, the MAXIMUM number that meet (1) and (2)? (2) seems to be a restriction that decreases the maximum, but doesn't affect the minimum. The minimum is still the same as the answers given above.

No; if one knight is in the corner and in check from another knight, he still has 1 other square he can escape to. However, in this puzzle, escaping to that square will put him in check again.

So you start with one knight, and you need at least 2 knights to cover the first. But then you need to cover the other two knights, etc etc.

[Guest]

No; if one knight is in the corner and in check from another knight, he still has 1 other square he can escape to. However, in this puzzle, escaping to that square will put him in check again.

So you start with one knight, and you need at least 2 knights to cover the first. But then you need to cover the other two knights, etc etc.

So the puzzle you are asking should be:

What is the minimum number of knights you can place knights on a chessboard (using their regular movement pattern) so they fullfill the following critera:

1 - Each knight is in check from another night

2 - No knight can escape from check ie. all knights are checkmated

edit: clarity

Edited August 28, 2008 by foolonthehill

[soop]

So the puzzle you are asking should be:

What is the minimum number of knights you can place knights on a chessboard (using their regular movement pattern) so they fullfill the following critera:

1 - Each knight is in check from another night

2 - No knight can escape from check ie. all knights are checkmated

edit: clarity

yeah - I just read my post - I ****ed it up again O_o.

Yes, your post is EXACTLY what I meant. Apologies to everyone who's read this so far - can we get a mod to tidy this up? I can't edit the first post anymore.

[Guest]

_X_XX_X_

_X____X_

X_X__X_X

_X_XX_X_

_X_XX_X_

X_X__X_X

_X____X_

_X_XX_X_

In total, there are 28 Knights, with at least 2 knights covering 52 squares, and the other 12 squares cannot be reached by any knight on a standard jump

[Guest]

I like this puzzle (mayve because I was too late to get any of the HoustonHokie's previous chess problems!)

I think you're going to need quite a lot of knights, because every one on a black square needs at least one knight on a white square to put him in check.

I have an answer for stalemate (attached but can't 'spoiler' it), but think the checkmate solution might cover pretty much every square

Edited August 28, 2008 by foolonthehill

[Guest]

I don't know chess, so forgive me if I'm wrong in labeling these, but the 3 Ks that seem out of place (A6, C8, and G6--assuming it goes up alphabetically and across increasing numerically). I assume one filled one necessary square, so the other two blocked off possible jumps made by the others. If that's so, which was the one that needed correcting? And if it needed correcting on one side, does it need it on the other side, or not because no extra 'wall space' exists on the left?

[soop]

I don't know chess, so forgive me if I'm wrong in labeling these, but the 3 Ks that seem out of place (A6, C8, and G6--assuming it goes up alphabetically and across increasing numerically). I assume one filled one necessary square, so the other two blocked off possible jumps made by the others. If that's so, which was the one that needed correcting? And if it needed correcting on one side, does it need it on the other side, or not because no extra 'wall space' exists on the left?

THat part is fine because (as you say) of the "wall". However there does need to be a knight on B3 in the picture otherwise C2 can escape.

[Guest]

I think the initial problem with my design was that I started off with the 4 knights in their actual chessboard positions and then built symmetrically from there

[soop]

I think the initial problem with my design was that I started off with the 4 knights in their actual chessboard positions and then built symmetrically from there

Try starting in the corner; that limits your initial cover, but it gets complicated fast.

[Guest]

However there does need to be a knight on B3 in the picture otherwise C2 can escape.

So there does. Thanks soop - I guess I should have spotted the symetry

Is there a better solution than Mumbles to the actual problem? At the moment, I can't beat it (and have too much 'real' work I should be doing instead!)

Edited August 28, 2008 by foolonthehill

[soop]

So there does. Thanks soop - I guess I should have spotted the symetry

Is there a better solution than Mumbles to the actual problem? At the moment, I can't beat it (and have too much 'real' work I should be doing instead!)

This is the sneaky thing about it; you start off pretty good, spacing them out as wisely as possible - but then there will be one little gap, and pretty soon you're flooded with knights. It's a real killer.

[Guest]

Try starting in the corner; that limits your initial cover, but it gets complicated fast.

Thanks--I had to use a color-coded excel spreadsheet to help me organize my thoughts

[Guest]

I'm new to the board, but this one intrigued me. As I understand, all knights must be in "check", and can not move out of "check"

I've come up with 24:

_ k k _ _ k k _

k _ _ _ _ _ _ k

k _ _ k k _ _ k

_ _ k _ _ k _ _

_ _ k _ _ k _ _

k _ _ k k _ _ k

k _ _ _ _ _ _ k

_ k k _ _ k k _

I don't know if this is the fewest possible, but it looks nice!

[soop]

I'm new to the board, but this one intrigued me. As I understand, all knights must be in "check", and can not move out of "check"

I've come up with 24:

_ k k _ _ k k _

k _ _ _ _ _ _ k

k _ _ k k _ _ k

_ _ k _ _ k _ _

_ _ k _ _ k _ _

k _ _ k k _ _ k

k _ _ _ _ _ _ k

_ k k _ _ k k _

I don't know if this is the fewest possible, but it looks nice!

That's beautiful! TBH, I never got anything as good as that. I think you're the winner! (Oh, I've checked, and I can see no flaws).

If anyone can do better, I'd be surprised - welcome to the boards, and what an entrance!

Ah, Just looking again, I've spotted an escape route, which makes it a tad uglier.

_ k k _ _ k k _

k _ _ _ _ _ _ k

k _ K k k K _ k

_ _ k _ _ k _ _

_ _ k _ _ k _ _

k _ K k k K _ k

k _ _ _ _ _ _ k

_ k k _ _ k k _

This way, the middle ones can't eascape to the outer edge. I think that's it. So 30 in total.

Edited August 28, 2008 by soop

[Guest]

Dang....I knew it was too good to be true. But thanks for the warm welcome nonetheless.

Here's another alternative to your revision:

_ k k _ _ k k _

k k _ _ _ _ k k

k _ _ k k _ _ k

_ _ k _ _ k _ _

_ _ k _ _ k _ _

k _ _ k k _ _ k

k k _ _ _ _ k k

_ k k _ _ k k _

Either way has 28, and either way works.

[Prime]

Need a clarification. Are we allowed to escape by capturing?

E.g., if there is a knight which is attaked by only one other knight, then that other knight escapes "check" by capturing the former.

[Guest]

Sorry, but I found another error. The top left corner is blank. If the knight 2 spaces below the blank jumps the knight just to the right of the blank, he will no longer be in check. This applies to all 4 corner locations. Remember that even if the space they jump to is occupied by another knight, a 3rd knight must also have the ability to jump that occupied space.

You must be in check after all possible jumps, not before them (hence the idea of checkmated)...Not exactly the same things

[Guest]

Sorry, but I found another error. The top left corner is blank. If the knight 2 spaces below the blank jumps the knight just to the right of the blank, he will no longer be in check. This applies to all 4 corner locations. Remember that even if the space they jump to is occupied by another knight, a 3rd knight must also have the ability to jump that occupied space.

You must be in check after all possible jumps, not before them (hence the idea of checkmated)...Not exactly the same things

OK... I see, so each knight must be double covered ( <_< my grand entrance is seeming less and less grand all the time)

So what if Soop and I combined answers?

_ k k _ _ k k _

k k _ _ _ _ k k

k _ k k k k _ k

_ _ k _ _ k _ _

_ _ k _ _ k _ _

k _ k k k k _ k

k k _ _ _ _ k k

_ k k _ _ k k _

Now we're at 32 total pieces (sadly, half of the board)

[Guest]

That's good, but my configuration has it in 28--good luck with beating it...it was the first solution I came up with so efficiency probably wasn't 100%

[Guest]

Hi

It should be 14 knights......

7 for black and 7 for white....

7 nights can cover the whole chessboard of one color. 7*2(for white) = 14 total.....

What is the correct answer?????