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Chess on an Infinite Plane. Chess game with no boundaries (infinite board), and Guard, Chancellor, and Hawk. () [All Comments] [Add Comment or Rating]
Ben Reiniger wrote on Sat, Feb 25, 2017 05:23 PM UTC:

Images seem to be broken.  You can upload images to this site (see the text near the bottom of the page), fixing any issues with Flickr.

Oh, but you resubmitted with new images. I've kept this one hidden in favor of that one. They are the same game, just with different images, right?


💡📝V. Reinhart wrote on Thu, Mar 9, 2017 01:07 AM UTC:

Thanks Ben,

I did update the images to the CVP site, and updated the links. It looks fine now. This same game was also released with Fergus' abstract icons, but I'd like to show the game with both pieces sets.

The game is already being played with the classic icons (this submission), and some people might prefer this style because it looks much more similar to the normal chess pieces.

Thanks for your support!


Aurelian Florea wrote on Mon, Jan 8, 2018 09:01 PM UTC:

In the comments section of chess with different armies Vickalan Reinhart (the author of CoaIP) stated " As far as I know, there is no upper bounds for the maximum size of a good board size, and even infinitelly-large boards are easily playable and fun. " I think the remark I am going to make is more appropriate here.

I just wanted to make the remark that the term "on an infinite plane" is appropriate as the game is not that infinite after all. It has a decent number of pieces and the effect of board largeness fades away far. It would be more "infinite" if maybe players would have pieces according to some rules  all over the board (with an 50% density maybe). For example white would have a knight on all white squares on even files, and black would have a knight on all black squares on odd files. That would be quite silly (almost all knights are undefended and attacked), but with work one can make more complex rules for more pieces.  From a mathematical point of view these games would be quite interesting but as of 2018 we don't have the technology or not even knowledge to tackle such complex problems :)! But we as cyborgs of the future will get there. Mark my words :)!


💡📝V. Reinhart wrote on Tue, Jan 9, 2018 02:28 AM UTC:

@Aurelian, that is a really interesting comment, and I love that idea. Like you say, an unbounded board can be populated with an infinite number of pieces in many ways. In addition to what you mention, and "on top" of the "normal" armies (near the kings), pawns can be populated for example on every 10th file and every 10th rank. So the density in half the plane is 1/10 x 1/10 = 1%. Similarly bishops and rooks can be populated with a density such as 0.1%, and queens populated on 0.05%.

So now you really have an "infinite" game with "infinitely many" pieces. But still - only one king of each color. As in the diagram, the White king starts on (5,1) and the Black king starts on (5,8).

This could have amazing and very interesting mathematical properties. Obviously, the strategy of the end-game cannot be "simplification" to just a few pieces! Players will need to learn to mate as other pieces slowly march inward, trying to replenish captured pieces.

Who is going to solve this type of game strategy? You said it best: "But we as cyborgs of the future will get there".  I agree with you 100%!!!


Aurelian Florea wrote on Tue, Jan 9, 2018 09:30 AM UTC:

Yes, that will obviously make it "more infinite". Still the kings are the point of atraction. But if you make more kings then you have two cases.

1. You have to mate one king to win. Which after a finite number of moves will often degrade in the previous situations as one king will become more vulnarable although each side could need to use plenty of moves for the defence of it's wekest king. And surellyanother win condition would be to simultanioulsy check 2 kings with an unattacked piece. Also winning. So 2 and in extremme conditions even 3 enemy kings could get involved but it is not a big difference in the end.

2. You have to mate all (which is probably a countable infinity, maybe the ordinal omega0). That is a whole other devil I don't know how to aproach yet.

You may also make different winning conditions like capture the flag, like bringing a piece to a certain area of the board withouht beeing caputed at the next turn.

Also you may make a condition to clear a certain area of enemy pieces.

So there is a whole world of possibilities out there.


Aurelian Florea wrote on Sun, Aug 6, 2023 10:18 AM UTC in reply to Aurelian Florea from Tue Jan 9 2018 09:30 AM:

While contemplating the discussion  here: https://www.chessvariants.com/index/listcomments.php?id=48814 about (5,x) pieces, with x=1..4, I remembered V. Reinhart's Huygens piece, and  I came up with one of my own piece on an infinite plane. A compound of all leapers (m,n) with m>n where m is any prime number strictly larger  than 1 and n are all the numbers from 1 to m-1. An then let all the leaps ride. The branches will not intersect with each other. While thinking this I came up with a compund leaper where the (m,n) pair is any ireducible fraction. I could not realize yet if this yields a rider where the branches don't interesect each other. Any other thoughts on this topic?


Bob Greenwade wrote on Sun, Aug 6, 2023 06:33 PM UTC in reply to Aurelian Florea from 10:18 AM:

It sounds a little (in principle, not in practice) like a piece I've been contemplating, the Root-N25 Leaper, which can leap to any space so long as the sum of the squares of the legs is a multiple of 25 (combine Root-25, Root-50, Root-100, etc., as far as is needed).

I'm not clear on what you mean by "intersect," though. If I take the word literally, two Rider paths will never intersect because they both go away from the starting point in different directions.


Aurelian Florea wrote on Mon, Aug 7, 2023 07:28 AM UTC in reply to Bob Greenwade from Sun Aug 6 06:33 PM:

I mean like the 4, 2 leaper intersects the nightrider!


H. G. Muller wrote on Mon, Aug 7, 2023 08:14 AM UTC in reply to Aurelian Florea from Sun Aug 6 10:18 AM:

I could not realize yet if this yields a rider where the branches don't interesect each other.

Isn't that obvious? To intersect they would need to have the same ratio of n and m. But there is only a single irreducible representation of n/m. (In other words, if n/m = n'/m' then nm' = n'm, and have the same factors. But since none of the factors in n is in m, they must all be in n', and none of the factors in n' is in m', and must be in n. So n = n'. This relies on the factorization of a number to be unique.)


Aurelian Florea wrote on Mon, Aug 7, 2023 09:16 AM UTC in reply to H. G. Muller from 08:14 AM:

Oh, yes. Silly me!


Bn Em wrote on Mon, Aug 14, 2023 09:05 PM UTC in reply to Aurelian Florea from Sun Aug 6 10:18 AM:

While thinking this I came up with a compound leaper where the (m,n) pair is any irreducible fraction

Iirc this is the Problemists' Wizard, also found in one of the later Man and Beasts. Indeed, its rider the Witch (as explained by H.G.) has only nonintersecting rides.


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