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Charles Gilman wrote on Wed, Sep 25, 2013 06:09 AM UTC:
I was intrigued by the idea in Daniil Frolov's latest variant of Queens and Nightriders as the basic linepieces, with Bisons as the short-range piece. It reminded me of something, and that is the relationship between the Glinsky and AltOrth interpretations of the hex geometry - although the derivation was the other way round. In this table illustrating the parallels, FIDE and AltOrth are collectively termed the "short" interpretations and Frolov virtual Octagonal and Glinsky as "long" ones.
square-cellhex-cell
Rook, BishopBasic linepieces in short interpretationForerook, Hincdrook
QueenTheir compound, the compound linepiece in the short interpretation but a basic linepiece in the long oneRook
KnightThe short-range piece in the short interpretationViceroy
NightriderIts rider, the other basic linepiece in the long interpretationUnicorn
AceriderThe compound linepiece in the long interpretationDuchess
BisonThe short-range piece in the long interpretationSennight
Well that made another idea dawn on me. A further development on AltOrth is Plattraum, which starts with the AltOrth linepieces but adds in the Unicorn as a third basic linepiece. There is then the potential for three two-way compounds - the Rook, Foreduchess, and Hindduchess - and the three-way Duchess just as cubic-cell boards have the potential for the two-way Queen, Duchess, and Governor and the three-way Empress. Thus it would be possible to have a square-cell Plattraum with the Rook, Bishop, and Nightrider as basic linepieces with the Queen, Marshrider, and Cardirider as two-wqy compounds and the Acerider as the three-way one - plus the Bison as a short-range piece.

Daniil Frolov wrote on Wed, Sep 25, 2013 09:14 AM UTC:
What an interesting observation i did not notice before, though i like
AltOrth chess and other twists with shape. It seems to me as being able to
open more paths of making curious chess interpretations.
The first idea that came in my head is to make a variant wich is
intermediate between Glinski and AltOrth hexchess - in same way as Betza's
Rectahex chess is intermediate between FIDE chess and my "Octagonal"
chess (hex of hex!)... But then i realized that such game would be
asymmetrical and ambiguous, just after noticing that difference between
numbers of directions of rook and forerook is odd.

Charles Gilman wrote on Sat, Nov 16, 2013 06:55 AM UTC:
I have a theory that every odd-SOLL leap on a hex board vcan be expressed as m diagnoal and n orthogonal steps at right angles in exactly one way, but cannot immediately see a way to prove it. Note that this is not the same as saying that every relevant SOLL can be expressed as 3m²+n² in exactly one way. For example, 49 is the SOLL of both the Heptagram (m=4, n=1) and the Settler (m=0, n=7) but as you can see, each has its own (unique) value of m and n. I have looked through several leapers without finding a counterexample. Does anyone know of either a counterexample or a proof of the theory?

Ben Reiniger wrote on Sat, Nov 16, 2013 09:37 PM UTC:
I don't understand the difference between your two statements.  What
exactly do you conjecture to be true?  Why does 49 not give a
counterexample?

Ben Reiniger wrote on Sat, Nov 16, 2013 11:31 PM UTC:
Do you mean something like multi-path?  Then the dababbah (thought of as a
1,1 diagonal-orthogonal leaper) fails, doesn't it?

Charles Gilman wrote on Sun, Nov 17, 2013 08:02 AM UTC:
The Dabbaba is not a counterexample as it has an even SOLL, namely 4, and my conjecture specificed odd SOLLs. The point about the pieces with SOLL 49 is that their leaps, though the same length, are to different destination cells relative to a starting cell. The following diagram shows this. Routes c and d lead to the same destinations but route b leads to a different one where:
@=statring cell
#=destination cell
a=intersection of b and c
b=7 orthogonal steps
c=1 orthogonal followed by 4 diagonal
d=4 diagonal folloewd by 1 orthogonal
 ___     ___     ___     ___     ___
/ . \___/ . \___/ . \___/ . \___/ # \
\___/ . \___/ . \___/ # \___/ . \___/
/ . \___/ . \___/ d \___/ . \___/ b \
\___/ . \___/ . \___/ . \___/ . \___/
/ . \___/ . \___/ . \___/ c \___/ b \
\___/ . \___/ . \___/ d \___/ . \___/
/ . \___/ . \___/ . \___/ . \___/ b \
\___/ # \___/ . \___/ . \___/ c \___/
/ . \___/ b \___/ . \___/ d \___/ b \
\___/ . \___/ b \___/ . \___/ . \___/
/ . \___/ . \___/ b \___/ . \___/ a \
\___/ . \___/ . \___/ b \___/ d \___/
/ # \___/ c \___/ c \___/ a \___/ a \
\___/ . \___/ . \___/ . \___/ a \___/
/ d \___/ d \___/ d \___/ d \___/ @ \
\___/ . \___/ . \___/ . \___/ . \___/

George Duke wrote on Sun, Nov 17, 2013 09:11 PM UTC:
To clarify for other than Charles and Ben, unit 1 is centre to centre. 
Each regular hexagon has six equilateral triangles, and each height is 1/2,
so each hexagon side is 1/root-3 by 30-60-90 triangle. That makes each
orthogonal step in hexagons the 1.0 and each diagonal step 3/root-3 seen by
inspection and counting.

[Examples with even SOLL irrelevantly: (0,10) and (5,5) have same LL and SOLL 10 and 100; also (0,12) and (6,6); also (0,6) and (3,3)]

What would be second case with odd SOLL? Many, (2,3) of 31 SOLL for one.

Ben Reiniger wrote on Sun, Nov 17, 2013 09:11 PM UTC:
So is your conjecture that in the hex-grid (or triangular grid depending on
how you look at it), there are no two right triangles that share a
hypotenuse with odd squared length?

EDIT: I should probably add "where the legs of the triangles are along the orthogonal and diagonal directions".

More edit: and to avoid the trivial swapping of the order of diagonal/orthogonal steps, make it "nonisomorphic triangles"

Ben Reiniger wrote on Sun, Nov 17, 2013 10:39 PM UTC:

I think the 4-coloring of the hex board can help to prove this. See
wikipedia's 4-coloring image

If a leap has odd SOLL, then in any right angle diagonal-orthogonal path--say m diagonal and n orthogonal as before--we must have that m and n have different parities (else 3m^2+n^2 is even). Then the starting and landing cells have different colors in the 4-coloring.

But each orthogonal-diagonal pair of directions at right angles involve exactly two colors, one of which is the starting cell's color. So traveling along distinct orthogonal-diagonal directions lands at distinct color cells.


Charles Gilman wrote on Mon, Nov 18, 2013 07:07 AM UTC:
"So is your conjecture that in the hex-grid (or triangular grid depending on how you look at it), there are no two right triangles that share a hypotenuse with odd squared length... where the legs of the triangles are along the orthogonal and diagonal directions?"

No, as I have already cited a counterexample to that conjecture if you include triangles of side zero. The orientation of the hypotenuse on the board matters as well as its length. I will have a think to see whether I can formulate the conjecture with purely numeric variables.

"I think the 4-coloring of the hex board can help to prove this. See wikipedia's 4-coloring image. If a leap has odd SOLL, then in any right angle diagonal-orthogonal path--say m diagonal and n orthogonal as before--we must have that m and n have different parities (else 3m^2+n^2 is even). Then the starting and landing cells have different colors in the 4-coloring. But each orthogonal-diagonal pair of directions at right angles involve exactly two colors, one of which is the starting cell's color. So traveling along distinct orthogonal-diagonal directions lands at distinct color cells."

I will have to think about this offline and reply at a later date.


Ben Reiniger wrote on Mon, Nov 18, 2013 03:11 PM UTC:
Charles: which example?  the 49?  Those triangles do not share a hypotenuse (aha, when I say "share a hypotenuse", I mean that geometrically, not just that they have the same length).  And, the orientation of the hypotenuse is determined by the choice of right-angled orthogonal-diagonal pair.

(I may still be misunderstanding your question, so the rephrasing and proof may yet be incorrect.)

Charles Gilman wrote on Tue, Nov 19, 2013 07:25 AM UTC:
"And, the orientation of the hypotenuse is determined by the choice of right-angled orthogonal-diagonal pair."

That is certainly part of what I was hoping someone could prove or disprove. It is self-evident for two orthogonals at right angles on a square-cell board, or even three on a cubic one. It is self-evident for a diagonal and an orhogonal with a 45° turn on a square-cell board. It is even self-evident for two orhogonals with a 60° turn on a hex board. It is however not only not self-evident for a diagonal and an orhogonal at right angles on a hex board, but untrue in the case of an even SOLL. Again taking m diagonal and orthogonal steps, m=1, n=1 gives the same destinations, not just leaps of the same length as by m=0, n=2. Likewise m=1, n=5 gives the same destinations as m=3, n=1. For odd-SOLL leaps I suspect from lack of counterexamples, and for prime-SOLL leaps I strongly suspect, that such duplicate vaules of m and n for the same destination are impossible, but cannot yet prove either. I have now checked all values of m up to 20 and n up to 40.


Joseph DiMuro wrote on Tue, Nov 19, 2013 04:28 PM UTC:
On a hex board, a dabbabah (a 2-square orthogonal leaper) can only reach
one-fourth of the spaces on the board.

Color a hex board in four colors, such that a dabbabah can only move to
hexes of the same color.

Claim: given two hexes, if there are two different paths of the type
Charles Gilman described between those hexes, then those hexes are the same
color. And every leap between two hexes of the same color is an even-SOLL
leap. Thus, Gilman is correct.

Proof left to the reader. :-D

Ben Reiniger wrote on Tue, Nov 19, 2013 04:39 PM UTC:
I am now more certain that I understand what you're saying, and that my proof from before works.  (Joseph has the same idea.)  Let me give some more details:

Given a cell, an orthogonal direction, and a diagonal direction at a right
angle to the orthogonal direction, the set of cells reachable from the
given cell in those directions has a rectangular geometry (it is a proper
subset of the entire board, consisting of all cells of two colors in the
4-coloring).

So, given starting and destination cells, there are at most three ways to
write the leap as a combination of orthogonal and diagonal (at right
angles) steps.  These three ways correspond to the three orthogonal (or
diagonal) directions.  But the 4-coloring has the property that these three
rectangular sub-geometries (not using that as a technical term) each
consist of the color of the starting cell and one other color, and these
three other colors are distinct among the three sub-geometries.

Furthermore, the 4-coloring has the following property: the destination
cell's color is the same as the starting cell's color if and only if the
number of diagonal steps and the number of orthogonal steps have the same
parity.  (This can be seen by noting that the 4-coloring induces the usual 2-coloring on any of the rectangular subgeometries.)

So, if the SOLL is odd, the target cell is of different color than the
starting cell, and so there is only one of these subgeometry paths.
If the SOLL is even, then the target cell is of the same color as the
starting cell, and there are three distinct subgeometry paths (except for
SOLL=0).  These three are distinct in that they use different
ortho-diagonal subgeometries, but they may have the same number of
orthogonal and diagonal steps.

For example, m=1,n=5; m=2,n=4; m=3,n=1 all have SOLL 28 and have a common
destination cell.  (Perhaps this is the smallest with all three distinct?)

Charles Gilman wrote on Wed, Nov 20, 2013 07:00 AM UTC:
I can see what you mean now. Intreerstingly while I've been offline I've noticed a pattern that explains why even-SOLL moves can be expressed in more ways than odd-SOLL ones. Taking m as the diagonal and n as the orthogonal, I looked at how the moves can be expressed as two orthogonal moves - p and q steps, say, where p>=q - with a 60° turn in between. Where 3m>n>m, p=2m and q=n-m, which translates into m=p/2 and n=q+p/2. Where n>3m, p=n-m and q=2m, which translates into m=q/2 and n=p+q/2. Where m>n, p=m+n and q=m-n, which translates into m=(p+q)/2 and n=(p-q)/2.

In the first case m must be even, in the second n must be even, and in the third m+n must be even. With an odd SOLL exactly one of these is the case and so only one of the three pairs of equations holds true. With an even SOLL all three are true and so all three pairs work, and not necessarily for the same values of m and n.

In the special case of 3m=n, p and q are equal and the first two pairs of equations are identical. In the special case of m=n, all equations except the second hold and q is zero and m and n both p/2. The missing equation may hold for different values of m and n if p and q are even.


Jörg Knappen wrote on Thu, Nov 21, 2013 08:54 AM UTC:
I think a got a proof for the hex geometry.

We orient the hexes such that there is a horizontal line of rook movement,
and denote that direction by 1. The other directions of rook movement are
denoted by \omega and (\omega-1) [the use of the letter \omega is
inspired by Eisenstein numbers]. The centre of a hex is given by a+b\omega
with a,b integer numbers.

First step is a drawing: When we go horizontally firs and vertically as a
hex bishop second, we can reach only one half of the hexes (a+2b\omega).
We repeat this for the other rook directions and mark the hexes
accordingly. They fall in two classes: (i) hexes which can be reached in
one way only (ii) hexes that can be reached in all three way.

The second class forms a grid described by 2a+2b\omega (both coordinates
must be even.

Finally we map these to rook and bishop moves. The path to a three-way
reachable hex (2a+2b\omega) using horizontal and vertical moves
(elementary vertical bishop step: (2\omega -1)) consists of b bishop steps
and b+2a rook steps. Therefore the number of rook and bishop steps are both
odd or both even, giving an even SOLL.

The other direction: Take r rook steps and s bishop steps and demand that
r+s is even. Then we go to r+s*(2\omega -1) = (r-s) +2s\omega. This is a
three-way reachable square again, because (r+s) even implies (r-s) even.

George Duke wrote on Sun, Nov 24, 2013 08:39 PM UTC:
Would the Rectahex board help or hinder visualizing Charles Gilman's
latest theorems?  http://www.chessvariants.org/diffmove.dir/rectahex.html.

We find: 'Hexagon -> Rectahex -> Square -> Triangle (all reversible)'!


Triangles: http://www.chessvariants.org/index/displaycomment.php?commentid=26710. 

 Squares: http://www.chessvariants.org/index/displaycomment.php?commentid=26708, 
http://www.chessvariants.org/index/displaycomment.php?commentid=26706.


This present topic of Chess Geometry is meaning Hexagonal chess geometry,
but Ralph Betza asks in 2003, "Is hexagonal chess really hexagonal or
merely a rectangular dream?" All the hex-board adjacencies are intact,
looking only at Squares once the Rectahex slide is performed.  There are
differences in some corners, but not edges which have all same adjacencies
cell-to-cell Hex and Rectahex.  However, LL and SOLL may be problematic
because of inconsistent perpendiculars from the skewing.
http://www.chessvariants.org/index/displaycomment.php?commentid=26724.

Charles Gilman wrote on Mon, Nov 25, 2013 06:52 AM UTC:
Thanks to all of you for your help clarifying the proof of my connjecture. I have credited you in my Man and Beast 14 update, and I also plan to mention it in pages 17 and 20 of the series. In the latter page I plan to drop the piece names Moses and Heracles in favour of seeing 90° orthogonal-hex diagonal alternators of this kind as "angle-" versions of 30° orthogonal-hex diagonal alternators with the same Even Move Direction. This still leaves those without a unique mapping, and I am thinking about reusing Moses and Heracles for two of them, but I am interested in other ideas for names of more of them.

Charles Gilman wrote on Mon, Jan 13, 2014 07:08 AM UTC:
While online yesterday morning I caught sight of an image of a genealogy showing a person and their ancestors only, on so small a scale that it could easily have been a Shogi variant camp with twice as many pieces on each rank in a pattern such as:
A-A-A-A-A-A-A-A
-B---C---C---B-
---D-------E---
-------F-------
where the letters are arbitary and no piece has a forward long-range orthogonal move - although B or C might have a forward long-range diagonal move or D or E a forward long-range Knightwise move. It was not however that idea for a square array that inspired me to think further. Instead it occurred to me to have a board with double the numbre of cells per rank, starting with 1. Initially I thought of going up to 16 and having two ranks that size before halving again back to 1, but then I began analysing what sort of cells this generates and I realised that by having just 9 cells on each middle rank I could make all the cells pentagonal. This resulted in the following array:
  ---------------
 |       k       |
  ---------------
 |   q   |   c   |
  ---------------
 | r | n | n | r |
  ---------------
 |p|p|p|p|p|p|p|p|
 -----------------
| | | | | | | | | |
 -----------------
| | | | | | | | | |
 -----------------
 |P|P|P|P|P|P|P|P|
  ---------------
 | R | N | N | R |
  ---------------
 |   Q   |   C   |
  ---------------
 |       K       |
  ---------------
The pieces are the Constrictor, Nadder, Rattlesnake, and Quetzalcoatl as defined in SerPent Chess, together with the Point of Wellisch hex Chess and the King. I rejected using the Boa as it is too weak. Points are promoted to Constrictor, Quetzalcoatl, or Mamba on entering the enemy camp. A Shogi variant would substitute Gold for rhe compound pieces, Silver for Nadder, and Waggle for Rattlesnake, with usual Shogi promotion.

Has anyone got any good ideas for a name for this third pentagonal geometry?


Daniil Frolov wrote on Mon, Jan 13, 2014 10:12 AM UTC:
I was thinking of something alike, while trying to figure out (not quite
succesful), how septagonal chess would look like. Game with progressive (or
wavy) number of cells per rank. Perhaps, even with cells, being
orthogonally-adjecent to three cells on next rank. While these three cells
could be orthogonally-adjecent to three further cells each (three different
for each (9), or external ones being shared by adjecent (7)), and so on.

Name? Progressively-shrinking pentagons, perhaps?

Ben Reiniger wrote on Tue, Jan 28, 2014 07:26 PM UTC:
Pritchard mentions in CECV a game with a similar topologically-pentagonal
but geometrically-"doubling" board.  It is called the Fourth Dimension,
but has no 4d characteristics.  It is played on a round board with
concentric ranks of 4, 8, 16, and 32 cells, a generic cell having one
neighbor in the next rank inward, two neighbors on its own rank, and two
neighbors in the next rank outward.  The pieces don't seem to be
particularly chesslike.

(Maybe this round version could be realized as almost a pentagonal tiling
of the hyperbolic plane?)

Charles Gilman wrote on Thu, Jan 30, 2014 07:13 AM UTC:
Taking together the comments
"Pritchard mentions in CECV a game with a similar topologically-pentagonal but geometrically-"doubling" board."
from this thread and
"Pritchard's CECV lists a game "Xyrixa Chess" by David Samuel c.1980 played on this same board (provided I'm reading correctly)."
from the .comments on Tetrahedral Chess, it seems that some Chessboards are more obvious to those who devise Chess variants than we realise when we think of them. Xyrixa is certainly a shorter word than Tetrahedral to describe the 3d geometry with 12 Rook and 6 Bishop directions, and if it is the older name for that geometry perhaps I should use it in place of Tetrahedral in Man and Beast. It would be interesting to know what Mark Thompson thinks. I will however retain the current name for Tetrahedral Shogi, not just because changing the display on search pages takes so long but because it was genuinely inspired by Mr. Thompson's game. Does Xyrixa have any prior meaning, or did Mr. Samuel coin it specifically for the variant?

Ben Reiniger wrote on Mon, Feb 3, 2014 05:03 PM UTC:
CECV doesn't suggest whether the word Xyrixa has some meaning.  Since we
now know that this geometry can be viewed as the rhombic dodecahedron
tiling, perhaps a name based on that is appropriate.  A clever name
lacking, I'll use RD for the rest of this message.

On RD geometry: it's been noted that it can be viewed as the standard 3D
cubic tiling, but using the cubic diagonal as the RD orthogonal (and
restricting to one color of the cubic 2-coloring, say black).  Charles's
RD bishop direction seems to correspond to the cubic dabbabah.  Perhaps
another interesting direction is the cubic triagonal, which must skip over
one white cell.  This direction corresponds to passing through a degree-3
vertex of the RD.  (The RD orthogonal passes through the faces, and
Charles's RD bishop direction passes through the degree-4 vertices.)  I'm
not quite sure, but I think passing through the edges of the RDs would
correspond to a (2,1,1)-leap in the cubic tiling.  I haven't thought yet
about: do these directions have meaning in the hex-plane parts of RD
geometry?

Charles Gilman wrote on Wed, Feb 5, 2014 07:24 AM UTC:
Ben Reiniger is quite correct, and what he describes is the basis of the dualities between hex or Tetrahedral (alias Xyrixa) and cubic pieces in pages 14 and 18 of Man and Beast. Taking out one Bishop binding of a cubic board converts a cubic piece whose coordinates are 2m Wazir steps at right angles to n Ferz ones into a Tetrahedral/Xyrixa piece whose coordinates are n Wazir steps at right angles to m Ferz ones, and an m+n:m:n leaper into a piece moving along a hex plane. The first gives dualities such as Lecturer/Elf, Effarig/Fencer, and Nurturer/Underscore while the seciond gives dualities such as Fortnight/Sennight, Sustainer/Student, and Votary/Overscore. Ferz/Wazir and Sexton/Viceroy fit both.

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