OT: Roman boardgame deciphered with help from AI

[Ajedrecista]

Hello:

Following my notation, b2 and c2 do not exist, just in case they appear in former posts. They would likely refer to a2 and d2, respectively.

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Assuming an empty board on the 10-spot variant, a piece can move from any spot to any spot in three moves at most:

Code: Select all

·—·—·—·  3
|/   \|
·—————·  2
|\   /|
·—·—·—·  1

a b c d

0123   1012   2101   3210
1  2   1  2   2  1   2  1
2233   2233   3322   3322

1122                 2211
0  1                 1  0
1122                 2211

2233   2233   3322   3322
1  2   1  2   2  1   2  1
0123   1012   2101   3210

With a higher mobility (less moves needed) of a2 and d2, as we already know, just the opposite of the corners.

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I toyed a little with Google free, online IA mode, explaining this exact variant. Then, inspired by this thread, I asked it to apply the Burnside's lemma, which I do not fully understand, so I copy the conclussions almost blindly: there could be 810 unique positions.

Code: Select all

N = (3150 + 30 + 30 + 30)/4 = 810

Where:

3150 is for identities: (10 over 4)*(6 over 2)
I guess that 10 spots and 4 pieces of the strong side, then 6=10-4 available spots and 2 pieces of the weak side.

30 for vertical reflections: (5 over 2)*(3 over 1)
Looks like 5 being the spots of one half of the board (left or right), then place one half of the pieces of the strong side (4/2=2), then 3=5-2 available spots in this half of the board and half of the pieces of the weak side (2/2=1).

30 for horizontal reflections: (5 over 2)*(3 over 1)
Probably 4 being the spots of one half (top or bottom) plus 1 of a2 or d2, with the same reasonment than for vertical reflections.

30 for 180º rotations: (5 over 2)*(3 over 1)
I could not understand why.

The IA also told that 200 of these unique positions where with both a2 and d2 empty; 160 with a2 and d2 occupied; and the remaining 450 with one occupied and one empty. I did not check manually, understandably.

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Inspired by the same thread than before, I asked the IA to apply the cycle index polynomial, which I also do not fully understand:

Code: Select all

Klein's group V4 (or C2 × C2) and order |G| = 4 according to the IA.

Identity: 10 fixed nodes.
10 cycles of length 1 (f_1;10)

Vertical reflections: swap left and right (a↔d and b↔c). 5 pairs (a1,d1), (b1,c1), (a2,d2), (a3,d3), (b3,c3).
5 cycles of length 2 (f_2;5)

Horizontal reflections: swap bottom and top (1↔3). 4 pairs (a1,a3), (b1,b3), (c1,c3), (d1,d3); plus 2 fixed nodes a2 and d2.
4 cycles of length 2 (f_2;4) and 2 cycles of length 1 (f_1;2), which result in (f_2;4)*(f_1;2)

180º rotation: the composition of both reflections. 5 pairs (a1,d3), (b1,c3), (c1,b3), (d1,a3), (a2,d2)
5 cycles of length 2 (f_2;5)

Polynomial:
Z(G) = (1/4) * [(f_1;10) + (f_2;5) + (f_2;4)*(f_1;2) + (f_2;5)]
Z(G) = (1/4) * [(f_1;10) + 2*(f_2;5) + (f_2;4)*(f_1;2)]

Taking a look to the thread of TalkChess, calling e to empty nodes, S to the pieces of the strong side and w to the pieces of the weak side, it should be:

Code: Select all

Z(G) = (1/4) * [(e+S+w)^10 + 2*(e^2+S^2+w^2)^5 + (e^2+S^2+w^2)^4 * (e+S+w)^2]

Expanding Z(G) and picking the coefficient of e^4 S^4 b^2, it is 810 indeed! Since this game does not allow captures, this is the only coefficient that matters and we have bounded the size of the game.

------------

I took some time to dig into the game without software analysis, i.e. without downloading the Ludii app, which in practice means that anyone who learns the game in five minutes by his/her own will get better strategies than me in some hours, such is life.

Thinking from the strong side POV, I initially thought that keeping a piece in one of the central spots (a2 or d2) forever was good for getting fast wins, as is the aim of the game; later, I changed my mind and while being very important, I let some incursions abandoning these spots in order to create zugzwangs to finally block the pieces of the other side. Given the small size of the board and how much crowded it is, I feel it is the same few maneouvres once and again, once and again... although I sometimes shuffles a bit because I have not solved the game.

One curious thing is that I came with a tactic once a piece is trapped and go for the other, that I leave only two spots for the weak side, blocking in the next move regardless of where it moves. I randomly called it 'pinza' (a Spanish word) and while toying with the IA and playing some games, the IA called it 'pinza' too without writing anything! I wrote in English and the IA answered in Spanish. It was funny, like great minds think alike, although the IA sometimes played illegal moves from time to time and I had to told why was wrong despite I described a precise topology of the board and rules.

I still think that once you can trap the first piece, trap it instead of search other moves in order to get faster wins, because it might not. Please remember that I do not master the game, so I might be wrong. Once you have a piece trapped, for example on d3, I analyzed some positions:

Code: Select all

w to move

·—·—S—w  3
|/   \|
w—————S  2
|\   /|
S—S—·—·  1

a b c d

     w      S
1. a2-a3  a1-a2  // Moving the piece of the rank where w moved.
2. a3-b3  a2-a3  // Creating the zugzwang.
3. b3-a2  a3-b3  // What I call the 'pinza', leaving only two corners available and blocking in the next move.
4. a2-a1  b3-a2  // Alternatively, 4. a2-a3  b1-a2

===============

w to move

·—·—S—w  3
|/   \|
w—————S  2
|\   /|
S—S—·—·  1

a b c d

     w      S
1. a2-b3  b1-a2  // Moving the piece of the rank where w moved.
2. b3-a3  a2-b3
3. a3-a2  a1-b1  // The 'pinza' again, creating the zugzwang and leaving only two corners available for blocking in the next move.
4. a2-a1  b3-a2  // Alternatively, 4. a2-a3  b1-a2

===============

S to move

·—·—S—w  3
|/   \|
w—————S  2
|\   /|
S—S—·—·  1

a b c d

     w      S
1.  ...   b1-c1
2. a2-b1  a1-a2  // Bad move by w, letting cornered faster.
3. b1-a1  c1-b1  // S wins.

===============

S to move

·—·—S—w  3
|/   \|
w—————S  2
|\   /|
S—S—·—·  1

a b c d

     w      S
1.  ...   b1-c1
2. a2-a3  a1-a2
3. a3-b3  a2-a3
4. b3-a2  c1-b1  // A variant of the 'pinza', explained below. S wins in the next move.
5. a2-b3  b1-a2  // Alternatively, 5. a2-a1  a3-a2

===============

S to move

·—·—S—w  3
|/   \|
w—————S  2
|\   /|
S—S—·—·  1

a b c d

     w      S
1.  ...   b1-c1
2. a2-b3  a1-a2
3. b2-a3  a2-b3
4. a3-a2  c1-b1  // The 'pinza' again. S wins in the next move.
5. a2-a1  b3-a2  // Alternatively, 5. a2-a3, b1-a2

The comment on the first strong side's move must serve as a rule of the thumb to help the stronger side and not waste moves.

There are positions of a double zugzwang or double 'pinza', with the two pieces of the weak side on the centre spots and all the corners empty (the double 'pinza'), with the weak side to move (the zugzwang or double zugzwang because one piece move and is trapped, and the weak sides faces other zugzwang again). Proof game:

Code: Select all

     w      S
1. a3-a2  d1-d2
2. d3-c3  d2-d3
3. a2-d2  b1-a2
4. c3-b3  d3-c3
5. b3-a3  a2-b3  // Setting the trap!
6. a3-a2  a1-b1  // Double zugzwang. The strong side blocks in 2 moves, winning the game in 8 moves.

w to move

·—S—S—·  3
|/   \|
w—————w  2
|\   /|
·—S—S—·  1

a b c d

The weak side has other moves after 5. ... a2-b3, of course. If 6. d2-d3, d1-d2 (traps the piece on d3); 7. a3-a2 (only move), a1-b2 (the 'pinza' again) and the strong side wins in the next move (8 moves again). If 6. d2-d1, there are more choices including a variant of the 'pinza':

Code: Select all

S to move

w—S—S—·  3
|/   \|
·—————·  2
|\   /|
S—·—S—w  1

a b c d

     w      S
6.  ...   c3-d2  // Traps the piece on d1.
7. a3-a2  a1-b2  // After the only move for w, the 'pinza' strikes again. S wins in 8 moves again.

     w      S
6.  ...   a1-a2  // Traps the piece on a3.
7. d1-d2  c3-d3  // After the only move for w, here appears a variant of the 'pinza' that does not leave two corners (only one) and blocks in the next move.
8. d2-c3  c1-d2  // Alternatively, 8. d2-d1  d3-d2

I played more than once this last blocking pattern against Ludii UCT online, not the typical pattern on two corners but this one on one file with w-S-w-S (or the symmetrical S-w-S-w) and the remaining S pieces on the centre spots.

If you look carefully, the weak side locked one piece of the strong side after 3. a2-d2 (contrary to the expected), but this could be considered greedy and triggered the variations shown above. My play on both sides might be suboptimal until the very end, of course.

I usually win Ludii UCT online in 7 to 9 moves, sometimes 10 if I get confused. Assuming 20 plies of play in this small board with low branching factor, I expect that a computer can solve the game and create DTM EGTB in few time. Please remember that there should be 810 unique positions only and, with reflections and rotations included, few thousands (3240). Optimal strategies shall be found easily, leaving this simple game with a status of more difficult than tic-tac-toe, but affordable to remember all the tricks, trap and shots to anyone who wants to take it seriously.

Regards from Spain.

Ajedrecista.

[Ajedrecista]

Hello:

I thought about the same board, but with different number of pieces that can make sense in a challenging, though small game. We now have #S = 4 pieces for the strong side and #w = 2 pieces for the weak side. I would say that #S > #w.

First of all, I take the cycle index polynomial Z(G) of my latest post and see how many unique positions are according to it. For the sake of completeness:

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#S #w   Unique positions
2 v 1           92
3 v 1          212
3 v 2          636
4 v 1          318
4 v 2          810
4 v 3         1056
5 v 1          318
5 v 2          636
5 v 3          636
5 v 4          318

Then, there are variants that I consider useless: if 4 v 2 is enough to win, is there a point on 4 v 1? I would say no, unless you want to prove the shortest optimal game. Other question arises: what initial setup must we choose in non-symmetrical scenarios like 3 v 2? I chose random setups, simply.

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2 v 1:

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w—·—·—·  3
|/   \|
·—————·  2
|\   /|
S—·—·—S  1

a b c d

I initially thought that the weak side could escape forever... I proved myself wrong in few time. Examples:

Code: Select all

     w      S
1. a3-a2  d1-d2
2. a2-a3  a1-a2
3. a3-b3  d2-c3
4. b3-a3  c3-b3

S wins in 4 moves.

==================

     w      S
1. a3-a2  d1-d2
2. a2-b1  a1-a2
3. a2-a1  d2-d1
4. a1-b1  d1-c1
5. b1-a1  c1-b1

S wins in 5 moves.

==================

     w      S
1. a3-a2  d1-d2
2. a2-b3  a1-a2
3. b3-a3  d2-d3
4. a3-b3  d3-c3
5. b3-a3  c3-b3

S wins in 5 moves.

==================

     w      S
1. a3-a2  d1-d2
2. a2-b3  a1-a2
3. b3-c3  a2-b3
4. c3-d3  b3-c3

S wins in 4 moves.

==================

     w      S
1. a3-b3  a1-a2
2. b3-a3  d1-d2
3. a3-b3  d2-c3
4. b3-a3  c3-b3

S wins in 4 moves.

==================

     w      S
1. a3-b3  a1-a2
2. b3-c3  d1-d2
3. c3-b3  d2-c3
4. b3-a3  c3-b3

S wins in 4 moves.

==================

     w      S
1. a3-b3  a1-a2
2. b3-c3  d1-d2
3. c3-d3  a2-a3
4. d3-c3  a3-b3
5. c3-d3  b3-c3

S wins in 5 moves.

My find is that the perfect game in 2 v 1 is won in 5 moves, while the wins in 4 moves are due to mistakes from the weak side. The strategy is clear for the strong side: to place a piece in a2 or d2 —avoiding to do so for the weak side— and manoeuvre with the other piece. The weak side can lose faster if goes far of the corner too early —a greedy, wrong strategy—; otherwise, the strong side must triangulate like in chess to avoid a zugzwang.

I find 3 v 1, 4 v 1 and 5 v 1 useless, following my criterium of the beginning of the post.

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3 v 2:

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w—·—·—w  3
|/   \|
·—————·  2
|\   /|
S—S—·—S  1

a b c d

With this setup, I think that the best opening move is 1. a3-a2, this is, avoiding two strong pieces to go to a2. If the initial setup had been the mirror (a piece on c1 instead of b1), then 1. d3-d2 would be the right choice for the same reason as before.

I did not find a winning strategy for the strong side. If the weak side blunders, there are positions where 3 pieces can lock 2, as seen in other post of this thread. For example a position with the strong pieces on a2, b1 and b3; and the weak pieces on a1 and a3. I show an example:

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      w      S
 1. a3-a2  d1-d2
 2. d3-c3  d2-d3
 3. c3-d2  b1-c1
 4. d2-c3  c1-d2
 5. c3-b3  d3-c3
 6. b3-a3  c3-b3
 7. a2-b1  a1-a2
 8. b1-a1  d2-d1  // 8. b1-a1? allows a manoeuvre like in 2 v 1; 8. b1-c1 and the game should be endless.
 9. a1-b1  d1-c1
10. b1-a1  c1-b1

S wins in 10 moves.

Unless I am missing something, 4 v 2 is needed and was widely analysed in this thread, because is the default variant. I find 5 v 2 useless, following my criterium of the beginning of the post.

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4 v 3:

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w—w—·—w  3
|/   \|
·—————·  2
|\   /|
S—S—S—S  1

a b c d

With this setup, I think that the best opening moves are 1. a3-a2 or 1. b3-a2 (1. a3-a2 better, explained in the proposed opening), because if a2 is not occupied, the strong side will do in the next move and will avoid two pieces of the weak side to move there later.

I did not find a winning strategy for the strong side. I propose this opening, with unclear eval:

Code: Select all

     w      S
1. a3-a2  d1-d2
2. b3-c3  c1-d1  // 2. b3-c3 does not allow the strong side to move into the weak side's half immediately.
3. c3-b3  d2-c3
4. d3-d2  b1-c1

·—w—S—·  3
|/   \|
w—————w 2
|\   /|
S—·—S—S  1

a b c d

w to move.
Unclear if S can force a win.

It is possible to lock 3 pieces with only 3: for example {a2, d2, d3} lock {a3, b3, c3}; {a2, c3, d2} lock {a3, b3, d3}; and {b1, b3, d2} lock {a1, a2, a3}.

There are positions where 4 pieces can lock 3. For example: {a1, b3, c1, d2} lock {a2, a3, d1}. Can the strong side force a win? I have not got an answer and claim 4 v 3 as variant of interest to research... though I found a fortress in 5 v 3 that can work here, depending on the initial setup of 4 v 3. Not in the 4 v 3 setup that I proposed, but yes with the weak pieces starting on {a3, b3, c3} —alternatively on {b3, c3, d3}—. Please see the reasoning at 5 v 3.

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5 v 3:

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w—w—·—w  3
|/   \|
·—————S  2
|\   /|
S—S—S—S  1

a b c d

5 v 3 appears here because 4 v 3 remains unresolved for me in terms of win/draw, let alone in terms of minimum number of moves in case of a forced win.

The board starts getting cramped. I purposedly avoided another initial setup like placing the piece on c3 instead of b3, for the immediate lose for the strong side with 1. a3-a2.

Getting back to the proposed initial setup, there is a short forced lose for the strong side:

Code: Select all

     w      S
1. a3-a2  d2-c3
2. d3-d2  c3-d3
3. b3-c3

S loses in 2 moves.

(Please remember the criterium of count only the moves of the strong side). We need to find other initial setup, for example:

Code: Select all

w—w—w—·  3
|/   \|
·—————S  2
|\   /|
S—S—S—S  1

a b c d

This is far more interesting! Here is a proposed game, surely far from perfect:

Code: Select all

     w      S
1. a3-a2  d2-d3
2. c3-d2  d3-c3
3. b3-a3  c3-b3
4. d2-d3  c1-d2
5. d3-c3  d2-d3
6. a2-d2  a1-a2  // 6. c3-d2?  d1-c1 (the 'pinza', as in 4 v 2) and S wins in 7 moves: {7. d2-c3  c1-d2} or {7. d2-d1  d3-d2}
7. d2-c1  d1-d2
8. c1-d1  b1-c1

S wins in 8 moves.

A win without obvious blunders is possible, but I might have blundered without knowing. What about a fortress, like in chess?

Code: Select all

     w      S
1. a3-a2  d2-d3
2. b3-a3  d1-d2  // 2. b3-a3! begins the idea of the fortress.
3. a3-b3  c1-d1  // 2. a3-b3! follows the key of the fortress.
4. b3-a3  b1-c1  // S tempts w to move 4. a2-b1, with unclear result. w just need to repeat b3-a3-b3-a3-... forever because S cannot move in between the w pieces.
5. a3-b3  a1-b1
6. b3-a3  b1-a1
7. a3-b3   ...

S draws.

Draw! Altough it heavily depends on the initial setup.

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5 v 4:

Code: Select all

w—w—w—w  3
|/   \|
·—————S  2
|\   /|
S—S—S—S  1

a b c d

Totally useless with this initial setup because there are immediate loses for the strong side with both 1. a3-a2 and 1. b3-a2. Other initial setups might be of interest, but such a crowded board with an only available square reminds me the 15 puzzle that Fischer was so good, which is not the spirit of this game (OK, this is a highly subjective opinion).

Regards from Spain.

Ajedrecista.

[Ajedrecista]

Hello:

I would like to see the AI digging into other old board games like senet, hounds and jackals, mehen, the Royal Game of Ur, ludus latrunculorum, XII scripta, alquerque and polis (unknown to me until right now and more difficult to investigate due to the absence of surviving boards), if they have not being investigated through AI yet, and see (new?) proposed rules and/or new conclusions. There is a list of some old board games here.

Quite a few of them are actually included in the Ludii game database. You can also play them against the computer. I don't know which ones they are actively researching, though.

There is a site that implements one of the variants of Ludus Coriovalli at Ludii Portal:

https://nunc.ch/domus-les-articles-du-b ... oriovalli/

The only 'but' is that this site forces to move the strong side first, opposite of what I am used to. The same tactics that weak side moving first apply, though the optimal opening moves might change a little. OTOH, I classify this AI as stronger than the UCT algorithm at Ludii. The AI of this Swiss site is divided in three levels.

I managed to win the strongest level in 8 moves of the strong side after few trials, changing my typical corner-to-centre move (a1-a2) to b1-a2 move in order to bring the 'pinza' setup at the right time. Remember, the strong side plays for the zugzwang and it heavily depends on who moves first. Here is a sample game:

Code: Select all

S starts.

w—·—·—w  3
|/   \|
·—————·  2
|\   /|
S—S—S—S  1

a b c d

     S      w
1. b1-a2  d3-d2
2. a2-b3  a3-a2
3. a1-b1  d2-d3
4. d1-d2  d3-c3
5. d2-d3  c3-d2
6. d3-c3  d2-d3
7. c1-d2  a2-a1
8. b3-a2

S wins in 8 moves.

·—·—S—w  3
|/   \|
S—————S  2
|\   /|
w—S—·—·  1

a b c d

I feel this AI to be very deterministic, so no variety expected, at least in the strongest level. This can be the cause of have been found optimal strategies in such a small game, but these are big words to say.

------------

One level up, this site writes about other ancient board games:

https://nunc.ch/domus-les-articles-du-b ... xantiques/

Aseb game (a kind of abbreviated Senet, but not the Royal Game of Ur) can also be played in this site against three levels of an AI. From an article of Wikipedia in French language:

[...]

Le jeu royal d'Ur, ou jeu des vingt cases, est joué avec un ensemble de pions sur un plateau richement décoré et date d'environ 2600 av. J.-C. C'est un jeu de parcours qui utilise des dés en os. Ce jeu est également connu et joué en Égypte sous le nom d'Aseb, avec un diagramme un peu différent mais toujours avec 20 cases. Un traité babylonien sur le jeu écrit sur une tablette d'argile montre que le jeu avait une signification astronomique et qu'il pouvait également être utilisé pour la divination.

[...]

Other games listed can be played online at Locus Ludi site. I won a game of Ludus latrunculorum (Séneca's variation) against an IA without knowing the rules!

Regards from Spain.

Ajedrecista.

[Ajedrecista]

Hello:

Yesterday, I deemed the AI at NUNC as stronger than the AI at Ludii Portal. Let us see if this is true with a game pair (error bars!), reversing sides.

First of all, I took the AI from here:

• Ludii (UCT) from here.

• Centurion level from here.

Secondly, the weak side starts at Ludii and the strong side starts at NUNC. To override that difference, I just triangulated manually at Ludii (a3-a2, d1-d2; a2-b3, d2-d1; b3-a3) and let both AI play to play from there. Here we go!

Code: Select all

S starts.

w—·—·—w  3
|/   \|
·—————·  2
|\   /|
S—S—S—S  1

a b c d

[S "NUNC (Centurion)"]
[w "Ludii (UCT)"]

      S      w
 1. a1-a2  a3-b3
 2. d1-d2  b3-a3
 3. d2-c3  d3-d2
 4. c3-b3  d2-c3
 5. c1-d2  c3-d3
 6. b3-c3  a3-b3
 7. b1-c1  b3-a3
 8. c3-b3  d3-c3
 9. c1-d1  c3-d3
10. b3-c3  a3-b3
11. a2-a3  b3-a2
12. d2-c1  a2-b1
13. c1-d2  b1-a1
14. d1-c1  a1-a2
15. a3-b3  a2-a3
16. b3-a2  a3-b3
17. c1-b1  b3-a3
18. c3-b3  d3-c3
19. b1-c1  c3-d3
20. c1-d1  d3-c3
21. d2-d3  c3-d2
22. d1-c1  d2-d1
23. d3-d2

S wins in 23 moves.

w—S—·—w·  3
|/   \|
S—————S  2
|\   /|
·—·—S—w  1

a b c d

Code: Select all

S starts.

w—·—·—w  3
|/   \|
·—————·  2
|\   /|
S—S—S—S  1

a b c d

[S "Ludii (UCT)"]
[w "NUNC (Centurion)"]

      S      w
 1. d1-d2  a3-a2
 2. d2-c3  d3-d2
 3. c3-b3  d2-d3
 4. b3-a3  a2-d2
 5. a3-b3  d2-c3
 6. c1-d2

S wins in 6 moves.

·—S—w—w  3
|/   \|
·—————S  2
|\   /|
S—S—·—·  1

a b c d

Ludii (UCT) won the match by a large margin of moves, 17.

NUNC (Centurion) was a total disappointment in this match: played horrible with the strong side, repeating and shuffling a lot; as well as with the weak side, trapping itself very early. This is the opposite of what I thought yesterday, when I played against NUNC (Centurion). Still, I should win either AI (Ludii or NUNC) with me playing the strong side in 8 moves at most, with enough practice.

However, I find the NUNC online GUI as more reliable than Ludii online GUI because Ludii sometimes freezes and one piece of the AI vanishes just after moving, only reappearing after the next move by the human.

------------

It is difficult to establish a rating system for this unbalanced game, so pairs of games reversing sides must be played, even more important than in chess. Other issue is to decide what limit of moves is sensible to claim a draw by insufficient knowledge of the strong side.

• The first tiebreak should be the percentage of won games. Unless big blunders where the weak side wins or the strong side does not know how to lock the weak side, this should be always 50%, needing the second tiebreak.

• The second tiebreak should be the difference in the number of moves, for example ±17 in the former match. Even better, averaging these differences with the number of matches if we have players with different number of matches.

Then, imagine a double Round-Robin like in recent Candidates Tournaments:

Code: Select all

Win:  a win playing the strong side.
Draw: the strong side can not lock the weak side in 50 moves.
Lose: a lose playing the strong side.

TB1: performance.
TB2: for a given player (n < 50 and m < 50):
       S wins  in  n moves counts as   -n   > -50
       S draws in 50 moves counts as  -50   = -50
       S loses in  n moves counts as -100+n < -50
       w wins  in  m moves counts as +100-m > +50
       w draws in 50 moves counts as  +50   = +50
       w loses in  m moves counts as   +m   < +50


A wins  B in  8 moves. \
                        ——— A 1.0-1.0 B →  +2 for A and  -2 for B
B wins  A in 10 moves. /


A wins  C in  7 moves. \
                        ——— A 1.5-0.5 C → +43 for A and -43 for C
C draws A in 50 moves. /


A wins  D in  7 moves. \
                        ——— A 2.0-0.0 D → +83 for A and -83 for D
D loses A in 10 moves. /


B wins  C in  9 moves. \
                        ——— B 1.5-0.5 C → +41 for B and -41 for C
C draws B in 50 moves. /


B wins  D in  8 moves. \
                        ——— B 2.0-0.0 D → +79 for B and -79 for D
D loses B in 13 moves. /


C draws D in 50 moves. \
                        ——— C 1.0-1.0 D →   0 for C and   0 for D
D draws C in 50 moves. /

Code: Select all

    TB1     TB2
A: 4.5/6   +128
B: 4.5/6   +118
C: 2.0/6   - 84
D: 1.0/6   -162

With a better pool of players, where the strong side always win within the limit of moves:

Code: Select all

Win:  a win playing the strong side.
Draw: the strong side can not lock the weak side in 50 moves.
Lose: a lose playing the strong side.

TB1: performance.
TB2: for a given player (n < 50 and m < 50):
       S wins  in  n moves counts as   -n   > -50
       S draws in 50 moves counts as  -50   = -50
       S loses in  n moves counts as -100+n < -50
       w wins  in  m moves counts as +100-m > +50
       w draws in 50 moves counts as  +50   = +50
       w loses in  m moves counts as   +m   < +50


A wins B in  8 moves. \
                        ——— A 1-1 B → +2 for A and -2 for B
B wins A in 10 moves. /


A wins C in  7 moves. \
                        ——— A 1-1 C → +4 for A and -4 for C
C wins A in 11 moves. /


A wins D in  7 moves. \
                        ——— A 1-1 D → +3 for A and -3 for D
D wins A in 10 moves. /


B wins C in  9 moves. \
                        ——— B 1-1 C → +6 for B and -6 for C
C wins B in 15 moves. /


B wins D in  8 moves. \
                        ——— B 1-1 D → +5 for B and -5 for D
D wins B in 13 moves. /


C wins D in 20 moves. \
                        ——— C 1-1 D → +4 for C and -4 for D
D wins C in 24 moves. /

Code: Select all

    TB1     TB2
A: 3.0/6    + 9
B: 3.0/6    + 9
C: 3.0/6    - 6
D: 3.0/6    -12

Another tiebreak could be direct encounter, thus A winning over B in the last tournament. An open question is the limit of 50 moves, which can be easily lowered to 40 or 30 and the tiebreaks of the difference of moves change accordingly.

Regards from Spain.

Ajedrecista.