ejsoon 写了： 用繁體中文翻譯以下文段：
I would like to answer some questions that came up regarding
the solution of Nine Men's Morris.
- How can a draw be reached?
A draw is reached through cyclical play, whereby neither
player can (or will) close a mill. Note that since the proof
was obtained using only the 9-9, 9-8 and 8-8 databases, neither
player need close more than one mill to reach a draw ,thereby
avoiding a loss with less than three stones.
- What is an optimal opening strategy (for humans)?
I must admit that I have not examined the opening moves
for any easily describable patterns.
However, the examination of mid- and endgame databases
has repeatedly shown optimal play to be beyond human
comprehension. Therefore, I very much doubt that any such
simple-to-describe strategy exists.
- Is a paper describing these results available?
I am now working on such a paper. I will send a copy to those
of you who have requested one as soon as it is completed.
There is however a paper describing some initial results
of the retrograde analysis phase of the project:
Gasser, R., "Applying Retrograde Analysis to Nine Men's
Morris", Heuristic Programming in Artificial
Intelligence 2: the Second Computer Olympiad, Levy, D.N.L.
and Beal, D.F. (Eds.), pp. 161-173, Ellis Horwood, London.
- Is the program available?
Making the program available is something of a problem
because of the databases it uses. In total the databases
amount to a few hundred MBytes.
There is of course a program available which plays without
any database (or with a few small db's). While this program
cannot claim to be invincible, it plays quite strongly, having
won a match against the British Champion.
This program was developed under the Smart Game Board and
therefore only runs on Macintosh computers. The program
is shareware and depending on the number of requests, I will
either make it available via ftp or snail mail.
Ralph Gasser (gasser@inf.ethz.ch)
Informatik, ETH
CH-8092 Zurich
Switzerland
Newsgroups: rec.games.abstract
From: colley@qucis.queensu.ca (Paul Colley)
Subject: Re: Nine Men's Morris: More about the draw
Organization: Computing & Information Science, Queen's University at Kingston
Date: Wed, 24 Nov 1993 15:41:54 GMT
In article <1993Nov24.103138.21488@neptune.inf.ethz.ch>
gasser@inf.ethz.ch (Ralph Gasser) writes:
>A draw is reached through cyclical play, whereby neither
>player can (or will) close a mill. Note that since the proof
>was obtained using only the 9-9, 9-8 and 8-8 databases, neither
>player need close more than one mill to reach a draw ,thereby
>avoiding a loss with less than three stones.
This appears to be a mistake. Er, let me rephrase that. "The evidence
supplied does not prove your conclusion."
In building the 9-9 and 9-8 databases, you would have used the smaller
data bases. Retrograde analysis starts from the smallest, and works
up.
Say some position in the 8-7 database is a draw. Then in the 8-8
database, a position where the player may close a mill and reduce to
the drawn 8-7 position gets marked as a draw (assuming there's no
winning move).
You don't need the 8-7 database in doing the opening search, because
you don't need to know why the particular 8-8 position is drawn. This
is NOT the same thing as
>neither player need close more than one mill to reach a draw
The fact that you didn't need the smaller databases in the opening
search just means that no player need close more than one mill *DURING
THE OPENING* to reach a draw. It doesn't mean the player doesn't need
to close a mill during the middle or end game.
The correct statement, based on the evidence you give, is "neither
player need close more than one mill, and only during the opening, to
reach a book draw." Well, now that the initial position of NMM is
solved, we could even say "neither player needs to make a move to reach
a book draw." Of course, if your opponent doesn't agree, you still
have to play out the book draw, and that (may) require the smaller
databases. Or may not require the smaller databases. I have nothing
to contradict your statement. I'm just asking for more support.
玩玩Chatgpt
Re: 玩玩Chatgpt
一个聊天机器人🤖
Re: 玩玩Chatgpt
用繁體中文翻譯以下文段：
I would like to answer some questions that came up regarding
the solution of Nine Men's Morris.
- How can a draw be reached?
A draw is reached through cyclical play, whereby neither
player can (or will) close a mill. Note that since the proof
was obtained using only the 9-9, 9-8 and 8-8 databases, neither
player need close more than one mill to reach a draw ,thereby
avoiding a loss with less than three stones.
- What is an optimal opening strategy (for humans)?
I must admit that I have not examined the opening moves
for any easily describable patterns.
However, the examination of mid- and endgame databases
has repeatedly shown optimal play to be beyond human
comprehension. Therefore, I very much doubt that any such
simple-to-describe strategy exists.
https://ejsoon.vip/
弈趣極光：享受思維樂趣
弈趣極光：享受思維樂趣
Re: 玩玩Chatgpt
我想要回答一些關於九子棋解法的問題。ejsoon 写了： 用繁體中文翻譯以下文段：
I would like to answer some questions that came up regarding
the solution of Nine Men's Morris.
- How can a draw be reached?
A draw is reached through cyclical play, whereby neither
player can (or will) close a mill. Note that since the proof
was obtained using only the 9-9, 9-8 and 8-8 databases, neither
player need close more than one mill to reach a draw ,thereby
avoiding a loss with less than three stones.
- What is an optimal opening strategy (for humans)?
I must admit that I have not examined the opening moves
for any easily describable patterns.
However, the examination of mid- and endgame databases
has repeatedly shown optimal play to be beyond human
comprehension. Therefore, I very much doubt that any such
simple-to-describe strategy exists.
- 如何達成平手？
平手是透過循環的遊戲過程達成的，在這種情況下，任何一方都無法（或不願意）關閉一個磨坊。請注意，由於證明僅使用 9-9、9-8 和 8-8 的資料庫獲得，因此任何一方都不需要關閉超過一個磨坊即可達成平手，從而避免少於三顆棋子時的失敗。
- 什麼是最佳的開局策略（對於人類來說）？
我必須承認，我尚未在開局走法中檢視任何易於描述的模式。
然而，對中局和殘局資料庫的檢視一再表明，最佳玩法超出了人類的理解範疇。因此，我非常懷疑是否存在任何易於描述的策略。
一个聊天机器人🤖
Re: 玩玩Chatgpt
用繁體中文翻譯以下英文：
For the rules of Nine Men’s Morris, see for example
Gasser’s work [6]. There are two points in the rules, for which
there is no consensus. The first question is what should happen
when two mills are closed with one move. The second is
whether a player should be allowed to take a stone from
a mill, when all the opponent’s stones are in mills. Our
implementation followed Gasser’s decisions: mill closure is
always followed by taking exactly one stone. Moreover, we
regard position repetitions as draws.
The main difference between Lasker Morris [7] and the
standard variant is that there are no distinct placement and
movement phases, i.e. the players can decide at every move
whether they want to place a stone on the board or move one
of their stones (as long as they have remaining stones to place).
The other difference is that players can place 10 stones instead
of 9.
Morabaraba differs from the standard variant in the graph
of the game board (see Fig. 3 and Fig. 4), and in the number
of stones the players can place, which is 12. There is also a
special rule: if the board becomes full (because neither player
closed a mill during the opening), the game ends in a draw.
There is no consensus about whether to use the rule or not, so
we implemented both versions. The state space of Morabaraba
is about four times bigger than that of the standard variant.
https://ejsoon.vip/
弈趣極光：享受思維樂趣
弈趣極光：享受思維樂趣
Re: 玩玩Chatgpt
九子棋的規則，請參考 Gasser 的著作 [6]。規則中有兩個方面尚未達成共識。第一個問題是，如果一步棋同時形成兩個磨坊，應該怎麼處理。第二個問題是，如果對手的所有棋子都在磨坊中，玩家是否可以從磨坊中取走棋子。我們的實作遵循 Gasser 的決定：形成磨坊後，必須取走一顆棋子。此外，我們將重複出現的局面視為和局。ejsoon 写了： 用繁體中文翻譯以下英文：
For the rules of Nine Men’s Morris, see for example
Gasser’s work [6]. There are two points in the rules, for which
there is no consensus. The first question is what should happen
when two mills are closed with one move. The second is
whether a player should be allowed to take a stone from
a mill, when all the opponent’s stones are in mills. Our
implementation followed Gasser’s decisions: mill closure is
always followed by taking exactly one stone. Moreover, we
regard position repetitions as draws.
The main difference between Lasker Morris [7] and the
standard variant is that there are no distinct placement and
movement phases, i.e. the players can decide at every move
whether they want to place a stone on the board or move one
of their stones (as long as they have remaining stones to place).
The other difference is that players can place 10 stones instead
of 9.
Morabaraba differs from the standard variant in the graph
of the game board (see Fig. 3 and Fig. 4), and in the number
of stones the players can place, which is 12. There is also a
special rule: if the board becomes full (because neither player
closed a mill during the opening), the game ends in a draw.
There is no consensus about whether to use the rule or not, so
we implemented both versions. The state space of Morabaraba
is about four times bigger than that of the standard variant.
拉斯克九子棋 [7] 與標準變體的主要區別在於，它沒有明確的放置和移動階段，也就是說，玩家可以在每一步決定是要在棋盤上放置棋子，還是移動其中一顆棋子（只要他們還有棋子可以放置）。另一個區別是，玩家可以放置 10 顆棋子，而不是 9 顆。
莫拉巴拉巴棋與標準變體的區別在於棋盤的圖形（參見圖 3 和圖 4），以及玩家可以放置的棋子數量，為 12 顆。還有一條特殊規則：如果棋盤被填滿（因為雙方玩家在開局階段都沒有形成磨坊），則比賽以平局結束。關於是否使用此規則，目前尚無共識，因此我們實作了兩個版本。莫拉巴拉巴棋的狀態空間大約是標準變體的四倍。
一个聊天机器人🤖
Re: 玩玩Chatgpt
用繁體中文翻譯以下文段：
Two points are subject to debate among Nine Men’s Morris enthusiasts. The
first hinges on the observation that in the opening it is possible to close two
mills simultaneously. Should the player then be allowed to remove one or two
opponent’s stones? Our implementation only allows one stone to be removed.
The second point concerns positions where the player to move has just closed a
mill, but all the opponent’s stones are also in mills. May she then remove a stone
or not? In our implementation she may remove any stone. It seems unlikely that
either of these rule variations affect the value of the game.
https://ejsoon.vip/
弈趣極光：享受思維樂趣
弈趣極光：享受思維樂趣
Re: 玩玩Chatgpt
九子棋愛好者之間有兩個爭論點。ejsoon 写了： 用繁體中文翻譯以下文段：
Two points are subject to debate among Nine Men’s Morris enthusiasts. The
first hinges on the observation that in the opening it is possible to close two
mills simultaneously. Should the player then be allowed to remove one or two
opponent’s stones? Our implementation only allows one stone to be removed.
The second point concerns positions where the player to move has just closed a
mill, but all the opponent’s stones are also in mills. May she then remove a stone
or not? In our implementation she may remove any stone. It seems unlikely that
either of these rule variations affect the value of the game.
第一個爭論點在於開局時有可能同時關閉兩個磨坊。玩家是否應該被允許移除一顆或兩顆對手的棋子？我們的實作只允許移除一顆棋子。
第二個爭論點涉及到玩家剛關閉一個磨坊，但對手的所有棋子也在磨坊中的情況。她是否可以移除一顆棋子？在我們的實作中，她可以移除任何棋子。這兩個規則變化的任一種似乎都不太可能影響遊戲的價值。
一个聊天机器人🤖
Re: 玩玩Chatgpt
用繁體中文翻譯以下文段：
After all stones have been placed, play proceeds to the midgame. Here a
player may slide one of her stones to an adjacent vacant point. If at any time
during the game a player succeeds in arranging three of her stones in a row—this
is known as closing a mill—she may remove any opponent’s stone that is not
part of a mill. In Figure 2, if White closes a mill in the opening by playing to
b6, she can now remove Black’s stone on a1, but not the one on d2.
As soon as a player has only three stones left, the endgame commences. When
it is her turn, the player with three stones may jump one of her stones to any
vacant point on the board.
https://ejsoon.vip/
弈趣極光：享受思維樂趣
弈趣極光：享受思維樂趣
Re: 玩玩Chatgpt
在所有棋子都放置好後，遊戲進入中盤階段。在此階段，玩家可以將自己的其中一顆棋子滑動到相鄰的空位。如果在遊戲中的任何時間點，玩家成功地將三顆自己的棋子排列成一排（稱為「成磨」），她就可以移除任何不屬於磨坊的對手棋子。在圖2中，如果白方在開局時下到b6，形成磨坊，她現在可以移除黑方在a1的棋子，但不能移除d2的棋子。ejsoon 写了： 用繁體中文翻譯以下文段：
After all stones have been placed, play proceeds to the midgame. Here a
player may slide one of her stones to an adjacent vacant point. If at any time
during the game a player succeeds in arranging three of her stones in a row—this
is known as closing a mill—she may remove any opponent’s stone that is not
part of a mill. In Figure 2, if White closes a mill in the opening by playing to
b6, she can now remove Black’s stone on a1, but not the one on d2.
As soon as a player has only three stones left, the endgame commences. When
it is her turn, the player with three stones may jump one of her stones to any
vacant point on the board.
一旦玩家只剩下三顆棋子，遊戲便進入終盤階段。輪到擁有三顆棋子的玩家時，她可以將其中一顆棋子跳到棋盤上的任何空位。
一个聊天机器人🤖
Re: 玩玩Chatgpt
用繁體中文翻譯以下文段：
Fig. 3: Game-theoretic values with DTWs of all possible
moves in the starting position of Morabaraba. The optimal
moves are marked with an “!”.
https://ejsoon.vip/
弈趣極光：享受思維樂趣
弈趣極光：享受思維樂趣
Re: 玩玩Chatgpt
圖 3：使用動態時間規整 (DTW) 計算出的 Morabaraba 遊戲起始位置所有可能移動的賽局理論值。最佳移動以 "!" 標記。ejsoon 写了： 用繁體中文翻譯以下文段：
Fig. 3: Game-theoretic values with DTWs of all possible
moves in the starting position of Morabaraba. The optimal
moves are marked with an “!”.
一个聊天机器人🤖
Re: 玩玩Chatgpt
用繁體中文翻譯以下文段：
Fig. 4: A game-theoretical draw position in the standard or
Lasker variant. White to move. Although White can preserve
his material advantage, he cannot win. This is one of the only
two such positions in the subspace of 8 white and 3 black
stones. The only move here that can keep this stone count
indefinitely is marked with an arrow.
https://ejsoon.vip/
弈趣極光：享受思維樂趣
弈趣極光：享受思維樂趣
Re: 玩玩Chatgpt
圖 4：標準或拉斯克變體中的博弈論平局局面。輪到白棋走。儘管白棋可以保持其物質優勢，但他無法獲勝。這是 8 個白棋和 3 個黑棋的子空間中僅有的兩個此類局面之一。這裡唯一可以無限期保持此棋子數量的走法用箭頭標記。ejsoon 写了： 用繁體中文翻譯以下文段：
Fig. 4: A game-theoretical draw position in the standard or
Lasker variant. White to move. Although White can preserve
his material advantage, he cannot win. This is one of the only
two such positions in the subspace of 8 white and 3 black
stones. The only move here that can keep this stone count
indefinitely is marked with an arrow.
一个聊天机器人🤖
Re: 玩玩Chatgpt
用繁體中文翻譯以下文段：
We strongly solved Morabaraba, which turned out to be
a win for the first player in 49 moves. We also calculated
extended strong solutions for Nine Men’s Morris, Lasker
Morris, and Morabaraba which provided some insights into
these games (and also confirmed the result of Stahlhacke [8]
that Lasker Morris is a draw).
Furthermore, we developed a multi-valued retrograde anal-
ysis. Then we modified this to have an algorithm which can
solve these games ultra-strongly. This means that the program
has a much higher chance to win a game of Nine Men’s
Morris or Lasker Morris against a fallible opponent instead
of just drawing it, compared to a program which uses only a
strong solution. This is important, because when a program is
playing Nine Men’s Morris based only on the strong solution,
it is surprisingly easy for the opponent to achieve a draw. The
algorithm classifies draws into subclasses based on a heuristic
value of the subspace that can be reached via perfect play. We
investigated two heuristics for the subspace values.
We compared our ultra-strong solution to the strong solution
by having them play against a heuristic (α-β) program, and
against human players, and found that it achieved wins more
often.
https://ejsoon.vip/
弈趣極光：享受思維樂趣
弈趣極光：享受思維樂趣
Re: 玩玩Chatgpt
我們強解了 Morabaraba，結果證明先手玩家可在 49 步內獲勝。我們也計算了九子棋、拉斯克棋和 Morabaraba 的擴展強解，這為我們提供了對這些遊戲的一些洞察（也證實了 Stahlhacke [8] 的結果，即拉斯克棋是平局）。ejsoon 写了： 用繁體中文翻譯以下文段：
We strongly solved Morabaraba, which turned out to be
a win for the first player in 49 moves. We also calculated
extended strong solutions for Nine Men’s Morris, Lasker
Morris, and Morabaraba which provided some insights into
these games (and also confirmed the result of Stahlhacke [8]
that Lasker Morris is a draw).
Furthermore, we developed a multi-valued retrograde anal-
ysis. Then we modified this to have an algorithm which can
solve these games ultra-strongly. This means that the program
has a much higher chance to win a game of Nine Men’s
Morris or Lasker Morris against a fallible opponent instead
of just drawing it, compared to a program which uses only a
strong solution. This is important, because when a program is
playing Nine Men’s Morris based only on the strong solution,
it is surprisingly easy for the opponent to achieve a draw. The
algorithm classifies draws into subclasses based on a heuristic
value of the subspace that can be reached via perfect play. We
investigated two heuristics for the subspace values.
We compared our ultra-strong solution to the strong solution
by having them play against a heuristic (α-β) program, and
against human players, and found that it achieved wins more
often.
此外，我們開發了一種多值逆向分析法。然後，我們對其進行修改，使其成為一種能夠超強解這些遊戲的算法。這意味著，與僅使用強解的程序相比，該程序在與容易犯錯的對手進行九子棋或拉斯克棋遊戲時，更有可能獲勝，而不僅僅是平局。這一點很重要，因為當一個程序僅基於強解來玩九子棋時，對手很容易就能達成平局。該算法根據可通過完美遊戲達成的子空間的啟發式值，將平局分為子類。我們研究了兩種用於子空間值的啟發式方法。
我們通過讓我們的超強解與啟發式（α-β）程序和人類玩家對戰，將其與強解進行比較，發現它更頻繁地取得勝利。
一个聊天机器人🤖
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