问题描述
我正在努力解决一个约束问题（它会因大值和/或如果我尝试优化而不是仅仅寻找任何解决方案而崩溃）。我已经根据之前一些问题的建议采取了一些措施来打破搜索空间，但它仍然停滞不前。是否有更多技术可以帮助我优化此计算？
%%% constants %%%
#const nRounds = 4.
#const nPlayers = 20.
#const nRooms = 4.
#const nDecks = 7.
player(1..nPlayers).
room(1..nRooms).
deck(1..nDecks).
writer(1,1;2,2;3,3;4,4).
% For reference - that's what I started with:
%nRounds { seat(Player,1..nRooms,1..nDecks) } nRounds :- player(Player).
% Now instead I'm using a few building blocks
% Each player shall only play nRounds decks
nRounds { what(Player,1..nDecks) } nRounds :- player(Player).
% Each player shall only play in up to nRounds rooms.
1 { where(Player,1..nRooms) } nRounds :- player(Player).
% For each deck,3 or 4 players can play in each room.
3 { who(1..nPlayers,Room,Deck) } 4 :- room(Room),deck(Deck).
% Putting it all together,hopefully,this leads to fewer combinations than the original monolithic choice rule.
{ seat(Player,Deck) } :- what(Player,Deck),where(Player,Room),who(Player,Deck).
% A player can only play a deck in a single room.
:- seat(Player,Room1,seat(Player,Room2,Room1 != Room2.
% A player must play nRounds decks overall.
:- player(Player),#count { Room,Deck: seat(Player,Deck) } != nRounds.
% Any deck in any room must be played by 3-4 players.
legal_player_count(3..4).
:- room(Room),deck(Deck),Players = #count { Player: seat(Player,Deck) },Players > 0,not legal_player_count(Players).
% Writers cannot play their own decks.
:- writer(Player,_,Deck).
% At least one non-playing player per room.
:- deck(Deck),Playing = #count { Player,Room: seat(Player,Rooms = #count { Room: seat(_,nPlayers - Playing < Rooms.
%:- room(R1),deck(D),room(R2),X = #sum { P: seat(P,R1,D) },Y = #sum { P: seat(P,R2,R1 > R2,X > Y.
#minimize { D: decks(D) }.
#show decks/1.
#show seat/3.
% #show common_games/3.
何时，或者如果，这变得可管理，我希望添加更多优化目标，以选择以下方面的最佳配置：
% Input points(P,R,D,X) to report points.
% winner(P,D) :- points(P,X),X = #max { Y : points(_,Y) }.
% Compute each player's rank based on each round:
% rank(P,R) :- points(P,winner(Winner,D),D_ = D - 1,% rank(P,D_,R_),% R = some_combination_of(X,P=Winner,R_).
% latest_rank(P,R) :- D = #max { DD: rank(P,DD,_) },rank(P,R).
% Total number of decks played throughout the night (for minimisation?)
decks(Decks) :- Decks = #count { Deck: seat(_,Deck) }.
% Total number of games played together by the same players (for minimisation)
% The total sum of this predicate is invariant
% Minimisation should took place by a superlinear value (e.g. square)
common_games(Player1,Player2,Games) :-
player(Player1),player(Player2),Player1 != Player2,Games = #count { Room,Deck:
seat(Player1,seat(Player2,Deck)
},Games > 0.
% For example:
% common_game_penalty(X) :- X = #sum { Y*Y,P1,P2 : common_games(P1,P2,Y) }.
% Another rank-based penalty needs to be added once the rank mechanics are there
% Then the 2 types of penalties need to be combined and / or passed to the optimiser
更新 - 问题描述
- P 玩家聚集在一起进行测验之夜。 D 甲板和 R 室是 可用播放。
- 每个房间只能容纳 3 或 4 名玩家（由于游戏规则，而不是空间）。
- 每个 Deck 最多播放一次，并同时在多个房间播放 - 所以从某种意义上说，Deck 是“Round”的同义词。
- 每位玩家最多只能玩同一副牌。
- 每位玩家在夜间只能玩 N 次（N 几乎是固定的，它是 4）。
- 所以如果在晚上玩 9 副套牌（即如果有很多玩家 现在），每个人将播放这 9 个中的 4 个。
- 因此，没有必要让每个玩家都参加每个“套牌/回合”。事实上，每副套牌都有一个作者，通常是其中一个玩家
- 当然，作者不能打自己的套牌，所以他们必须在那轮比赛中待在外面。此外，对于每一副牌/轮， 必须有人阅读每个房间的问题，所以如果 16 名玩家 现在有4个房间，16个玩家不可能 玩。可以有 4 个房间，每个房间有 3 个玩家（并且 剩下的 4 名玩家读出问题）或拥有 3 个房间，其中 4 个 每位玩家（3 名玩家朗读问题，1 名玩家 观看）。
希望这能消除混乱，如果没有，我可以尝试提供更详细的示例，但基本上，假设您有 4 个房间和 30 个玩家：
- 您选择 16 名将参加比赛，另外 4 名将宣读问题
- 然后你有 16 人玩了他们的 1/4 套牌/轮，还有 14 人仍然在 0/4
- 那么您可以让其他 14 人玩（每个房间 4、4、3、3 名玩家）或继续最大化房间效用，以便在第二轮之后每个人至少玩一次，2/30 名玩家已经玩了 2/4 场比赛。
- 然后你继续挑选一些人，直到每个人都玩了 4 副套牌/轮。
附言你有两个回合的概念 - 一个是个人层面，每个人都有 4 个可以玩，另一个是联盟级别，有一定数量的套牌>4，每个套牌在在场的每个人眼中都被认为是“一轮”。据我了解，这是关于设置的最令人困惑的一点，我在开始时没有很好地说明。
解决方法
我已经用你的新规范重写了编码，没有进行太多优化来解决问题。
备注： 我认为“阅读问题”的人是作家？ 我保证每个房间有 1 位作家可用，但我没有说出它的名字。
#const nPlayers = 20.
#const nRooms = 4.
#const nDecks = 6.
player(1..nPlayers).
room(1..nRooms).
deck(1..nDecks).
% player P plays in room R in round D
{plays(P,R,D)} :- deck(D),room(R),player(P).
% a player may only play in a single room each round
:- player(P),deck(D),1 < #sum {1,R : plays(P,D)}.
% not more than 4 players per room
:- deck(D),4 < #sum {1,P : plays(P,D)}.
% not less than 3 players per room
:- deck(D),3 > #sum {1,D)}.
plays(P,D) :- plays(P,D).
% at least one writer per room (we need at least one player not playing for each room,we do not care who does it)
:- deck(D),nRooms > #sum {1,P : not plays(P,D),player(P)}.
% each player only plays 4 times during the night
:- player(P),not 4 = #sum {1,D : plays(P,D)}.
#show plays/3.
%%% shortcut if too many decks are used,each player can only play 4 times but at least 3 players have to play in a room (currently there is no conecpt of an empty room)
:- 3*nRooms*nDecks > nPlayers*4.
请注意，我添加了最后一个约束，因为您的初始配置无法解决（每个玩家必须恰好玩 4 轮，我们有 20 个玩家，这是 80 个单独的游戏。考虑到至少 3 个玩家必须在一个房间，我们有 4 个房间和 7 个甲板，这是 3 * 4 * 7 = 84，我们至少需要玩 84 场个人游戏）。你也可以计算我认为的甲板数量。