SET游戏赔率模拟（MATLAB）

Question

我最近发现了很棒的卡片 - SET。简而言之，有81张卡片具有以下四个特征：符号（椭圆形，波浪形或菱形），颜色（红色，紫色或绿色），数字（一，二或三）和着色（实心，条纹或开放）。任务是找到（从选定的12张卡片）一套3张卡片，其中每张卡片上的四个特征中的每一个都是相同的，或者每张卡片上的所有特征都是不同的（没有2 + 1组合）。

我在MATLAB中对其进行编码以找到解决方案，并估计随机选择卡片中设置一组的几率。

这是我估算赔率的代码：

%% initialization
K = 12; % cards to draw
NF = 4; % number of features (usually 3 or 4)
setallcards = unique(nchoosek(repmat(1:3,1,NF),NF),'rows'); % all cards: rows - cards, columns - features
setallcomb = nchoosek(1:K,3); % index of all combinations of K cards by 3

%% test
tic
NIter=1e2; % number of test iterations
setexists = 0; % test results holder
% C = progress('init'); % if you have progress function from FileExchange
for d = 1:NIter
% C = progress(C,d/NIter);    

% cards for current test
setdrawncardidx = randi(size(setallcards,1),K,1);
setdrawncards = setallcards(setdrawncardidx,:);

% find all sets in current test iteration
for setcombidx = 1:size(setallcomb,1)
    setcomb = setdrawncards(setallcomb(setcombidx,:),:);
    if all(arrayfun(@(x) numel(unique(setcomb(:,x))), 1:NF)~=2) % test one combination
        setexists = setexists + 1;
        break % to find only the first set
    end
end
end
fprintf('Set:NoSet = %g:%g = %g:1\n', setexists, NIter-setexists, setexists/(NIter-setexists))
toc

100-1000次迭代很快，但要小心多了。在我的家用电脑上进行一百万次迭代大约需要15个小时。无论如何，有12张卡和4个功能，我有13：1有一套。这实际上是一个问题。说明书说这个数字应该是33：1。它最近由Peter Norvig确认。他提供了Python代码，但我还没有测试它。

那你能找到错误吗？欢迎任何关于性能改进的评论。

Answer 1

这是一个矢量化版本，可以在大约一分钟内计算1M手。我得到了大约28：1，因此找到“所有不同的”套装可能仍然会有一些东西。我的猜测是，这也是您的解决方案遇到的问题。

%# initialization
K = 12; %# cards to draw
NF = 4; %# number of features (this is hard-coded to 4)
nIter = 100000; %# number of iterations

%# each card has four features. This means that a card can be represented
%# by a coordinate in 4D space. A set is a full row, column, etc in 4D
%# space. We can even parallelize the iterations, at least as long as we
%# have RAM (each hand costs 81 bytes)
%# make card space - one dimension per feature, plus one for the iterations
cardSpace = false(3,3,3,3,nIter);

%# To draw cards, we put K trues into each cardSpace. I can't think of a
%# good, fast way to draw exactly K cards that doesn't involve calling
%# unique
for i=1:nIter
    shuffle = randperm(81) + (i-1) * 81;
    cardSpace(shuffle(1:K)) = true;
end

%# to test, all we have to do is check whether there is any row, column,
%# with all 1's
isEqual = squeeze(any(any(any(all(cardSpace,1),2),3),4) | ...
    any(any(any(all(cardSpace,2),1),3),4) | ...
    any(any(any(all(cardSpace,3),2),1),4) | ...
    any(any(any(all(cardSpace,4),2),3),1));
%# to get a set of 3 cards where all symbols are different, we require that
%# no 'sub-volume' is completely empty - there may be something wrong with this
%# but since my test looked ok, I'm not going to investigate on Friday night
isDifferent = squeeze(~any(all(all(all(~cardSpace,1),2),3),4) & ...
    ~any(all(all(all(~cardSpace,1),2),4),3) & ...
    ~any(all(all(all(~cardSpace,1),3),4),2) & ...
    ~any(all(all(all(~cardSpace,4),2),3),1));

isSet = isEqual | isDifferent;

%# find the odds
fprintf('odds are %5.2f:1\n',sum(isSet)/(nIter-sum(isSet)))

Answer 2

在查看代码之前，我解决了编写自己的实现的问题。我的第一次尝试与您已经拥有的非常相似：）

%# some parameters
NUM_ITER = 100000;  %# number of simulations to run
DRAW_SZ = 12;       %# number of cards we are dealing
SET_SZ = 3;         %# number of cards in a set
FEAT_NUM = 4;       %# number of features (symbol,color,number,shading)
FEAT_SZ = 3;        %# number of values per feature (eg: red/purple/green, ...)

%# cards features
features = {
    'oval' 'squiggle' 'diamond' ;    %# symbol
    'red' 'purple' 'green' ;         %# color
    'one' 'two' 'three' ;            %# number
    'solid' 'striped' 'open'         %# shading
};
fIdx = arrayfun(@(k) grp2idx(features(k,:)), 1:FEAT_NUM, 'UniformOutput',0);

%# list of all cards. Each card: [symbol,color,number,shading]
[W X Y Z] = ndgrid(fIdx{:});
cards = [W(:) X(:) Y(:) Z(:)];

%# all possible sets: choose 3 from 12
setsInd = nchoosek(1:DRAW_SZ,SET_SZ);

%# count number of valid sets in random draws of 12 cards
counterValidSet = 0;
for i=1:NUM_ITER
    %# pick 12 cards
    ord = randperm( size(cards,1) );
    cardsDrawn = cards(ord(1:DRAW_SZ),:);

    %# check for valid sets: features are all the same or all different
    for s=1:size(setsInd,1)
        %# set of 3 cards
        set = cardsDrawn(setsInd(s,:),:);

        %# check if set is valid
        count = arrayfun(@(k) numel(unique(set(:,k))), 1:FEAT_NUM);
        isValid = (count==1|count==3);

        %# increment counter
        if isValid
            counterValidSet = counterValidSet + 1;
            break           %# break early if found valid set among candidates
        end
    end
end

%# ratio of found-to-notfound
fprintf('Size=%d, Set=%d, NoSet=%d, Set:NoSet=%g\n', ...
    DRAW_SZ, counterValidSet, (NUM_ITER-counterValidSet), ...
    counterValidSet/(NUM_ITER-counterValidSet))

在使用Profiler发现热点之后，可以主要通过尽可能早地中断循环来进行一些改进。主要的瓶颈是调用UNIQUE函数。我们检查有效集合的上面两行可以改写为：

%# check if set is valid
isValid = true;
for k=1:FEAT_NUM
    count = numel(unique(set(:,k)));
    if count~=1 && count~=3
        isValid = false;
        break   %# break early if one of the features doesnt meet conditions
    end
end

不幸的是，对于较大的模拟，模拟仍然很慢。因此，我的下一个解决方案是矢量化版本，其中对于每次迭代，我们从12张抽取牌的手中构建所有可能的3张牌组的单个矩阵。对于所有这些候选集，我们使用逻辑向量来指示存在的特征，从而避免调用UNIQUE / NUMEL（我们希望在集合的每张卡上的所有相同或全部不同的特征）。

我承认代码现在可读性较差，难以遵循（因此我发布了两个版本进行比较）。原因是我试图尽可能地优化代码，以便每个迭代循环都完全矢量化。这是最终的代码：

%# some parameters
NUM_ITER = 100000;  %# number of simulations to run
DRAW_SZ = 12;       %# number of cards we are dealing
SET_SZ = 3;         %# number of cards in a set
FEAT_NUM = 4;       %# number of features (symbol,color,number,shading)
FEAT_SZ = 3;        %# number of values per feature (eg: red/purple/green, ...)

%# cards features
features = {
    'oval' 'squiggle' 'diamond' ;    %# symbol
    'red' 'purple' 'green' ;         %# color
    'one' 'two' 'three' ;            %# number
    'solid' 'striped' 'open'         %# shading
};
fIdx = arrayfun(@(k) grp2idx(features(k,:)), 1:FEAT_NUM, 'UniformOutput',0);

%# list of all cards. Each card: [symbol,color,number,shading]
[W X Y Z] = ndgrid(fIdx{:});
cards = [W(:) X(:) Y(:) Z(:)];

%# all possible sets: choose 3 from 12
setsInd = nchoosek(1:DRAW_SZ,SET_SZ);

%# optimizations: some calculations taken out of the loop
ss = setsInd(:);
set_sz2 = numel(ss)*FEAT_NUM/SET_SZ;
col = repmat(1:set_sz2,SET_SZ,1);
col = FEAT_SZ.*(col(:)-1);
M = false(FEAT_SZ,set_sz2);

%# progress indication
%#hWait = waitbar(0./NUM_ITER, 'Simulation...');

%# count number of valid sets in random draws of 12 cards
counterValidSet = 0;
for i=1:NUM_ITER
    %# update progress
    %#waitbar(i./NUM_ITER, hWait);

    %# pick 12 cards
    ord = randperm( size(cards,1) );
    cardsDrawn = cards(ord(1:DRAW_SZ),:);

    %# put all possible sets of 3 cards next to each other
    set = reshape(cardsDrawn(ss,:)',[],SET_SZ)';
    set = set(:);

    %# check for valid sets: features are all the same or all different
    M(:) = false;            %# if using PARFOR, it will complain about this
    M(set+col) = true;
    isValid = all(reshape(sum(M)~=2,FEAT_NUM,[]));

    %# increment counter if there is at least one valid set in all candidates
    if any(isValid)
        counterValidSet = counterValidSet + 1;
    end
end

%# ratio of found-to-notfound
fprintf('Size=%d, Set=%d, NoSet=%d, Set:NoSet=%g\n', ...
    DRAW_SZ, counterValidSet, (NUM_ITER-counterValidSet), ...
    counterValidSet/(NUM_ITER-counterValidSet))

%# close progress bar
%#close(hWait)

如果您有并行处理工具箱，则可以使用并行PARFOR轻松替换普通FOR循环（您可能希望再次在循环内移动矩阵M的初始化：替换{{1}与M(:) = false;）

以下是50000次模拟的一些示例输出（PARFOR与2个本地实例池一起使用）：

M = false(FEAT_SZ,set_sz2);

并且有一百万次迭代（12次为PARFOR，15次为非PARFOR）：

» tic, SET_game2, toc
Size=12, Set=48376, NoSet=1624, Set:NoSet=29.7882
Elapsed time is 5.653933 seconds.

» tic, SET_game2, toc
Size=15, Set=49981, NoSet=19, Set:NoSet=2630.58
Elapsed time is 9.414917 seconds.

比值比与Peter Norvig报告的结果一致。

Answer 3

我发现了我的错误。感谢Jonas对RANDPERM的暗示。

我使用RANDI随机抽取K卡，但即使在12张牌中也有大约50％的机会获得重复。当我用randperm替换这一行时，我得到了33.8：1，10000次迭代，非常接近指令书中的数字。

setdrawncardidx = randperm(81);
setdrawncardidx = setdrawncardidx(1:K);

无论如何，看到问题的其他方法会很有趣。

Answer 4

我确定我对这些赔率的计算有问题，因为其他几个人已通过模拟确认它接近33：1，如说明书所示，但是以下逻辑有什么问题？

对于12张随机卡，有三种卡的220种可能组合（12！/（9！3！）= 220）。三张牌的每个组合都有1/79的机会成为一套，所以有三张任意牌的78/79几率不是一套。因此，如果你检查了所有220个组合并且每个组合都有78/79的机会，那么你没有找到检查所有可能组合的集合的机会将是78/79提升到220次幂，或0.0606，这是约。 17：1赔率。

我一定错过了什么......？

克里斯托弗