如何找到对组的所有组合

Question

我想生成 3 男 2 女对的所有组合。有很多配对示例（例如参见this），但没有一个涉及配对集。

例如，如果我有：

Men   = {'M1', 'M2'};
Women = {'W1', 'W2', 'W3'};

我想要的结果是以下几组：

(M1, W1), (M2, W2)
(M1, W1), (M2, W3)
(M1, W2), (M2, W1)
(M1, W2), (M2, W3)
(M1, W3), (M2, W1)
(M1, W3), (M2, W2)

谢谢。

Answer 1

其实很简单。要填充一组 k 对，您需要 k 个男人和 k 个女人，所以让我们找出所有可能的 k 组合em> 男人和 k 女人优先：

%// Find all possible combinations of sets of k pairs of men and women
k = 2;
idx_m = nchoosek(1:numel(Men), k);             % // Indices of men
idx_w = nchoosek(1:numel(Women), k);           % // Indices of women
idx_w = reshape(idx_w(:, perms(1:k)), [], k);  % // All permutations

然后让我们构建 k 个男人和 k 个女人的集合的所有可能组合：

[idx_comb_w, idx_comb_m] = find(ones(size(idx_w , 1), size(idx_m , 1)));
idx = sortrows([idx_m(idx_comb_m(:), :), idx_w(idx_comb_w(:), :)]);
idx = idx(:, reshape(1:size(idx, 2), k, [])'); %'// Rearrange in pairs

生成的矩阵idx 包含集合中男性和女性的索引（第一列为男性，第二列为女性，第三列为男性，第四列为女性，依此类推...）。

示例

Men = {'M1', 'M2'};
Women = {'W1', 'W2', 'W3'};

%// Find all possible combinations of sets of k pairs of men and women
k = 2;
idx_m = nchoosek(1:numel(Men), k);
idx_w = nchoosek(1:numel(Women), k);
idx_w = reshape(idx_w(:, perms(1:k)), [], k);
[idx_comb_w, idx_comb_m] = find(ones(size(idx_w , 1), size(idx_m , 1)));

%// Construct pairs from indices and print sets nicely
idx = sortrows([idx_m(idx_comb_m(:), :), idx_w(idx_comb_w(:), :)]);
idx = idx(:, reshape(1:size(idx, 2), k, [])');

%// Obtain actual sets
sets = cell(size(idx));
sets(:, 1:2:end) = Men(idx(:, 1:2:end));
sets(:, 2:2:end) = Women(idx(:, 2:2:end));

%// Print sets nicely
sets_t = sets';
fprintf([repmat('(%s, %s), ', 1, k - 1), '(%s, %s)\n'], sets_t{:})

这里生成的数组sets 已经适应了包含来自Men 和Women 的实际值。结果是：

(M1, W1), (M2, W2)
(M1, W1), (M2, W3)
(M1, W2), (M2, W1)
(M1, W2), (M2, W3)
(M1, W3), (M2, W1)
(M1, W3), (M2, W2)

Answer 2

这个例子正在使用FEX file allcomb

men = {'M1', 'M2', 'M3'};
women = {'W1', 'W2'};
allPeople = [men, women];
%// Play with vector index because allcomb doesn't work with cell
[~, id_men] = ismember(men,allPeople);
[~, id_women] = ismember(women,allPeople);


%// give all combinations for men/women
setOfMenWomen = allcomb(id_men,id_women);
%// give all combinations of pairs
nComb = size(setOfMenWomen,1);
setOfPair = nchoosek(1:nComb,2);
%// give all combinations for men/women/men/women
setOfPairMenWomen = cell2mat(arrayfun(@(id) setOfMenWomen(id,:), setOfPair, 'UniformOutput', 0));
%// get ids of set of pairs with the same men twice
isDoubleMen = setOfPairMenWomen(:,1) == setOfPairMenWomen(:,3);
%// get ids of set of pairs with the same women twice
isDoubleWomen = setOfPairMenWomen(:,2) == setOfPairMenWomen(:,4);
%// We don't want to have set of pairs with twice same men or same women
cond = isDoubleWomen | isDoubleMen;
setOfPairMenWomen(cond,:) = [];

%//results :
allPeople(setOfPairMenWomen)

Answer 3

我看不出你的问题到底是什么。您是否在 MATLAB 中搜索算法或实现？我会尝试用数学方法来回答。

您可以通过将其表述为graph 问题来解决您的问题：您的一组人就是一组顶点。通过将所有男人与所有女人联系起来形成边缘。现在找到（最大）matchings 的集合。

这在一定程度上取决于您到底想要什么：例如，您是否希望元素 ((M1, W1)) 成为您解决方案的一部分？然后您正在搜索匹配项（请参阅Hosoya index 了解解决方案集中的元素数量）。如果您只想要无法添加进一步配对的元素，那么您只考虑最大匹配。

如果您将问题扩展到通用的非性别对，这可能会有所帮助：http://en.wikipedia.org/wiki/Telephone_number_(mathematics)