我正在使用Matlab处理代码，我需要在其中找到覆盖引用列表所有元素所需的最少数字列表（在一些给定列表中）。
例如，假设我的参考列表是
X = [0 1 2 3 4 5 6 7 8 9]
我有一组给定的列表如下：
A = [0 1 3 5 6 7 9]
B = [0 1 2 3 4]
C = [5 6 7 8 9]
D = [1 2 3 4]
E = [1 5 7 8]
涵盖X中每个元素所需的最少数量的列表是2（B和C），但是，如果我最初只搜索列表，涵盖了大多数元素（A），然后尝试查找将涵盖其余元素的其他列表，我最终将使用至少3列表。编写可以搜索所需的最少数量的列表的代码的最佳方法是什么（它会给我输出B和C）？任何帮助都将非常感激...即使只是一个概念性解释（而非实际代码）如何最好地解决这个问题将是一个巨大的帮助！
答案 0 :(得分：4)
下面是一个可能的算法，说明如何解决问题。代码本身应该是不言自明的，但我们的想法是我们测试所有可能的列表组合，直到找到有效的列表（因此我们没有遇到您描述的问题，我们错误地根据它们的长度选择列表）。 / p>
function varargout = q36323802
R = [0 1 2 3 4 5 6 7 8 9]; %// Reference List
L = {... // As per Dan's suggestion:
[0 1 3 5 6 7 9]
[0 1 2 3 4]
[5 6 7 8 9]
[1 2 3 4]
[1 5 7 8]
};
out = []; %// Initialize output
%% // Brute-force approach:
nLists = numel(L);
for indN = 1:nLists
setCombinationsToCheck = nchoosek(1:nLists,indN);
for indC = 1:size(setCombinationsToCheck,1)
u = unique(cat(2,L{setCombinationsToCheck(indC,:)}));
if all(ismember(R,u))
out = setCombinationsToCheck(indC,:);
disp(['The minimum number of required sets is ' num2str(indN) ...
', and their indices are: ' num2str(out)]);
return;
end
end
end
disp('No amount of lists found to cover the reference.');
if nargout > 0
varargout{1} = out;
end
对于您的示例，输出为：
The minimum number of required sets is 2, and their indices are: 2 3
n-1中不使用长度为n的列表来执行一些冗余计算，这些列表已在先前的迭代中找到（适用时）。在这种情况下，递归解决方案可能有效。感谢 Adiel 提出了一些改进建议，感谢Amro找到了一些错误！
我也试图构建一个递归求解器。现在它找到了一个解决方案，但它不够通用（实际上问题是它只返回第一个结果，不一定是最好的结果）。这种方法背后的原因是我们可以将您的问题视为树搜索问题，因此我们可以使用搜索/寻路算法（请参阅BFS，DFS，IDS等）。我认为下面的算法最接近DFS。和以前一样，这应该主要说明解决问题的方法。
function q36323802_DFS(R,L)
%% //Input checking:
if nargin < 2 || isempty(L)
L = {... // As per Dan's suggestion:
[0 1 3 5 6 7 9]
[0 1 2 3 4]
[5 6 7 8 9]
[1 2 3 4]
[1 5 7 8]
};
end
if nargin < 1 || isempty(R)
R = [0 1 2 3 4 5 6 7 8 9]; %// Reference List
end
%% // Algorithm (DFS: breadth-first search):
out = DFS_search(R,L,0);
if isempty(out)
disp('No amount of lists found to cover the reference.');
else
disp(['The minimum number of required sets is ' num2str(numel(out)) ...
', and their indices are: ' num2str(out)]);
end
end
function out = DFS_search(R,L,depth)
%// Check to see if we should stop:
if isempty(R) || isempty(L)
% // Backtrack here?
out = [];
return;
end
if isnan(R)
out = [];
return;
end
nLists = numel(L);
reducedR = cellfun(@(R,L)setdiff(R,L),repmat({R},[nLists,1]),L,'UniformOutput',false)';
%'// We consider a case where the reduction had no effect as "hopeless" and
%// "drop" it.
isFullCoverage = cellfun(@isempty,reducedR);
isHopeless = cellfun(@(R)all(isnan(R)),reducedR) | cellfun(@(rR)isequal(rR,R),reducedR);
reducedR(isHopeless) = deal({NaN});
if all(isHopeless) && ~any(isFullCoverage)
out = [];
return
end
if any(isFullCoverage) %// Check current "breadth level"
out = find(isFullCoverage,1,'first');
return
else
for indB = 1:nLists
out = DFS_search(reducedR{indB},L,depth+1);
if ~isempty(out)
out = [indB out]; %#ok
%// TODO: test if one of the sets is covered by the others and remove it
%// from the list "out".
%// Also, keep track of the best path and only return (finally) if shortest
return
end
end
end
end
答案 1 :(得分：1)
通过Amro对Dev-iL＆＃39; 1 st 方法的类似解决方案：
function varargout = q36323802A
R = [0 1 2 3 4 5 6 7 8 9];
names = {'A' 'B' 'C' 'D' 'E'};
L = {...
[0 1 3 5 6 7 9]
[0 1 2 3 4]
[5 6 7 8 9]
[1 2 3 4]
[1 5 7 8]
};
N = numel(L);
%// powerset of L: set of all subsets (excluding empty set)
powerset = cell(1,N);
for k=1:N
sets = nchoosek(1:N, k);
powerset{k} = num2cell(sets,2);
end
powerset = cat(1, powerset{:});
%// for each possible subset, check if it covers the target R
mask = false(size(powerset));
for i=1:numel(powerset)
elems = unique([L{powerset{i}}]);
mask(i) = all(ismember(R, elems));
end
if ~any(mask), error('cant cover target'); end
%// from candidates, choose the one with least amount of sets
candidates = powerset(mask);
len = cellfun(@numel, candidates);
[~,idx] = min(len);
out = candidates{idx};
varargout{1} = names(out);