Matlab 矩阵/向量的线性组合

B是一个[1x8]矩阵，它也可以被视为两部分，如下所示：

B = -1 -1 0 0   0 0 1 1

在这里，上半部分可以有一个、两个、三个或四个

-1

，下半部分应该有相同数量的

1

。它应该以线性组合的方式进行

例如，如果上半部分有两个

-1

，则可以将它们放置在

4中，选择2=6个
方式，并且对于每一个
1
方式，将有6个方式将两个
1
放置在下半部分。所以系统总共有6*6=36路。i、 e.如果上半部分有两个
-1
，则B的36个不同值

我该怎么做呢？
您可以先生成所有可能的1和0排列，然后扔掉多余的排列

%# make permutations using dec2bin (start from 17 since it's the first solution)
allB = str2double(num2cell(dec2bin(17:255)));

%# change sign in the first half, then check that the total is ok
allB(:,1:4) = - allB(:,1:4);
allB = allB(sum(allB,2)==0,:);

allB
的每一行都是
B
的一个可能值，下面是另一个解决方案：

%# generate all possible version of first half
h1 = num2cell(-(dec2bin(1:15)-'0'),2);

%# generate all possible version of second half
h2 = arrayfun(@(i) unique(perms([zeros(1,4-i) ones(1,i)]),'rows'), (1:4)', 'UniformOutput',false);

%'# number of 1s in each row of h1
n = -cellfun(@sum, h1);

%# get final result by combining h1 and h2
B = cellfun(@(a,b) [repmat(a,size(b,1),1) b], h1, h2(n), 'UniformOutput',false);
B = cell2mat(B);

结果是：

B =
     0     0     0    -1     0     0     0     1
     0     0     0    -1     0     0     1     0
     0     0     0    -1     0     1     0     0
     0     0     0    -1     1     0     0     0
     0     0    -1     0     0     0     0     1
     0     0    -1     0     0     0     1     0
     0     0    -1     0     0     1     0     0
     0     0    -1     0     1     0     0     0
     0     0    -1    -1     0     0     1     1
     0     0    -1    -1     0     1     0     1
     0     0    -1    -1     0     1     1     0
     0     0    -1    -1     1     0     0     1
     0     0    -1    -1     1     0     1     0
     0     0    -1    -1     1     1     0     0
     0    -1     0     0     0     0     0     1
     0    -1     0     0     0     0     1     0
     0    -1     0     0     0     1     0     0
     0    -1     0     0     1     0     0     0
     0    -1     0    -1     0     0     1     1
     0    -1     0    -1     0     1     0     1
     0    -1     0    -1     0     1     1     0
     0    -1     0    -1     1     0     0     1
     0    -1     0    -1     1     0     1     0
     0    -1     0    -1     1     1     0     0
     0    -1    -1     0     0     0     1     1
     0    -1    -1     0     0     1     0     1
     0    -1    -1     0     0     1     1     0
     0    -1    -1     0     1     0     0     1
     0    -1    -1     0     1     0     1     0
     0    -1    -1     0     1     1     0     0
     0    -1    -1    -1     0     1     1     1
     0    -1    -1    -1     1     0     1     1
     0    -1    -1    -1     1     1     0     1
     0    -1    -1    -1     1     1     1     0
    -1     0     0     0     0     0     0     1
    -1     0     0     0     0     0     1     0
    -1     0     0     0     0     1     0     0
    -1     0     0     0     1     0     0     0
    -1     0     0    -1     0     0     1     1
    -1     0     0    -1     0     1     0     1
    -1     0     0    -1     0     1     1     0
    -1     0     0    -1     1     0     0     1
    -1     0     0    -1     1     0     1     0
    -1     0     0    -1     1     1     0     0
    -1     0    -1     0     0     0     1     1
    -1     0    -1     0     0     1     0     1
    -1     0    -1     0     0     1     1     0
    -1     0    -1     0     1     0     0     1
    -1     0    -1     0     1     0     1     0
    -1     0    -1     0     1     1     0     0
    -1     0    -1    -1     0     1     1     1
    -1     0    -1    -1     1     0     1     1
    -1     0    -1    -1     1     1     0     1
    -1     0    -1    -1     1     1     1     0
    -1    -1     0     0     0     0     1     1
    -1    -1     0     0     0     1     0     1
    -1    -1     0     0     0     1     1     0
    -1    -1     0     0     1     0     0     1
    -1    -1     0     0     1     0     1     0
    -1    -1     0     0     1     1     0     0
    -1    -1     0    -1     0     1     1     1
    -1    -1     0    -1     1     0     1     1
    -1    -1     0    -1     1     1     0     1
    -1    -1     0    -1     1     1     1     0
    -1    -1    -1     0     0     1     1     1
    -1    -1    -1     0     1     0     1     1
    -1    -1    -1     0     1     1     0     1
    -1    -1    -1     0     1     1     1     0
    -1    -1    -1    -1     1     1     1     1

你的意思是要根据你的定义产生所有可能的（即16+36+16+1=69）向量吗？你的解决方案也是正确的。你也可以看看我类似的另一个问题吗。在+1的情况下，通过消除来进行肯定更容易