B 是一个 [1x8] 矩阵，也可以被视为如下两半：
B = -1 -1 0 0 0 0 1 1
这里前半部分可以有一个、两个、三个或四个-1，后半部分应该有相同数量的1。应该是线性组合。
例如，如果前半部分有两个-1，则可以将它们放在4 choose 2 = 6的方式中，并且对于它们中的每一个，将有6种方式来放置两个1下半场。所以系统总共有 6*6 = 36 种方式。即如果前半部分有两个-1，则 B 有 36 个不同的值。
我该怎么做？
【问题讨论】：
您可以先生成所有可能的 1 和 0 排列，然后丢弃多余的排列。
%# 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 的可能值
【讨论】：
这是另一个解决方案：
%# 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：
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
【讨论】：