考虑到之前的评论，我的代码会生成向量的随机排列并检查所有这些组合是否都在排列中。如果找到某个条件，则会生成新的排列：
vector = [1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18];
not_valid = 1;
while not_valid == 1
randpermutation = vector(randperm(18))
not_valid = 0;
for i = 1:length(randpermutation)-1
if randpermutation(i) == 1 && randpermutation(i+1) == 7 || randpermutation(i) == 7 && randpermutation(i+1) == 1
not_valid = 1;
end
if randpermutation(i) == 1 && randpermutation(i+1) == 8 || randpermutation(i) == 8 && randpermutation(i+1) == 1
not_valid = 1;
end
if randpermutation(i) == 1 && randpermutation(i+1) == 17 || randpermutation(i) == 17 && randpermutation(i+1) == 1
not_valid = 1;
end
if randpermutation(i) == 1 && randpermutation(i+1) == 18 || randpermutation(i) == 18 && randpermutation(i+1) == 1
not_valid = 1;
end
if randpermutation(i) == 2 && randpermutation(i+1) == 7 || randpermutation(i) == 7 && randpermutation(i+1) == 2
not_valid = 1;
end
if randpermutation(i) == 2 && randpermutation(i+1) == 8 || randpermutation(i) == 8 && randpermutation(i+1) == 2
not_valid = 1;
end
if randpermutation(i) == 2 && randpermutation(i+1) == 17 || randpermutation(i) == 17 && randpermutation(i+1) == 2
not_valid = 1;
end
if randpermutation(i) == 2 && randpermutation(i+1) == 18 || randpermutation(i) == 18 && randpermutation(i+1) == 2
not_valid = 1;
end
if randpermutation(i) == 2 && randpermutation(i+1) == 6 || randpermutation(i) == 6 && randpermutation(i+1) == 2
not_valid = 1;
end
end
end
感谢您的回答，我曾想过类似的事情，我将验证它在计算复杂度方面的性能，因为我将生成 10,000 到 100,000 个排列的矩阵。