因此，想法是有四种情况：HH，LL，LH和HL。每个都基于标准的普通RNG（randn（n，1））。窃听者进行测量并将其记录下来；其中一个与四个都相同。我正在尝试编写一个程序，以计算窃听者破解案件所花费的时间，偏差在1％以内。例如，HH的第一点可能与LL的第一点相同。这就是为什么窃听者需要继续进行测量，直到消除可能性并仅保留一种情况。每个案例有1000分：

n = 1000;

我的方法是，将所测案例与这四个案例之间的差异：

diffHH = M - HH;
diffLL = M - LL;
diffHL = M - HL;
diffLH = M - LH;

然后，我编写了四个for循环来检查每种情况：

for guesses = 1:1:n
    if (abs(diffHH(guesses)) <= 0.01) 
        if (abs(diffHH(guesses)) <= 0.01) && (abs(diffLL(guesses)) <= 0.01)
            continue;
        elseif (abs(diffHH(guesses)) <= 0.01)&& (abs(diffHL(guesses)) <= 0.01)
            continue;
        elseif (abs(diffHH(guesses)) <= 0.01) && (abs(diffLL(guesses)) <= 0.01)
            continue;
        else
            disp('Eve guesses HH');
            break;
        end
    end
end
for guesses = 1:1:n
    if (abs(diffLL(guesses)) <= 0.01) 
        if (abs(diffLL(guesses)) <= 0.01) && (abs(diffHH(guesses)) <= 0.01)
            continue;
        elseif (abs(diffLL(guesses)) <= 0.01)&& (abs(diffHL(guesses)) <= 0.01)
            continue;
        elseif (abs(diffLL(guesses)) <= 0.01) && (abs(diffLH(guesses)) <= 0.01)
            continue;
        else
            disp('Eve guesses LL');
            break;
        end
    end
end
for guesses = 1:1:n
    if (abs(diffHL(guesses)) <= 0.01) 
        if (abs(diffHL(guesses)) <= 0.01) && (abs(diffHH(guesses)) <= 0.01)
            continue;
        elseif (abs(diffHL(guesses)) <= 0.01)&& (abs(diffLL(guesses)) <= 0.01)
            continue;
        elseif (abs(diffHL(guesses)) <= 0.01) && (abs(diffLH(guesses)) <= 0.01)
            continue;
        else
            disp('Eve guesses HL');
            break;
        end
    end
end
for guesses = 1:1:n
    if (abs(diffLH(guesses)) <= 0.01) 
        if (abs(diffLH(guesses)) <= 0.01) && (abs(diffHH(guesses)) <= 0.01)
            continue;
        elseif (abs(diffLH(guesses)) <= 0.01)&& (abs(diffLL(guesses)) <= 0.01)
            continue;
        elseif (abs(diffLH(guesses)) <= 0.01) && (abs(diffHL(guesses)) <= 0.01)
            continue;
        else
            disp('Eve guesses LH');
            break;
        end
    end
end

猜测值似乎一直上升到1000，而且不应该；四种情况截然不同。我到底想念什么？