随机16位字母大小写加数字不可能这么快被破解啊,随机16位字母大小写加数字是16*log(62)/log(2)=95 bits of entropy,可用以下命令计算:
$ echo'16*l(62)/l(2)' | bc -l
95.26714096619000334174
95 bits of entropy虽然不如RSA2048的112 bits of entropy但可比RSA1024的80 bits of entropy高多了,见https://crypto.stackexchange.com/questions/8687/security-strength-of-rsa-in-relation-with-the-modulus-size/8692#8692,一直有传言说RSA1024被量子计算机破解但从未有实证,反正我是不信,80 bits of entropy的RSA1024都从未有实证被破解,更别说你随机16位字母大小写加数字是95 bits of entropy,10分钟内被破解?不可能
随机16位字母大小写加数字不可能这么快被破解啊,随机16位字母大小写加数字是16*log(62)/log(2)=95 bits of entropy,可用以下命令计算:
$ echo'16*l(62)/l(2)' | bc -l
95.26714096619000334174
95 bits of entropy虽然不如RSA2048的112 bits of entropy但可比RSA1024的80 bits of entropy高多了,见https://crypto.stackexchange.com/questions/8687/security-strength-of-rsa-in-relation-with-the-modulus-size/8692#8692,一直有传言说RSA1024被量子计算机破解但从未有实证,反正我是不信,80 bits of entropy的RSA1024都从未有实证被破解,更别说你随机16位字母大小写加数字是95 bits of entropy,10分钟内被破解?不可能
随机16位字母大小写加数字不可能这么快被破解啊,随机16位字母大小写加数字是16*log(62)/log(2)=95 bits of entropy,可用以下命令计算:
$ echo'16*l(62)/l(2)' | bc -l
95.26714096619000334174
95 bits of entropy虽然不如RSA2048的112 bits of entropy但可比RSA1024的80 bits of entropy高多了,见https://crypto.stackexchange.com/questions/8687/security-strength-of-rsa-in-relation-with-the-modulus-size/8692#8692,一直有传言说RSA1024被量子计算机破解但从未有实证,反正我是不信,80 bits of entropy的RSA1024都从未有实证被破解,更别说你随机16位字母大小写加数字是95 bits of entropy,10分钟内被破解?不可能
On average, an attacker will have to try half the possible number of passwords before finding the correct one.
更多见https://www.nodeseek.com/post-750965-2#15:
@domingokkk #9 发布于5/29/2026, 5:10:15 AM
@sv6cr #5 很特殊的符号不支持
没有特殊符号不代表密码不强,如果是随机生成的密码,只要计算出来的bits of entropy高于你认为的强密码该有的bits of entropy那么这密码就是强密码
我认为高于RSA1024的bits of entropy的密码就是强密码,RSA1024的bits of entropy是80,之所以我选择高于RSA1024的bits of entropy是因为几年前有过量子计算机已经能破解RSA1024的传闻
那么如果我只用纯数字和大小写字母而不用符号,纯数字和大小写字母有62个字符,通过以下计算
$ echo 'l(2^80)/l(62)' | bc -l
13.43590231656335538018
可以得出我密码只需要至少14位随机生成的纯数字和大小写字母就是强密码
l()是log()
80是bits of entropy
62是纯数字和大小写字母有62个字符
根据这个2012年的网站文章25-GPU cluster cracks every standard Windows password in <6 hours当时此文章使用的显卡阵列能做到350 billion(3500亿)次猜测每秒,我们可以计算14位随机生成的纯数字和大小写字母的密码需要此显卡阵列计算几年才能破解:
$ echo '62^14/2/(350*10^9)/3600/24/365' | bc -l
561796.47000514273538356454
RSA1024 about 80 bits entropy. Recently there's rumors (possiblely false) that quantum computer is able to crack RSA1024 now. Although possiblely false, It shows people's expectation and technology trend, so possible higher than it would be better.
another result: if I want to use only [:alnum:], which has 62 chars; and if I want bits of entropy of at least 80 (because RSA1024); then I calculate with l(2^80)/l(62)=13.43590231656335538018, so I need at least 14 chars; with 14 chars, I got 14*l(62)/l(2)=83.35874834541625292402 this test shows bits of entropy good. But for pass cli I prefer longer password better, so I prefer pass cli without specify length which will use default length PASSWORD_STORE_CHARACTER_SET='[:alnum:]' pass generate -i only_alnum