tag:blogger.com,1999:blog-7266528187680728229.post7074021731854902292..comments2024-03-07T22:27:00.958-08:00Comments on The Bug Charmer: A note on password mathStevenhttp://www.blogger.com/profile/08515783026293944881noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-7266528187680728229.post-26476886036771815922012-11-18T02:44:09.169-08:002012-11-18T02:44:09.169-08:00that was an excellent pdf, thanks, only now i foun...that was an excellent pdf, thanks, only now i found the time to read it....if you got more like that, shoot!:)micheehttps://www.blogger.com/profile/05937230965190972040noreply@blogger.comtag:blogger.com,1999:blog-7266528187680728229.post-5387447002533969002012-11-17T06:06:01.848-08:002012-11-17T06:06:01.848-08:00You definitely need to subtract 85^8. The 85^8 co...You definitely need to subtract 85^8. The 85^8 comes about because there are 85^8 passwords that contain no digits in any of the 8 positions. You need to remove these to ensure that the remaining passwords all contain at least one digit. If you look at the link in the previous comment, it shows how to calculate a similar problem where there are 36^6 possible 6 character alphanumeric passwords and 26^6 of those contain no digits so that (36^6 - 26^6) contain at least one digit.Stevenhttps://www.blogger.com/profile/08515783026293944881noreply@blogger.comtag:blogger.com,1999:blog-7266528187680728229.post-40515980731956484202012-11-17T02:40:12.304-08:002012-11-17T02:40:12.304-08:00i'm glad you like my question, it is indeed a ...i'm glad you like my question, it is indeed a tough one i think...i thought about it after reading some stuff on passwords math, including this post.....<br />I think you're wrong with something though. You've started by substracting 85^8 ,meaning passwords with no digits. <br /><br />Instead i think you should have substracted <br /><br />85 x 95^7, meaning passwords with no digits in just one position, or better yet, what if you subtract from the start like this:<br /><br />95^8 - 85 x 69 x 69 x 95 ^ 5 (this would be the "complementary of 10 x 26 x 26 x 95^5 we have started with") = final answer ? <br /><br />micheehttps://www.blogger.com/profile/05937230965190972040noreply@blogger.comtag:blogger.com,1999:blog-7266528187680728229.post-56928460198837401862012-11-16T20:57:39.055-08:002012-11-16T20:57:39.055-08:00That's actually a complicated question. To st...That's actually a complicated question. To start, let's be clear that there are 95^8 possible 8 character passwords using all printable ASCII characters. To get the answer you are looking for, we need to subtract out the passwords that contain no digits, no upper case, or no lower case letters.<br /><br />If any readers with a good understanding of probability want to double-check my math here, I'd appreciate it.<br /><br />There are 85^8 passwords with no digits. There are 69^8 passwords with no upper case letters and, similarly, 69^8 passwords with no lower case letters.<br /><br />The naive (and wrong) way to calculate the number of passwords with at least 1 digit, at least 1 upper case, and at least 1 lower case letters would be:<br /><br />95^8 - 85^8 - 69^8 - 69^8 = 2,881,702,313,642,720<br /><br />The problem is that there is some overlap between the groups. For instance, some of the passwords with no digits also contain no lower case letters. So, among the 85^8 passwords with no digits and the 69^8 passwords with no lower case letters, there are 59^8 passwords with no lower case or digits and we counted them twice.<br /><br />So, taking the initial (wrong) answer, we need to add back the number of passwords with no digits or lower case, no digits or upper case and no upper or lower case.<br /><br />There are 59^8 passwords with no digits or lower case<br />There are 59^8 passwords with no digits or upper case<br />There are 43^8 passwords with no upper or lower case<br /><br />This gets us to 95^8 - 85^8 - 69^8 - 69^8 + 59^8 + 59^8 + 43^8. <br /><br />We're still not done, every one of the three groups we added or subtracted from 95^8 includes the 33^8 passwords that consist of no upper or lower case or digits (only special characters) and all three of the groups we added back also included them so we need to subtract them again:<br /><br />We now have:<br /><br />95^8 - 85^8 - 69^8 - 69^8 + 59^8 + 59^8 + 43^8 - 33^8 = 3,185,644,980,510,720. <br /><br />There are some notes on password math worth reading here:<br /><br />http://people.math.gatech.edu/~heitsch/Teaching/F06/passwd.pdf<br /><br /><br /><br />Stevenhttps://www.blogger.com/profile/08515783026293944881noreply@blogger.comtag:blogger.com,1999:blog-7266528187680728229.post-66998510505383982502012-11-16T09:36:01.896-08:002012-11-16T09:36:01.896-08:00no, my questions is like this:
what if you're ...no, my questions is like this:<br />what if you're forced to use 1 digit, 1 lower, 1 upper, the other 5 are not restricted<br />so the final number would be smth. like:<br /><br />10 x 26 x 26 x 95 x 95 x 95 x 95 x 95 .....is this correct?micheehttps://www.blogger.com/profile/05937230965190972040noreply@blogger.comtag:blogger.com,1999:blog-7266528187680728229.post-63178952059160573922012-11-16T08:21:14.430-08:002012-11-16T08:21:14.430-08:00This comment has been removed by the author.Stevenhttps://www.blogger.com/profile/08515783026293944881noreply@blogger.comtag:blogger.com,1999:blog-7266528187680728229.post-17312893200666477982012-11-16T01:46:41.576-08:002012-11-16T01:46:41.576-08:00what if you are forced to use 3 charsets in a pass...what if you are forced to use 3 charsets in a password of 8? how many passwords would then be? i'm having some problems on that.....thanks.micheehttps://www.blogger.com/profile/05937230965190972040noreply@blogger.com