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Are randomly-generated passwords starting with “a” less secure?


Using a Cryptovariable with an API Designed for Human-Generated PasswordsHow secure are “pattern” passwords?Single randomly generated passphrase for both identity and authenticationHow come a controlledly generated password is more secure than a purely randomly generated oneWhat makes one random strong password more resistant to a brute force search than anotherIs randomly generating passwords from an assortment of dictionary words cryptographically secure?How to brute-force a somewhat remembered aescrypt password?Protecting Passwords Using Short Hash LengthCracking passwords via hashcat or rainbow tables. Which is faster?Are partial passwords a security improvement over full passwords?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2















I generated a password recently for a new account and the first three characters were "aa1".



After exhausting all other attacks, a cracker would start brute forcing. On the assumption they'd start from "a", my password of "aa1" would be cracked faster than, say, "ba1", and that faster than, say, "za1".



This password was very long so this question is more theoretical than practical. (Unless password lengths are limited, of course...) Are my assumptions right about brute-forcing and passwords?










share|improve this question






















  • brute-force would probably be parallelized, so if you have 26 running attack instances, then the 1st starting lowercase letter does not change a thing, but the question is pretty interesting : )

    – Xenos
    8 hours ago











  • It depends on the total length and character set used, the more the marrier. You must make sure a rainbow attack is not viable, you should check out this rainbow table calculator tobtu.com/rtcalc.php

    – mootmoot
    6 hours ago

















2















I generated a password recently for a new account and the first three characters were "aa1".



After exhausting all other attacks, a cracker would start brute forcing. On the assumption they'd start from "a", my password of "aa1" would be cracked faster than, say, "ba1", and that faster than, say, "za1".



This password was very long so this question is more theoretical than practical. (Unless password lengths are limited, of course...) Are my assumptions right about brute-forcing and passwords?










share|improve this question






















  • brute-force would probably be parallelized, so if you have 26 running attack instances, then the 1st starting lowercase letter does not change a thing, but the question is pretty interesting : )

    – Xenos
    8 hours ago











  • It depends on the total length and character set used, the more the marrier. You must make sure a rainbow attack is not viable, you should check out this rainbow table calculator tobtu.com/rtcalc.php

    – mootmoot
    6 hours ago













2












2








2








I generated a password recently for a new account and the first three characters were "aa1".



After exhausting all other attacks, a cracker would start brute forcing. On the assumption they'd start from "a", my password of "aa1" would be cracked faster than, say, "ba1", and that faster than, say, "za1".



This password was very long so this question is more theoretical than practical. (Unless password lengths are limited, of course...) Are my assumptions right about brute-forcing and passwords?










share|improve this question














I generated a password recently for a new account and the first three characters were "aa1".



After exhausting all other attacks, a cracker would start brute forcing. On the assumption they'd start from "a", my password of "aa1" would be cracked faster than, say, "ba1", and that faster than, say, "za1".



This password was very long so this question is more theoretical than practical. (Unless password lengths are limited, of course...) Are my assumptions right about brute-forcing and passwords?







passwords password-cracking






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 9 hours ago









gnulynnuxgnulynnux

254 bronze badges




254 bronze badges












  • brute-force would probably be parallelized, so if you have 26 running attack instances, then the 1st starting lowercase letter does not change a thing, but the question is pretty interesting : )

    – Xenos
    8 hours ago











  • It depends on the total length and character set used, the more the marrier. You must make sure a rainbow attack is not viable, you should check out this rainbow table calculator tobtu.com/rtcalc.php

    – mootmoot
    6 hours ago

















  • brute-force would probably be parallelized, so if you have 26 running attack instances, then the 1st starting lowercase letter does not change a thing, but the question is pretty interesting : )

    – Xenos
    8 hours ago











  • It depends on the total length and character set used, the more the marrier. You must make sure a rainbow attack is not viable, you should check out this rainbow table calculator tobtu.com/rtcalc.php

    – mootmoot
    6 hours ago
















brute-force would probably be parallelized, so if you have 26 running attack instances, then the 1st starting lowercase letter does not change a thing, but the question is pretty interesting : )

– Xenos
8 hours ago





brute-force would probably be parallelized, so if you have 26 running attack instances, then the 1st starting lowercase letter does not change a thing, but the question is pretty interesting : )

– Xenos
8 hours ago













It depends on the total length and character set used, the more the marrier. You must make sure a rainbow attack is not viable, you should check out this rainbow table calculator tobtu.com/rtcalc.php

– mootmoot
6 hours ago





It depends on the total length and character set used, the more the marrier. You must make sure a rainbow attack is not viable, you should check out this rainbow table calculator tobtu.com/rtcalc.php

– mootmoot
6 hours ago










2 Answers
2






active

oldest

votes


















6














It would seem that it depends on how exactly the attacker is going to bruteforce your password. However, my opinion is that in the end it doesn't matter.



A serious attacker will never start from the beginning in alphanumeric order, from aaaaaaaa to 99999999, unless they know they can do that in a reasonable time. If that's going to take them a thousand years, why should they use that method, knowing they will necessarily have to stop at, say, cccccccc? But if the attacker knows that they can try all the possibilities in a reasonable time, then it doesn't matter whether your password is among the first combinations or among the last, because in the end they will find it anyway (in a reasonable time).



Most passwords are still weak (say, your dog's name, plus maybe your date of birth, etc.) and the attackers don't like wasting too much time, let alone years to crack passwords. So what attackers normally do is use dictionaries and patterns. They will first try passwords like: pass123, 123pass, john90, john91, John92, JOHN93, suckmyd1ck, l1nux4dm1n, etc. If every attempt with dictionaries and patterns fails, they might move on and assume that the password looks truly random. How long will it take to try all the possible passwords? If that can be done in a reasonable time, they might try them all (for example from aaaaaaaa to 99999999). Otherwise if the attacker assumes that they will never be able to try them all, they might try to bruteforce the password with some random guesses (random strings, not ordered): 12hrisn589sjlf, 9f2jcvew85hdye, otnwc739vhe82b, etc. If the attacker is lucky they might find the password, sooner or later. However if the password is too strong, such that it would take them too many years to guess it, they had better give up or think of an alternative attack (phishing, shoulder surfing, keyloggers, etc.)






share|improve this answer























  • Indeed, I think that's what it really comes down to - the hacker is well aware of what their has rate is, and therefore what passwords they can crack in a reasonable amount of time. Bruteforcing the full search space is either possible in a reasonable amount of time or it isn't, in which case they won't bother and it won't matter.

    – Conor Mancone
    6 hours ago


















6














Entropy is really the biggest concern, and entropy is determined by the amount of randomness in your password generation process.



Let's use an 18 character password as an example. We'll look at alphanumeric characters only (62 possible characters). This gives:



log2(62**18) = 107 bits of entropy



If you decide to always convert the first 3 characters to Z to make it harder to bruteforce then you have effectively removed 3 random characters from your password, leaving you with:



log2(62**15) = 89 bits of entropy



Which is about a factor of 1,000,000 weaker. Of course, both are still absolutely and completely impossible to bruteforce, so if someone is trying to brute force your password I wouldn't worry. However, it's worth emphasizing how impossible it is to brute force such a password (just in case my links didn't convince you). A top of the line MD5 hashing rig can try 200 billion hashes per second. It's blazing fast (which is why MD5 isn't supposed to be used for passwords). Even your weaker password (when you replaced the first three characters with z) has 7.6e26 possible combinations. At 200 billion hashes per second it will only take 120 MILLION years to try all password possibilities if the password was hashed with MD5. I don't think you have anything to worry about.



So, is it better to start with "higher" characters to make it harder for someone to brute force your password? Maybe, maybe not. It depends on how someone actually tries to crack the password (and there aren't any guarantees about what direction they might work in). Either way though, if you use a long random password, there is absolutely zero chance that it will get hacked. So I wouldn't worry about it. When it comes to passwords length is king.






share|improve this answer

























  • This is a good answer. It may be improved by a brief discussion on Kerckhoff's Principle.

    – Adonalsium
    8 hours ago











  • More than 256 bits of entropy with only 18 chars and no symbols? That's surprising, I thought. Because in fact the calculation is wrong: it's not log2(18^62), but log2(62^18), which is 107 bits of entropy. Your general point (length is king) is good, but unfortunately the rest of your post is based on wrong calculations.

    – reed
    7 hours ago












  • @reed lol, this isn't the first time I've gotten those backwards...

    – Conor Mancone
    7 hours ago











  • @reed fortunately the correct numbers don't change the actual answer - it's just a lot less dramatic.

    – Conor Mancone
    7 hours ago






  • 1





    @Adonalsium I think that could be a good addition - I'll add something in when I have more time later.

    – Conor Mancone
    7 hours ago













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6














It would seem that it depends on how exactly the attacker is going to bruteforce your password. However, my opinion is that in the end it doesn't matter.



A serious attacker will never start from the beginning in alphanumeric order, from aaaaaaaa to 99999999, unless they know they can do that in a reasonable time. If that's going to take them a thousand years, why should they use that method, knowing they will necessarily have to stop at, say, cccccccc? But if the attacker knows that they can try all the possibilities in a reasonable time, then it doesn't matter whether your password is among the first combinations or among the last, because in the end they will find it anyway (in a reasonable time).



Most passwords are still weak (say, your dog's name, plus maybe your date of birth, etc.) and the attackers don't like wasting too much time, let alone years to crack passwords. So what attackers normally do is use dictionaries and patterns. They will first try passwords like: pass123, 123pass, john90, john91, John92, JOHN93, suckmyd1ck, l1nux4dm1n, etc. If every attempt with dictionaries and patterns fails, they might move on and assume that the password looks truly random. How long will it take to try all the possible passwords? If that can be done in a reasonable time, they might try them all (for example from aaaaaaaa to 99999999). Otherwise if the attacker assumes that they will never be able to try them all, they might try to bruteforce the password with some random guesses (random strings, not ordered): 12hrisn589sjlf, 9f2jcvew85hdye, otnwc739vhe82b, etc. If the attacker is lucky they might find the password, sooner or later. However if the password is too strong, such that it would take them too many years to guess it, they had better give up or think of an alternative attack (phishing, shoulder surfing, keyloggers, etc.)






share|improve this answer























  • Indeed, I think that's what it really comes down to - the hacker is well aware of what their has rate is, and therefore what passwords they can crack in a reasonable amount of time. Bruteforcing the full search space is either possible in a reasonable amount of time or it isn't, in which case they won't bother and it won't matter.

    – Conor Mancone
    6 hours ago















6














It would seem that it depends on how exactly the attacker is going to bruteforce your password. However, my opinion is that in the end it doesn't matter.



A serious attacker will never start from the beginning in alphanumeric order, from aaaaaaaa to 99999999, unless they know they can do that in a reasonable time. If that's going to take them a thousand years, why should they use that method, knowing they will necessarily have to stop at, say, cccccccc? But if the attacker knows that they can try all the possibilities in a reasonable time, then it doesn't matter whether your password is among the first combinations or among the last, because in the end they will find it anyway (in a reasonable time).



Most passwords are still weak (say, your dog's name, plus maybe your date of birth, etc.) and the attackers don't like wasting too much time, let alone years to crack passwords. So what attackers normally do is use dictionaries and patterns. They will first try passwords like: pass123, 123pass, john90, john91, John92, JOHN93, suckmyd1ck, l1nux4dm1n, etc. If every attempt with dictionaries and patterns fails, they might move on and assume that the password looks truly random. How long will it take to try all the possible passwords? If that can be done in a reasonable time, they might try them all (for example from aaaaaaaa to 99999999). Otherwise if the attacker assumes that they will never be able to try them all, they might try to bruteforce the password with some random guesses (random strings, not ordered): 12hrisn589sjlf, 9f2jcvew85hdye, otnwc739vhe82b, etc. If the attacker is lucky they might find the password, sooner or later. However if the password is too strong, such that it would take them too many years to guess it, they had better give up or think of an alternative attack (phishing, shoulder surfing, keyloggers, etc.)






share|improve this answer























  • Indeed, I think that's what it really comes down to - the hacker is well aware of what their has rate is, and therefore what passwords they can crack in a reasonable amount of time. Bruteforcing the full search space is either possible in a reasonable amount of time or it isn't, in which case they won't bother and it won't matter.

    – Conor Mancone
    6 hours ago













6












6








6







It would seem that it depends on how exactly the attacker is going to bruteforce your password. However, my opinion is that in the end it doesn't matter.



A serious attacker will never start from the beginning in alphanumeric order, from aaaaaaaa to 99999999, unless they know they can do that in a reasonable time. If that's going to take them a thousand years, why should they use that method, knowing they will necessarily have to stop at, say, cccccccc? But if the attacker knows that they can try all the possibilities in a reasonable time, then it doesn't matter whether your password is among the first combinations or among the last, because in the end they will find it anyway (in a reasonable time).



Most passwords are still weak (say, your dog's name, plus maybe your date of birth, etc.) and the attackers don't like wasting too much time, let alone years to crack passwords. So what attackers normally do is use dictionaries and patterns. They will first try passwords like: pass123, 123pass, john90, john91, John92, JOHN93, suckmyd1ck, l1nux4dm1n, etc. If every attempt with dictionaries and patterns fails, they might move on and assume that the password looks truly random. How long will it take to try all the possible passwords? If that can be done in a reasonable time, they might try them all (for example from aaaaaaaa to 99999999). Otherwise if the attacker assumes that they will never be able to try them all, they might try to bruteforce the password with some random guesses (random strings, not ordered): 12hrisn589sjlf, 9f2jcvew85hdye, otnwc739vhe82b, etc. If the attacker is lucky they might find the password, sooner or later. However if the password is too strong, such that it would take them too many years to guess it, they had better give up or think of an alternative attack (phishing, shoulder surfing, keyloggers, etc.)






share|improve this answer













It would seem that it depends on how exactly the attacker is going to bruteforce your password. However, my opinion is that in the end it doesn't matter.



A serious attacker will never start from the beginning in alphanumeric order, from aaaaaaaa to 99999999, unless they know they can do that in a reasonable time. If that's going to take them a thousand years, why should they use that method, knowing they will necessarily have to stop at, say, cccccccc? But if the attacker knows that they can try all the possibilities in a reasonable time, then it doesn't matter whether your password is among the first combinations or among the last, because in the end they will find it anyway (in a reasonable time).



Most passwords are still weak (say, your dog's name, plus maybe your date of birth, etc.) and the attackers don't like wasting too much time, let alone years to crack passwords. So what attackers normally do is use dictionaries and patterns. They will first try passwords like: pass123, 123pass, john90, john91, John92, JOHN93, suckmyd1ck, l1nux4dm1n, etc. If every attempt with dictionaries and patterns fails, they might move on and assume that the password looks truly random. How long will it take to try all the possible passwords? If that can be done in a reasonable time, they might try them all (for example from aaaaaaaa to 99999999). Otherwise if the attacker assumes that they will never be able to try them all, they might try to bruteforce the password with some random guesses (random strings, not ordered): 12hrisn589sjlf, 9f2jcvew85hdye, otnwc739vhe82b, etc. If the attacker is lucky they might find the password, sooner or later. However if the password is too strong, such that it would take them too many years to guess it, they had better give up or think of an alternative attack (phishing, shoulder surfing, keyloggers, etc.)







share|improve this answer












share|improve this answer



share|improve this answer










answered 7 hours ago









reedreed

4,5383 gold badges13 silver badges34 bronze badges




4,5383 gold badges13 silver badges34 bronze badges












  • Indeed, I think that's what it really comes down to - the hacker is well aware of what their has rate is, and therefore what passwords they can crack in a reasonable amount of time. Bruteforcing the full search space is either possible in a reasonable amount of time or it isn't, in which case they won't bother and it won't matter.

    – Conor Mancone
    6 hours ago

















  • Indeed, I think that's what it really comes down to - the hacker is well aware of what their has rate is, and therefore what passwords they can crack in a reasonable amount of time. Bruteforcing the full search space is either possible in a reasonable amount of time or it isn't, in which case they won't bother and it won't matter.

    – Conor Mancone
    6 hours ago
















Indeed, I think that's what it really comes down to - the hacker is well aware of what their has rate is, and therefore what passwords they can crack in a reasonable amount of time. Bruteforcing the full search space is either possible in a reasonable amount of time or it isn't, in which case they won't bother and it won't matter.

– Conor Mancone
6 hours ago





Indeed, I think that's what it really comes down to - the hacker is well aware of what their has rate is, and therefore what passwords they can crack in a reasonable amount of time. Bruteforcing the full search space is either possible in a reasonable amount of time or it isn't, in which case they won't bother and it won't matter.

– Conor Mancone
6 hours ago













6














Entropy is really the biggest concern, and entropy is determined by the amount of randomness in your password generation process.



Let's use an 18 character password as an example. We'll look at alphanumeric characters only (62 possible characters). This gives:



log2(62**18) = 107 bits of entropy



If you decide to always convert the first 3 characters to Z to make it harder to bruteforce then you have effectively removed 3 random characters from your password, leaving you with:



log2(62**15) = 89 bits of entropy



Which is about a factor of 1,000,000 weaker. Of course, both are still absolutely and completely impossible to bruteforce, so if someone is trying to brute force your password I wouldn't worry. However, it's worth emphasizing how impossible it is to brute force such a password (just in case my links didn't convince you). A top of the line MD5 hashing rig can try 200 billion hashes per second. It's blazing fast (which is why MD5 isn't supposed to be used for passwords). Even your weaker password (when you replaced the first three characters with z) has 7.6e26 possible combinations. At 200 billion hashes per second it will only take 120 MILLION years to try all password possibilities if the password was hashed with MD5. I don't think you have anything to worry about.



So, is it better to start with "higher" characters to make it harder for someone to brute force your password? Maybe, maybe not. It depends on how someone actually tries to crack the password (and there aren't any guarantees about what direction they might work in). Either way though, if you use a long random password, there is absolutely zero chance that it will get hacked. So I wouldn't worry about it. When it comes to passwords length is king.






share|improve this answer

























  • This is a good answer. It may be improved by a brief discussion on Kerckhoff's Principle.

    – Adonalsium
    8 hours ago











  • More than 256 bits of entropy with only 18 chars and no symbols? That's surprising, I thought. Because in fact the calculation is wrong: it's not log2(18^62), but log2(62^18), which is 107 bits of entropy. Your general point (length is king) is good, but unfortunately the rest of your post is based on wrong calculations.

    – reed
    7 hours ago












  • @reed lol, this isn't the first time I've gotten those backwards...

    – Conor Mancone
    7 hours ago











  • @reed fortunately the correct numbers don't change the actual answer - it's just a lot less dramatic.

    – Conor Mancone
    7 hours ago






  • 1





    @Adonalsium I think that could be a good addition - I'll add something in when I have more time later.

    – Conor Mancone
    7 hours ago















6














Entropy is really the biggest concern, and entropy is determined by the amount of randomness in your password generation process.



Let's use an 18 character password as an example. We'll look at alphanumeric characters only (62 possible characters). This gives:



log2(62**18) = 107 bits of entropy



If you decide to always convert the first 3 characters to Z to make it harder to bruteforce then you have effectively removed 3 random characters from your password, leaving you with:



log2(62**15) = 89 bits of entropy



Which is about a factor of 1,000,000 weaker. Of course, both are still absolutely and completely impossible to bruteforce, so if someone is trying to brute force your password I wouldn't worry. However, it's worth emphasizing how impossible it is to brute force such a password (just in case my links didn't convince you). A top of the line MD5 hashing rig can try 200 billion hashes per second. It's blazing fast (which is why MD5 isn't supposed to be used for passwords). Even your weaker password (when you replaced the first three characters with z) has 7.6e26 possible combinations. At 200 billion hashes per second it will only take 120 MILLION years to try all password possibilities if the password was hashed with MD5. I don't think you have anything to worry about.



So, is it better to start with "higher" characters to make it harder for someone to brute force your password? Maybe, maybe not. It depends on how someone actually tries to crack the password (and there aren't any guarantees about what direction they might work in). Either way though, if you use a long random password, there is absolutely zero chance that it will get hacked. So I wouldn't worry about it. When it comes to passwords length is king.






share|improve this answer

























  • This is a good answer. It may be improved by a brief discussion on Kerckhoff's Principle.

    – Adonalsium
    8 hours ago











  • More than 256 bits of entropy with only 18 chars and no symbols? That's surprising, I thought. Because in fact the calculation is wrong: it's not log2(18^62), but log2(62^18), which is 107 bits of entropy. Your general point (length is king) is good, but unfortunately the rest of your post is based on wrong calculations.

    – reed
    7 hours ago












  • @reed lol, this isn't the first time I've gotten those backwards...

    – Conor Mancone
    7 hours ago











  • @reed fortunately the correct numbers don't change the actual answer - it's just a lot less dramatic.

    – Conor Mancone
    7 hours ago






  • 1





    @Adonalsium I think that could be a good addition - I'll add something in when I have more time later.

    – Conor Mancone
    7 hours ago













6












6








6







Entropy is really the biggest concern, and entropy is determined by the amount of randomness in your password generation process.



Let's use an 18 character password as an example. We'll look at alphanumeric characters only (62 possible characters). This gives:



log2(62**18) = 107 bits of entropy



If you decide to always convert the first 3 characters to Z to make it harder to bruteforce then you have effectively removed 3 random characters from your password, leaving you with:



log2(62**15) = 89 bits of entropy



Which is about a factor of 1,000,000 weaker. Of course, both are still absolutely and completely impossible to bruteforce, so if someone is trying to brute force your password I wouldn't worry. However, it's worth emphasizing how impossible it is to brute force such a password (just in case my links didn't convince you). A top of the line MD5 hashing rig can try 200 billion hashes per second. It's blazing fast (which is why MD5 isn't supposed to be used for passwords). Even your weaker password (when you replaced the first three characters with z) has 7.6e26 possible combinations. At 200 billion hashes per second it will only take 120 MILLION years to try all password possibilities if the password was hashed with MD5. I don't think you have anything to worry about.



So, is it better to start with "higher" characters to make it harder for someone to brute force your password? Maybe, maybe not. It depends on how someone actually tries to crack the password (and there aren't any guarantees about what direction they might work in). Either way though, if you use a long random password, there is absolutely zero chance that it will get hacked. So I wouldn't worry about it. When it comes to passwords length is king.






share|improve this answer















Entropy is really the biggest concern, and entropy is determined by the amount of randomness in your password generation process.



Let's use an 18 character password as an example. We'll look at alphanumeric characters only (62 possible characters). This gives:



log2(62**18) = 107 bits of entropy



If you decide to always convert the first 3 characters to Z to make it harder to bruteforce then you have effectively removed 3 random characters from your password, leaving you with:



log2(62**15) = 89 bits of entropy



Which is about a factor of 1,000,000 weaker. Of course, both are still absolutely and completely impossible to bruteforce, so if someone is trying to brute force your password I wouldn't worry. However, it's worth emphasizing how impossible it is to brute force such a password (just in case my links didn't convince you). A top of the line MD5 hashing rig can try 200 billion hashes per second. It's blazing fast (which is why MD5 isn't supposed to be used for passwords). Even your weaker password (when you replaced the first three characters with z) has 7.6e26 possible combinations. At 200 billion hashes per second it will only take 120 MILLION years to try all password possibilities if the password was hashed with MD5. I don't think you have anything to worry about.



So, is it better to start with "higher" characters to make it harder for someone to brute force your password? Maybe, maybe not. It depends on how someone actually tries to crack the password (and there aren't any guarantees about what direction they might work in). Either way though, if you use a long random password, there is absolutely zero chance that it will get hacked. So I wouldn't worry about it. When it comes to passwords length is king.







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edited 7 hours ago

























answered 8 hours ago









Conor ManconeConor Mancone

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  • This is a good answer. It may be improved by a brief discussion on Kerckhoff's Principle.

    – Adonalsium
    8 hours ago











  • More than 256 bits of entropy with only 18 chars and no symbols? That's surprising, I thought. Because in fact the calculation is wrong: it's not log2(18^62), but log2(62^18), which is 107 bits of entropy. Your general point (length is king) is good, but unfortunately the rest of your post is based on wrong calculations.

    – reed
    7 hours ago












  • @reed lol, this isn't the first time I've gotten those backwards...

    – Conor Mancone
    7 hours ago











  • @reed fortunately the correct numbers don't change the actual answer - it's just a lot less dramatic.

    – Conor Mancone
    7 hours ago






  • 1





    @Adonalsium I think that could be a good addition - I'll add something in when I have more time later.

    – Conor Mancone
    7 hours ago

















  • This is a good answer. It may be improved by a brief discussion on Kerckhoff's Principle.

    – Adonalsium
    8 hours ago











  • More than 256 bits of entropy with only 18 chars and no symbols? That's surprising, I thought. Because in fact the calculation is wrong: it's not log2(18^62), but log2(62^18), which is 107 bits of entropy. Your general point (length is king) is good, but unfortunately the rest of your post is based on wrong calculations.

    – reed
    7 hours ago












  • @reed lol, this isn't the first time I've gotten those backwards...

    – Conor Mancone
    7 hours ago











  • @reed fortunately the correct numbers don't change the actual answer - it's just a lot less dramatic.

    – Conor Mancone
    7 hours ago






  • 1





    @Adonalsium I think that could be a good addition - I'll add something in when I have more time later.

    – Conor Mancone
    7 hours ago
















This is a good answer. It may be improved by a brief discussion on Kerckhoff's Principle.

– Adonalsium
8 hours ago





This is a good answer. It may be improved by a brief discussion on Kerckhoff's Principle.

– Adonalsium
8 hours ago













More than 256 bits of entropy with only 18 chars and no symbols? That's surprising, I thought. Because in fact the calculation is wrong: it's not log2(18^62), but log2(62^18), which is 107 bits of entropy. Your general point (length is king) is good, but unfortunately the rest of your post is based on wrong calculations.

– reed
7 hours ago






More than 256 bits of entropy with only 18 chars and no symbols? That's surprising, I thought. Because in fact the calculation is wrong: it's not log2(18^62), but log2(62^18), which is 107 bits of entropy. Your general point (length is king) is good, but unfortunately the rest of your post is based on wrong calculations.

– reed
7 hours ago














@reed lol, this isn't the first time I've gotten those backwards...

– Conor Mancone
7 hours ago





@reed lol, this isn't the first time I've gotten those backwards...

– Conor Mancone
7 hours ago













@reed fortunately the correct numbers don't change the actual answer - it's just a lot less dramatic.

– Conor Mancone
7 hours ago





@reed fortunately the correct numbers don't change the actual answer - it's just a lot less dramatic.

– Conor Mancone
7 hours ago




1




1





@Adonalsium I think that could be a good addition - I'll add something in when I have more time later.

– Conor Mancone
7 hours ago





@Adonalsium I think that could be a good addition - I'll add something in when I have more time later.

– Conor Mancone
7 hours ago

















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