Mixing
64 bits password = 13 characters?
Clash Royale CLAN TAG#URR8PPP
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1
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If we use upper and lower case letters, and 10 digits, we get
approximately 6 bits per character. Then, strings of 13 characters
should work.
I saw above explanation in the material, but I cannot understand how 13 characters are appeared. How can I compute?
passwords
asked 5 hours ago
baeharam
112
add a comment |Â
up vote
1
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favorite
If we use upper and lower case letters, and 10 digits, we get
approximately 6 bits per character. Then, strings of 13 characters
should work.
I saw above explanation in the material, but I cannot understand how 13 characters are appeared. How can I compute?
passwords
asked 5 hours ago
baeharam
112
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If we use upper and lower case letters, and 10 digits, we get
approximately 6 bits per character. Then, strings of 13 characters
should work.
I saw above explanation in the material, but I cannot understand how 13 characters are appeared. How can I compute?
passwords
asked 5 hours ago
baeharam
112
If we use upper and lower case letters, and 10 digits, we get
approximately 6 bits per character. Then, strings of 13 characters
should work.
I saw above explanation in the material, but I cannot understand how 13 characters are appeared. How can I compute?
passwords
passwords
asked 5 hours ago
baeharam
112
asked 5 hours ago
baeharam
112
asked 5 hours ago
baeharam
112
asked 5 hours ago
baeharam
112
asked 5 hours ago
baeharam
112
112
add a comment |Â
add a comment |Â
2 Answers
2
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oldest
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up vote
2
down vote
It's likely an error. The number of possible passwords of length $l$ from a character set size of $n$ is $n^l$. The number 13 probably came from a calculation for a character set with only one case. A set of $36^12$ passwords is smaller than a set of $2^64$. You need one additional character to get 64 bits of information.
$$36^12 < 2^64 < 36^13$$
A 13 character long password from a 62 element character set is equivalent to
$$log_2(62^13) = 13 * log_2(62) approx 77.4 text(bits)$$
The actual minimum number of characters for that character set is
$$lceillog_62(2^64)rceil = lceil64 * log_62(2)rceil = 11 text(characters)$$
And for that character set there are $log_262 approx 5.95$ bits of information per character. You can get the same number $11 = lceil 64 / ~5.95 rceil$.
A 13 character string* is sufficient to get a password with at least 64 bit strength. However that's satisfied by any number of characters no less than 11, making 13 characters unnecessarily long.
* Randomly selected from a uniform distribution
answered 4 hours ago
Future Security
1,5741117
add a comment |Â
up vote
1
down vote
Suppose the number of a set $S$ is denoted as $Card(S)$ (which is sort for cardinality if you didn't know), then
$Card($uppercase-letters$)+Card($lowercase-letters$)+Card($digits$)
=26+26+10=62$
To represent an element from a set of 62, you need a string of at least 6 bits, this is because $2^6 = 64 ge 62$
However I don't understand the title where it says "64 bits password = 13 characters" because $ 6 cdot 13 = 78 $ which is way bigger here, than 64.
To achieve at least 64-bit entropy, you need to uniformly draw $ 64 over log_262 approx 11.00$ characters (10.749 to be more exact).
answered 4 hours ago
DannyNiu
840324
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
It's likely an error. The number of possible passwords of length $l$ from a character set size of $n$ is $n^l$. The number 13 probably came from a calculation for a character set with only one case. A set of $36^12$ passwords is smaller than a set of $2^64$. You need one additional character to get 64 bits of information.
$$36^12 < 2^64 < 36^13$$
A 13 character long password from a 62 element character set is equivalent to
$$log_2(62^13) = 13 * log_2(62) approx 77.4 text(bits)$$
The actual minimum number of characters for that character set is
$$lceillog_62(2^64)rceil = lceil64 * log_62(2)rceil = 11 text(characters)$$
And for that character set there are $log_262 approx 5.95$ bits of information per character. You can get the same number $11 = lceil 64 / ~5.95 rceil$.
A 13 character string* is sufficient to get a password with at least 64 bit strength. However that's satisfied by any number of characters no less than 11, making 13 characters unnecessarily long.
* Randomly selected from a uniform distribution
answered 4 hours ago
Future Security
1,5741117
add a comment |Â
up vote
2
down vote
It's likely an error. The number of possible passwords of length $l$ from a character set size of $n$ is $n^l$. The number 13 probably came from a calculation for a character set with only one case. A set of $36^12$ passwords is smaller than a set of $2^64$. You need one additional character to get 64 bits of information.
$$36^12 < 2^64 < 36^13$$
A 13 character long password from a 62 element character set is equivalent to
$$log_2(62^13) = 13 * log_2(62) approx 77.4 text(bits)$$
The actual minimum number of characters for that character set is
$$lceillog_62(2^64)rceil = lceil64 * log_62(2)rceil = 11 text(characters)$$
And for that character set there are $log_262 approx 5.95$ bits of information per character. You can get the same number $11 = lceil 64 / ~5.95 rceil$.
A 13 character string* is sufficient to get a password with at least 64 bit strength. However that's satisfied by any number of characters no less than 11, making 13 characters unnecessarily long.
* Randomly selected from a uniform distribution
answered 4 hours ago
Future Security
1,5741117
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It's likely an error. The number of possible passwords of length $l$ from a character set size of $n$ is $n^l$. The number 13 probably came from a calculation for a character set with only one case. A set of $36^12$ passwords is smaller than a set of $2^64$. You need one additional character to get 64 bits of information.
$$36^12 < 2^64 < 36^13$$
A 13 character long password from a 62 element character set is equivalent to
$$log_2(62^13) = 13 * log_2(62) approx 77.4 text(bits)$$
The actual minimum number of characters for that character set is
$$lceillog_62(2^64)rceil = lceil64 * log_62(2)rceil = 11 text(characters)$$
And for that character set there are $log_262 approx 5.95$ bits of information per character. You can get the same number $11 = lceil 64 / ~5.95 rceil$.
A 13 character string* is sufficient to get a password with at least 64 bit strength. However that's satisfied by any number of characters no less than 11, making 13 characters unnecessarily long.
* Randomly selected from a uniform distribution
answered 4 hours ago
Future Security
1,5741117
It's likely an error. The number of possible passwords of length $l$ from a character set size of $n$ is $n^l$. The number 13 probably came from a calculation for a character set with only one case. A set of $36^12$ passwords is smaller than a set of $2^64$. You need one additional character to get 64 bits of information.
$$36^12 < 2^64 < 36^13$$
A 13 character long password from a 62 element character set is equivalent to
$$log_2(62^13) = 13 * log_2(62) approx 77.4 text(bits)$$
The actual minimum number of characters for that character set is
$$lceillog_62(2^64)rceil = lceil64 * log_62(2)rceil = 11 text(characters)$$
And for that character set there are $log_262 approx 5.95$ bits of information per character. You can get the same number $11 = lceil 64 / ~5.95 rceil$.
A 13 character string* is sufficient to get a password with at least 64 bit strength. However that's satisfied by any number of characters no less than 11, making 13 characters unnecessarily long.
* Randomly selected from a uniform distribution
answered 4 hours ago
Future Security
1,5741117
edited 4 hours ago
answered 4 hours ago
Future Security
1,5741117
answered 4 hours ago
Future Security
1,5741117
answered 4 hours ago
Future Security
1,5741117
1,5741117
add a comment |Â
add a comment |Â
up vote
1
down vote
Suppose the number of a set $S$ is denoted as $Card(S)$ (which is sort for cardinality if you didn't know), then
$Card($uppercase-letters$)+Card($lowercase-letters$)+Card($digits$)
=26+26+10=62$
To represent an element from a set of 62, you need a string of at least 6 bits, this is because $2^6 = 64 ge 62$
However I don't understand the title where it says "64 bits password = 13 characters" because $ 6 cdot 13 = 78 $ which is way bigger here, than 64.
To achieve at least 64-bit entropy, you need to uniformly draw $ 64 over log_262 approx 11.00$ characters (10.749 to be more exact).
answered 4 hours ago
DannyNiu
840324
add a comment |Â
up vote
1
down vote
Suppose the number of a set $S$ is denoted as $Card(S)$ (which is sort for cardinality if you didn't know), then
$Card($uppercase-letters$)+Card($lowercase-letters$)+Card($digits$)
=26+26+10=62$
To represent an element from a set of 62, you need a string of at least 6 bits, this is because $2^6 = 64 ge 62$
However I don't understand the title where it says "64 bits password = 13 characters" because $ 6 cdot 13 = 78 $ which is way bigger here, than 64.
To achieve at least 64-bit entropy, you need to uniformly draw $ 64 over log_262 approx 11.00$ characters (10.749 to be more exact).
answered 4 hours ago
DannyNiu
840324
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Suppose the number of a set $S$ is denoted as $Card(S)$ (which is sort for cardinality if you didn't know), then
$Card($uppercase-letters$)+Card($lowercase-letters$)+Card($digits$)
=26+26+10=62$
To represent an element from a set of 62, you need a string of at least 6 bits, this is because $2^6 = 64 ge 62$
However I don't understand the title where it says "64 bits password = 13 characters" because $ 6 cdot 13 = 78 $ which is way bigger here, than 64.
To achieve at least 64-bit entropy, you need to uniformly draw $ 64 over log_262 approx 11.00$ characters (10.749 to be more exact).
answered 4 hours ago
DannyNiu
840324
Suppose the number of a set $S$ is denoted as $Card(S)$ (which is sort for cardinality if you didn't know), then
$Card($uppercase-letters$)+Card($lowercase-letters$)+Card($digits$)
=26+26+10=62$
To represent an element from a set of 62, you need a string of at least 6 bits, this is because $2^6 = 64 ge 62$
However I don't understand the title where it says "64 bits password = 13 characters" because $ 6 cdot 13 = 78 $ which is way bigger here, than 64.
To achieve at least 64-bit entropy, you need to uniformly draw $ 64 over log_262 approx 11.00$ characters (10.749 to be more exact).
answered 4 hours ago
DannyNiu
840324
answered 4 hours ago
DannyNiu
840324
answered 4 hours ago
DannyNiu
840324
answered 4 hours ago
DannyNiu
840324
840324
add a comment |Â
add a comment |Â
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