```There is an aspect of SQRL which has not been tempered by reason:

In designing the user password needed to unlock the password system (SQRL)
we have been guided by opinion, not facts. I hope to solve that here.

All should become clear in the "(3)++++++" heading, if not, then sorry, I am
not able to make it simpler.

++++++++++++++
In any calculation for password length or strength there is a basic
function, the logarithm, which is very tricky to understand and really
manage well.

There are two aspects in which the logarithm function is used:
1.- To calculate the number of binary bits equivalent to password strength.
2.- To calculate the number of characters needed to reach a set bit length.

If you do not understand the two lines above don't despair, let's make it
simple, as simple as possible:

(1)++++++
The strength of a password could be understood by the length of time that is
needed to break it, and that it is proportional to the number of trials
needed to make a full brute-force attack (try every possible option). So if
we want a password that live for 1000 years, and 1000 passwords could be
tested by second, we will need to perform:
1000 trials/second * 1000 years * (60*60*24*365.24) seconds/year =
31556736000000 trials = 3.16E13 trials

Here comes the first application of the log function. That number could be
converted to equivalent entropy bits by taking the "binary base" (base 2)
logarithm of it:
log(31556736000000 trials ; 2) = log(3.16E13 trials;2) = 44.84301 bits.

(2)++++++
Now, the strength of a password is a function of the number of symbols used
at each position and the number of positions. A decimal number has 10
symbols (0-9) and as many positions as needed to express the number. A 5
digit number has 10^5 (10 raised to the 5 power) options, or to say the same
in a different way: 100,000 possible values (0-99999). A 5 position base64
string will have 64^5 values, or 1,073,741,824

Here comes the second use of the logarithm function. If we take the
logarithm of such number, the exponent becomes a multiplicand, and we get:
a = b^n    ===>>  log(a;2) = n * log(b;2)
Setting t as the number of trials and s the number of symbols, we could
write:
t = s^n    ===>>  log(t;2) = n * log(s;2)
Then the number of needed positions (n) becomes:
n = log(t;2) / log(s;2)

That brings all the complexity to just one function, that function could be
written in a cell in any worksheet program.

(3)++++++
Bringing both items together, we could ask: what is the length required for
a password that lasts 1000 years at 1000 tests/second in ASCII (95 symbols):

Trials    Time     Total  equivalent      symb
per sec.  Years    Trials    entropy        95
1000      1000     3.16E13     44.84        10.36

sec/year is: 60 sec/min * 60 min/hr * 24 hr/day * 365.24 day/year = 31536000

(4)++++++
As the Enscrypt could be made to last 1 sec, we could write this table:
(rounding up to the nearest integer the number of characters):

Trials    Time     Total        equivalent      symb
per sec.  Years    Trials        entropy      95  64  52  32  16  10
1         100       3155673600  31,55        5   6   6   7   8  10
1       1.000      31556736000  34,87        6   6   7   7   9  11
1      10.000     315567360000  38,19        6   7   7   8  10  12
1     100.000    3155673600000  41,52        7   7   8   9  11  13
1   1.000.000   31556736000000  44,84        7   8   8   9  12  14
1  10.000.000  315567360000000  48,16        8   9   9  10  13  15
1 100.000.000 3155673600000000  51,48        8   9  10  11  13  16

The numbers may seem small, but are accurate (AFAICT).
We could include a factor of 1000, and still get:
Trials      Time     Total        equivalent      symb
per sec.    Years    Trials        entropy      95  64  52  32  16  10
1000         100    3,16E+012   41,52          7   7   8   9  11  13
1000       1.000    3,16E+013   44,84          7   8   8   9  12  14
1000      10.000    3,16E+014   48,16          8   9   9  10  13  15
1000     100.000    3,16E+015   51,49          8   9  10  11  13  16
1000   1.000.000    3,16E+016   54,81          9  10  10  11  14  17
1000  10.000.000    3,16E+017   58,13          9  10  11  12  15  18
1000 100.000.000    3,16E+018   61,45         10  11  11  13  16  19

So, for example, a 10 character upper and lower case (52 symbols) should
resist for 100 thousand years without any problem.

++++++++++++++

Note: the Haystack page use a more complex formula to calculate the "Search
Space" or number of total trials, exactly : b*((b^n)-1)/(b-1) (sum of a
geometric series) b=character set, n=number of positions. Not far from this
calculation anyway.

--
Mark Cross @ 02/10/2014 11:15 p.m.
Eagles may soar, but weasels don't get sucked into jet engines.

```
 0
Mark
2/11/2014 3:17:50 AM
grc.sqrl 459 articles. 0 followers.

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Mark Cross was heard to say :

> There is an aspect of SQRL which has not been tempered by reason:
>

Probably the idea is better understood with a worksheet, so here is one from
LibreOffice. It has been modified to fit the space contains of this server
but should work.

If anyone could confirm it works, it will be appreciated, thanks.

--
Mark Cross @ 02/11/2014 8:06 p.m.
The difference between greatness and mediocrity is often how an individual
views a mistake... — Nelson Boswell

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YXRpb25zMi9pbWFnZXMvQml0bWFwcy9QSwECFAAUAAAIAAB3AExEAAAAAAAAAAAAAAAAGgAAAAAA
AAAAAAAAAAAAGAAAAAAAAAAAAAAAAAAoGQAAQ29uZmlndXJhdGlvbnMyL2Zsb2F0ZXIvUEsBAhQA
ZW51YmFyL1BLAQIUABQAAAgIAHcATER3W33yEwEAAOEDAAAVAAAAAAAAAAAAAAAAAJQZAABNRVRB

--nextPart1945325.py4hMg1fq9--

```
 0
Mark
2/12/2014 12:09:44 AM
```This is an OpenPGP/MIME signed message (RFC 4880 and 3156)
--sU7wtwaRJLHrBtba2ELWOmN9XpOetneDB
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: quoted-printable

Hi,

Am 12.02.2014 01:09, schrieb Mark Cross:
> If anyone could confirm it works, it will be appreciated, thanks.

What do you mean by "works"? It opens up fine with LibreOffice 4.1.5 ...

Best regards,
Karol Babioch

--sU7wtwaRJLHrBtba2ELWOmN9XpOetneDB
Content-Type: application/pgp-signature; name="signature.asc"
Content-Description: OpenPGP digital signature
Content-Disposition: attachment; filename="signature.asc"

-----BEGIN PGP SIGNATURE-----
Version: GnuPG v2.0.22 (GNU/Linux)

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--sU7wtwaRJLHrBtba2ELWOmN9XpOetneDB--
```
 0
Karol
2/12/2014 1:22:21 AM
```Karol Babioch was heard to say :

> Hi,
>
> Am 12.02.2014 01:09, schrieb Mark Cross:
>> If anyone could confirm it works, it will be appreciated, thanks.
>
> What do you mean by "works"? It opens up fine with LibreOffice 4.1.5 ...

Just that !!

And that there are the correct formulas in the cells and it is editable and
you could play with the numbers.

The blue cells are intended to be modified as needed !!

But otherwise, just that is a correct worksheet file.
The point being that I had to erase several components of the internal
compressed original file to make it small enough for this news server,
I was not sure it would work (open) correctly in some other computer.

Thanks !!

--
Mark Cross @ 02/11/2014 9:25 p.m.
I'm a nobody, nobody is perfect, therefore I'm perfect.

```
 0
Mark
2/12/2014 1:30:12 AM
```On 2/10/2014 10:17 PM, Mark Cross wrote:
> There is an aspect of SQRL which has not been tempered by reason:
>
>
> In designing the user password needed to unlock the password system (SQRL)
> we have been guided by opinion, not facts. I hope to solve that here.

http://www.GRC.com/groups/sqrl:5001

There is still room for some conjecture, simply because we don't yet
know what advances will be made at performing Scrypt.  Yes, there are
indeed physical hardware limits to searching vast areas of memory, *but*
can we predict with any accuracy what the state-of-the-art Scrypt speed
will be in 10 years?  What about 20?  We can't imagine what we can't
imagine.

I will say this, however: if people obtain the technology to do 100
trillion guesses per second on passwords EnScrypted for 1 second or
longer, it's possible that our fundamental encryption technologies may
be at a higher risk of failure than our EnScrypted passwords/passcodes.

In all honesty, we don't need to resist *all* future attacks; just those
we can envision plus some reasonable safety margin.  If a paradigm shift
occurs due to unforeseeable technological advances (redundant, I know),
then everyone will simply create new identities on a hardened SQRL5
platform, just as they will stop using the cracked TLS1.5 in favor of
something else someday.  Yes, SQRL will be much more entrenched than an
SSL version (though history suggests otherwise), but we cannot plan for
all futures and trying to do so is counter-productive.

Yes, we want to be well-protected from unknown attacks, but we also
don't want to "waste" entropy that merely clogs up usability while
providing no *relevant* security whatsoever.  Loss or theft, not brute
force, are the biggest threats to SQRL Access Codes *and even* user

This is why SQRL has been so difficult to design, IMO; our dependence
upon first-party responsibility means we *must* eventually hand over
responsibility to those parties.  The website authentication part was
fairly straightforward right up until that pesky *first-party* started
asking for revocation privileges!  Try as we might, we cannot create a
user-proof identity solution.  If users want to be responsible for their
own online identities, then they must backup their keys in a secure
location.

--

RobAllen
_____________________________________________________
```
 0
RobAllen
2/12/2014 3:40:59 AM
```Worked fine here. This spreadsheet is COOL!

Disclosure: --Not a dev. Just an occasional lurker.--

effectiveness of a password you might be considering. From this, I've
learned that when you can use some means to put the brakes on the rate
an attacker can make trials, the effectiveness of a given password
length goes up dramatically. So a really long and or complex password is
necessary only when your attacker will be more or less unbounded in
their ability to bring conceivable computing resources to bear on a
brute-force effort (e.g. a website from which where an attacker

Take from the table in your original post, the /seemingly/ weak password
of length-5 from the full ASCII set, with the equivalent entropy of
31.55 bits. If you had the power of today's Bitcoin hashing network
(crunching 22606 Tera-hashes/sec), you could crack that flimsy thing on
the order of (2^31.55/22606E+12) = 139 ns. But, slow the trials rate to
1/sec, and you need on the order of 3 Gs (100 years) to complete a full
brute force.

So what you're saying is: since SQRL eliminates the need for the website
to keep something like an encrypted password database, and the design of
the system incorporates a method for fixing relative upper-bounds on the
achievable trials rate, it's not as important that a user of SQRL fret
too much about the length of the password safeguarding their ID key.

Have I got that?
Good demo.

--
OpenPGP 0x06b4de19 (http://keyserver.pgp.com; http://keys.gnupg.net:11371)
```
 0
Andrew
2/18/2014 8:44:01 AM
```Andrew Skretvedt was heard to say :

> Worked fine here. This spreadsheet is COOL!

Thanks, I am happy you like it.

> Like the Haystacks page, your spreadsheet brings great visibility to the
> effectiveness of a password you might be considering. From this, I've
> learned that when you can use some means to put the brakes on the rate
> an attacker can make trials, the effectiveness of a given password
> length goes up dramatically. So a really long and or complex password is
> necessary only when your attacker will be more or less unbounded in
> their ability to bring conceivable computing resources to bear on a
> brute-force effort (e.g. a website from which where an attacker
> exfiltrated the encrypted password database).

Your description is excellent and exactly the point I was trying to get tru.
Thanks for sharing your thoughs on the matter.

> Take from the table in your original post, the /seemingly/ weak password
> of length-5 from the full ASCII set, with the equivalent entropy of
> 31.55 bits. If you had the power of today's Bitcoin hashing network
> (crunching 22606 Tera-hashes/sec), you could crack that flimsy thing on
> the order of (2^31.55/22606E+12) = 139 ns. But, slow the trials rate to
> 1/sec, and you need on the order of 3 Gs (100 years) to complete a full
> brute force.

Yes!

> So what you're saying is: since SQRL eliminates the need for the website
> to keep something like an encrypted password database, and the design of
> the system incorporates a method for fixing relative upper-bounds on the
> achievable trials rate, it's not as important that a user of SQRL fret
> too much about the length of the password safeguarding their ID key.
>
> Have I got that?

Yes, you have get it very well. :-)

> Good demo.

Thanks again.

--
Mark Cross @ 02/18/2014 6:20 p.m.
Everybody wants to go to heaven, but nobody wants to die.

```
 0
Mark
2/18/2014 10:23:10 PM

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