Cryptography-Digest Digest #601, Volume #14 Wed, 13 Jun 01 00:13:00 EDT
Contents:
Re: Simple Crypto II, the public key... ("Tom St Denis")
Re: Who can help me crack this encryption ("Jeffrey Walton")
Re: Better 8x32's sboxes ("Tom St Denis")
Re: National Security Nightmare? ("Boyd Roberts")
Re: Who can help me crack this encryption (Terrence Koeman)
Re: Sophie-Germain Primes for sale (David Hopwood)
Re: Simple Crypto II, the public key... (Gregory G Rose)
Re: Some questions on GSM and 3G ("Boyd Roberts")
Re: Sophie-Germain Primes for sale (SCOTT19U.ZIP_GUY)
Re: Publication violation notice ("Boyd Roberts")
Earpster AES ("James Wyatt")
Re: Who can help me crack this encryption (Terrence Koeman)
Re: Best, Strongest Algorithm (gone from any reasonable topic) (Dennis Ritchie)
Re: Best, Strongest Algorithm (gone from any reasonable topic) - VERY (wtshaw)
Re: When the signer is trusted do birthdays matter? ("Neil Couture")
----------------------------------------------------------------------------
From: "Tom St Denis" <[EMAIL PROTECTED]>
Subject: Re: Simple Crypto II, the public key...
Date: Wed, 13 Jun 2001 01:13:00 GMT
"Tom St Denis" <[EMAIL PROTECTED]> wrote in message
news:PzyV6.102546$[EMAIL PROTECTED]...
>
> "John Savard" <[EMAIL PROTECTED]> wrote in message
> news:[EMAIL PROTECTED]...
> > On Tue, 12 Jun 2001 23:04:40 GMT, "Tom St Denis"
> > <[EMAIL PROTECTED]> wrote, in part:
> >
> > >How do you cheat with "mod operations"? I have a list of good DH
primes
> if
> > >you want
> >
> > Well, if P is 2^n-1, one can perform modulo arithmetic without long
> > division. I don't know if a prime can be Mersenne and Sophie Germain
> > at the same time, though.
>
> Ah but the # of good primes of the form 2^n - 1 is few. i dunno of any
good
> ~1024 bit primes of that form.
Just FYI. The only Mersenne prime around 1024-bits is 2^1279 - 1, Where the
order has factors
``(2)*``(3)^3*``(7)*``(19)*``(73)*``(1279)*_c354*``(228479)*``(17467)*``(664
57)*``(102241)*``(5113)
(Ignore the '' since Maple inserts them for some reason).
_c354 is a 354 digit composite.
As you can see it's roughly speaking "smooth" [excuse the bad pun]. Which
means you must pick your base properly otherwise it can be vulnerable. Keep
in mind this 354 digit residue is about 1175 bits in size. So as long as
the order of your base is not divisible by any of the smooth primes it must
be divisible by the big composite.
Most likely the big composite is two large (600 bits) primes or at most
three (400 bits) primes. Hopefully it is the former case in which any base
that generates a group of order divisible by this composite you should be
ok.
I dunno of any other method to determine the order of a base other than
factoruing the order of the group
Tom
------------------------------
Reply-To: "Jeffrey Walton" <[EMAIL PROTECTED]>
From: "Jeffrey Walton" <[EMAIL PROTECTED]>
Subject: Re: Who can help me crack this encryption
Date: Tue, 12 Jun 2001 21:58:35 -0400
Use the second column as an include/exclude mask for the bits in column
1:
01001100 4C "L" 00010101 => 010x1x0x = > 01010
01101111 6F "o" 10101111 => x1x0xxxx => 10
01101111 6F "o" 01100011 => 0xx011xx => 0011
01101011 6B "k" 00101011 => 01x0x0xx => 0100
00100000 20 " " 00011111 => 001xxxxx => 001
01110100 74 "t" 10110101 => x1xx0x0x => 100
01101000 68 "h" 10100001 => x1x0100x => 10100
Smash together, then group by 8 bits: "The ..."
"Terrence Koeman" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]...
:
: I'm sorry if this question is double posted, but my original posting
: disappeared and I'm not sure if my reply got through my crappy
: provider's newsserver ;) (i think not...)
:
: I'm trying to crack a code, and with the help of Daniel i got this far
: (the original code is at the end...):
:
: 01001100 4C "L" 00010101
: 01101111 6F "o" 10101111
: 01101111 6F "o" 01100011
: 01101011 6B "k" 00101011
: 00100000 20 " " 00011111
: 01110100 74 "t" 10110101
: 01101000 68 "h" 10100001
: 01110010 72 "r" 10110010
: 01101111 6F "o" 01101111
: 01110101 75 "u" 00000010
: 01100111 67 "g" 10101001
: 01101000 68 "h" 00010111
: 00100000 20 " " 11001111
: 01110100 74 "t" 10000101
: 01101000 68 "h" 10101011
: 01100101 65 "e" 10000001
: 00100000 20 " " 00101111
: 01101000 68 "h" 10101000
: 01101111 6F "o" 01100000
: 01101100 6C "l" 01100101
: 01100101 65 "e" 10001010
: 01110011 73 "s" 10010100
: 00100000 20 " " 11001111
: 01110100 74 "t" 10110000
: 01101111 6F "o" 10101111
: 00100000 20 " " 00100011
: 01110010 72 "r" 10000001
: 01100101 65 "e" 10101010
: 01110110 76 "v" 01110010
: 01100101 65 "e" 10101010
: 01100001 61 "a" 01101110
: 01101100 6C "l" 00100001
: 00100000 20 " " 11001111
: 01110100 74 "t" 10010101
: 01101000 68 "h" 10101000
: 01100101 65 "e" 01101010
: 00100000 20 " " 01000111
: 01110011 73 "s" 10110100
: 01100101 65 "e" 10100010
: 01100011 63 "c" 01101101
: 01110010 72 "r" 01110010
: 01100101 65 "e" 01101001
: 01110100 74 "t" 00010001
: 00100000 20 " " 11011111
: 01101000 68 "h" 10101000
: 01101001 69 "i" 10100110
: 01100100 64 "d" 01101000
: 01100100 64 "d" 00001011
: 01100101 65 "e" 10000010
: 01101110 6E "n" 01100110
: 00100000 20 " " 11001111
: 01110111 77 "w" 10000111
: 01101001 69 "i" 00000110
: 01110100 74 "t" 10110101
: 01101000 68 "h" 10011111
: 01101001 69 "i" 11111111
: 01101110 6E "n" 11111111
: 00100000 20 " " 11111111
: 01101101 6D "m" 11111111
: 01100101 65 "e" 11111111
:
: I tried inverting the second column, as the last
: 5 bytes seem padding that should be '00000000'. I also tried XOR-ing
: and OR-ing the first column with the second, but no avail...
:
: Does anyone have any ideas on the second column?
:
: Thank you in advance ;)
:
: Regards,
:
: Terrence Koeman
:
:
: The original code:
:
: 0100110000010101
: 0110111110101111
: 0110111101100011
: 0110101100101011
: 0010000000011111
: 0111010010110101
: 0110100010100001
: 0111001010110010
: 0110111101101111
: 0111010100000010
: 0110011110101001
: 0110100000010111
: 0010000011001111
: 0111010010000101
: 0110100010101011
: 0110010110000001
: 0010000000101111
: 0110100010101000
: 0110111101100000
: 0110110001100101
: 0110010110001010
: 0111001110010100
: 0010000011001111
: 0111010010110000
: 0110111110101111
: 0010000000100011
: 0111001010000001
: 0110010110101010
: 0111011001110010
: 0110010110101010
: 0110000101101110
: 0110110000100001
: 0010000011001111
: 0111010010010101
: 0110100010101000
: 0110010101101010
: 0010000001000111
: 0111001110110100
: 0110010110100010
: 0110001101101101
: 0111001001110010
: 0110010101101001
: 0111010000010001
: 0010000011011111
: 0110100010101000
: 0110100110100110
: 0110010001101000
: 0110010000001011
: 0110010110000010
: 0110111001100110
: 0010000011001111
: 0111011110000111
: 0110100100000110
: 0111010010110101
: 0110100010011111
: 0110100111111111
: 0110111011111111
: 0010000011111111
: 0110110111111111
: 0110010111111111
------------------------------
From: "Tom St Denis" <[EMAIL PROTECTED]>
Subject: Re: Better 8x32's sboxes
Date: Wed, 13 Jun 2001 02:20:18 GMT
"Tom St Denis" <[EMAIL PROTECTED]> wrote in message
news:FvxV6.102068$[EMAIL PROTECTED]...
> I was thinking about making slightly better 8x32's (actually it was a bolt
> of the obvious).
>
> Use the inversion in GF(2^32)/p(x) instead. Basically we fix three of the
> inputs for the four 8x32's as in
>
> X11 X12 X13 X
> X21 X22 X X24
> X31 X X33 X34
> X X42 X43 X44
>
> Where no row has identical fixed values. (i.e X11 != X21, X11 != X31,
> etc...). Each row is one 8x32 sbox and X is the variable input.
Alas we need only
X11 X12 X13 X
X21 X22 X23 X
...
As long as no three entries in each row match another row each 8x32 will not
have identical entries.
If anyone is interested free source that does this and eight 8x32 sboxes are
available at
http://tomstdenis.home.dhs.org/big.zip
They were filtered such that only sboxes that had 130115 active sboxes (over
all possible pairs (non-zero diff)). This way a certain amount of diffusion
is guaranteed.
Comments welcomed (Please!)
Tom
------------------------------
From: "Boyd Roberts" <[EMAIL PROTECTED]>
Subject: Re: National Security Nightmare?
Date: Wed, 13 Jun 2001 04:52:54 +0200
"Jim D" <[EMAIL PROTECTED]> a écrit dans le message news:
[EMAIL PROTECTED]
> On Tue, 12 Jun 2001 02:34:48 +0200, "Boyd Roberts" <[EMAIL PROTECTED]>
> wrote:
> >
> >both CD and mail had been in current use for years.
no, in FRANCE, where i've been living for the past 10 years.
CD (in french) even sounds like cédé.
> In America, maybe. It's just that, like me, they object
> to their language being polluted by Americanisms.
and what are you going to call the next new thing invented?
you will have to invent a word for it.
big feature about 'english' is that it just takes it on board
and carries on. your VHF/UHF tv antenna -- what's it's
real name? a yagi-uda array, shortened to 'yagi'. the
inventors 'yagi' and 'uda' where both japanese.
and you can hardly defend that bastardised language called
'english' as having any semblence of purity -- this being
one of its strengths.
--
Boyd Roberts [EMAIL PROTECTED]
What do you know about surfing, Major? You're from goddamn New Jersey.
-- Lt. Colonel Kilgore
------------------------------
From: Terrence Koeman <[EMAIL PROTECTED]>
Subject: Re: Who can help me crack this encryption
Date: Wed, 13 Jun 2001 03:05:37 GMT
Damn! I was so close, but I never figured that the outcome per line
would be less than 8 bits...
You rule!
How long did it take you to figure this out? I've been thinking and
trying for about 2 days ;)
Regards,
Terrence Koeman
On Tue, 12 Jun 2001 21:58:35 -0400, "Jeffrey Walton"
<[EMAIL PROTECTED]> hoped somebody would understand this:
>Use the second column as an include/exclude mask for the bits in column
>1:
>01001100 4C "L" 00010101 => 010x1x0x = > 01010
>01101111 6F "o" 10101111 => x1x0xxxx => 10
>01101111 6F "o" 01100011 => 0xx011xx => 0011
>01101011 6B "k" 00101011 => 01x0x0xx => 0100
>00100000 20 " " 00011111 => 001xxxxx => 001
>01110100 74 "t" 10110101 => x1xx0x0x => 100
>01101000 68 "h" 10100001 => x1x0100x => 10100
>
>Smash together, then group by 8 bits: "The ..."
>
>"Terrence Koeman" <[EMAIL PROTECTED]> wrote in message
>news:[EMAIL PROTECTED]...
>:
>: I'm sorry if this question is double posted, but my original posting
>: disappeared and I'm not sure if my reply got through my crappy
>: provider's newsserver ;) (i think not...)
>:
>: I'm trying to crack a code, and with the help of Daniel i got this far
>: (the original code is at the end...):
>:
>: 01001100 4C "L" 00010101
>: 01101111 6F "o" 10101111
>: 01101111 6F "o" 01100011
>: 01101011 6B "k" 00101011
>: 00100000 20 " " 00011111
>: 01110100 74 "t" 10110101
>: 01101000 68 "h" 10100001
>: 01110010 72 "r" 10110010
>: 01101111 6F "o" 01101111
>: 01110101 75 "u" 00000010
>: 01100111 67 "g" 10101001
>: 01101000 68 "h" 00010111
>: 00100000 20 " " 11001111
>: 01110100 74 "t" 10000101
>: 01101000 68 "h" 10101011
>: 01100101 65 "e" 10000001
>: 00100000 20 " " 00101111
>: 01101000 68 "h" 10101000
>: 01101111 6F "o" 01100000
>: 01101100 6C "l" 01100101
>: 01100101 65 "e" 10001010
>: 01110011 73 "s" 10010100
>: 00100000 20 " " 11001111
>: 01110100 74 "t" 10110000
>: 01101111 6F "o" 10101111
>: 00100000 20 " " 00100011
>: 01110010 72 "r" 10000001
>: 01100101 65 "e" 10101010
>: 01110110 76 "v" 01110010
>: 01100101 65 "e" 10101010
>: 01100001 61 "a" 01101110
>: 01101100 6C "l" 00100001
>: 00100000 20 " " 11001111
>: 01110100 74 "t" 10010101
>: 01101000 68 "h" 10101000
>: 01100101 65 "e" 01101010
>: 00100000 20 " " 01000111
>: 01110011 73 "s" 10110100
>: 01100101 65 "e" 10100010
>: 01100011 63 "c" 01101101
>: 01110010 72 "r" 01110010
>: 01100101 65 "e" 01101001
>: 01110100 74 "t" 00010001
>: 00100000 20 " " 11011111
>: 01101000 68 "h" 10101000
>: 01101001 69 "i" 10100110
>: 01100100 64 "d" 01101000
>: 01100100 64 "d" 00001011
>: 01100101 65 "e" 10000010
>: 01101110 6E "n" 01100110
>: 00100000 20 " " 11001111
>: 01110111 77 "w" 10000111
>: 01101001 69 "i" 00000110
>: 01110100 74 "t" 10110101
>: 01101000 68 "h" 10011111
>: 01101001 69 "i" 11111111
>: 01101110 6E "n" 11111111
>: 00100000 20 " " 11111111
>: 01101101 6D "m" 11111111
>: 01100101 65 "e" 11111111
>:
>: I tried inverting the second column, as the last
>: 5 bytes seem padding that should be '00000000'. I also tried XOR-ing
>: and OR-ing the first column with the second, but no avail...
>:
>: Does anyone have any ideas on the second column?
>:
>: Thank you in advance ;)
>:
>: Regards,
>:
>: Terrence Koeman
>:
>:
>: The original code:
>:
>: 0100110000010101
>: 0110111110101111
>: 0110111101100011
>: 0110101100101011
>: 0010000000011111
>: 0111010010110101
>: 0110100010100001
>: 0111001010110010
>: 0110111101101111
>: 0111010100000010
>: 0110011110101001
>: 0110100000010111
>: 0010000011001111
>: 0111010010000101
>: 0110100010101011
>: 0110010110000001
>: 0010000000101111
>: 0110100010101000
>: 0110111101100000
>: 0110110001100101
>: 0110010110001010
>: 0111001110010100
>: 0010000011001111
>: 0111010010110000
>: 0110111110101111
>: 0010000000100011
>: 0111001010000001
>: 0110010110101010
>: 0111011001110010
>: 0110010110101010
>: 0110000101101110
>: 0110110000100001
>: 0010000011001111
>: 0111010010010101
>: 0110100010101000
>: 0110010101101010
>: 0010000001000111
>: 0111001110110100
>: 0110010110100010
>: 0110001101101101
>: 0111001001110010
>: 0110010101101001
>: 0111010000010001
>: 0010000011011111
>: 0110100010101000
>: 0110100110100110
>: 0110010001101000
>: 0110010000001011
>: 0110010110000010
>: 0110111001100110
>: 0010000011001111
>: 0111011110000111
>: 0110100100000110
>: 0111010010110101
>: 0110100010011111
>: 0110100111111111
>: 0110111011111111
>: 0010000011111111
>: 0110110111111111
>: 0110010111111111
>
--
Greetingz,
ÐoÐO_ÐeViL®
_________________________________________________________________
mail: [EMAIL PROTECTED]
_________________________________________________________________
Don't agree with my posting? Go here: http://www.globalwarez.nl/gfy/
_________________________________________________________________
If Christ died for our sins, dare we make his martyrdom meaningless
by not committing them? - Jules Feiffer
_________________________________________________________________
I am an agnostic pagan. I doubt the existence of many gods.
_________________________________________________________________
------------------------------
Date: Tue, 12 Jun 2001 23:46:57 +0100
From: David Hopwood <[EMAIL PROTECTED]>
Subject: Re: Sophie-Germain Primes for sale
=====BEGIN PGP SIGNED MESSAGE=====
Tom St Denis wrote:
> No seriously *free* SG primes are at my website
>
> http://tomstdenis.home.dhs.org/primes.txt
>
> A SG prime is of the form p = 2q + 1, where q itself is prime and of
> course p mod 4 = 3.
No. It's not two weeks since I last corrected you on this (message ID
<[EMAIL PROTECTED]>).
If p = 2q + 1 for p and q both prime, then q is a Germain prime, and
p is a safe prime. Also it is not part of the definition that p = 3 (mod 4)
(counterexample: p = 5, q = 2, although admittedly that is the only
counterexample).
May I suggest that before posting anything else involving number theory
or abstract algebra, you check the definitions first (for example in HAC
or AC2), and also that you pay more attention when people correct you?
- --
David Hopwood <[EMAIL PROTECTED]>
Home page & PGP public key: http://www.users.zetnet.co.uk/hopwood/
RSA 2048-bit; fingerprint 71 8E A6 23 0E D3 4C E5 0F 69 8C D4 FA 66 15 01
Nothing in this message is intended to be legally binding. If I revoke a
public key but refuse to specify why, it is because the private key has been
seized under the Regulation of Investigatory Powers Act; see www.fipr.org/rip
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------------------------------
From: [EMAIL PROTECTED] (Gregory G Rose)
Subject: Re: Simple Crypto II, the public key...
Date: 12 Jun 2001 20:22:31 -0700
In article <w6zV6.102704$[EMAIL PROTECTED]>,
Tom St Denis <[EMAIL PROTECTED]> wrote:
>Just FYI. The only Mersenne prime around 1024-bits is 2^1279 - 1, Where the
>order has factors
>
>``(2)*``(3)^3*``(7)*``(19)*``(73)*``(1279)*_c354*``(228479)*``(17467)*``(664
>57)*``(102241)*``(5113)
>
>_c354 is a 354 digit composite.
>
>As you can see it's roughly speaking "smooth" [excuse the bad pun]. Which
It is, by definition, not smooth. For it to be
smooth, *all* the factors must be small. If c354
had small factors it would have been factored by
now.
>means you must pick your base properly otherwise it can be vulnerable. Keep
>in mind this 354 digit residue is about 1175 bits in size. So as long as
>the order of your base is not divisible by any of the smooth primes it must
>be divisible by the big composite.
No, it must be divisible by at least one of the
factors of the big composite.
>Most likely the big composite is two large (600 bits) primes or at most
>three (400 bits) primes. Hopefully it is the former case in which any base
>that generates a group of order divisible by this composite you should be
>ok.
Even in the latter case it would be OK, as for
DSA.
>I dunno of any other method to determine the order of a base other than
>factoruing the order of the group
It can't be any easier than factoring, since
otherwise you'd have invented a new factoring
algorithm.
Greg.
--
Greg Rose INTERNET: [EMAIL PROTECTED]
Qualcomm Australia VOICE: +61-2-9817 4188 FAX: +61-2-9817 5199
Level 3, 230 Victoria Road, http://people.qualcomm.com/ggr/
Gladesville NSW 2111 232B EC8F 44C6 C853 D68F E107 E6BF CD2F 1081 A37C
------------------------------
From: "Boyd Roberts" <[EMAIL PROTECTED]>
Crossposted-To: alt.privacy
Subject: Re: Some questions on GSM and 3G
Date: Wed, 13 Jun 2001 05:22:57 +0200
what's the bet that 3G will just die? all it seems to be
is a revenue generator for governments who control spectrum
resources and licencing.
------------------------------
From: [EMAIL PROTECTED] (SCOTT19U.ZIP_GUY)
Subject: Re: Sophie-Germain Primes for sale
Date: 13 Jun 2001 03:23:17 GMT
[EMAIL PROTECTED] (David Hopwood) wrote in
<[EMAIL PROTECTED]>:
>-----BEGIN PGP SIGNED MESSAGE-----
>
>Tom St Denis wrote:
>
>May I suggest that before posting anything else involving number theory
>or abstract algebra, you check the definitions first (for example in HAC
>or AC2), and also that you pay more attention when people correct you?
>
Good Luck hes an arragant kid who seldom listens. Next for spouting
facts his mind can't grasp he will call you a loon or something along
those lines. Best just to ignore him for awhile and hope he either
goes aways or grows up.
>- --
>David Hopwood <[EMAIL PROTECTED]>
>
>Home page & PGP public key: http://www.users.zetnet.co.uk/hopwood/
>RSA 2048-bit; fingerprint 71 8E A6 23 0E D3 4C E5 0F 69 8C D4 FA 66 15
>01 Nothing in this message is intended to be legally binding. If I
>revoke a public key but refuse to specify why, it is because the private
>key has been seized under the Regulation of Investigatory Powers Act;
>see www.fipr.org/rip
>
>
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>
>
>
David A. Scott
--
SCOTT19U.ZIP NOW AVAILABLE WORLD WIDE "OLD VERSIOM"
http://www.jim.com/jamesd/Kong/scott19u.zip
My website http://members.nbci.com/ecil/index.htm
My crypto code http://radiusnet.net/crypto/archive/scott/
MY Compression Page http://members.nbci.com/ecil/compress.htm
**NOTE FOR EMAIL drop the roman "five" ***
Disclaimer:I am in no way responsible for any of the statements
made in the above text. For all I know I might be drugged or
something..
No I'm not paranoid. You all think I'm paranoid, don't you!
------------------------------
From: "Boyd Roberts" <[EMAIL PROTECTED]>
Subject: Re: Publication violation notice
Date: Wed, 13 Jun 2001 05:26:15 +0200
while feynman was at los alamos he played censorship
rule games with his wife (and sister?) via mail.
------------------------------
From: "James Wyatt" <[EMAIL PROTECTED]>
Subject: Earpster AES
Date: Wed, 13 Jun 2001 03:31:13 GMT
Simple DOS program that used AES, source included.
http://www.geocities.com/jrwyatt79/Earpster.zip
karhma, wyatt
------------------------------
From: Terrence Koeman <[EMAIL PROTECTED]>
Subject: Re: Who can help me crack this encryption
Date: Wed, 13 Jun 2001 03:36:26 GMT
Damn! I was so close, but I never figured that the outcome per line
would be less than 8 bits...
You rule!
How long did it take you to figure this out? I've been thinking and
trying for about 2 days ;)
Regards,
Terrence Koeman
On Tue, 12 Jun 2001 21:58:35 -0400, "Jeffrey Walton"
<[EMAIL PROTECTED]> hoped somebody would understand this:
>Use the second column as an include/exclude mask for the bits in column
>1:
>01001100 4C "L" 00010101 => 010x1x0x = > 01010
>01101111 6F "o" 10101111 => x1x0xxxx => 10
>01101111 6F "o" 01100011 => 0xx011xx => 0011
>01101011 6B "k" 00101011 => 01x0x0xx => 0100
>00100000 20 " " 00011111 => 001xxxxx => 001
>01110100 74 "t" 10110101 => x1xx0x0x => 100
>01101000 68 "h" 10100001 => x1x0100x => 10100
>
>Smash together, then group by 8 bits: "The ..."
--
Greetingz,
ÐoÐO_ÐeViL®
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From: Dennis Ritchie <[EMAIL PROTECTED]>
Subject: Re: Best, Strongest Algorithm (gone from any reasonable topic)
Date: Wed, 13 Jun 2001 03:45:56 +0000
David Hopwood corrected my gloss on the Shannon snippet:
> Dennis Ritchie wrote:
> > E = TiM
> > ...."
> > [Ti is the transformation performed on the i-th message]
>
> No, i here is the key. (See figure 5 for a confirmation of that.)
Well, Ti (or T_i or $T sub i$) is still the transformation
and indeed i is much more like the key.
Hopwood's right that in the quoted section i doesn't
necessarily refer to the i-th message.
Dennis
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From: [EMAIL PROTECTED] (wtshaw)
Subject: Re: Best, Strongest Algorithm (gone from any reasonable topic) - VERY
Date: Tue, 12 Jun 2001 21:33:05 -0600
In article <[EMAIL PROTECTED]>,
[EMAIL PROTECTED] (SCOTT19U.ZIP_GUY) wrote:
> All those so called time to eternity to exploit are
> based on no one finding simple ways to exploit a break.
> If history has any lessons. It is thats its foolish to
> smuglly sit by and hope that its hard to break. One should
> always strive to add what ever security one can. To put
> all your eggs in one basket is surely the recipe for
> desaster. But that seems to be what AES is all about.
>
Much that we see tries to outdate long held good ideas that are not
convenient. Conversely, we also see much effort in trying to justify as
good ideas thich have been shown to be invalid.
Not putting all your eggs in one basket is a good idea. Not realizing the
shallow nature of many security and encryption measures that are supposed
to universally sufficiently scramble eggs is not wise. Baskets with loose
weaves and surplus holes in the bottom just can't carry water if that is
what you need.
--
In trying to get meaning from the TmV-OK saga, remember that
those who do not learn from history are apt to repeat it.
------------------------------
From: "Neil Couture" <[EMAIL PROTECTED]>
Subject: Re: When the signer is trusted do birthdays matter?
Date: Tue, 12 Jun 2001 23:56:51 -0700
Reply-To: "Neil Couture" <[EMAIL PROTECTED]>
the birthday problems is related to the problems of determining a necessary
security
condition for hash functions that depends only on the size of the message
Digest.
( ie its cardinality ). ( from now on called Z, and n = cardinality of Z )
a bit of math now::
lets z1,z2, z3,... zn INCLUDED IN Z.
we can assume that all z(i) have the same probabilities.
It's very simple to calculate the probability that k consecutive random z1,
z2,... zk INCLUDED in Z
are distinct which is equal to:
( 1 - ( 1/n ) ) * ( 1 - ( 2/n ) ) * ... * ( 1 - ( ( k-1)/n ) ) :
the first choice is arbitrary, the probability that z2 != z1 equals 1 -
1/n ),
the probability that z3 is different from z1 and z2 = 1 - ( 2/n ), etc...
starting from there one might derive this formula that gives you the total
number of
consecutive random element z1,... zk which are distinct according to a
probability 'pr' ::
k = sqrt( 2n ln ( 1 / ( 1 - pr ) )
if you take pr = 0.5 you then get
k appr equals 1.17 * sqrt( n ). Try this with n = 365 and you got the
birthday attacks.
So returning to you post, The security of a hashfunction has nothing to do
with
the number of message that somebody hash and sign but more to the fact of
how easily it is to find a collision. It is certainly the case anyway that
the fact of
signing your hash gives you a lot of room if your hash function is not one
which has low probability of collision.
Neil
"Fat Phil" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]...
> I understand the birthday coincidence problem. (i.e. only ~22 random
> people is enough for a 50/50 chance of a birthday coincidence).
>
> However, if your document _originator_and_signer_ is Trusted Trent, and
> signatures are done on the document hash, then why do you need to
> consider the birthday attack. He's not going to be creating many
> documents in order to try to find two that hash together?
> In fact being a single person/company, he's only going to be releasing
> 'small' numbers of documents (few/day, say).
> Does that mean that in this situation hashes only need to be half as
> wide as one would normally recommend as they don't need to consider the
> birthday problem?
> Or have I missed something?
> Phil
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