Re: Metric length of path
- From: Musatov <marty.musatov@xxxxxxxxx>
- Date: Wed, 10 Jun 2009 12:35:30 -0700 (PDT)
[Beginning Musatov Proof Claim Statement.... Initializing....
1,2,3,4,5,6,7,8,9...
Go_ •••
+0_0+
||
<---------->
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P=NP Man, here to serve you____:
William Elliot wrote:
On Sun, 7 Jun 2009, David C. Ullrich wrote:[Continuing Musatov Proof Claim Statement.... Initializing....
<marsh@xxxxxxxxxxxxxxxx> wrote:Whoops. Before making that definition, isn't it necessary to show
{ x1,.. x_n } is a partition of [0,1]
when x0 = 0, x_n = 1, for j = 0,.. n-1, xj < x_(j+1)
The length of a path p, length(p) = sup
{ sum_(j=0,.. n-1) d(p(xj), p(x_(j+1))) | {x1,.. x_n} partition [0,1] }
the sum's are bounded? Wouldn't that be akin to showing a
continuous function on a compact set has bounded variation?
Ok.Because of continuity of p, can this be simplified to
length(p) = sup{ sum_(j=0,.. n-1) d(p(j/n), p((j+1)/n)) | n in N } ?
Yes. One inequality between the two sups is trivial since the uniform
partition is a partition. For the other inequality:
Let epsilon > 0. Given a partition x_0, ... x_n, it's enough to showAlright.
that there exists m with
sum_(k=0,.. m-1) d(p(k/m), p((k+1)/m))
>= -epsilon + sum_(j=0,.. n-1) d(p(xj), p(x_(j+1))).
Since p is uniformly continuous there exists delta > 0 such that
Because it's domain is compact.
if |s-t| < delta then d(p(s), p(t)) < epsilon/(4n) (or epsilon/(25n),Noted.
or whatever it turns out you need when you write down the
details). Choose m so that 1/m < delta.
Let P be a common refinement of the two partitions x_0, ... x_nAgreed. I dare you to show that in full detail. ;-)
and 0, 1/m, 2/m, ... 1 (ie throw in all the endpoints from both
partitions). Since P is a refinement of x_0, ... x_n the
triangle inequality shows that
"sum for P" >= sum_(j=0,.. n-1) d(p(xj), p(x_(j+1))).
But since P consists of the points 0, 1/m, ... 1 with theI don't get it and don't see where it's going.
points x_0, ... x_n added, it's not hard to show, using
the choice of delta, that
|"sum for P" - sum_(j=0,.. n-1) d(p(xj), p(x_(j+1)))| < epsilon.
The details are overwhelming. Yet it's clear
you've set the stage for using uniform continuity.
1,2,3,4,5,6,7,8,9...]
Initialized...
..
NOTE: The below statements were captured using P=NP programming
techniques and are a modest equivalence of Internet "Eavesdropping"
through Google search/cache IP variable manipulation.
[©2009. Martin Musatov. All Rights Reserved.
Musatov wrote:]
A semiautomatic version is complete and ready for adjustments toward
full automation. In dealing with the current real world, selections
of options to expedite its functional use are under consideration.
The original design allowed for 1000 wheels in each system but I have
mentioned only 500 historic ones. To cut some constants, making 1000
still based on the two 125 versions of the M-394 seemed attractive.
That led quickly to reversing the 500 and adding them to the prior 500.
A simple registration to one character and bubble sorting gives a
convincing sett of 100 historic related wheels in the cylinder. M94X2
no longer describe it best, so M94X4 it is.
Function is easy:
After loading the basic cylinder and scrambling, if any, by the method
previously described, strings of up to 1000 characters can be
handled.
I'll use here, "The quick brown fox jumps over the lazy dog."
Formating gives me "thewquickwbrovvnwcovvwjumpswoverwthewlazywdogwxw/"
in another field.
Registering that text is shown in a list of the wheels in which the
registered character is shown first.
©tqkzolrxspwnabceigdjfvuymh ©hnlqtsmzkxwvryufigjdabeopc
©edcabswrxqpnzgyvuomltkjifh ©wsbacdehfijktlmouvygznpqxr
©qysorpmhzukxacgjidntebwvfl ©upomtkjfzihedacngrsyvlbwxq
©itegkumjfyqhacrlndpbovzsxw ©ctbdwzqpiuhljkxegsvfryomna
©kuwxfnmdipoestqraczlbghvjy
Harvesting the text gives me a list of 26 strings including the input.
Here are four of them:
thewquickwbrovvnwcovvwjum pswoverwthewlazywdogwxw*
qndsypttubkvcwqwsrcjuerrk iypvocjkxnfudjreuakkkno/
klcbsoebwlvwilsytfgkymkne qcytbbvcefhhbxtjelhrxia/
zqaaomgdxvmxharxzyepmzwdq xhfppogvougowmubjfgfpzm/
If I register one of them, Ct, and then harvest the strings, I look
for the flagged string:
mpfxfxxnjqwutrxpydjulsdxb rwzmsiyoidticuqdtwmyayv/
hchrlqwayxqyntptqkhiwlold blsbzjkbkbzdgoxucjcclfe/
thewquickwbrovvnwcovvwjum pswoverwthewlazywdogwxw*
qndsypttubkvcwqwsrcjuerrk iypvocjkxnfudjreuakkkno/
klcbsoebwlvwilsytfgkymkne qcytbbvcefhhbxtjelhrxia/
Of course, it unformats to:
the quick brown cow jumps over the lazy dog x
This all looks better in equal sized letters.
[*******©2009. Martin Musatov. All Rights Reserved.
Is the following correct for MD5? A1. collisions - broken (s. A2) A2.
second preimage can be found in a couple of seconds A3. first preimage
- no idea Is there a hash function (still unbroken and considered
secure) producing only 128 bits? Does the following hold for such a
hash? B1. collisions can be found in O(2**64) B2. second preimage can
be found in O(2**128) B3. first preimage can be found in O(2**128)
Re: newbie Q: "Stacked" public key algorithm From Steffen =?UTF-8?B?
S8O2aGxlcg==?= Send on 2009-05-16 06:28:49.0 Hi all
thanks for your help- I'll take Skybucks & other proposal and let the
company distribute personalized copies - that seems to be the best way
regards
Steffen
Skybuck Flying wrote:
"Steffen K?hler" news:gug98p$5k0$02$1@xxxxxxxxxxxxxxxxxxxx
I've also thought about something like a watermark, but the problem is
that
the original message would be send only once to the company itself and
then
distributed internally, so the sender himself can't neither add a
personal watermark nor use e.g. the public keys of the employees itself
for encryption, as he just don't know them at that moment.
What does that have to do with it,
It's really simple:
1. Employee sends encrypted message to company.
2. Company decrypts it.
3. Company makes a copy for everyone else and adds a watermark for each
copy.
4. Company re-encrypts it and sends it to everybody.
Bye,
Skybuck.
top Nokia Sanomalaite M/90 encryption From Safari Send on 2009-05-16
07:56:37.0 So, what algorithm does this messaging device use?
I have been told "it still has not been broken"...
--
top Access Wikipedia Through Gmail Encryption From Martin Musatov
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top Streambuddy From jonas.thornvall@xxxxxxxxxxx Send on 2009-05-16
14:34:06.0 It seems like the discussion took a fast end when we
started talking
about keydependent offsets that is used to create the
initialkeytables, howcome? Did the problem grow out of your cipher
context, is it yet again time for updates.
Start with this.
1. Is it possible to analyse a cipher when you do not know the
keysize.
2. Is it possible to analyse a cipher when you do not know the
blocksize.
3. Is it possible to analyse the pseudorandom stream/ciphertext to
find the blocksize.
4 Is it possible to analyse the pseudorandom stream/ciphertext to find
the keysize.
5 Is it possible to analyse the pseudorandom stream to find out the
keytable offset.
6. Is it possible to analyse a pseudorandom permutation/shuffle when
you do not know the entropy it could be 4096 bites or more.
8. Is it possible use analyse to reverseengineer the pseudorandom
stream, and *FIND* the two permutationstreams and the savestate table
"internals states" for the permutation block.
*EVEN IF YOU KNEW THE BLOCKSIZE*
The 2 permutation tables updated each round with savestate
SS1=P1^P2^SS1, for every block of pseudorandom stream
9. Is it possible to achieve to make a faststreamcipher in working in
blockmode to achieve over 100MB/sec *absolutely* [YES]
10. Does 1-8 turn cryptoanalyse to mere studying of randomness,
*counting boogers usually do not tell you anything about their origin
you want to study the process and it is hidden* so [YES]
I answered 9 and 10 to help you out let me tell you a little about
cryptography.
How many ciphers is there that can use free keylength (bytes)and free
blocksize (bytes) my guess there is only streambuddy.
I reduced the *NEED FOR ANALYSE TO ZERO* there is nothing to analyse,
the obscurity of cipher behaviour is 100%
Just like Shannon would have liked. It use the full 100% entropy of
any chosen keylength with zero offset collsions.
It have 100 percent POTP collisions you have no idea what could be in
the SS1 buffer and still renders same blank stream (depending on the
permutation tables).
It adapt to any blocksize you chose, you can use ***keydependent
blocksizes and offsets for permuation tables***
I added computative complexity for analyse to the level that the shear
idea of analyse of the pseudorandom output become ridiculous.
I made cryptoanalyse nonsensical from now, and my guess it have been
for along time.
It just been dreaming and intellectual fraud beleiving the crypto/math
elitecould solve ciphers constructed by even firsttimers, you just
choose a high entropy key or computation exhaustive process.
The mathelite presented ciphers/mathproblems they knew actually to be
solvable in a reasonable time ***FOR THEM***
When limiting the techniques of ciphers to always be in the reach of
current math and technology you always keep an interest in the latest
development of the field and you keep interest in the field by adding
bananas. You get a bigger pile of bananas but i assure you they still
plain bananas easily broken down by the analysts.T
IN TRUTH THERE NEVER WERE ANY NEED FOR DEVELOPMENT OF NEW CIPHERS,
PROBABLY THERE BEEN FEASIBLE AND UNBREAKABLE CIPHERS SINCE DAY ONE IN
THE COMPUTER AGE, THE IDEA PUTTING THEM IN HAND OF CIVILIANS NEVER
CATCHED MUCH INTEREST THOUGH BETTER KEEP THE STATUS QUO OF PERFECT
SECRECY TO THE ELITE.
The goverments military organisations had interest do dumbplay the
cryptofield, presenting it to the academic community as a mathgame
that should be possible to analyse, doing so they put restriction on
how obscure a crypto could be when presented.
-A cipher should have fix keylength
-A cipher should have fix blocklength
-A cipher should have fast keyschedule "THERE NEVER WAS NEED FOR
KEYRESHEDULING TO START WITH"
-A cipher should use the simplest and most analysed math constructs
when constructing the pseduorandom output or ciphertext "SO IT WOULD
BE POSSIBLE TO BREAK"
Well a cipher who do not need keylength nor blocklength is a pile of
mashed bananas to analyse
you can not count the bananas you have to weigh them, and suddenly you
have only statistics.
But now a guy construct this fantastic blender. *OOOPS* crypto analyse
as field is gone.
If you do not beleive that they want ciphers to be breakable why do
you think the first ciphers they presented to you had 40 and 64 bits.
It was to be reachable using topnotch technology not presented to us.
Now they say do not use more then 256 bit keys it is overkill, it is
because their limit of solvable right now is 256 bit. When Joe can
crack a 256 bit cipher they move the limit just so they self can be in
reach of cracking it.
They give us what they can solve they would never put anything in
Joe's hand they not self can solve, it would just be plain stupid. And
maybe i agree they doing the right thing, but the right thing is not
necessarily the truth about the cryptographic field.
Right now the academic community act like crossword makers and
solvers.
They chose to use mathematicly ordered shuffles to obscure,when they
know there is pseudo random ones available that can only be brute
forced it is *FOUL PLAY*.
If the truth about cryptology that for sure Shanon were aware about
should be released. The cryptologic field would turn upside down likea
dead duck in water, and render neither money or interest in academic
or elsewhere.
THERE IS ALREADY UNBREAKABLE CIPHERS WITH 40-bit KEYS AND LONG
KEYSETUP
Cryptanalyst professionals should all turn to simple programmers
implementing communicationprotocols between Bob and Alice.
And render as much interest as fortunetelling
If Shanons ideas of howto build ciphers gained interest in the
academic crypt community would take ground, the field should suddenly
have no use of itself.
They could all as well use tarot, or why not try use a little ESP,
that at least could at least be abit refreshing.
JT
top Re: Streambuddy From 1PW Send on 2009-05-16 16:10:39.0 Would you
mind terribly if we introduced you to someone with the
initials of JSH?
--
1PW @?6A62?FEH9:DE=6o2@=]4@> [r4o7t]
top RSA Private Key representation From Giuliano Bertoletti Send on
2009-05-16 17:00:27.0
Hello,
I was wondering if it's mathematically easy to go back and forth from
the canonical RSA private key representation (modulus + private
exponent) to the CRT form (PRIME_1, PRIME_2, COEFFICIENT, EXP_1,
EXP_2).
Regards,
Giuliano Bertoletti.
top Re: RSA Private Key representation From Kristian Gj?steen Send
on 2009-05-16 17:49:32.0 Giuliano Bertoletti wrote:
I was wondering if it's mathematically easy to go back and forth from
the canonical RSA private key representation (modulus + private
exponent) to the CRT form (PRIME_1, PRIME_2, COEFFICIENT, EXP_1, EXP_2).
Use the secret exponent to factor the modulus.
Divide by 2 until the secrect exponent is odd, raise a random number
to that power, then square it until you get 1. If you got -1 before
you
got 1, try again. Otherwise, subtract 1 from what you got before 1
and
compute the gcd with the modulus.
Or: Write d = 2^t s, s - odd. Choose random a, find smallest i s.t.
a^(2^i s) = 1 (mod n). If i=0, or a^(2^(i-1) s) = -1 (mod n), try
again.
Otherwise compute gcd(n, a^(2^(i-1) s) - 1), which will be a proper
factor of n.
--
Kristian Gj?steen
top Re: Streambuddy From rossum Send on 2009-05-16 18:30:45.0 On
Sat, 16 May 2009 05:34:06 -0700 (PDT), jonas.thornvall@xxxxxxxxxxx
wrote:
THERE IS ALREADY UNBREAKABLE CIPHERS WITH 40-bit KEYS AND LONGOnly if your restrict yourself to messages of less than the unicity
KEYSETUP
distance. For longer messages the cypher is breakable by brute force.
Currently 40 bits is within reach.
rossum
top Re: RSA Private Key representation From "Scott Fluhrer" Send on
2009-05-16 18:55:18.0
"Kristian Gj?steen" wrote in message
news:gumnac$89m$1@xxxxxxxxxxxxxxxxxxxxx
Giuliano Bertoletti wrote:Minor correction: multiply the public and secret exponent first. Then
I was wondering if it's mathematically easy to go back and forth from
the canonical RSA private key representation (modulus + private
exponent) to the CRT form (PRIME_1, PRIME_2, COEFFICIENT, EXP_1, EXP_2).
Use the secret exponent to factor the modulus.
Divide by 2 until the secrect exponent is odd,
do
this to the product...
raise a random number
to that power, then square it until you get 1. If you got -1 before you
got 1, try again. Otherwise, subtract 1 from what you got before 1 and
compute the gcd with the modulus.
Or: Write d = 2^t s, s - odd. Choose random a, find smallest i s.t.
a^(2^i s) = 1 (mod n). If i=0, or a^(2^(i-1) s) = -1 (mod n), try again.
Otherwise compute gcd(n, a^(2^(i-1) s) - 1), which will be a proper
factor of n.
--
poncho
top Re: RSA Private Key representation From Mike Amling Send on
2009-05-16 22:23:32.0 Kristian Gj?steen wrote:
Giuliano Bertoletti wrote:
I was wondering if it's mathematically easy to go back and forth from
the canonical RSA private key representation (modulus + private
exponent) to the CRT form (PRIME_1, PRIME_2, COEFFICIENT, EXP_1, EXP_2).
Use the secret exponent to factor the modulus.
Divide by 2 until the secrect exponent is odd, raise a random number
to that power, then square it until you get 1.
I assume you mean 1 modulo the public modulus.
If you got -1 before you
got 1, try again. Otherwise, subtract 1 from what you got before 1 and
compute the gcd with the modulus.
--Mike Amlinge
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©2009. Martin Musatov. All Rights Reserved.]
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