Re: new idea for random number generation
On 8/4/17, Randydandy <randydandy@randydandy.de> wrote:
Would not a coin flip with the exact same initial parameters (height, force, deterministic air currents, and striking surface) have the same result? [...] The problem with your algorithm is that there are no parameters that would be unknown. Only if you get randomness from another source as parameters which makes this pseudo random.
Okay, I should know this, but who came up with the idea of using random-number generators for ciphers anyway? Most XOR ciphers (which is unbreakable if, for example, you use a key as long as the text) use a key as a source, AFAIK. Anyone have a reference? Now, chaotic functions are known mathematically to be "topologically complete" (I believe that's the term I learned) -- that means they will cover every number in the range. An amazing feat: consider if you're using 2^63 (which will fit in ONE world) to express your real number in the logistic function. Your chance of landing on a already visited number is 9.22^308 -- beyond the scale of indexing every atom in the universe. So, in theory you have both an UNBREAKABLE and PRACTICAL cipher method: use the logistic function to make a pseudo-random number generator, think of unique words that you can transform into your seed values, and communicate those "humane" keys to your recipient. How can you possibly break this (apart from psychological tactics of guessing the words used to create the seed values)? Marxos
On 8/4/17, \0xDynamite <dreamingforward@gmail.com> wrote:
On 8/4/17, Randydandy <randydandy@randydandy.de> wrote:
Would not a coin flip with the exact same initial parameters (height, force, deterministic air currents, and striking surface) have the same result? [...] The problem with your algorithm is that there are no parameters that would be unknown. Only if you get randomness from another source as parameters which makes this pseudo random.
Okay, I should know this, but who came up with the idea of using random-number generators for ciphers anyway? Most XOR ciphers (which is unbreakable if, for example, you use a key as long as the text) use a key as a source, AFAIK. Anyone have a reference?
Sorry, I meant "use a relatively [short, simple] key as a source" since it is unwieldy to supply a long key. In theory, I would suppose a key that is log(base 26) + (length of impracticality of brute force keyspace search) of your text to be "good enough" to use. marxos
Now, chaotic functions are known mathematically to be "topologically complete" (I believe that's the term I learned) -- that means they will cover every number in the range. An amazing feat: consider if you're using 2^63 (which will fit in ONE world) to express your real number in the logistic function. Your chance of landing on a already visited number is 9.22^308 -- beyond the scale of indexing every atom in the universe.
So, in theory you have both an UNBREAKABLE and PRACTICAL cipher method: use the logistic function to make a pseudo-random number generator, think of unique words that you can transform into your seed values, and communicate those "humane" keys to your recipient.
How can you possibly break this (apart from psychological tactics of guessing the words used to create the seed values)?
Marxos
"Vernam cipher" Sent from Yahoo Mail on Android On Fri, Aug 4, 2017 at 14:17, \0xDynamite<dreamingforward@gmail.com> wrote: On 8/4/17, Randydandy <randydandy@randydandy.de> wrote:
Would not a coin flip with the exact same initial parameters (height, force, deterministic air currents, and striking surface) have the same result? [...] The problem with your algorithm is that there are no parameters that would be unknown. Only if you get randomness from another source as parameters which makes this pseudo random.
Okay, I should know this, but who came up with the idea of using random-number generators for ciphers anyway? Most XOR ciphers (which is unbreakable if, for example, you use a key as long as the text) use a key as a source, AFAIK. Anyone have a reference? Now, chaotic functions are known mathematically to be "topologically complete" (I believe that's the term I learned) -- that means they will cover every number in the range. An amazing feat: consider if you're using 2^63 (which will fit in ONE world) to express your real number in the logistic function. Your chance of landing on a already visited number is 9.22^308 -- beyond the scale of indexing every atom in the universe. So, in theory you have both an UNBREAKABLE and PRACTICAL cipher method: use the logistic function to make a pseudo-random number generator, think of unique words that you can transform into your seed values, and communicate those "humane" keys to your recipient. How can you possibly break this (apart from psychological tactics of guessing the words used to create the seed values)? Marxos
On 8/4/17, jim bell <jdb10987@yahoo.com> wrote:
"Vernam cipher"
YES! That is the cipher. I can't believe I forgot the name of it. Geez, I'm just re-realizing all of my undergrad training. AFAICT, it is UNBREAKABLE if the keysize is at least half the size of your plaintext. Marxos
On Sunday, September 24, 2017, 3:25:10 PM PDT, \0xDynamite <dreamingforward@gmail.com> wrote: On 8/4/17, jim bell <jdb10987@yahoo.com> wrote:
"Vernam cipher"
YES! That is the cipher. I can't believe I forgot the name of it. Geez, I'm just re-realizing all of my undergrad training.
AFAICT, it is UNBREAKABLE if the keysize is at least half the size of your plaintext.>Marxos
My understanding is that the keysize ought to be as long as the message to be encrypted. And yes, it is unbreakable... UNLESS you try to reuse the keys!!! Google-search the word "Venona" to learn more. Jim Bell
On 09/25/2017 02:38 AM, jim bell wrote:
On Sunday, September 24, 2017, 3:25:10 PM PDT, \0xDynamite <dreamingforward@gmail.com> wrote:
On 8/4/17, jim bell <jdb10987@yahoo.com <mailto:jdb10987@yahoo.com>> wrote:
"Vernam cipher"
YES! That is the cipher. I can't believe I forgot the name of it. Geez, I'm just re-realizing all of my undergrad training.
AFAICT, it is UNBREAKABLE if the keysize is at least half the size of your plaintext. Marxos
My understanding is that the keysize ought to be as long as the message to be encrypted.
And yes, it is unbreakable... UNLESS you try to reuse the keys!!! Google-search the word "Venona" to learn more.
Jim Bell
Very very easy to implement too, though getting quality key material is a bit of a hassle. I have considered filming rough water then taking the low bits and repacking them. I wonder if low bits from the sounds in a chicken coop would work too. --- Marin
On Monday, September 25, 2017, 7:48:23 AM PDT, Marina Brown <catskillmarina@gmail.com> wrote: On 09/25/2017 02:38 AM, jim bell wrote:
On Sunday, September 24, 2017, 3:25:10 PM PDT, \0xDynamite <dreamingforward@gmail.com> wrote:
On 8/4/17, jim bell <jdb10987@yahoo.com <mailto:jdb10987@yahoo.com>> wrote:
"Vernam cipher"
YES! That is the cipher. I can't believe I forgot the name of it. Geez, I'm just re-realizing all of my undergrad training.
AFAICT, it is UNBREAKABLE if the keysize is at least half the size of your plaintext. Marxos
My understanding is that the keysize ought to be as long as the message to be encrypted.
And yes, it is unbreakable... UNLESS you try to reuse the keys!!! Google-search the word "Venona" to learn more.
Jim Bell
Very very easy to implement too, though getting quality key material is a bit of a hassle.
75 years ago, that was true. Today, not. A year ago, I saw what looked like a good implementation of a random-number generator chip. This might have been it: http://www.fdk.com/cyber-e/pi_ic_rpg100.html × 250 kbps is plenty for many applications, especially since the generator could run continuously, with the output stored away and used as needed.
I have considered filming rough water then taking the low bits and repacking them.
In a continuously windy area, a camera aimed at a tree with leaves would work, too.
I wonder if low bits from the sounds in a chicken coop would work too.
"Cluck you!!" B^) Jim Bell
On Mon, Sep 25, 2017 at 10:49:10AM -0400, Marina Brown wrote:
On 09/25/2017 02:38 AM, jim bell wrote:
On Sunday, September 24, 2017, 3:25:10 PM PDT, \0xDynamite <dreamingforward@gmail.com> wrote:
On 8/4/17, jim bell <jdb10987@yahoo.com <mailto:jdb10987@yahoo.com>> wrote:
"Vernam cipher"
YES! That is the cipher. I can't believe I forgot the name of it. Geez, I'm just re-realizing all of my undergrad training.
AFAICT, it is UNBREAKABLE if the keysize is at least half the size of your plaintext. Marxos
My understanding is that the keysize ought to be as long as the message to be encrypted.
And yes, it is unbreakable... UNLESS you try to reuse the keys!!! Google-search the word "Venona" to learn more.
Jim Bell
Very very easy to implement too, though getting quality key material is a bit of a hassle.
I have considered filming rough water then taking the low bits and repacking them.
I wonder if low bits from the sounds in a chicken coop would work too.
Out here's a ways in the boonies, we grab dem chicken low bits in the morning', add milk, scramble up some brekky for the yunguns - best with cheese, tomaytoes, and onions too when we got 'em; Pa likes fresh CIA bacon whupass with 'is...
I guess the CIA's whupass has inserted itself into the cypherpunks email chain again. Zenaan On Tue, Sep 26, 2017 at 07:51:52AM +1000, Zenaan Harkness wrote:
On Mon, Sep 25, 2017 at 10:49:10AM -0400, Marina Brown wrote:
On 09/25/2017 02:38 AM, jim bell wrote:
On Sunday, September 24, 2017, 3:25:10 PM PDT, \0xDynamite <dreamingforward@gmail.com> wrote:
On 8/4/17, jim bell <jdb10987@yahoo.com <mailto:jdb10987@yahoo.com>> wrote:
"Vernam cipher"
YES! That is the cipher. I can't believe I forgot the name of it. Geez, I'm just re-realizing all of my undergrad training.
AFAICT, it is UNBREAKABLE if the keysize is at least half the size of your plaintext. Marxos
My understanding is that the keysize ought to be as long as the message to be encrypted.
And yes, it is unbreakable... UNLESS you try to reuse the keys!!! Google-search the word "Venona" to learn more.
Jim Bell
Very very easy to implement too, though getting quality key material is a bit of a hassle.
I have considered filming rough water then taking the low bits and repacking them.
I wonder if low bits from the sounds in a chicken coop would work too.
Out here's a ways in the boonies, we grab dem chicken low bits in the morning', add milk, scramble up some brekky for the yunguns - best with cheese, tomaytoes, and onions too when we got 'em; Pa likes fresh CIA bacon whupass with 'is...
There are plenty of good online references for the issues raised in this thread. I'll give two for each, a more-or-less standard reference & a Citizendium article that I wrote much of & that may be easier reading. Of course Wikipedia has both as well, and various crypto texts have deeper detail. Random numbers: http://www.ietf.org/rfc/rfc4086.txt http://en.citizendium.org/wiki/Random_number_generator One-time pads: http://www.ranum.com/security/computer_security/papers/otp-faq/ http://en.citizendium.org/wiki/One-time_pad
AFAICT, it is UNBREAKABLE if the keysize is at least half the size of your plaintext.
My understanding is that the keysize ought to be as long as the message to be encrypted.
Well if they don't know the key length, then using somewhere between 1/3 and 2/3 should make it unbreakable.
And yes, it is unbreakable... UNLESS you try to reuse the keys!!!
Hmm, but THEY don't know if you re-used them. So it would take how many ciphertexts messages to figure that out? Marxos
On Tuesday, September 26, 2017, 8:14:35 AM PDT, \0xDynamite <dreamingforward@gmail.com> wrote:
AFAICT, it is UNBREAKABLE if the keysize is at least half the size of your plaintext.
My understanding is that the keysize ought to be as long as the message to be encrypted.
Well if they don't know the key length, then using somewhere between 1/3 and 2/3 should make it unbreakable.
No, using the full length is necessary. Using a key length less than the full length of the message is very dangerous.
And yes, it is unbreakable... UNLESS you try to reuse the keys!!!
Hmm, but THEY don't know if you re-used them. So it would take how many ciphertexts messages to figure that out?
You presumably haven't Google-searched "Venona" yet. I believe it took a few thousand. See: https://en.wikipedia.org/wiki/Venona_project Jim Bell
From that article:
"Decryption[edit] This message traffic, which was encrypted with a one-time pad system, was stored and analyzed in relative secrecy by hundreds of cryptanalysts over a 40-year period starting in the early 1940s. Due to a serious blunder on the part of the Soviets, some of this traffic was vulnerable to cryptanalysis. The Soviet company that manufactured the one-time pads produced around 35,000 pages of duplicate key numbers, as a result of pressures brought about by the German advance on Moscow during World War II. The duplication—which undermines the security of a one-time system—was discovered and attempts to lessen its impact were made by sending the duplicates to widely separated users.[16] Despite this, the reuse was detected by cryptologists in the US."
Well if they don't know the key length, then using somewhere between 1/3 and 2/3 should make it unbreakable.
No, using the full length is necessary. Using a key length less than the full length of the message is very dangerous.
Hmm, as far as I'm aware, you can't do cryptanalysis on 2 characters. Marxos
On Wednesday, September 27, 2017, 7:24:22 AM PDT, \0xDynamite <dreamingforward@gmail.com> wrote:
Well if they don't know the key length, then using somewhere between 1/3 and 2/3 should make it unbreakable.
No, using the full length is necessary. Using a key length less than the full length of the message is very dangerous.
Hmm, as far as I'm aware, you can't do cryptanalysis on 2 characters. Marxos A factoid which is irrelevant in this context. If you only wanted to encrypt two characters, once, you could use ROT-13 and nobody would be the wiser. Jim Bell
Hmm, as far as I'm aware, you can't do cryptanalysis on 2 characters. Marxos A factoid which is irrelevant in this context. If you only wanted to encrypt two characters, once, you could use ROT-13 and nobody would be the wiser.
Not a factoid. I'm making the claim that you can't decrypt it. I don't want to do the proof, because I can deduce it, with Pretty Good Certainty (TM). If you have a key approx 1/2 the size of the text, AND you do NOT know the length of the key, you have essentially two characters to do cryptanalysis. Now the key may be a small enough dictionary to do more analysis, and since you know that each set of the same modulus is using the same key character, but without knowing the length of the key, it's going to be next to impossible. There are many false leads (key-text combinations which result in what seems like valid plaintext) and the search space becomes too large. Marxos
On Wednesday, September 27, 2017, 11:51:22 AM PDT, \0xDynamite <dreamingforward@gmail.com> wrote:
Hmm, as far as I'm aware, you can't do cryptanalysis on 2 characters. Marxos A factoid which is irrelevant in this context. If you only wanted to encrypt two characters, once, you could use ROT-13 and nobody would be the wiser.
ot a factoid. I'm making the claim that you can't decrypt it. I don't want to do the proof, because I can deduce it, with Pretty Good Certainty (TM). If you have a key approx 1/2 the size of the text, AND you do NOT know the length of the key, you have essentially two characters to do cryptanalysis. Now the key may be a small enough dictionary to do more analysis, and since you know that each set of the same modulus is using the same key character, but without knowing the length of the key, it's going to be next to impossible. There are many false leads (key-text combinations which result in what seems like valid plaintext) and the search space becomes too large.>Marxos
You've added the silly limiting condition that you are referring only to messages with two characters, which has virtually nothing to do with most messages sent by OTP. And, I observe if indeed you are only referring to two character messages, then the One Time Pad with half that number of characters, ONE character, amounts to the same thing as ROT-13, except where the number "13" is allowed to vary between 0 and 25. Highly UNinteresting. Jim Bell
Hmm, as far as I'm aware, you can't do cryptanalysis on 2 characters. Marxos A factoid which is irrelevant in this context. If you only wanted to encrypt two characters, once, you could use ROT-13 and nobody would be the wiser.
ot a factoid. I'm making the claim that you can't decrypt it. I don't want to do the proof, because I can deduce it, with Pretty Good Certainty (TM). If you have a key approx 1/2 the size of the text, AND you do NOT know the length of the key, you have essentially two characters to do cryptanalysis. Now the key may be a small enough dictionary to do more analysis, and since you know that each set of the same modulus is using the same key character, but without knowing the length of the key, it's going to be next to impossible. There are many false leads (key-text combinations which result in what seems like valid plaintext) and the search space becomes too large.>
You've added the silly limiting condition that you are referring only to messages with two characters, which has virtually nothing to do with most messages sent by OTP.
No, I'm saying if the key is about half the text -- THAT ALL YOU GET TO ANALYZE.
And, I observe if indeed you are only referring to two character messages, then the One Time Pad with half that number of characters, ONE character, amounts to the same thing as ROT-13, except where the number "13" is allowed to vary between 0 and 25.
Perhaps you didn't see my followup, but you still don't get it: there are many false positives: decrypts that lead to semi-valid texts WHICH YOU HAVE NO WAY TO DETERMINE VALIDITY (given no futher data). You don't know if the key is 1 char or 2 characters, so there is a complete set of valid decrypts which may or may not be the original and you have no way to determine that. marxos
On Sep 26, 2017, at 11:14 AM, \0xDynamite <dreamingforward@gmail.com> wrote:
AFAICT, it is UNBREAKABLE if the keysize is at least half the size of your plaintext.
My understanding is that the keysize ought to be as long as the message to be encrypted.
Well if they don't know the key length, then using somewhere between 1/3 and 2/3 should make it unbreakable.
It's not a one time pad, and not secure, if the key is smaller than the clear text.
And yes, it is unbreakable... UNLESS you try to reuse the keys!!!
Hmm, but THEY don't know if you re-used them. So it would take how many ciphertexts messages to figure that out?
Marxos
I've mentioned this before in regards to OTP, and I think Jim just mentioned it - https://en.m.wikipedia.org/wiki/Venona_project
On Sep 26, 2017, at 11:14 AM, \0xDynamite <dreamingforward@gmail.com> wrote:
I've mentioned this before in regards to OTP, and I think Jim just mentioned it -
The only commercially viable one-time pad is found in the feminine hygiene section of your local supermarket. perl -pe 's/^\s+//g' *.py
On Sep 26, 2017, at 10:29 PM, alan@clueserver.org wrote:
On Sep 26, 2017, at 11:14 AM, \0xDynamite <dreamingforward@gmail.com> wrote:
I've mentioned this before in regards to OTP, and I think Jim just mentioned it -
The only commercially viable one-time pad is found in the feminine hygiene section of your local supermarket.
Yeah, venona decrypts were from days when encrypts were made with pencil, paper, radio. But spy agencies fuck everything up. I don't imagine OTP are used by spooks at all anymore ... or maybe i'm wrong?
perl -pe 's/^\s+//g' *.py
Not a fan of pythons indentation-based syntax eh? ;)
On Wed, 27 Sep 2017 03:20:03 -0400 John Newman <jnn@synfin.org> wrote:
Yeah, venona decrypts were from days when encrypts were made with pencil, paper, radio. But spy agencies fuck everything up. I don't imagine OTP are used by spooks at all anymore ... or maybe i'm wrong?
I imagine they are used by anyobody with half a brain. Furthermore the use of such technique means that internet surveillance is pretty pointless when directed at any even half-valuable 'target'.
perl -pe 's/^\s+//g' *.py
Not a fan of pythons indentation-based syntax eh? ;)
participants (8)
-
\0xDynamite -
alan＠clueserver.org -
jim bell -
John Newman -
juan -
Marina Brown -
Sandy Harris -
Zenaan Harkness