Author Topic: How cryptography shows that quantum mechanics is incomplete?  (Read 2279 times)

Re: How cryptography shows that quantum mechanics is incomplete?
« Reply #15 on: May 30, 2019, 04:55:20 PM »
Soon we'll have quantum cryptography, which will be inherently unbreakable.
God Not Found
"There is a sucker born-again every minute." - C. Spellman

Offline Baruch

Re: How cryptography shows that quantum mechanics is incomplete?
« Reply #16 on: May 31, 2019, 12:09:29 AM »
Soon we'll have quantum cryptography, which will be inherently unbreakable.


Correct.  The government will have it, you and I will not.  But there probably aren't to many organizations, other than national ones, who need that kind of protection.  Corporations and individuals can deal with breakable encryption, just like the old wiretap days of rotary phones.
Ha’át’íísh baa naniná?
Azee’ ła’ish nanídį́į́h?
Táadoo ánít’iní.
What are you doing?
Are you taking any medications?
Don't do that.

Offline Cavebear

Re: How cryptography shows that quantum mechanics is incomplete?
« Reply #17 on: May 31, 2019, 11:36:42 AM »

Correct.  The government will have it, you and I will not.  But there probably aren't to many organizations, other than national ones, who need that kind of protection.  Corporations and individuals can deal with breakable encryption, just like the old wiretap days of rotary phones.

Any "unbreakable" password will be broken.  Ita quod umquam fuerit non habet.
Atheist born, atheist bred.  And when I die, atheist dead!

Offline Baruch

Re: How cryptography shows that quantum mechanics is incomplete?
« Reply #18 on: May 31, 2019, 10:00:07 PM »
Any "unbreakable" password will be broken.  Ita quod umquam fuerit non habet.

Not if, but when.  Decoding Enigma from 1940, in 1960 is too late. It only has to be sufficiently hard to break.  In self education, I have come to understand this principle (known by many others) in my own experience.

Please reverse engineer  my challenge or ...
Ha’át’íísh baa naniná?
Azee’ ła’ish nanídį́į́h?
Táadoo ánít’iní.
What are you doing?
Are you taking any medications?
Don't do that.

Re: How cryptography shows that quantum mechanics is incomplete?
« Reply #19 on: June 01, 2019, 11:29:57 AM »
An idiot tried to rob a bank after putting lemon juice all over his face, thinking that because it could be used to make invisible ink it would also make his face invisible!

Why a bank robber thought covering himself in lemon juice would help him get away with it


:-P :-D

Is this a joke ?








Voicjit - yes, that is a string run thru "my" system.  Gets harder the longer the message, usually it gets easier to break with length, because you get better statistics.  But statistics won't help much.  I had a profession hacker try to break it with his best tools (I gave him only the crypt text, nothing else) and he failed ;-)

Basically decryption is looking for correlation patterns.  If you work against that in particular, you can make a system as hard as you like (but that isn't enough to make it commercially viable).  Basically you first run statistics on individual characters, then on doubles, then on triples until you get to the total length of the message.  So the first set of stats have N values equal to the total length, and the final stat has just one value.  That is looking at letters just sequentially (there are other choices, like every other letter).

THE0RAIN0IN0SPAIN0FALLS0MAINLY0ON0THE0PLAIN
L5CRAGQIPVPVW9MTD827I6Q1ROVO0C8K51XJ416JXN0

What can I say about your challenge ? It's important to know what we already know before do a cryptanalysis work because what we know will help us.

1.We know the plaintext (In blue) and the cipher text (In grey).
2.Each of these have 43 characters.
3.We don't know the cryptosystem used but its author say us it's a symmetric block cipher.
4.We don't know the number of blocks used.
5.We don't know the size of a block.
6.We don't know the mode of operation used (CBC , CFB , CCM , CTR , CTS , CWC , EAX , ECB , GCM , IACBC , IAPM , KFB , LRW , OCB , OFB , PCBC , SGCM , TCB , XCBC , XEX , XTS etc...).
7.We don't know the initialization vector.
8.Does the initialization vector use a cryptographic nonce ?
9.Does the cryptosystem use a key schedule ?
10.How deep is avalanche effect ? (Refer to confusion and diffusion)
11.Does it use a keystream ?
12.Does it use a whitening transformation ?
13.Does it use a one way-compression function ?
14.Does it use a one way function ?
15.Does it use a P-Box ?
16.Does it use a S-Box ?
17.How product cipher is made if there are one ?
18.What is architecture used (Feistel network , Lai–Massey scheme , Substitution-permutation network etc...) ?
19.We don't know the size of the key but we can suppose it does only use Latin characters or / and numbers. If it does use Latin characters we can suppose there are no case sensitivity.
20.Does this cipher have bijective , injective or subjective functions ?
21.How many rounds used ?

If we find the answer to one or more of these questions we can resolve the problem more easily.



Frequency analysis of the plaintext (Unuseful for cryptanalysis but can help to understand more about how the cipher operate)  :
There are 8 numbers and this is still the digit 0.
There are 35 letters but many of these are repetitions.
T = Appear 1 time as the first character and first letter , Appear a second time as the thirtieth-five character and twentieth-seven letter
H = Appear 1 time as the second character and second letter , Appear a second time as the thirtieth-six character and twentieth-eight letter
E = Appear 1 time as the third character and third letter , Appear a second time as the thirtieth-six  character and twentieth-nine letter
0 = Appear 8 times (Fourth character , Ninth character , Twelfth character , Eighteenth character , twentieth-four character , thirtieth-one character , thirteenth-four character , thirteenth-eight character)
R = Appear 1 time as the fifth character and fourth letter
A = Appear 1 time as the sixth character and fifth letter , Appear a second time as the fifteenth character and twelfth letter , Appear a third time as the twentieth character and sixteenth letter , Appear a fourth time as the twentieth-six character and twentieth letter , Appear a fifth time as the fortieth-one character and fortieth-two letter
I = Appear 1 time as the seventh character and sixth letter , Appear a second time as the tenth character and eight letter , Appear a third time as the sixteenth character and thirteenth letter , Appear a fourth time as the twentieth-seven character and twentieth-one letter , Appear a fifth time as the fortieth-two character and fortieth-three letter
N = Appear 1 time as the eighth character and seventh letter , Appear a second time as the eleventh character and ninth letter , Appear a third time as the seventeenth character and fourteenth letter , Appear a fourth time as the twentieth-eight character and twentieth-two letter , Appear a fifth time as the thirtieth-three character and twentieth-six letter , Appear a sixth time as the fortieth-three character and fortieth-four letter
S = Appear 1 time as the thirteenth character and tenth letter , Appear a second time as the twentieth-three character and nineteenth letter
P = Appear 1 time as the fourteenth character and eleventh letter , Appear a second time as the thirtieth-nine character and fortieth letter
F = Appear 1 time as the nineteenth character and fifteenth letter
L =  Appear 1 time as the twentieth-one character and seventeenth letter , Appear a second time as the twentieth-two character and eighteenth letter , Appear a third time as the twentieth-nine character and twentieth-three , Appear a fourth time as the fortieth character and fortieth one letter
M = Appear 1 time as the twentieth-five character and nineteenth letter
Y = Appear 1 time as the thirtieth character and twentieth-four letter
O = Appear 1 time as the thirtieth-two character and twentieth-five letter

15 characters are used in the plaintext (1 digit and 14 letters) and the characters that appear more often are the next : 0NAILTHESPRFMYO



Frequency analysis of the ciphertext (Unuseful for cryptanalysis but can help to understand more about how the cipher operate)  :
There are 9 numbers but many of these are repetitions.
There are 34 letters but many of these are repetitions.

L = Appear 1 time as the first character and first letter
5 = Appear 1 time as the second character and first digit , Appear a second time as the thirtieth-three character and tenth digit
C = Appear 1 time as the third character and second letter , Appear a second time as the thirtieth character and twentieth-two letter
R = Appear 1 time as the fourth character and third letter , Appear a second time as the twentieth-five character and eighteen letter
A =  Appear 1 time as the fifth character and fourth letter
G = Appear 1 time as the sixth character and fifth letter
Q = Appear 1 time as the seventh character and sixth letter , Appear a second time as the twentieth-three character and seventeenth letter
I = Appear 1 time as the eight character and seventh letter , Appear a second time as the twentieth-one character and sixteenth letter
P = Appear 1 time as the ninth character and eight letter , Appear a second time as the eleventh character and tenth letter
V = Appear 1 time as the tenth character and ninth letter , Appear a second time as the twelfth character and eleventh letter , Appear a third time as the twentieth-seven character and twentieth letter , Appear a fourth time as the twentieth-seven character and twentieth letter
W = Appear 1 time as the thirteenth character and twelfth letter
9 = Appear 1 time as the fourteenth character and second digit
M = Appear 1 time as the fifteenth character and thirteenth letter
T = Appear 1 time as the sixteenth character and fourteenth letter
D = Appear 1 time as the seventeenth character and fifteenth letter
8 = Appear 1 time as the eighteenth character and third digit , Appear a second time as the thirtieth-one character and ninth digit
2 = Appear 1 time as the nineteenth character and fourth digit
7 = Appear 1 time as the twentieth character and fifth digit
6 = Appear 1 time as the twentieth-two character and sixth digit , Appear a second time as the thirtieth-nine character and fourteenth digit
1 = Appear 1 time as the twentieth-four character and seventh digit , Appear a second time as the thirtieth-four character and eleventh digit , Appear a third time as the thirtieth-eight character and thirteenth digit
O = Appear 1 time as the twentieth-six character and nineteenth letter , Appear a second time as the twentieth-eight character and twentieth-one letter
0 = Appear 1 time as the twentieth-nine character and eight digit , Appear a second time as the fortieth-three character and fifteenth digit
K = Appear 1 time as the thirtieth-two character and twentieth-three letter
X = Appear 1 time as the thirtieth-five character and twentieth-four letter , Appear a second time as the fortieth-one character and twentieth-six letter
J = Appear a second time as the thirtieth-six character and twentieth-five letter , Appear a second time as the fortieth character and twentieth-five letter
4 = Appear 1 time as the thirtieth-seven character and twelfth digit
N = Appear 1 time as the fortieth-two character and twentieth-seven letter

27 characters are used in the ciphertext (9 digits and 18 letters) and the characters that appear more often are the next : V15CRQIP86O0XJLAGW9MTD27K4N

Who can say if my frequency analysis of the plaintext and ciphertext is accurate ?

Offline Baruch

Re: How cryptography shows that quantum mechanics is incomplete?
« Reply #20 on: June 01, 2019, 06:50:02 PM »
It would appear we have a professional in our midst!!

There are exactly 36 possible characters, 0-9 and A-Z

The arithmetic is modular (of course).  Aside from the key itself, it is mod36 (of course).

The key (almost pure random) is not mod36.  That would create unnecessary "depth" via aliasing.

The key is muxed with the message, by base36 addition, but dynamically, not statically.

There is in addition, when converting character back and forth with base36 … a constant re-lettering vector (sticker per Enigma) for all 36 characters.

The initial offset is simply the effect of adding the first key character to the first message character, there is no other offset.
Ha’át’íísh baa naniná?
Azee’ ła’ish nanídį́į́h?
Táadoo ánít’iní.
What are you doing?
Are you taking any medications?
Don't do that.

Re: How cryptography shows that quantum mechanics is incomplete?
« Reply #21 on: June 03, 2019, 01:40:19 PM »
It would appear we have a professional in our midst!!

There are exactly 36 possible characters, 0-9 and A-Z

The arithmetic is modular (of course).  Aside from the key itself, it is mod36 (of course).

The key (almost pure random) is not mod36.  That would create unnecessary "depth" via aliasing.

The key is muxed with the message, by base36 addition, but dynamically, not statically.

There is in addition, when converting character back and forth with base36 … a constant re-lettering vector (sticker per Enigma) for all 36 characters.

The initial offset is simply the effect of adding the first key character to the first message character, there is no other offset.

Why do you say I'm a professional ?
I know only a few things in crypto. I don't know the mathematical details in each concept.

Offline Baruch

Re: How cryptography shows that quantum mechanics is incomplete?
« Reply #22 on: June 03, 2019, 03:00:50 PM »
Why do you say I'm a professional ?
I know only a few things in crypto. I don't know the mathematical details in each concept.

Take it as praise.
Ha’át’íísh baa naniná?
Azee’ ła’ish nanídį́į́h?
Táadoo ánít’iní.
What are you doing?
Are you taking any medications?
Don't do that.