Title: **How cryptography shows that quantum mechanics is incomplete?**

Post by:**Baruch** on **March 23, 2019, 05:34:11 AM**

Post by:

In quantum mechanics, information is neither created nor destroyed (barring black holes). See my argument after the film, on why the title is a rhetorical question.

https://www.youtube.com/watch?v=HF-9Dy6iB_4

In a cryptography situation, if you know the key, you can encrypt a message, and decrypt it too. The total process involves conservation of information. And QM supports this total model. But you really encrypted a message using a random key (in the sense that you don't know what it is) then you won't be able to decrypt the message, only encrypt it. If you look at the Shannon Information Entropy for a classical system, or the Von Neumann Information Entropy for a quantum system, you will see that the entropy (and thus the information) changes going from the plain text to the crypt text. The missing information is restored when decrypted by someone with the correct key. With asymmetric key exchange, you do have a situation where there is a public key and a private key. I encrypt my email with my private key (which subtly matches the public key), and using a third party key holder, you can use my public key to decrypt my email. Without knowing my private key. Again, a more sophisticated situation, but because the plain text is restored from the crypt text, information is conserved over the total process.

https://www.youtube.com/watch?v=HF-9Dy6iB_4

In a cryptography situation, if you know the key, you can encrypt a message, and decrypt it too. The total process involves conservation of information. And QM supports this total model. But you really encrypted a message using a random key (in the sense that you don't know what it is) then you won't be able to decrypt the message, only encrypt it. If you look at the Shannon Information Entropy for a classical system, or the Von Neumann Information Entropy for a quantum system, you will see that the entropy (and thus the information) changes going from the plain text to the crypt text. The missing information is restored when decrypted by someone with the correct key. With asymmetric key exchange, you do have a situation where there is a public key and a private key. I encrypt my email with my private key (which subtly matches the public key), and using a third party key holder, you can use my public key to decrypt my email. Without knowing my private key. Again, a more sophisticated situation, but because the plain text is restored from the crypt text, information is conserved over the total process.

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**SoldierofFortune** on **March 23, 2019, 03:59:40 PM**

Post by:

Firstly i should express that i havent got even a clue about about what you are talking about?

What do you mean by information? it is something like "the sun rises in the east" or "birds fly, fish swim"...

if the information you mean in the universe is this, the mission for us is to discover that

What do you mean by information? it is something like "the sun rises in the east" or "birds fly, fish swim"...

if the information you mean in the universe is this, the mission for us is to discover that

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**Baruch** on **March 23, 2019, 06:22:15 PM**

Post by:

Quote from: SoldierofFortune on March 23, 2019, 03:59:40 PM

Firstly i should express that i havent got even a clue about about what you are talking about?

What do you mean by information? it is something like "the sun rises in the east" or "birds fly, fish swim"...

if the information you mean in the universe is this, the mission for us is to discover that

Well, the OP is a rhetorical question ... turns out, within the limits outlined, cryptography is consistent with quantum mechanics being complete.

I will give you an elementary lesson ...

Shannon Entropy is for classical systems ...

https://www.youtube.com/watch?v=R4OlXb9aTvQ

This whole series (this is part 12) is the best intro to Information Theory ever. It really helped when I first studied this 4 years ago.

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**Cavebear** on **April 07, 2019, 04:24:48 PM**

Post by:

Quote from: Baruch on March 23, 2019, 05:34:11 AM

In quantum mechanics, information is neither created nor destroyed (barring black holes). See my argument after the film, on why the title is a rhetorical question.

In a cryptography situation, if you know the key, you can encrypt a message, and decrypt it too. The total process involves conservation of information. And QM supports this total model. But you really encrypted a message using a random key (in the sense that you don't know what it is) then you won't be able to decrypt the message, only encrypt it. If you look at the Shannon Information Entropy for a classical system, or the Von Neumann Information Entropy for a quantum system, you will see that the entropy (and thus the information) changes going from the plain text to the crypt text. The missing information is restored when decrypted by someone with the correct key. With asymmetric key exchange, you do have a situation where there is a public key and a private key. I encrypt my email with my private key (which subtly matches the public key), and using a third party key holder, you can use my public key to decrypt my email. Without knowing my private key. Again, a more sophisticated situation, but because the plain text is restored from the crypt text, information is conserved over the total process.

Did you just say that you are falsifying who you are using 3rd party encryption?

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**Baruch** on **April 07, 2019, 04:40:08 PM**

Post by:

Nope. That is a different discussion. Here is how PKI keys are used asymmetrically. You know a reliable third party key holder. You create a key pair, one private, the other public. You send your public key, along with your verified ID to the third party key holder. Other people do the same. So if I want to exchange encrypted mail with another party, I encrypt my outgoing meal with someone else's public key. The receiving party gets my public key from the third party key holder, as well. Once the receiving party gets my email, they have their private key all to themselves, and use it to decrypt my message, that was encrypted with their public key. The other party, sending encrypted mail to me, does the same thing symmetrically. They encrypt their email using my public key, and when I receive it, I use my private key, that only I have, to decrypt the message sent to me. Basically blind sharing of keys.

Unfortunately the majority of asymmetrical key exchange, is based on the factoring of dual prime numbers (an integer made up of just two large primes multiplied together). Quantum computing is uniquely able to factor dual prime numbers. Quantum computing is an analog/digital hybrid, useful for certain hard problems, but not all, and not practical for ordinary computing. Once you have factored the large prime, you can read an asymmetrical PKI (which might be 256 bits long).

Unfortunately the majority of asymmetrical key exchange, is based on the factoring of dual prime numbers (an integer made up of just two large primes multiplied together). Quantum computing is uniquely able to factor dual prime numbers. Quantum computing is an analog/digital hybrid, useful for certain hard problems, but not all, and not practical for ordinary computing. Once you have factored the large prime, you can read an asymmetrical PKI (which might be 256 bits long).

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**Cavebear** on **May 26, 2019, 08:25:55 AM**

Post by:

Quote from: Baruch on April 07, 2019, 04:40:08 PM

Nope. That is a different discussion. Here is how PKI keys are used asymmetrically. You know a reliable third party key holder. You create a key pair, one private, the other public. You send your public key, along with your verified ID to the third party key holder. Other people do the same. So if I want to exchange encrypted mail with another party, I encrypt my outgoing meal with someone else's public key. The receiving party gets my public key from the third party key holder, as well. Once the receiving party gets my email, they have their private key all to themselves, and use it to decrypt my message, that was encrypted with their public key. The other party, sending encrypted mail to me, does the same thing symmetrically. They encrypt their email using my public key, and when I receive it, I use my private key, that only I have, to decrypt the message sent to me. Basically blind sharing of keys.

Unfortunately the majority of asymmetrical key exchange, is based on the factoring of dual prime numbers (an integer made up of just two large primes multiplied together). Quantum computing is uniquely able to factor dual prime numbers. Quantum computing is an analog/digital hybrid, useful for certain hard problems, but not all, and not practical for ordinary computing. Once you have factored the large prime, you can read an asymmetrical PKI (which might be 256 bits long).

Thank you for mansplaining the dual prime numbers multiples. GAGobvious!

I'll wait for the practical application.

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**Baruch** on **May 26, 2019, 10:27:44 AM**

Post by:

Quote from: Cavebear on May 26, 2019, 08:25:55 AM

Thank you for mansplaining the dual prime numbers multiples. GAGobvious!

I'll wait for the practical application.

Supposedly the dual prime numbers technique (used to public key distribution) is vulnerable to quantum computers. Not everything is. But this is a hugely important target. It is yet to be seen that quantum computing can be used in a practical way to break typical public keys.

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**Cavebear** on **May 26, 2019, 11:43:57 PM**

Post by:

Quote from: Baruch on May 26, 2019, 10:27:44 AM

Supposedly the dual prime numbers technique (used to public key distribution) is vulnerable to quantum computers. Not everything is. But this is a hugely important target. It is yet to be seen that quantum computing can be used in a practical way to break typical public keys.

I agree that quantum computers could make passwords obsolete. I'll go mostly offline then. But there will be other ways to protect stuff, so maybe not.

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**Baruch** on **May 27, 2019, 12:06:53 AM**

Post by:

Quote from: Cavebear on May 26, 2019, 11:43:57 PM

I agree that quantum computers could make passwords obsolete. I'll go mostly offline then. But there will be other ways to protect stuff, so maybe not.

I hear lemon juice is a useful invisible ink ;-)

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**viocjit** on **May 27, 2019, 01:32:39 PM**

Post by:

It does exist solutions against cryptanalysis with a quantum computer.

Officially , it doesn't exist one with a sufficient calculation power to get a RSA private key with a size of 4096 bit**(Maximum size for the majority of RSA implementation)** from the public key.

Shor's algorithm can be theoretically used with a quantum computer**(With a sufficient calculation power)** to break RSA and others asymetric cipher based on integer factorization like Benaloh , Blumâ€"Goldwasser , DamgÃ¥rdâ€"Jurik , Goldwasserâ€"Micali , Naccacheâ€"Stern , Okamotoâ€"Uchiyama , Rabin , Schmidt-Samoa etc...

We speak often of RSA when we are speaking of public-key cryptography but we forget RSA isn't the one PK cipher in the world.

We have many solutions against quantum computers.

A list of solutions against quantum cryptanalysis : https://en.wikipedia.org/wiki/Post-Quantum_Cryptography_Standardization

One-time pad is an unpractical solution against quantum computers to resist to these but if we use it under some conditions**(Never use the same key , Key size is the same than message send , The generation of key must to be aleatory)** it can't be broke : https://en.wikipedia.org/wiki/One-time_pad

McEliece cryptosystem is immune against Shor's algorithm but I doubt it will still resist to quantum computers as I'm certain new algorithms will appear : https://en.wikipedia.org/wiki/McEliece_cryptosystem

Benaloh : https://en.wikipedia.org/wiki/Benaloh_cryptosystem

Blumâ€"Goldwasser : https://en.wikipedia.org/wiki/Blum%E2%80%93Goldwasser_cryptosystem

DamgÃ¥rdâ€"Jurik : https://en.wikipedia.org/wiki/Damg%C3%A5rd%E2%80%93Jurik_cryptosystem

Goldwasserâ€"Micali : https://en.wikipedia.org/wiki/Goldwasser%E2%80%93Micali_cryptosystem

Integer Factorization : https://en.wikipedia.org/wiki/Integer_factorization

Naccacheâ€"Stern : https://en.wikipedia.org/wiki/Naccache%E2%80%93Stern_cryptosystem

Post-quantum cryptography : https://en.wikipedia.org/wiki/Post-quantum_cryptography

Rabin : https://en.wikipedia.org/wiki/Rabin_cryptosystem

RSA : https://en.wikipedia.org/wiki/RSA_(cryptosystem)

Schmidt-Samoa : https://en.wikipedia.org/wiki/Schmidt-Samoa_cryptosystem

Shor's algorithm : https://en.wikipedia.org/wiki/Shor%27s_algorithm

Okamotoâ€"Uchiyama : https://en.wikipedia.org/wiki/Okamoto%E2%80%93Uchiyama_cryptosystem

Officially , it doesn't exist one with a sufficient calculation power to get a RSA private key with a size of 4096 bit

Shor's algorithm can be theoretically used with a quantum computer

We speak often of RSA when we are speaking of public-key cryptography but we forget RSA isn't the one PK cipher in the world.

We have many solutions against quantum computers.

A list of solutions against quantum cryptanalysis : https://en.wikipedia.org/wiki/Post-Quantum_Cryptography_Standardization

One-time pad is an unpractical solution against quantum computers to resist to these but if we use it under some conditions

McEliece cryptosystem is immune against Shor's algorithm but I doubt it will still resist to quantum computers as I'm certain new algorithms will appear : https://en.wikipedia.org/wiki/McEliece_cryptosystem

Benaloh : https://en.wikipedia.org/wiki/Benaloh_cryptosystem

Blumâ€"Goldwasser : https://en.wikipedia.org/wiki/Blum%E2%80%93Goldwasser_cryptosystem

DamgÃ¥rdâ€"Jurik : https://en.wikipedia.org/wiki/Damg%C3%A5rd%E2%80%93Jurik_cryptosystem

Goldwasserâ€"Micali : https://en.wikipedia.org/wiki/Goldwasser%E2%80%93Micali_cryptosystem

Integer Factorization : https://en.wikipedia.org/wiki/Integer_factorization

Naccacheâ€"Stern : https://en.wikipedia.org/wiki/Naccache%E2%80%93Stern_cryptosystem

Post-quantum cryptography : https://en.wikipedia.org/wiki/Post-quantum_cryptography

Rabin : https://en.wikipedia.org/wiki/Rabin_cryptosystem

RSA : https://en.wikipedia.org/wiki/RSA_(cryptosystem)

Schmidt-Samoa : https://en.wikipedia.org/wiki/Schmidt-Samoa_cryptosystem

Shor's algorithm : https://en.wikipedia.org/wiki/Shor%27s_algorithm

Okamotoâ€"Uchiyama : https://en.wikipedia.org/wiki/Okamoto%E2%80%93Uchiyama_cryptosystem

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**Baruch** on **May 27, 2019, 01:48:25 PM**

Post by:

Ah, someone who is more than a recreational mathematician ;-)

With circular functions, you have the truly periodic and the almost periodic. The truly periodic corresponds to a pure timbre, the simplest being a single pure tone (sine wave). If the ratio of at least one irregular component among the purely periodic ones, meet the number theory criteria of not being rational, not even algebraic eg square root of two, but is transcendental "pi/e/etc" ... then even after an infinite expansion of digits, the odd component will never align (harmonize). Such is the theory of almost-periodic functions, and how it becomes harder and harder to derive an accurate fourier analysis of the non-harmonics. Such is "percussive" sound, which can be partly harmonic (tympany) or nearly pure noise (true random number).

Consider not a true one time pad, but one that is arbitrarily close to a one time pad. Then the matter of being hard to decrypt is simply a parameter, that one can extend as necessary, as decryption gets better (an arms race). What if you have a key that isn't periodic (has a recurring depth), but is algebraic or even transcendental. In that case you can extend the key system indefinitely, without running into recurrence (which weakens the encryption). You can't as a practical matter use a true non-rational number, one approximates a non-rational number with a close enough rational number eg you use however many digits of "e" as necessary to meet the message size. The true key being incommensurable, while the practical key is a many digit rational approximation.

So what quantum computer can calculate the last digit of square root of two, or of "pi"? There is no free lunch on general problem solving, because of the "halting" problem. But specific classes of problems can be solved in closed form. We will have to wait to see, a practical demonstration that the multiplication of two large prime numbers is easy to factor or not. There are always surprises. To what extent is mathematics actually empirical not analytical?

With circular functions, you have the truly periodic and the almost periodic. The truly periodic corresponds to a pure timbre, the simplest being a single pure tone (sine wave). If the ratio of at least one irregular component among the purely periodic ones, meet the number theory criteria of not being rational, not even algebraic eg square root of two, but is transcendental "pi/e/etc" ... then even after an infinite expansion of digits, the odd component will never align (harmonize). Such is the theory of almost-periodic functions, and how it becomes harder and harder to derive an accurate fourier analysis of the non-harmonics. Such is "percussive" sound, which can be partly harmonic (tympany) or nearly pure noise (true random number).

Consider not a true one time pad, but one that is arbitrarily close to a one time pad. Then the matter of being hard to decrypt is simply a parameter, that one can extend as necessary, as decryption gets better (an arms race). What if you have a key that isn't periodic (has a recurring depth), but is algebraic or even transcendental. In that case you can extend the key system indefinitely, without running into recurrence (which weakens the encryption). You can't as a practical matter use a true non-rational number, one approximates a non-rational number with a close enough rational number eg you use however many digits of "e" as necessary to meet the message size. The true key being incommensurable, while the practical key is a many digit rational approximation.

So what quantum computer can calculate the last digit of square root of two, or of "pi"? There is no free lunch on general problem solving, because of the "halting" problem. But specific classes of problems can be solved in closed form. We will have to wait to see, a practical demonstration that the multiplication of two large prime numbers is easy to factor or not. There are always surprises. To what extent is mathematics actually empirical not analytical?

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**Baruch** on **May 27, 2019, 02:13:23 PM**

Post by:

A challenge ...

One can try to decrypt a known crypt message with a known method but unknown key (Bletchley Park), but how about when you have the plain text and crypt text but not the other two components ...

Please reverse engineer the following plain text plus crypt text for unknown method/key ...

THE0RAIN0IN0SPAIN0FALLS0MAINLY0ON0THE0PLAIN

L5CRAGQIPVPVW9MTD827I6Q1ROVO0C8K51XJ416JXN0

This is a symmetric key block encryption (single key, but encrypt method is inverse of decrypt method). Same number of characters in plain text and crypt text. Same symbol table used for both. One could start with a statistical analysis. However a simple symbol frequency analysis would fail, even with a longer message, because a given plain text symbol isn't converted to the same crypt text symbol, but varies with position in the chain ...

This is generated from a nearly periodic one time pad (one time in that it can be set to recompute via computer clock over whatever short period you need to make a new key. Enigma was changed once per day at midnight. The method can be shown statistically to be as good as AES with message lengths greater than 40 characters (shorter messages are easier to guess, which is to say effective depth is less). For example, a one character message, in a 36 symbol alphabet (as used in this example) can be correctly found with at most 36 guesses, and on average 18 guesses. With 40 character messages, given the nearly random key generation, you need worst case 36^40~1.8x10^62 guesses, or on average half that. Some claim, with favorable problems, a quantum computer at best can achieve the average on every try (which would be quite an accomplishment) but still insufficient to break my 40 character message.

One can try to decrypt a known crypt message with a known method but unknown key (Bletchley Park), but how about when you have the plain text and crypt text but not the other two components ...

Please reverse engineer the following plain text plus crypt text for unknown method/key ...

THE0RAIN0IN0SPAIN0FALLS0MAINLY0ON0THE0PLAIN

L5CRAGQIPVPVW9MTD827I6Q1ROVO0C8K51XJ416JXN0

This is a symmetric key block encryption (single key, but encrypt method is inverse of decrypt method). Same number of characters in plain text and crypt text. Same symbol table used for both. One could start with a statistical analysis. However a simple symbol frequency analysis would fail, even with a longer message, because a given plain text symbol isn't converted to the same crypt text symbol, but varies with position in the chain ...

This is generated from a nearly periodic one time pad (one time in that it can be set to recompute via computer clock over whatever short period you need to make a new key. Enigma was changed once per day at midnight. The method can be shown statistically to be as good as AES with message lengths greater than 40 characters (shorter messages are easier to guess, which is to say effective depth is less). For example, a one character message, in a 36 symbol alphabet (as used in this example) can be correctly found with at most 36 guesses, and on average 18 guesses. With 40 character messages, given the nearly random key generation, you need worst case 36^40~1.8x10^62 guesses, or on average half that. Some claim, with favorable problems, a quantum computer at best can achieve the average on every try (which would be quite an accomplishment) but still insufficient to break my 40 character message.

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**viocjit** on **May 30, 2019, 11:04:52 AM**

Post by:

Quote from: Baruch on May 27, 2019, 01:48:25 PM

Ah, someone who is more than a recreational mathematician ;-)

With circular functions, you have the truly periodic and the almost periodic. The truly periodic corresponds to a pure timbre, the simplest being a single pure tone (sine wave). If the ratio of at least one irregular component among the purely periodic ones, meet the number theory criteria of not being rational, not even algebraic eg square root of two, but is transcendental "pi/e/etc" ... then even after an infinite expansion of digits, the odd component will never align (harmonize). Such is the theory of almost-periodic functions, and how it becomes harder and harder to derive an accurate fourier analysis of the non-harmonics. Such is "percussive" sound, which can be partly harmonic (tympany) or nearly pure noise (true random number).

Consider not a true one time pad, but one that is arbitrarily close to a one time pad. Then the matter of being hard to decrypt is simply a parameter, that one can extend as necessary, as decryption gets better (an arms race). What if you have a key that isn't periodic (has a recurring depth), but is algebraic or even transcendental. In that case you can extend the key system indefinitely, without running into recurrence (which weakens the encryption). You can't as a practical matter use a true non-rational number, one approximates a non-rational number with a close enough rational number eg you use however many digits of "e" as necessary to meet the message size. The true key being incommensurable, while the practical key is a many digit rational approximation.

So what quantum computer can calculate the last digit of square root of two, or of "pi"? There is no free lunch on general problem solving, because of the "halting" problem. But specific classes of problems can be solved in closed form. We will have to wait to see, a practical demonstration that the multiplication of two large prime numbers is easy to factor or not. There are always surprises. To what extent is mathematics actually empirical not analytical?

I'm not a Mathematician. Not even a recreational one.

I think quantum computers will maybe be able a day to break a cryptosystem like this.

We will certainly discover new mathematical algorithms

But before ask if quantum computers will be able to do so a moment or another.

I think we must use a sufficient encryption process until the time the information we want to kept won't be necessary to be secret.

An info that must to be a secret indefinitely

What was said in this meeting can be fully secret , partially secret or not secret.

If this meeting is about the secret project of companies for the next five years. Information must to be protected for the next five years.

If this meeting is about public and secret projects of companies for the next ten years. Information about public project can be released fully or partially released and info about secret projects must to be protected for the next ten years. The same for non-released info on public projects.

If this meeting is about public projects for the next two years. Information can be released fully or partially and non-released info must to be protected for the next two years.

Even if a project is public it doesn't means all details are public because some of these need to be secret in some situations.

If a project is secret. The public must not known its existence and therefore the content should not be known to the public.

Few secrets need to be kept indefinitely.

An industrial process can need to be a forever a secret but the fact a video game is being created don't need to be kept indefinitely as the game will be released unless the project is abandoned.

If this ciphertext and plain text is really from you ? I can't find it over Internet. I can suppose you're the author of this.

THE0RAIN0IN0SPAIN0FALLS0MAINLY0ON0THE0PLAIN

L5CRAGQIPVPVW9MTD827I6Q1ROVO0C8K51XJ416JXN0

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**Unbeliever** on **May 30, 2019, 01:23:08 PM**

Post by:

Quote from: Baruch on May 27, 2019, 12:06:53 AM

I hear lemon juice is a useful invisible ink ;-)

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!

:-P :-D

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**Baruch** on **May 30, 2019, 04:46:42 PM**

Post by:

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).

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).

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**Unbeliever** on **May 30, 2019, 04:55:20 PM**

Post by:

Soon we'll have **quantum cryptography** (https://en.wikipedia.org/wiki/Quantum_cryptography), which will be inherently unbreakable.

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**Baruch** on **May 31, 2019, 12:09:29 AM**

Post by:

Quote from: Unbeliever on May 30, 2019, 04:55:20 PM

Soon we'll havequantum cryptography(https://en.wikipedia.org/wiki/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.

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**Cavebear** on **May 31, 2019, 11:36:42 AM**

Post by:

Quote from: Baruch on May 31, 2019, 12:09:29 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.

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**Baruch** on **May 31, 2019, 10:00:07 PM**

Post by:

Quote from: Cavebear on May 31, 2019, 11:36:42 AM

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 ...

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**viocjit** on **June 01, 2019, 11:29:57 AM**

Post by:

Quote from: Unbeliever on May 30, 2019, 01:23:08 PM

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(https://qz.com/986221/what-know-it-alls-dont-know-or-the-illusion-of-competence/)

:-P :-D

Is this a joke ?

Quote from: Baruch on May 30, 2019, 04:46:42 PM

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

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

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 ?

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

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

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

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

Frequency analysis of the ciphertext

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

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

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**Baruch** on **June 01, 2019, 06:50:02 PM**

Post by:

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 â€¦ 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.

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 â€¦ 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.

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**viocjit** on **June 03, 2019, 01:40:19 PM**

Post by:

Quote from: Baruch 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 â€¦ 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.

Title: **Re: How cryptography shows that quantum mechanics is incomplete?**

Post by:**Baruch** on **June 03, 2019, 03:00:50 PM**

Post by:

Quote from: viocjit on June 03, 2019, 01:40:19 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.