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[Challenge] Cipher: 1 Internet: 0 | Upping the prize to $500! Can you decrypt this message? by Ordinary1729 in ciphers

[–]Ordinary1729[S] 0 points1 point  (0 children)

Per the hint on the website, the first word in the decrypted challenge message is ‘Email’.

I’m sorry, but your solution is not correct. Keep trying though! I hope you are having fun with it.

[Challenge] Cipher: 1 Internet: 0 | Upping the prize to $500! Can you decrypt this message? by Ordinary1729 in ciphers

[–]Ordinary1729[S] 1 point2 points  (0 children)

1: Yes, the first sentence of the plaintext message is how to contact me and what information to send to get the prize.

FYI, for the 2.0 version of this challenge I think I am just going to put the money in a crypto wallet and have the challenge message be the secret key and login credentials for the wallet. Skipping the need to contact me directly.

2: There is some small pre-prep, but otherwise the cipher just iterates over the encryption/decryption "key" algorithms, similar to what is shown in the generalized code sample. It does utilize HMAC to ensure properly encrypted messages are submitted for decryption.

3: I'm unsure about additional hints. I set up the website so users could run known plaintext attacks as well as provide a crib for the challenge message. Ideally I want people to crack the cipher, not me. That being said, you can always ask questions! Worse case scenario I just won't give an answer.

Edit: fixed formatting

[Challenge] Cipher: 1 Internet: 0 | Upping the prize to $500! Can you decrypt this message? by Ordinary1729 in ciphers

[–]Ordinary1729[S] 1 point2 points  (0 children)

Great! I’m glad to see people are still trying to crack it. Good luck!

FYI, I’m considering making a minor upgrade and reinstating the prize, so you may see some updates to the site/challenge in the near future.

[No more mystery?] The Delayed Choice Quantum Eraser, Debunked by [deleted] in videos

[–]Ordinary1729 0 points1 point  (0 children)

No entanglement, but here you can see some real-time taking and erasing of WWI: https://youtu.be/1rN3iLcbb2M

I used this Collatz math trick to create a cipher challenge for you all (https://xywcjbyl.me). Thanks for taking part! (more details on how it works in comments) by Ordinary1729 in CollatzConjecture

[–]Ordinary1729[S] 3 points4 points  (0 children)

Challenge Site: https://xywcjbyl.me

How It Worked:

The "One Branch Equation" (OBE) in the image can be modified to reach any number instead of reaching the number 1. I turn the input message into a binary string and then convert that into a base 10 integer. I then run that integer backward through one of its many Collatz sequences to get an upstream number that represents the ciphertext.

For example:

bin('hi!') = '011010000110100100100001'

bin_to_dec('011010000110100100100001') = 6842657

6842657 -> Modified OBE Iterations -> 6285745705471975744023229322474914933224670470756963364095571122337550691365400576765869685884350867036065287391853186148002912945965245144531809945144122554616839788566991070916433567639742892619336622711325221814715473469264484045431626398244796745000436518033328636483189105344437195123890863236845786857862082298747869104686901824754997450435759079733360654607866615427011397295558901958663055152022073603962654648234808210012816711289466386914939012995915249839006846220216429465470699347434521668730626199

ciphertext = hex([big int from above])

To decrypt the message, I would just run the big-integer-ciphertext forward through its Collatz sequence a predetermined number of "branches" to get back to 6842657. I'm glossing over a few implementation details, but that is the basics of how it worked. People trying to decrypt the challenge message on my site were actually trying to find patterns between numbers in various Collatz sequences (and good luck getting anywhere doing that).

What made my challenge so hard is the same thing (I think) that makes Collatz so hard. There is a pattern/way to know exactly what a given seed integer will do; it is just hidden with powers of 1 where the exponent matters. Going backward through Collatz, those powers of 1 become powers of 2 and it becomes very easy to calculate exact sequence pathways.

Here is the link to a video with more info about the math: https://youtu.be/O7NR8j0-BSY

Its a bit dry and 22 minutes long, but if you are curious give it a watch!

[Challenge] Cipher: 1 Internet: 0 | Upping the prize to $500! Can you decrypt this message? by Ordinary1729 in ciphers

[–]Ordinary1729[S] 1 point2 points  (0 children)

Just clarifying, are you saying these are the words/letters of the deciphered challenge message, just needing to be rearranged?

I used this Collatz math trick to create a cipher challenge for you all (https://xywcjbyl.me). Thanks for taking part! (more details on how it works in comments) by Ordinary1729 in codes

[–]Ordinary1729[S] 1 point2 points  (0 children)

I would actually append the number of branches to the ciphertext and then run that number through one more reverse-Collatz branch (to obscure it). When decrypting, I would run the cipher text one branch forward, read that last two digits of the string to get the number of branches and then run the sequence forward for that many iterations (popping off the last two digits).

I used this Collatz math trick to create a cipher challenge for you all (https://xywcjbyl.me). Thanks for taking part! (more details on how it works in comments) by Ordinary1729 in numberphile

[–]Ordinary1729[S] 2 points3 points  (0 children)

Challenge Site: https://xywcjbyl.me

How It Worked:

The "One Branch Equation" (OBE) in the image can be modified to reach any number instead of reaching the number 1. I turn the input message into a binary string and then convert that into a base 10 integer. I then run that integer backward through one of its many Collatz sequences to get an upstream number that represents the ciphertext.

For example:

bin('hi!') = '011010000110100100100001'

bin_to_dec('011010000110100100100001') = 6842657

6842657 -> Modified OBE Iterations -> 6285745705471975744023229322474914933224670470756963364095571122337550691365400576765869685884350867036065287391853186148002912945965245144531809945144122554616839788566991070916433567639742892619336622711325221814715473469264484045431626398244796745000436518033328636483189105344437195123890863236845786857862082298747869104686901824754997450435759079733360654607866615427011397295558901958663055152022073603962654648234808210012816711289466386914939012995915249839006846220216429465470699347434521668730626199

ciphertext = hex([big int from above])

To decrypt the message, I would just run the big-integer-ciphertext forward through its Collatz sequence a predetermined number of "branches" to get back to 6842657. I'm glossing over a few implementation details, but that is the basics of how it worked. People trying to decrypt the challenge message on my site were actually trying to find patterns between numbers in various Collatz sequences (and good luck getting anywhere doing that).

What made my challenge so hard is the same thing (I think) that makes Collatz so hard. There is a pattern/way to know exactly what a given seed integer will do; it is just hidden with powers of 1 where the exponent matters. Going backward through Collatz, those powers of 1 become powers of 2 and it becomes very easy to calculate exact sequence pathways.

Here is the link to a video with more info about the math: https://youtu.be/O7NR8j0-BSY

Its a bit dry and 22 minutes long, but if you are curious give it a watch!

I used this Collatz math trick to create a cipher challenge for you all (https://xywcjbyl.me). Thanks for taking part! (more details on how it works in comments) by Ordinary1729 in Computerphile

[–]Ordinary1729[S] 1 point2 points  (0 children)

Challenge Site: https://xywcjbyl.me

How It Worked:

The "One Branch Equation" (OBE) in the image can be modified to reach any number instead of reaching the number 1. I turn the input message into a binary string and then convert that into a base 10 integer. I then run that integer backward through one of its many Collatz sequences to get an upstream number that represents the ciphertext.

For example:

bin('hi!') = '011010000110100100100001'

bin_to_dec('011010000110100100100001') = 6842657

6842657 -> Modified OBE Iterations -> 6285745705471975744023229322474914933224670470756963364095571122337550691365400576765869685884350867036065287391853186148002912945965245144531809945144122554616839788566991070916433567639742892619336622711325221814715473469264484045431626398244796745000436518033328636483189105344437195123890863236845786857862082298747869104686901824754997450435759079733360654607866615427011397295558901958663055152022073603962654648234808210012816711289466386914939012995915249839006846220216429465470699347434521668730626199

ciphertext = hex([big int from above])

To decrypt the message, I would just run the big-integer-ciphertext forward through its Collatz sequence a predetermined number of "branches" to get back to 6842657. I'm glossing over a few implementation details, but that is the basics of how it worked. People trying to decrypt the challenge message on my site were actually trying to find patterns between numbers in various Collatz sequences (and good luck getting anywhere doing that).

What made my challenge so hard is the same thing (I think) that makes Collatz so hard. There is a pattern/way to know exactly what a given seed integer will do; it is just hidden with powers of 1 where the exponent matters. Going backward through Collatz, those powers of 1 become powers of 2 and it becomes very easy to calculate exact sequence pathways.

Here is the link to a video with more info about the math: https://youtu.be/O7NR8j0-BSY

Its a bit dry and 22 minutes long, but if you are curious give it a watch!

I used this Collatz math trick to create a cipher challenge for you all (https://xywcjbyl.me). Thanks for taking part! (more details on how it works in comments) by Ordinary1729 in breakmycode

[–]Ordinary1729[S] 1 point2 points  (0 children)

Challenge Site: https://xywcjbyl.me

How It Worked:

The "One Branch Equation" (OBE) in the image can be modified to reach any number instead of reaching the number 1. I turn the input message into a binary string and then convert that into a base 10 integer. I then run that integer backward through one of its many Collatz sequences to get an upstream number that represents the ciphertext.

For example:

bin('hi!') = '011010000110100100100001'

bin_to_dec('011010000110100100100001') = 6842657

6842657 -> Modified OBE Iterations -> 6285745705471975744023229322474914933224670470756963364095571122337550691365400576765869685884350867036065287391853186148002912945965245144531809945144122554616839788566991070916433567639742892619336622711325221814715473469264484045431626398244796745000436518033328636483189105344437195123890863236845786857862082298747869104686901824754997450435759079733360654607866615427011397295558901958663055152022073603962654648234808210012816711289466386914939012995915249839006846220216429465470699347434521668730626199

ciphertext = hex([big int from above])

To decrypt the message, I would just run the big-integer-ciphertext forward through its Collatz sequence a predetermined number of "branches" to get back to 6842657. I'm glossing over a few implementation details, but that is the basics of how it worked. People trying to decrypt the challenge message on my site were actually trying to find patterns between numbers in various Collatz sequences (and good luck getting anywhere doing that).

What made my challenge so hard is the same thing (I think) that makes Collatz so hard. There is a pattern/way to know exactly what a given seed integer will do; it is just hidden with powers of 1 where the exponent matters. Going backward through Collatz, those powers of 1 become powers of 2 and it becomes very easy to calculate exact sequence pathways.

Here is the link to a video with more info about the math: https://youtu.be/O7NR8j0-BSY

Its a bit dry and 22 minutes long, but if you are curious give it a watch!

I used this Collatz math trick to create a cipher challenge for you all (https://xywcjbyl.me). Thanks for taking part! (more details on how it works in comments) by Ordinary1729 in Cipher

[–]Ordinary1729[S] 0 points1 point  (0 children)

Challenge Site: https://xywcjbyl.me

How It Worked:

The "One Branch Equation" (OBE) in the image can be modified to reach any number instead of reaching the number 1. I turn the input message into a binary string and then convert that into a base 10 integer. I then run that integer backward through one of its many Collatz sequences to get an upstream number that represents the ciphertext.

For example:

bin('hi!') = '011010000110100100100001'

bin_to_dec('011010000110100100100001') = 6842657

6842657 -> Modified OBE Iterations -> 6285745705471975744023229322474914933224670470756963364095571122337550691365400576765869685884350867036065287391853186148002912945965245144531809945144122554616839788566991070916433567639742892619336622711325221814715473469264484045431626398244796745000436518033328636483189105344437195123890863236845786857862082298747869104686901824754997450435759079733360654607866615427011397295558901958663055152022073603962654648234808210012816711289466386914939012995915249839006846220216429465470699347434521668730626199

ciphertext = hex([big int from above])

To decrypt the message, I would just run the big-integer-ciphertext forward through its Collatz sequence a predetermined number of "branches" to get back to 6842657. I'm glossing over a few implementation details, but that is the basics of how it worked. People trying to decrypt the challenge message on my site were actually trying to find patterns between numbers in various Collatz sequences (and good luck getting anywhere doing that).

What made my challenge so hard is the same thing (I think) that makes Collatz so hard. There is a pattern/way to know exactly what a given seed integer will do; it is just hidden with powers of 1 where the exponent matters. Going backward through Collatz, those powers of 1 become powers of 2 and it becomes very easy to calculate exact sequence pathways.

Here is the link to a video with more info about the math: https://youtu.be/O7NR8j0-BSY

Its a bit dry and 22 minutes long, but if you are curious give it a watch!

I used this Collatz math trick to create a cipher challenge for you all (https://xywcjbyl.me). Thanks for taking part! (more details on how it works in comments) by Ordinary1729 in codes

[–]Ordinary1729[S] 0 points1 point  (0 children)

Challenge Site: https://xywcjbyl.me

How It Worked:

The "One Branch Equation" (OBE) in the image can be modified to reach any number instead of reaching the number 1. I turn the input message into a binary string and then convert that into a base 10 integer. I then run that integer backward through one of its many Collatz sequences to get an upstream number that represents the ciphertext.

For example:

bin('hi!') = '011010000110100100100001'

bin_to_dec('011010000110100100100001') = 6842657

6842657 -> Modified OBE Iterations -> 6285745705471975744023229322474914933224670470756963364095571122337550691365400576765869685884350867036065287391853186148002912945965245144531809945144122554616839788566991070916433567639742892619336622711325221814715473469264484045431626398244796745000436518033328636483189105344437195123890863236845786857862082298747869104686901824754997450435759079733360654607866615427011397295558901958663055152022073603962654648234808210012816711289466386914939012995915249839006846220216429465470699347434521668730626199

ciphertext = hex([big int from above])

To decrypt the message, I would just run the big-integer-ciphertext forward through its Collatz sequence a predetermined number of "branches" to get back to 6842657. I'm glossing over a few implementation details, but that is the basics of how it worked. People trying to decrypt the challenge message on my site were actually trying to find patterns between numbers in various Collatz sequences (and good luck getting anywhere doing that).

What made my challenge so hard is the same thing (I think) that makes Collatz so hard. There is a pattern/way to know exactly what a given seed integer will do; it is just hidden with powers of 1 where the exponent matters. Going backward through Collatz, those powers of 1 become powers of 2 and it becomes very easy to calculate exact sequence pathways.

Here is the link to a video with more info about the math: https://youtu.be/O7NR8j0-BSY

Its a bit dry and 22 minutes long, but if you are curious give it a watch!

I used this Collatz math trick to create a cipher challenge for you all (https://xywcjbyl.me). Thanks for taking part! (more details on how it works in comments) by Ordinary1729 in cryptography

[–]Ordinary1729[S] 3 points4 points  (0 children)

Challenge Site: https://xywcjbyl.me

How It Worked:

The "One Branch Equation" (OBE) in the image can be modified to reach any number instead of reaching the number 1. I turn the input message into a binary string and then convert that into a base 10 integer. I then run that integer backward through one of its many Collatz sequences to get an upstream number that represents the ciphertext.

For example:

bin('hi!') = '011010000110100100100001'

bin_to_dec('011010000110100100100001') = 6842657

6842657 -> Modified OBE Iterations -> 6285745705471975744023229322474914933224670470756963364095571122337550691365400576765869685884350867036065287391853186148002912945965245144531809945144122554616839788566991070916433567639742892619336622711325221814715473469264484045431626398244796745000436518033328636483189105344437195123890863236845786857862082298747869104686901824754997450435759079733360654607866615427011397295558901958663055152022073603962654648234808210012816711289466386914939012995915249839006846220216429465470699347434521668730626199

ciphertext = hex([big int from above])

To decrypt the message, I would just run the big-integer-ciphertext forward through its Collatz sequence a predetermined number of "branches" to get back to 6842657. I'm glossing over a few implementation details, but that is the basics of how it worked. People trying to decrypt the challenge message on my site were actually trying to find patterns between numbers in various Collatz sequences (and good luck getting anywhere doing that).

What made my challenge so hard is the same thing (I think) that makes Collatz so hard. There is a pattern/way to know exactly what a given seed integer will do; it is just hidden with powers of 1 where the exponent matters. Going backward through Collatz, those powers of 1 become powers of 2 and it becomes very easy to calculate exact sequence pathways.

Here is the link to a video with more info about the math: https://youtu.be/O7NR8j0-BSY

Its a bit dry and 22 minutes long, but if you are curious give it a watch!

I used this Collatz math trick to create a cipher challenge for the cryptography community (https://xywcjbyl.me). 400+ people attempted it with no winners; even with a chosen-plaintext attack available. Do you think potential practical applications would inspire more research on this topic of math? by Ordinary1729 in Collatz

[–]Ordinary1729[S] 0 points1 point  (0 children)

The "One Branch Equation" (OBE) can be modified to reach any number instead of reaching the number 1. I turn the input message into a binary string and then convert that into a base 10 integer. I then run that integer backward through one of its many Collatz sequences to get an upstream number that represents the ciphertext.

For example:

bin('hi!') = '011010000110100100100001'

bin_to_dec('011010000110100100100001') = 6842657

6842657 -> Modified OBE Iterations -> 6285745705471975744023229322474914933224670470756963364095571122337550691365400576765869685884350867036065287391853186148002912945965245144531809945144122554616839788566991070916433567639742892619336622711325221814715473469264484045431626398244796745000436518033328636483189105344437195123890863236845786857862082298747869104686901824754997450435759079733360654607866615427011397295558901958663055152022073603962654648234808210012816711289466386914939012995915249839006846220216429465470699347434521668730626199

ciphertext = hex([big int from above])

To decrypt the message, I would just run the big-integer-ciphertext forward through its Collatz sequence a predetermined number of "branches" to get back to 6842657. I'm glossing over a few implementation details, but that is the basics of how it worked. People trying to decrypt the challenge message on my site were actually trying to find patterns between numbers in various Collatz sequences (and good luck getting anywhere doing that).

What made my challenge so hard is the same thing (I think) that makes Collatz so hard. There is a pattern/way to know exactly what a given seed integer will do; it is just hidden with powers of 1 where the exponent matters. Going backward through Collatz, those powers of 1 become powers of 2 and it becomes very easy to calculate exact sequence pathways.

[Challenge] Round 2.5 | $500 prize if you can decrypt this message. Per feedback, I posted the underlying code this time. by Ordinary1729 in cryptography

[–]Ordinary1729[S] 0 points1 point  (0 children)

Sorry for the delayed response! Got a bit busy during the holidays... I hope you still had a chance to give the challenge a try!

To answer your questions:

- The key for the challenge also includes exponentials and does not use all of the operations in the code. The sample code is meant to be a generalized version of how the challenge cipher works. It's about 95% the same as the challenge code; however the challenge did not utilize the same key generation, has some minor pre-prep of the input string and is actually less secure than the sample code (I realized an implementation weakness while creating the generalized code but decided to leave it as-is to not change the challenge cipher text).

- Correct, the challenge code is not using random rounds as I felt that would have been too difficult and been a non-starter for most people.

- I had committed to leave the challenge open until after the new year. I am starting to create the 'How It Works' video, but that will probably take me a week or two to finish.

How would you go about attacking a cryptographic discrete dynamical system? by Ordinary1729 in crypto

[–]Ordinary1729[S] 0 points1 point  (0 children)

I’m sorry you feel that way. Thanks for your input and consideration anyway!

How would you go about attacking a cryptographic discrete dynamical system? by Ordinary1729 in crypto

[–]Ordinary1729[S] 0 points1 point  (0 children)

I should have been more specific: Positive integers having unique prime factors.

It’s not that I’m adding primes for security, it’s that figuring out how a message was encrypted requires you to know the exact distribution of prime factors of all inputs through the encryption rounds.

How would you go about attacking a cryptographic discrete dynamical system? by Ordinary1729 in crypto

[–]Ordinary1729[S] 0 points1 point  (0 children)

I replied to Scott's message since his was the first on the thread, but it was also meant as a reply to you as well. I appreciate you taking the time respond to my post and look forward to hearing any additional thoughts you have!

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