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Is it possible to determine the symmetric encryption method used by output size?

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2

$begingroup$

I'm attempting to identify the method of encryption for a black-box symmetric encryptor that produces blocks of output that are 4 bytes in length (e.g. small inputs fit in 16 bytes, then 20 bytes and 24 bytes as more input characters are added).

It's symmetric encryption and the value is always the same for the same input text. Is it possible to determine which method of encryption is used? I'm assuming it's a block cipher as a result of the blocks of output it produces.

block-cipher symmetric blocksize

share|improve this question

edited 4 hours ago

jSherz

asked 4 hours ago

jSherzjSherz

1114

New contributor

jSherz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related question on security.SE: security.stackexchange.com/questions/38797/… (I thought I remembered seeing a duplicate or at least a very similar question before on this site as well, but if so, I can't find it.)
  $endgroup$
  – Ilmari Karonen
  3 hours ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

I'm attempting to identify the method of encryption for a black-box symmetric encryptor that produces blocks of output that are 4 bytes in length (e.g. small inputs fit in 16 bytes, then 20 bytes and 24 bytes as more input characters are added).

It's symmetric encryption and the value is always the same for the same input text. Is it possible to determine which method of encryption is used? I'm assuming it's a block cipher as a result of the blocks of output it produces.

block-cipher symmetric blocksize

share|improve this question

edited 4 hours ago

jSherz

asked 4 hours ago

jSherzjSherz

1114

New contributor

jSherz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related question on security.SE: security.stackexchange.com/questions/38797/… (I thought I remembered seeing a duplicate or at least a very similar question before on this site as well, but if so, I can't find it.)
  $endgroup$
  – Ilmari Karonen
  3 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

I'm attempting to identify the method of encryption for a black-box symmetric encryptor that produces blocks of output that are 4 bytes in length (e.g. small inputs fit in 16 bytes, then 20 bytes and 24 bytes as more input characters are added).

It's symmetric encryption and the value is always the same for the same input text. Is it possible to determine which method of encryption is used? I'm assuming it's a block cipher as a result of the blocks of output it produces.

block-cipher symmetric blocksize

share|improve this question

edited 4 hours ago

jSherz

asked 4 hours ago

jSherzjSherz

1114

New contributor

jSherz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited 4 hours ago

jSherz

asked 4 hours ago

jSherzjSherz

1114

New contributor

jSherz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 4 hours ago

jSherz

asked 4 hours ago

jSherzjSherz

1114

New contributor

jSherz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 4 hours ago

jSherzjSherz

1114

New contributor

jSherz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 4 hours ago

jSherzjSherz

1114

2 Answers
2

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oldest

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2

$begingroup$

Simply put: No.

Without knowing other details, you cannot be sure. That being said, in the case you described, the black box uses 4 bytes blocks, which is rather uncommon with modern block ciphers. AES e.g. uses 128bits (16bytes), Blowfish uses 64bits (8bytes). 4byte block ciphers are very uncommon now. Even DES, which is quite outdated and old uses 8 bytes. The only block cipher used that has 32bit block size and comes to my mind is RC5.

So as you can see, you can make an educated guess. But given just the ciphertext, this does not inform you at all about encryption used. This is one of the key aspects of cryptography btw, encrypted data should look as much as random data as possible.

You cannot even be sure that this is a block cipher. Could be a stream cipher with padding. So if the only thing you know is: "I have a box. I feed it data and it spits out data in chunks of 4 bytes" - then you know nothin', j Sherz

share|improve this answer

edited 3 hours ago

answered 3 hours ago

michnovkamichnovka

19629

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Re: "This is one of the key aspects of cryptography btw, encrypted data should look as much as random data as possible": I don't really agree with this. From a security standpoint, it would be perfectly fine for an encryption scheme to always produce output that begins with, say, This message encrypted with ...; see https://... for details. In practice, however, encryption schemes don't generally do this; instead, a secure protocol will typically provide a separate (metadata) field to indicate the encryption scheme in use.
  $endgroup$
  – ruakh
  6 mins ago
  

  
  
  
  

add a comment | 

0

$begingroup$

You can try to determine the block size by modifying each byte of the input data, starting from the last one.

A typical block chipher implementation will mix the change into all other bytes in the block, and usually also to all following blocks, but not to blocks that come before it.

So if you observe e.g.:

 Plaintext      Encrypted

 0000 0000      2348 1234

 0000 0001      2348 4292

 0000 0010      2348 9823

 0000 0100      2348 2149

 0000 1000      2348 6785

 0001 0000      8173 1437

you could be quite certain that it is a block chipher with 4-byte blocks.

For comparison, in typical applications of a stream chipher, only a single byte would usually change. If it is a stream cipher, you can obtain the keystream by encoding 000...0000, because it sounds like it doesn't have a random initialization vector.

share

answered 9 mins ago

jpajpa

1614

$endgroup$

add a comment | 

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2

$begingroup$

Simply put: No.

Without knowing other details, you cannot be sure. That being said, in the case you described, the black box uses 4 bytes blocks, which is rather uncommon with modern block ciphers. AES e.g. uses 128bits (16bytes), Blowfish uses 64bits (8bytes). 4byte block ciphers are very uncommon now. Even DES, which is quite outdated and old uses 8 bytes. The only block cipher used that has 32bit block size and comes to my mind is RC5.

So as you can see, you can make an educated guess. But given just the ciphertext, this does not inform you at all about encryption used. This is one of the key aspects of cryptography btw, encrypted data should look as much as random data as possible.

You cannot even be sure that this is a block cipher. Could be a stream cipher with padding. So if the only thing you know is: "I have a box. I feed it data and it spits out data in chunks of 4 bytes" - then you know nothin', j Sherz

share|improve this answer

edited 3 hours ago

answered 3 hours ago

michnovkamichnovka

19629

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Re: "This is one of the key aspects of cryptography btw, encrypted data should look as much as random data as possible": I don't really agree with this. From a security standpoint, it would be perfectly fine for an encryption scheme to always produce output that begins with, say, This message encrypted with ...; see https://... for details. In practice, however, encryption schemes don't generally do this; instead, a secure protocol will typically provide a separate (metadata) field to indicate the encryption scheme in use.
  $endgroup$
  – ruakh
  6 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Simply put: No.

Without knowing other details, you cannot be sure. That being said, in the case you described, the black box uses 4 bytes blocks, which is rather uncommon with modern block ciphers. AES e.g. uses 128bits (16bytes), Blowfish uses 64bits (8bytes). 4byte block ciphers are very uncommon now. Even DES, which is quite outdated and old uses 8 bytes. The only block cipher used that has 32bit block size and comes to my mind is RC5.

So as you can see, you can make an educated guess. But given just the ciphertext, this does not inform you at all about encryption used. This is one of the key aspects of cryptography btw, encrypted data should look as much as random data as possible.

You cannot even be sure that this is a block cipher. Could be a stream cipher with padding. So if the only thing you know is: "I have a box. I feed it data and it spits out data in chunks of 4 bytes" - then you know nothin', j Sherz

share|improve this answer

edited 3 hours ago

answered 3 hours ago

michnovkamichnovka

19629

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Re: "This is one of the key aspects of cryptography btw, encrypted data should look as much as random data as possible": I don't really agree with this. From a security standpoint, it would be perfectly fine for an encryption scheme to always produce output that begins with, say, This message encrypted with ...; see https://... for details. In practice, however, encryption schemes don't generally do this; instead, a secure protocol will typically provide a separate (metadata) field to indicate the encryption scheme in use.
  $endgroup$
  – ruakh
  6 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

Simply put: No.

Without knowing other details, you cannot be sure. That being said, in the case you described, the black box uses 4 bytes blocks, which is rather uncommon with modern block ciphers. AES e.g. uses 128bits (16bytes), Blowfish uses 64bits (8bytes). 4byte block ciphers are very uncommon now. Even DES, which is quite outdated and old uses 8 bytes. The only block cipher used that has 32bit block size and comes to my mind is RC5.

So as you can see, you can make an educated guess. But given just the ciphertext, this does not inform you at all about encryption used. This is one of the key aspects of cryptography btw, encrypted data should look as much as random data as possible.

You cannot even be sure that this is a block cipher. Could be a stream cipher with padding. So if the only thing you know is: "I have a box. I feed it data and it spits out data in chunks of 4 bytes" - then you know nothin', j Sherz

share|improve this answer

edited 3 hours ago

answered 3 hours ago

michnovkamichnovka

19629

$endgroup$

share|improve this answer

edited 3 hours ago

answered 3 hours ago

michnovkamichnovka

19629

answered 3 hours ago

michnovkamichnovka

19629

answered 3 hours ago

michnovkamichnovka

19629

0

$begingroup$

You can try to determine the block size by modifying each byte of the input data, starting from the last one.

A typical block chipher implementation will mix the change into all other bytes in the block, and usually also to all following blocks, but not to blocks that come before it.

So if you observe e.g.:

 Plaintext      Encrypted

 0000 0000      2348 1234

 0000 0001      2348 4292

 0000 0010      2348 9823

 0000 0100      2348 2149

 0000 1000      2348 6785

 0001 0000      8173 1437

you could be quite certain that it is a block chipher with 4-byte blocks.

For comparison, in typical applications of a stream chipher, only a single byte would usually change. If it is a stream cipher, you can obtain the keystream by encoding 000...0000, because it sounds like it doesn't have a random initialization vector.

share

answered 9 mins ago

jpajpa

1614

$endgroup$

add a comment | 

0

$begingroup$

You can try to determine the block size by modifying each byte of the input data, starting from the last one.

A typical block chipher implementation will mix the change into all other bytes in the block, and usually also to all following blocks, but not to blocks that come before it.

So if you observe e.g.:

 Plaintext      Encrypted

 0000 0000      2348 1234

 0000 0001      2348 4292

 0000 0010      2348 9823

 0000 0100      2348 2149

 0000 1000      2348 6785

 0001 0000      8173 1437

you could be quite certain that it is a block chipher with 4-byte blocks.

For comparison, in typical applications of a stream chipher, only a single byte would usually change. If it is a stream cipher, you can obtain the keystream by encoding 000...0000, because it sounds like it doesn't have a random initialization vector.

share

answered 9 mins ago

jpajpa

1614

$endgroup$

add a comment | 

$begingroup$

You can try to determine the block size by modifying each byte of the input data, starting from the last one.

A typical block chipher implementation will mix the change into all other bytes in the block, and usually also to all following blocks, but not to blocks that come before it.

So if you observe e.g.:

 Plaintext      Encrypted

 0000 0000      2348 1234

 0000 0001      2348 4292

 0000 0010      2348 9823

 0000 0100      2348 2149

 0000 1000      2348 6785

 0001 0000      8173 1437

you could be quite certain that it is a block chipher with 4-byte blocks.

For comparison, in typical applications of a stream chipher, only a single byte would usually change. If it is a stream cipher, you can obtain the keystream by encoding 000...0000, because it sounds like it doesn't have a random initialization vector.

share

answered 9 mins ago

jpajpa

1614

$endgroup$

 Plaintext      Encrypted

 0000 0000      2348 1234

 0000 0001      2348 4292

 0000 0010      2348 9823

 0000 0100      2348 2149

 0000 1000      2348 6785

 0001 0000      8173 1437

share

answered 9 mins ago

jpajpa

1614

answered 9 mins ago

jpajpa

1614

answered 9 mins ago

jpajpa

1614