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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 Apr 27 at 16:16

jSherz

asked Apr 27 at 15:37

jSherzjSherz

1135

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 27 at 16:26
  
  

  
  
  
  

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 Apr 27 at 16:16

jSherz

asked Apr 27 at 15:37

jSherzjSherz

1135

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 27 at 16:26
  
  

  
  
  
  

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 Apr 27 at 16:16

jSherz

asked Apr 27 at 15:37

jSherzjSherz

1135

$endgroup$

share|improve this question

edited Apr 27 at 16:16

jSherz

asked Apr 27 at 15:37

jSherzjSherz

1135

share|improve this question

edited Apr 27 at 16:16

jSherz

asked Apr 27 at 15:37

jSherzjSherz

1135

asked Apr 27 at 15:37

jSherzjSherz

1135

asked Apr 27 at 15:37

jSherzjSherz

1135

2 Answers
2

active

oldest

votes

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 Apr 27 at 16:41

answered Apr 27 at 16:31

michnovkamichnovka

22139

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 27 at 20:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ruakh but This message encrypted with ...; see https://... for details is not the encrypted data, it's just a header. The actual encrypted data should still be indistinguishable from random data.
  $endgroup$
  – marcelm
  Apr 27 at 22:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @marcelm: You can choose to make that distinction, but there's no need to; if the ciphertext requires a certain fixed prefix, then that prefix is essentially part of the ciphertext. If you'd like a different example: there's no problem with an encryption scheme where every Nth byte is a checksum, or one where the ciphertext only uses ASCII printable characters (for compatibility with MIME). There is literally no need for encrypted data to be "indistinguishable from random data". All that matters is that it be impossible to guess anything about the plaintext (other than its length).
  $endgroup$
  – ruakh
  Apr 27 at 23:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't believe there's any block cipher which has a default block size that small. Even for RC5, 32 bits was the lowest it could go (default was 128, I think). But a few like SIMON and SPECK support 32-bit blocks. The cipher with the smallest default block size I know of is PRINTcipher, which has 48-bit blocks, IIRC.
  $endgroup$
  – forest
  Apr 28 at 8:44
  

  
  
  
  

add a comment | 

3

$begingroup$

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

A typical block cipher 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 cipher with 4-byte blocks.

For comparison, in typical applications of a stream cipher, 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|improve this answer

edited Apr 28 at 6:50

answered Apr 27 at 20:13

jpajpa

1914

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For block ciphers, that will depend entirely on the mode of operation.
  $endgroup$
  – Ilmari Karonen
  Apr 27 at 21:03
  

  
  
  
  

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 Apr 27 at 16:41

answered Apr 27 at 16:31

michnovkamichnovka

22139

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 27 at 20:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ruakh but This message encrypted with ...; see https://... for details is not the encrypted data, it's just a header. The actual encrypted data should still be indistinguishable from random data.
  $endgroup$
  – marcelm
  Apr 27 at 22:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @marcelm: You can choose to make that distinction, but there's no need to; if the ciphertext requires a certain fixed prefix, then that prefix is essentially part of the ciphertext. If you'd like a different example: there's no problem with an encryption scheme where every Nth byte is a checksum, or one where the ciphertext only uses ASCII printable characters (for compatibility with MIME). There is literally no need for encrypted data to be "indistinguishable from random data". All that matters is that it be impossible to guess anything about the plaintext (other than its length).
  $endgroup$
  – ruakh
  Apr 27 at 23:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't believe there's any block cipher which has a default block size that small. Even for RC5, 32 bits was the lowest it could go (default was 128, I think). But a few like SIMON and SPECK support 32-bit blocks. The cipher with the smallest default block size I know of is PRINTcipher, which has 48-bit blocks, IIRC.
  $endgroup$
  – forest
  Apr 28 at 8:44
  

  
  
  
  

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 Apr 27 at 16:41

answered Apr 27 at 16:31

michnovkamichnovka

22139

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 27 at 20:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ruakh but This message encrypted with ...; see https://... for details is not the encrypted data, it's just a header. The actual encrypted data should still be indistinguishable from random data.
  $endgroup$
  – marcelm
  Apr 27 at 22:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @marcelm: You can choose to make that distinction, but there's no need to; if the ciphertext requires a certain fixed prefix, then that prefix is essentially part of the ciphertext. If you'd like a different example: there's no problem with an encryption scheme where every Nth byte is a checksum, or one where the ciphertext only uses ASCII printable characters (for compatibility with MIME). There is literally no need for encrypted data to be "indistinguishable from random data". All that matters is that it be impossible to guess anything about the plaintext (other than its length).
  $endgroup$
  – ruakh
  Apr 27 at 23:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't believe there's any block cipher which has a default block size that small. Even for RC5, 32 bits was the lowest it could go (default was 128, I think). But a few like SIMON and SPECK support 32-bit blocks. The cipher with the smallest default block size I know of is PRINTcipher, which has 48-bit blocks, IIRC.
  $endgroup$
  – forest
  Apr 28 at 8:44
  

  
  
  
  

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 Apr 27 at 16:41

answered Apr 27 at 16:31

michnovkamichnovka

22139

$endgroup$

share|improve this answer

edited Apr 27 at 16:41

answered Apr 27 at 16:31

michnovkamichnovka

22139

answered Apr 27 at 16:31

michnovkamichnovka

22139

answered Apr 27 at 16:31

michnovkamichnovka

22139

3

$begingroup$

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

A typical block cipher 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 cipher with 4-byte blocks.

For comparison, in typical applications of a stream cipher, 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|improve this answer

edited Apr 28 at 6:50

answered Apr 27 at 20:13

jpajpa

1914

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For block ciphers, that will depend entirely on the mode of operation.
  $endgroup$
  – Ilmari Karonen
  Apr 27 at 21:03
  

  
  
  
  

add a comment | 

3

$begingroup$

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

A typical block cipher 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 cipher with 4-byte blocks.

For comparison, in typical applications of a stream cipher, 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|improve this answer

edited Apr 28 at 6:50

answered Apr 27 at 20:13

jpajpa

1914

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For block ciphers, that will depend entirely on the mode of operation.
  $endgroup$
  – Ilmari Karonen
  Apr 27 at 21:03
  

  
  
  
  

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 cipher 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 cipher with 4-byte blocks.

For comparison, in typical applications of a stream cipher, 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|improve this answer

edited Apr 28 at 6:50

answered Apr 27 at 20:13

jpajpa

1914

$endgroup$

 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|improve this answer

edited Apr 28 at 6:50

answered Apr 27 at 20:13

jpajpa

1914

answered Apr 27 at 20:13

jpajpa

1914

answered Apr 27 at 20:13

jpajpa

1914