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Why does AES have exactly 10 rounds for a 128-bit key, 12 for 192 bits and 14 for a 256-bit key size?

Increase number of rounds for SPN and Feistel ciphersIs AES-256 weaker than 192 and 128 bit versions?What is the security loss from reducing Rijndael to 128 bits block size from 256 bits?Can Poly1305-AES be used with AES-256?What are the constraints on using GCM with a tag size of 96 and 128 bits?AES - What is the advantage of a 256-bit key with a 128-bit block cipher?AES function with 128 bit key and 128 bit input size - does it have perfect secrecy?Replacing a block cipher's key schedule with a stream cipherA Lightweight Matrix Suggestion for MixColumns State of AESOCB-AES: Ambiguous definition of “Encipher” in RFC document

23

4

$begingroup$

I was reading about the AES algorithm to be used in one of our projects and found that the exact number of rounds is fixed in AES for specific key sizes:

$$
beginarray
hline
beginarrayc textbfKey Size \ left(textbitsright) endarray
&beginarrayc textbfRounds \ left(textnumberright) endarray \ hline
128 & 10 \ hline
192 & 12 \ hline
256 & 14 \ hline
endarray
$$

Why these specific numbers of rounds only?

aes

share|improve this question

edited Mar 21 at 23:58

Nat

2081411

asked Mar 21 at 19:36

kapilkapil

11816

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that AES is a subset of the Rijndael cipher. The same number of rounds are applicable for Rijndael, but there are more options available depending on key size and block size (AES has just one block size: 128 bits and 3 key sizes, Rijndael has 3 block sizes and 5 key sizes, and therefore 15 combinations of both, rather than just the 3 for AES).
  $endgroup$
  – Maarten Bodewes♦
  Mar 24 at 13:32
  

  
  
  
  

add a comment | 

23

4

$begingroup$

I was reading about the AES algorithm to be used in one of our projects and found that the exact number of rounds is fixed in AES for specific key sizes:

$$
beginarray
hline
beginarrayc textbfKey Size \ left(textbitsright) endarray
&beginarrayc textbfRounds \ left(textnumberright) endarray \ hline
128 & 10 \ hline
192 & 12 \ hline
256 & 14 \ hline
endarray
$$

Why these specific numbers of rounds only?

aes

share|improve this question

edited Mar 21 at 23:58

Nat

2081411

asked Mar 21 at 19:36

kapilkapil

11816

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that AES is a subset of the Rijndael cipher. The same number of rounds are applicable for Rijndael, but there are more options available depending on key size and block size (AES has just one block size: 128 bits and 3 key sizes, Rijndael has 3 block sizes and 5 key sizes, and therefore 15 combinations of both, rather than just the 3 for AES).
  $endgroup$
  – Maarten Bodewes♦
  Mar 24 at 13:32
  

  
  
  
  

add a comment | 

$begingroup$

I was reading about the AES algorithm to be used in one of our projects and found that the exact number of rounds is fixed in AES for specific key sizes:

$$
beginarray
hline
beginarrayc textbfKey Size \ left(textbitsright) endarray
&beginarrayc textbfRounds \ left(textnumberright) endarray \ hline
128 & 10 \ hline
192 & 12 \ hline
256 & 14 \ hline
endarray
$$

Why these specific numbers of rounds only?

aes

share|improve this question

edited Mar 21 at 23:58

Nat

2081411

asked Mar 21 at 19:36

kapilkapil

11816

New contributor

kapil 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 Mar 21 at 23:58

Nat

2081411

asked Mar 21 at 19:36

kapilkapil

11816

New contributor

kapil 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 Mar 21 at 23:58

Nat

2081411

asked Mar 21 at 19:36

kapilkapil

11816

New contributor

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

edited Mar 21 at 23:58

Nat

2081411

edited Mar 21 at 23:58

Nat

2081411

asked Mar 21 at 19:36

kapilkapil

11816

New contributor

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

asked Mar 21 at 19:36

kapilkapil

11816

1 Answer
1

active

oldest

votes

31

$begingroup$

Why these specific number of rounds only?

Because AES is a standard; AES is an acronym for "Advanced Encryption Standard".

The standard specifies these specific number of rounds to ensure that different implementations are interoperable.

Why not more or less?

The reason these specific numbers of rounds were chosen was a choice of the designers. They did a lot of math to determine that these were the sweet spot between sufficient security and optimal performance.

Less might be insecure, and more might be slower with no benefit.

To quote the above book (from Section 3.5 The Number of Rounds):

For Rijndael versions with a longer key, the number of rounds was raised by one for every additional 32 bits in the cipher key. This was done for the following reasons:

1. One of the main objectives is the absence of shortcut attacks, i.e. attacks that are more efficient than an exhaustive key search. Since the workload of an exhaustive key search grows with the key length, shortcut attacks can afford to be less efficient for longer keys.




2. (Partially) known-key and related-key attacks exploit the knowledge of cipher key bits or the ability to apply different cipher keys. If the cipher key grows, the range of possibilities available to the cryptanalyst increases.

share|improve this answer

edited Mar 21 at 21:20

puzzlepalace

2,9101133

answered Mar 21 at 19:51

Ella Rose♦Ella Rose

16.7k44482

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That quote only explains why with longer keys the number of rounds is higher. It does not explain why exactly the 128 bit version uses 10 rounds. The reason for the 10 rounds (which I could misrebemmer since it has been almost 20 years) is as follows: The security against all known attacks was analyzed and 6 rounds was found to be enough against attacks known at the time. It takes 2 rounds to achieve a full avalanche effect in AES, so 10 rounds corresponds to enough rounds for a full avalanche effect before and after the 6 rounds needed for security against known attacks.
  $endgroup$
  – kasperd
  Mar 22 at 10:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kasperd I think it was 6 rounds at the time + 2 rounds because attacks only get better + 2 rounds for full avalanche.
  $endgroup$
  – Martin Bonner
  Mar 22 at 12:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MartinBonner The way the paper described it was "so it can be thought of as padding the vulnerable 6 rounds with two full diffusion steps" or something along those lines, as kasperd says.
  $endgroup$
  – forest
  Mar 22 at 22:44
  
  

  
  
  
  

add a comment | 

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31

$begingroup$

Why these specific number of rounds only?

Because AES is a standard; AES is an acronym for "Advanced Encryption Standard".

The standard specifies these specific number of rounds to ensure that different implementations are interoperable.

Why not more or less?

The reason these specific numbers of rounds were chosen was a choice of the designers. They did a lot of math to determine that these were the sweet spot between sufficient security and optimal performance.

Less might be insecure, and more might be slower with no benefit.

To quote the above book (from Section 3.5 The Number of Rounds):

For Rijndael versions with a longer key, the number of rounds was raised by one for every additional 32 bits in the cipher key. This was done for the following reasons:

1. One of the main objectives is the absence of shortcut attacks, i.e. attacks that are more efficient than an exhaustive key search. Since the workload of an exhaustive key search grows with the key length, shortcut attacks can afford to be less efficient for longer keys.




2. (Partially) known-key and related-key attacks exploit the knowledge of cipher key bits or the ability to apply different cipher keys. If the cipher key grows, the range of possibilities available to the cryptanalyst increases.

share|improve this answer

edited Mar 21 at 21:20

puzzlepalace

2,9101133

answered Mar 21 at 19:51

Ella Rose♦Ella Rose

16.7k44482

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That quote only explains why with longer keys the number of rounds is higher. It does not explain why exactly the 128 bit version uses 10 rounds. The reason for the 10 rounds (which I could misrebemmer since it has been almost 20 years) is as follows: The security against all known attacks was analyzed and 6 rounds was found to be enough against attacks known at the time. It takes 2 rounds to achieve a full avalanche effect in AES, so 10 rounds corresponds to enough rounds for a full avalanche effect before and after the 6 rounds needed for security against known attacks.
  $endgroup$
  – kasperd
  Mar 22 at 10:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kasperd I think it was 6 rounds at the time + 2 rounds because attacks only get better + 2 rounds for full avalanche.
  $endgroup$
  – Martin Bonner
  Mar 22 at 12:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MartinBonner The way the paper described it was "so it can be thought of as padding the vulnerable 6 rounds with two full diffusion steps" or something along those lines, as kasperd says.
  $endgroup$
  – forest
  Mar 22 at 22:44
  
  

  
  
  
  

add a comment | 

31

$begingroup$

Why these specific number of rounds only?

Because AES is a standard; AES is an acronym for "Advanced Encryption Standard".

The standard specifies these specific number of rounds to ensure that different implementations are interoperable.

Why not more or less?

The reason these specific numbers of rounds were chosen was a choice of the designers. They did a lot of math to determine that these were the sweet spot between sufficient security and optimal performance.

Less might be insecure, and more might be slower with no benefit.

To quote the above book (from Section 3.5 The Number of Rounds):

For Rijndael versions with a longer key, the number of rounds was raised by one for every additional 32 bits in the cipher key. This was done for the following reasons:

1. One of the main objectives is the absence of shortcut attacks, i.e. attacks that are more efficient than an exhaustive key search. Since the workload of an exhaustive key search grows with the key length, shortcut attacks can afford to be less efficient for longer keys.




2. (Partially) known-key and related-key attacks exploit the knowledge of cipher key bits or the ability to apply different cipher keys. If the cipher key grows, the range of possibilities available to the cryptanalyst increases.

share|improve this answer

edited Mar 21 at 21:20

puzzlepalace

2,9101133

answered Mar 21 at 19:51

Ella Rose♦Ella Rose

16.7k44482

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That quote only explains why with longer keys the number of rounds is higher. It does not explain why exactly the 128 bit version uses 10 rounds. The reason for the 10 rounds (which I could misrebemmer since it has been almost 20 years) is as follows: The security against all known attacks was analyzed and 6 rounds was found to be enough against attacks known at the time. It takes 2 rounds to achieve a full avalanche effect in AES, so 10 rounds corresponds to enough rounds for a full avalanche effect before and after the 6 rounds needed for security against known attacks.
  $endgroup$
  – kasperd
  Mar 22 at 10:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kasperd I think it was 6 rounds at the time + 2 rounds because attacks only get better + 2 rounds for full avalanche.
  $endgroup$
  – Martin Bonner
  Mar 22 at 12:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MartinBonner The way the paper described it was "so it can be thought of as padding the vulnerable 6 rounds with two full diffusion steps" or something along those lines, as kasperd says.
  $endgroup$
  – forest
  Mar 22 at 22:44
  
  

  
  
  
  

add a comment | 

$begingroup$

Why these specific number of rounds only?

Because AES is a standard; AES is an acronym for "Advanced Encryption Standard".

The standard specifies these specific number of rounds to ensure that different implementations are interoperable.

Why not more or less?

The reason these specific numbers of rounds were chosen was a choice of the designers. They did a lot of math to determine that these were the sweet spot between sufficient security and optimal performance.

Less might be insecure, and more might be slower with no benefit.

To quote the above book (from Section 3.5 The Number of Rounds):

For Rijndael versions with a longer key, the number of rounds was raised by one for every additional 32 bits in the cipher key. This was done for the following reasons:

1. One of the main objectives is the absence of shortcut attacks, i.e. attacks that are more efficient than an exhaustive key search. Since the workload of an exhaustive key search grows with the key length, shortcut attacks can afford to be less efficient for longer keys.




2. (Partially) known-key and related-key attacks exploit the knowledge of cipher key bits or the ability to apply different cipher keys. If the cipher key grows, the range of possibilities available to the cryptanalyst increases.

share|improve this answer

edited Mar 21 at 21:20

puzzlepalace

2,9101133

answered Mar 21 at 19:51

Ella Rose♦Ella Rose

16.7k44482

$endgroup$

share|improve this answer

edited Mar 21 at 21:20

puzzlepalace

2,9101133

answered Mar 21 at 19:51

Ella Rose♦Ella Rose

16.7k44482

edited Mar 21 at 21:20

puzzlepalace

2,9101133

edited Mar 21 at 21:20

puzzlepalace

2,9101133

answered Mar 21 at 19:51

Ella Rose♦Ella Rose

16.7k44482

answered Mar 21 at 19:51

Ella Rose♦Ella Rose

16.7k44482