Why does AES have exactly 10 rounds for a 128-bit key, 12 for 192 bits and 14 for a 256-bit key size? ...
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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?
The 2019 Stack Overflow Developer Survey Results Are InIncrease 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
$begingroup$
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:
$$
begin{array}{|c|c|}
hline
begin{array}{c} textbf{Key Size} \ left(text{bits}right) end{array}
&begin{array}{c} textbf{Rounds} \ left(text{number}right) end{array} \ hline
128 & 10 \ hline
192 & 12 \ hline
256 & 14 \ hline
end{array}
$$
Why these specific numbers of rounds only?
aes
edited Mar 21 at 23:58
Nat
2081411
asked Mar 21 at 19:36
kapilkapil
11816
$endgroup$
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:
$$
begin{array}{|c|c|}
hline
begin{array}{c} textbf{Key Size} \ left(text{bits}right) end{array}
&begin{array}{c} textbf{Rounds} \ left(text{number}right) end{array} \ hline
128 & 10 \ hline
192 & 12 \ hline
256 & 14 \ hline
end{array}
$$
Why these specific numbers of rounds only?
aes
edited Mar 21 at 23:58
Nat
2081411
asked Mar 21 at 19:36
kapilkapil
11816
$endgroup$
$begingroup$
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:
$$
begin{array}{|c|c|}
hline
begin{array}{c} textbf{Key Size} \ left(text{bits}right) end{array}
&begin{array}{c} textbf{Rounds} \ left(text{number}right) end{array} \ hline
128 & 10 \ hline
192 & 12 \ hline
256 & 14 \ hline
end{array}
$$
Why these specific numbers of rounds only?
aes
edited Mar 21 at 23:58
Nat
2081411
asked Mar 21 at 19:36
kapilkapil
11816
$endgroup$
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:
$$
begin{array}{|c|c|}
hline
begin{array}{c} textbf{Key Size} \ left(text{bits}right) end{array}
&begin{array}{c} textbf{Rounds} \ left(text{number}right) end{array} \ hline
128 & 10 \ hline
192 & 12 \ hline
256 & 14 \ hline
end{array}
$$
Why these specific numbers of rounds only?
aes
aes
edited Mar 21 at 23:58
Nat
2081411
asked Mar 21 at 19:36
kapilkapil
11816
edited Mar 21 at 23:58
Nat
2081411
asked Mar 21 at 19:36
kapilkapil
11816
edited Mar 21 at 23:58
Nat
2081411
edited Mar 21 at 23:58
Nat
2081411
edited Mar 21 at 23:58
Nat
2081411
2081411
asked Mar 21 at 19:36
kapilkapil
11816
asked Mar 21 at 19:36
kapilkapil
11816
asked Mar 21 at 19:36
kapilkapil
11816
11816
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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 |
1 Answer
1
active
oldest
votes
$begingroup$
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:
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.
(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.
edited Mar 21 at 21:20
puzzlepalace
2,9601133
answered Mar 21 at 19:51
Ella Rose♦Ella Rose
17k44483
$endgroup$
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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$begingroup$
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:
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.
(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.
edited Mar 21 at 21:20
puzzlepalace
2,9601133
answered Mar 21 at 19:51
Ella Rose♦Ella Rose
17k44483
$endgroup$
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:
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.
(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.
edited Mar 21 at 21:20
puzzlepalace
2,9601133
answered Mar 21 at 19:51
Ella Rose♦Ella Rose
17k44483
$endgroup$
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:
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.
(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.
edited Mar 21 at 21:20
puzzlepalace
2,9601133
answered Mar 21 at 19:51
Ella Rose♦Ella Rose
17k44483
$endgroup$
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:
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.
(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.
edited Mar 21 at 21:20
puzzlepalace
2,9601133
answered Mar 21 at 19:51
Ella Rose♦Ella Rose
17k44483
edited Mar 21 at 21:20
puzzlepalace
2,9601133
edited Mar 21 at 21:20
puzzlepalace
2,9601133
edited Mar 21 at 21:20
puzzlepalace
2,9601133
2,9601133
answered Mar 21 at 19:51
Ella Rose♦Ella Rose
17k44483
answered Mar 21 at 19:51
Ella Rose♦Ella Rose
17k44483
answered Mar 21 at 19:51
Ella Rose♦Ella Rose
17k44483
17k44483
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 |
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
1
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
$begingroup$
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
$begingroup$
@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
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
$begingroup$
@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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$begingroup$
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