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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?

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8

1

$begingroup$

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 34 mins ago

Nat

2081411

asked 4 hours ago

kapilkapil

432

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.

$endgroup$

add a comment | 

8

1

$begingroup$

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 34 mins ago

Nat

2081411

asked 4 hours ago

kapilkapil

432

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.

$endgroup$

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 34 mins ago

Nat

2081411

asked 4 hours ago

kapilkapil

432

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 34 mins ago

Nat

2081411

asked 4 hours ago

kapilkapil

432

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 34 mins ago

Nat

2081411

asked 4 hours ago

kapilkapil

432

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 34 mins ago

Nat

2081411

edited 34 mins ago

Nat

2081411

asked 4 hours ago

kapilkapil

432

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 4 hours ago

kapilkapil

432

1 Answer
1

active

oldest

votes

10

$begingroup$

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 3 hours ago

puzzlepalace

2,8701133

answered 4 hours ago

Ella Rose♦Ella Rose

16.5k44382

$endgroup$

add a comment | 

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10

$begingroup$

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 3 hours ago

puzzlepalace

2,8701133

answered 4 hours ago

Ella Rose♦Ella Rose

16.5k44382

$endgroup$

add a comment | 

10

$begingroup$

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 3 hours ago

puzzlepalace

2,8701133

answered 4 hours ago

Ella Rose♦Ella Rose

16.5k44382

$endgroup$

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 3 hours ago

puzzlepalace

2,8701133

answered 4 hours ago

Ella Rose♦Ella Rose

16.5k44382

$endgroup$

share|improve this answer

edited 3 hours ago

puzzlepalace

2,8701133

answered 4 hours ago

Ella Rose♦Ella Rose

16.5k44382

edited 3 hours ago

puzzlepalace

2,8701133

edited 3 hours ago

puzzlepalace

2,8701133

answered 4 hours ago

Ella Rose♦Ella Rose

16.5k44382

answered 4 hours ago

Ella Rose♦Ella Rose

16.5k44382