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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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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













8












$begingroup$


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:



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



Why these specific numbers of rounds only?










share|improve this question









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.







$endgroup$
















    8












    $begingroup$


    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:



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



    Why these specific numbers of rounds only?










    share|improve this question









    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.







    $endgroup$














      8












      8








      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:



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



      Why these specific numbers of rounds only?










      share|improve this question









      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.







      $endgroup$




      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:



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



      Why these specific numbers of rounds only?







      aes






      share|improve this question









      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









      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




      share|improve this question








      edited 48 mins ago









      Nat

      2081411




      2081411






      New contributor




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      Check out our Code of Conduct.









      asked 5 hours ago









      kapilkapil

      432




      432




      New contributor




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      New contributor





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




















          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











          $endgroup$












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            1 Answer
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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            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











            $endgroup$

















              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











              $endgroup$















                10












                10








                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











                $endgroup$



                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:



                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














                share|improve this answer



                share|improve this answer








                edited 3 hours ago









                puzzlepalace

                2,8701133




                2,8701133










                answered 4 hours ago









                Ella RoseElla Rose

                16.5k44382




                16.5k44382




















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