Symetric key encryption is not Authentication?











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Is there any way to show CCA-secure symmetric-key encryption does not have to be an authenticated encryption?



I know you can create a setup that isn't authentication, but how will you go about it?






authenticated-encryption chosen-ciphertext-attack







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    up vote
    2
    down vote

    favorite












    Is there any way to show CCA-secure symmetric-key encryption does not have to be an authenticated encryption?



    I know you can create a setup that isn't authentication, but how will you go about it?






    authenticated-encryption chosen-ciphertext-attack







    share|improve this question









    New contributor




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






















      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Is there any way to show CCA-secure symmetric-key encryption does not have to be an authenticated encryption?



      I know you can create a setup that isn't authentication, but how will you go about it?






      authenticated-encryption chosen-ciphertext-attack







      share|improve this question









      New contributor




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











      Is there any way to show CCA-secure symmetric-key encryption does not have to be an authenticated encryption?



      I know you can create a setup that isn't authentication, but how will you go about it?






      authenticated-encryption chosen-ciphertext-attack




      authenticated-encryption chosen-ciphertext-attack






      share|improve this question









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      Uhntiss is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|improve this question









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      share|improve this question




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      edited 4 hours ago









      kelalaka

      4,84121837




      4,84121837






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      asked 6 hours ago









      Uhntiss

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          1 Answer
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          Consider a CCA2 secure secret key encryption scheme $mathcal{E}$ with an exponentially large randomness space. Turn this into an encryption scheme $mathcal{E'}$ where we extend the randomness space of $mathcal{E}$ with one more element. Let's think of it as $0$ (we can get there by rearranging the randomness space). For randomness $0$ the encryption in $mathcal{E'}$ becomes the identity for every element.



          Now, I claim $mathcal{E'}$ is still CCA secure as the probability that for the challenge query randomness 0 is chosen is negligible. However, $mathcal{E'}$ is not a secure MAC as everybody knows a valid tag for any message.






          share|improve this answer





















          • Two comments/nitpicks: 1. You don't need to add an additional randomness value, you can just repurpose an existing one (e.g. the all 0 string). 2. For that scheme to be be complete you need to mark from which mode of the scheme the ciphertext resulted. You can do that e.g. by prefixing actual encryptions with 0 and the identity with 1.
            – Maeher
            1 hour ago











          Your Answer





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          up vote
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          down vote













          Consider a CCA2 secure secret key encryption scheme $mathcal{E}$ with an exponentially large randomness space. Turn this into an encryption scheme $mathcal{E'}$ where we extend the randomness space of $mathcal{E}$ with one more element. Let's think of it as $0$ (we can get there by rearranging the randomness space). For randomness $0$ the encryption in $mathcal{E'}$ becomes the identity for every element.



          Now, I claim $mathcal{E'}$ is still CCA secure as the probability that for the challenge query randomness 0 is chosen is negligible. However, $mathcal{E'}$ is not a secure MAC as everybody knows a valid tag for any message.






          share|improve this answer





















          • Two comments/nitpicks: 1. You don't need to add an additional randomness value, you can just repurpose an existing one (e.g. the all 0 string). 2. For that scheme to be be complete you need to mark from which mode of the scheme the ciphertext resulted. You can do that e.g. by prefixing actual encryptions with 0 and the identity with 1.
            – Maeher
            1 hour ago















          up vote
          2
          down vote













          Consider a CCA2 secure secret key encryption scheme $mathcal{E}$ with an exponentially large randomness space. Turn this into an encryption scheme $mathcal{E'}$ where we extend the randomness space of $mathcal{E}$ with one more element. Let's think of it as $0$ (we can get there by rearranging the randomness space). For randomness $0$ the encryption in $mathcal{E'}$ becomes the identity for every element.



          Now, I claim $mathcal{E'}$ is still CCA secure as the probability that for the challenge query randomness 0 is chosen is negligible. However, $mathcal{E'}$ is not a secure MAC as everybody knows a valid tag for any message.






          share|improve this answer





















          • Two comments/nitpicks: 1. You don't need to add an additional randomness value, you can just repurpose an existing one (e.g. the all 0 string). 2. For that scheme to be be complete you need to mark from which mode of the scheme the ciphertext resulted. You can do that e.g. by prefixing actual encryptions with 0 and the identity with 1.
            – Maeher
            1 hour ago













          up vote
          2
          down vote










          up vote
          2
          down vote









          Consider a CCA2 secure secret key encryption scheme $mathcal{E}$ with an exponentially large randomness space. Turn this into an encryption scheme $mathcal{E'}$ where we extend the randomness space of $mathcal{E}$ with one more element. Let's think of it as $0$ (we can get there by rearranging the randomness space). For randomness $0$ the encryption in $mathcal{E'}$ becomes the identity for every element.



          Now, I claim $mathcal{E'}$ is still CCA secure as the probability that for the challenge query randomness 0 is chosen is negligible. However, $mathcal{E'}$ is not a secure MAC as everybody knows a valid tag for any message.






          share|improve this answer












          Consider a CCA2 secure secret key encryption scheme $mathcal{E}$ with an exponentially large randomness space. Turn this into an encryption scheme $mathcal{E'}$ where we extend the randomness space of $mathcal{E}$ with one more element. Let's think of it as $0$ (we can get there by rearranging the randomness space). For randomness $0$ the encryption in $mathcal{E'}$ becomes the identity for every element.



          Now, I claim $mathcal{E'}$ is still CCA secure as the probability that for the challenge query randomness 0 is chosen is negligible. However, $mathcal{E'}$ is not a secure MAC as everybody knows a valid tag for any message.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 3 hours ago









          mephisto

          2,2471226




          2,2471226












          • Two comments/nitpicks: 1. You don't need to add an additional randomness value, you can just repurpose an existing one (e.g. the all 0 string). 2. For that scheme to be be complete you need to mark from which mode of the scheme the ciphertext resulted. You can do that e.g. by prefixing actual encryptions with 0 and the identity with 1.
            – Maeher
            1 hour ago


















          • Two comments/nitpicks: 1. You don't need to add an additional randomness value, you can just repurpose an existing one (e.g. the all 0 string). 2. For that scheme to be be complete you need to mark from which mode of the scheme the ciphertext resulted. You can do that e.g. by prefixing actual encryptions with 0 and the identity with 1.
            – Maeher
            1 hour ago
















          Two comments/nitpicks: 1. You don't need to add an additional randomness value, you can just repurpose an existing one (e.g. the all 0 string). 2. For that scheme to be be complete you need to mark from which mode of the scheme the ciphertext resulted. You can do that e.g. by prefixing actual encryptions with 0 and the identity with 1.
          – Maeher
          1 hour ago




          Two comments/nitpicks: 1. You don't need to add an additional randomness value, you can just repurpose an existing one (e.g. the all 0 string). 2. For that scheme to be be complete you need to mark from which mode of the scheme the ciphertext resulted. You can do that e.g. by prefixing actual encryptions with 0 and the identity with 1.
          – Maeher
          1 hour ago










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