With DES, is ECB encryption mode equivalent to CBC with null IV for short data blocks?

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I got to this by experimentation, but can someone maybe provide an explanation for this?






cbc des ecb







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    up vote
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    down vote

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    I got to this by experimentation, but can someone maybe provide an explanation for this?






    cbc des ecb







    share|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I got to this by experimentation, but can someone maybe provide an explanation for this?






      cbc des ecb







      share|improve this question













      I got to this by experimentation, but can someone maybe provide an explanation for this?






      cbc des ecb




      cbc des ecb






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      asked 4 hours ago









      bbozo

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          • The CBC mode processed as $c_i = E_k(p_i oplus c_i-1)$ with $c_0 = IV$


          • The ECB mode processed as $c_i = E_k(p_i)$


          If you set the first block with $c_o = IV = 0$ in the CBC mode, than it is calculated as $c_1 = E_k(p_1)$. This is exactly ECB mode.



          The next blocks, however, will not be equal;



          in ECB mode $c_i = E_k(p_i)$ whereas in CBC mode $c_i = E_k(p_i oplus c_i-1) neq E_k(p_i)$ for $1 < i < m$ where $m$ is the number of blocks.



          Equality is only valid for first block.






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          • Exactly as I suspected, thank you for a more formal explanation <3
            – bbozo
            19 mins ago










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          active

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          • The CBC mode processed as $c_i = E_k(p_i oplus c_i-1)$ with $c_0 = IV$


          • The ECB mode processed as $c_i = E_k(p_i)$


          If you set the first block with $c_o = IV = 0$ in the CBC mode, than it is calculated as $c_1 = E_k(p_1)$. This is exactly ECB mode.



          The next blocks, however, will not be equal;



          in ECB mode $c_i = E_k(p_i)$ whereas in CBC mode $c_i = E_k(p_i oplus c_i-1) neq E_k(p_i)$ for $1 < i < m$ where $m$ is the number of blocks.



          Equality is only valid for first block.






          share|improve this answer






















          • Exactly as I suspected, thank you for a more formal explanation <3
            – bbozo
            19 mins ago














          up vote
          2
          down vote



          accepted










          • The CBC mode processed as $c_i = E_k(p_i oplus c_i-1)$ with $c_0 = IV$


          • The ECB mode processed as $c_i = E_k(p_i)$


          If you set the first block with $c_o = IV = 0$ in the CBC mode, than it is calculated as $c_1 = E_k(p_1)$. This is exactly ECB mode.



          The next blocks, however, will not be equal;



          in ECB mode $c_i = E_k(p_i)$ whereas in CBC mode $c_i = E_k(p_i oplus c_i-1) neq E_k(p_i)$ for $1 < i < m$ where $m$ is the number of blocks.



          Equality is only valid for first block.






          share|improve this answer






















          • Exactly as I suspected, thank you for a more formal explanation <3
            – bbozo
            19 mins ago












          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          • The CBC mode processed as $c_i = E_k(p_i oplus c_i-1)$ with $c_0 = IV$


          • The ECB mode processed as $c_i = E_k(p_i)$


          If you set the first block with $c_o = IV = 0$ in the CBC mode, than it is calculated as $c_1 = E_k(p_1)$. This is exactly ECB mode.



          The next blocks, however, will not be equal;



          in ECB mode $c_i = E_k(p_i)$ whereas in CBC mode $c_i = E_k(p_i oplus c_i-1) neq E_k(p_i)$ for $1 < i < m$ where $m$ is the number of blocks.



          Equality is only valid for first block.






          share|improve this answer














          • The CBC mode processed as $c_i = E_k(p_i oplus c_i-1)$ with $c_0 = IV$


          • The ECB mode processed as $c_i = E_k(p_i)$


          If you set the first block with $c_o = IV = 0$ in the CBC mode, than it is calculated as $c_1 = E_k(p_1)$. This is exactly ECB mode.



          The next blocks, however, will not be equal;



          in ECB mode $c_i = E_k(p_i)$ whereas in CBC mode $c_i = E_k(p_i oplus c_i-1) neq E_k(p_i)$ for $1 < i < m$ where $m$ is the number of blocks.



          Equality is only valid for first block.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 40 mins ago

























          answered 3 hours ago









          kelalaka

          1,086217




          1,086217











          • Exactly as I suspected, thank you for a more formal explanation <3
            – bbozo
            19 mins ago
















          • Exactly as I suspected, thank you for a more formal explanation <3
            – bbozo
            19 mins ago















          Exactly as I suspected, thank you for a more formal explanation <3
          – bbozo
          19 mins ago




          Exactly as I suspected, thank you for a more formal explanation <3
          – bbozo
          19 mins ago

















           

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