double encryption - One Time Pad












1












$begingroup$


Can the security of the encryption system be improved by using double encryption with two randomly chosen keys?



By double encryption I mean using two keys K1 and K2, each 20-bit long, to obtain ciphertext $C = ( M ⊕ K_1 ) ⊕ K_2$.










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  • 1




    $begingroup$
    An ideal one time pad already has perfect security, you can't just get twice as perfect. The second layer doesn't add much.
    $endgroup$
    – Natanael
    5 hours ago






  • 1




    $begingroup$
    I don't see how enumerating the key values makes any difference for the one-time pad. Even a single bit key for a single bit message would be secure.
    $endgroup$
    – Maarten Bodewes
    3 hours ago


















1












$begingroup$


Can the security of the encryption system be improved by using double encryption with two randomly chosen keys?



By double encryption I mean using two keys K1 and K2, each 20-bit long, to obtain ciphertext $C = ( M ⊕ K_1 ) ⊕ K_2$.










share|improve this question









New contributor




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







$endgroup$








  • 1




    $begingroup$
    An ideal one time pad already has perfect security, you can't just get twice as perfect. The second layer doesn't add much.
    $endgroup$
    – Natanael
    5 hours ago






  • 1




    $begingroup$
    I don't see how enumerating the key values makes any difference for the one-time pad. Even a single bit key for a single bit message would be secure.
    $endgroup$
    – Maarten Bodewes
    3 hours ago
















1












1








1





$begingroup$


Can the security of the encryption system be improved by using double encryption with two randomly chosen keys?



By double encryption I mean using two keys K1 and K2, each 20-bit long, to obtain ciphertext $C = ( M ⊕ K_1 ) ⊕ K_2$.










share|improve this question









New contributor




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







$endgroup$




Can the security of the encryption system be improved by using double encryption with two randomly chosen keys?



By double encryption I mean using two keys K1 and K2, each 20-bit long, to obtain ciphertext $C = ( M ⊕ K_1 ) ⊕ K_2$.







one-time-pad multiple-encryption






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share|improve this question




share|improve this question








edited 4 hours ago









Ella Rose

17k44483




17k44483






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asked 6 hours ago









MinaMina

61




61




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Check out our Code of Conduct.








  • 1




    $begingroup$
    An ideal one time pad already has perfect security, you can't just get twice as perfect. The second layer doesn't add much.
    $endgroup$
    – Natanael
    5 hours ago






  • 1




    $begingroup$
    I don't see how enumerating the key values makes any difference for the one-time pad. Even a single bit key for a single bit message would be secure.
    $endgroup$
    – Maarten Bodewes
    3 hours ago
















  • 1




    $begingroup$
    An ideal one time pad already has perfect security, you can't just get twice as perfect. The second layer doesn't add much.
    $endgroup$
    – Natanael
    5 hours ago






  • 1




    $begingroup$
    I don't see how enumerating the key values makes any difference for the one-time pad. Even a single bit key for a single bit message would be secure.
    $endgroup$
    – Maarten Bodewes
    3 hours ago










1




1




$begingroup$
An ideal one time pad already has perfect security, you can't just get twice as perfect. The second layer doesn't add much.
$endgroup$
– Natanael
5 hours ago




$begingroup$
An ideal one time pad already has perfect security, you can't just get twice as perfect. The second layer doesn't add much.
$endgroup$
– Natanael
5 hours ago




1




1




$begingroup$
I don't see how enumerating the key values makes any difference for the one-time pad. Even a single bit key for a single bit message would be secure.
$endgroup$
– Maarten Bodewes
3 hours ago






$begingroup$
I don't see how enumerating the key values makes any difference for the one-time pad. Even a single bit key for a single bit message would be secure.
$endgroup$
– Maarten Bodewes
3 hours ago












3 Answers
3






active

oldest

votes


















4












$begingroup$

The answer is we cannot improve the security of the one-time pad in this manner. Intuitively the reason is that the double one-time pad is just a less efficient one time pad.



The security of the traditional xor-based one-time pad is requires that the key $K$ is chosen uniformly at random for each message and that the key is at least as large as the message.



Because the xor operation is associative, we could re-write the double one-time pad as $C = M oplus(K_1 oplus K_2)$ or $ C = M oplus K$, where $K = K_1 oplus K_2$. Now if $K_1$ and $K_2$ are uniform, it is easy to show that $K$ is also uniform. Assuming the key and the message have length $n$



$P_K[k] = sum_{x in {0,1}^n}P_{K_1,K_2}[k_1 = x, k_2 = k oplus k_1] = frac{1}{2^n}$



In other words $K_1 oplus K_2$ could be simply replaced by a single uniform key, therefore $K_1$ is enough. Moreover the double one-time pad is inefficient because it requires two xor operations.





Regarding cascade encryption, Maurer and Massey showed that cascade encryption is as strong as the first cipher. Matthew Green's blog post is a nice (easy to read) summary of multi-encryption security.






share|improve this answer











$endgroup$





















    2












    $begingroup$

    For the one-time-pad the answer is no, since it already achieves information-theoretic security (meaning that the ciphertext is statistically independent from the plaintext).
    So, applying it twice doesn’t add any extra security (and may actually worsen security if the second key is not independent from the first).






    share|improve this answer








    New contributor




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






    $endgroup$





















      0












      $begingroup$

      The intention of a good cipher is to remove all orders and bit arrangements in a plaintext and produce output ciphertext in which there is no distinguishable orders and arrangements obtainable.I mean that the ciphertext should be a uniform distribution.



      By considering this remark, if we have a good encrypted output(means plaintext which is encrypted with a good encryption algorithm ex AES),we can not find bit orders on it and re-encrypting the same output is ineffective and useless. In other words, imposing more security affections on the second encrypted ciphertext based on using the same cipher with another key is not remarkable, However we can not give a global rule for this.



      An obvious example of this cipher types is 2DES (or even 3DES). This cipher, encrypts a plaintext with 2 different keys. Again the security of ciphertext rely on the DES structure. but we should consider that the whole system(2DES or 3DES) is vulnerable to meet-in-the-middle attack and in this situation and also the key space is not the sum of keys( k1+k2).



      Therefore using ciphers in this manner is not more common. but if we want to encrypt a plaintext with two different ciphers with different keys, the whole structure gives more stamina to ciphertext against cryptanalysis.



      And about One-time pad encryption, this cipher has a perfect security, so imposing another encryption with different key is waste of time and resources.






      share|improve this answer











      $endgroup$













      • $begingroup$
        I'm not convinced that reencrypting the output of AES cannot strengthen the ciphertext in case weaknesses are found. For 2DES: even 2DES is more secure than single DES. It is not as strong as it should be given the doubling of the key size (which is actually also true for 3DES, only less so). So I'm finding myself disagreeing with a large portion of the answer (and agreeing with the last two sections).
        $endgroup$
        – Maarten Bodewes
        4 hours ago












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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

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      4












      $begingroup$

      The answer is we cannot improve the security of the one-time pad in this manner. Intuitively the reason is that the double one-time pad is just a less efficient one time pad.



      The security of the traditional xor-based one-time pad is requires that the key $K$ is chosen uniformly at random for each message and that the key is at least as large as the message.



      Because the xor operation is associative, we could re-write the double one-time pad as $C = M oplus(K_1 oplus K_2)$ or $ C = M oplus K$, where $K = K_1 oplus K_2$. Now if $K_1$ and $K_2$ are uniform, it is easy to show that $K$ is also uniform. Assuming the key and the message have length $n$



      $P_K[k] = sum_{x in {0,1}^n}P_{K_1,K_2}[k_1 = x, k_2 = k oplus k_1] = frac{1}{2^n}$



      In other words $K_1 oplus K_2$ could be simply replaced by a single uniform key, therefore $K_1$ is enough. Moreover the double one-time pad is inefficient because it requires two xor operations.





      Regarding cascade encryption, Maurer and Massey showed that cascade encryption is as strong as the first cipher. Matthew Green's blog post is a nice (easy to read) summary of multi-encryption security.






      share|improve this answer











      $endgroup$


















        4












        $begingroup$

        The answer is we cannot improve the security of the one-time pad in this manner. Intuitively the reason is that the double one-time pad is just a less efficient one time pad.



        The security of the traditional xor-based one-time pad is requires that the key $K$ is chosen uniformly at random for each message and that the key is at least as large as the message.



        Because the xor operation is associative, we could re-write the double one-time pad as $C = M oplus(K_1 oplus K_2)$ or $ C = M oplus K$, where $K = K_1 oplus K_2$. Now if $K_1$ and $K_2$ are uniform, it is easy to show that $K$ is also uniform. Assuming the key and the message have length $n$



        $P_K[k] = sum_{x in {0,1}^n}P_{K_1,K_2}[k_1 = x, k_2 = k oplus k_1] = frac{1}{2^n}$



        In other words $K_1 oplus K_2$ could be simply replaced by a single uniform key, therefore $K_1$ is enough. Moreover the double one-time pad is inefficient because it requires two xor operations.





        Regarding cascade encryption, Maurer and Massey showed that cascade encryption is as strong as the first cipher. Matthew Green's blog post is a nice (easy to read) summary of multi-encryption security.






        share|improve this answer











        $endgroup$
















          4












          4








          4





          $begingroup$

          The answer is we cannot improve the security of the one-time pad in this manner. Intuitively the reason is that the double one-time pad is just a less efficient one time pad.



          The security of the traditional xor-based one-time pad is requires that the key $K$ is chosen uniformly at random for each message and that the key is at least as large as the message.



          Because the xor operation is associative, we could re-write the double one-time pad as $C = M oplus(K_1 oplus K_2)$ or $ C = M oplus K$, where $K = K_1 oplus K_2$. Now if $K_1$ and $K_2$ are uniform, it is easy to show that $K$ is also uniform. Assuming the key and the message have length $n$



          $P_K[k] = sum_{x in {0,1}^n}P_{K_1,K_2}[k_1 = x, k_2 = k oplus k_1] = frac{1}{2^n}$



          In other words $K_1 oplus K_2$ could be simply replaced by a single uniform key, therefore $K_1$ is enough. Moreover the double one-time pad is inefficient because it requires two xor operations.





          Regarding cascade encryption, Maurer and Massey showed that cascade encryption is as strong as the first cipher. Matthew Green's blog post is a nice (easy to read) summary of multi-encryption security.






          share|improve this answer











          $endgroup$



          The answer is we cannot improve the security of the one-time pad in this manner. Intuitively the reason is that the double one-time pad is just a less efficient one time pad.



          The security of the traditional xor-based one-time pad is requires that the key $K$ is chosen uniformly at random for each message and that the key is at least as large as the message.



          Because the xor operation is associative, we could re-write the double one-time pad as $C = M oplus(K_1 oplus K_2)$ or $ C = M oplus K$, where $K = K_1 oplus K_2$. Now if $K_1$ and $K_2$ are uniform, it is easy to show that $K$ is also uniform. Assuming the key and the message have length $n$



          $P_K[k] = sum_{x in {0,1}^n}P_{K_1,K_2}[k_1 = x, k_2 = k oplus k_1] = frac{1}{2^n}$



          In other words $K_1 oplus K_2$ could be simply replaced by a single uniform key, therefore $K_1$ is enough. Moreover the double one-time pad is inefficient because it requires two xor operations.





          Regarding cascade encryption, Maurer and Massey showed that cascade encryption is as strong as the first cipher. Matthew Green's blog post is a nice (easy to read) summary of multi-encryption security.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 4 hours ago

























          answered 4 hours ago









          Marc IlungaMarc Ilunga

          37817




          37817























              2












              $begingroup$

              For the one-time-pad the answer is no, since it already achieves information-theoretic security (meaning that the ciphertext is statistically independent from the plaintext).
              So, applying it twice doesn’t add any extra security (and may actually worsen security if the second key is not independent from the first).






              share|improve this answer








              New contributor




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






              $endgroup$


















                2












                $begingroup$

                For the one-time-pad the answer is no, since it already achieves information-theoretic security (meaning that the ciphertext is statistically independent from the plaintext).
                So, applying it twice doesn’t add any extra security (and may actually worsen security if the second key is not independent from the first).






                share|improve this answer








                New contributor




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






                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  For the one-time-pad the answer is no, since it already achieves information-theoretic security (meaning that the ciphertext is statistically independent from the plaintext).
                  So, applying it twice doesn’t add any extra security (and may actually worsen security if the second key is not independent from the first).






                  share|improve this answer








                  New contributor




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






                  $endgroup$



                  For the one-time-pad the answer is no, since it already achieves information-theoretic security (meaning that the ciphertext is statistically independent from the plaintext).
                  So, applying it twice doesn’t add any extra security (and may actually worsen security if the second key is not independent from the first).







                  share|improve this answer








                  New contributor




                  guilhermemtr 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 answer



                  share|improve this answer






                  New contributor




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









                  answered 4 hours ago









                  guilhermemtrguilhermemtr

                  1214




                  1214




                  New contributor




                  guilhermemtr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                  New contributor





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






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























                      0












                      $begingroup$

                      The intention of a good cipher is to remove all orders and bit arrangements in a plaintext and produce output ciphertext in which there is no distinguishable orders and arrangements obtainable.I mean that the ciphertext should be a uniform distribution.



                      By considering this remark, if we have a good encrypted output(means plaintext which is encrypted with a good encryption algorithm ex AES),we can not find bit orders on it and re-encrypting the same output is ineffective and useless. In other words, imposing more security affections on the second encrypted ciphertext based on using the same cipher with another key is not remarkable, However we can not give a global rule for this.



                      An obvious example of this cipher types is 2DES (or even 3DES). This cipher, encrypts a plaintext with 2 different keys. Again the security of ciphertext rely on the DES structure. but we should consider that the whole system(2DES or 3DES) is vulnerable to meet-in-the-middle attack and in this situation and also the key space is not the sum of keys( k1+k2).



                      Therefore using ciphers in this manner is not more common. but if we want to encrypt a plaintext with two different ciphers with different keys, the whole structure gives more stamina to ciphertext against cryptanalysis.



                      And about One-time pad encryption, this cipher has a perfect security, so imposing another encryption with different key is waste of time and resources.






                      share|improve this answer











                      $endgroup$













                      • $begingroup$
                        I'm not convinced that reencrypting the output of AES cannot strengthen the ciphertext in case weaknesses are found. For 2DES: even 2DES is more secure than single DES. It is not as strong as it should be given the doubling of the key size (which is actually also true for 3DES, only less so). So I'm finding myself disagreeing with a large portion of the answer (and agreeing with the last two sections).
                        $endgroup$
                        – Maarten Bodewes
                        4 hours ago
















                      0












                      $begingroup$

                      The intention of a good cipher is to remove all orders and bit arrangements in a plaintext and produce output ciphertext in which there is no distinguishable orders and arrangements obtainable.I mean that the ciphertext should be a uniform distribution.



                      By considering this remark, if we have a good encrypted output(means plaintext which is encrypted with a good encryption algorithm ex AES),we can not find bit orders on it and re-encrypting the same output is ineffective and useless. In other words, imposing more security affections on the second encrypted ciphertext based on using the same cipher with another key is not remarkable, However we can not give a global rule for this.



                      An obvious example of this cipher types is 2DES (or even 3DES). This cipher, encrypts a plaintext with 2 different keys. Again the security of ciphertext rely on the DES structure. but we should consider that the whole system(2DES or 3DES) is vulnerable to meet-in-the-middle attack and in this situation and also the key space is not the sum of keys( k1+k2).



                      Therefore using ciphers in this manner is not more common. but if we want to encrypt a plaintext with two different ciphers with different keys, the whole structure gives more stamina to ciphertext against cryptanalysis.



                      And about One-time pad encryption, this cipher has a perfect security, so imposing another encryption with different key is waste of time and resources.






                      share|improve this answer











                      $endgroup$













                      • $begingroup$
                        I'm not convinced that reencrypting the output of AES cannot strengthen the ciphertext in case weaknesses are found. For 2DES: even 2DES is more secure than single DES. It is not as strong as it should be given the doubling of the key size (which is actually also true for 3DES, only less so). So I'm finding myself disagreeing with a large portion of the answer (and agreeing with the last two sections).
                        $endgroup$
                        – Maarten Bodewes
                        4 hours ago














                      0












                      0








                      0





                      $begingroup$

                      The intention of a good cipher is to remove all orders and bit arrangements in a plaintext and produce output ciphertext in which there is no distinguishable orders and arrangements obtainable.I mean that the ciphertext should be a uniform distribution.



                      By considering this remark, if we have a good encrypted output(means plaintext which is encrypted with a good encryption algorithm ex AES),we can not find bit orders on it and re-encrypting the same output is ineffective and useless. In other words, imposing more security affections on the second encrypted ciphertext based on using the same cipher with another key is not remarkable, However we can not give a global rule for this.



                      An obvious example of this cipher types is 2DES (or even 3DES). This cipher, encrypts a plaintext with 2 different keys. Again the security of ciphertext rely on the DES structure. but we should consider that the whole system(2DES or 3DES) is vulnerable to meet-in-the-middle attack and in this situation and also the key space is not the sum of keys( k1+k2).



                      Therefore using ciphers in this manner is not more common. but if we want to encrypt a plaintext with two different ciphers with different keys, the whole structure gives more stamina to ciphertext against cryptanalysis.



                      And about One-time pad encryption, this cipher has a perfect security, so imposing another encryption with different key is waste of time and resources.






                      share|improve this answer











                      $endgroup$



                      The intention of a good cipher is to remove all orders and bit arrangements in a plaintext and produce output ciphertext in which there is no distinguishable orders and arrangements obtainable.I mean that the ciphertext should be a uniform distribution.



                      By considering this remark, if we have a good encrypted output(means plaintext which is encrypted with a good encryption algorithm ex AES),we can not find bit orders on it and re-encrypting the same output is ineffective and useless. In other words, imposing more security affections on the second encrypted ciphertext based on using the same cipher with another key is not remarkable, However we can not give a global rule for this.



                      An obvious example of this cipher types is 2DES (or even 3DES). This cipher, encrypts a plaintext with 2 different keys. Again the security of ciphertext rely on the DES structure. but we should consider that the whole system(2DES or 3DES) is vulnerable to meet-in-the-middle attack and in this situation and also the key space is not the sum of keys( k1+k2).



                      Therefore using ciphers in this manner is not more common. but if we want to encrypt a plaintext with two different ciphers with different keys, the whole structure gives more stamina to ciphertext against cryptanalysis.



                      And about One-time pad encryption, this cipher has a perfect security, so imposing another encryption with different key is waste of time and resources.







                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited 5 hours ago

























                      answered 5 hours ago









                      Arsalan VahiArsalan Vahi

                      1169




                      1169












                      • $begingroup$
                        I'm not convinced that reencrypting the output of AES cannot strengthen the ciphertext in case weaknesses are found. For 2DES: even 2DES is more secure than single DES. It is not as strong as it should be given the doubling of the key size (which is actually also true for 3DES, only less so). So I'm finding myself disagreeing with a large portion of the answer (and agreeing with the last two sections).
                        $endgroup$
                        – Maarten Bodewes
                        4 hours ago


















                      • $begingroup$
                        I'm not convinced that reencrypting the output of AES cannot strengthen the ciphertext in case weaknesses are found. For 2DES: even 2DES is more secure than single DES. It is not as strong as it should be given the doubling of the key size (which is actually also true for 3DES, only less so). So I'm finding myself disagreeing with a large portion of the answer (and agreeing with the last two sections).
                        $endgroup$
                        – Maarten Bodewes
                        4 hours ago
















                      $begingroup$
                      I'm not convinced that reencrypting the output of AES cannot strengthen the ciphertext in case weaknesses are found. For 2DES: even 2DES is more secure than single DES. It is not as strong as it should be given the doubling of the key size (which is actually also true for 3DES, only less so). So I'm finding myself disagreeing with a large portion of the answer (and agreeing with the last two sections).
                      $endgroup$
                      – Maarten Bodewes
                      4 hours ago




                      $begingroup$
                      I'm not convinced that reencrypting the output of AES cannot strengthen the ciphertext in case weaknesses are found. For 2DES: even 2DES is more secure than single DES. It is not as strong as it should be given the doubling of the key size (which is actually also true for 3DES, only less so). So I'm finding myself disagreeing with a large portion of the answer (and agreeing with the last two sections).
                      $endgroup$
                      – Maarten Bodewes
                      4 hours ago










                      Mina is a new contributor. Be nice, and check out our Code of Conduct.










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