double encryption - One Time Pad The 2019 Stack Overflow Developer Survey Results Are...

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double encryption - One Time Pad

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double encryption - One Time Pad



The 2019 Stack Overflow Developer Survey Results Are In“Padless” One-time-Pad encryptionHave these compositions of block ciphers the same security?One-time pad mistake in the transmission?OTP same message encrypted twice (with different keys)Breaking One Time Pad with CCAOne-time pad using RSA and Diffie-Hellman functionsDoes sending the same message with 2 different keys in OTP leak information?One Time Pad Alphabet SizeSafely using ciphers that take small key sizesIs one time pad cipher reuse and random key secure?












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









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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
    1 hour ago






  • 1




    $begingroup$
    Hi. Is the size of the 20 part within your control? 20 bits = 1 million values. That can easily be enumerated. A hundred keys won't prevent that. And what Natanael says.
    $endgroup$
    – Paul Uszak
    1 hour 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
    19 mins 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
    1 hour ago






  • 1




    $begingroup$
    Hi. Is the size of the 20 part within your control? 20 bits = 1 million values. That can easily be enumerated. A hundred keys won't prevent that. And what Natanael says.
    $endgroup$
    – Paul Uszak
    1 hour 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
    19 mins 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






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.











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.









share|improve this question




share|improve this question








edited 58 mins ago









Ella Rose

17k44483




17k44483






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asked 3 hours ago









MinaMina

61




61




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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.






Mina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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
    1 hour ago






  • 1




    $begingroup$
    Hi. Is the size of the 20 part within your control? 20 bits = 1 million values. That can easily be enumerated. A hundred keys won't prevent that. And what Natanael says.
    $endgroup$
    – Paul Uszak
    1 hour 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
    19 mins 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
    1 hour ago






  • 1




    $begingroup$
    Hi. Is the size of the 20 part within your control? 20 bits = 1 million values. That can easily be enumerated. A hundred keys won't prevent that. And what Natanael says.
    $endgroup$
    – Paul Uszak
    1 hour 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
    19 mins 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
1 hour 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
1 hour ago




1




1




$begingroup$
Hi. Is the size of the 20 part within your control? 20 bits = 1 million values. That can easily be enumerated. A hundred keys won't prevent that. And what Natanael says.
$endgroup$
– Paul Uszak
1 hour ago




$begingroup$
Hi. Is the size of the 20 part within your control? 20 bits = 1 million values. That can easily be enumerated. A hundred keys won't prevent that. And what Natanael says.
$endgroup$
– Paul Uszak
1 hour 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
19 mins 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
19 mins ago












3 Answers
3






active

oldest

votes


















3












$begingroup$

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











$endgroup$





















    1












    $begingroup$

    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
        21 mins ago












      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      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











      $endgroup$


















        3












        $begingroup$

        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











        $endgroup$
















          3












          3








          3





          $begingroup$

          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 28 mins ago

























          answered 40 mins ago









          Marc IlungaMarc Ilunga

          36617




          36617























              1












              $begingroup$

              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.






              $endgroup$


















                1












                $begingroup$

                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.






                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  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 1 hour ago









                  guilhermemtrguilhermemtr

                  1114




                  1114




                  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
                        21 mins 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
                        21 mins 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.







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                      edited 1 hour ago

























                      answered 1 hour ago









                      Arsalan VahiArsalan Vahi

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                      • $begingroup$
                        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
                        21 mins 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
                        21 mins 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
                      21 mins 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
                      21 mins ago










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