Does adding complexity mean a more secure cipher? The 2019 Stack Overflow Developer Survey Results Are InEasy explanation of “IND-” security notions?Is this hand cipher any more secure than the Vigenère cipher?What does ''latency'' really mean when a block cipher is partially unrolled?RSA: how does it work and how is it more secure than symmetric systemsComplexity of attacks on affine cipherIs my protocol that uses hybrid cryptography and AES-GCM secure?What does 0…0 and 1…1 meanDoes AAD make GCM encryption more secure?What does (block cipher) decryption parallelizable mean?What does “$<!!<!!<$” mean?What does fullstop mean in this context?

Is it possible for absolutely everyone to attain enlightenment?

How to obtain a position of last non-zero element

Why doesn't shell automatically fix "useless use of cat"?

What information about me do stores get via my credit card?

Are spiders unable to hurt humans, especially very small spiders?

Dropping list elements from nested list after evaluation

What could be the right powersource for 15 seconds lifespan disposable giant chainsaw?

Can we generate random numbers using irrational numbers like π and e?

How do PCB vias affect signal quality?

How much of the clove should I use when using big garlic heads?

Worn-tile Scrabble

Cooking pasta in a water boiler

Button changing its text & action. Good or terrible?

Can there be female White Walkers?

How can I define good in a religion that claims no moral authority?

What's the name of these plastic connectors

Did the UK government pay "millions and millions of dollars" to try to snag Julian Assange?

How can I add encounters in the Lost Mine of Phandelver campaign without giving PCs too much XP?

Is it correct to say the Neural Networks are an alternative way of performing Maximum Likelihood Estimation? if not, why?

If I score a critical hit on an 18 or higher, what are my chances of getting a critical hit if I roll 3d20?

Why didn't the Event Horizon Telescope team mention Sagittarius A*?

Get name of standard action overriden in Visualforce contorller

Inverse Relationship Between Precision and Recall

Why don't hard Brexiteers insist on a hard border to prevent illegal immigration after Brexit?



Does adding complexity mean a more secure cipher?



The 2019 Stack Overflow Developer Survey Results Are InEasy explanation of “IND-” security notions?Is this hand cipher any more secure than the Vigenère cipher?What does ''latency'' really mean when a block cipher is partially unrolled?RSA: how does it work and how is it more secure than symmetric systemsComplexity of attacks on affine cipherIs my protocol that uses hybrid cryptography and AES-GCM secure?What does 0…0 and 1…1 meanDoes AAD make GCM encryption more secure?What does (block cipher) decryption parallelizable mean?What does “$<!!<!!<$” mean?What does fullstop mean in this context?










4












$begingroup$


I have a cryptography workshop question I'm having trouble with as follows;



Person A creates a cipher $E_k(m)$ which produces a ciphertext from message "m" using key "k". The function inside E is kept secret but the length of $E_K(m)$ is known.



Person B recommends "increasing" security of the cipher by instead doing :



$(E_k(m) oplus m) || (E_k(m) oplus 1111...11)$



Does this in fact increase security of the cipher or increase new problems.



My thinking is, depending on the function within E, xoring the output of the cipher with the plaintext message could expose the key, meaning the extra complexity is for nothing. Am I on the right track, or am I missing something?



I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.



Any steers in the right direction would be greatly appreciated, I'm more than happy to do the research myself just unsure what specifically to look for.



Unfortunately the above context is all I have been provided for this question.










share|improve this question











$endgroup$



migrated from stackoverflow.com 7 hours ago


This question came from our site for professional and enthusiast programmers.






















    4












    $begingroup$


    I have a cryptography workshop question I'm having trouble with as follows;



    Person A creates a cipher $E_k(m)$ which produces a ciphertext from message "m" using key "k". The function inside E is kept secret but the length of $E_K(m)$ is known.



    Person B recommends "increasing" security of the cipher by instead doing :



    $(E_k(m) oplus m) || (E_k(m) oplus 1111...11)$



    Does this in fact increase security of the cipher or increase new problems.



    My thinking is, depending on the function within E, xoring the output of the cipher with the plaintext message could expose the key, meaning the extra complexity is for nothing. Am I on the right track, or am I missing something?



    I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.



    Any steers in the right direction would be greatly appreciated, I'm more than happy to do the research myself just unsure what specifically to look for.



    Unfortunately the above context is all I have been provided for this question.










    share|improve this question











    $endgroup$



    migrated from stackoverflow.com 7 hours ago


    This question came from our site for professional and enthusiast programmers.




















      4












      4








      4





      $begingroup$


      I have a cryptography workshop question I'm having trouble with as follows;



      Person A creates a cipher $E_k(m)$ which produces a ciphertext from message "m" using key "k". The function inside E is kept secret but the length of $E_K(m)$ is known.



      Person B recommends "increasing" security of the cipher by instead doing :



      $(E_k(m) oplus m) || (E_k(m) oplus 1111...11)$



      Does this in fact increase security of the cipher or increase new problems.



      My thinking is, depending on the function within E, xoring the output of the cipher with the plaintext message could expose the key, meaning the extra complexity is for nothing. Am I on the right track, or am I missing something?



      I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.



      Any steers in the right direction would be greatly appreciated, I'm more than happy to do the research myself just unsure what specifically to look for.



      Unfortunately the above context is all I have been provided for this question.










      share|improve this question











      $endgroup$




      I have a cryptography workshop question I'm having trouble with as follows;



      Person A creates a cipher $E_k(m)$ which produces a ciphertext from message "m" using key "k". The function inside E is kept secret but the length of $E_K(m)$ is known.



      Person B recommends "increasing" security of the cipher by instead doing :



      $(E_k(m) oplus m) || (E_k(m) oplus 1111...11)$



      Does this in fact increase security of the cipher or increase new problems.



      My thinking is, depending on the function within E, xoring the output of the cipher with the plaintext message could expose the key, meaning the extra complexity is for nothing. Am I on the right track, or am I missing something?



      I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.



      Any steers in the right direction would be greatly appreciated, I'm more than happy to do the research myself just unsure what specifically to look for.



      Unfortunately the above context is all I have been provided for this question.







      encryption






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 6 hours ago









      Ella Rose

      17k44483




      17k44483










      asked 7 hours ago









      melloncolliemelloncollie

      235




      235




      migrated from stackoverflow.com 7 hours ago


      This question came from our site for professional and enthusiast programmers.









      migrated from stackoverflow.com 7 hours ago


      This question came from our site for professional and enthusiast programmers.






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$


          xoring the output of the cipher with the plaintext message




          Xoring the message into the ciphertext removes the ability to decrypt the ciphertext.



          If all you have is $k, c = E_k(m) oplus m$, then you need to know $m$ in order to strip the external $m$ off of $E_k(m)$ before you can apply $m = D_k(E_k(m))$; Basically, you would need to know the message in order to "decrypt" the message, but since you already know the message, then there would be no knowledge gained from "decrypting".



          $(E_k(m)oplus m)||(E_k(m) oplus 1111...11)$



          The previous section was striked out, because there was more to the suggestion than simply xoring the message into the ciphertext.



          In fact, the complete suggestion is far, far worse then simply implying the inability to decrypt a ciphertext: Anyone can decrypt a ciphertext from this scheme without requiring the key.



          $$c = (E_k(m) oplus m) || E_k(m) oplus 1111dots 11)\c_texta = E_k(m) oplus m\c_textb = E_k(m) oplus 1111 dots 11\c' = c_textb oplus 1111dots11\m = c_texta oplus c'$$



          The value $1111dots11$ is known to all, so anyone can compute $$E_k(m) = E_k(m) oplus 1111dots11 oplus 1111dots11$$



          So $E_k(m)$ is effectively public knowledge, so again anyone can compute $$m = E_k(m) oplus m oplus E_k(m)$$



          This scheme is completely broken.




          I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.




          See the previous section - "encryption" is an invertible process: You have to be able to undo the transformation so that you can retrieve the plaintext from any given ciphertext.



          It is completely broken, so you won't find anything similar to it (other than situations where people asked this same question and found it to be broken)




          Does this in fact increase security of the cipher or increase new problems.




          It's a lot easier to determine whether or not doing x or y will increase/decrease security once you have a clear goal of what it means to be secure. This is a context dependent notion.



          It helps to list:



          • What you have

          • What your adversary can do

          • What you want to accomplish (in very specific terms).

          If you don't know where to start, then look for the standard notions of security that cryptographers have already worked out for the context of interest (e.g. indistinguishability for ciphers)






          share|improve this answer











          $endgroup$








          • 1




            $begingroup$
            In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
            $endgroup$
            – Marc Ilunga
            6 hours ago











          • $begingroup$
            @MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
            $endgroup$
            – Ella Rose
            6 hours ago










          • $begingroup$
            Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
            $endgroup$
            – melloncollie
            6 hours ago


















          3












          $begingroup$

          This is indeed a example of complexity not adding security and actually weakening it.



          The second encryption can be written as $c = c_1|| c_2$, where $c_1 = E_k(m) oplus m$ and $c_2 = E_k(m) oplus 111ldots11$.



          Now observe that $m' = c_1 oplus c_2 = m oplus 111ldots11$.
          And we can easily get $m$ as $m' oplus 111ldots 11$






          share|improve this answer









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function ()
            return StackExchange.using("mathjaxEditing", function ()
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            );
            );
            , "mathjax-editing");

            StackExchange.ready(function()
            var channelOptions =
            tags: "".split(" "),
            id: "281"
            ;
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function()
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled)
            StackExchange.using("snippets", function()
            createEditor();
            );

            else
            createEditor();

            );

            function createEditor()
            StackExchange.prepareEditor(
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: false,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: null,
            bindNavPrevention: true,
            postfix: "",
            imageUploader:
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            ,
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            );



            );













            draft saved

            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcrypto.stackexchange.com%2fquestions%2f68705%2fdoes-adding-complexity-mean-a-more-secure-cipher%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$


            xoring the output of the cipher with the plaintext message




            Xoring the message into the ciphertext removes the ability to decrypt the ciphertext.



            If all you have is $k, c = E_k(m) oplus m$, then you need to know $m$ in order to strip the external $m$ off of $E_k(m)$ before you can apply $m = D_k(E_k(m))$; Basically, you would need to know the message in order to "decrypt" the message, but since you already know the message, then there would be no knowledge gained from "decrypting".



            $(E_k(m)oplus m)||(E_k(m) oplus 1111...11)$



            The previous section was striked out, because there was more to the suggestion than simply xoring the message into the ciphertext.



            In fact, the complete suggestion is far, far worse then simply implying the inability to decrypt a ciphertext: Anyone can decrypt a ciphertext from this scheme without requiring the key.



            $$c = (E_k(m) oplus m) || E_k(m) oplus 1111dots 11)\c_texta = E_k(m) oplus m\c_textb = E_k(m) oplus 1111 dots 11\c' = c_textb oplus 1111dots11\m = c_texta oplus c'$$



            The value $1111dots11$ is known to all, so anyone can compute $$E_k(m) = E_k(m) oplus 1111dots11 oplus 1111dots11$$



            So $E_k(m)$ is effectively public knowledge, so again anyone can compute $$m = E_k(m) oplus m oplus E_k(m)$$



            This scheme is completely broken.




            I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.




            See the previous section - "encryption" is an invertible process: You have to be able to undo the transformation so that you can retrieve the plaintext from any given ciphertext.



            It is completely broken, so you won't find anything similar to it (other than situations where people asked this same question and found it to be broken)




            Does this in fact increase security of the cipher or increase new problems.




            It's a lot easier to determine whether or not doing x or y will increase/decrease security once you have a clear goal of what it means to be secure. This is a context dependent notion.



            It helps to list:



            • What you have

            • What your adversary can do

            • What you want to accomplish (in very specific terms).

            If you don't know where to start, then look for the standard notions of security that cryptographers have already worked out for the context of interest (e.g. indistinguishability for ciphers)






            share|improve this answer











            $endgroup$








            • 1




              $begingroup$
              In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
              $endgroup$
              – Marc Ilunga
              6 hours ago











            • $begingroup$
              @MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
              $endgroup$
              – Ella Rose
              6 hours ago










            • $begingroup$
              Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
              $endgroup$
              – melloncollie
              6 hours ago















            3












            $begingroup$


            xoring the output of the cipher with the plaintext message




            Xoring the message into the ciphertext removes the ability to decrypt the ciphertext.



            If all you have is $k, c = E_k(m) oplus m$, then you need to know $m$ in order to strip the external $m$ off of $E_k(m)$ before you can apply $m = D_k(E_k(m))$; Basically, you would need to know the message in order to "decrypt" the message, but since you already know the message, then there would be no knowledge gained from "decrypting".



            $(E_k(m)oplus m)||(E_k(m) oplus 1111...11)$



            The previous section was striked out, because there was more to the suggestion than simply xoring the message into the ciphertext.



            In fact, the complete suggestion is far, far worse then simply implying the inability to decrypt a ciphertext: Anyone can decrypt a ciphertext from this scheme without requiring the key.



            $$c = (E_k(m) oplus m) || E_k(m) oplus 1111dots 11)\c_texta = E_k(m) oplus m\c_textb = E_k(m) oplus 1111 dots 11\c' = c_textb oplus 1111dots11\m = c_texta oplus c'$$



            The value $1111dots11$ is known to all, so anyone can compute $$E_k(m) = E_k(m) oplus 1111dots11 oplus 1111dots11$$



            So $E_k(m)$ is effectively public knowledge, so again anyone can compute $$m = E_k(m) oplus m oplus E_k(m)$$



            This scheme is completely broken.




            I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.




            See the previous section - "encryption" is an invertible process: You have to be able to undo the transformation so that you can retrieve the plaintext from any given ciphertext.



            It is completely broken, so you won't find anything similar to it (other than situations where people asked this same question and found it to be broken)




            Does this in fact increase security of the cipher or increase new problems.




            It's a lot easier to determine whether or not doing x or y will increase/decrease security once you have a clear goal of what it means to be secure. This is a context dependent notion.



            It helps to list:



            • What you have

            • What your adversary can do

            • What you want to accomplish (in very specific terms).

            If you don't know where to start, then look for the standard notions of security that cryptographers have already worked out for the context of interest (e.g. indistinguishability for ciphers)






            share|improve this answer











            $endgroup$








            • 1




              $begingroup$
              In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
              $endgroup$
              – Marc Ilunga
              6 hours ago











            • $begingroup$
              @MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
              $endgroup$
              – Ella Rose
              6 hours ago










            • $begingroup$
              Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
              $endgroup$
              – melloncollie
              6 hours ago













            3












            3








            3





            $begingroup$


            xoring the output of the cipher with the plaintext message




            Xoring the message into the ciphertext removes the ability to decrypt the ciphertext.



            If all you have is $k, c = E_k(m) oplus m$, then you need to know $m$ in order to strip the external $m$ off of $E_k(m)$ before you can apply $m = D_k(E_k(m))$; Basically, you would need to know the message in order to "decrypt" the message, but since you already know the message, then there would be no knowledge gained from "decrypting".



            $(E_k(m)oplus m)||(E_k(m) oplus 1111...11)$



            The previous section was striked out, because there was more to the suggestion than simply xoring the message into the ciphertext.



            In fact, the complete suggestion is far, far worse then simply implying the inability to decrypt a ciphertext: Anyone can decrypt a ciphertext from this scheme without requiring the key.



            $$c = (E_k(m) oplus m) || E_k(m) oplus 1111dots 11)\c_texta = E_k(m) oplus m\c_textb = E_k(m) oplus 1111 dots 11\c' = c_textb oplus 1111dots11\m = c_texta oplus c'$$



            The value $1111dots11$ is known to all, so anyone can compute $$E_k(m) = E_k(m) oplus 1111dots11 oplus 1111dots11$$



            So $E_k(m)$ is effectively public knowledge, so again anyone can compute $$m = E_k(m) oplus m oplus E_k(m)$$



            This scheme is completely broken.




            I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.




            See the previous section - "encryption" is an invertible process: You have to be able to undo the transformation so that you can retrieve the plaintext from any given ciphertext.



            It is completely broken, so you won't find anything similar to it (other than situations where people asked this same question and found it to be broken)




            Does this in fact increase security of the cipher or increase new problems.




            It's a lot easier to determine whether or not doing x or y will increase/decrease security once you have a clear goal of what it means to be secure. This is a context dependent notion.



            It helps to list:



            • What you have

            • What your adversary can do

            • What you want to accomplish (in very specific terms).

            If you don't know where to start, then look for the standard notions of security that cryptographers have already worked out for the context of interest (e.g. indistinguishability for ciphers)






            share|improve this answer











            $endgroup$




            xoring the output of the cipher with the plaintext message




            Xoring the message into the ciphertext removes the ability to decrypt the ciphertext.



            If all you have is $k, c = E_k(m) oplus m$, then you need to know $m$ in order to strip the external $m$ off of $E_k(m)$ before you can apply $m = D_k(E_k(m))$; Basically, you would need to know the message in order to "decrypt" the message, but since you already know the message, then there would be no knowledge gained from "decrypting".



            $(E_k(m)oplus m)||(E_k(m) oplus 1111...11)$



            The previous section was striked out, because there was more to the suggestion than simply xoring the message into the ciphertext.



            In fact, the complete suggestion is far, far worse then simply implying the inability to decrypt a ciphertext: Anyone can decrypt a ciphertext from this scheme without requiring the key.



            $$c = (E_k(m) oplus m) || E_k(m) oplus 1111dots 11)\c_texta = E_k(m) oplus m\c_textb = E_k(m) oplus 1111 dots 11\c' = c_textb oplus 1111dots11\m = c_texta oplus c'$$



            The value $1111dots11$ is known to all, so anyone can compute $$E_k(m) = E_k(m) oplus 1111dots11 oplus 1111dots11$$



            So $E_k(m)$ is effectively public knowledge, so again anyone can compute $$m = E_k(m) oplus m oplus E_k(m)$$



            This scheme is completely broken.




            I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.




            See the previous section - "encryption" is an invertible process: You have to be able to undo the transformation so that you can retrieve the plaintext from any given ciphertext.



            It is completely broken, so you won't find anything similar to it (other than situations where people asked this same question and found it to be broken)




            Does this in fact increase security of the cipher or increase new problems.




            It's a lot easier to determine whether or not doing x or y will increase/decrease security once you have a clear goal of what it means to be secure. This is a context dependent notion.



            It helps to list:



            • What you have

            • What your adversary can do

            • What you want to accomplish (in very specific terms).

            If you don't know where to start, then look for the standard notions of security that cryptographers have already worked out for the context of interest (e.g. indistinguishability for ciphers)







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 6 hours ago

























            answered 6 hours ago









            Ella RoseElla Rose

            17k44483




            17k44483







            • 1




              $begingroup$
              In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
              $endgroup$
              – Marc Ilunga
              6 hours ago











            • $begingroup$
              @MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
              $endgroup$
              – Ella Rose
              6 hours ago










            • $begingroup$
              Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
              $endgroup$
              – melloncollie
              6 hours ago












            • 1




              $begingroup$
              In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
              $endgroup$
              – Marc Ilunga
              6 hours ago











            • $begingroup$
              @MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
              $endgroup$
              – Ella Rose
              6 hours ago










            • $begingroup$
              Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
              $endgroup$
              – melloncollie
              6 hours ago







            1




            1




            $begingroup$
            In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
            $endgroup$
            – Marc Ilunga
            6 hours ago





            $begingroup$
            In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
            $endgroup$
            – Marc Ilunga
            6 hours ago













            $begingroup$
            @MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
            $endgroup$
            – Ella Rose
            6 hours ago




            $begingroup$
            @MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
            $endgroup$
            – Ella Rose
            6 hours ago












            $begingroup$
            Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
            $endgroup$
            – melloncollie
            6 hours ago




            $begingroup$
            Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
            $endgroup$
            – melloncollie
            6 hours ago











            3












            $begingroup$

            This is indeed a example of complexity not adding security and actually weakening it.



            The second encryption can be written as $c = c_1|| c_2$, where $c_1 = E_k(m) oplus m$ and $c_2 = E_k(m) oplus 111ldots11$.



            Now observe that $m' = c_1 oplus c_2 = m oplus 111ldots11$.
            And we can easily get $m$ as $m' oplus 111ldots 11$






            share|improve this answer









            $endgroup$

















              3












              $begingroup$

              This is indeed a example of complexity not adding security and actually weakening it.



              The second encryption can be written as $c = c_1|| c_2$, where $c_1 = E_k(m) oplus m$ and $c_2 = E_k(m) oplus 111ldots11$.



              Now observe that $m' = c_1 oplus c_2 = m oplus 111ldots11$.
              And we can easily get $m$ as $m' oplus 111ldots 11$






              share|improve this answer









              $endgroup$















                3












                3








                3





                $begingroup$

                This is indeed a example of complexity not adding security and actually weakening it.



                The second encryption can be written as $c = c_1|| c_2$, where $c_1 = E_k(m) oplus m$ and $c_2 = E_k(m) oplus 111ldots11$.



                Now observe that $m' = c_1 oplus c_2 = m oplus 111ldots11$.
                And we can easily get $m$ as $m' oplus 111ldots 11$






                share|improve this answer









                $endgroup$



                This is indeed a example of complexity not adding security and actually weakening it.



                The second encryption can be written as $c = c_1|| c_2$, where $c_1 = E_k(m) oplus m$ and $c_2 = E_k(m) oplus 111ldots11$.



                Now observe that $m' = c_1 oplus c_2 = m oplus 111ldots11$.
                And we can easily get $m$ as $m' oplus 111ldots 11$







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 6 hours ago









                Marc IlungaMarc Ilunga

                40817




                40817



























                    draft saved

                    draft discarded
















































                    Thanks for contributing an answer to Cryptography Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid


                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.

                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcrypto.stackexchange.com%2fquestions%2f68705%2fdoes-adding-complexity-mean-a-more-secure-cipher%23new-answer', 'question_page');

                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Bett Inhaltsverzeichnis Geschichte | Bettformen | Bettgrößen | Andere Bezeichnungen | Bettenmangel | Betten in der bildenden Kunst | Schlafmedizinische Gesichtspunkte | Siehe auch | Literatur | Weblinks | Einzelnachweise | NavigationsmenüBett, Bettstatt, BettstelleCommons: BettBabybetten: Anwendung, Ausstattungsmerkmale und VergleichskriterienWasserbetten. Vorurteile im TestHapfnNursch10.1007/s11818-012-0584-74006250-8AKS4329276-8

                    Luksemburg Sisukord Nimi | Asend | Loodus | Riigikord | Haldusjaotus | Rahvastik | Riigikaitse | Majandus | Taristu | Ajalugu | Eesti ja Luksemburgi suhted | Haridus | Kultuur | Vaata ka | Viited | Välislingid | Navigeerimismenüü50° N, 6° EÜlevaade Luksemburgi kaitsealadest.Luksemburgi rahvaarv. Statistikaamet.World Bank'i andmebaasÜlevaade Luksemburgi loodusest.Ülevaade Luksemburgi metsadest.Guy Colling. "Red List of the Vascular Plants of Luxembourg." Travaux scientifiques du Musée national d’histoire naturelle Luxembourg. 2005.Luxembourg’s biodiversity at risk.Maailma kahepaiksete andmebaas.Denis Lepage. "Luxembourg." Avibase.Ülevaade temperatuuridest. Luksemburgi meteoroloogiateenistus.Ülevaade Luksemburgist. Euroopa Liidu esinduse koduleht.Système politique. TerritoireÜlevaade Luksemburgi rahvastikust. Luksemburgi statistikaamet.Luksemburgi rahvastik. Luksemburgi statistikaamet.The World FactbookMonique Borsenberger, Paul Dickes. "Religions au Luxembourg. Quelle évolution entre 1999-2008". Luksemburgi statistikaamet. 2011.Luksemburgi peapiiskopkond. Catholic-Hierarchy.Luksemburgi armee koduleht.Luksemburgi armee relvastus.Eesti Välisministeerium.Luksemburgi rahvastik. Luksemburgi statistikaamet.Luksemburgi Eesti Seltsi koduleht.Helen Eelrand. "Raadio, mis muutis maailma." Eesti Päevaleht. 13. märts 2004.Ülevaade Luksemburgi haridussüsteemist.Ülevaade Luksemburgi keskkoolidest.Luksemburgr

                    Valle di Casies Indice Geografia fisica | Origini del nome | Storia | Società | Amministrazione | Sport | Note | Bibliografia | Voci correlate | Altri progetti | Collegamenti esterni | Menu di navigazione46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)Sito istituzionaleAstat Censimento della popolazione 2011 - Determinazione della consistenza dei tre gruppi linguistici della Provincia Autonoma di Bolzano-Alto Adige - giugno 2012Numeri e fattiValle di CasiesDato IstatTabella dei gradi/giorno dei Comuni italiani raggruppati per Regione e Provincia26 agosto 1993, n. 412Heraldry of the World: GsiesStatistiche I.StatValCasies.comWikimedia CommonsWikimedia CommonsValle di CasiesSito ufficialeValle di CasiesMM14870458910042978-6