Hill cipher: why can the cipher key matrix’s determinant not share common factors with the modulus?












0














Background



The Hill cipher works by:




  1. defining a letter-to-number substitution table/list/pattern/etc.;

  2. encoding a cypher-word into a column vector $u$ whose components are determined by the said list;

  3. multiplying $u$ with a random “key” matrix $K$ of corresponding dimension, rendering another column vector $z$;

  4. converting $z$ into a cypher-word.


Decryption works by:




  1. considering the number of characters in the list to be $m$;

  2. multiplying $z$ by $K^{-1}$ to give another column vector $w$;

  3. attaining $u$ via the relationship $uequiv wpmod m$ where all components of $u$ are in the list; and

  4. obtaining the original word using the list.


There is a caveat: In addition to $K$ needing to be invertible, the determinant of the key $det K$ must share no common factors with the modulus $m$.



Personal investigation



In order to understand why this condition is required, I created my own example in which I use the pattern



$$begin{array}{ccccccccc} text{a} & text b & text c & text d & text e & text f & text g & cdots & text z \ 0 & 1 & 2 & 3 & 4 & 5 & 6 & cdots & 25 end{array}$$



and the key $$K = pmatrix{2&1\2&2}$$ to encrypt the word $$text{ab}mapsto pmatrix{0\1}=u$$ (like the muscle), in which I know $det K=2$ shares a factor with the modulus $26$.



The cipher-word is attained by



$$z=Ku=pmatrix{2&1\2&2}pmatrix{0\1}=pmatrix{1\2}mapstotext{bc}$$



The original word is obtained by



$$K^{-1}z=pmatrix{1&-1/2\-1&1}pmatrix{1\2}=pmatrix{0\1}mapstotext{ab}$$



as originally planned. . . .



Question



At first I thought there was a possibility that my cipher working was due to the matrix components being small numbers, but after working out several additional examples, it seems that no matter what key I choose, it works flawlessly if the key matrix is invertible.



Is it even necessary that the key matrix not share factors with the modulus? If so, why?






linear-algebra matrices cryptography







share|cite|improve this question


















  • 1




    In your example, what is $-frac 12cdot ybmod{26}$ when $y$ is odd? Upon closer look, whatever you encode by multiplying with $K$, will have even second letter! This means that you reach only $26cdot 13$ code words instead of $26cdot 26$. Thus you cannot regain your originally possible $26cdot 26$ clear texts
    – Hagen von Eitzen
    Nov 21 '18 at 21:48












  • The key matrix is not invertible in the ring $mathbb{Z}_{26}$ that you choose to work in.
    – Henno Brandsma
    Nov 22 '18 at 4:45
















0














Background



The Hill cipher works by:




  1. defining a letter-to-number substitution table/list/pattern/etc.;

  2. encoding a cypher-word into a column vector $u$ whose components are determined by the said list;

  3. multiplying $u$ with a random “key” matrix $K$ of corresponding dimension, rendering another column vector $z$;

  4. converting $z$ into a cypher-word.


Decryption works by:




  1. considering the number of characters in the list to be $m$;

  2. multiplying $z$ by $K^{-1}$ to give another column vector $w$;

  3. attaining $u$ via the relationship $uequiv wpmod m$ where all components of $u$ are in the list; and

  4. obtaining the original word using the list.


There is a caveat: In addition to $K$ needing to be invertible, the determinant of the key $det K$ must share no common factors with the modulus $m$.



Personal investigation



In order to understand why this condition is required, I created my own example in which I use the pattern



$$begin{array}{ccccccccc} text{a} & text b & text c & text d & text e & text f & text g & cdots & text z \ 0 & 1 & 2 & 3 & 4 & 5 & 6 & cdots & 25 end{array}$$



and the key $$K = pmatrix{2&1\2&2}$$ to encrypt the word $$text{ab}mapsto pmatrix{0\1}=u$$ (like the muscle), in which I know $det K=2$ shares a factor with the modulus $26$.



The cipher-word is attained by



$$z=Ku=pmatrix{2&1\2&2}pmatrix{0\1}=pmatrix{1\2}mapstotext{bc}$$



The original word is obtained by



$$K^{-1}z=pmatrix{1&-1/2\-1&1}pmatrix{1\2}=pmatrix{0\1}mapstotext{ab}$$



as originally planned. . . .



Question



At first I thought there was a possibility that my cipher working was due to the matrix components being small numbers, but after working out several additional examples, it seems that no matter what key I choose, it works flawlessly if the key matrix is invertible.



Is it even necessary that the key matrix not share factors with the modulus? If so, why?






linear-algebra matrices cryptography







share|cite|improve this question


















  • 1




    In your example, what is $-frac 12cdot ybmod{26}$ when $y$ is odd? Upon closer look, whatever you encode by multiplying with $K$, will have even second letter! This means that you reach only $26cdot 13$ code words instead of $26cdot 26$. Thus you cannot regain your originally possible $26cdot 26$ clear texts
    – Hagen von Eitzen
    Nov 21 '18 at 21:48












  • The key matrix is not invertible in the ring $mathbb{Z}_{26}$ that you choose to work in.
    – Henno Brandsma
    Nov 22 '18 at 4:45














0












0








0







Background



The Hill cipher works by:




  1. defining a letter-to-number substitution table/list/pattern/etc.;

  2. encoding a cypher-word into a column vector $u$ whose components are determined by the said list;

  3. multiplying $u$ with a random “key” matrix $K$ of corresponding dimension, rendering another column vector $z$;

  4. converting $z$ into a cypher-word.


Decryption works by:




  1. considering the number of characters in the list to be $m$;

  2. multiplying $z$ by $K^{-1}$ to give another column vector $w$;

  3. attaining $u$ via the relationship $uequiv wpmod m$ where all components of $u$ are in the list; and

  4. obtaining the original word using the list.


There is a caveat: In addition to $K$ needing to be invertible, the determinant of the key $det K$ must share no common factors with the modulus $m$.



Personal investigation



In order to understand why this condition is required, I created my own example in which I use the pattern



$$begin{array}{ccccccccc} text{a} & text b & text c & text d & text e & text f & text g & cdots & text z \ 0 & 1 & 2 & 3 & 4 & 5 & 6 & cdots & 25 end{array}$$



and the key $$K = pmatrix{2&1\2&2}$$ to encrypt the word $$text{ab}mapsto pmatrix{0\1}=u$$ (like the muscle), in which I know $det K=2$ shares a factor with the modulus $26$.



The cipher-word is attained by



$$z=Ku=pmatrix{2&1\2&2}pmatrix{0\1}=pmatrix{1\2}mapstotext{bc}$$



The original word is obtained by



$$K^{-1}z=pmatrix{1&-1/2\-1&1}pmatrix{1\2}=pmatrix{0\1}mapstotext{ab}$$



as originally planned. . . .



Question



At first I thought there was a possibility that my cipher working was due to the matrix components being small numbers, but after working out several additional examples, it seems that no matter what key I choose, it works flawlessly if the key matrix is invertible.



Is it even necessary that the key matrix not share factors with the modulus? If so, why?






linear-algebra matrices cryptography







share|cite|improve this question













Background



The Hill cipher works by:




  1. defining a letter-to-number substitution table/list/pattern/etc.;

  2. encoding a cypher-word into a column vector $u$ whose components are determined by the said list;

  3. multiplying $u$ with a random “key” matrix $K$ of corresponding dimension, rendering another column vector $z$;

  4. converting $z$ into a cypher-word.


Decryption works by:




  1. considering the number of characters in the list to be $m$;

  2. multiplying $z$ by $K^{-1}$ to give another column vector $w$;

  3. attaining $u$ via the relationship $uequiv wpmod m$ where all components of $u$ are in the list; and

  4. obtaining the original word using the list.


There is a caveat: In addition to $K$ needing to be invertible, the determinant of the key $det K$ must share no common factors with the modulus $m$.



Personal investigation



In order to understand why this condition is required, I created my own example in which I use the pattern



$$begin{array}{ccccccccc} text{a} & text b & text c & text d & text e & text f & text g & cdots & text z \ 0 & 1 & 2 & 3 & 4 & 5 & 6 & cdots & 25 end{array}$$



and the key $$K = pmatrix{2&1\2&2}$$ to encrypt the word $$text{ab}mapsto pmatrix{0\1}=u$$ (like the muscle), in which I know $det K=2$ shares a factor with the modulus $26$.



The cipher-word is attained by



$$z=Ku=pmatrix{2&1\2&2}pmatrix{0\1}=pmatrix{1\2}mapstotext{bc}$$



The original word is obtained by



$$K^{-1}z=pmatrix{1&-1/2\-1&1}pmatrix{1\2}=pmatrix{0\1}mapstotext{ab}$$



as originally planned. . . .



Question



At first I thought there was a possibility that my cipher working was due to the matrix components being small numbers, but after working out several additional examples, it seems that no matter what key I choose, it works flawlessly if the key matrix is invertible.



Is it even necessary that the key matrix not share factors with the modulus? If so, why?






linear-algebra matrices cryptography




linear-algebra matrices cryptography






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 21 '18 at 21:44









Chase Ryan TaylorChase Ryan Taylor

4,38021530




4,38021530








  • 1




    In your example, what is $-frac 12cdot ybmod{26}$ when $y$ is odd? Upon closer look, whatever you encode by multiplying with $K$, will have even second letter! This means that you reach only $26cdot 13$ code words instead of $26cdot 26$. Thus you cannot regain your originally possible $26cdot 26$ clear texts
    – Hagen von Eitzen
    Nov 21 '18 at 21:48












  • The key matrix is not invertible in the ring $mathbb{Z}_{26}$ that you choose to work in.
    – Henno Brandsma
    Nov 22 '18 at 4:45














  • 1




    In your example, what is $-frac 12cdot ybmod{26}$ when $y$ is odd? Upon closer look, whatever you encode by multiplying with $K$, will have even second letter! This means that you reach only $26cdot 13$ code words instead of $26cdot 26$. Thus you cannot regain your originally possible $26cdot 26$ clear texts
    – Hagen von Eitzen
    Nov 21 '18 at 21:48












  • The key matrix is not invertible in the ring $mathbb{Z}_{26}$ that you choose to work in.
    – Henno Brandsma
    Nov 22 '18 at 4:45








1




1




In your example, what is $-frac 12cdot ybmod{26}$ when $y$ is odd? Upon closer look, whatever you encode by multiplying with $K$, will have even second letter! This means that you reach only $26cdot 13$ code words instead of $26cdot 26$. Thus you cannot regain your originally possible $26cdot 26$ clear texts
– Hagen von Eitzen
Nov 21 '18 at 21:48






In your example, what is $-frac 12cdot ybmod{26}$ when $y$ is odd? Upon closer look, whatever you encode by multiplying with $K$, will have even second letter! This means that you reach only $26cdot 13$ code words instead of $26cdot 26$. Thus you cannot regain your originally possible $26cdot 26$ clear texts
– Hagen von Eitzen
Nov 21 '18 at 21:48














The key matrix is not invertible in the ring $mathbb{Z}_{26}$ that you choose to work in.
– Henno Brandsma
Nov 22 '18 at 4:45




The key matrix is not invertible in the ring $mathbb{Z}_{26}$ that you choose to work in.
– Henno Brandsma
Nov 22 '18 at 4:45










1 Answer
1






active

oldest

votes


















1














The matrix $K^{-1}$ that you propose is only the inverse over the rational numbers, not in the ring $mathbb{Z}_{26}$ that you are working over (the characters are $0$ to $25$). The determinant is not coprime with $n$ so has no inverse in the ring. This means that you don't always get the right result with your "pseudoinverse"



To see where things go wrong concretely in your example:



$an$ is encrypted to $na$ but applying your inverse you'd get $nn$ as the decrypt.
In fact anyone receiving $na$ as a ciphertext cannot tell whether to decrypt it to $an$ or $nn$. Both plaintexts give that same ciphertext. Encryption is thus not 1-1 and hence cannot be invertible.



Other such cases (it goes wrong in half of the $26^2$ pairs):



$ao rightarrow oc rightarrow no$ (Last step is your pseudo inverse)



$es rightarrow as rightarrow rs$



$do rightarrow ui rightarrow qo$ etc. Write a program to generate all such pairs (I did).



You can see that the requirement is truly essential.






share|cite|improve this answer























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    The matrix $K^{-1}$ that you propose is only the inverse over the rational numbers, not in the ring $mathbb{Z}_{26}$ that you are working over (the characters are $0$ to $25$). The determinant is not coprime with $n$ so has no inverse in the ring. This means that you don't always get the right result with your "pseudoinverse"



    To see where things go wrong concretely in your example:



    $an$ is encrypted to $na$ but applying your inverse you'd get $nn$ as the decrypt.
    In fact anyone receiving $na$ as a ciphertext cannot tell whether to decrypt it to $an$ or $nn$. Both plaintexts give that same ciphertext. Encryption is thus not 1-1 and hence cannot be invertible.



    Other such cases (it goes wrong in half of the $26^2$ pairs):



    $ao rightarrow oc rightarrow no$ (Last step is your pseudo inverse)



    $es rightarrow as rightarrow rs$



    $do rightarrow ui rightarrow qo$ etc. Write a program to generate all such pairs (I did).



    You can see that the requirement is truly essential.






    share|cite|improve this answer




























      1














      The matrix $K^{-1}$ that you propose is only the inverse over the rational numbers, not in the ring $mathbb{Z}_{26}$ that you are working over (the characters are $0$ to $25$). The determinant is not coprime with $n$ so has no inverse in the ring. This means that you don't always get the right result with your "pseudoinverse"



      To see where things go wrong concretely in your example:



      $an$ is encrypted to $na$ but applying your inverse you'd get $nn$ as the decrypt.
      In fact anyone receiving $na$ as a ciphertext cannot tell whether to decrypt it to $an$ or $nn$. Both plaintexts give that same ciphertext. Encryption is thus not 1-1 and hence cannot be invertible.



      Other such cases (it goes wrong in half of the $26^2$ pairs):



      $ao rightarrow oc rightarrow no$ (Last step is your pseudo inverse)



      $es rightarrow as rightarrow rs$



      $do rightarrow ui rightarrow qo$ etc. Write a program to generate all such pairs (I did).



      You can see that the requirement is truly essential.






      share|cite|improve this answer


























        1












        1








        1






        The matrix $K^{-1}$ that you propose is only the inverse over the rational numbers, not in the ring $mathbb{Z}_{26}$ that you are working over (the characters are $0$ to $25$). The determinant is not coprime with $n$ so has no inverse in the ring. This means that you don't always get the right result with your "pseudoinverse"



        To see where things go wrong concretely in your example:



        $an$ is encrypted to $na$ but applying your inverse you'd get $nn$ as the decrypt.
        In fact anyone receiving $na$ as a ciphertext cannot tell whether to decrypt it to $an$ or $nn$. Both plaintexts give that same ciphertext. Encryption is thus not 1-1 and hence cannot be invertible.



        Other such cases (it goes wrong in half of the $26^2$ pairs):



        $ao rightarrow oc rightarrow no$ (Last step is your pseudo inverse)



        $es rightarrow as rightarrow rs$



        $do rightarrow ui rightarrow qo$ etc. Write a program to generate all such pairs (I did).



        You can see that the requirement is truly essential.






        share|cite|improve this answer














        The matrix $K^{-1}$ that you propose is only the inverse over the rational numbers, not in the ring $mathbb{Z}_{26}$ that you are working over (the characters are $0$ to $25$). The determinant is not coprime with $n$ so has no inverse in the ring. This means that you don't always get the right result with your "pseudoinverse"



        To see where things go wrong concretely in your example:



        $an$ is encrypted to $na$ but applying your inverse you'd get $nn$ as the decrypt.
        In fact anyone receiving $na$ as a ciphertext cannot tell whether to decrypt it to $an$ or $nn$. Both plaintexts give that same ciphertext. Encryption is thus not 1-1 and hence cannot be invertible.



        Other such cases (it goes wrong in half of the $26^2$ pairs):



        $ao rightarrow oc rightarrow no$ (Last step is your pseudo inverse)



        $es rightarrow as rightarrow rs$



        $do rightarrow ui rightarrow qo$ etc. Write a program to generate all such pairs (I did).



        You can see that the requirement is truly essential.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 23 '18 at 5:13

























        answered Nov 22 '18 at 22:32









        Henno BrandsmaHenno Brandsma

        105k347114




        105k347114






























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