Naïve RSA decryption in Python












1












$begingroup$


I am making a code with basic RSA encryption/decryption. My professor wants me to speed up this function but it is already so simple and I am lost. Any ideas?



def decrypt(kenc,d,n):            

    kdec=(kenc**d)%n

    return kdec





python performance homework cryptography







share|improve this question











$endgroup$

















    1












    $begingroup$


    I am making a code with basic RSA encryption/decryption. My professor wants me to speed up this function but it is already so simple and I am lost. Any ideas?



    def decrypt(kenc,d,n):            
    
    kdec=(kenc**d)%n
    
    return kdec





    python performance homework cryptography







    share|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I am making a code with basic RSA encryption/decryption. My professor wants me to speed up this function but it is already so simple and I am lost. Any ideas?



      def decrypt(kenc,d,n):            
      
    kdec=(kenc**d)%n
      
    return kdec





      python performance homework cryptography







      share|improve this question











      $endgroup$




      I am making a code with basic RSA encryption/decryption. My professor wants me to speed up this function but it is already so simple and I am lost. Any ideas?



      def decrypt(kenc,d,n):            
      
    kdec=(kenc**d)%n
      
    return kdec





      python performance homework cryptography




      python performance homework cryptography






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 2 hours ago









      200_success

      130k17154419




      130k17154419










      asked 6 hours ago









      Chad TChad T

      211




      211






















          1 Answer
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          4












          $begingroup$

          Simple does not mean fast, so you cannot judge performance based on how simple the implementation looks. Usually the most efficient way to perform a non-trivial task is not also the simplest way to do it. In this case though, there is a much more efficient solution that is about equally simple, and is probably sufficient.



          There is a serious problem with this implementation: it computes kenc**d.



          kenc**d is in general a very big number that takes a long time to compute, and then it takes a long time again to reduce it modulo n. For example, trying it out with 1024bit RSA (the lowest setting!):



          import Crypto
          
from Crypto.PublicKey import RSA
          
from Crypto import Random
          

          
random_generator = Random.new().read
          
key = RSA.generate(1024, random_generator)
          

          
def decrypt(kenc,d,n):
          
  kdec=(kenc**d)%n
          
  return kdec
          

          
(ciphertext,) = key.encrypt(42, 0)
          
print(decrypt(ciphertext, key.d, key.n))


          This does not finish in a reasonable time. Estimating the size of kenc**d, it is expected to be up to (and usually close to) 1024*1024 = 1048576 bits (both kenc and d are 1024 bit numbers), that will certainly fit on a computer these days, but that's still a very big number and calculations on such large numbers take a lot of time, especially multiplication and remainder.



          There is a simple remedy: use modular exponentiation, which keeps the size of the numbers that it is working with low throughout the whole calculation by reducing modulo n as it goes along. You could implement it yourself, but Python handily provides a built-in function for this: pow(x, e, n)



          So decrypt can be written as:



          def decrypt(kenc, d, n):
          
  return pow(kenc, d, n)


          With that change, the code above decodes the message quickly.



          Further improvements are possible, but more complicated, and won't be drop-in replacements.






          share|improve this answer









          $endgroup$













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            $begingroup$

            Simple does not mean fast, so you cannot judge performance based on how simple the implementation looks. Usually the most efficient way to perform a non-trivial task is not also the simplest way to do it. In this case though, there is a much more efficient solution that is about equally simple, and is probably sufficient.



            There is a serious problem with this implementation: it computes kenc**d.



            kenc**d is in general a very big number that takes a long time to compute, and then it takes a long time again to reduce it modulo n. For example, trying it out with 1024bit RSA (the lowest setting!):



            import Crypto
            
from Crypto.PublicKey import RSA
            
from Crypto import Random
            

            
random_generator = Random.new().read
            
key = RSA.generate(1024, random_generator)
            

            
def decrypt(kenc,d,n):
            
  kdec=(kenc**d)%n
            
  return kdec
            

            
(ciphertext,) = key.encrypt(42, 0)
            
print(decrypt(ciphertext, key.d, key.n))


            This does not finish in a reasonable time. Estimating the size of kenc**d, it is expected to be up to (and usually close to) 1024*1024 = 1048576 bits (both kenc and d are 1024 bit numbers), that will certainly fit on a computer these days, but that's still a very big number and calculations on such large numbers take a lot of time, especially multiplication and remainder.



            There is a simple remedy: use modular exponentiation, which keeps the size of the numbers that it is working with low throughout the whole calculation by reducing modulo n as it goes along. You could implement it yourself, but Python handily provides a built-in function for this: pow(x, e, n)



            So decrypt can be written as:



            def decrypt(kenc, d, n):
            
  return pow(kenc, d, n)


            With that change, the code above decodes the message quickly.



            Further improvements are possible, but more complicated, and won't be drop-in replacements.






            share|improve this answer









            $endgroup$


















              4












              $begingroup$

              Simple does not mean fast, so you cannot judge performance based on how simple the implementation looks. Usually the most efficient way to perform a non-trivial task is not also the simplest way to do it. In this case though, there is a much more efficient solution that is about equally simple, and is probably sufficient.



              There is a serious problem with this implementation: it computes kenc**d.



              kenc**d is in general a very big number that takes a long time to compute, and then it takes a long time again to reduce it modulo n. For example, trying it out with 1024bit RSA (the lowest setting!):



              import Crypto
              
from Crypto.PublicKey import RSA
              
from Crypto import Random
              

              
random_generator = Random.new().read
              
key = RSA.generate(1024, random_generator)
              

              
def decrypt(kenc,d,n):
              
  kdec=(kenc**d)%n
              
  return kdec
              

              
(ciphertext,) = key.encrypt(42, 0)
              
print(decrypt(ciphertext, key.d, key.n))


              This does not finish in a reasonable time. Estimating the size of kenc**d, it is expected to be up to (and usually close to) 1024*1024 = 1048576 bits (both kenc and d are 1024 bit numbers), that will certainly fit on a computer these days, but that's still a very big number and calculations on such large numbers take a lot of time, especially multiplication and remainder.



              There is a simple remedy: use modular exponentiation, which keeps the size of the numbers that it is working with low throughout the whole calculation by reducing modulo n as it goes along. You could implement it yourself, but Python handily provides a built-in function for this: pow(x, e, n)



              So decrypt can be written as:



              def decrypt(kenc, d, n):
              
  return pow(kenc, d, n)


              With that change, the code above decodes the message quickly.



              Further improvements are possible, but more complicated, and won't be drop-in replacements.






              share|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                Simple does not mean fast, so you cannot judge performance based on how simple the implementation looks. Usually the most efficient way to perform a non-trivial task is not also the simplest way to do it. In this case though, there is a much more efficient solution that is about equally simple, and is probably sufficient.



                There is a serious problem with this implementation: it computes kenc**d.



                kenc**d is in general a very big number that takes a long time to compute, and then it takes a long time again to reduce it modulo n. For example, trying it out with 1024bit RSA (the lowest setting!):



                import Crypto
                
from Crypto.PublicKey import RSA
                
from Crypto import Random
                

                
random_generator = Random.new().read
                
key = RSA.generate(1024, random_generator)
                

                
def decrypt(kenc,d,n):
                
  kdec=(kenc**d)%n
                
  return kdec
                

                
(ciphertext,) = key.encrypt(42, 0)
                
print(decrypt(ciphertext, key.d, key.n))


                This does not finish in a reasonable time. Estimating the size of kenc**d, it is expected to be up to (and usually close to) 1024*1024 = 1048576 bits (both kenc and d are 1024 bit numbers), that will certainly fit on a computer these days, but that's still a very big number and calculations on such large numbers take a lot of time, especially multiplication and remainder.



                There is a simple remedy: use modular exponentiation, which keeps the size of the numbers that it is working with low throughout the whole calculation by reducing modulo n as it goes along. You could implement it yourself, but Python handily provides a built-in function for this: pow(x, e, n)



                So decrypt can be written as:



                def decrypt(kenc, d, n):
                
  return pow(kenc, d, n)


                With that change, the code above decodes the message quickly.



                Further improvements are possible, but more complicated, and won't be drop-in replacements.






                share|improve this answer









                $endgroup$



                Simple does not mean fast, so you cannot judge performance based on how simple the implementation looks. Usually the most efficient way to perform a non-trivial task is not also the simplest way to do it. In this case though, there is a much more efficient solution that is about equally simple, and is probably sufficient.



                There is a serious problem with this implementation: it computes kenc**d.



                kenc**d is in general a very big number that takes a long time to compute, and then it takes a long time again to reduce it modulo n. For example, trying it out with 1024bit RSA (the lowest setting!):



                import Crypto
                
from Crypto.PublicKey import RSA
                
from Crypto import Random
                

                
random_generator = Random.new().read
                
key = RSA.generate(1024, random_generator)
                

                
def decrypt(kenc,d,n):
                
  kdec=(kenc**d)%n
                
  return kdec
                

                
(ciphertext,) = key.encrypt(42, 0)
                
print(decrypt(ciphertext, key.d, key.n))


                This does not finish in a reasonable time. Estimating the size of kenc**d, it is expected to be up to (and usually close to) 1024*1024 = 1048576 bits (both kenc and d are 1024 bit numbers), that will certainly fit on a computer these days, but that's still a very big number and calculations on such large numbers take a lot of time, especially multiplication and remainder.



                There is a simple remedy: use modular exponentiation, which keeps the size of the numbers that it is working with low throughout the whole calculation by reducing modulo n as it goes along. You could implement it yourself, but Python handily provides a built-in function for this: pow(x, e, n)



                So decrypt can be written as:



                def decrypt(kenc, d, n):
                
  return pow(kenc, d, n)


                With that change, the code above decodes the message quickly.



                Further improvements are possible, but more complicated, and won't be drop-in replacements.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 4 hours ago









                haroldharold

                1,32867




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