Naïve RSA decryption in Python












3












$begingroup$


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




    $begingroup$
    You can trivially make the code simpler/shorter and also (minimally) more efficient by removing the unnecessary variable assignment. But of course this isn’t what your professor meant.
    $endgroup$
    – Konrad Rudolph
    2 days ago


















3












$begingroup$


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




    $begingroup$
    You can trivially make the code simpler/shorter and also (minimally) more efficient by removing the unnecessary variable assignment. But of course this isn’t what your professor meant.
    $endgroup$
    – Konrad Rudolph
    2 days ago
















3












3








3





$begingroup$


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 days ago









200_success

130k17154419




130k17154419










asked 2 days ago









Chad TChad T

311




311








  • 1




    $begingroup$
    You can trivially make the code simpler/shorter and also (minimally) more efficient by removing the unnecessary variable assignment. But of course this isn’t what your professor meant.
    $endgroup$
    – Konrad Rudolph
    2 days ago
















  • 1




    $begingroup$
    You can trivially make the code simpler/shorter and also (minimally) more efficient by removing the unnecessary variable assignment. But of course this isn’t what your professor meant.
    $endgroup$
    – Konrad Rudolph
    2 days ago










1




1




$begingroup$
You can trivially make the code simpler/shorter and also (minimally) more efficient by removing the unnecessary variable assignment. But of course this isn’t what your professor meant.
$endgroup$
– Konrad Rudolph
2 days ago






$begingroup$
You can trivially make the code simpler/shorter and also (minimally) more efficient by removing the unnecessary variable assignment. But of course this isn’t what your professor meant.
$endgroup$
– Konrad Rudolph
2 days ago












1 Answer
1






active

oldest

votes


















12












$begingroup$

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









$endgroup$









  • 1




    $begingroup$
    And of course, it's even faster not to have decrypt at all when all it does is call pow.
    $endgroup$
    – MSalters
    2 days ago










  • $begingroup$
    Keeping decrypt would be a good idea. If you were actually implementing RSA, you would also want PKCS or something there, and the performance penalty of a func call will be small compared to the time to call pow.
    $endgroup$
    – Oscar Smith
    2 days ago






  • 1




    $begingroup$
    Also note that pow probably doesn't exhibit timing independent of base and exponent ("constant-time") allowing for timing side-channel attacks which one may want / need to defend against.
    $endgroup$
    – SEJPM
    2 days ago











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1 Answer
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1 Answer
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active

oldest

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12












$begingroup$

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









$endgroup$









  • 1




    $begingroup$
    And of course, it's even faster not to have decrypt at all when all it does is call pow.
    $endgroup$
    – MSalters
    2 days ago










  • $begingroup$
    Keeping decrypt would be a good idea. If you were actually implementing RSA, you would also want PKCS or something there, and the performance penalty of a func call will be small compared to the time to call pow.
    $endgroup$
    – Oscar Smith
    2 days ago






  • 1




    $begingroup$
    Also note that pow probably doesn't exhibit timing independent of base and exponent ("constant-time") allowing for timing side-channel attacks which one may want / need to defend against.
    $endgroup$
    – SEJPM
    2 days ago
















12












$begingroup$

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









$endgroup$









  • 1




    $begingroup$
    And of course, it's even faster not to have decrypt at all when all it does is call pow.
    $endgroup$
    – MSalters
    2 days ago










  • $begingroup$
    Keeping decrypt would be a good idea. If you were actually implementing RSA, you would also want PKCS or something there, and the performance penalty of a func call will be small compared to the time to call pow.
    $endgroup$
    – Oscar Smith
    2 days ago






  • 1




    $begingroup$
    Also note that pow probably doesn't exhibit timing independent of base and exponent ("constant-time") allowing for timing side-channel attacks which one may want / need to defend against.
    $endgroup$
    – SEJPM
    2 days ago














12












12








12





$begingroup$

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 2 days ago









haroldharold

1,40868




1,40868








  • 1




    $begingroup$
    And of course, it's even faster not to have decrypt at all when all it does is call pow.
    $endgroup$
    – MSalters
    2 days ago










  • $begingroup$
    Keeping decrypt would be a good idea. If you were actually implementing RSA, you would also want PKCS or something there, and the performance penalty of a func call will be small compared to the time to call pow.
    $endgroup$
    – Oscar Smith
    2 days ago






  • 1




    $begingroup$
    Also note that pow probably doesn't exhibit timing independent of base and exponent ("constant-time") allowing for timing side-channel attacks which one may want / need to defend against.
    $endgroup$
    – SEJPM
    2 days ago














  • 1




    $begingroup$
    And of course, it's even faster not to have decrypt at all when all it does is call pow.
    $endgroup$
    – MSalters
    2 days ago










  • $begingroup$
    Keeping decrypt would be a good idea. If you were actually implementing RSA, you would also want PKCS or something there, and the performance penalty of a func call will be small compared to the time to call pow.
    $endgroup$
    – Oscar Smith
    2 days ago






  • 1




    $begingroup$
    Also note that pow probably doesn't exhibit timing independent of base and exponent ("constant-time") allowing for timing side-channel attacks which one may want / need to defend against.
    $endgroup$
    – SEJPM
    2 days ago








1




1




$begingroup$
And of course, it's even faster not to have decrypt at all when all it does is call pow.
$endgroup$
– MSalters
2 days ago




$begingroup$
And of course, it's even faster not to have decrypt at all when all it does is call pow.
$endgroup$
– MSalters
2 days ago












$begingroup$
Keeping decrypt would be a good idea. If you were actually implementing RSA, you would also want PKCS or something there, and the performance penalty of a func call will be small compared to the time to call pow.
$endgroup$
– Oscar Smith
2 days ago




$begingroup$
Keeping decrypt would be a good idea. If you were actually implementing RSA, you would also want PKCS or something there, and the performance penalty of a func call will be small compared to the time to call pow.
$endgroup$
– Oscar Smith
2 days ago




1




1




$begingroup$
Also note that pow probably doesn't exhibit timing independent of base and exponent ("constant-time") allowing for timing side-channel attacks which one may want / need to defend against.
$endgroup$
– SEJPM
2 days ago




$begingroup$
Also note that pow probably doesn't exhibit timing independent of base and exponent ("constant-time") allowing for timing side-channel attacks which one may want / need to defend against.
$endgroup$
– SEJPM
2 days ago


















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