Naïve RSA decryption in PythonIs my Encryption Module Secure?Encoding and decoding small strings of textC#...
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Naïve RSA decryption in Python
Is my Encryption Module Secure?Encoding and decoding small strings of textC# AES + RSA Encryption ImplementationEncryption in C#My images have secrets A.K.A. the making of aesthetic passwords V.2ISO mode encryption/decryption with BouncyCastleEncryption/decryption by matrix multiplication in CC++ RSA ImplementationGenerate two random primes and find their totientPython RSA/DSA File Cryptography, Key Generation, Key Protection
$begingroup$
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
edited 2 hours ago
200_success
130k17154419
asked 6 hours ago
Chad TChad T
211
$endgroup$
add a comment |
$begingroup$
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
edited 2 hours ago
200_success
130k17154419
asked 6 hours ago
Chad TChad T
211
$endgroup$
add a comment |
$begingroup$
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
edited 2 hours ago
200_success
130k17154419
asked 6 hours ago
Chad TChad T
211
$endgroup$
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
edited 2 hours ago
200_success
130k17154419
asked 6 hours ago
Chad TChad T
211
edited 2 hours ago
200_success
130k17154419
asked 6 hours ago
Chad TChad T
211
edited 2 hours ago
200_success
130k17154419
edited 2 hours ago
200_success
130k17154419
edited 2 hours ago
200_success
130k17154419
130k17154419
asked 6 hours ago
Chad TChad T
211
asked 6 hours ago
Chad TChad T
211
asked 6 hours ago
Chad TChad T
211
211
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered 4 hours ago
haroldharold
1,32867
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 4 hours ago
haroldharold
1,32867
$endgroup$
add a comment |
$begingroup$
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.
answered 4 hours ago
haroldharold
1,32867
$endgroup$
add a comment |
$begingroup$
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.
answered 4 hours ago
haroldharold
1,32867
$endgroup$
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.
answered 4 hours ago
haroldharold
1,32867
answered 4 hours ago
haroldharold
1,32867
answered 4 hours ago
haroldharold
1,32867
answered 4 hours ago
haroldharold
1,32867
1,32867
add a comment |
add a comment |
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