Insecure private-key encryption
$begingroup$
I am learning about crytography and the differences between the public and private keys encryption examples and I was wondering if it is possible to have a private key encryption scheme that is completely insecure? I have not been able to find what that could look like but I'm curious? Or is this not possible? Is the encryption scheme always secure with private keys?
encryption symmetric
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migrated from stackoverflow.com 7 hours ago
This question came from our site for professional and enthusiast programmers.
add a comment |
$begingroup$
I am learning about crytography and the differences between the public and private keys encryption examples and I was wondering if it is possible to have a private key encryption scheme that is completely insecure? I have not been able to find what that could look like but I'm curious? Or is this not possible? Is the encryption scheme always secure with private keys?
encryption symmetric
$endgroup$
migrated from stackoverflow.com 7 hours ago
This question came from our site for professional and enthusiast programmers.
add a comment |
$begingroup$
I am learning about crytography and the differences between the public and private keys encryption examples and I was wondering if it is possible to have a private key encryption scheme that is completely insecure? I have not been able to find what that could look like but I'm curious? Or is this not possible? Is the encryption scheme always secure with private keys?
encryption symmetric
$endgroup$
I am learning about crytography and the differences between the public and private keys encryption examples and I was wondering if it is possible to have a private key encryption scheme that is completely insecure? I have not been able to find what that could look like but I'm curious? Or is this not possible? Is the encryption scheme always secure with private keys?
encryption symmetric
encryption symmetric
edited 5 hours ago
Ilmari Karonen
35k373138
35k373138
asked 12 hours ago
droidnoob
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.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Any encryption scheme can be insecure if the key space is small enough.
For example, you could perform encryption with an 8 bit RSA key. For a key of that size, it's trivial to determine the private key given the public key.
$endgroup$
$begingroup$
would this be true even with a small message space? Or that wouldn't factor in?
$endgroup$
– droidnoob
12 hours ago
1
$begingroup$
@droidnoob If your key is insecure nothing else matters. For example given an 8 bit RSA public key with exponent 3 modulus 187, it's trival to factor the modulus into 11 and 17, calculate lamda(187) = lcm(11-1,17-1) = 80, then calculate the private exponent d = 3^-1 mod 80 = 27.
$endgroup$
– dbush
11 hours ago
$begingroup$
thanks for reply. can I ask you one more? What if the key is completely secure and protected, would there be any way to have a completely secure key but insecure scheme?
$endgroup$
– droidnoob
11 hours ago
2
$begingroup$
@droidnoob If you mean a poorly designed encryption algorithm, then yes.
$endgroup$
– dbush
11 hours ago
2
$begingroup$
You could derive a subkey in your scheme that is much smaller than the input key, e.g. by hashing it and taking the leftmost bits. The original, large key could be completely secure - protected by the cryptographic hash - but you have again an encryption scheme with a small key space. Presto.
$endgroup$
– Maarten Bodewes♦
10 hours ago
|
show 1 more comment
$begingroup$
What if the key is completely secure and protected, would there be any way to have a completely secure key but insecure scheme?
Yes, certainly. For example, consider the following encryption scheme:
The key is a 256 bit (or, heck, 512 or 1024 bit if you want) string chosen uniformly at random by a cryptographically secure true random number generator, stored securely in a locked vault deep underground, with multiple armed guards watching the entrance 24/7.
The encryption method doesn't use the key for anything, and instead encrypts the data using rot13.
The decryption method is the same as the encryption method.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Any encryption scheme can be insecure if the key space is small enough.
For example, you could perform encryption with an 8 bit RSA key. For a key of that size, it's trivial to determine the private key given the public key.
$endgroup$
$begingroup$
would this be true even with a small message space? Or that wouldn't factor in?
$endgroup$
– droidnoob
12 hours ago
1
$begingroup$
@droidnoob If your key is insecure nothing else matters. For example given an 8 bit RSA public key with exponent 3 modulus 187, it's trival to factor the modulus into 11 and 17, calculate lamda(187) = lcm(11-1,17-1) = 80, then calculate the private exponent d = 3^-1 mod 80 = 27.
$endgroup$
– dbush
11 hours ago
$begingroup$
thanks for reply. can I ask you one more? What if the key is completely secure and protected, would there be any way to have a completely secure key but insecure scheme?
$endgroup$
– droidnoob
11 hours ago
2
$begingroup$
@droidnoob If you mean a poorly designed encryption algorithm, then yes.
$endgroup$
– dbush
11 hours ago
2
$begingroup$
You could derive a subkey in your scheme that is much smaller than the input key, e.g. by hashing it and taking the leftmost bits. The original, large key could be completely secure - protected by the cryptographic hash - but you have again an encryption scheme with a small key space. Presto.
$endgroup$
– Maarten Bodewes♦
10 hours ago
|
show 1 more comment
$begingroup$
Any encryption scheme can be insecure if the key space is small enough.
For example, you could perform encryption with an 8 bit RSA key. For a key of that size, it's trivial to determine the private key given the public key.
$endgroup$
$begingroup$
would this be true even with a small message space? Or that wouldn't factor in?
$endgroup$
– droidnoob
12 hours ago
1
$begingroup$
@droidnoob If your key is insecure nothing else matters. For example given an 8 bit RSA public key with exponent 3 modulus 187, it's trival to factor the modulus into 11 and 17, calculate lamda(187) = lcm(11-1,17-1) = 80, then calculate the private exponent d = 3^-1 mod 80 = 27.
$endgroup$
– dbush
11 hours ago
$begingroup$
thanks for reply. can I ask you one more? What if the key is completely secure and protected, would there be any way to have a completely secure key but insecure scheme?
$endgroup$
– droidnoob
11 hours ago
2
$begingroup$
@droidnoob If you mean a poorly designed encryption algorithm, then yes.
$endgroup$
– dbush
11 hours ago
2
$begingroup$
You could derive a subkey in your scheme that is much smaller than the input key, e.g. by hashing it and taking the leftmost bits. The original, large key could be completely secure - protected by the cryptographic hash - but you have again an encryption scheme with a small key space. Presto.
$endgroup$
– Maarten Bodewes♦
10 hours ago
|
show 1 more comment
$begingroup$
Any encryption scheme can be insecure if the key space is small enough.
For example, you could perform encryption with an 8 bit RSA key. For a key of that size, it's trivial to determine the private key given the public key.
$endgroup$
Any encryption scheme can be insecure if the key space is small enough.
For example, you could perform encryption with an 8 bit RSA key. For a key of that size, it's trivial to determine the private key given the public key.
answered 12 hours ago
dbushdbush
22615
22615
$begingroup$
would this be true even with a small message space? Or that wouldn't factor in?
$endgroup$
– droidnoob
12 hours ago
1
$begingroup$
@droidnoob If your key is insecure nothing else matters. For example given an 8 bit RSA public key with exponent 3 modulus 187, it's trival to factor the modulus into 11 and 17, calculate lamda(187) = lcm(11-1,17-1) = 80, then calculate the private exponent d = 3^-1 mod 80 = 27.
$endgroup$
– dbush
11 hours ago
$begingroup$
thanks for reply. can I ask you one more? What if the key is completely secure and protected, would there be any way to have a completely secure key but insecure scheme?
$endgroup$
– droidnoob
11 hours ago
2
$begingroup$
@droidnoob If you mean a poorly designed encryption algorithm, then yes.
$endgroup$
– dbush
11 hours ago
2
$begingroup$
You could derive a subkey in your scheme that is much smaller than the input key, e.g. by hashing it and taking the leftmost bits. The original, large key could be completely secure - protected by the cryptographic hash - but you have again an encryption scheme with a small key space. Presto.
$endgroup$
– Maarten Bodewes♦
10 hours ago
|
show 1 more comment
$begingroup$
would this be true even with a small message space? Or that wouldn't factor in?
$endgroup$
– droidnoob
12 hours ago
1
$begingroup$
@droidnoob If your key is insecure nothing else matters. For example given an 8 bit RSA public key with exponent 3 modulus 187, it's trival to factor the modulus into 11 and 17, calculate lamda(187) = lcm(11-1,17-1) = 80, then calculate the private exponent d = 3^-1 mod 80 = 27.
$endgroup$
– dbush
11 hours ago
$begingroup$
thanks for reply. can I ask you one more? What if the key is completely secure and protected, would there be any way to have a completely secure key but insecure scheme?
$endgroup$
– droidnoob
11 hours ago
2
$begingroup$
@droidnoob If you mean a poorly designed encryption algorithm, then yes.
$endgroup$
– dbush
11 hours ago
2
$begingroup$
You could derive a subkey in your scheme that is much smaller than the input key, e.g. by hashing it and taking the leftmost bits. The original, large key could be completely secure - protected by the cryptographic hash - but you have again an encryption scheme with a small key space. Presto.
$endgroup$
– Maarten Bodewes♦
10 hours ago
$begingroup$
would this be true even with a small message space? Or that wouldn't factor in?
$endgroup$
– droidnoob
12 hours ago
$begingroup$
would this be true even with a small message space? Or that wouldn't factor in?
$endgroup$
– droidnoob
12 hours ago
1
1
$begingroup$
@droidnoob If your key is insecure nothing else matters. For example given an 8 bit RSA public key with exponent 3 modulus 187, it's trival to factor the modulus into 11 and 17, calculate lamda(187) = lcm(11-1,17-1) = 80, then calculate the private exponent d = 3^-1 mod 80 = 27.
$endgroup$
– dbush
11 hours ago
$begingroup$
@droidnoob If your key is insecure nothing else matters. For example given an 8 bit RSA public key with exponent 3 modulus 187, it's trival to factor the modulus into 11 and 17, calculate lamda(187) = lcm(11-1,17-1) = 80, then calculate the private exponent d = 3^-1 mod 80 = 27.
$endgroup$
– dbush
11 hours ago
$begingroup$
thanks for reply. can I ask you one more? What if the key is completely secure and protected, would there be any way to have a completely secure key but insecure scheme?
$endgroup$
– droidnoob
11 hours ago
$begingroup$
thanks for reply. can I ask you one more? What if the key is completely secure and protected, would there be any way to have a completely secure key but insecure scheme?
$endgroup$
– droidnoob
11 hours ago
2
2
$begingroup$
@droidnoob If you mean a poorly designed encryption algorithm, then yes.
$endgroup$
– dbush
11 hours ago
$begingroup$
@droidnoob If you mean a poorly designed encryption algorithm, then yes.
$endgroup$
– dbush
11 hours ago
2
2
$begingroup$
You could derive a subkey in your scheme that is much smaller than the input key, e.g. by hashing it and taking the leftmost bits. The original, large key could be completely secure - protected by the cryptographic hash - but you have again an encryption scheme with a small key space. Presto.
$endgroup$
– Maarten Bodewes♦
10 hours ago
$begingroup$
You could derive a subkey in your scheme that is much smaller than the input key, e.g. by hashing it and taking the leftmost bits. The original, large key could be completely secure - protected by the cryptographic hash - but you have again an encryption scheme with a small key space. Presto.
$endgroup$
– Maarten Bodewes♦
10 hours ago
|
show 1 more comment
$begingroup$
What if the key is completely secure and protected, would there be any way to have a completely secure key but insecure scheme?
Yes, certainly. For example, consider the following encryption scheme:
The key is a 256 bit (or, heck, 512 or 1024 bit if you want) string chosen uniformly at random by a cryptographically secure true random number generator, stored securely in a locked vault deep underground, with multiple armed guards watching the entrance 24/7.
The encryption method doesn't use the key for anything, and instead encrypts the data using rot13.
The decryption method is the same as the encryption method.
$endgroup$
add a comment |
$begingroup$
What if the key is completely secure and protected, would there be any way to have a completely secure key but insecure scheme?
Yes, certainly. For example, consider the following encryption scheme:
The key is a 256 bit (or, heck, 512 or 1024 bit if you want) string chosen uniformly at random by a cryptographically secure true random number generator, stored securely in a locked vault deep underground, with multiple armed guards watching the entrance 24/7.
The encryption method doesn't use the key for anything, and instead encrypts the data using rot13.
The decryption method is the same as the encryption method.
$endgroup$
add a comment |
$begingroup$
What if the key is completely secure and protected, would there be any way to have a completely secure key but insecure scheme?
Yes, certainly. For example, consider the following encryption scheme:
The key is a 256 bit (or, heck, 512 or 1024 bit if you want) string chosen uniformly at random by a cryptographically secure true random number generator, stored securely in a locked vault deep underground, with multiple armed guards watching the entrance 24/7.
The encryption method doesn't use the key for anything, and instead encrypts the data using rot13.
The decryption method is the same as the encryption method.
$endgroup$
What if the key is completely secure and protected, would there be any way to have a completely secure key but insecure scheme?
Yes, certainly. For example, consider the following encryption scheme:
The key is a 256 bit (or, heck, 512 or 1024 bit if you want) string chosen uniformly at random by a cryptographically secure true random number generator, stored securely in a locked vault deep underground, with multiple armed guards watching the entrance 24/7.
The encryption method doesn't use the key for anything, and instead encrypts the data using rot13.
The decryption method is the same as the encryption method.
answered 4 hours ago
Ilmari KaronenIlmari Karonen
35k373138
35k373138
add a comment |
add a comment |
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