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I like crypto. Need I say more?


Dec
16
comment Is there a general method to crack this type of fractionating cipher?
What happens if e->121 and d->21. I think your methods are good, but like any method for cryptanalysis, it is not perfect. +1 though.
Dec
16
comment Advantages of bilinear map
Compared to what?
Dec
16
comment Is there a general method to crack this type of fractionating cipher?
On your ciphertext only attack do you mean given say 154153 run the analysis on 1,5,4,1,5,3; 15,41,53; 154,153? Or do you mean to run it on 1,5,4,1,5,3; 15,54,41,14,43; 154,541,415,153? If the former, I don't think it will work.
Dec
11
revised Do test vectors ensure a cipher is free of backdoors?
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Dec
11
revised Do test vectors ensure a cipher is free of backdoors?
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Dec
11
answered Do test vectors ensure a cipher is free of backdoors?
Dec
10
answered How is an AES Encrypted Message Decrypted with Random Padding?
Dec
8
comment How does this happen in RSA malleability?
RSA works in the multiplicative group of integers $\mathbb{Z}_n$ or the integers mod n. In this group $594\equiv 34$. I'd suggest reading up on abstract algebra, groups in particular.
Dec
8
comment How does this happen in RSA malleability?
What is 594 mod 35?
Dec
8
awarded  Popular Question
Dec
5
comment Get RSA PlainText without Knowing Private Key
@aligajani you call rsa_crack with a different ciphertext each time. Where the new ciphertext was obtained by multiplying by $i^e$
Dec
5
comment Get RSA PlainText without Knowing Private Key
@AliGajani you say why isn't c2 written as c2=.m^e mod n. What is m? In writing c2 the way I did I simply wanted to show that given a ciphertext you could compute another ciphertext with is related to the original in a predictable manner. In this case it was a multiply by 2. It could have been multiply by m too and that would have been fine.
Dec
5
comment Get RSA PlainText without Knowing Private Key
@AliGajani the output of RSA-Crack() would be $m_1\cdot i$. So to get $m_1$, you'd have to divide by $i$. Remember the goal is to get $m_1$ from a function which works with 0.01 probability.
Dec
5
comment How hard is to invert the function that computes the middle-bits of (x^2)?
cleaned up some comments that were no longer needed
Dec
5
revised How hard is to invert the function that computes the middle-bits of (x^2)?
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Dec
4
reviewed Approve suggested edit on Is a simple stream cipher “partially homomorphic” if no integrity check is applied?
Dec
4
revised How hard is to invert the function that computes the middle-bits of (x^2)?
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Dec
4
revised How hard is to invert the function that computes the middle-bits of (x^2)?
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Dec
4
comment Is a simple stream cipher “partially homomorphic” if no integrity check is applied?
@Polynomial Most homomorphic ciphers allow you to add and/or multiply in plaintext values (in addition to ciphertext values). I just believe that to be called a homomorphic cipher it should securely be able to combine ciphertexts.
Dec
4
comment Is a simple stream cipher “partially homomorphic” if no integrity check is applied?
@poncho, not sure, but I sure don't want to use software that spits out multiple ciphertexts encrypted with the same keystream (whether it spits out 2 or 3 or more).