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Results tagged with blinding
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Technique mainly employed for factorization based cryptosystems to hide information. Usually blinding involves using a multiplicative homomorphic property of the cipher, it may be used to realize blind-signatures. It's also commonly used to protect cryptosystems like RSA and Rabin-Williams against timing-attacks.
4
votes
Is multiplicative blinding less secure than additive?
Whether multiplicative blinding is drastically worse than additive blinding depends fairly strongly on:
The ring are you doing the blinding in
Whether you need to blind the value 0
The second is a … fairly obvious, as multiplicative blinding doesn't disguise 0 at all. …
- 151k
1
vote
Accepted
One time blinding threshold encryption?
Notes:
Blinding works because the blinded value $C^2S^e = (P^2S)^e \pmod N$, and as $P^2S < N$, RSA decryption results in the integer $P^2S$. … It should be obvious why blinding an already blinded value doesn't work.
Obviously, we need to specify the sizes of $N$ and $T$. …
- 151k
1
vote
Accepted
How much risk is there for RSA blinding random number not being relatively prime to N
That being said, how much of a security risk does the random number used for blinding not being relatively prime to N pose? …
- 151k
3
votes
Accepted
Encryption using blinded key
Would it be a requirement that, given $f(k)$ and $c$, it is hard to rederive $m$?
If so, then what you are effectively asking for is a public key encryption system; your public key is $f(k)$, and your …
- 151k
7
votes
Accepted
Data Leakage and Data Switching
Given $v_1$ and $v_2$, can the server learn anything about $a$ and $b$?
Yes, they can (with high probability) determine whether $a = b$; if $v_2 = 0$, then either $r_1 = 0$ or $a = b$; given that …
- 151k
3
votes
Accepted
Protocol for Sharing Blinding Factor with RSA
First of all, you need to understand what's the point of key blinding in the first place. … through the above, $R$ is used only during the decrypt operation (or the signature operation); the original encryptor (the guy who created $C = P^e$ in the first place) need not know it (or even if RSA blinding …
- 151k