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awarded | Critic |
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Proper Way to Encrypt Data with Two Keys? You are describing your proposed solution instead of describing your problem/goal. |
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Apr 24 |
answered | Attacking historical ciphers methodology |
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Apr 24 |
revised |
Using the output of a stream cipher, how to guarantee the integrity of 4 bytes of data? added 451 characters in body |
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Apr 24 |
answered | Using the output of a stream cipher, how to guarantee the integrity of 4 bytes of data? |
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Apr 2 |
comment |
How to calculate y value from ((y*y) mod prime) efficiently it's the other way around: You can square any value, but not every value has a square root. If you have $y^2$, then $\pm y$ will be the two square roots. If you just try to calculate the root of a random value, there might not be one. |
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Mar 25 |
comment |
Perfectly hiding / binding commitment scheme I did not say "Alice needs to be computationally unbound to decomit any value" (for perfect hiding). Perfect hiding just implies, that decomitment of any value is possible. And this is equal to "(unbounded) Bob can learn no information from the commitment alone.", which is what you describe with "Bob's view is independent of Alice's value" |
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Mar 25 |
answered | Where can I begin to study the math behind modern cryptography? |
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Mar 25 |
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Security of Deterministic Encryption Scheme Ehm, I am not sure where you're going with this, and maybe you should specify which information should be retrievable by whom, and who has the private key(s). |
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Mar 25 |
awarded | Editor |
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Mar 25 |
revised |
How to calculate y value from ((y*y) mod prime) efficiently added 468 characters in body |
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Mar 22 |
comment |
Perfectly hiding / binding commitment scheme Uhm, yes... that's the same coin from the other side. Decomitment to ANY value is possible. If Alice is comp. unbound, she can decomit anything. If Bob is unbound, he still can not figure out Alice input. |
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Mar 22 |
answered | How to calculate y value from ((y*y) mod prime) efficiently |
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Mar 22 |
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How would one crack a weak but unknown encryption protocol? This is it. From even a couple of ciphertexts you wont get much. almost anything is secure against ciphertext only attacks. The next step is knowledge about the plaintext (distribution, common patterns, etc), and then the known plaintext attack. At this point most historical algorithms cracked (Enigma etc.). |
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Mar 22 |
answered | Security of Deterministic Encryption Scheme |
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Mar 22 |
answered | Perfectly hiding / binding commitment scheme |
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Mar 22 |
answered | Comparison: complexity measures vs. security |
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Dec 21 |
comment |
Why the following attack in common modulus RSA works? "Computing the RSA Secret Key Is Deterministic Polynomial Time Equivalent to Factoring" from May explains, how to get $(p,q)$ from $(e,d)$ in poly time. This means, $u_1$ will just factorize $N$ and calculate the private key of $u_2$. |
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Dec 21 |
answered | What would the Internet be like without public-key cryptography? |
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Nov 27 |
comment |
Cracking WWII-era codes - code found on a pigeon's leg in Surrey One time pad is: non-repeating and truly random. A Stream cipher with a repeating short sequence is called Vigenere cipher, and was broken about 150 years ago. The alternative with text passages of books as keystream can be seen as a variant... a very unsafe one at that. Words in any written language have certain characteristics, which makes them insecure. |