# poncho

less info
reputation
12463
bio website location age member for 3 years, 2 months seen 27 mins ago profile views 473

# 2,387 Actions

 Apr7 comment How could Fully Homomorphic Encryption support power operations? @D.W.: in my opinion, the answer needed to be sketched out in greater detail. If sashank didn't mean "addition == XOR" and instead he meant that + and $\times$ can be considered to be made up of XOR and AND gates, and we can use the embedded gates in a universal way, well, that doesn't obviously follow, After all, + and - can also be considered combinations of XOR and AND gates, however that pair is not a universal set, hence we cannot use the embedded gates from those two operators in a universal way. Apr7 comment How could Fully Homomorphic Encryption support power operations? While the conclusions are correct, it turns out that the details are a bit more complicated than "ADD==XOR". The problem is that if we want to limit are internal data to the values $(0,1)$, the simple computation $A+B$ may result a result 2 which is outside of that. One way to obtain an universal gate is to compute the NAND function by encrypting the values $1$ and $-1$, and computing $NAND(A,B) = 1 + (-1 \times A \times B)$, where $+$ and $\times$ are the FHE operations. Apr6 awarded Necromancer Apr5 comment Definition of a Statistical Test An extremely low false positive rate; when I attempt to estimate the probability of a false positive of a truly random long string, I get something on the order of $10^{-88}$ Apr4 answered HMAC-SHA1 vs HMAC-SHA256 Apr3 comment TLS Key Block calculation - What is a PRF? @Eddie: you are correct, except that PRFs in general do not produce arbitrary long outputs; that is a special property of the PRF that TLS uses Apr2 answered TLS Key Block calculation - What is a PRF? Apr2 comment What is h in this RSA variant? Well, given a public key $(N,e)$ and plaintext/ciphertext pairs $(P,C)$, an attacker can replace it with an RSA key $(N,e)$ and plaintext/ciphertext pairs $(P,eC)$; any attack on this system would immediately imply an attack on the RSA system. We believe the RSA system to be secure (because, while $e$ was chosen so that $e^{-1}$ is small, there's no known weakness there), and hence this system is secure. Apr1 comment What is h in this RSA variant? Actually, these authors just copied the 2KGEA algorithm from their reference [3]: iasir.net/IJSWSpapers/IJSWS13-272.pdf - however, that paper doesn't answer any of the above questions. Apr1 revised What is h in this RSA variant? deleted 5 characters in body Apr1 answered What is h in this RSA variant? Apr1 comment Demonstrating Diffie-Hellman key exchange using only p, A, B; There are 50976 generators for $Z/104933$, and that's assuming that, by generator, the question meant a generator for the entire group, and not the generator for the subgroup we'll be doing DH in. So, 5 is a generator; how do we know that that's the generator that's meant, and not any of the 50975 other generators that exist? Apr1 comment Demonstrating Diffie-Hellman key exchange using only p, A, B; No, there is not enough information given to let you derive $g$. Now, if they said "use the smallest value that generates the group for $g$", then you could find that, and then use that. However, remember that any value of $g$ will work within the protocol (although some choices, such as $g=1$, do have some security issues, of course, in this toy example, there aren't any secure choices). Also, in practice, we generally don't use values of $g$ which generate the entire group, and so you are told to use such a value, that's something contrary to common practice. Apr1 comment Demonstrating Diffie-Hellman key exchange using only p, A, B; I believe you'll also need the value of $g$; you cannot derive it with the information you have Mar31 comment Can a commutative block cipher be indistinguishable from a random permutation, for fixed key? @fgrieu: actually, it sounds like what you really ask about are known plaintext attacks; given a random set of $(x, P(x))$ pairs, can we distinguish a random $P$ from $E(k)$? Given the Pohlig-Hellman cipher, we can distinguish it in this model; is that a fundamental property of all commutative ciphers, or is that just a side effect of how Pohlig-Hellman achieves commutativity and another commutative cipher might be secure in this model? Mar31 answered Can a commutative block cipher be indistinguishable from a random permutation, for fixed key? Mar29 awarded rsa Mar27 answered Is it secure to choose d in a RSA key pair? Mar26 comment m ∈ Zn \Z*n, RSA works but not secure You still have the wrong value for $\phi(p)$. $\phi(p)$ is defined to be the number of values between 1 and $p-1$ which is relatively prime to $p$. If $p$ is prime, how many of the values between 1 and $p-1$ are relatively prime to $p$? Mar26 reviewed Approve suggested edit on Encryption of numeric value using playfair
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8093727827072144, "perplexity": 816.8478470098344}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414119646180.24/warc/CC-MAIN-20141024030046-00316-ip-10-16-133-185.ec2.internal.warc.gz"}