phin | relied upon by important projects | Runtime Evironment library
kandi X-RAY | phin Summary
kandi X-RAY | phin Summary
Phin is relied upon by important projects and large companies. The hundreds of contributors at Less, for example, depend on Phin as part of their development process. Also, Phin is very lightweight. To compare to other libraries, see Phin vs. the Competition.
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Currently covering the most popular Java, JavaScript and Python libraries. See a Sample of phin
phin Key Features
phin Examples and Code Snippets
import https from 'https';
import { CookieJar } from 'tough-cookie';
import { HttpsCookieAgent } from 'http-cookie-agent';
const jar = new CookieJar();
const agent = new HttpsCookieAgent({ jar });
https.get('https://example.com', { agent }, (res) =
Community Discussions
Trending Discussions on phin
QUESTION
I have two Lists (CalculateDensityPorosity() and Neuton) and I wanna do Math calculation for each element of this lists; I am in a class so I cannot use "foreach"
the code that I tried is:
...ANSWER
Answered 2019-Aug-24 at 15:21Your example:
QUESTION
I’m implementing the RSA cryptosystem for a school project. I’m doing the project in C on Linux and I’m using the GNU MP library for most of the mathematical operations.
For some reason I keep getting the same cyphertext for different public keys with the same message, so either I don’t understand RSA, or there’s something wrong with my code but I can’t figure it out.
Here’s the code in question:
...ANSWER
Answered 2018-May-10 at 16:0648 to the 3rd is always 110592. 110592 mod (big number) is 110592.
The RSA primitives are not a secure algorithm. In order to effectively hide the input for encryption you need to guarantee that the "message" value raised to the power of the public exponent actually exceeds the modulus.
The way that this is accomplished is called padding. All four padding families for RSA set a bit in the second-most-significant byte to guarantee that a) any exponentiation of that value will exceed the modulus and b) that the padded value itself does not exceed the modulus (which is why it's not the most-significant-byte). Then you apply the RSA primitive to the padded message instead of the original message, and now a secure system can be born.
(4 families? RSA PKCS#1 v1.5 encryption, RSA PKCS#1 v1.5 signature, RSA-OAEP, RSA-PSS)
QUESTION
How do we generate the nth term of a Fibonacci sequence which has starting values other than 0 and 1. That is, a user gives in two starting values say, 1 and 4, then the code generates the fibonacci based on user input. For example, 1 and 4 will give 1,4,5,9,13,23... Coming up with this quite easy with tabulation or recursion but I tried googling for a general formula (to save running time) and I fell on this one:
G(a, b, n) = ( (a(√5 – 1) + 2b) Phin + (a(√5 + 1) – 2b) ( –phi)n ) / (2√5)
Please note in above formula that 'Phin' is Phi to the power n. (Phi**n)
where a and b are starting values, phi, as u guessed is (1+√5)/2 and n the nth value to be gotten.
I tried implementing the above formula in python, but does not give me expected output, (values are not what am expecting). The site on which I found this formula, found here has a built-in generator for the sequence, this generator works as expected, but my code does not. Can anyone spot where I went wrong ?
When I do fibon(1,4,3) it produces 13.260990336999413. But answer should be 9; 1,4,5,9
G(a, b, n) = ( (a(√5 – 1) + 2b) Phin + (a(√5 + 1) – 2b) ( –phi)n ) / (2√5)
...ANSWER
Answered 2018-May-05 at 13:41You wrote
QUESTION
So I am trying to create a very simple RSA encryption/decryption program where I only encrypt/decrypt int numbers. All works fine except for one problem. Sometimes my decrypted number (message) does not match my original number (the message I needed to encrypt and decrypt). This seems to happen whenever my inputted number (message) is close to my the number in my 'n' variable (n=p*q where p and q are prime numbers). I have browsed Stackoverflow a bit now and have found out that RSA algoritms cannot properly decrypt messages that are greater than 'n'. But in my case it fails to decrypt messages that are close to 'n'. If n=35 and my input number (the number to encrypt/decrypt) is 32, the program does not properly decrypt it back to 32 despite 32 being lower than 35 (works for 31, 30 ... though). Why?
Code:
...ANSWER
Answered 2017-Dec-16 at 17:02Solved it! I made two little mistakes in my code. First one was that I assumed that the "^" symbol meant "to the power of" in Python (the correct symbol is "**") and the second one was that I forgot to add "mod n" to the line in my decrypt() function (so return M**(d) % n
instead of return M^(d)
).
QUESTION
I am trying to implement the RSA Blind digital signature scheme, using the BigInteger
class for generating large prime numbers. Samantha generates the public key, the private key, chooses a message, masks it, then signs it and then Victor verifies the signature.
Problem: As long as I use the modular exponentiation method modPow from the BigInteger
class, everything works perfectly (the verification algorithm returns true everytime). However, I have built a custom class where I have implemented several algebraic algorithms on my own; when I switch the modPow call with my modExp method, I keep getting false returns from the verification algorithm (about 50-60 % of the time), even though I should not. If instead of using large, random integers, I set small, hardcoded numbers for testing purposes, I get the correct result.
Question: As a consequence, I am pretty sure that my modExp method is the problem, however I can't seem to find out did I do wrong, even after changing the algorithm several times. What is the problem?
My code so far:
RSA_test() -- Method used for the precomputation step and testing
...ANSWER
Answered 2017-Nov-01 at 18:53You don't handle negative exponents correctly, except to check that gcd(base, modulus) == 1
. The following snippet shows one correct way to do it.
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