phin | relied upon by important projects | Runtime Evironment library

 by   ethanent JavaScript Version: 3.7.0 License: MIT

kandi X-RAY | phin Summary

kandi X-RAY | phin Summary

phin is a JavaScript library typically used in Server, Runtime Evironment, Nodejs, Express.js, NPM applications. phin has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. You can download it from GitHub, Maven.

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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            kandi-support Support

              phin has a low active ecosystem.
              It has 532 star(s) with 35 fork(s). There are 8 watchers for this library.
              OutlinedDot
              It had no major release in the last 12 months.
              There are 13 open issues and 36 have been closed. On average issues are closed in 95 days. There are 7 open pull requests and 0 closed requests.
              It has a neutral sentiment in the developer community.
              The latest version of phin is 3.7.0

            kandi-Quality Quality

              phin has 0 bugs and 0 code smells.

            kandi-Security Security

              phin has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
              phin code analysis shows 0 unresolved vulnerabilities.
              There are 0 security hotspots that need review.

            kandi-License License

              phin is licensed under the MIT License. This license is Permissive.
              Permissive licenses have the least restrictions, and you can use them in most projects.

            kandi-Reuse Reuse

              phin releases are not available. You will need to build from source code and install.
              Deployable package is available in Maven.
              Installation instructions are not available. Examples and code snippets are available.
              phin saves you 430 person hours of effort in developing the same functionality from scratch.
              It has 1019 lines of code, 0 functions and 9 files.
              It has low code complexity. Code complexity directly impacts maintainability of the code.

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            phin Key Features

            No Key Features are available at this moment for phin.

            phin Examples and Code Snippets

            Using with HTTP clients
            TypeScriptdot img1Lines of Code : 121dot img1License : Permissive (MIT)
            copy iconCopy
            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

            QUESTION

            Math calculation for two lists
            Asked 2019-Aug-24 at 15:21

            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 15:21

            QUESTION

            Textbook RSA Same cyphertext with different keys
            Asked 2018-May-29 at 18:31

            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 16:06

            48 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)

            Source https://stackoverflow.com/questions/50274852

            QUESTION

            Getting nth value of a G-Series (General Fibonnaci Sequence)
            Asked 2018-May-05 at 13:41

            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 13:41

            QUESTION

            Python - RSA decryption does not return original message (very simple, short program)
            Asked 2017-Dec-16 at 17:02

            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 17:02

            Solved 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)).

            Source https://stackoverflow.com/questions/47759060

            QUESTION

            RSA digital signature failing
            Asked 2017-Nov-01 at 18:53

            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 18:53

            You don't handle negative exponents correctly, except to check that gcd(base, modulus) == 1. The following snippet shows one correct way to do it.

            Source https://stackoverflow.com/questions/47061259

            Community Discussions, Code Snippets contain sources that include Stack Exchange Network

            Vulnerabilities

            No vulnerabilities reported

            Install phin

            You can download it from GitHub, Maven.

            Support

            There's a lot more which can be done with the Phin library. See the Phin documentation.
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            Install
          • npm

            npm i phin

          • CLONE
          • HTTPS

            https://github.com/ethanent/phin.git

          • CLI

            gh repo clone ethanent/phin

          • sshUrl

            git@github.com:ethanent/phin.git

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