kdf | kdf password strengthener | Security library
kandi X-RAY | kdf Summary
kandi X-RAY | kdf Summary
Strengthen passwords with up to 24 bits security. algortithm = sha224( 2^24 * sha512 ). requires: xclip (for clipboard access) - sudo apt-get install xclip. Compatibel with: python2/3 and Tails. Run: python kdf.py -h.
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- Derives a base value from a base
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kdf Key Features
kdf Examples and Code Snippets
Lines of Code : 31
License : No License
Copyfrom Crypto.Cipher import AES
import scrypt, os, binascii
def encrypt_AES_GCM(msg, password):
kdfSalt = os.urandom(16)
secretKey = scrypt.hash(password, kdfSalt, N=16384, r=8, p=1, buflen=32)
aesCipher = AES.new(secretKey, AES.MODE_GCM)
Community Discussions
Trending Discussions on kdf
QUESTION
I am making a program which encrypts and decrypts texts. I am using Python 3.7 and cryptography.fernet library. I want to enter some information about my program's encryption standard to the GitHub page but I didn't understand which encryption does Fernet uses.
Here is my sample code which I am used in my project. I want to encrypt with 256-bit (AES-256) key but the key which this code generates is longer than 32 characters. It's 44 characters. But in official web site of cryptography library it says this code generates 128-bit key. What is the name of this 44 character (352-bit) key? Or is there any way for 256-bit symmetric encryption without PyCrypto?
...ANSWER
Answered 2021-May-03 at 13:26It is well written in the documentation;
Fernet is built on top of a number of standard cryptographic primitives. Specifically it uses:
- AES in CBC mode with a 128-bit key for encryption; using PKCS7 padding.
- HMAC using SHA256 for authentication.
- Initialization vectors are generated using os.urandom().
For complete details consult the specification.
Therefore you cannot use AES-256 with Fernet
- Cryptography.io library has other modes too, in the hazardous material layer including CBC, CTR, and GCM for AES-256, too.
PyCrypto can use a wide range of mode of operations for AES-256 including CBC, CTR, GCM, SIV, and OCB
Not clear how you get 44 bytes, here is the way to get the 32-bytes;
QUESTION
I'm using (CryptoKit) to use AES-GCM to encrypt some data and authenticate it as well.
However, I was wondering how I would generate an AES-GCM key from a plain text password. Normally, you would use a KDF function for that, like PBKDF2.
In CryptoKit, there is a HKDF class which does about what I want: https://developer.apple.com/documentation/cryptokit/hkdf
However, I am wondering what KDF algorithm the DeriveKey function uses. Does it use PBKDF2? Does it use bcrypt? If so, how do I specify settings, or are the settings automatically determined?
ANSWER
Answered 2021-May-29 at 21:03HKDF is defined in RFC5869. It is intended to generate keys from some cryptographically secure "keying material" (IKM). It is not intended for stretching a human-generated password. As discussed in section 4 Applications of HKDF:
On the other hand, it is anticipated that some applications will not be able to use HKDF "as-is" due to specific operational requirements, or will be able to use it but without the full benefits of the scheme. One significant example is the derivation of cryptographic keys from a source of low entropy, such as a user's password. The extract step in HKDF can concentrate existing entropy but cannot amplify entropy. In the case of password-based KDFs, a main goal is to slow down dictionary attacks using two ingredients: a salt value, and the intentional slowing of the key derivation computation. HKDF naturally accommodates the use of salt; however, a slowing down mechanism is not part of this specification. Applications interested in a password-based KDF should consider whether, for example, [PKCS5] meets their needs better than HKDF.
I don't believe that CryptoKit offers a PBKDF of any kind (PBKDF2, scrypt, bcrypt, argon2). It's a very limited framework (I have yet to find a situation where it was useful). You will likely need to continue to use CommonCrypto for this, or implement it yourself (or use something like CryptoSwift, which I believe implements several).
QUESTION
Recently my lambda code stopped working. I am no longer able to create connection to Snowflake with sqlalchemy. See error stack below.
...ANSWER
Answered 2021-Jan-13 at 19:26For completeness, moving the answer from @Clement in a comment to an answer:
This error can happen when loading the oscrypto (libcrypto) if the memory usage is too high. The OOM state cascades upward.
QUESTION
I tried to install the https://pypi.org/project/keyrings.google-artifactregistry-auth/ package, but installation failed because it claims that Rust is required to install:
This package requires Rust >=1.41.0.
How can I install this? Do I need to install Rust?
Full output is here:
...ANSWER
Answered 2021-May-24 at 18:59The issue is that your pip version is too old to install one of this project's subdependencies, cryptography, which is using newer features.
Upgrading pip with the following will make it possible to install this package:
QUESTION
I am trying to add a new column to my existing Koalas dataframe. But the values turn into NaN's as soon as the new column is added. I am not sure what's going on here, could anyone give me some pointers?
Here's the code:
...ANSWER
Answered 2021-May-24 at 08:00You need to match the index when adding a new column:
QUESTION
I have previously used the following code on a Pandas dataframe to successfully generate a pivot table containing both column and row totals like in Excel:
...ANSWER
Answered 2021-May-06 at 01:51Since koalas.pivot_table() doesn't support margins, I would just add the totals manually:
QUESTION
I use the AES method to encrypt a sentance called from a txt file. I used the GCM Mode and created a specific key too. Everything is working (the code is below).
...ANSWER
Answered 2021-Apr-29 at 17:00PyCryptodome has a good documentation. The GCM example there uses JSON for concatenating/separating nonce, ciphertext, and tag, but the principle is the same and can easily be applied to your code.
Since you are using the implicitly derived nonce, be aware that PyCryptodome applies a 16 bytes nonce. Note, however, that the recommendation for GCM is a 12 bytes nonce (s. here, Note section).
The following decryption example uses a key and ciphertext created with the code you posted for encryption:
QUESTION
I have a Dockerfile, docker-compose.yml, requirements.txt defined below for a django project. The Dockerfile uses python:3.8.3-alpine and the docker-compose.yml have a db service that uses postgres:12.0-alpine image. I made sure that the dependencies are defined in the Dockerfile as required. However, it seems to me that django-allauth require extra dependencies. I have tried for days to fix this issue but get an error that says
This package requires Rust >=1.41.0.
ERROR: Failed building wheel for cryptography. I haved pasted the whole error for reference sake. Any help will be much appreciated. Thanks in advance.
ANSWER
Answered 2021-Apr-02 at 11:31django-allauth requires cryptography which now requires Rust to compile. You could try updating your Dockerfile with the newer python release, i.e. FROM python:3.8.8-alpine, which might let it fetch the prebuilt binary for cryptography.
If that doesn't work you need to add the Rust dependencies so it can compile the package.
QUESTION
I am trying to sign another public key using gpg --sign-key which then errors with signing failed: No secret key
Full output:
...ANSWER
Answered 2021-Apr-09 at 09:12After some time I was able to figure it out.
gpg --sign-key only work if you have the private key of the masterkey, meaning I had to sign the public key on my offline machine with the master private key. Subkeys don't work with gpg --sign-key
QUESTION
I am validating the input text with specific text "ABC 1234" in a sentence using javascript regex expression. My desired text can be occurred at begin, middle or end of the sentence or line. the text can be upper or lower case.
My desired validation: ex:
- "hdkjd dldld f abc 1234" - valid
- "jkdlfd df kdkd 333 abc 1234 dkdk dfd 444 d" - valid
- "dkdkf f fd kkd dfabc1234" - valid
- "kdkdfjf dkf dfk . fd,f kd abc 1234,kdf" - valid
- "kdkdfjf dkf dfk . fd,f kd abc 1234.kdf" - valid
- "kdkdfjf dkf dfk . fd,f kd abc 1234kdf" - valid
- "kdkdfjf dkf dfk . fd,f kd abc 1234, 5" - valid
- "kdkdfjf dkf dfk . fd,f kd abc 12345kdf" - invalid
- "kdkdfjf dkf dfk . fd,f kd abc 12345" - invalid
I have tried below expression, but its failing if there is any digit after 1234.
...ANSWER
Answered 2021-Mar-29 at 23:18Use a negative lookahead after 4 that disallows a digit there.
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Install kdf
You can use kdf like any standard Python library. You will need to make sure that you have a development environment consisting of a Python distribution including header files, a compiler, pip, and git installed. Make sure that your pip, setuptools, and wheel are up to date. When using pip it is generally recommended to install packages in a virtual environment to avoid changes to the system.
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