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Asymmetric XML encryption (in .NET)

I am trying to decrypt XML EncryptedData in .NET Framework:

As I understand it, I need to implement these steps:

recreate the public key stored in the OriginatorKeyInfo node using the x, y parameters of PublicKey and the specified curve, the result is publicKey
KeyAgreement operation (algorithm: ECDH-ES) using my private key and publicKey from step 1, the result is sharedKey
Key derivation function operation (algorithm: ConcatKDF) using sharedKey, result is wrappingKey
do Key Unwrap (algorithm kw-aes256) in EncryptedKey> CipherData> CipherValue using unwrappingKey, the result is encryptionKey
decrypt (algorithm aes256-gcm) in EncryptedData> CipherData> CipherValue using encryptionKey

I am currently in step 1. Using jose-jwt I can create a public key with EccKey.New(x, y, CngKeyUsages.KeyAgreement), but I don't understand why there is no curve name parameter?
If anyone knows how to make this work or some examples (bouncy castle?) Please comment.

encryption: using Chacha20 with a nonce NULL

I am new to ChaCha20. From the RFC –

The entrances to ChaCha20 are:

o A 256-bit key, treated as a 32-bit eight-bit concatenation.
endian integers.

o A 96-bit nonce, treated as a concatenation of three small 32-bit bits
endian integers.

o A 32-bit block count parameter, treated as a 32-bit little endian
whole.

Now looking at Apple's CryptoKit, the counter is not provided as an input option. They implement the counter under the hood. For the sake of this question, suppose the counter is implemented correctly.

The key is set randomly for the encryption session (which is network data encryption).

Now if I use a nonce NULL, the counter is still used. It will be OK?

Is the maximum number of messages that can be encrypted in this way before changing the key is 2 ^ 32? or is it the maximum number of bytes?

encryption: GnuPG decryption does not request a passphrase

Do you store the secret key somewhere and use it (I also stored my secret key in the GnuPG key chain, use it)?

GnuPG only uses keys from its keychain, so should be there to use it.

How can I force the system to ask for the passphrase every time?

Earlier versions of GnuPG used the gpg-agent, which caches the passphrase for a specified time. Use the option --no-use-agent or add a line no-use-agent to ~/.gnupg/gpg.conf to avoid using the agent.

For the latest versions (v2.1 +), disable password caching for the agent by creating ~/.gnupg/gpg-agent.conf and adding the following lines:

default-cache-ttl 1
max-cache-ttl 1

Restart the agent with:

echo RELOADAGENT | gpg-connect-agent

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Let the variable x be the position of the alphabet in the list of alphabets sorted alphabetically Let the variable y be the position of the alphabet in the entered word For example, if the user enters 'abcd', the program would find x and y for a, b, c and d one by one Then, it would find the variable z = 26-x+y and z would be the position of the alphabet in the code in the list of alphabets In abcd : for a - x = 1, y = 1, so, z = 26-1+1 = 26, so coded alphabet = 26th alphabet = z Similarly, all of a,b,c and d will have coded alphabets as 'z' So, 'abcd' will be coded as 'zzzz'

alphabets =('a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z') # Defining all sorts of functions def position(tofind,source): size = len(source) for i in range(0,size,1): if (source(i) == tofind): p = i + 1 break return p def encrypt(a): output = () sizea = len(a) for i in range(0,sizea,1): x = i+1 y = position(a(i),alphabets) z = 26-x+y if (z>26): z = z % 26 element = alphabets(z-1) output.append(element) return output def converttolist(text): size = len(text) l = () for i in range(0,size,1): l.append(text(i)) return l # The main program print () print ("NOTE : Please enter all the alphabets in lowercase...") print () given = str(input("Please enter the word to be coded : ")) givenlist = converttolist(given) outputlist = encrypt(givenlist) print () print ("The code for ",given," is :-") outputlistlength = len(outputlist) for i in range(0,outputlistlength,1): print (outputlist(i),end = "")

>>> fscrypt status filesystems supporting encryption: 2 filesystems with fscrypt metadata: 2 MOUNTPOINT DEVICE FILESYSTEM ENCRYPTION FSCRYPT / /dev/sda2 ext4 supported Yes /mnt/hd1 /dev/sdb2 ext4 supported Yes >>> fscrypt status /mnt/hd1/ ext4 filesystem "/mnt/hd1" has 1 protector and 1 policy PROTECTOR LINKED DESCRIPTION 7db5baf4xxxxxxxx No login protector for user1 POLICY UNLOCKED PROTECTORS 0000ed45xxxxxxxx No 7db5baf4xxxxxxxx >>> fscrypt unlock /mnt/hd1/home/user1 fscrypt unlock: user keyring not linked into session keyring This is usually the result of a bad PAM configuration. Either correct the problem in your PAM stack, enable pam_keyinit.so, or run "keyctl link @u @s". >>> fscrypt unlock /mnt/hd1/home/user1 --unlock-with=/mnt/hd1:7db5baf4xxxxxxxx fscrypt unlock: user keyring not linked into session keyring This is usually the result of a bad PAM configuration. Either correct the problem in your PAM stack, enable pam_keyinit.so, or run "keyctl link @u @s".

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Tag: encryption

Asymmetric XML encryption (in .NET)

encryption: using Chacha20 with a nonce NULL

encryption: GnuPG decryption does not request a passphrase

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