PROTECTED CONTENT: END-TO-END PGP ENCRYPTION FOR DRUPAL

Protected Content
Rodrigo Панчиняк Fernandes
DRUPALCAMP KYIV 2019 – Kyiv, Ucrain
25/05/2019

Keyring generation
• User types in a password
• Password is concatenated with a hash used as a salt specific to
the user in the current enveiroment
• OpenPGPjs uses username, email address and random
numbers to create users public and private key
• Private key is symmetrically encrypted (aes256) with the salted
password

Password (and salt) are not submitted (proc/keyring/add)
Public key and aes256-encrypted private key are submitted via hidden fields.

Encryption of a file
• Recipient user needs to have registered encryption keys
• Sender user does not need to have registered encryption keys
(there is no signing)
• It is possible to encrypt the same file using multiple public
keys
• Mary can encrypt to herself as long as Mary has registered
keys
• Public keys of recipient users are loaded by their UIDs in a CSV
argument at /proc/add/<UIDs-CSV>

Text and binary files are treated the same way (binary Blob)
(/proc/new/<UIDs-CSV>)
Original file will never be submitted or leave the browser. When the user is able to press “Save”
the file has been encrypted with the public keys of the users identified by their UIDs.

Cipher text (ASCII Armored format) is ready to be submitted
File was read as a FileReader() object and converted to Uint8Array. A ReadableStream() is started with
it and composes together with recipient(s) pubkey(s) the “options” needed at
openpgp.encrypt(options)

Once the cipher text has been stored user gets its exclusive access link
Only the recipients will be allowed to access the link by Drupal permission and only the holder of the
decryption key will be able to decrypt the ciphertext. Metadata as filename and size are stored in the
database as plain text and are used to name and check the file created at the end of the decryption
process

Password must be typed in by the recipient user
Password is never submitted. If it fails an error message comes from OpenPGPjs library.

Wrong password
Wrong password fails to decrypt a cipher text.
Decryption (or fail on doing it) is triggered on “Get it” link

Decryption of a file
• Drupal checks if the current user is recipient of the ciphertext requested
• If this is the case, ciphertext is attached to the form at /proc/<pid>
• Ciphertext and encrypted privkey of the current user/recipient is
loaded in by js via Drupal.settings and becomes available for decryption
• First the privkey is decrypted in a promise
• Then the privkey composes together with the cipher text the “options”
needed at openpgp.decrypt(options)
• A new Blob (type octet-binary) object is created with the plaintext
• The size is checked against the database and if it matches, the link
becomes available according to the original file name.

Bonus 1: asymetric cyptography in a
metaphor
• The double locked box
• Alice wants to send a letter to Mary
• Alice puts the letter in a box. She locks it and only she has the key. Sehe
sends the box to Mary.
• Mary does not open it. Instead, Mary adds her own locker and sends the
now double locked box back to Alice.
• Alice removes her own locker and sends the box back to Mary.
• Mary opens the box removing her own locker

Bonus 2: rsa in 10 steps (by Simon Singh, The
Code Book)
• 1. Alice picks two giant prime numbers, p and q. The primes
• should be enormous, but for simplicity we assume that Alice
• chooses p = 17, q = 11. She must keep these numbers secret.
• 2. Alice multiplies them together to get another number, N. In this case N
= 187. She now picks another number e, and in this case she chooses e = 7
(e and (p – 1) ✕ (q – 1) should be relatively prime, but this is a
technicality).

Code Book)
• 3. Alice can now publish e and N in something akin to a telephone
directory. Since these two numbers are necessary for encryption, they
must be available to anybody who might want to encrypt a message to
Alice. Together these numbers are called the public key. (As well as being
part of Alice’s public key, e could also be part of everybody else’s public
key. However, everybody must have a different value of N, which depends
on their choice of p and q.)

Code Book)
• 4. To encrypt a message, the message must first be converted into a
number, M. For example, a word is changed into ASCII binary digits, and
the binary digits can be considered as a decimal number. M is then
encrypted to give the ciphertext, C, according to the formula C = M e
(mod N ).

Code Book)
• 5. Imagine that Bob wants to send Alice a simple kiss: just the letter X. In
ASCII this is represented by 1011000, which is equivalent to 88 in decimal.
So, M = 88.
• 6. To encrypt this message, Bob looks up Alice’s public key, and discovers
that N = 187 and e = 7. This provides him with the encryption formula
required to encrypt messages to Alice. With M = 88, the formula gives C =
88 7 (mod 187).

Code Book)
• 7. Working this out directly on a calculator is tough, because the display
cannot cope with such large numbers. However, there is a trick for
calculating exponentials in modular arithmetic. We know that since 7 = 4
+ 2 + 1
Bob now sends the ciphertext, C =
11, to Alice.

Code Book)
• 8. We know that exponentials in modular arithmetic are one-way functions,
so it is very difficult to work backward from C = 11 and recover the original
message, M. Hence, Eve cannot decipher the message.

Code Book)
• 9. However, Alice can decipher the message because she has some special
information: she knows the values of p and q. She calculates a special
number, d, the decryption key,otherwise known as her private key. The
number d is calculated according to the following formula:

Code Book)
• 10. To decrypt the message, Alice uses this formula:

PROTECTED CONTENT: END-TO-END PGP ENCRYPTION FOR DRUPAL

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PROTECTED CONTENT: END-TO-END PGP ENCRYPTION FOR DRUPAL