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+# Copyright (c) 2006, Mathieu Fenniak
+# Copyright (c) 2007, Ashish Kulkarni <kulkarni.ashish@gmail.com>
+#
+# All rights reserved.
+#
+# Redistribution and use in source and binary forms, with or without
+# modification, are permitted provided that the following conditions are
+# met:
+#
+# * Redistributions of source code must retain the above copyright notice,
+# this list of conditions and the following disclaimer.
+# * Redistributions in binary form must reproduce the above copyright notice,
+# this list of conditions and the following disclaimer in the documentation
+# and/or other materials provided with the distribution.
+# * The name of the author may not be used to endorse or promote products
+# derived from this software without specific prior written permission.
+#
+# THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
+# AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
+# IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
+# ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT OWNER OR CONTRIBUTORS BE
+# LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR
+# CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF
+# SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS
+# INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN
+# CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
+# ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE
+# POSSIBILITY OF SUCH DAMAGE.
+
+"""Anything related to encryption / decryption."""
+
+import struct
+from hashlib import md5
+from typing import Tuple, Union
+
+from ._utils import b_, ord_, str_
+from .generic import ByteStringObject
+
+try:
+ from typing import Literal # type: ignore[attr-defined]
+except ImportError:
+ # PEP 586 introduced typing.Literal with Python 3.8
+ # For older Python versions, the backport typing_extensions is necessary:
+ from typing_extensions import Literal # type: ignore[misc]
+
+# ref: pdf1.8 spec section 3.5.2 algorithm 3.2
+_encryption_padding = (
+ b"\x28\xbf\x4e\x5e\x4e\x75\x8a\x41\x64\x00\x4e\x56"
+ b"\xff\xfa\x01\x08\x2e\x2e\x00\xb6\xd0\x68\x3e\x80\x2f\x0c"
+ b"\xa9\xfe\x64\x53\x69\x7a"
+)
+
+
+def _alg32(
+ password: str,
+ rev: Literal[2, 3, 4],
+ keylen: int,
+ owner_entry: ByteStringObject,
+ p_entry: int,
+ id1_entry: ByteStringObject,
+ metadata_encrypt: bool = True,
+) -> bytes:
+ """
+ Implementation of algorithm 3.2 of the PDF standard security handler.
+
+ See section 3.5.2 of the PDF 1.6 reference.
+ """
+ # 1. Pad or truncate the password string to exactly 32 bytes. If the
+ # password string is more than 32 bytes long, use only its first 32 bytes;
+ # if it is less than 32 bytes long, pad it by appending the required number
+ # of additional bytes from the beginning of the padding string
+ # (_encryption_padding).
+ password_bytes = b_((str_(password) + str_(_encryption_padding))[:32])
+ # 2. Initialize the MD5 hash function and pass the result of step 1 as
+ # input to this function.
+ m = md5(password_bytes)
+ # 3. Pass the value of the encryption dictionary's /O entry to the MD5 hash
+ # function.
+ m.update(owner_entry.original_bytes)
+ # 4. Treat the value of the /P entry as an unsigned 4-byte integer and pass
+ # these bytes to the MD5 hash function, low-order byte first.
+ p_entry_bytes = struct.pack("<i", p_entry)
+ m.update(p_entry_bytes)
+ # 5. Pass the first element of the file's file identifier array to the MD5
+ # hash function.
+ m.update(id1_entry.original_bytes)
+ # 6. (Revision 3 or greater) If document metadata is not being encrypted,
+ # pass 4 bytes with the value 0xFFFFFFFF to the MD5 hash function.
+ if rev >= 3 and not metadata_encrypt:
+ m.update(b"\xff\xff\xff\xff")
+ # 7. Finish the hash.
+ md5_hash = m.digest()
+ # 8. (Revision 3 or greater) Do the following 50 times: Take the output
+ # from the previous MD5 hash and pass the first n bytes of the output as
+ # input into a new MD5 hash, where n is the number of bytes of the
+ # encryption key as defined by the value of the encryption dictionary's
+ # /Length entry.
+ if rev >= 3:
+ for _ in range(50):
+ md5_hash = md5(md5_hash[:keylen]).digest()
+ # 9. Set the encryption key to the first n bytes of the output from the
+ # final MD5 hash, where n is always 5 for revision 2 but, for revision 3 or
+ # greater, depends on the value of the encryption dictionary's /Length
+ # entry.
+ return md5_hash[:keylen]
+
+
+def _alg33(
+ owner_password: str, user_password: str, rev: Literal[2, 3, 4], keylen: int
+) -> bytes:
+ """
+ Implementation of algorithm 3.3 of the PDF standard security handler,
+ section 3.5.2 of the PDF 1.6 reference.
+ """
+ # steps 1 - 4
+ key = _alg33_1(owner_password, rev, keylen)
+ # 5. Pad or truncate the user password string as described in step 1 of
+ # algorithm 3.2.
+ user_password_bytes = b_((user_password + str_(_encryption_padding))[:32])
+ # 6. Encrypt the result of step 5, using an RC4 encryption function with
+ # the encryption key obtained in step 4.
+ val = RC4_encrypt(key, user_password_bytes)
+ # 7. (Revision 3 or greater) Do the following 19 times: Take the output
+ # from the previous invocation of the RC4 function and pass it as input to
+ # a new invocation of the function; use an encryption key generated by
+ # taking each byte of the encryption key obtained in step 4 and performing
+ # an XOR operation between that byte and the single-byte value of the
+ # iteration counter (from 1 to 19).
+ if rev >= 3:
+ for i in range(1, 20):
+ new_key = ""
+ for key_char in key:
+ new_key += chr(ord_(key_char) ^ i)
+ val = RC4_encrypt(new_key, val)
+ # 8. Store the output from the final invocation of the RC4 as the value of
+ # the /O entry in the encryption dictionary.
+ return val
+
+
+def _alg33_1(password: str, rev: Literal[2, 3, 4], keylen: int) -> bytes:
+ """Steps 1-4 of algorithm 3.3"""
+ # 1. Pad or truncate the owner password string as described in step 1 of
+ # algorithm 3.2. If there is no owner password, use the user password
+ # instead.
+ password_bytes = b_((password + str_(_encryption_padding))[:32])
+ # 2. Initialize the MD5 hash function and pass the result of step 1 as
+ # input to this function.
+ m = md5(password_bytes)
+ # 3. (Revision 3 or greater) Do the following 50 times: Take the output
+ # from the previous MD5 hash and pass it as input into a new MD5 hash.
+ md5_hash = m.digest()
+ if rev >= 3:
+ for _ in range(50):
+ md5_hash = md5(md5_hash).digest()
+ # 4. Create an RC4 encryption key using the first n bytes of the output
+ # from the final MD5 hash, where n is always 5 for revision 2 but, for
+ # revision 3 or greater, depends on the value of the encryption
+ # dictionary's /Length entry.
+ key = md5_hash[:keylen]
+ return key
+
+
+def _alg34(
+ password: str,
+ owner_entry: ByteStringObject,
+ p_entry: int,
+ id1_entry: ByteStringObject,
+) -> Tuple[bytes, bytes]:
+ """
+ Implementation of algorithm 3.4 of the PDF standard security handler.
+
+ See section 3.5.2 of the PDF 1.6 reference.
+ """
+ # 1. Create an encryption key based on the user password string, as
+ # described in algorithm 3.2.
+ rev: Literal[2] = 2
+ keylen = 5
+ key = _alg32(password, rev, keylen, owner_entry, p_entry, id1_entry)
+ # 2. Encrypt the 32-byte padding string shown in step 1 of algorithm 3.2,
+ # using an RC4 encryption function with the encryption key from the
+ # preceding step.
+ U = RC4_encrypt(key, _encryption_padding)
+ # 3. Store the result of step 2 as the value of the /U entry in the
+ # encryption dictionary.
+ return U, key
+
+
+def _alg35(
+ password: str,
+ rev: Literal[2, 3, 4],
+ keylen: int,
+ owner_entry: ByteStringObject,
+ p_entry: int,
+ id1_entry: ByteStringObject,
+ metadata_encrypt: bool,
+) -> Tuple[bytes, bytes]:
+ """
+ Implementation of algorithm 3.4 of the PDF standard security handler.
+
+ See section 3.5.2 of the PDF 1.6 reference.
+ """
+ # 1. Create an encryption key based on the user password string, as
+ # described in Algorithm 3.2.
+ key = _alg32(password, rev, keylen, owner_entry, p_entry, id1_entry)
+ # 2. Initialize the MD5 hash function and pass the 32-byte padding string
+ # shown in step 1 of Algorithm 3.2 as input to this function.
+ m = md5()
+ m.update(_encryption_padding)
+ # 3. Pass the first element of the file's file identifier array (the value
+ # of the ID entry in the document's trailer dictionary; see Table 3.13 on
+ # page 73) to the hash function and finish the hash. (See implementation
+ # note 25 in Appendix H.)
+ m.update(id1_entry.original_bytes)
+ md5_hash = m.digest()
+ # 4. Encrypt the 16-byte result of the hash, using an RC4 encryption
+ # function with the encryption key from step 1.
+ val = RC4_encrypt(key, md5_hash)
+ # 5. Do the following 19 times: Take the output from the previous
+ # invocation of the RC4 function and pass it as input to a new invocation
+ # of the function; use an encryption key generated by taking each byte of
+ # the original encryption key (obtained in step 2) and performing an XOR
+ # operation between that byte and the single-byte value of the iteration
+ # counter (from 1 to 19).
+ for i in range(1, 20):
+ new_key = b""
+ for k in key:
+ new_key += b_(chr(ord_(k) ^ i))
+ val = RC4_encrypt(new_key, val)
+ # 6. Append 16 bytes of arbitrary padding to the output from the final
+ # invocation of the RC4 function and store the 32-byte result as the value
+ # of the U entry in the encryption dictionary.
+ # (implementer note: I don't know what "arbitrary padding" is supposed to
+ # mean, so I have used null bytes. This seems to match a few other
+ # people's implementations)
+ return val + (b"\x00" * 16), key
+
+
+def RC4_encrypt(key: Union[str, bytes], plaintext: bytes) -> bytes: # TODO
+ S = list(range(256))
+ j = 0
+ for i in range(256):
+ j = (j + S[i] + ord_(key[i % len(key)])) % 256
+ S[i], S[j] = S[j], S[i]
+ i, j = 0, 0
+ retval = []
+ for plaintext_char in plaintext:
+ i = (i + 1) % 256
+ j = (j + S[i]) % 256
+ S[i], S[j] = S[j], S[i]
+ t = S[(S[i] + S[j]) % 256]
+ retval.append(b_(chr(ord_(plaintext_char) ^ t)))
+ return b"".join(retval)