From: Harald_Krusekamp@MS.maus.de (Harald Krusekamp)
Newsgroups: alt.security.pgp
Subject: Breaking WordPerfect passwords
From: gnu@hoptoad.uucp (John Gilmore)
Newsgroups: comp.os.msdos.apps,sci.crypt
Subject: Word Perfect "locked document encryption" is trivial to break
One thing that came up at Crypto '90 was a short paper from Ms. Helen
Bergen at Queensland U. in Australia. She noticed the 'locked
document' commands in Word Perfect, used by all the secretaries in her
dept., and looked to see how strong it was. It turned out that the
MSDOS DEBUG command and an envelope for scratch paper are enough for
anyone to decode both a document and the key used for it! Word Perfect
Corp. didn't care about her results (letter reproduced below), but I
thought that some Word Perfect losers, I mean users, here on the net
might want to know.
You should consider WP locked documents like ROT13: fine to keep the
text garbled until you type a command, useless for keeping things private.
John Gilmore
From:
Date: Mon, 27 Aug 90 10:28 +1000
To: cygint!gnu
Dear John,
Here is the letter and a copy of the Latex source of my paper. It
will be published in CRYPTOLOGIA in the near future. Thanks for your
interest,
Regards,
Helen Bergen
Quote from letter received from WordPerfect Pacific:
Thankyou for the copy of your paper entitled "File Security in
WordPerfect 5.0". I sent a copy of the paper to WordPerfect Corporation
in the USA and recently received a reply from them.
They confirmed that people have written programs to break the password.
However, WordPerfect Corporation does not have such a program and
therefore has no way of breaking it. They also pointed out that very
few users would know how to write such a program.
It is possible that the manual may be amended in a future edition to
clarify the protection that a password gives. They recommend that
anyone concerned about security may want to take higher precautions
than the password protection.
Thankyou for your interest in WordPerfect
FILE SECURITY IN WORDPERFECT 5.0
H.A. Bergen, School of Computing Science
W.J. Caelli, Information Security Research Centre
Faculty of Information Technology
Queensland University of Technology
G.P.O. Box 2434, Brisbane, Q 4001, AUSTRALIA
Abstract
Cryptanalysis of files encrypted with the 'locked document'
option of the word processing package WordPerfect V5.0, is shown to be
remarkably simple. The encryption key and the plaintext are easily
recovered in a ciphertext only attack. File security is thus
compromised and is not in accord with the claim by the manufacturer
that: "If you forget the password, there is absolutely no way to
retrieve the document".
Keywords
Cryptanalysis, WordPerfect.
Introduction
WordPerfect is one of the most popular word processing packages in use
today. It has a 'locked document' option which aims at protection of a
WordPerfect file from unauthorised access. The manual states "You can
protect or lock your documents with a password so that no one will be
able to retrieve or print the file without knowing the password - not
even you". The manual also claims that "If you forget the password,
there is absolutely no way to retrieve the document" [1].
This option is used to 'add' a password to an existing or newly created
WordPerfect file. The file is then encrypted using the password as the
cryptographic key, and is stored on disk. Any subsequent retrieval or
printing of the file via WordPerfect requires the entry of the correct
password. With the increasing use of distributed file systems and
sharing of data, this option might appear to be an attractive means of
protecting sensitive files, particularly where they may reside on a
shared network server. It is easily implemented without the expense
and installation of another software protection/encryption package.
The encryption algorithm used in the WordPerfect 4.2 version, however,
was successfully cryptanalysed by Bennett [2]. He concluded that the
encryption system was unsatisfactory for protection of sensitive
documents.
The present study extends this work to an investigation of the security
of the WordPerfect 5.0 encryption system on both the IBM PC and DEC VAX
systems as well as WordPerfect 5.1 on the IBM PC.
WordPerfect Files
WordPerfect version 5.0 was used on an IBM-PC and other compatible
systems to create various files consisting of original documents and
their associated ciphertext with different passwords. The DOS utility
DEBUG was used to display the content of the files in hexadecimal
notation.
The WordPerfect files were created on three different systems. By this
we mean, three different licenced copies of WordPerfect running on
different Personal Computers with different printers. An example from
just one of these systems has been given in detail.
Version 4.2 format
Files created under 4.2 contain just the ASCII representation of the
character text. Printer definitions and setup parameters are in
separate files and are used only when the file is to be printed.
For example, a file may contain zeros (ASCII code 30 hex) and new line
characters (these are converted to the ASCII line feed character, 0A
hex). The plaintext file in hexadecimal would be
30 30 30 30 30 30 30 30 30 30 0A
30 30 30 30 30 30 30 30 30 30 0A
30 30 30 30 30 30 30 30 30 30
The corresponding ciphertext file with a key value of the ASCII letter A is
FE FF 61 61 41 00
73 72 75 74 77 76 79 78 7B 7A 47
7C 7F 7E 61 60 63 62 65 64 67 5C
69 68 6B 6A 6D 6C 6F 6E 51 50
Encrypted files contain an extra 6 bytes, shown in the first line of
the above. The first 4 bytes are constant for all keys and are used by
the WordPerfect program to determine whether the file is plaintext or
ciphertext. The latter 2 bytes contain a checksum derived from the
key, as described by Bennett [2].
For example,
FE FF 61 61 43 00 key = C
FE FF 61 61 71 C0 key = AA
Version 5.0 format
Files created under version 5.0 are stored in a different format. With
the default WordPerfect format, the file contains the document text
appended to printer setup information. There are other options to save
the file in DOS text format or in 4.2 format, and in these the printer
information is omitted. For example, a document containing 32
characters of text is saved in 5.0 format as a file of approximately
600 - 1000 bytes (depending on the particular printer system) and in
4.2 or DOS format as a file of 32 bytes.
The locked document option in version 5.0 allows encryption of files
only in WordPerfect format, the one containing all the printer
information.
-
All version 5.0 files, original and encrypted forms have
the same four characters in byte positions 0 - 3 :
FF 57 50 43 (HEX)
or W P C (ASCII)
These codes were unchanged for files created on three different
systems, i.e., three different licenced copies of WordPerfect 5.0 on
three different PC's using different printers.
- BYTES 4 - 7 are related to the offset address of the text, ie. the
start of the document text.
- BYTES 8 - 11 are constant for all files:
01 0A 00 00
- ENCRYPTED TEXT STARTS HERE
- BYTES 12 - 15 are constant for a plaintext file:
00 00 00 00
For an encrypted file, however, bytes 12 and 13 in the above contain a
checksum related to the key value used. This checksum appears to be
the same as that used in the 4.2 version [2].
- BYTES 16 - 21 were constant for files prepared on the three
different systems and contained:
FB FF 05 00 32 00
- BYTES 22 - 31. Of these 10 bytes, 22, 23, 28 are file and system
dependent, but bytes 24, 26, 29, 30, 31 are constant with value 00.
- BYTES 32 - 39 were constant for files prepared on the three
different systems and contained
42 00 00 00 02 00 56 00
- BYTES 40 - 47. Of these 8 bytes, 42, 46, 47 are file and system
dependent, but bytes 40, 41, 43, 44, 45 are constant with value 00.
- The remaining bytes of the printer header information are
dependent on the particular hardware and printer in use. These might
change according to the printer setup values.
- The remaining bytes are document text. The offset address in
bytes 4 - 5 gives the start of the document text. This is dependent on
the size of the printer information and this can obviously change from
one system to another.
- Other systems may well have different printers or setup parameters
which change some of the bytes that we found to be constant. In general
though, there will be a reasonable number of constant known plaintext
bytes.
Analysis
The encryption algorithm was found to be the same as that used in the
4.2 version [2]. The main differences between version 4.2 and 5.0 are
in the file formats.
Bytes 0 - 15 of the original and encrypted files contain some useful
information. The offset address in bytes 4-5 gives the starting point
of the document text. The checksum of the key in the encrypted file is
in bytes 12 - 13. This gives the key directly if the key is a single
character.
The encryption of the file starts at byte number 16, so all the printer
information as well as the document is encrypted.
The Encryption Algorithm
- Firstly, the ciphertext is XORed with an ascending sequence of
bytes based on the sequence in Hexadecimal :
02 03 04 ... 79 7A 7B ... FD FE FF 00 01 02 03 ...
Note that the sequence repeats from 00 not 02 after reaching FF. The
keylength determines the starting point of the sequence to be used, ie.
starting point = keylength + 1
For example, for key = QWERTY the starting point of the ascending
sequence would be at position 6 in the sequence giving a starting value
of 07.
- Secondly, the resulting text is XORed with the key characters in
blocks of key length, to restore the original plaintext. This type of
polyalphabetic substitution is called a Vigenere cipher. The analysis
of Vigenere ciphers is well known and covered in the standard
cryptography literature e.g. [3,4,5].
Plan of attack
In the 4.2 version, the only text encrypted was that contained in the
actual document. This is unknown plaintext. In version 5.0, however,
the printer information as well as the document text is encrypted. We
have identified bytes 16 - 21, 24 - 27, 29 - 41, 43 - 45 as being
constant for a particular system (as defined earlier, a particular
licenced copy of WordPerfect on a particular PC and printer), and they
do not change markedly from one system to another.
So we have the ideal situation of known plaintext for a reasonable
number of bytes. This can greatly simplify our attack as it makes it
possible to recover the actual key. Then it is trivial to recover the
plaintext by using WordPerfect to retrieve the file using the
recovered key as the ''password''. Alternatively, a program could be
written to do this as the encryption/decryption algorithm is known. We
outline a strategy with the following example from one particular
system:
Document text consists of three lines of ten ASCII zeros each. The
size of the original file and the encrypted file is 651 bytes.
0000000000
0000000000
0000000000
Plaintext file contains in hexadecimal (for a particular printer):
BYTES 0-15 FF 57 50 43 6B 02 00 00-01 0A 00 00 00 00 00 00
16-31 FB FF 05 00 32 00 2D 02-00 00 07 00 11 00 00 00
32-47 42 00 00 00 02 00 56 00-00 00 53 00 00 00 0C 00
........
619-623 30 30 30 30 30
624-639 30 30 30 30 30 0A 30 30-30 30 30 30 30 30 30 30
640-650 0A 30 30 30 30 30 30 30-30 30 30
Ciphertext file contains in hexadecimal:
BYTES 0-15 FF 57 50 43 6B 02 00 00-01 0A 00 00 6E 50 00 00
16-31 B0 B4 42 41 7E 47 6C 46-53 53 58 59 45 5F 59 4C
32-47 19 5B 57 51 5E 57 07 74-63 63 3C 69 64 6F 65 7C
.......
619-623 19 14 1F 19 0C
624-639 1B 1B 17 11 1C 2D 11 14-03 03 0F 09 04 0F 09 1C
640-650 31 0B 07 01 0C 07 01 E4-F3 F3 FF
We will illustrate a known ciphertext only attack, even though we
obviously know the exact plaintext in this particular example. So we
assume that we have a ciphertext file produced on some other hardware
system using a different licenced copy of WordPerfect. As explained
earlier, we can be confident that a substantial portion of text is
common to all systems. Thus to summarise, the known plaintext we have
is
BYTES 16 - 21 known
BYTES 22 - 31 known except for 22, 23, 28
BYTES 31 - 39 known
BYTES 40 - 47 known except for 42, 46, 47
- Firstly, look at bytes in positions 12 - 15 in the ciphertext file
above which contain the checksum of the key. If the key is one
character, it will be evident in byte number 12. For longer keys bytes
12 and 13 are probably non zero. In this example the checksum is 6E 50
which implies a key size greater then 1.
- Now we consider bytes 16 - 47. For byte number 16, we will try to
deduce the key character used. To do this, choose a keylength starting
with likely values say, 4 to 10 characters. Then XOR the plaintext
characters with the ascending sequence (in the algorithm section)
starting with position keylength which has the value keylength + 1.
Then XOR that result with the associated ciphertext and the key
character should result. For example,
keylength of 4 keylength of 8
plaintext FB 1111 1011 FB 1111 1011
sequence 05 0000 0101 09 0000 1001
xor --------- ---------
1111 1110 1111 0010
cipher B0 1011 0000 B0 1011 0000
xor --------- ---------
0100 1110 0100 0010
=> key 4 E 4 2
Thus we get the following table:
keylength starting possible
sequence key character
4 05 4E
5 06 4D
6 07 4C
7 08 43
8 09 42
9 0A 41
10 0B 40
- Now for a keylength of 4, byte 16 gives a possible key character
of 4E. Bytes 20, 24, 28 ... must also have been created from the same
key character, so we deduce a potential key character for these other
bytes to see if it is also 4E. It turns out that the other potential
key characters are not 4E.
- So we take the next possible key length, 5. Deduce the key
character for bytes 21, 26 .. to see if they match the value for byte
number 16 for that keylength, which is 4D. They do not.
- When a match is obtained for the first key character, deduce the
key characters for the remaining positions. We show the full analysis
for bytes 16 - 31 for a keylength of 8. As we stated earlier, bytes at
positions 22, 23 and 28 are unknown, and we signify these as ??.
BYTES 16 - 23
plaintext FB FF 05 00 32 00 ?? ??
sequence 09 0A 0B 0C 0D 0E 0F 10
xor ------------------------------
F2 F5 0E 0C 3F 0C ?? ??
cipher B0 B4 42 41 7E 47 6C 46
xor ------------------------------
=> key 42 41 4C 4D 41 49 ?? ??
BYTES 24 - 31
plaintext 00 00 07 00 ?? 00 00 00
sequence 11 12 13 14 15 16 17 18
xor ------------------------------
11 12 14 14 ?? 16 17 18
cipher 53 53 58 59 45 5F 59 4C
xor -------------------------------
=> key 42 41 4C 4D ?? 49 4E 54
- The repeating sequence for the key is obvious, even with three
unknown bytes at positions 22, 23 and 28, and so the key characters
are:
42 41 4C 4D 41 49 4E 54
B A L M A I N T
- Further checks on the key could be done using the known bytes from
32-47, if the repeating pattern of the key characters is ambiguous.
- In general, the probability of deducing the key bytes is dependent
on the keylength. Some definitions relating to the key byte are
useful:
- Known: the key byte may be determined at two or more
different positions which correspond to known plaintext.
- Possible: the key byte may be determined at only one position.
- Unknown: the key byte may not be determined as there is no
overlap of this byte with known plaintext.
In summary, for a keylength of 1-9, the key bytes are all known and
thus all of the key may always be deduced. For a keylength of 10-13,
15-17, there is a small proportion of possible to known key bytes. Thus
all the key may be deduced with a high probability. Keys with
keylengths of 14, 18-24 contain one, two or three unknown key bytes and
an increasingly high proportion of possible to known key bytes. At
least five bytes of the key may always be determined.
- Retrieve the plaintext using WordPerfect with the
key as the password. This is the easiest way to decrypt
the document text.
- If no access to WordPerfect is available, then it is
straightforward to recover the plaintext with a short C
program which implements the decryption algorithm as described
previously. This has been done successfully.
Conclusion
The encryption key is easily recovered in an apparent known ciphertext
only attack, as the system provides enough known plaintext in the
printer information regardless of the document plaintext. The
analysis, as shown, can literally be done on the back of a (large)
envelope.
The analysis may be slightly more difficult where the physical system
on which the files were prepared is completely unknown and vastly
different to any system we have encountered, as this may reduce the
amount of known plaintext. In these situations, statistical analysis
based on the characteristic frequencies of characters in a language is
used to decipher text files. This is a standard method which is
straightforward although a program may have to be written.
In summary, the cryptanalysis of files encrypted with the 'locked
document' option in WordPerfect version 5.0 is remarkably simple. The
inclusion of portions of known plaintext in the encrypted file is a
fatal flaw in the system, since it provides a mechanism of attack in
which the key can be recovered by hand, and document plaintext easily
retrieved. All of the key can easily be recovered for keylengths of
1-13 and 15-17, far in excess of commonly used passwords of 8
characters. A high proportion of the key can be deduced for keylengths
of 14 and 18-24. The cipher used is too weak, providing little or no
protection.
If the attacker has knowledge of any other unencrypted file from the
same system, the analysis is made even more simple. We stress that
both the key and the plaintext can be recovered, independent of
the content of the plaintext.
The worst problem is that it may give a false sense of security. For
example, an attacker may decrypt a document, modify it and re-encrypt
so that the originator is unaware of the alterations. We conclude that
the file security is not consistent with claims made by the
manufacturer and is not sufficent to protect sensitive documents from
anything but the most naive attack.
References
- WORDPERFECT CORPORATION (1989): WordPerfect for IBM Personal
Computers.
- BENNETT, J (1987): Analysis of the encryption algorithm
used in the WordPerfect Word Processing Program,
Cryptologia, Vol XI. No 4. pp 206-210.
- KONHEIM, A G (1981): Cryptography, A Primer, Wiley.
- DENNING, D E (1981): Cryptography and Data Security,
Addison Wesley.
- CARROLL, J and Robbins, L E (1989): Computer Cryptanalysis
of Product Ciphers, Cryptologia, Vol XIII. No 4. pp 303-326.
Biographical
Helen Bergen is a Lecturer in the School of Computing Science, Faculty
of Information Technology, at the Queensland University of Technology.
Her research interests within the Information Security Research Centre,
Faculty of Information Technology, include cryptology and the
application of supercomputers.
Bill Caelli is Director of the Information Security Research Centre
within the Faculty of Information Technology at the Queensland
University of Technology. He is also Technical Director and Founder of
ERACOM Pty. Ltd., a manufacturer of cryptographic equipment. His
research interests lie in the development and application of
cryptographic systems to enhance security, control and management of
computer and data network systems.
John Gilmore {sun,pacbell,uunet,pyramid}!hoptoad!gnu gnu@toad.com
The Gutenberg Bible is printed on hemp (marijuana) paper. So was the July 2,
1776 draft of the Declaration of Independence. Why can't we grow it now?
MfG Harald
Kryptographie - Verschiedenes
Meine Homepage
UNIX-AG Homepage
Impressum