ZIP Attacks with Reduced Known-Plaintext

Michael Stay

AccessData Corporation

2500 N. University Ave. Ste. 200

Provo, UT 84606

[email protected]

published by +Tsehp March 2001

Abstract. In [BK94] Biham and Kocher demonstrated that the PKZIP stream cipher was weak and presented an attack requiring thirteen bytes of plaintext.The deflate algorithm ďzippersĒ now use to compress the plaintext before encryption makes it difficult to get known plaintext.We consider the problem of reducing the amount of known plaintext by finding other ways to filter key guesses.In most cases we can reduce the amount of known plaintext from the archived file to two or three bytes, depending on the zipper used and the number of files in the archive.For the most popular zippers on the Internet, there is a fast attack that does not require any information about the files in the archive.


PKZIP is a compression / archival program created by Phil Katz.Katz had the foresight to document his file format completely in the file APPNOTE.TXT, distributed with every copy of PKZIP; there are now literally hundreds of ďzipperĒ programs available, and the ZIP file format has become a de facto standard on the Internet.


In [BK94] Biham and Kocher demonstrated that the PKZIP stream cipher was weak and presented an attack requiring thirteen bytes of plaintext.Eight bytes of the plaintext must be contiguous and all of the bytes must be the text that was encrypted, which is usually compressed plaintext.[K92] shows that the compression method used at the time, implode, produces many predictable bytes suitable for mounting their attack.


Most zippers available today only implement one of the compression methods defined in APPNOTE.TXT, called deflate.This is in part due to [BK94]ís results, and in part to the better compression ratio it provides.Deflate uses Huffman coding followed by a variant of Lempel-Ziv.Once the dictionary reaches a certain size, the process starts over.Itís easy to see that the codes for any of the plaintext depend on a great deal of surrounding plaintext, and that thereís almost no way to get it right unless you have the beginning of the file.The difficulty of getting known plaintext was one reason Phil Zimmerman decided to use it in PGP [PGP98].Practically speaking, if one has enough of the original file to get the thirteen bytes of plaintext required for the attack in [BK94], one has enough to break the encryption almost instantly.


Without the original file, all is not lost; we have the fileís type as indicated by its extension, and we have its size.We have at least one byte of known plaintext used for filtering incorrect passwords.Zippers also encrypt output from a pseudorandom number generator that may be vulnerable to attack.


It is the authorís opinion that the only reason the PKZIP cipher has held up so well in light of [BK94] is due to the deflate algorithm and the difficulty of getting enough plaintext.This paper treats the question of how far we can reduce the plaintext requirement and still break the cipher in a reasonable amount of time.


1.1The PKZIP Stream Cipher

The PKZIP stream cipher was designed by Roger Schaffely and is fully described in the file APPNOTE.TXT found in most PKZIP distributions.The internal state of the cipher consists of three 32-bit words: key0, key1, and key2.These values are initialized to 0x12345678, 0x23456789, and 0x34567890, respectively.The internal state is updated by mixing in the next plaintext byte; we follow [BK94] in our description, combining the decrypt_byte() and update_keys() functions described in APPNOTE.TXT into a single function:


unsigned char PKZIP_stream_byte (unsigned char pt)


†††† unsigned short temp;

†††† key0 = crc32 (key0, pt);

†††† key1 = (key1+(key0 & 0xFF)) * 0x08088405 + 1;

†††† key2 = crc32 (key2, key1 >> 24);

†††† temp = (key2 & 0xFFFC) | 2;

†††† return ( (temp * (temp ^ 1)) >> 8)&0xFF;



where ^ is XOR, | is OR, & is AND, and >> is right shift.


For the purposes of this paper, we define crc32() to be


#define crc32(crc, b) ((crc >> 8) ^ crctab[(crc & 0xFF) ^ b])


The old CRC state is shifted right eight bits and XORed with the 32-bit entry of a byte-indexed table to produce the new state.The index is the low byte of the old state XORed with b.The function is linear; that is,

crctab[x ^ y] = crctab[x] ^ crctab[y].

The cipher is keyed by encrypting the userís password and throwing away the corresponding stream bytes. The stream bytes produced after this point are XORed with the plaintext to produce the ciphertext.


The crux of all of our attacks is the fact that there is almost no diffusion in the internal state.Of the ninety-six bits of internal state, eight bits of key0 affect key1; eight bits of key1 affect key2; and fourteen bits of key2 affect the output stream byte.

1.2Encrypted File Format

Zippers must prepend twelve bytes to the beginning of the file to be encrypted.The ZIP file format specifies that the first eleven should be random and that the last should be the low byte of the archived fileís CRC.The entire CRC is stored in plaintext, and this byte serves as a password filter.Some zippers, like InfoZIP [IZ] and WinZip [WZ], store ten random bytes and the low two bytes of the CRC.


We assume that deflate was the algorithm used to compress the underlying data. The author did a crude statistical test on a few hundred files of varying types and sizes and found that given the file type, as indicated by its extension, and its size, one can guess about the first two and a half bytes of the compressed file.Since the checksum bytes are at the very end of the prepended header, we can use them to augment the plaintext from the file in mounting our attacks.

2Biham and Kocherís Attack

For completeness, we review [BK94]ís results.We begin with some terminology.Bits are numbered from right to left: bit 0 is the onesí place, bit 1 is the twosí place, bit 2 is the foursí place, etc.Let pi be the ith known-plaintext byte, i = 1, 2, 3, ...Let si be the ith stream byte.Let key0i, key1i, and key2i be the value of key0, key1, and key2 after processing pi.Note that s1 is a function of the random header and the password; it is independent of the plaintext.In general, bits 2 through 15 of key2i determine si+1.


Their attack proceeds as follows:


XOR ciphertext and known plaintext to get known stream bytes s1 through s13.

Guess 22 bits of key213.

This guess combined with s13 is enough to fill in eight more bits of key213, for a total of thirty.s12 provides enough information to derive 30 bits of key212 and the most significant of key113.In general, each stream byte si allows us to calculate thirty bits of key2i-1 and the most significant byte ofkey1i.

We continue using stream bytes to make a list of the most significant bytes of key113 through key18.

For each list, we find 216 possibilities for the low 24 bits of key113 through key19 by calculating the low byte of (key1i+LSB(key0i+1)) such that you get the right high byte of key1i+1.

From each of the 216 lists of complete key1ís, derive the low bytes of key013 through key010.

Once we have the low bytes of key010, key011, key012, and key013, we can use our knowledge of the plaintext bytes to invert the CRC function, since itís linear, and find the complete internal state at one point along the encryption.

Once we have the complete internal state, we can decrypt backwards as far as we want; we decrypt the ciphertext corresponding to p1 through p5 and filter out wrong keys.


We can break a file with work equivalent to encrypting around 238 bytes and negligible memory.We need a total of thirteen bytes of known plaintext: eight for the attack, and five to filter the 238 keys that remain.

2.1Minor Improvement in the Amount of Plaintext Required

[BK94] throws away six bits in key17.By using them, we can reduce the plaintext requirement to twelve bytes at the cost of increasing the work factor by four.

2.2More Files in the Archive

If we have more than one file in the archive, we can make the reasonable assumption that they were encrypted with the same password.ďZippersĒ encrypt at least one check byte into every encrypted file to verify that the user entered the correct password. Once we have the complete internal state of the cipher, we can run it backwards to the beginning of the file and read out key0, key1, and key2.Since this state is the same at the beginning of each file (it only depends on the password), we can decrypt the check byte in each file and use it to filter with instead of known plaintext from a single file.This also works if the files are in different archives, but have the same password.


Since the attack is so fast, we can afford to guess a couple of stream bytes.If we guess N of them, we get 240 + 8N keys to filter; this increases the amount of work and the number of additional files we need, but reduces the amount of known plaintext required from a single file.


Consider the N=1 case: if the file was created in a zipper with two checksum bytes, we can break the file with work equivalent to encrypting 248 bytes.We need two checksum bytes followed by only four more known plaintext bytes in one file, and three other files in the archive (six check bytes) to filter the 248 possible keys.


If there is only one checksum byte, we can break the file with the same amount of work, but we need seven files in the archive and five bytes of known plaintext in addition to the checksum byte.

3Divide and Conquer

The limited diffusion of the internal state prompts us to ask how much of the state we need to guess to process one byte.If it is small enough, we can guess it and filter out keys that wonít work with our known stream bytes, then proceed to the next part.


It turns out that we can get by with as few as 23 bits.Note that we donít need to guess 16 bits of key00 to calculate the low byte of key01: if we distribute the XOR in the definition of crc32(), we see that we only need to guess 8 bits of crc32(key00, 0):


LSB(key01) = LSB(crc32(key00, p1))

††††††††† = LSB(crc32(key00, 0)) ^ LSB(crctab[p1]).


Now we distribute the multiplication across the addition in the next step:

MSB(key11) = MSB( (key10 + LSB(key01)) * 0x08088405 + 1 )

(A)††††† = MSB(LSB(key01) * 0x08088405) +

(B)†††††††† MSB(key10 * 0x08088405) + possible carry bit.


We separate the equation into parts we know (A) and parts we need to guess (B), and find we need to guess nine bits, including a possible carry bit.Note that since we know the low bits of (LSB(key01) * 0x08088405),the carry bit will usually give us more than one bit of information in the form of an upper or lower bound on the rest of (key10 * 0x08088405) that we havenít guessed yet.


Given a stream byte si+1, we can find sixty four values for bits 2..15 of key2i.Itís easy to see why: fourteen bits of key2i produce eight bits of si+1, so there are six left over.We can create a table of 256 x 64 bytes such that given si+1 and bits 10..15 of key2i, we can look up bits 2..9 of key2i. We call this the preimage table.


We guess bits 10..15 of crc32(key20, 0) and use s2, the preimage table, and crctab[MSB(key11)] to find bits 2..9 of crc32(key20, 0).We end up with 223 key guesses.


To find the next part of the internal state, we have to guess about the same amount. This guess is not illustrated, but we basically guess about eight more bits of information in each of the three keys.The only complicated part is separating what we know about key1 from what we donít.


We guess bits 8..15 of crc32(key00, 0) directly; the next guess involving key1 is a little more complicated:


MSB(key12) = MSB( (key11 + LSB(key02)) * 0x08088405 + 1 )


†††††† †††† = MSB( LSB(key02) * 0x08088405 ) +

†††††† †††††† MSB( key11 * 0x08088405 ) + possible carry bit



†††††† †††† = MSB( LSB(key02) * 0x08088405 ) +

†††††† †††††† MSB( (key10 + LSB(key01)) * 0xD4652819 ) +

possible carry bit


(A)††††††† = MSB( LSB(key02) * 0x08088405 ) +

(A)††† †††† MSB( LSB(key01) * 0xD4652819 ) +

(B)††† †††† MSB( key10 * 0xD4652819 ) + possible carry bit.


Again, (A) is known and we have to guess (B) nine bits, including a possible carry bit.The carry bit establishes an upper or lower bound on (key10 * 0xD4652819).We end this filter by guessing bits 16..23 and bits 0..1 of crc32(key20, 0) and calculating a stream byte.We have guessed 27 more bits, but the output byte has to match s3, so we expect 223+27-8 = 242 key guesses to pass this filter.


At this point, we have guessed 24 bits of crc32(key20, 0) and we know s1.From this we can calculate, on average, one full value of key20.There are also only around 213 possibilities for key10 due to the restrictions from the carry bits.So the third stage consists of guessing bits 16..23 of crc32(key00, 0) and running through the 213 possible values for key10.We expect 242+13+8-8 = 255 key pieces to pass this filter.


Finally, we guess the last eight bits of key00 and we have a complete internal state.We will have 263 complete keys to filter with other bytes, whether they are in the archived file or in checksum bytes in other files.The cost is approximately the same as encrypting 263 bytes under the stream cipher.The plaintext requirement is four bytes total; at least one of these may come from the fileís own check byte(s).


This is 128 times faster than guessing three stream bytes and using [BK94].

4Random Number Generation

InfoZIP is a cross-platform freeware zipper distribution.Because the C source code is readily available and is free, it forms the basis of most non-PKZIP zippers, including the very popular WinZip and NetZip*.


APPNOTE.TXT does not specify how to generate the prepended random bytes; it only says that they are used to scramble the internal state of the cipher and are discarded after decryption.InfoZIP implements it as follows:


srand(time(NULL) ^ getpid())

For each file in the order they are stored,

Generate ten random bytes by calling rand() ten times and discarding all but the high eight bits of each return value.

Initialize the cipher with the password.

Encrypt the ten random bytes.

Append the low two bytes of the checksum.

Reinitialize the cipher with the password and encrypt the twelve-byte header and the compressed file.


rand() is usually implemented as a truncated linear congruential generator.WinZip and NetZip use Microsoft Visual C++ís implementation, which has a 31-bit seed:


††† unsigned long seed;


void srand( unsigned long s ) { seed = s; }


unsigned short rand()


†††††† seed = 0x343FD * seed + 0x269EC3;

†††††† return ( ( seed >> 16) & 0x7FFF );



Let ri, j be the jth random byte in the ith archived file; i, j = 1, 2, 3, ... Note that the internal state of the cipher is the same both times ri, 1 is encrypted.Since XOR is its own inverse, ri, 1 is decrypted for all i.Also, every r1, j reveals the high eight bits of the internal state of the random number generator.


Since rand() is linear, we can compute two new constants for a generator such that it outputs every tenth output of the original.We know the upper eight bits of the generator, so we guess the low 23 bits and start generating every tenth output and comparing them to the leaked bytes.Four archived files suffice to uniquely determine the seed that was used in the random number generator, and therefore every ri,j.


Let us emphasize that we do not have known plaintext at this point, in the sense that [BK94] requires, because the plaintext was encrypted twice, and we do not know the actual values of the stream bytes.What we can derive is the XOR of the stream bytes in the first and second encryption.

5Parallel Divide and Conquer

We can adapt the divide-and-conquer algorithm from section 3 to use this information.Once we know the ďrandomĒ headers, we can exploit the fact that the internal state was the same at the beginning of each embedded file and filter guesses with multiple known plaintext bytes in parallel, instead of being restricted to one byte as in section 3.


Let si,j,k be the jth stream byte of the kth encryption of the bytes in the ith archived file; i, j = 1, 2, 3, ...; k = 1, 2.We guess the same 23 bits as in section 3, but since we donít know the actual value of s1,2,1, we have to guess it, too.It is equivalent, and more convenient, to guess bits 2..9 of crc32(key20, 0).Now we have a prediction for s1,2,1, and can derive s1,2,2.We donít have any information at all about si,1,1, since itís the same as si,1,2 and cancels out.We guess it, and check to see that the second encryption spits out s1,2,2.We have to guess a carry bit for the second encryption, too, so of the 223+8+8+1 = 240 key guesses, we expect 240-8 = 232 key pieces to pass this filter on this file.


We want to filter out all but the correct guess at this stage; fortunately, we know that the state we are trying to guess was the same at the start of each encryption.We have an eight-bit value to filter with in each file, si,2,1 XOR si,2,2, but we also guess two carry bits, so with four more files in the archive, we can reduce the number of false positives to around 232 - 6*4 = 256.Note that we now have ten carry bits putting restrictions on key10 instead of just one.


We continue to the next byte of each file.This time we guess the same 26 bits as in section 3 plus two carry bits, one for each encryption.With five files, we have 30 bits to filter with.We expect that 28+26+2-30 = 26 = 64 key guesses survive the second stage.Total work for this stage is 28+26+2 = 236 byte encryptions.


At this point we can derive key20 as before.Due to all the carry bit restrictions, we only have on the order of 28 possible key10ís.We guess eight more bits of crc32(key10, 0) and run through all the remaining key10ís.Since we arenít guessing carry bits any more, we have 40 bits to filter with.26+16-40 < 1, and we expect that only the correct guess survives.Finally, we guess the last eight bits of crc32(key10, 0) and only the correct guess survives.


Experimentally, we have found that a key guess passes the second stage only if our guess for si,1,1 is correct.This usually occurs about one quarter of the way through the first 40-bit keyspace.After that, we only try one value for si,1,1 instead of 256 and the rest of the attack takes at most a few minutes.


The work done in the first stage dwarfs the rest of the work needed.The total work is therefore about the same as encrypting 239 bytes.Cracking a file created with this kind of weak PRNG usually takes about two hours on a 500 MHz Pentium II.One can then take the three keys and use [BK94]ís second algorithm to derive a password, if one desires, although the three keys suffice to decrypt the files.


The PKZIP stream cipher is very weak.The deflate algorithm makes it harder to get plaintext, but in most cases we can reduce the plaintext requirement to the point where one can guess enough plaintext based on file type and size alone.The most popular zippers on the internet are also susceptible to an attack that runs in two hours on a single PC based on known plaintext provided by the application and independent of the archived files themselves.


[BK94] Biham, Eli and Paul Kocher. ďA Known Plaintext Attack on the PKZIP Stream Cipher.Ē Fast Software Encryption 2, Proceedings of the Leuven Workshop, LNCS 1008, December 1994.







[K92] Kocher, Paul. ZIPCRACK 2.00 Documentation. 1992.


[PGP98] Userís Guide, Version 6.0.Network Associates, Inc., 1998.p.145.