Security weaknesses of VTun

[Jerome Etienne <jme () off net>]

Hello,

the following text describes security flaws in vtund. It includes a
description of the security based on the source and lists the 
possible attacks. An attacker can modify packets, replay them,
learn pattern of the plain text or easily guess low-entropy password.

comments are welcome

ps: version in .ps, .pdf and .html can be found in http://www.off.net/~jme


                           Security analysis of VTun

                           Jerome Etienne jme () off net

Abstract

   This text is a security analysis of VTun. It includes a description of the
   security (see section 2) based on the source and lists the possible
   attacks (see section 3). An attacker can modify packets, replay them,
   learn pattern of the plain text or easily guess low-entropy password.

1  Introduction

   From the man page, "VTun provides the method for creating Virtual Tunnels
   over TCP/IP networks and allows to shape, compress, encrypt traffic in
   that tunnels." The analyze has been done on VTun version 2.5b1 which has
   been found at http://vtun.sourceforge.net.

   From the FAQ, "VTun doesn't try to be the MOST secure tunneling software
   in the world, it tries to be ... secure enough instead." In my opinion, it
   is a rather dangerous statement as the definition of the 'enough' entirely
   depends on the user and not on the designer or implementor. This text aims
   to clarify the security provided by VTun.

2  Security description

   The security has been analyzed from the source as the distribution doesn't
   contain any detailed description.

  2.1  packet forwarding

   The forwarded packets are encrypted with blowfish in ECB using MD5( user
   password ) as encryption key (see lfd_encrypt.c). As ECB requires the
   cipher text to be block aligned and blowfish has 64bit blocks, the packet
   is 64bit aligned. The pad is zeros prepended to the packet and the first
   byte of the packet is its length.

  2.2  Connection establishment

   During the connection Establishment, the client authenticates itself to
   the server with a challenge/response scheme (i.e. a simple way to
   authenticate without sending passwords in clear) based on a user password.
   The challenge is 16bytes of random (see VTUN_CHAL_SIZE) chosen by the
   server. They are encrypted with a key equal to MD5( user password ). The
   server sends the encrypted challenge to the client, the client decrypts it
   and replies it.

   The above explanation assumes the HAVE_SSL is defined. If it isn't, the
   authentication is very insecure because the challenges is just XOR-ed with
   the password, and the challenge is based on rand() output which is known
   as easily predicable.

3  Vulnerabilities

   This section explain how an attacker can modify packets, replay them,
   learn pattern of the plain text or easily guess low-entropy password.

  3.1  forwarded packet aren't authenticated

   The forwarded packet aren't authenticated, so an attacker can modify them
   without being detected. The aim of encryption is to make the data
   unreadable for anybody who doesn't know the key. It doesn't prevent an
   attacker from modifying the data. People assume that an attacker won't do
   it because the attacker wouldn't be able to choose the resulting clear
   text. But this section shows that the attacker can choose the resulting
   clear text to some extends and that modifying the cypher text data may be
   interesting even if the attacker ignores the result.

    3.1.1  To insert random data

   If the attacker modifies the cipher text without choosing the resulting
   clear text, it will likely produce random data. The legitimate user won't
   detect the modification and will use them as if they were valid. As they
   likely appears random, it will result of a Denial of Service (aka DoS).

    3.1.2  To insert chosen data

   The encryption mode used by encrypted loop device is ECB[oST81]. ECB
   allows cut/past attacks i.e. the attacker can cut encrypted data from one
   part of a packet and paste them anywhere in another packet. As both data
   sections have been encrypted by the same key, the clear text won't be
   completely random data.

   This lack of authentication isn't a ECB flaw. Authentication isn't
   considered a aim of the encryption mode, so most modes (e.g. CBC, CFB,
   OFB) doesn't authenticate the data. To use another mode would be flawed in
   the same way except if they explicitly protect against forgery. Recently
   some modes including authentication popped up to speed up the encryption /
   authentication couple but as far as i know they are all patented.

  3.2  Easy dictionary attacks

   The authentication is based on a secret key chosen by the user. The key is
   trivially derived from the user password by computing MD5( user password
   ).

   Unfortunately, users often choose low-entropy passwords because those are
   easier to remember, even if it is a bad behavior from a security point of
   view. This allows attackers to try dictionary attacks i.e. to try likely
   password (e.g. jack the ripper). This weakness isn't inherently a VTun
   weakness as the password choice depends on the users. He may choose a
   random password (e.g. /dev/random output) and so won't be vulnerable.

   When the security ultimately relies on a low-entropy password chosen by a
   user, dictionary attacks can't be stopped but they can be made
   sufficiently harder to be impractical (e.g. salt, key derivation
   sufficiently slow). VTun doesn't use those tricks.

  3.3  No anti-replay protection

   VTun doesn't include any protection against packet replay, so an attacker
   who eavesdrops the encrypted packets can successfully replay them later as
   the destination will consider them as legitimate. They can be replayed
   inside the same tunnel or in another instance the tunnel. The attacker can
   even replay them to the source: a packet from A to B can be send to A
   which will accept it.

  3.4  Usage of ECB

   VTun uses blowfish with ECB but ECB doesn't hide the patterns inside the
   plain text. A given plain text block will produce the same cipher text
   block independently of the packet in which it is and of its location
   inside them. The attacker can recognize the repetition of identical cipher
   text blocks and obtain informations on the plain text.

4  Conclusion

   This text describes vulnerabilities of VTun security. An attacker can
   modify packets, replay them, learn pattern of the plain text or easily
   guess low-entropy password. All those attacks are independent and can be
   combined to perform even stronger attacks.

References

   [oST81]
           National Institute of Standards and Technology. implementing and
           using the nbs data encryption standard. Federal information
           processing standards fips74, April 1981.