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TUCoPS :: Unix :: General :: unix4989.htm

VTun VPN Crypto Analysis white paper



10th Jan 2002 [SBWID-4989]
COMMAND

	VTun VPN Crypto Analysis white paper

SYSTEMS AFFECTED

	All versions

PROBLEM

	Jerome Etienne [http://www.off.net/~jme] wrote :
	

	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.

SOLUTION

	See recommandations


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