什么是 WEP 共享密钥认证算法 [英] What is the WEP shared-key authentication algorithm
问题描述
我最近在读一本名为 802.11 无线网络权威指南(第二版)的书.我发现自己无法理解 WEP 共享密钥认证的算法.
I'm reading a book named 802.11 Wireless Networks The Definitive Guide(second edition) recently. I find myself unable to understand the algorithm of WEP shared-key authentication.
在这本书的第 8.3 章共享密钥认证的遗产"一节中，它说
In the book, chapter 8.3, section "The legacy of shared-key authentication", it says
第三帧是移动站对挑战的响应.为了证明它在网络上是允许的，移动站构造了一个包含三个信息元素的管理帧:认证算法标识符、序列号 3 和质询文本.在发送帧之前，移动台使用 WEP (BUT HOW???) 处理帧.将帧标识为认证帧的报头被保留，但信息元素被 WEP 隐藏.
The third frame is the mobile station's response to the challenge. To prove that it is allowed on the network, the mobile station constructs a management frame with three information elements: the Authentication Algorithm Identifier, a Sequence Number of 3, and the Challenge Text. Before transmitting the frame, the mobile station processes the frame with WEP (BUT HOW???). The header identifying the frame as an authentication frame is preserved, but the information elements are hidden by WEP.
所以，我想问一下这里的好心人.
So, I'd like to ask the kind community here.
这是我使用 Tamosoft Commview for wifi 6.3 捕获的 WEP 身份验证会话数据包示例.
Here is my example WEP auth session packets captured with Tamosoft Commview for wifi 6.3.
- AP MAC:000E.2E7C.52A9 (Edimax)
- Wifi 客户端:0020.4A96.23C7 (Lantronix WiPort)
- WEP 密钥是 437B7A57F6762CC7271EBB16FC
你可以在这里找到我的数据包捕获:http://down.nlscan.com/misc/WEP128-shared-key-success-1.ncf
You can find my packet capture here: http://down.nlscan.com/misc/WEP128-shared-key-success-1.ncf
Packet #55,#57,#59,#61 是 WEP 认证数据包.#59是第三帧".
Packet #55,#57,#59,#61 is the WEP authentication packets. #59 is "the third frame".
============================================================================
Packet #55, Direction: Pass-through, Time:16:11:42.634285, Size: 30
Wireless Packet Info
Signal level: 100%
Rate: 2.0 Mbps
Band: 802.11g
Channel: 11 - 2462 MHz
802.11
Frame Control: 0x00B0 (176)
Protocol version: 0
To DS: 0
From DS: 0
More Fragments: 0
Retry: 0
Power Management: 0
More Data: 0
Protected Frame: 0
Order: 0
Type: 0 - Management
Subtype: 11 - Authentication
Duration: 0x0102 (258)
Destination Address: 00:0E:2E:7C:52:A9
Source Address: 00:20:4A:96:23:C7
BSS ID: 00:0E:2E:7C:52:A9
Fragment Number: 0x0000 (0)
Sequence Number: 0x000E (14)
Authentication
Algorithm Number: 0x0001 (1) - Shared Key
Transaction Sequence Number: 0x0001 (1)
Status Code: 0x0000 (0) - Successful
Raw Data:
0x0000 B0 00 02 01 00 0E 2E 7C-52 A9 00 20 4A 96 23 C7 °......|R©. J–#Ç
0x0010 00 0E 2E 7C 52 A9 E0 00-01 00 01 00 00 00 ...|R©à.......
============================================================================
Packet #57, Direction: Pass-through, Time:16:11:42.638429, Size: 160
Wireless Packet Info
Signal level: 100%
Rate: 1.0 Mbps
Band: 802.11g
Channel: 11 - 2462 MHz
802.11
Frame Control: 0x00B0 (176)
Protocol version: 0
To DS: 0
From DS: 0
More Fragments: 0
Retry: 0
Power Management: 0
More Data: 0
Protected Frame: 0
Order: 0
Type: 0 - Management
Subtype: 11 - Authentication
Duration: 0x013A (314)
Destination Address: 00:20:4A:96:23:C7
Source Address: 00:0E:2E:7C:52:A9
BSS ID: 00:0E:2E:7C:52:A9
Fragment Number: 0x0000 (0)
Sequence Number: 0x0343 (835)
Authentication
Algorithm Number: 0x0001 (1) - Shared Key
Transaction Sequence Number: 0x0002 (2)
Status Code: 0x0000 (0) - Successful
Challenge text: 28 B8 9B EC 79 C1 AC B6 24 AD 54 A5 5A 96 EE 24 3E 25 F2 D5 B8 11 1C 2F E9 8D 2B A2 63 EA 3D 1F 40 6E 8C 3D 2C 7E 37 E9 5C 9C F4 0E F2 9C 50 88 21 DA 35 09 97 AE E3 BA 4E 56 77 9A B4 B1 F2 34 E9 AD
Raw Data:
0x0000 B0 00 3A 01 00 20 4A 96-23 C7 00 0E 2E 7C 52 A9 °.:.. J–#Ç...|R©
0x0010 00 0E 2E 7C 52 A9 30 34-01 00 02 00 00 00 10 80 ...|R©04.......€
0x0020 28 B8 9B EC 79 C1 AC B6-24 AD 54 A5 5A 96 EE 24 (¸›ìyÁ¬¶$T¥Z–î$
0x0030 3E 25 F2 D5 B8 11 1C 2F-E9 8D 2B A2 63 EA 3D 1F >%òÕ¸../é+¢cê=.
0x0040 40 6E 8C 3D 2C 7E 37 E9-5C 9C F4 0E F2 9C 50 88 @nŒ=,~7é\œô.òœPˆ
0x0050 21 DA 35 09 97 AE E3 BA-4E 56 77 9A B4 B1 F2 34 !Ú5.—®ãºNVwš´±ò4
0x0060 E9 AD 8D 98 05 28 A1 AD-3F DA 66 05 60 66 EA 24 é˜.(¡?Úf.`fê$
0x0070 02 DA 14 AC 66 CD DC E6-93 A8 79 23 70 87 39 44 .Ú.¬fÍÜæ"¨y#p‡9D
0x0080 17 4E 0F AC A2 CA 9F 84-5F 94 66 3C 04 AB 86 8E .N.¬¢ÊŸ„_"f<.«†Ž
0x0090 99 78 AB C9 E9 C0 91 95-9E 52 B1 7C 6B 22 63 C0 ™x«ÉéÀ‘•žR±|k"cÀ
============================================================================
Packet #59, Direction: Pass-through, Time:16:11:42.639825, Size: 168
Wireless Packet Info
Signal level: 100%
Rate: 2.0 Mbps
Band: 802.11g
Channel: 11 - 2462 MHz
802.11
Frame Control: 0x40B0 (16560)
Protocol version: 0
To DS: 0
From DS: 0
More Fragments: 0
Retry: 0
Power Management: 0
More Data: 0
Protected Frame: 1
Order: 0
Type: 0 - Management
Subtype: 11 - Authentication
Duration: 0x0102 (258)
Destination Address: 00:0E:2E:7C:52:A9
Source Address: 00:20:4A:96:23:C7
BSS ID: 00:0E:2E:7C:52:A9
Fragment Number: 0x0000 (0)
Sequence Number: 0x000F (15)
Authentication
Algorithm Number: 0x1300 (4864) - Reserved
Transaction Sequence Number: 0x00F6 (246)
Status Code: 0xB4BA (46266) - Reserved
Raw Data:
0x0000 B0 40 02 01 00 0E 2E 7C-52 A9 00 20 4A 96 23 C7 °@.....|R©. J–#Ç
0x0010 00 0E 2E 7C 52 A9 F0 00-00 13 F6 00 BA B4 A9 F5 ...|R©ð...ö.º´©õ
0x0020 77 E9 5D 1F A2 B2 CE 3A-AD 1E FD 31 EA 55 90 B8 wé].¢²Î:.ý1êU¸
0x0030 56 F6 EF 81 CE C5 95 B6-9B 2F C4 77 BD E0 DD 73 VöïÎÅ•¶›/Äw½àÝs
0x0040 C6 C8 CE F6 0B 3F 0E 8D-08 15 93 5C 26 6E DA 17 ÆÈÎö.?..."\&nÚ.
0x0050 83 34 A2 53 51 65 3C AE-7A 5C A5 EA 04 97 6E F0 ƒ4¢SQe<®z\¥ê.—nð
0x0060 53 02 02 91 08 51 87 8E-83 38 CD 23 35 E7 56 1B S..‘.Q‡Žƒ8Í#5çV.
0x0070 1D A8 52 8F E1 D4 21 FD-46 41 65 AD 26 AB 74 3D .¨RáÔ!ýFAe&«t=
0x0080 E0 13 12 66 F5 C1 67 B3-71 7F 83 77 A0 34 16 55 à..fõÁg³qƒw 4.U
0x0090 25 96 31 01 A0 9C D9 13-1E 7C E6 8F 15 8D 8A 7B %–1. œÙ..|æ.Š{
0x00A0 8E 6B 65 97 74 0B 23 71- Žke—t.#q
============================================================================
Packet #61, Direction: Pass-through, Time:16:11:42.640916, Size: 30
Wireless Packet Info
Signal level: 100%
Rate: 1.0 Mbps
Band: 802.11g
Channel: 11 - 2462 MHz
802.11
Frame Control: 0x00B0 (176)
Protocol version: 0
To DS: 0
From DS: 0
More Fragments: 0
Retry: 0
Power Management: 0
More Data: 0
Protected Frame: 0
Order: 0
Type: 0 - Management
Subtype: 11 - Authentication
Duration: 0x013A (314)
Destination Address: 00:20:4A:96:23:C7
Source Address: 00:0E:2E:7C:52:A9
BSS ID: 00:0E:2E:7C:52:A9
Fragment Number: 0x0000 (0)
Sequence Number: 0x0344 (836)
Authentication
Algorithm Number: 0x0001 (1) - Shared Key
Transaction Sequence Number: 0x0004 (4)
Status Code: 0x0000 (0) - Successful
Raw Data:
0x0000 B0 00 3A 01 00 20 4A 96-23 C7 00 0E 2E 7C 52 A9 °.:.. J–#Ç...|R©
0x0010 00 0E 2E 7C 52 A9 40 34-01 00 04 00 00 00 ...|R©@4......
============================================================================
我从书中知道 RC4 的工作原理，并且我编写了一个 Python 程序来验证 WEP 如何加密 802.11 数据包.
I know, from the book, how RC4 works and I have written a python program to verify how WEP encrypts a 802.11 packet.
剩下的问题是我无法弄清楚 WEP 身份验证算法是如何工作的(#59 是如何计算的.
The remaining issue is I just cannot figure out how WEP authentication algorithm works(how #59 is calculated.
等待您的慷慨帮助.
推荐答案
我的推测是，当以明文形式发送质询时，移动站选择一个随机 IV 并使用预共享的 WEP 密钥，它使用RC4 当 IV 用作密钥的前 3 个字节时.这 3 个字节通过信道以明文形式发送.然后，正如本书后面提到的，问题从这里开始.任何正在侦听通信的人都可以窃听并找出挑战和密文，将这两者进行异或，攻击者获得密钥流并请求新的挑战，并使用与合法用户相同的 IV 和相同的密钥流对其进行加密.然后他可以简单地进行身份验证，尽管他仍然不知道 WEP 密钥 :)
my speculation says that when the challenge is sent in cleartext, the mobile station picks a random IV and using the pre-shared WEP key, it encrypts the challenge using RC4 when IV is used as the first 3 bytes of the key. These 3 bytes are sent in clear over the channel. Then, the problem starts here as mentioned later in the book. Anyone who is listening to the communication, can eavesdrop and find out the challenge and the ciphertext, xoring these two, the attacker gets the keystream and asks for a new challenge and encrypts it with the same IV as the legitimate user and the same keystream. He can then simply authenticate, though he still does not know the WEP key :)
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