1 00:00:02,420 --> 00:00:08,870 Let's move on that the security measures for ensuring data confidentiality as we've said the first security 2 00:00:08,870 --> 00:00:17,410 measure developed was what or what key is indirectly used not only for secure data but also to authenticate 3 00:00:17,410 --> 00:00:25,780 computers one access point doesn't encrypt frames it can't decrypt if the frames were encrypted with 4 00:00:25,780 --> 00:00:28,770 an incorrect Webcke it will be rejected. 5 00:00:30,040 --> 00:00:34,040 What are the flaws of the web technology. 6 00:00:34,090 --> 00:00:39,020 One of the fundamental rules of cryptography is that one should not be used frequently. 7 00:00:40,210 --> 00:00:43,540 The more often you use the same key the less secure it becomes 8 00:00:47,610 --> 00:00:53,040 the standard doesn't define any solutions for automating key generation. 9 00:00:53,100 --> 00:00:59,810 If you wanted to change a key you'd have to do this manually and separately for each computer as a result 10 00:00:59,810 --> 00:01:00,610 of this. 11 00:01:00,620 --> 00:01:02,000 These are rarely changed 12 00:01:06,400 --> 00:01:09,030 the encryption used in the standard is weak. 13 00:01:09,310 --> 00:01:14,710 The process can be seen in the picture below. 14 00:01:14,870 --> 00:01:19,040 It's necessary to calculate the checks some of the packet first. 15 00:01:19,170 --> 00:01:23,990 It's an ordinary CRC checksum not a cryptographic hash function. 16 00:01:24,000 --> 00:01:30,310 This means that the well protected packets are vulnerable to tamper and attacker can modify the contents 17 00:01:30,310 --> 00:01:33,540 of the frame and forward it by calculating the correct checksum. 18 00:01:35,650 --> 00:01:43,440 Checksum is hazarded Tuason packet we've mentioned already that this is not secured in any way. 19 00:01:43,520 --> 00:01:48,790 RC for a stream cypher is used for encryption. 20 00:01:48,820 --> 00:01:55,220 We'll talk about this cipher at the end of all modules in secure use of the RC for algorithm is based 21 00:01:55,220 --> 00:02:01,730 on the rule that the key shouldn't be used more than once to encrypt the same data. 22 00:02:01,750 --> 00:02:07,870 If this rule is flouted working the ciphertext is trivially easy. 23 00:02:07,950 --> 00:02:12,070 Or C 4 is based on X or operations. 24 00:02:12,120 --> 00:02:15,560 Let's leave aside cryptographic details for a moment. 25 00:02:15,750 --> 00:02:22,940 The same data is exchanged repeatedly over the network this transmission doesn't relate to people visiting 26 00:02:22,940 --> 00:02:24,910 the same web site. 27 00:02:25,080 --> 00:02:27,420 It's about packet headers having a set structure 28 00:02:29,840 --> 00:02:32,810 predicting field values of the headers is very easy. 29 00:02:33,880 --> 00:02:41,330 Given those randomisation was required to improve security in this case randomisation is provided by 30 00:02:41,330 --> 00:02:44,020 the presence of an initialization vector. 31 00:02:44,210 --> 00:02:50,960 Ivy unfortunately it's just 24 bit 24 bit. 32 00:02:50,960 --> 00:02:55,730 ITV is too small this doesn't end the list of problems though. 33 00:02:55,890 --> 00:02:57,400 We'll return to that soon. 34 00:03:00,710 --> 00:03:09,460 When an I.V. is used data stream has to be mapped to the keystream the exclusive or ex-SO are is used 35 00:03:09,460 --> 00:03:10,610 for this purpose. 36 00:03:12,230 --> 00:03:17,630 The end result is the ciphertext or the encrypted version of a packet that was sent to an access point 37 00:03:20,000 --> 00:03:29,180 length depends on what version it can be a 64 bit or even 256 bit long even if the key is really long 38 00:03:29,540 --> 00:03:32,290 you can ensure the security of a network on its own. 39 00:03:35,760 --> 00:03:40,820 This is due to a small number of initialization vectors. 40 00:03:40,900 --> 00:03:46,630 There's only 16 million seven hundred and seventy seven thousand two hundred sixteen available IVI values 41 00:03:46,630 --> 00:03:49,950 in the pool. 42 00:03:49,950 --> 00:03:55,890 The problem is that the risk of repeated values the risk of collision is much higher than it would seem 43 00:03:55,890 --> 00:03:57,090 at first glance. 44 00:03:58,670 --> 00:04:03,980 It's enough to capture more or less 5000 packets for the risk of finding two packets encrypted with 45 00:04:03,980 --> 00:04:06,340 the same IP to be over 50 percent. 46 00:04:08,530 --> 00:04:14,530 And having the same vector equals having the same key because as we said WEP doesn't support dynamic 47 00:04:14,720 --> 00:04:17,570 change. 48 00:04:17,600 --> 00:04:23,510 This means that we've managed to capture data encrypted with the same key. 49 00:04:23,550 --> 00:04:29,540 All it takes now to extract nearly full information on the key is to combine the ciphertext using ex-SO 50 00:04:29,540 --> 00:04:29,970 or 51 00:04:32,410 --> 00:04:37,250 implementation of this week security mechanism is even worse. 52 00:04:37,420 --> 00:04:44,000 The standard doesn't specify the entities responsible for generating Ivey's this has to be done by a 53 00:04:44,000 --> 00:04:54,300 network card how standard doesn't say initially network card manufacturers use the following system. 54 00:04:55,190 --> 00:05:02,010 One was for the Wii for encrypting the first packet two was for encrypting the second packet three for 55 00:05:02,010 --> 00:05:06,620 the third it was highly predictable. 56 00:05:06,660 --> 00:05:15,820 This means that there can be attacks that consist of assuming a specific value of a vector. 57 00:05:15,830 --> 00:05:20,650 You can also encrypt data but not encrypt headers. 58 00:05:20,670 --> 00:05:25,740 In that case an attacker is not only able to capture physical addresses of the clan he's eavesdropping 59 00:05:25,740 --> 00:05:26,540 on. 60 00:05:26,700 --> 00:05:29,590 You can also modify them. 61 00:05:29,710 --> 00:05:41,370 You can control the packets transmitted in this network. 62 00:05:41,400 --> 00:05:47,630 We've talked about the potential problems connected to adding up checksum to packets this issue will 63 00:05:47,630 --> 00:05:51,720 be noticed for example by korek another's. 64 00:05:51,820 --> 00:05:57,190 If we had a checksum to an encrypted packet there's a one to one relationship between the checks some 65 00:05:57,190 --> 00:05:59,020 bits and the ciphertext 66 00:06:01,590 --> 00:06:06,940 by modifying any checksum bit can determine the validity of any part of the packet. 67 00:06:08,400 --> 00:06:10,900 The checksums will or won't match accordingly. 68 00:06:13,540 --> 00:06:18,820 An access point rejects the packets that have an incorrect checksum. 69 00:06:18,830 --> 00:06:24,100 This is a basic mechanism for maintaining reliability of data sent over wireless networks where there's 70 00:06:24,140 --> 00:06:26,790 a lot of interference. 71 00:06:26,870 --> 00:06:33,490 You can modify a ciphertext byte by byte modify the checksum and see if an access point rejects the 72 00:06:33,490 --> 00:06:36,480 packet. 73 00:06:36,480 --> 00:06:40,770 This will help you make sure whether or not you have correctly cracked the valid combination for the 74 00:06:40,770 --> 00:06:42,090 key for each part. 75 00:06:46,900 --> 00:06:55,110 The RC for algorithm is vulnerable to revealing certain relationships between the key NIV. 76 00:06:55,210 --> 00:06:57,800 How can all of these problems be solved. 77 00:06:57,820 --> 00:07:02,390 Don't use WEP to start with look at the alternative solutions.