Network Working Group A. Kato
Request for Comments: 5528 NTT Software Corporation
Category: Informational M. Kanda
NTT
S. Kanno
NTT Software Corporation
April 2009
Network Working Group A. Kato
Request for Comments: 5528 NTT Software Corporation
Category: Informational M. Kanda
NTT
S. Kanno
NTT Software Corporation
April 2009
Camellia Counter Mode and Camellia Counter with CBC-MAC Mode Algorithms
Camellia计数器模式和采用CBC-MAC模式算法的Camellia计数器
Status of This Memo
关于下段备忘
This memo provides information for the Internet community. It does not specify an Internet standard of any kind. Distribution of this memo is unlimited.
本备忘录为互联网社区提供信息。它没有规定任何类型的互联网标准。本备忘录的分发不受限制。
Copyright Notice
版权公告
Copyright (c) 2009 IETF Trust and the persons identified as the document authors. All rights reserved.
版权所有（c）2009 IETF信托基金和确定为文件作者的人员。版权所有。
This document is subject to BCP 78 and the IETF Trust's Legal Provisions Relating to IETF Documents in effect on the date of publication of this document (http://trustee.ietf.org/license-info). Please review these documents carefully, as they describe your rights and restrictions with respect to this document.
本文件受BCP 78和IETF信托在本文件出版之日生效的与IETF文件有关的法律规定的约束(http://trustee.ietf.org/license-info). 请仔细阅读这些文件，因为它们描述了您对本文件的权利和限制。
Abstract
摘要
This document describes the algorithms and presents test vectors for the Camellia block cipher algorithm in Counter mode (CTR) and Counter with Cipher Block Chaining MAC mode (CCM). The purpose of this document is to make the Camellia-CTR and Camellia-CCM algorithm conveniently available to the Internet Community.
本文描述了计数器模式（CTR）下的Camellia分组密码算法和密码分组链MAC模式（CCM）下的计数器的算法，并给出了测试向量。本文档的目的是使Camellia CTR和Camellia CCM算法方便地提供给互联网社区。
Table of Contents
目录
1. Introduction ....................................................2
1.1. Terminology ................................................3
2. The Camellia Cipher Algorithm ...................................3
2.1. Key Size ...................................................3
2.2. Weak Keys ..................................................3
2.3. Block Size and Padding .....................................3
2.4. Performance ................................................4
3. Modes of Operation ..............................................4
4. Test Vectors ....................................................4
4.1. Camellia-CTR ...............................................4
4.2. Camellia-CCM ...............................................7
5. Security Considerations ........................................20
6. Acknowledgments ................................................20
7. References .....................................................20
7.1. Normative References ......................................20
7.2. Informative References ....................................20
1. Introduction ....................................................2
1.1. Terminology ................................................3
2. The Camellia Cipher Algorithm ...................................3
2.1. Key Size ...................................................3
2.2. Weak Keys ..................................................3
2.3. Block Size and Padding .....................................3
2.4. Performance ................................................4
3. Modes of Operation ..............................................4
4. Test Vectors ....................................................4
4.1. Camellia-CTR ...............................................4
4.2. Camellia-CCM ...............................................7
5. Security Considerations ........................................20
6. Acknowledgments ................................................20
7. References .....................................................20
7.1. Normative References ......................................20
7.2. Informative References ....................................20
This document describes the use of the Camellia block cipher algorithm in Counter (CTR) mode and Counter with CBC-MAC (CCM) mode.
本文档描述了在计数器（CTR）模式和计数器与CBC-MAC（CCM）模式下使用Camellia分组密码算法。
Camellia is a symmetric cipher with a Feistel structure. Camellia was developed jointly by NTT and Mitsubishi Electric Corporation in 2000. It was designed to withstand all known cryptanalytic attacks, and it has been scrutinized by worldwide cryptographic experts. Camellia is suitable for implementation in software and hardware, offering encryption speed in software and hardware implementations that is comparable to Advanced Encryption Standard (AES) [5].
Camellia是一种具有Feistel结构的对称密码。茶花由NTT和三菱电机公司于2000年联合开发。它被设计来抵抗所有已知的密码分析攻击，并且已经被全世界的密码专家仔细检查过。Camellia适合在软件和硬件中实现，在软件和硬件实现中提供与高级加密标准（AES）相当的加密速度[5]。
Camellia supports 128-bit block size and 128-, 192-, and 256-bit key lengths, i.e., the same interface specifications as the AES. Therefore, it is easy to implement Camellia-based algorithms by replacing the AES block of AES-based algorithms with a Camellia block.
Camellia支持128位块大小和128、192和256位密钥长度，即与AES相同的接口规范。因此，通过使用Camellia块替换基于AES的算法的AES块，可以很容易地实现基于Camellia的算法。
Camellia already has been adopted by the IETF and other international standardization organizations; in particular, the IETF has published specifications for the use of Camellia with IPsec [6], TLS [7], Secure/Multipurpose Internet Mail Extensions (S/MIME) [8], and XML Securiy [9]. Camellia is one of the three ISO/IEC international standard [10] 128-bit block ciphers (Camellia, AES, and Super Effective and Efficient Delivery (SEED)). Camellia was selected as a recommended cryptographic primitive by the EU NESSIE (New European Schemes for Signatures, Integrity and Encryption) project [11] and
Camellia已经被IETF和其他国际标准化组织采用；特别是，IETF已经发布了使用Camellia与IPsec[6]、TLS[7]、安全/多用途Internet邮件扩展（S/MIME）[8]和XML安全[9]的规范。Camellia是三种ISO/IEC国际标准[10]128位分组密码（Camellia、AES和超高效和高效传递（SEED））之一。Camellia被欧盟NESSIE（新欧洲签名、完整性和加密方案）项目选为推荐的加密原语[11]和
was included in the list of cryptographic techniques for Japanese e-Government systems that was selected by the Japanese CRYPTREC (Cryptography Research and Evaluation Committees) [12].
被列入日本CRYPTREC（密码学研究和评估委员会）选定的日本电子政务系统密码技术清单[12]。
Since optimized source code is provided under several open source licenses [13], Camellia has also been adopted by several open source projects (OpenSSL, FreeBSD, Linux, and Firefox).
由于优化的源代码是在几个开源许可下提供的[13]，Camellia也被几个开源项目（OpenSSL、FreeBSD、Linux和Firefox）采用。
The algorithm specification and object identifiers are described in [1].
[1]中描述了算法规范和对象标识符。
The Camellia web site [14] contains a wealth of information about Camellia, including detailed specification, security analysis, performance figures, reference implementation, optimized implementation, test vectors (TVs), and intellectual property information.
Camellia网站[14]包含了大量关于Camellia的信息，包括详细的规范、安全性分析、性能数据、参考实现、优化实现、测试向量（TVs）和知识产权信息。
The key words "MUST", "MUST NOT", "REQUIRED", "SHALL", "SHALL NOT", "SHOULD", "SHOULD NOT", "RECOMMENDED", "MAY", and "OPTIONAL" in this document are to be interpreted as described in RFC 2119 [2].
本文件中的关键词“必须”、“不得”、“要求”、“应”、“不应”、“应”、“不应”、“建议”、“可”和“可选”应按照RFC 2119[2]中所述进行解释。
All multi-octet values in this document are encoded and represented in network byte order, i.e., most significant octet first.
本文件中的所有多个八位字节值均按网络字节顺序编码和表示，即最重要的八位字节优先。
All symmetric block cipher algorithms share common characteristics and variables, including mode, key size, weak keys, and block size. The following sections contain descriptions of the relevant characteristics of Camellia.
所有对称分组密码算法都具有共同的特征和变量，包括模式、密钥大小、弱密钥和块大小。以下各节介绍了山茶花的相关特性。
Camellia supports three key sizes: 128 bits, 192 bits, and 256 bits. The default key size is 128 bits, and all implementations MUST support this key size. Implementations MAY also support key sizes of 192 bits and 256 bits.
Camellia支持三种密钥大小：128位、192位和256位。默认密钥大小为128位，所有实现都必须支持此密钥大小。实现还可以支持192位和256位的密钥大小。
At the time of writing this document, there are no known weak keys for Camellia.
在编写本文档时，没有已知的Camellia的弱键。
Camellia uses a block size of 16 octets (128 bits).
Camellia使用16个八位字节（128位）的块大小。
Padding is required by the algorithm to maintain a 16-octet (128-bit) block size. Padding MUST be added such that the data to be encrypted has a length that is a multiple of 16 octets.
算法需要填充以保持16个八位组（128位）的块大小。必须添加填充，以便要加密的数据的长度为16个八位字节的倍数。
Because of the algorithm-specific padding requirement, no additional padding is required to ensure that the ciphertext terminates on a 4-octet boundary (i.e., maintaining a 16-octet block size guarantees that the Encapsulating Security Payload (ESP) Pad Length and Next Header fields will be right aligned within a 4-octet word). Additional padding MAY be included as long as the 16-octet block size is maintained.
由于算法特定的填充要求，不需要额外的填充来确保密文在4个八位字节的边界上终止（即，保持16个八位字节的块大小可以保证封装安全有效负载（ESP）填充长度和下一个报头字段将在4个八位字节的字内右对齐）。只要保持16个八位组块的大小，就可以包括额外的填充。
Performance figures for Camellia are available at [14]. The NESSIE project has reported on the performance of optimized implementations independently [11].
茶花的性能数据见[14]。NESSIE项目独立报告了优化实现的性能[11]。
Camellia Counter (Camellia-CTR) mode and Camellia Counter with CBC-MAC (Camellia-CCM) mode are based on [3][15][4].
茶花计数器（茶花CTR）模式和带CBC-MAC（茶花CCM）模式的茶花计数器基于[3][15][4]。
CTR mode [3] behaves like a stream cipher, but is based on a block cipher primitive (that is, CTR mode operation of a block cipher results in a stream cipher).
CTR模式[3]的行为类似于流密码，但基于分组密码原语（即，分组密码的CTR模式操作产生流密码）。
CCM mode [15][4] is a generic authenticate-and-encrypt block cipher mode. In this specification, CCM is used with the Camellia [1] block cipher.
CCM模式[15][4]是一种通用的身份验证和加密分组密码模式。在本规范中，CCM与Camellia[1]分组密码一起使用。
This section contains nine TVs, which can be used to confirm that an implementation has correctly implemented Camellia-CTR. The first three TVs use Camellia with a 128-bit key; the next three TVs use Camellia with a 192-bit key; and the last three TVs use Camellia with a 256-bit key.
本节包含九个TV，可用于确认实现已正确实现Camellia CTR。前三台电视使用带128位密钥的茶花；接下来的三台电视使用带有192位钥匙的茶花；最后三台电视使用带256位键的茶花。
TV #1: Encrypting 16 octets using Camellia-CTR with 128-bit key Camellia Key : AE 68 52 F8 12 10 67 CC 4B F7 A5 76 55 77 F3 9E Camellia-CTR IV : 00 00 00 00 00 00 00 00 Nonce : 00 00 00 30 Plaintext : 53 69 6E 67 6C 65 20 62 6C 6F 63 6B 20 6D 73 67 Counter Block (1): 00 00 00 30 00 00 00 00 00 00 00 00 00 00 00 01 Key Stream (1): 83 F4 AC FD EE 71 41 F8 4C E8 1F 1D FB 72 78 58 Ciphertext : D0 9D C2 9A 82 14 61 9A 20 87 7C 76 DB 1F 0B 3F
TV#1：使用Camellia CTR加密16个八位字节，使用128位密钥Camellia密钥：AE 68 52 F8 12 10 67 CC 4B F7 A5 76 55 77 F3 9E Camellia CTR IV:00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 30明文：53 69 6E 67 6C 6F 63 6B 20 6D 73 67计数器块（1）：00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 01密钥流（1）：83 F4 AC FD EE 71 41 F8 4C E8 1F 1D FB 72 78 58密文：D0 9D C2 9A 82 14 61 9A 20 87 7C 76 DB 1F 0B 3F
TV #2: Encrypting 32 octets using Camellia-CTR with 128-bit key Camellia Key : 7E 24 06 78 17 FA E0 D7 43 D6 CE 1F 32 53 91 63 Camellia-CTR IV : C0 54 3B 59 DA 48 D9 0B Nonce : 00 6C B6 DB Plaintext : 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F : 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F Counter Block (1): 00 6C B6 DB C0 54 3B 59 DA 48 D9 0B 00 00 00 01 Key Stream (1): DB F2 C5 8E C4 86 90 D3 D2 75 9A 7C 69 B6 C5 4B Counter Block (2): 00 6C B6 DB C0 54 3B 59 DA 48 D9 0B 00 00 00 02 Key Stream (2): 3B 9F 9C 1C 25 E5 CA B0 34 6D 0D F8 4F 7D FE 57 Ciphertext : DB F3 C7 8D C0 83 96 D4 DA 7C 90 77 65 BB CB 44 : 2B 8E 8E 0F 31 F0 DC A7 2C 74 17 E3 53 60 E0 48
电视#2：使用带128位密钥的Camellia CTR加密32个八位字节Camellia密钥：7E 24 06 78 17 FA E0 D7 43 D6 CE 1F 32 53 91 63 Camellia CTR IV:C0 54 3B 59 DA 48 D9 0B当前值：00 6C B6 DB明文：00 01 02 03 04 05 07 08 09 0A 0B 0C 0D 0E：10 11 12 13 15 16 18 19 1A 1C 1D 1F计数器块（1）：00 6C B6 DB C0 54 3B 59 DA 48 D9 0B 00 00 01密钥流（1）：DB F2 C5 8E C4 86 90 D3 D2 75 9A 7C 69 B6 C5 4B计数器块（2）：00 6C B6 DB C0 54 3B 59 DA 48 D9 0B 00 00 02密钥流（2）：3B 9F 9C 1C 25 E5 CA B0 34 6D 0D F8 4F 7D FE 57密文：DB F3 C7 8D C0 83 96 D4 DA 7C 90 77 65 BB CB 44:2B 8E 8E 0F 31 F0 DC A7 2C 74 17 E3 53 60 E0 48
TV #3: Encrypting 36 octets using Camellia-CTR with 128-bit key
Camellia Key : 76 91 BE 03 5E 50 20 A8 AC 6E 61 85 29 F9 A0 DC
Camellia-CTR IV : 27 77 7F 3F 4A 17 86 F0
Nonce : 00 E0 01 7B
Plaintext : 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
: 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
: 20 21 22 23
Counter Block (1): 00 E0 01 7B 27 77 7F 3F 4A 17 86 F0 00 00 00 01
Key Stream (1): B1 9C 1D CE CF 70 ED 8F 27 8D 96 E9 41 88 C1 7C
Counter Block (2): 00 E0 01 7B 27 77 7F 3F 4A 17 86 F0 00 00 00 02
Key Stream (2): 8C F7 59 38 48 88 65 E6 57 34 47 86 D2 85 97 D2
Counter Block (3): 00 E0 01 7B 27 77 7F 3F 4A 17 86 F0 00 00 00 03
Key Stream (3): FF 71 A4 B5 D8 86 12 53 6A 9D 10 A1 13 0F 14 F8
Ciphertext : B1 9D 1F CD CB 75 EB 88 2F 84 9C E2 4D 85 CF 73
: 9C E6 4B 2B 5C 9D 73 F1 4F 2D 5D 9D CE 98 89 CD
: DF 50 86 96
TV #3: Encrypting 36 octets using Camellia-CTR with 128-bit key
Camellia Key : 76 91 BE 03 5E 50 20 A8 AC 6E 61 85 29 F9 A0 DC
Camellia-CTR IV : 27 77 7F 3F 4A 17 86 F0
Nonce : 00 E0 01 7B
Plaintext : 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
: 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
: 20 21 22 23
Counter Block (1): 00 E0 01 7B 27 77 7F 3F 4A 17 86 F0 00 00 00 01
Key Stream (1): B1 9C 1D CE CF 70 ED 8F 27 8D 96 E9 41 88 C1 7C
Counter Block (2): 00 E0 01 7B 27 77 7F 3F 4A 17 86 F0 00 00 00 02
Key Stream (2): 8C F7 59 38 48 88 65 E6 57 34 47 86 D2 85 97 D2
Counter Block (3): 00 E0 01 7B 27 77 7F 3F 4A 17 86 F0 00 00 00 03
Key Stream (3): FF 71 A4 B5 D8 86 12 53 6A 9D 10 A1 13 0F 14 F8
Ciphertext : B1 9D 1F CD CB 75 EB 88 2F 84 9C E2 4D 85 CF 73
: 9C E6 4B 2B 5C 9D 73 F1 4F 2D 5D 9D CE 98 89 CD
: DF 50 86 96
TV #4: Encrypting 16 octets using Camellia-CTR with 192-bit key Camellia Key : 16 AF 5B 14 5F C9 F5 79 C1 75 F9 3E 3B FB 0E ED : 86 3D 06 CC FD B7 85 15 Camellia-CTR IV : 36 73 3C 14 7D 6D 93 CB Nonce : 00 00 00 48 Plaintext : 53 69 6E 67 6C 65 20 62 6C 6F 63 6B 20 6D 73 67 Counter Block (1): 00 00 00 48 36 73 3C 14 7D 6D 93 CB 00 00 00 01 Key Stream (1): 70 10 57 F9 E6 E8 0B 49 7A 1F 4C AC AB F3 E5 F1 Ciphertext : 23 79 39 9E 8A 8D 2B 2B 16 70 2F C7 8B 9E 96 96
TV#4:使用Camellia CTR和192位密钥加密16个八位字节Camellia密钥：16 AF 5B 14 5F C9 F5 79 C1 75 F9 3E 3B FB 0E ED:86 3D 06 CC FD B7 85 15 Camellia CTR IV:36 73 3C 14 7D 6D 93 CB Nonce:00 00 00 00 48明文：53 69 6E 67 6C 65 62 6C 6F 63 6B 20 6D 73 67计数器块（1）：00 00 00 48 36 73 3C 14 7D 6D 93 CB 00 00 01密钥流（1）：70 10 57 F9 E6 E8 0B 49 7A 1F 4C AC AB F3 E5 F1密文：23 79 39 9E 8A 8D 2B 16 70 2F C7 8B 9E 96
TV #5: Encrypting 32 octets using Camellia-CTR with 192-bit key
Camellia Key : 7C 5C B2 40 1B 3D C3 3C 19 E7 34 08 19 E0 F6 9C
: 67 8C 3D B8 E6 F6 A9 1A
Camellia-CTR IV : 02 0C 6E AD C2 CB 50 0D
Nonce : 00 96 B0 3B
Plaintext : 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
: 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
Counter Block (1): 00 96 B0 3B 02 0C 6E AD C2 CB 50 0D 00 00 00 01
Key Stream (1): 7D EE 36 F4 A1 D5 E2 12 6F 42 75 F7 A2 6A C9 52
Counter Block (2): 00 96 B0 3B 02 0C 6E AD C2 CB 50 0D 00 00 00 02
Key Stream (2): C0 09 AA 7C E6 25 47 F7 4E 20 30 82 EF 47 52 F2
Ciphertext : 7D EF 34 F7 A5 D0 E4 15 67 4B 7F FC AE 67 C7 5D
: D0 18 B8 6F F2 30 51 E0 56 39 2A 99 F3 5A 4C ED
TV #5: Encrypting 32 octets using Camellia-CTR with 192-bit key
Camellia Key : 7C 5C B2 40 1B 3D C3 3C 19 E7 34 08 19 E0 F6 9C
: 67 8C 3D B8 E6 F6 A9 1A
Camellia-CTR IV : 02 0C 6E AD C2 CB 50 0D
Nonce : 00 96 B0 3B
Plaintext : 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
: 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
Counter Block (1): 00 96 B0 3B 02 0C 6E AD C2 CB 50 0D 00 00 00 01
Key Stream (1): 7D EE 36 F4 A1 D5 E2 12 6F 42 75 F7 A2 6A C9 52
Counter Block (2): 00 96 B0 3B 02 0C 6E AD C2 CB 50 0D 00 00 00 02
Key Stream (2): C0 09 AA 7C E6 25 47 F7 4E 20 30 82 EF 47 52 F2
Ciphertext : 7D EF 34 F7 A5 D0 E4 15 67 4B 7F FC AE 67 C7 5D
: D0 18 B8 6F F2 30 51 E0 56 39 2A 99 F3 5A 4C ED
TV #6: Encrypting 36 octets using Camellia-CTR with 192-bit key
Camellia Key : 02 BF 39 1E E8 EC B1 59 B9 59 61 7B 09 65 27 9B
: F5 9B 60 A7 86 D3 E0 FE
Camellia-CTR IV : 5C BD 60 27 8D CC 09 12
Nonce : 00 07 BD FD
Plaintext : 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
: 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
: 20 21 22 23
Counter Block (1): 00 07 BD FD 5C BD 60 27 8D CC 09 12 00 00 00 01
Key Stream (1): 57 11 E7 55 E5 4D 7C 27 BD A5 04 78 FD 93 40 77
Counter Block (2): 00 07 BD FD 5C BD 60 27 8D CC 09 12 00 00 00 02
Key Stream (2): 66 E2 6D CF 85 A4 F9 5A 55 B4 F2 FD 7A BB 53 11
Counter Block (3): 00 07 BD FD 5C BD 60 27 8D CC 09 12 00 00 00 03
Key Stream (3): F5 76 89 74 63 52 A8 C5 1E 82 DE 66 C3 9F 38 34
Ciphertext : 57 10 E5 56 E1 48 7A 20 B5 AC 0E 73 F1 9E 4E 78
: 76 F3 7F DC 91 B1 EF 4D 4D AD E8 E6 66 A6 4D 0E
: D5 57 AB 57
TV #6: Encrypting 36 octets using Camellia-CTR with 192-bit key
Camellia Key : 02 BF 39 1E E8 EC B1 59 B9 59 61 7B 09 65 27 9B
: F5 9B 60 A7 86 D3 E0 FE
Camellia-CTR IV : 5C BD 60 27 8D CC 09 12
Nonce : 00 07 BD FD
Plaintext : 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
: 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
: 20 21 22 23
Counter Block (1): 00 07 BD FD 5C BD 60 27 8D CC 09 12 00 00 00 01
Key Stream (1): 57 11 E7 55 E5 4D 7C 27 BD A5 04 78 FD 93 40 77
Counter Block (2): 00 07 BD FD 5C BD 60 27 8D CC 09 12 00 00 00 02
Key Stream (2): 66 E2 6D CF 85 A4 F9 5A 55 B4 F2 FD 7A BB 53 11
Counter Block (3): 00 07 BD FD 5C BD 60 27 8D CC 09 12 00 00 00 03
Key Stream (3): F5 76 89 74 63 52 A8 C5 1E 82 DE 66 C3 9F 38 34
Ciphertext : 57 10 E5 56 E1 48 7A 20 B5 AC 0E 73 F1 9E 4E 78
: 76 F3 7F DC 91 B1 EF 4D 4D AD E8 E6 66 A6 4D 0E
: D5 57 AB 57
TV #7: Encrypting 16 octets using Camellia-CTR with 256-bit key Camellia Key : 77 6B EF F2 85 1D B0 6F 4C 8A 05 42 C8 69 6F 6C : 6A 81 AF 1E EC 96 B4 D3 7F C1 D6 89 E6 C1 C1 04 Camellia-CTR IV : DB 56 72 C9 7A A8 F0 B2 Nonce : 00 00 00 60 Plaintext : 53 69 6E 67 6C 65 20 62 6C 6F 63 6B 20 6D 73 67 Counter Block (1): 00 00 00 60 DB 56 72 C9 7A A8 F0 B2 00 00 00 01 Key Stream (1): 67 68 97 AF 48 1B DF AC D1 06 F7 1A 6C 76 C8 76 Ciphertext : 34 01 F9 C8 24 7E FF CE BD 69 94 71 4C 1B BB 11
电视#7：使用带256位密钥的Camellia CTR加密16个八位字节Camellia密钥：77 6B EF F2 85 1D B0 6F 4C 8A 05 42 C8 69 6F 6C:6A 81 AF 1E EC 96 B4 D3 7F C1 D6 89 E6 C1 C1 04 Camellia CTR IV:DB 56 72 C9 7A A8 F0 B2 Nonce:00 00 00 00 60明文：53 69 6E 67 6C 65 20 62 6C 63 6B 20 6D 73 67计数器块（1）：00 00 00 60 DB 56 72 C9 7A A8 F0 B2 00 00 00 01密钥流（1）：67 68 97 AF 48 1B DF AC D1 06 F7 1A 6C 76 C8 76密文：34 01 F9 C8 24 7E FF CE BD 69 94 71 4C 1B BB 11
TV #8: Encrypting 32 octets using Camellia-CTR with 256-bit key
Camellia Key : F6 D6 6D 6B D5 2D 59 BB 07 96 36 58 79 EF F8 86
: C6 6D D5 1A 5B 6A 99 74 4B 50 59 0C 87 A2 38 84
Camellia-CTR IV : C1 58 5E F1 5A 43 D8 75
Nonce : 00 FA AC 24
Plaintext : 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
: 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
Counter Block (1): 00 FA AC 24 C1 58 5E F1 5A 43 D8 75 00 00 00 01
Key Stream (1): D6 C2 01 91 20 6A 7E 0F A0 35 21 29 A4 8E 90 4A
Counter Block (2): 00 FA AC 24 C1 58 5E F1 5A 43 D8 75 00 00 00 02
Key Stream (2): F5 0D C6 99 08 CA 56 79 A4 85 D8 C8 B7 9E 5F 17
Ciphertext : D6 C3 03 92 24 6F 78 08 A8 3C 2B 22 A8 83 9E 45
: E5 1C D4 8A 1C DF 40 6E BC 9C C2 D3 AB 83 41 08
TV #8: Encrypting 32 octets using Camellia-CTR with 256-bit key
Camellia Key : F6 D6 6D 6B D5 2D 59 BB 07 96 36 58 79 EF F8 86
: C6 6D D5 1A 5B 6A 99 74 4B 50 59 0C 87 A2 38 84
Camellia-CTR IV : C1 58 5E F1 5A 43 D8 75
Nonce : 00 FA AC 24
Plaintext : 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
: 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
Counter Block (1): 00 FA AC 24 C1 58 5E F1 5A 43 D8 75 00 00 00 01
Key Stream (1): D6 C2 01 91 20 6A 7E 0F A0 35 21 29 A4 8E 90 4A
Counter Block (2): 00 FA AC 24 C1 58 5E F1 5A 43 D8 75 00 00 00 02
Key Stream (2): F5 0D C6 99 08 CA 56 79 A4 85 D8 C8 B7 9E 5F 17
Ciphertext : D6 C3 03 92 24 6F 78 08 A8 3C 2B 22 A8 83 9E 45
: E5 1C D4 8A 1C DF 40 6E BC 9C C2 D3 AB 83 41 08
TV #9: Encrypting 36 octets using Camellia-CTR with 256-bit key
Camellia Key : FF 7A 61 7C E6 91 48 E4 F1 72 6E 2F 43 58 1D E2
: AA 62 D9 F8 05 53 2E DF F1 EE D6 87 FB 54 15 3D
Camellia-CTR IV : 51 A5 1D 70 A1 C1 11 48
Nonce : 00 1C C5 B7
Plaintext : 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
: 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
: 20 21 22 23
Counter Block (1): 00 1C C5 B7 51 A5 1D 70 A1 C1 11 48 00 00 00 01
Key Stream (1): A4 DB 21 FF E2 A0 F9 AD 65 6D A4 91 0A 5F AA 23
Counter Block (2): 00 1C C5 B7 51 A5 1D 70 A1 C1 11 48 00 00 00 02
Key Stream (2): C1 70 B1 58 71 EC 71 88 6D D9 05 0B 03 6C 39 70
Counter Block (3): 00 1C C5 B7 51 A5 1D 70 A1 C1 11 48 00 00 00 03
Key Stream (3): 35 CE 2F AE 90 78 B3 72 F5 76 12 39 1F 8B AF BF
Ciphertext : A4 DA 23 FC E6 A5 FF AA 6D 64 AE 9A 06 52 A4 2C
: D1 61 A3 4B 65 F9 67 9F 75 C0 1F 10 1F 71 27 6F
: 15 EF 0D 8D
TV #9: Encrypting 36 octets using Camellia-CTR with 256-bit key
Camellia Key : FF 7A 61 7C E6 91 48 E4 F1 72 6E 2F 43 58 1D E2
: AA 62 D9 F8 05 53 2E DF F1 EE D6 87 FB 54 15 3D
Camellia-CTR IV : 51 A5 1D 70 A1 C1 11 48
Nonce : 00 1C C5 B7
Plaintext : 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
: 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
: 20 21 22 23
Counter Block (1): 00 1C C5 B7 51 A5 1D 70 A1 C1 11 48 00 00 00 01
Key Stream (1): A4 DB 21 FF E2 A0 F9 AD 65 6D A4 91 0A 5F AA 23
Counter Block (2): 00 1C C5 B7 51 A5 1D 70 A1 C1 11 48 00 00 00 02
Key Stream (2): C1 70 B1 58 71 EC 71 88 6D D9 05 0B 03 6C 39 70
Counter Block (3): 00 1C C5 B7 51 A5 1D 70 A1 C1 11 48 00 00 00 03
Key Stream (3): 35 CE 2F AE 90 78 B3 72 F5 76 12 39 1F 8B AF BF
Ciphertext : A4 DA 23 FC E6 A5 FF AA 6D 64 AE 9A 06 52 A4 2C
: D1 61 A3 4B 65 F9 67 9F 75 C0 1F 10 1F 71 27 6F
: 15 EF 0D 8D
This section contains twenty four TVs, which can be used to confirm that an implementation has correctly implemented Camellia-CCM. In each of these TVs, the least significant sixteen bits of the counter block is used for the block counter, and the nonce is 13 octets. Some of the TVs include an eight octet authentication value, and others include a ten octet authentication value.
本节包含24个TV，可用于确认实现已正确实现Camellia CCM。在每个tv中，计数器块的最低有效16位用于块计数器，nonce为13个八位字节。一些tv包括八个八位字节的认证值，而其他tv包括十个八位字节的认证值。
=============== Packet Vector #1 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 03 02 01 00 A0 A1 A2 A3 A4 A5
Total packet length = 31. [Input (8 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E
CBC IV in: 59 00 00 00 03 02 01 00 A0 A1 A2 A3 A4 A5 00 17
CBC IV out:D4 DB CD 92 A8 96 41 56 1D 0D BB D0 D5 7F 7E 1D
After xor: D4 D3 CD 93 AA 95 45 53 1B 0A BB D0 D5 7F 7E 1D [hdr]
After CAM: BD 84 03 80 73 59 37 B7 CE F5 E4 BA 1B 18 54 DC
After xor: B5 8D 09 8B 7F 54 39 B8 DE E4 F6 A9 0F 0D 42 CB [msg]
After CAM: CE 21 82 9C F6 F2 4D A2 CB 35 D1 FD 81 27 63 EC
After xor: D6 38 98 87 EA EF 53 A2 CB 35 D1 FD 81 27 63 EC [msg]
After CAM: 20 11 FE E2 53 B1 A7 DB 02 77 FA 37 6D 78 EE 10
MIC tag : 20 11 FE E2 53 B1 A7 DB
CTR Start: 01 00 00 00 03 02 01 00 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: B2 7A 7B 8E EB 14 3F 0B 82 E2 98 4C 06 44 CC 42
CTR[0002]: E2 E2 D3 52 98 97 13 45 D1 63 22 90 E7 F8 15 4A
CTR[MIC ]: DC BF 30 96 38 8C 1E 76
Total packet length = 39. [Encrypted]
00 01 02 03 04 05 06 07 BA 73 71 85 E7 19 31 04
92 F3 8A 5F 12 51 DA 55 FA FB C9 49 84 8A 0D FC
AE CE 74 6B 3D B9 AD
=============== Packet Vector #1 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 03 02 01 00 A0 A1 A2 A3 A4 A5
Total packet length = 31. [Input (8 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E
CBC IV in: 59 00 00 00 03 02 01 00 A0 A1 A2 A3 A4 A5 00 17
CBC IV out:D4 DB CD 92 A8 96 41 56 1D 0D BB D0 D5 7F 7E 1D
After xor: D4 D3 CD 93 AA 95 45 53 1B 0A BB D0 D5 7F 7E 1D [hdr]
After CAM: BD 84 03 80 73 59 37 B7 CE F5 E4 BA 1B 18 54 DC
After xor: B5 8D 09 8B 7F 54 39 B8 DE E4 F6 A9 0F 0D 42 CB [msg]
After CAM: CE 21 82 9C F6 F2 4D A2 CB 35 D1 FD 81 27 63 EC
After xor: D6 38 98 87 EA EF 53 A2 CB 35 D1 FD 81 27 63 EC [msg]
After CAM: 20 11 FE E2 53 B1 A7 DB 02 77 FA 37 6D 78 EE 10
MIC tag : 20 11 FE E2 53 B1 A7 DB
CTR Start: 01 00 00 00 03 02 01 00 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: B2 7A 7B 8E EB 14 3F 0B 82 E2 98 4C 06 44 CC 42
CTR[0002]: E2 E2 D3 52 98 97 13 45 D1 63 22 90 E7 F8 15 4A
CTR[MIC ]: DC BF 30 96 38 8C 1E 76
Total packet length = 39. [Encrypted]
00 01 02 03 04 05 06 07 BA 73 71 85 E7 19 31 04
92 F3 8A 5F 12 51 DA 55 FA FB C9 49 84 8A 0D FC
AE CE 74 6B 3D B9 AD
=============== Packet Vector #2 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 04 03 02 01 A0 A1 A2 A3 A4 A5
Total packet length = 32. [Input (8 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
CBC IV in: 59 00 00 00 04 03 02 01 A0 A1 A2 A3 A4 A5 00 18
CBC IV out:07 0B 22 50 8A 24 3C DD 5B BA 54 DB 60 52 88 06
After xor: 07 03 22 51 88 27 38 D8 5D BD 54 DB 60 52 88 06 [hdr]
After CAM: 10 FD C2 F2 90 4A 9F 96 B0 4F 62 A4 A1 A9 31 1E
After xor: 18 F4 C8 F9 9C 47 91 99 A0 5E 70 B7 B5 BC 27 09 [msg]
After CAM: E4 C8 82 02 89 55 5C 15 CE 7F E4 60 B1 B9 5A 08
After xor: FC D1 98 19 95 48 42 0A CE 7F E4 60 B1 B9 5A 08 [msg]
After CAM: D2 96 BA 4F 83 DE B5 DF A2 19 08 F7 47 4E 3C 40
MIC tag : D2 96 BA 4F 83 DE B5 DF
CTR Start: 01 00 00 00 04 03 02 01 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: 55 2C 6E B4 82 A2 EF D6 85 37 FE 12 79 0E E6 55
CTR[0002]: 54 E2 C8 D6 7E 99 91 2C F2 8A D7 8E 83 04 10 36
CTR[MIC ]: B2 24 93 12 71 9C 36 37
Total packet length = 40. [Encrypted]
00 01 02 03 04 05 06 07 5D 25 64 BF 8E AF E1 D9
95 26 EC 01 6D 1B F0 42 4C FB D2 CD 62 84 8F 33
60 B2 29 5D F2 42 83 E8
=============== Packet Vector #2 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 04 03 02 01 A0 A1 A2 A3 A4 A5
Total packet length = 32. [Input (8 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
CBC IV in: 59 00 00 00 04 03 02 01 A0 A1 A2 A3 A4 A5 00 18
CBC IV out:07 0B 22 50 8A 24 3C DD 5B BA 54 DB 60 52 88 06
After xor: 07 03 22 51 88 27 38 D8 5D BD 54 DB 60 52 88 06 [hdr]
After CAM: 10 FD C2 F2 90 4A 9F 96 B0 4F 62 A4 A1 A9 31 1E
After xor: 18 F4 C8 F9 9C 47 91 99 A0 5E 70 B7 B5 BC 27 09 [msg]
After CAM: E4 C8 82 02 89 55 5C 15 CE 7F E4 60 B1 B9 5A 08
After xor: FC D1 98 19 95 48 42 0A CE 7F E4 60 B1 B9 5A 08 [msg]
After CAM: D2 96 BA 4F 83 DE B5 DF A2 19 08 F7 47 4E 3C 40
MIC tag : D2 96 BA 4F 83 DE B5 DF
CTR Start: 01 00 00 00 04 03 02 01 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: 55 2C 6E B4 82 A2 EF D6 85 37 FE 12 79 0E E6 55
CTR[0002]: 54 E2 C8 D6 7E 99 91 2C F2 8A D7 8E 83 04 10 36
CTR[MIC ]: B2 24 93 12 71 9C 36 37
Total packet length = 40. [Encrypted]
00 01 02 03 04 05 06 07 5D 25 64 BF 8E AF E1 D9
95 26 EC 01 6D 1B F0 42 4C FB D2 CD 62 84 8F 33
60 B2 29 5D F2 42 83 E8
=============== Packet Vector #3 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 05 04 03 02 A0 A1 A2 A3 A4 A5
Total packet length = 33. [Input (8 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
20
CBC IV in: 59 00 00 00 05 04 03 02 A0 A1 A2 A3 A4 A5 00 19
CBC IV out:6F 69 15 DF A6 A0 DF 24 84 A7 37 88 A3 65 F9 2E
After xor: 6F 61 15 DE A4 A3 DB 21 82 A0 37 88 A3 65 F9 2E [hdr]
After CAM: 59 5D 99 48 79 04 DA C9 13 93 36 C9 11 A8 09 1D
After xor: 51 54 93 43 75 09 D4 C6 03 82 24 DA 05 BD 1F 0A [msg]
After CAM: 1A 43 D7 19 65 43 97 C1 43 6F 4F 11 A7 6C 6B ED
After xor: 02 5A CD 02 79 5E 89 DE 63 6F 4F 11 A7 6C 6B ED [msg]
After CAM: 30 0B 06 8A A0 D1 4D C5 9E 44 22 84 82 45 42 0B
MIC tag : 30 0B 06 8A A0 D1 4D C5
CTR Start: 01 00 00 00 05 04 03 02 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: 89 FF 69 DD CB 75 76 18 E9 31 24 1B AD 97 BB 02
CTR[0002]: C4 32 A7 9C CB 4B E9 8D 24 A8 F0 AB C6 87 16 11
CTR[MIC ]: C5 5A D0 E2 8F F2 E7 83
Total packet length = 41. [Encrypted]
00 01 02 03 04 05 06 07 81 F6 63 D6 C7 78 78 17
F9 20 36 08 B9 82 AD 15 DC 2B BD 87 D7 56 F7 92
04 F5 51 D6 68 2F 23 AA 46
=============== Packet Vector #3 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 05 04 03 02 A0 A1 A2 A3 A4 A5
Total packet length = 33. [Input (8 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
20
CBC IV in: 59 00 00 00 05 04 03 02 A0 A1 A2 A3 A4 A5 00 19
CBC IV out:6F 69 15 DF A6 A0 DF 24 84 A7 37 88 A3 65 F9 2E
After xor: 6F 61 15 DE A4 A3 DB 21 82 A0 37 88 A3 65 F9 2E [hdr]
After CAM: 59 5D 99 48 79 04 DA C9 13 93 36 C9 11 A8 09 1D
After xor: 51 54 93 43 75 09 D4 C6 03 82 24 DA 05 BD 1F 0A [msg]
After CAM: 1A 43 D7 19 65 43 97 C1 43 6F 4F 11 A7 6C 6B ED
After xor: 02 5A CD 02 79 5E 89 DE 63 6F 4F 11 A7 6C 6B ED [msg]
After CAM: 30 0B 06 8A A0 D1 4D C5 9E 44 22 84 82 45 42 0B
MIC tag : 30 0B 06 8A A0 D1 4D C5
CTR Start: 01 00 00 00 05 04 03 02 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: 89 FF 69 DD CB 75 76 18 E9 31 24 1B AD 97 BB 02
CTR[0002]: C4 32 A7 9C CB 4B E9 8D 24 A8 F0 AB C6 87 16 11
CTR[MIC ]: C5 5A D0 E2 8F F2 E7 83
Total packet length = 41. [Encrypted]
00 01 02 03 04 05 06 07 81 F6 63 D6 C7 78 78 17
F9 20 36 08 B9 82 AD 15 DC 2B BD 87 D7 56 F7 92
04 F5 51 D6 68 2F 23 AA 46
=============== Packet Vector #4 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 06 05 04 03 A0 A1 A2 A3 A4 A5
Total packet length = 31. [Input (12 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E
CBC IV in: 59 00 00 00 06 05 04 03 A0 A1 A2 A3 A4 A5 00 13
CBC IV out:F5 51 CF 6C 7C F7 D4 0B 2B 76 F1 6B 57 F0 19 FE
After xor: F5 5D CF 6D 7E F4 D0 0E 2D 71 F9 62 5D FB 19 FE [hdr]
After CAM: 02 2B 21 1B EB 97 02 3B F8 10 7D CC 62 14 E5 7C
After xor: 0E 26 2F 14 FB 86 10 28 EC 05 6B DB 7A 0D FF 67 [msg]
After CAM: 48 14 A4 2D 31 25 1C 37 19 C5 6F DD 5A 37 81 42
After xor: 54 09 BA 2D 31 25 1C 37 19 C5 6F DD 5A 37 81 42 [msg]
After CAM: CF 85 25 D2 80 D5 F0 09 53 2C 9D 43 4E F3 04 47
MIC tag : CF 85 25 D2 80 D5 F0 09
CTR Start: 01 00 00 00 06 05 04 03 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: C6 E2 10 8D 62 00 A2 9C 6F CC 19 1F DF 6B 92 DB
CTR[0002]: 6C B9 BE EE 1E A2 E9 B3 2D D6 C2 9A E8 26 D5 C2
CTR[MIC ]: 44 BF B6 E8 E3 31 67 A9
Total packet length = 39. [Encrypted]
00 01 02 03 04 05 06 07 08 09 0A 0B CA EF 1E 82
72 11 B0 8F 7B D9 0F 08 C7 72 88 C0 70 A4 A0 8B
3A 93 3A 63 E4 97 A0
=============== Packet Vector #4 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 06 05 04 03 A0 A1 A2 A3 A4 A5
Total packet length = 31. [Input (12 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E
CBC IV in: 59 00 00 00 06 05 04 03 A0 A1 A2 A3 A4 A5 00 13
CBC IV out:F5 51 CF 6C 7C F7 D4 0B 2B 76 F1 6B 57 F0 19 FE
After xor: F5 5D CF 6D 7E F4 D0 0E 2D 71 F9 62 5D FB 19 FE [hdr]
After CAM: 02 2B 21 1B EB 97 02 3B F8 10 7D CC 62 14 E5 7C
After xor: 0E 26 2F 14 FB 86 10 28 EC 05 6B DB 7A 0D FF 67 [msg]
After CAM: 48 14 A4 2D 31 25 1C 37 19 C5 6F DD 5A 37 81 42
After xor: 54 09 BA 2D 31 25 1C 37 19 C5 6F DD 5A 37 81 42 [msg]
After CAM: CF 85 25 D2 80 D5 F0 09 53 2C 9D 43 4E F3 04 47
MIC tag : CF 85 25 D2 80 D5 F0 09
CTR Start: 01 00 00 00 06 05 04 03 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: C6 E2 10 8D 62 00 A2 9C 6F CC 19 1F DF 6B 92 DB
CTR[0002]: 6C B9 BE EE 1E A2 E9 B3 2D D6 C2 9A E8 26 D5 C2
CTR[MIC ]: 44 BF B6 E8 E3 31 67 A9
Total packet length = 39. [Encrypted]
00 01 02 03 04 05 06 07 08 09 0A 0B CA EF 1E 82
72 11 B0 8F 7B D9 0F 08 C7 72 88 C0 70 A4 A0 8B
3A 93 3A 63 E4 97 A0
=============== Packet Vector #5 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 07 06 05 04 A0 A1 A2 A3 A4 A5
Total packet length = 32. [Input (12 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
CBC IV in: 59 00 00 00 07 06 05 04 A0 A1 A2 A3 A4 A5 00 14
CBC IV out:73 72 9D 76 7A BD B9 82 60 3A 12 7B EF 26 FB 80
After xor: 73 7E 9D 77 78 BE BD 87 66 3D 1A 72 E5 2D FB 80 [hdr]
After CAM: E1 B7 A6 72 E2 5C 87 75 91 21 22 A4 07 13 CD 5B
After xor: ED BA A8 7D F2 4D 95 66 85 34 34 B3 1F 0A D7 40 [msg]
After CAM: 13 2F 58 D9 5D 0F 95 B8 90 BF 6F 1D 31 84 54 C7
After xor: 0F 32 46 C6 5D 0F 95 B8 90 BF 6F 1D 31 84 54 C7 [msg]
After CAM: 47 8F 1E B0 71 24 8B 13 AF C8 C8 44 E6 0F 88 B6
MIC tag : 47 8F 1E B0 71 24 8B 13
CTR Start: 01 00 00 00 07 06 05 04 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: 26 DE B4 D6 5F D4 3C 81 AA 56 98 95 64 09 39 A2
CTR[0002]: 76 97 69 3A 21 13 0C 39 2E 4E EB BF 48 7B 24 BE
CTR[MIC ]: C8 2E 65 17 82 15 50 1A
Total packet length = 40. [Encrypted]
00 01 02 03 04 05 06 07 08 09 0A 0B 2A D3 BA D9
4F C5 2E 92 BE 43 8E 82 7C 10 23 B9 6A 8A 77 25
8F A1 7B A7 F3 31 DB 09
=============== Packet Vector #5 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 07 06 05 04 A0 A1 A2 A3 A4 A5
Total packet length = 32. [Input (12 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
CBC IV in: 59 00 00 00 07 06 05 04 A0 A1 A2 A3 A4 A5 00 14
CBC IV out:73 72 9D 76 7A BD B9 82 60 3A 12 7B EF 26 FB 80
After xor: 73 7E 9D 77 78 BE BD 87 66 3D 1A 72 E5 2D FB 80 [hdr]
After CAM: E1 B7 A6 72 E2 5C 87 75 91 21 22 A4 07 13 CD 5B
After xor: ED BA A8 7D F2 4D 95 66 85 34 34 B3 1F 0A D7 40 [msg]
After CAM: 13 2F 58 D9 5D 0F 95 B8 90 BF 6F 1D 31 84 54 C7
After xor: 0F 32 46 C6 5D 0F 95 B8 90 BF 6F 1D 31 84 54 C7 [msg]
After CAM: 47 8F 1E B0 71 24 8B 13 AF C8 C8 44 E6 0F 88 B6
MIC tag : 47 8F 1E B0 71 24 8B 13
CTR Start: 01 00 00 00 07 06 05 04 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: 26 DE B4 D6 5F D4 3C 81 AA 56 98 95 64 09 39 A2
CTR[0002]: 76 97 69 3A 21 13 0C 39 2E 4E EB BF 48 7B 24 BE
CTR[MIC ]: C8 2E 65 17 82 15 50 1A
Total packet length = 40. [Encrypted]
00 01 02 03 04 05 06 07 08 09 0A 0B 2A D3 BA D9
4F C5 2E 92 BE 43 8E 82 7C 10 23 B9 6A 8A 77 25
8F A1 7B A7 F3 31 DB 09
=============== Packet Vector #6 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 08 07 06 05 A0 A1 A2 A3 A4 A5
Total packet length = 33. [Input (12 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
20
CBC IV in: 59 00 00 00 08 07 06 05 A0 A1 A2 A3 A4 A5 00 15
CBC IV out:EB 59 05 CC 3F 52 61 10 26 24 75 93 DD B9 A0 F4
After xor: EB 55 05 CD 3D 51 65 15 20 23 7D 9A D7 B2 A0 F4 [hdr]
After CAM: 18 A9 AE A4 3D D2 A9 11 6C 0A E5 4F 40 D1 4D 9F
After xor: 14 A4 A0 AB 2D C3 BB 02 78 1F F3 58 58 C8 57 84 [msg]
After CAM: FA C4 13 18 98 54 1B 54 93 9C 64 B8 CB FD 5B 18
After xor: E6 D9 0D 07 B8 54 1B 54 93 9C 64 B8 CB FD 5B 18 [msg]
After CAM: 49 E6 E8 ED 32 FB CA 2F 2E 55 CD AF D0 F2 B3 05
MIC tag : 49 E6 E8 ED 32 FB CA 2F
CTR Start: 01 00 00 00 08 07 06 05 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: F2 A8 46 04 B5 2E BA C0 D7 51 34 BD D6 54 FC 64
CTR[0002]: E6 26 A9 24 8B E6 86 CB 92 D6 FB FC 2E F2 91 98
CTR[MIC ]: E2 D0 49 03 7D 1B 34 07
Total packet length = 41. [Encrypted]
00 01 02 03 04 05 06 07 08 09 0A 0B FE A5 48 0B
A5 3F A8 D3 C3 44 22 AA CE 4D E6 7F FA 3B B7 3B
AB AB 36 A1 EE 4F E0 FE 28
=============== Packet Vector #6 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 08 07 06 05 A0 A1 A2 A3 A4 A5
Total packet length = 33. [Input (12 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
20
CBC IV in: 59 00 00 00 08 07 06 05 A0 A1 A2 A3 A4 A5 00 15
CBC IV out:EB 59 05 CC 3F 52 61 10 26 24 75 93 DD B9 A0 F4
After xor: EB 55 05 CD 3D 51 65 15 20 23 7D 9A D7 B2 A0 F4 [hdr]
After CAM: 18 A9 AE A4 3D D2 A9 11 6C 0A E5 4F 40 D1 4D 9F
After xor: 14 A4 A0 AB 2D C3 BB 02 78 1F F3 58 58 C8 57 84 [msg]
After CAM: FA C4 13 18 98 54 1B 54 93 9C 64 B8 CB FD 5B 18
After xor: E6 D9 0D 07 B8 54 1B 54 93 9C 64 B8 CB FD 5B 18 [msg]
After CAM: 49 E6 E8 ED 32 FB CA 2F 2E 55 CD AF D0 F2 B3 05
MIC tag : 49 E6 E8 ED 32 FB CA 2F
CTR Start: 01 00 00 00 08 07 06 05 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: F2 A8 46 04 B5 2E BA C0 D7 51 34 BD D6 54 FC 64
CTR[0002]: E6 26 A9 24 8B E6 86 CB 92 D6 FB FC 2E F2 91 98
CTR[MIC ]: E2 D0 49 03 7D 1B 34 07
Total packet length = 41. [Encrypted]
00 01 02 03 04 05 06 07 08 09 0A 0B FE A5 48 0B
A5 3F A8 D3 C3 44 22 AA CE 4D E6 7F FA 3B B7 3B
AB AB 36 A1 EE 4F E0 FE 28
=============== Packet Vector #7 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 09 08 07 06 A0 A1 A2 A3 A4 A5
Total packet length = 31. [Input (8 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E
CBC IV in: 61 00 00 00 09 08 07 06 A0 A1 A2 A3 A4 A5 00 17
CBC IV out:AC F1 5D 79 99 1A 15 BF 5C DC F6 C4 45 AE 1F CB
After xor: AC F9 5D 78 9B 19 11 BA 5A DB F6 C4 45 AE 1F CB [hdr]
After CAM: E9 C0 AC FD C7 E8 E7 1D FA E8 8B 66 95 9E 01 45
After xor: E1 C9 A6 F6 CB E5 E9 12 EA F9 99 75 81 8B 17 52 [msg]
After CAM: 9C FF ED 72 09 A6 7D 2A 48 B7 29 BF D8 BE 39 59
After xor: 84 E6 F7 69 15 BB 63 2A 48 B7 29 BF D8 BE 39 59 [msg]
After CAM: 4F 41 FA DE B2 58 F3 32 54 0A 55 7A 80 4A A3 F5
MIC tag : 4F 41 FA DE B2 58 F3 32 54 0A
CTR Start: 01 00 00 00 09 08 07 06 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: 5C 5A 2A 2D E9 41 1F 95 9D 27 CB FF 7A 0B CF 63
CTR[0002]: 0E D1 6A 97 57 41 32 4F 33 1B 4A 42 B1 4A 54 63
CTR[MIC ]: E3 EE 59 62 7D 22 BD 8D C1 79
Total packet length = 41. [Encrypted]
00 01 02 03 04 05 06 07 54 53 20 26 E5 4C 11 9A
8D 36 D9 EC 6E 1E D9 74 16 C8 70 8C 4B 5C 2C AC
AF A3 BC CF 7A 4E BF 95 73
=============== Packet Vector #7 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 09 08 07 06 A0 A1 A2 A3 A4 A5
Total packet length = 31. [Input (8 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E
CBC IV in: 61 00 00 00 09 08 07 06 A0 A1 A2 A3 A4 A5 00 17
CBC IV out:AC F1 5D 79 99 1A 15 BF 5C DC F6 C4 45 AE 1F CB
After xor: AC F9 5D 78 9B 19 11 BA 5A DB F6 C4 45 AE 1F CB [hdr]
After CAM: E9 C0 AC FD C7 E8 E7 1D FA E8 8B 66 95 9E 01 45
After xor: E1 C9 A6 F6 CB E5 E9 12 EA F9 99 75 81 8B 17 52 [msg]
After CAM: 9C FF ED 72 09 A6 7D 2A 48 B7 29 BF D8 BE 39 59
After xor: 84 E6 F7 69 15 BB 63 2A 48 B7 29 BF D8 BE 39 59 [msg]
After CAM: 4F 41 FA DE B2 58 F3 32 54 0A 55 7A 80 4A A3 F5
MIC tag : 4F 41 FA DE B2 58 F3 32 54 0A
CTR Start: 01 00 00 00 09 08 07 06 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: 5C 5A 2A 2D E9 41 1F 95 9D 27 CB FF 7A 0B CF 63
CTR[0002]: 0E D1 6A 97 57 41 32 4F 33 1B 4A 42 B1 4A 54 63
CTR[MIC ]: E3 EE 59 62 7D 22 BD 8D C1 79
Total packet length = 41. [Encrypted]
00 01 02 03 04 05 06 07 54 53 20 26 E5 4C 11 9A
8D 36 D9 EC 6E 1E D9 74 16 C8 70 8C 4B 5C 2C AC
AF A3 BC CF 7A 4E BF 95 73
=============== Packet Vector #8 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 0A 09 08 07 A0 A1 A2 A3 A4 A5
Total packet length = 32. [Input (8 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
CBC IV in: 61 00 00 00 0A 09 08 07 A0 A1 A2 A3 A4 A5 00 18
CBC IV out:AD CA 1C 1D 45 E7 E2 62 58 D5 DA 46 D8 2F 69 3A
After xor: AD C2 1C 1C 47 E4 E6 67 5E D2 DA 46 D8 2F 69 3A [hdr]
After CAM: FA DE 0E B4 3E CA C1 E9 69 BB 8C A4 7C 0D 80 8F
After xor: F2 D7 04 BF 32 C7 CF E6 79 AA 9E B7 68 18 96 98 [msg]
After CAM: D2 87 35 C2 D0 E4 AE 4E BC C2 99 FF B3 77 F8 A1
After xor: CA 9E 2F D9 CC F9 B0 51 BC C2 99 FF B3 77 F8 A1 [msg]
After CAM: BD F6 FB 55 9E 90 C0 E7 DF 4B 0C 37 DC 42 32 A2
MIC tag : BD F6 FB 55 9E 90 C0 E7 DF 4B
CTR Start: 01 00 00 00 0A 09 08 07 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: 82 D8 91 0B 16 8A DF 47 E4 C8 39 FC 20 47 4A DB
CTR[0002]: FB BF 26 7E 0E BB EB 6A 07 4E 29 CF 3D 12 E6 DB
CTR[MIC ]: CE 7E 1F C4 A0 61 87 E6 2B 0A
Total packet length = 42. [Encrypted]
00 01 02 03 04 05 06 07 8A D1 9B 00 1A 87 D1 48
F4 D9 2B EF 34 52 5C CC E3 A6 3C 65 12 A6 F5 75
73 88 E4 91 3E F1 47 01 F4 41
=============== Packet Vector #8 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 0A 09 08 07 A0 A1 A2 A3 A4 A5
Total packet length = 32. [Input (8 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
CBC IV in: 61 00 00 00 0A 09 08 07 A0 A1 A2 A3 A4 A5 00 18
CBC IV out:AD CA 1C 1D 45 E7 E2 62 58 D5 DA 46 D8 2F 69 3A
After xor: AD C2 1C 1C 47 E4 E6 67 5E D2 DA 46 D8 2F 69 3A [hdr]
After CAM: FA DE 0E B4 3E CA C1 E9 69 BB 8C A4 7C 0D 80 8F
After xor: F2 D7 04 BF 32 C7 CF E6 79 AA 9E B7 68 18 96 98 [msg]
After CAM: D2 87 35 C2 D0 E4 AE 4E BC C2 99 FF B3 77 F8 A1
After xor: CA 9E 2F D9 CC F9 B0 51 BC C2 99 FF B3 77 F8 A1 [msg]
After CAM: BD F6 FB 55 9E 90 C0 E7 DF 4B 0C 37 DC 42 32 A2
MIC tag : BD F6 FB 55 9E 90 C0 E7 DF 4B
CTR Start: 01 00 00 00 0A 09 08 07 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: 82 D8 91 0B 16 8A DF 47 E4 C8 39 FC 20 47 4A DB
CTR[0002]: FB BF 26 7E 0E BB EB 6A 07 4E 29 CF 3D 12 E6 DB
CTR[MIC ]: CE 7E 1F C4 A0 61 87 E6 2B 0A
Total packet length = 42. [Encrypted]
00 01 02 03 04 05 06 07 8A D1 9B 00 1A 87 D1 48
F4 D9 2B EF 34 52 5C CC E3 A6 3C 65 12 A6 F5 75
73 88 E4 91 3E F1 47 01 F4 41
=============== Packet Vector #9 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 0B 0A 09 08 A0 A1 A2 A3 A4 A5
Total packet length = 33. [Input (8 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
20
CBC IV in: 61 00 00 00 0B 0A 09 08 A0 A1 A2 A3 A4 A5 00 19
CBC IV out:D0 A9 A5 94 00 63 86 40 11 0D DB 40 CA F8 4A 9C
After xor: D0 A1 A5 95 02 60 82 45 17 0A DB 40 CA F8 4A 9C [hdr]
After CAM: 7B CA 4E 2D 79 82 0D 1E 15 22 DD E8 37 B9 B1 F0
After xor: 73 C3 44 26 75 8F 03 11 05 33 CF FB 23 AC A7 E7 [msg]
After CAM: 6B 75 9F 83 C0 8F 56 64 F2 FA D5 7F 67 01 B8 21
After xor: 73 6C 85 98 DC 92 48 7B D2 FA D5 7F 67 01 B8 21 [msg]
After CAM: 7D B7 BE FF 72 F3 26 74 9E 20 07 28 1E 5B 1A 8A
MIC tag : 7D B7 BE FF 72 F3 26 74 9E 20
CTR Start: 01 00 00 00 0B 0A 09 08 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: 55 B9 87 69 4C 73 60 3E C6 1E 8E B1 D2 11 62 36
CTR[0002]: 82 D9 A4 4B DC C9 BB 68 A7 FE 15 A5 19 51 57 87
CTR[MIC ]: E9 61 5C CF BF D6 EF 8A 21 A7
Total packet length = 43. [Encrypted]
00 01 02 03 04 05 06 07 5D B0 8D 62 40 7E 6E 31
D6 0F 9C A2 C6 04 74 21 9A C0 BE 50 C0 D4 A5 77
87 94 D6 E2 30 CD 25 C9 FE BF 87
=============== Packet Vector #9 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 0B 0A 09 08 A0 A1 A2 A3 A4 A5
Total packet length = 33. [Input (8 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
20
CBC IV in: 61 00 00 00 0B 0A 09 08 A0 A1 A2 A3 A4 A5 00 19
CBC IV out:D0 A9 A5 94 00 63 86 40 11 0D DB 40 CA F8 4A 9C
After xor: D0 A1 A5 95 02 60 82 45 17 0A DB 40 CA F8 4A 9C [hdr]
After CAM: 7B CA 4E 2D 79 82 0D 1E 15 22 DD E8 37 B9 B1 F0
After xor: 73 C3 44 26 75 8F 03 11 05 33 CF FB 23 AC A7 E7 [msg]
After CAM: 6B 75 9F 83 C0 8F 56 64 F2 FA D5 7F 67 01 B8 21
After xor: 73 6C 85 98 DC 92 48 7B D2 FA D5 7F 67 01 B8 21 [msg]
After CAM: 7D B7 BE FF 72 F3 26 74 9E 20 07 28 1E 5B 1A 8A
MIC tag : 7D B7 BE FF 72 F3 26 74 9E 20
CTR Start: 01 00 00 00 0B 0A 09 08 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: 55 B9 87 69 4C 73 60 3E C6 1E 8E B1 D2 11 62 36
CTR[0002]: 82 D9 A4 4B DC C9 BB 68 A7 FE 15 A5 19 51 57 87
CTR[MIC ]: E9 61 5C CF BF D6 EF 8A 21 A7
Total packet length = 43. [Encrypted]
00 01 02 03 04 05 06 07 5D B0 8D 62 40 7E 6E 31
D6 0F 9C A2 C6 04 74 21 9A C0 BE 50 C0 D4 A5 77
87 94 D6 E2 30 CD 25 C9 FE BF 87
=============== Packet Vector #10 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 0C 0B 0A 09 A0 A1 A2 A3 A4 A5
Total packet length = 31. [Input (12 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E
CBC IV in: 61 00 00 00 0C 0B 0A 09 A0 A1 A2 A3 A4 A5 00 13
CBC IV out:B1 85 73 A3 1C 6F EC 01 90 E3 CE 94 27 11 04 B9
After xor: B1 89 73 A2 1E 6C E8 04 96 E4 C6 9D 2D 1A 04 B9 [hdr]
After CAM: A6 AD EA 9C FA 3F 76 78 4C 17 8A F3 DC 69 F0 82
After xor: AA A0 E4 93 EA 2E 64 6B 58 02 9C E4 C4 70 EA 99 [msg]
After CAM: 35 50 B7 27 78 F8 C6 BF 02 4B 65 60 05 C0 E1 ED
After xor: 29 4D A9 27 78 F8 C6 BF 02 4B 65 60 05 C0 E1 ED [msg]
After CAM: 3D B5 A6 E6 85 AF 1C 58 80 B0 32 2E 01 74 91 FC
MIC tag : 3D B5 A6 E6 85 AF 1C 58 80 B0
CTR Start: 01 00 00 00 0C 0B 0A 09 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: D7 1C 82 C1 D1 A9 64 0F 93 69 CE 81 22 7E CC E8
CTR[0002]: A7 A1 42 44 32 4E 69 FE 4C D0 36 65 A5 31 0B AB
CTR[MIC ]: ED 27 3F 0D 94 5C 0E AA B2 87
Total packet length = 41. [Encrypted]
00 01 02 03 04 05 06 07 08 09 0A 0B DB 11 8C CE
C1 B8 76 1C 87 7C D8 96 3A 67 D6 F3 BB BC 5C D0
92 99 EB 11 F3 12 F2 32 37
=============== Packet Vector #10 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 0C 0B 0A 09 A0 A1 A2 A3 A4 A5
Total packet length = 31. [Input (12 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E
CBC IV in: 61 00 00 00 0C 0B 0A 09 A0 A1 A2 A3 A4 A5 00 13
CBC IV out:B1 85 73 A3 1C 6F EC 01 90 E3 CE 94 27 11 04 B9
After xor: B1 89 73 A2 1E 6C E8 04 96 E4 C6 9D 2D 1A 04 B9 [hdr]
After CAM: A6 AD EA 9C FA 3F 76 78 4C 17 8A F3 DC 69 F0 82
After xor: AA A0 E4 93 EA 2E 64 6B 58 02 9C E4 C4 70 EA 99 [msg]
After CAM: 35 50 B7 27 78 F8 C6 BF 02 4B 65 60 05 C0 E1 ED
After xor: 29 4D A9 27 78 F8 C6 BF 02 4B 65 60 05 C0 E1 ED [msg]
After CAM: 3D B5 A6 E6 85 AF 1C 58 80 B0 32 2E 01 74 91 FC
MIC tag : 3D B5 A6 E6 85 AF 1C 58 80 B0
CTR Start: 01 00 00 00 0C 0B 0A 09 A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: D7 1C 82 C1 D1 A9 64 0F 93 69 CE 81 22 7E CC E8
CTR[0002]: A7 A1 42 44 32 4E 69 FE 4C D0 36 65 A5 31 0B AB
CTR[MIC ]: ED 27 3F 0D 94 5C 0E AA B2 87
Total packet length = 41. [Encrypted]
00 01 02 03 04 05 06 07 08 09 0A 0B DB 11 8C CE
C1 B8 76 1C 87 7C D8 96 3A 67 D6 F3 BB BC 5C D0
92 99 EB 11 F3 12 F2 32 37
=============== Packet Vector #11 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 0D 0C 0B 0A A0 A1 A2 A3 A4 A5
Total packet length = 32. [Input (12 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
CBC IV in: 61 00 00 00 0D 0C 0B 0A A0 A1 A2 A3 A4 A5 00 14
CBC IV out:45 DF B5 07 6F BB 10 EA F1 15 15 AD 21 4F B0 0E
After xor: 45 D3 B5 06 6D B8 14 EF F7 12 1D A4 2B 44 B0 0E [hdr]
After CAM: 17 52 F9 6D DD BC 5B 1C 1E EB 80 FC F6 10 AC 03
After xor: 1B 5F F7 62 CD AD 49 0F 0A FE 96 EB EE 09 B6 18 [msg]
After CAM: BE F0 A0 B9 EC 94 B6 B3 E8 EC 1B 82 14 14 09 87
After xor: A2 ED BE A6 EC 94 B6 B3 E8 EC 1B 82 14 14 09 87 [msg]
After CAM: 70 16 E4 F9 C4 2C 30 10 84 BF EC 69 34 89 91 FD
MIC tag : 70 16 E4 F9 C4 2C 30 10 84 BF
CTR Start: 01 00 00 00 0D 0C 0B 0A A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: 70 C5 33 82 D4 80 11 41 4F 5D 2B D2 D2 67 B3 B0
CTR[0002]: 9D 36 6E 49 39 C5 16 76 5C 1C 25 12 81 79 94 70
CTR[MIC ]: 77 8B 4B 03 1E 3A FC DF A8 F1
Total packet length = 42. [Encrypted]
00 01 02 03 04 05 06 07 08 09 0A 0B 7C C8 3D 8D
C4 91 03 52 5B 48 3D C5 CA 7E A9 AB 81 2B 70 56
07 9D AF FA DA 16 CC CF 2C 4E
=============== Packet Vector #11 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 0D 0C 0B 0A A0 A1 A2 A3 A4 A5
Total packet length = 32. [Input (12 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
CBC IV in: 61 00 00 00 0D 0C 0B 0A A0 A1 A2 A3 A4 A5 00 14
CBC IV out:45 DF B5 07 6F BB 10 EA F1 15 15 AD 21 4F B0 0E
After xor: 45 D3 B5 06 6D B8 14 EF F7 12 1D A4 2B 44 B0 0E [hdr]
After CAM: 17 52 F9 6D DD BC 5B 1C 1E EB 80 FC F6 10 AC 03
After xor: 1B 5F F7 62 CD AD 49 0F 0A FE 96 EB EE 09 B6 18 [msg]
After CAM: BE F0 A0 B9 EC 94 B6 B3 E8 EC 1B 82 14 14 09 87
After xor: A2 ED BE A6 EC 94 B6 B3 E8 EC 1B 82 14 14 09 87 [msg]
After CAM: 70 16 E4 F9 C4 2C 30 10 84 BF EC 69 34 89 91 FD
MIC tag : 70 16 E4 F9 C4 2C 30 10 84 BF
CTR Start: 01 00 00 00 0D 0C 0B 0A A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: 70 C5 33 82 D4 80 11 41 4F 5D 2B D2 D2 67 B3 B0
CTR[0002]: 9D 36 6E 49 39 C5 16 76 5C 1C 25 12 81 79 94 70
CTR[MIC ]: 77 8B 4B 03 1E 3A FC DF A8 F1
Total packet length = 42. [Encrypted]
00 01 02 03 04 05 06 07 08 09 0A 0B 7C C8 3D 8D
C4 91 03 52 5B 48 3D C5 CA 7E A9 AB 81 2B 70 56
07 9D AF FA DA 16 CC CF 2C 4E
=============== Packet Vector #12 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 0E 0D 0C 0B A0 A1 A2 A3 A4 A5
Total packet length = 33. [Input (12 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
20
CBC IV in: 61 00 00 00 0E 0D 0C 0B A0 A1 A2 A3 A4 A5 00 15
CBC IV out:81 E4 EB 1E 50 A9 70 CE 18 CA 1A 4B 68 39 80 2E
After xor: 81 E8 EB 1F 52 AA 74 CB 1E CD 12 42 62 32 80 2E [hdr]
After CAM: 04 AB D9 62 34 B9 8F 32 8C 0F 08 3F 3D 87 9D 57
After xor: 08 A6 D7 6D 24 A8 9D 21 98 1A 1E 28 25 9E 87 4C [msg]
After CAM: BD A2 EA CB 3A DA 6A E7 9F BB C2 2C E6 4C 98 89
After xor: A1 BF F4 D4 1A DA 6A E7 9F BB C2 2C E6 4C 98 89 [msg]
After CAM: B6 FC E1 46 D3 EA DC 91 E0 AB 10 AD D8 55 E7 03
MIC tag : B6 FC E1 46 D3 EA DC 91 E0 AB
CTR Start: 01 00 00 00 0E 0D 0C 0B A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: 20 DE 55 87 30 C3 2C 69 B7 44 A6 FE 37 DE 89 7C
CTR[0002]: 3F 96 32 D8 68 6D C2 B5 22 97 42 27 EB F9 26 5E
CTR[MIC ]: 7D 45 AD 6F 94 93 E1 F5 4F DE
Total packet length = 43. [Encrypted]
00 01 02 03 04 05 06 07 08 09 0A 0B 2C D3 5B 88
20 D2 3E 7A A3 51 B0 E9 2F C7 93 67 23 8B 2C C7
48 CB B9 4C 29 47 79 3D 64 AF 75
=============== Packet Vector #12 ==================
CAM Key: C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF
Nonce = 00 00 00 0E 0D 0C 0B A0 A1 A2 A3 A4 A5
Total packet length = 33. [Input (12 cleartext header octets)]
00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
20
CBC IV in: 61 00 00 00 0E 0D 0C 0B A0 A1 A2 A3 A4 A5 00 15
CBC IV out:81 E4 EB 1E 50 A9 70 CE 18 CA 1A 4B 68 39 80 2E
After xor: 81 E8 EB 1F 52 AA 74 CB 1E CD 12 42 62 32 80 2E [hdr]
After CAM: 04 AB D9 62 34 B9 8F 32 8C 0F 08 3F 3D 87 9D 57
After xor: 08 A6 D7 6D 24 A8 9D 21 98 1A 1E 28 25 9E 87 4C [msg]
After CAM: BD A2 EA CB 3A DA 6A E7 9F BB C2 2C E6 4C 98 89
After xor: A1 BF F4 D4 1A DA 6A E7 9F BB C2 2C E6 4C 98 89 [msg]
After CAM: B6 FC E1 46 D3 EA DC 91 E0 AB 10 AD D8 55 E7 03
MIC tag : B6 FC E1 46 D3 EA DC 91 E0 AB
CTR Start: 01 00 00 00 0E 0D 0C 0B A0 A1 A2 A3 A4 A5 00 01
CTR[0001]: 20 DE 55 87 30 C3 2C 69 B7 44 A6 FE 37 DE 89 7C
CTR[0002]: 3F 96 32 D8 68 6D C2 B5 22 97 42 27 EB F9 26 5E
CTR[MIC ]: 7D 45 AD 6F 94 93 E1 F5 4F DE
Total packet length = 43. [Encrypted]
00 01 02 03 04 05 06 07 08 09 0A 0B 2C D3 5B 88
20 D2 3E 7A A3 51 B0 E9 2F C7 93 67 23 8B 2C C7
48 CB B9 4C 29 47 79 3D 64 AF 75
=============== Packet Vector #13 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 A9 70 11 0E 19 27 B1 60 B6 A3 1C 1C
Total packet length = 31. [Input (8 cleartext header octets)]
6B 7F 46 45 07 FA E4 96 C6 B5 F3 E6 CA 23 11 AE
F7 47 2B 20 3E 73 5E A5 61 AD B1 7D 56 C5 A3
CBC IV in: 59 00 A9 70 11 0E 19 27 B1 60 B6 A3 1C 1C 00 17
CBC IV out:D7 24 B0 0F B1 87 04 C6 C1 4E 90 37 AA F2 F1 F9
After xor: D7 2C DB 70 F7 C2 03 3C 25 D8 90 37 AA F2 F1 F9 [hdr]
After CAM: 9B 13 6D E3 D9 9F C3 6D 7D 0D B7 D8 A1 BF E9 BD
After xor: 5D A6 9E 05 13 BC D2 C3 8A 4A 9C F8 9F CC B7 18 [msg]
After CAM: F8 BF 25 7D 23 F8 D9 B5 82 E6 C9 3E C8 9B 85 73
After xor: 99 12 94 00 75 3D 7A B5 82 E6 C9 3E C8 9B 85 73 [msg]
After CAM: D9 D6 62 21 6D B2 CA FD 1F C6 FE 9D 2C AF 5B 69
MIC tag : D9 D6 62 21 6D B2 CA FD
CTR Start: 01 00 A9 70 11 0E 19 27 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 62 80 24 C1 FE AE CC 8C 67 38 55 98 CB 8E E5 E8
CTR[0002]: F2 30 17 2F 1B 71 55 9F 8B CE 79 E5 13 01 FC 6A
CTR[MIC ]: 9C 8E A2 0C 48 03 ED 13
Total packet length = 39. [Encrypted]
6B 7F 46 45 07 FA E4 96 A4 35 D7 27 34 8D DD 22
90 7F 7E B8 F5 FD BB 4D 93 9D A6 52 4D B4 F6 45
58 C0 2D 25 B1 27 EE
=============== Packet Vector #13 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 A9 70 11 0E 19 27 B1 60 B6 A3 1C 1C
Total packet length = 31. [Input (8 cleartext header octets)]
6B 7F 46 45 07 FA E4 96 C6 B5 F3 E6 CA 23 11 AE
F7 47 2B 20 3E 73 5E A5 61 AD B1 7D 56 C5 A3
CBC IV in: 59 00 A9 70 11 0E 19 27 B1 60 B6 A3 1C 1C 00 17
CBC IV out:D7 24 B0 0F B1 87 04 C6 C1 4E 90 37 AA F2 F1 F9
After xor: D7 2C DB 70 F7 C2 03 3C 25 D8 90 37 AA F2 F1 F9 [hdr]
After CAM: 9B 13 6D E3 D9 9F C3 6D 7D 0D B7 D8 A1 BF E9 BD
After xor: 5D A6 9E 05 13 BC D2 C3 8A 4A 9C F8 9F CC B7 18 [msg]
After CAM: F8 BF 25 7D 23 F8 D9 B5 82 E6 C9 3E C8 9B 85 73
After xor: 99 12 94 00 75 3D 7A B5 82 E6 C9 3E C8 9B 85 73 [msg]
After CAM: D9 D6 62 21 6D B2 CA FD 1F C6 FE 9D 2C AF 5B 69
MIC tag : D9 D6 62 21 6D B2 CA FD
CTR Start: 01 00 A9 70 11 0E 19 27 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 62 80 24 C1 FE AE CC 8C 67 38 55 98 CB 8E E5 E8
CTR[0002]: F2 30 17 2F 1B 71 55 9F 8B CE 79 E5 13 01 FC 6A
CTR[MIC ]: 9C 8E A2 0C 48 03 ED 13
Total packet length = 39. [Encrypted]
6B 7F 46 45 07 FA E4 96 A4 35 D7 27 34 8D DD 22
90 7F 7E B8 F5 FD BB 4D 93 9D A6 52 4D B4 F6 45
58 C0 2D 25 B1 27 EE
=============== Packet Vector #14 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 83 CD 8C E0 CB 42 B1 60 B6 A3 1C 1C
Total packet length = 32. [Input (8 cleartext header octets)]
98 66 05 B4 3D F1 5D E7 01 F6 CE 67 64 C5 74 48
3B B0 2E 6B BF 1E 0A BD 26 A2 25 72 B4 D8 0E E7
CBC IV in: 59 00 83 CD 8C E0 CB 42 B1 60 B6 A3 1C 1C 00 18
CBC IV out:A0 8A 29 78 36 23 1D 84 96 76 93 FF 0A 4C 92 7A
After xor: A0 82 B1 1E 33 97 20 75 CB 91 93 FF 0A 4C 92 7A [hdr]
After CAM: 8C F5 F4 23 BF 09 1C 74 CD 47 00 C1 32 5D 5C 92
After xor: 8D 03 3A 44 DB CC 68 3C F6 F7 2E AA 8D 43 56 2F [msg]
After CAM: 69 DA 48 24 41 1E AC 8E A9 0A CD 8B DD 00 2B 9A
After xor: 4F 78 6D 56 F5 C6 A2 69 A9 0A CD 8B DD 00 2B 9A [msg]
After CAM: C2 03 3B 08 6D B3 CB 3B 2C C8 5D E7 76 A1 C0 44
MIC tag : C2 03 3B 08 6D B3 CB 3B
CTR Start: 01 00 83 CD 8C E0 CB 42 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 8B 16 9C 37 EB 7B BE DB 15 84 41 6E 5F C2 07 46
CTR[0002]: E9 31 BB DD 4E E6 56 9B 68 95 13 5F AB A4 DF EF
CTR[MIC ]: 44 7E 55 14 25 C3 F3 3D
Total packet length = 40. [Encrypted]
98 66 05 B4 3D F1 5D E7 8A E0 52 50 8F BE CA 93
2E 34 6F 05 E0 DC 0D FB CF 93 9E AF FA 3E 58 7C
86 7D 6E 1C 48 70 38 06
=============== Packet Vector #14 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 83 CD 8C E0 CB 42 B1 60 B6 A3 1C 1C
Total packet length = 32. [Input (8 cleartext header octets)]
98 66 05 B4 3D F1 5D E7 01 F6 CE 67 64 C5 74 48
3B B0 2E 6B BF 1E 0A BD 26 A2 25 72 B4 D8 0E E7
CBC IV in: 59 00 83 CD 8C E0 CB 42 B1 60 B6 A3 1C 1C 00 18
CBC IV out:A0 8A 29 78 36 23 1D 84 96 76 93 FF 0A 4C 92 7A
After xor: A0 82 B1 1E 33 97 20 75 CB 91 93 FF 0A 4C 92 7A [hdr]
After CAM: 8C F5 F4 23 BF 09 1C 74 CD 47 00 C1 32 5D 5C 92
After xor: 8D 03 3A 44 DB CC 68 3C F6 F7 2E AA 8D 43 56 2F [msg]
After CAM: 69 DA 48 24 41 1E AC 8E A9 0A CD 8B DD 00 2B 9A
After xor: 4F 78 6D 56 F5 C6 A2 69 A9 0A CD 8B DD 00 2B 9A [msg]
After CAM: C2 03 3B 08 6D B3 CB 3B 2C C8 5D E7 76 A1 C0 44
MIC tag : C2 03 3B 08 6D B3 CB 3B
CTR Start: 01 00 83 CD 8C E0 CB 42 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 8B 16 9C 37 EB 7B BE DB 15 84 41 6E 5F C2 07 46
CTR[0002]: E9 31 BB DD 4E E6 56 9B 68 95 13 5F AB A4 DF EF
CTR[MIC ]: 44 7E 55 14 25 C3 F3 3D
Total packet length = 40. [Encrypted]
98 66 05 B4 3D F1 5D E7 8A E0 52 50 8F BE CA 93
2E 34 6F 05 E0 DC 0D FB CF 93 9E AF FA 3E 58 7C
86 7D 6E 1C 48 70 38 06
=============== Packet Vector #15 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 5F 54 95 0B 18 F2 B1 60 B6 A3 1C 1C
Total packet length = 33. [Input (8 cleartext header octets)]
48 F2 E7 E1 A7 67 1A 51 CD F1 D8 40 6F C2 E9 01
49 53 89 70 05 FB FB 8B A5 72 76 F9 24 04 60 8E
08
CBC IV in: 59 00 5F 54 95 0B 18 F2 B1 60 B6 A3 1C 1C 00 19
CBC IV out:76 74 53 37 95 23 3C F0 EB 77 CE 93 73 06 99 A8
After xor: 76 7C 1B C5 72 C2 9B 97 F1 26 CE 93 73 06 99 A8 [hdr]
After CAM: EF 79 8B 70 34 E4 D5 6B 57 3A F9 44 F0 AF D6 9A
After xor: 22 88 53 30 5B 26 3C 6A 1E 69 70 34 F5 54 2D 11 [msg]
After CAM: 63 BF 4E 10 01 79 38 0B E4 EC C1 39 B2 B4 3B 8C
After xor: C6 CD 38 E9 25 7D 58 85 EC EC C1 39 B2 B4 3B 8C [msg]
After CAM: 39 E1 0E FA BD 2F 43 00 50 9E E7 EB A4 FF 6B 8F
MIC tag : 39 E1 0E FA BD 2F 43 00
CTR Start: 01 00 5F 54 95 0B 18 F2 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: C5 47 A6 A2 73 49 1B 6F 0E 6D C9 F5 9C 12 3B 08
CTR[0002]: C8 18 86 42 3C DB 35 C8 64 4D 8C 4C 58 01 47 27
CTR[MIC ]: 91 E9 76 5D 2D 68 2E E5
Total packet length = 41. [Encrypted]
48 F2 E7 E1 A7 67 1A 51 08 B6 7E E2 1C 8B F2 6E
47 3E 40 85 99 E9 C0 83 6D 6A F0 BB 18 DF 55 46
6C A8 08 78 A7 90 47 6D E5
=============== Packet Vector #15 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 5F 54 95 0B 18 F2 B1 60 B6 A3 1C 1C
Total packet length = 33. [Input (8 cleartext header octets)]
48 F2 E7 E1 A7 67 1A 51 CD F1 D8 40 6F C2 E9 01
49 53 89 70 05 FB FB 8B A5 72 76 F9 24 04 60 8E
08
CBC IV in: 59 00 5F 54 95 0B 18 F2 B1 60 B6 A3 1C 1C 00 19
CBC IV out:76 74 53 37 95 23 3C F0 EB 77 CE 93 73 06 99 A8
After xor: 76 7C 1B C5 72 C2 9B 97 F1 26 CE 93 73 06 99 A8 [hdr]
After CAM: EF 79 8B 70 34 E4 D5 6B 57 3A F9 44 F0 AF D6 9A
After xor: 22 88 53 30 5B 26 3C 6A 1E 69 70 34 F5 54 2D 11 [msg]
After CAM: 63 BF 4E 10 01 79 38 0B E4 EC C1 39 B2 B4 3B 8C
After xor: C6 CD 38 E9 25 7D 58 85 EC EC C1 39 B2 B4 3B 8C [msg]
After CAM: 39 E1 0E FA BD 2F 43 00 50 9E E7 EB A4 FF 6B 8F
MIC tag : 39 E1 0E FA BD 2F 43 00
CTR Start: 01 00 5F 54 95 0B 18 F2 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: C5 47 A6 A2 73 49 1B 6F 0E 6D C9 F5 9C 12 3B 08
CTR[0002]: C8 18 86 42 3C DB 35 C8 64 4D 8C 4C 58 01 47 27
CTR[MIC ]: 91 E9 76 5D 2D 68 2E E5
Total packet length = 41. [Encrypted]
48 F2 E7 E1 A7 67 1A 51 08 B6 7E E2 1C 8B F2 6E
47 3E 40 85 99 E9 C0 83 6D 6A F0 BB 18 DF 55 46
6C A8 08 78 A7 90 47 6D E5
=============== Packet Vector #16 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 EC 60 08 63 31 9A B1 60 B6 A3 1C 1C
Total packet length = 31. [Input (12 cleartext header octets)]
DE 97 DF 3B 8C BD 6D 8E 50 30 DA 4C B0 05 DC FA
0B 59 18 14 26 A9 61 68 5A 99 3D 8C 43 18 5B
CBC IV in: 59 00 EC 60 08 63 31 9A B1 60 B6 A3 1C 1C 00 13
CBC IV out:78 EE 05 5A 88 48 E3 5B 8A 45 46 8F 35 4F 0C A2
After xor: 78 E2 DB CD 57 73 6F E6 E7 CB 16 BF EF 03 0C A2 [hdr]
After CAM: A9 C6 7F 15 00 1A C6 92 81 67 BD EC DF D2 35 C9
After xor: 19 C3 A3 EF 0B 43 DE 86 A7 CE DC 84 85 4B 08 45 [msg]
After CAM: 7C A8 9C 90 46 42 4B E2 4D 96 DF CF BA 12 FD 18
After xor: 3F B0 C7 90 46 42 4B E2 4D 96 DF CF BA 12 FD 18 [msg]
After CAM: 89 C7 B4 E8 A4 24 8C 6C 52 ED 34 50 E3 53 AD F5
MIC tag : 89 C7 B4 E8 A4 24 8C 6C
CTR Start: 01 00 EC 60 08 63 31 9A B1 60 B6 A3 1C 1C 00 01
CTR[0001]: D3 B2 57 B3 6C E8 86 CF 91 9A AC 79 4E 6F 73 3E
CTR[0002]: 65 10 C8 72 39 AF 0F 52 9F D0 A4 DF 54 BF D6 EB
CTR[MIC ]: E1 04 E0 6A 29 B1 80 A9
Total packet length = 39. [Encrypted]
DE 97 DF 3B 8C BD 6D 8E 50 30 DA 4C 63 B7 8B 49
67 B1 9E DB B7 33 CD 11 14 F6 4E B2 26 08 93 68
C3 54 82 8D 95 0C C5
=============== Packet Vector #16 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 EC 60 08 63 31 9A B1 60 B6 A3 1C 1C
Total packet length = 31. [Input (12 cleartext header octets)]
DE 97 DF 3B 8C BD 6D 8E 50 30 DA 4C B0 05 DC FA
0B 59 18 14 26 A9 61 68 5A 99 3D 8C 43 18 5B
CBC IV in: 59 00 EC 60 08 63 31 9A B1 60 B6 A3 1C 1C 00 13
CBC IV out:78 EE 05 5A 88 48 E3 5B 8A 45 46 8F 35 4F 0C A2
After xor: 78 E2 DB CD 57 73 6F E6 E7 CB 16 BF EF 03 0C A2 [hdr]
After CAM: A9 C6 7F 15 00 1A C6 92 81 67 BD EC DF D2 35 C9
After xor: 19 C3 A3 EF 0B 43 DE 86 A7 CE DC 84 85 4B 08 45 [msg]
After CAM: 7C A8 9C 90 46 42 4B E2 4D 96 DF CF BA 12 FD 18
After xor: 3F B0 C7 90 46 42 4B E2 4D 96 DF CF BA 12 FD 18 [msg]
After CAM: 89 C7 B4 E8 A4 24 8C 6C 52 ED 34 50 E3 53 AD F5
MIC tag : 89 C7 B4 E8 A4 24 8C 6C
CTR Start: 01 00 EC 60 08 63 31 9A B1 60 B6 A3 1C 1C 00 01
CTR[0001]: D3 B2 57 B3 6C E8 86 CF 91 9A AC 79 4E 6F 73 3E
CTR[0002]: 65 10 C8 72 39 AF 0F 52 9F D0 A4 DF 54 BF D6 EB
CTR[MIC ]: E1 04 E0 6A 29 B1 80 A9
Total packet length = 39. [Encrypted]
DE 97 DF 3B 8C BD 6D 8E 50 30 DA 4C 63 B7 8B 49
67 B1 9E DB B7 33 CD 11 14 F6 4E B2 26 08 93 68
C3 54 82 8D 95 0C C5
=============== Packet Vector #17 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 60 CF F1 A3 1E A1 B1 60 B6 A3 1C 1C
Total packet length = 32. [Input (12 cleartext header octets)]
A5 EE 93 E4 57 DF 05 46 6E 78 2D CF 2E 20 21 12
98 10 5F 12 9D 5E D9 5B 93 F7 2D 30 B2 FA CC D7
CBC IV in: 59 00 60 CF F1 A3 1E A1 B1 60 B6 A3 1C 1C 00 14
CBC IV out:C3 34 69 7D 11 38 73 06 BD 34 E2 10 1F 66 17 E8
After xor: C3 38 CC 93 82 DC 24 D9 B8 72 8C 68 32 A9 17 E8 [hdr]
After CAM: 43 6F 37 74 AB 94 3B 41 EA AD 00 CA C3 99 13 7B
After xor: 6D 4F 16 66 33 84 64 53 77 F3 D9 91 50 6E 3E 4B [msg]
After CAM: 2D 28 FB 62 DA 06 97 A7 4C D4 31 B8 B5 AE AE EE
After xor: 9F D2 37 B5 DA 06 97 A7 4C D4 31 B8 B5 AE AE EE [msg]
After CAM: F3 DE 10 CD 91 4D B1 B6 CC 37 F0 A2 4A 5A B7 A1
MIC tag : F3 DE 10 CD 91 4D B1 B6
CTR Start: 01 00 60 CF F1 A3 1E A1 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 25 E6 9A F0 30 A9 56 E6 FF C0 3F 87 87 7A 89 74
CTR[0002]: A2 1B 46 23 76 A2 1E DD F2 AC 4B EC 42 95 3D D3
CTR[MIC ]: C2 99 28 FF E7 BB DB 29
Total packet length = 40. [Encrypted]
A5 EE 93 E4 57 DF 05 46 6E 78 2D CF 0B C6 BB E2
A8 B9 09 F4 62 9E E6 DC 14 8D A4 44 10 E1 8A F4
31 47 38 32 76 F6 6A 9F
=============== Packet Vector #17 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 60 CF F1 A3 1E A1 B1 60 B6 A3 1C 1C
Total packet length = 32. [Input (12 cleartext header octets)]
A5 EE 93 E4 57 DF 05 46 6E 78 2D CF 2E 20 21 12
98 10 5F 12 9D 5E D9 5B 93 F7 2D 30 B2 FA CC D7
CBC IV in: 59 00 60 CF F1 A3 1E A1 B1 60 B6 A3 1C 1C 00 14
CBC IV out:C3 34 69 7D 11 38 73 06 BD 34 E2 10 1F 66 17 E8
After xor: C3 38 CC 93 82 DC 24 D9 B8 72 8C 68 32 A9 17 E8 [hdr]
After CAM: 43 6F 37 74 AB 94 3B 41 EA AD 00 CA C3 99 13 7B
After xor: 6D 4F 16 66 33 84 64 53 77 F3 D9 91 50 6E 3E 4B [msg]
After CAM: 2D 28 FB 62 DA 06 97 A7 4C D4 31 B8 B5 AE AE EE
After xor: 9F D2 37 B5 DA 06 97 A7 4C D4 31 B8 B5 AE AE EE [msg]
After CAM: F3 DE 10 CD 91 4D B1 B6 CC 37 F0 A2 4A 5A B7 A1
MIC tag : F3 DE 10 CD 91 4D B1 B6
CTR Start: 01 00 60 CF F1 A3 1E A1 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 25 E6 9A F0 30 A9 56 E6 FF C0 3F 87 87 7A 89 74
CTR[0002]: A2 1B 46 23 76 A2 1E DD F2 AC 4B EC 42 95 3D D3
CTR[MIC ]: C2 99 28 FF E7 BB DB 29
Total packet length = 40. [Encrypted]
A5 EE 93 E4 57 DF 05 46 6E 78 2D CF 0B C6 BB E2
A8 B9 09 F4 62 9E E6 DC 14 8D A4 44 10 E1 8A F4
31 47 38 32 76 F6 6A 9F
=============== Packet Vector #18 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 0F 85 CD 99 5C 97 B1 60 B6 A3 1C 1C
Total packet length = 33. [Input (12 cleartext header octets)]
24 AA 1B F9 A5 CD 87 61 82 A2 50 74 26 45 94 1E
75 63 2D 34 91 AF 0F C0 C9 87 6C 3B E4 AA 74 68
C9
CBC IV in: 59 00 0F 85 CD 99 5C 97 B1 60 B6 A3 1C 1C 00 15
CBC IV out:72 0A 46 75 0F 40 59 53 F2 3B D2 1F 6A 11 60 F6
After xor: 72 06 62 DF 14 B9 FC 9E 75 5A 50 BD 3A 65 60 F6 [hdr]
After CAM: 67 73 A0 FD D5 7E D3 5E E8 24 06 D0 A1 8B 0E 18
After xor: 41 36 34 E3 A0 1D FE 6A 79 8B 09 10 68 0C 62 23 [msg]
After CAM: BB 1E D8 9F 60 29 D0 99 09 14 06 A5 E3 8B 72 7B
After xor: 5F B4 AC F7 A9 29 D0 99 09 14 06 A5 E3 8B 72 7B [msg]
After CAM: 3E 4F 40 73 D1 31 E9 B8 02 C8 99 BC FD AC 19 4B
MIC tag : 3E 4F 40 73 D1 31 E9 B8
CTR Start: 01 00 0F 85 CD 99 5C 97 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 04 6F 42 2C 8F 52 FB 9B 06 A3 3B 9F B7 F0 A6 00
CTR[0002]: 34 76 51 DB 89 10 FB E6 73 E8 56 6E DB 66 47 5D
CTR[MIC ]: 9F EC 93 6C 5C 7A AD 0F
Total packet length = 41. [Encrypted]
24 AA 1B F9 A5 CD 87 61 82 A2 50 74 22 2A D6 32
FA 31 D6 AF 97 0C 34 5F 7E 77 CA 3B D0 DC 25 B3
40 A1 A3 D3 1F 8D 4B 44 B7
=============== Packet Vector #18 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 0F 85 CD 99 5C 97 B1 60 B6 A3 1C 1C
Total packet length = 33. [Input (12 cleartext header octets)]
24 AA 1B F9 A5 CD 87 61 82 A2 50 74 26 45 94 1E
75 63 2D 34 91 AF 0F C0 C9 87 6C 3B E4 AA 74 68
C9
CBC IV in: 59 00 0F 85 CD 99 5C 97 B1 60 B6 A3 1C 1C 00 15
CBC IV out:72 0A 46 75 0F 40 59 53 F2 3B D2 1F 6A 11 60 F6
After xor: 72 06 62 DF 14 B9 FC 9E 75 5A 50 BD 3A 65 60 F6 [hdr]
After CAM: 67 73 A0 FD D5 7E D3 5E E8 24 06 D0 A1 8B 0E 18
After xor: 41 36 34 E3 A0 1D FE 6A 79 8B 09 10 68 0C 62 23 [msg]
After CAM: BB 1E D8 9F 60 29 D0 99 09 14 06 A5 E3 8B 72 7B
After xor: 5F B4 AC F7 A9 29 D0 99 09 14 06 A5 E3 8B 72 7B [msg]
After CAM: 3E 4F 40 73 D1 31 E9 B8 02 C8 99 BC FD AC 19 4B
MIC tag : 3E 4F 40 73 D1 31 E9 B8
CTR Start: 01 00 0F 85 CD 99 5C 97 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 04 6F 42 2C 8F 52 FB 9B 06 A3 3B 9F B7 F0 A6 00
CTR[0002]: 34 76 51 DB 89 10 FB E6 73 E8 56 6E DB 66 47 5D
CTR[MIC ]: 9F EC 93 6C 5C 7A AD 0F
Total packet length = 41. [Encrypted]
24 AA 1B F9 A5 CD 87 61 82 A2 50 74 22 2A D6 32
FA 31 D6 AF 97 0C 34 5F 7E 77 CA 3B D0 DC 25 B3
40 A1 A3 D3 1F 8D 4B 44 B7
=============== Packet Vector #19 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 C2 9B 2C AA C4 CD B1 60 B6 A3 1C 1C
Total packet length = 31. [Input (8 cleartext header octets)]
69 19 46 B9 CA 07 BE 87 07 01 35 A6 43 7C 9D B1
20 CD 61 D8 F6 C3 9C 3E A1 25 FD 95 A0 D2 3D
CBC IV in: 61 00 C2 9B 2C AA C4 CD B1 60 B6 A3 1C 1C 00 17
CBC IV out:74 AD F8 04 05 2A 48 E7 46 97 38 D5 BA A1 27 79
After xor: 74 A5 91 1D 43 93 82 E0 F8 10 38 D5 BA A1 27 79 [hdr]
After CAM: BD C3 B1 41 1C 64 C8 B3 A9 DC 6A 94 78 97 88 E2
After xor: BA C2 84 E7 5F 18 55 02 89 11 0B 4C 8E 54 14 DC [msg]
After CAM: 7D 6C 8A BF AD 68 48 D8 C5 FB CD 1E AF F2 44 99
After xor: DC 49 77 2A 0D BA 75 D8 C5 FB CD 1E AF F2 44 99 [msg]
After CAM: 19 99 AB 92 5E 30 46 96 3D EF FB 1B 4C 87 F7 76
MIC tag : 19 99 AB 92 5E 30 46 96 3D EF
CTR Start: 01 00 C2 9B 2C AA C4 CD B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 02 B9 D4 1F 87 E0 60 E7 EF DE 6B 7E D3 DE 5E D2
CTR[0002]: 61 49 31 C5 2F 34 AA 47 A3 E4 D3 2C 0B 36 41 C6
CTR[MIC ]: B9 9F C6 C5 96 7B AA 8E 1A 87
Total packet length = 41. [Encrypted]
69 19 46 B9 CA 07 BE 87 05 B8 E1 B9 C4 9C FD 56
CF 13 0A A6 25 1D C2 EC C0 6C CC 50 8F E6 97 A0
06 6D 57 C8 4B EC 18 27 68
=============== Packet Vector #19 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 C2 9B 2C AA C4 CD B1 60 B6 A3 1C 1C
Total packet length = 31. [Input (8 cleartext header octets)]
69 19 46 B9 CA 07 BE 87 07 01 35 A6 43 7C 9D B1
20 CD 61 D8 F6 C3 9C 3E A1 25 FD 95 A0 D2 3D
CBC IV in: 61 00 C2 9B 2C AA C4 CD B1 60 B6 A3 1C 1C 00 17
CBC IV out:74 AD F8 04 05 2A 48 E7 46 97 38 D5 BA A1 27 79
After xor: 74 A5 91 1D 43 93 82 E0 F8 10 38 D5 BA A1 27 79 [hdr]
After CAM: BD C3 B1 41 1C 64 C8 B3 A9 DC 6A 94 78 97 88 E2
After xor: BA C2 84 E7 5F 18 55 02 89 11 0B 4C 8E 54 14 DC [msg]
After CAM: 7D 6C 8A BF AD 68 48 D8 C5 FB CD 1E AF F2 44 99
After xor: DC 49 77 2A 0D BA 75 D8 C5 FB CD 1E AF F2 44 99 [msg]
After CAM: 19 99 AB 92 5E 30 46 96 3D EF FB 1B 4C 87 F7 76
MIC tag : 19 99 AB 92 5E 30 46 96 3D EF
CTR Start: 01 00 C2 9B 2C AA C4 CD B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 02 B9 D4 1F 87 E0 60 E7 EF DE 6B 7E D3 DE 5E D2
CTR[0002]: 61 49 31 C5 2F 34 AA 47 A3 E4 D3 2C 0B 36 41 C6
CTR[MIC ]: B9 9F C6 C5 96 7B AA 8E 1A 87
Total packet length = 41. [Encrypted]
69 19 46 B9 CA 07 BE 87 05 B8 E1 B9 C4 9C FD 56
CF 13 0A A6 25 1D C2 EC C0 6C CC 50 8F E6 97 A0
06 6D 57 C8 4B EC 18 27 68
=============== Packet Vector #20 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 2C 6B 75 95 EE 62 B1 60 B6 A3 1C 1C
Total packet length = 32. [Input (8 cleartext header octets)]
D0 C5 4E CB 84 62 7D C4 C8 C0 88 0E 6C 63 6E 20
09 3D D6 59 42 17 D2 E1 88 77 DB 26 4E 71 A5 CC
CBC IV in: 61 00 2C 6B 75 95 EE 62 B1 60 B6 A3 1C 1C 00 18
CBC IV out:35 A9 48 70 F9 B0 C7 85 FB 32 1A D1 3C 8C A4 9A
After xor: 35 A1 98 B5 B7 7B 43 E7 86 F6 1A D1 3C 8C A4 9A [hdr]
After CAM: 0A 3C E3 0F AC 09 DC 5C 00 10 5C 69 AC 19 F7 19
After xor: C2 FC 6B 01 C0 6A B2 7C 09 2D 8A 30 EE 0E 25 F8 [msg]
After CAM: 61 CD 80 D0 72 E6 84 E1 BF E1 4A 00 27 2A 4D 96
After xor: E9 BA 5B F6 3C 97 21 2D BF E1 4A 00 27 2A 4D 96 [msg]
After CAM: E5 F9 F2 AB 47 FD 7B 8D 6F 72 F4 72 74 D7 69 BB
MIC tag : E5 F9 F2 AB 47 FD 7B 8D 6F 72
CTR Start: 01 00 2C 6B 75 95 EE 62 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 9C 0E 31 66 B2 81 58 31 5E 63 16 5A 9D BD CE 35
CTR[0002]: 00 3E 66 D3 E0 5F 7E A7 EF C8 9A 5F DD 39 E3 54
CTR[MIC ]: 9A 5E 87 1A 17 10 38 0E AA DB
Total packet length = 42. [Encrypted]
D0 C5 4E CB 84 62 7D C4 54 CE B9 68 DE E2 36 11
57 5E C0 03 DF AA 1C D4 88 49 BD F5 AE 2E DB 6B
7F A7 75 B1 50 ED 43 83 C5 A9
=============== Packet Vector #20 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 2C 6B 75 95 EE 62 B1 60 B6 A3 1C 1C
Total packet length = 32. [Input (8 cleartext header octets)]
D0 C5 4E CB 84 62 7D C4 C8 C0 88 0E 6C 63 6E 20
09 3D D6 59 42 17 D2 E1 88 77 DB 26 4E 71 A5 CC
CBC IV in: 61 00 2C 6B 75 95 EE 62 B1 60 B6 A3 1C 1C 00 18
CBC IV out:35 A9 48 70 F9 B0 C7 85 FB 32 1A D1 3C 8C A4 9A
After xor: 35 A1 98 B5 B7 7B 43 E7 86 F6 1A D1 3C 8C A4 9A [hdr]
After CAM: 0A 3C E3 0F AC 09 DC 5C 00 10 5C 69 AC 19 F7 19
After xor: C2 FC 6B 01 C0 6A B2 7C 09 2D 8A 30 EE 0E 25 F8 [msg]
After CAM: 61 CD 80 D0 72 E6 84 E1 BF E1 4A 00 27 2A 4D 96
After xor: E9 BA 5B F6 3C 97 21 2D BF E1 4A 00 27 2A 4D 96 [msg]
After CAM: E5 F9 F2 AB 47 FD 7B 8D 6F 72 F4 72 74 D7 69 BB
MIC tag : E5 F9 F2 AB 47 FD 7B 8D 6F 72
CTR Start: 01 00 2C 6B 75 95 EE 62 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 9C 0E 31 66 B2 81 58 31 5E 63 16 5A 9D BD CE 35
CTR[0002]: 00 3E 66 D3 E0 5F 7E A7 EF C8 9A 5F DD 39 E3 54
CTR[MIC ]: 9A 5E 87 1A 17 10 38 0E AA DB
Total packet length = 42. [Encrypted]
D0 C5 4E CB 84 62 7D C4 54 CE B9 68 DE E2 36 11
57 5E C0 03 DF AA 1C D4 88 49 BD F5 AE 2E DB 6B
7F A7 75 B1 50 ED 43 83 C5 A9
=============== Packet Vector #21 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 C5 3C D4 C2 AA 24 B1 60 B6 A3 1C 1C
Total packet length = 33. [Input (8 cleartext header octets)]
E2 85 E0 E4 80 8C DA 3D F7 5D AA 07 10 C4 E6 42
97 79 4D C2 B7 D2 A2 07 57 B1 AA 4E 44 80 02 FF
AB
CBC IV in: 61 00 C5 3C D4 C2 AA 24 B1 60 B6 A3 1C 1C 00 19
CBC IV out:2A 3C 23 B2 43 F5 1C 35 F7 79 5A CB 3B 20 21 2F
After xor: 2A 34 C1 37 A3 11 9C B9 2D 44 5A CB 3B 20 21 2F [hdr]
After CAM: A1 7E AD 4C EE AB 51 21 1D 2A 32 F2 D4 45 A6 D6
After xor: 56 23 07 4B FE 6F B7 63 8A 53 7F 30 63 97 04 D1 [msg]
After CAM: A9 A1 32 55 8F C6 9B 98 A9 CC 23 96 FE CA 84 EB
After xor: FE 10 98 1B CB 46 99 67 02 CC 23 96 FE CA 84 EB [msg]
After CAM: 6A 5E 04 42 D1 A5 7E 17 9A 6C 8B 56 F7 19 80 C5
MIC tag : 6A 5E 04 42 D1 A5 7E 17 9A 6C
CTR Start: 01 00 C5 3C D4 C2 AA 24 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 46 1D EF 41 AF A2 94 52 5D 51 AE CB 04 49 74 CD
CTR[0002]: 29 2E 62 66 1B 66 9A 2B 97 72 6B 77 32 A8 DC 35
CTR[MIC ]: B8 54 06 A2 6C 6F 93 37 8A BF
Total packet length = 43. [Encrypted]
E2 85 E0 E4 80 8C DA 3D B1 40 45 46 BF 66 72 10
CA 28 E3 09 B3 9B D6 CA 7E 9F C8 28 5F E6 98 D4
3C D2 0A 02 E0 BD CA ED 20 10 D3
=============== Packet Vector #21 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 C5 3C D4 C2 AA 24 B1 60 B6 A3 1C 1C
Total packet length = 33. [Input (8 cleartext header octets)]
E2 85 E0 E4 80 8C DA 3D F7 5D AA 07 10 C4 E6 42
97 79 4D C2 B7 D2 A2 07 57 B1 AA 4E 44 80 02 FF
AB
CBC IV in: 61 00 C5 3C D4 C2 AA 24 B1 60 B6 A3 1C 1C 00 19
CBC IV out:2A 3C 23 B2 43 F5 1C 35 F7 79 5A CB 3B 20 21 2F
After xor: 2A 34 C1 37 A3 11 9C B9 2D 44 5A CB 3B 20 21 2F [hdr]
After CAM: A1 7E AD 4C EE AB 51 21 1D 2A 32 F2 D4 45 A6 D6
After xor: 56 23 07 4B FE 6F B7 63 8A 53 7F 30 63 97 04 D1 [msg]
After CAM: A9 A1 32 55 8F C6 9B 98 A9 CC 23 96 FE CA 84 EB
After xor: FE 10 98 1B CB 46 99 67 02 CC 23 96 FE CA 84 EB [msg]
After CAM: 6A 5E 04 42 D1 A5 7E 17 9A 6C 8B 56 F7 19 80 C5
MIC tag : 6A 5E 04 42 D1 A5 7E 17 9A 6C
CTR Start: 01 00 C5 3C D4 C2 AA 24 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 46 1D EF 41 AF A2 94 52 5D 51 AE CB 04 49 74 CD
CTR[0002]: 29 2E 62 66 1B 66 9A 2B 97 72 6B 77 32 A8 DC 35
CTR[MIC ]: B8 54 06 A2 6C 6F 93 37 8A BF
Total packet length = 43. [Encrypted]
E2 85 E0 E4 80 8C DA 3D B1 40 45 46 BF 66 72 10
CA 28 E3 09 B3 9B D6 CA 7E 9F C8 28 5F E6 98 D4
3C D2 0A 02 E0 BD CA ED 20 10 D3
=============== Packet Vector #22 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 BE E9 26 7F BA DC B1 60 B6 A3 1C 1C
Total packet length = 31. [Input (12 cleartext header octets)]
6C AE F9 94 11 41 57 0D 7C 81 34 05 C2 38 82 2F
AC 5F 98 FF 92 94 05 B0 AD 12 7A 4E 41 85 4E
CBC IV in: 61 00 BE E9 26 7F BA DC B1 60 B6 A3 1C 1C 00 13
CBC IV out:20 60 6A D1 E1 A0 84 52 2F A3 8B F4 88 1D D6 8B
After xor: 20 6C 06 7F 18 34 95 13 78 AE F7 75 BC 18 D6 8B [hdr]
After CAM: 71 FD FF E7 D9 C8 95 75 D3 EC 0B 7E 7B 8B BE E7
After xor: B3 C5 7D C8 75 97 0D 8A 41 78 0E CE D6 99 C4 A9 [msg]
After CAM: CA AD 93 9C 59 BA 40 AA 1A 0B 88 1B EE 3D 3C 65
After xor: 8B 28 DD 9C 59 BA 40 AA 1A 0B 88 1B EE 3D 3C 65 [msg]
After CAM: DC 48 8F AA 9C 75 E7 03 17 56 C2 C7 48 48 8D 1B
MIC tag : DC 48 8F AA 9C 75 E7 03 17 56
CTR Start: 01 00 BE E9 26 7F BA DC B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 56 F0 17 B3 BD 09 02 D6 EA A5 A2 91 AD 4A 2D E5
CTR[0002]: 20 3D 34 21 EF 5B F8 FC 7B 21 5C 76 7B A5 21 A6
CTR[MIC ]: F1 A2 86 9C 2A 9E B8 61 48 0B
Total packet length = 41. [Encrypted]
6C AE F9 94 11 41 57 0D 7C 81 34 05 94 C8 95 9C
11 56 9A 29 78 31 A7 21 00 58 57 AB 61 B8 7A 2D
EA 09 36 B6 EB 5F 62 5F 5D
=============== Packet Vector #22 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 BE E9 26 7F BA DC B1 60 B6 A3 1C 1C
Total packet length = 31. [Input (12 cleartext header octets)]
6C AE F9 94 11 41 57 0D 7C 81 34 05 C2 38 82 2F
AC 5F 98 FF 92 94 05 B0 AD 12 7A 4E 41 85 4E
CBC IV in: 61 00 BE E9 26 7F BA DC B1 60 B6 A3 1C 1C 00 13
CBC IV out:20 60 6A D1 E1 A0 84 52 2F A3 8B F4 88 1D D6 8B
After xor: 20 6C 06 7F 18 34 95 13 78 AE F7 75 BC 18 D6 8B [hdr]
After CAM: 71 FD FF E7 D9 C8 95 75 D3 EC 0B 7E 7B 8B BE E7
After xor: B3 C5 7D C8 75 97 0D 8A 41 78 0E CE D6 99 C4 A9 [msg]
After CAM: CA AD 93 9C 59 BA 40 AA 1A 0B 88 1B EE 3D 3C 65
After xor: 8B 28 DD 9C 59 BA 40 AA 1A 0B 88 1B EE 3D 3C 65 [msg]
After CAM: DC 48 8F AA 9C 75 E7 03 17 56 C2 C7 48 48 8D 1B
MIC tag : DC 48 8F AA 9C 75 E7 03 17 56
CTR Start: 01 00 BE E9 26 7F BA DC B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 56 F0 17 B3 BD 09 02 D6 EA A5 A2 91 AD 4A 2D E5
CTR[0002]: 20 3D 34 21 EF 5B F8 FC 7B 21 5C 76 7B A5 21 A6
CTR[MIC ]: F1 A2 86 9C 2A 9E B8 61 48 0B
Total packet length = 41. [Encrypted]
6C AE F9 94 11 41 57 0D 7C 81 34 05 94 C8 95 9C
11 56 9A 29 78 31 A7 21 00 58 57 AB 61 B8 7A 2D
EA 09 36 B6 EB 5F 62 5F 5D
=============== Packet Vector #23 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 DF A8 B1 24 50 07 B1 60 B6 A3 1C 1C
Total packet length = 32. [Input (12 cleartext header octets)]
36 A5 2C F1 6B 19 A2 03 7A B7 01 1E 4D BF 3E 77
4A D2 45 E5 D5 89 1F 9D 1C 32 A0 AE 02 2C 85 D7
CBC IV in: 61 00 DF A8 B1 24 50 07 B1 60 B6 A3 1C 1C 00 14
CBC IV out:78 FD B6 AF 61 9E 1C 8D 82 41 17 A8 73 60 1B 70
After xor: 78 F1 80 0A 4D 6F 77 94 20 42 6D 1F 72 7E 1B 70 [hdr]
After CAM: 62 2E 28 65 92 43 DB 82 88 79 09 1E A7 24 54 67
After xor: 2F 91 16 12 D8 91 9E 67 5D F0 16 83 BB 16 F4 C9 [msg]
After CAM: 95 0E 52 08 FF 16 70 8C 1E D9 BB 06 3E 1E 41 CF
After xor: 97 22 D7 DF FF 16 70 8C 1E D9 BB 06 3E 1E 41 CF [msg]
After CAM: BA CD 51 FC 77 F4 02 8D 47 D5 7D 54 7D 46 33 4B
MIC tag : BA CD 51 FC 77 F4 02 8D 47 D5
CTR Start: 01 00 DF A8 B1 24 50 07 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 15 D6 DD DD 98 96 39 91 35 75 1A 64 B8 D8 D4 F9
CTR[0002]: 7D 61 6D 1D EB 92 00 2B 6F FA AB 53 BC AF 69 89
CTR[MIC ]: 33 E9 27 BE E1 59 06 9C DB 32
Total packet length = 42. [Encrypted]
36 A5 2C F1 6B 19 A2 03 7A B7 01 1E 58 69 E3 AA
D2 44 7C 74 E0 FC 05 F9 A4 EA 74 57 7F 4D E8 CA
89 24 76 42 96 AD 04 11 9C E7
=============== Packet Vector #23 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 DF A8 B1 24 50 07 B1 60 B6 A3 1C 1C
Total packet length = 32. [Input (12 cleartext header octets)]
36 A5 2C F1 6B 19 A2 03 7A B7 01 1E 4D BF 3E 77
4A D2 45 E5 D5 89 1F 9D 1C 32 A0 AE 02 2C 85 D7
CBC IV in: 61 00 DF A8 B1 24 50 07 B1 60 B6 A3 1C 1C 00 14
CBC IV out:78 FD B6 AF 61 9E 1C 8D 82 41 17 A8 73 60 1B 70
After xor: 78 F1 80 0A 4D 6F 77 94 20 42 6D 1F 72 7E 1B 70 [hdr]
After CAM: 62 2E 28 65 92 43 DB 82 88 79 09 1E A7 24 54 67
After xor: 2F 91 16 12 D8 91 9E 67 5D F0 16 83 BB 16 F4 C9 [msg]
After CAM: 95 0E 52 08 FF 16 70 8C 1E D9 BB 06 3E 1E 41 CF
After xor: 97 22 D7 DF FF 16 70 8C 1E D9 BB 06 3E 1E 41 CF [msg]
After CAM: BA CD 51 FC 77 F4 02 8D 47 D5 7D 54 7D 46 33 4B
MIC tag : BA CD 51 FC 77 F4 02 8D 47 D5
CTR Start: 01 00 DF A8 B1 24 50 07 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: 15 D6 DD DD 98 96 39 91 35 75 1A 64 B8 D8 D4 F9
CTR[0002]: 7D 61 6D 1D EB 92 00 2B 6F FA AB 53 BC AF 69 89
CTR[MIC ]: 33 E9 27 BE E1 59 06 9C DB 32
Total packet length = 42. [Encrypted]
36 A5 2C F1 6B 19 A2 03 7A B7 01 1E 58 69 E3 AA
D2 44 7C 74 E0 FC 05 F9 A4 EA 74 57 7F 4D E8 CA
89 24 76 42 96 AD 04 11 9C E7
=============== Packet Vector #24 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 3B 8F D8 D3 A9 37 B1 60 B6 A3 1C 1C
Total packet length = 33. [Input (12 cleartext header octets)]
A4 D4 99 F7 84 19 72 8C 19 17 8B 0C 9D C9 ED AE
2F F5 DF 86 36 E8 C6 DE 0E ED 55 F7 86 7E 33 33
7D
CBC IV in: 61 00 3B 8F D8 D3 A9 37 B1 60 B6 A3 1C 1C 00 15
CBC IV out:84 E6 CF DD 6A 37 68 5D E6 71 AD 54 B3 BE FE B9
After xor: 84 EA 6B 09 F3 C0 EC 44 94 FD B4 43 38 B2 FE B9 [hdr]
After CAM: C5 0F A0 62 20 18 F1 21 0E BC 3D 2E 47 B7 B8 C3
After xor: 58 C6 4D CC 0F ED 2E A7 38 54 FB F0 49 5A ED 34 [msg]
After CAM: C4 6F 6D C3 17 3C 2A 7A 81 FC 2D DA 7F B7 C6 60
After xor: 42 11 5E F0 6A 3C 2A 7A 81 FC 2D DA 7F B7 C6 60 [msg]
After CAM: DF AB 2E 76 B0 67 50 B3 7C DD 9A AC F3 79 17 71
MIC tag : DF AB 2E 76 B0 67 50 B3 7C DD
CTR Start: 01 00 3B 8F D8 D3 A9 37 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: D6 D0 6C F8 16 CE D0 F1 A0 E0 AC 71 BA B9 AD 34
CTR[0002]: 76 4A FF 9A 1B F8 55 1F 68 54 39 0A EE 37 24 28
CTR[MIC ]: 4B F4 31 B8 17 86 4B 5D 16 F2
Total packet length = 43. [Encrypted]
A4 D4 99 F7 84 19 72 8C 19 17 8B 0C 4B 19 81 56
39 3B 0F 77 96 08 6A AF B4 54 F8 C3 F0 34 CC A9
66 94 5F 1F CE A7 E1 1B EE 6A 2F
=============== Packet Vector #24 ==================
CAM Key: D7 5C 27 78 07 8C A9 3D 97 1F 96 FD E7 20 F4 CD
Nonce = 00 3B 8F D8 D3 A9 37 B1 60 B6 A3 1C 1C
Total packet length = 33. [Input (12 cleartext header octets)]
A4 D4 99 F7 84 19 72 8C 19 17 8B 0C 9D C9 ED AE
2F F5 DF 86 36 E8 C6 DE 0E ED 55 F7 86 7E 33 33
7D
CBC IV in: 61 00 3B 8F D8 D3 A9 37 B1 60 B6 A3 1C 1C 00 15
CBC IV out:84 E6 CF DD 6A 37 68 5D E6 71 AD 54 B3 BE FE B9
After xor: 84 EA 6B 09 F3 C0 EC 44 94 FD B4 43 38 B2 FE B9 [hdr]
After CAM: C5 0F A0 62 20 18 F1 21 0E BC 3D 2E 47 B7 B8 C3
After xor: 58 C6 4D CC 0F ED 2E A7 38 54 FB F0 49 5A ED 34 [msg]
After CAM: C4 6F 6D C3 17 3C 2A 7A 81 FC 2D DA 7F B7 C6 60
After xor: 42 11 5E F0 6A 3C 2A 7A 81 FC 2D DA 7F B7 C6 60 [msg]
After CAM: DF AB 2E 76 B0 67 50 B3 7C DD 9A AC F3 79 17 71
MIC tag : DF AB 2E 76 B0 67 50 B3 7C DD
CTR Start: 01 00 3B 8F D8 D3 A9 37 B1 60 B6 A3 1C 1C 00 01
CTR[0001]: D6 D0 6C F8 16 CE D0 F1 A0 E0 AC 71 BA B9 AD 34
CTR[0002]: 76 4A FF 9A 1B F8 55 1F 68 54 39 0A EE 37 24 28
CTR[MIC ]: 4B F4 31 B8 17 86 4B 5D 16 F2
Total packet length = 43. [Encrypted]
A4 D4 99 F7 84 19 72 8C 19 17 8B 0C 4B 19 81 56
39 3B 0F 77 96 08 6A AF B4 54 F8 C3 F0 34 CC A9
66 94 5F 1F CE A7 E1 1B EE 6A 2F
Camellia-CTR and Camellia-CCM employ CTR mode for confidentiality. For the security of CTR mode, refer to the Security Considerations of [16].
Camellia CTR和Camellia CCM采用CTR模式进行保密。有关CTR模式的安全性，请参阅[16]中的安全注意事项。
Thanks to Rui Hodai for comments and suggestions. Special thanks to Alfred Hoenes for several very detailed reviews and suggestions.
感谢芮和岱的评论和建议。特别感谢阿尔弗雷德·霍恩斯（Alfred Hoenes）几次非常详细的评论和建议。
[1] Matsui, M., Nakajima, J., and S. Moriai, "A Description of the Camellia Encryption Algorithm", RFC 3713, April 2004.
[1] Matsui，M.，Nakajima，J.，和S.Moraii，“茶花加密算法的描述”，RFC 3713，2004年4月。
[2] Bradner, S., "Key words for use in RFCs to Indicate Requirement Levels", BCP 14, RFC 2119, March 1997.
[2] Bradner，S.，“RFC中用于表示需求水平的关键词”，BCP 14，RFC 2119，1997年3月。
[3] Dworkin, M., "Recommendation for Block Cipher Modes of Operation - Methods and Techniques", NIST Special Publication 800-38A, December 2001, <http://csrc.nist.gov/ publications/nistpubs/800-38a/sp800-38a.pdf>.
[3] Dworkin，M.“分组密码操作模式的建议-方法和技术”，NIST特别出版物800-38A，2001年12月<http://csrc.nist.gov/ 出版物/nistpubs/800-38a/sp800-38a.pdf>。
[4] National Institute of Standards and Technology, "Recommendation for Block Cipher Modes Operation : The CCM Mode for Authentication and Confidentiality", May 2004, <http:// csrc.nist.gov/publications/nistpubs/800-38C/SP800-38C.pdf>.
[4] 国家标准与技术研究所，“分组密码模式操作建议：认证和保密的CCM模式”，2004年5月，<http://csrc.nist.gov/publications/nistpubs/800-38C/SP800-38C.pdf>。
[5] National Institute of Standards and Technology, "Advanced Encryption Standard (AES)", FIPS PUB 197, November 2001, <http://csrc.nist.gov/publications/fips/fips197/fips-197.pdf>.
[5] 国家标准与技术研究所，“高级加密标准（AES）”，FIPS PUB 197，2001年11月<http://csrc.nist.gov/publications/fips/fips197/fips-197.pdf>.
[6] Kato, A., Moriai, S., and M. Kanda, "The Camellia Cipher Algorithm and Its Use With IPsec", RFC 4312, December 2005.
[6] Kato，A.，Moraii，S.，和M.Kanda，“茶花密码算法及其与IPsec的使用”，RFC 4312，2005年12月。
[7] Moriai, S., Kato, A., and M. Kanda, "Addition of Camellia Cipher Suites to Transport Layer Security (TLS)", RFC 4132, July 2005.
[7] Moraii，S.，Kato，A.，和M.Kanda，“将茶花密码套件添加到传输层安全性（TLS）”，RFC 4132，2005年7月。
[8] Moriai, S. and A. Kato, "Use of the Camellia Encryption Algorithm in Cryptographic Message Syntax (CMS)", RFC 3657, January 2004.
[8] Moraii，S.和A.Kato，“密码学消息语法（CMS）中茶花加密算法的使用”，RFC 3657，2004年1月。
[9] Eastlake, D., "Additional XML Security Uniform Resource Identifiers (URIs)", RFC 4051, April 2005.
[9] Eastlake，D.，“附加XML安全统一资源标识符（URI）”，RFC 4051，2005年4月。
[10] International Organization for Standardization, "Information technology - Security techniques - Encryption algorithms - Part 3: Block ciphers", ISO/IEC 18033-3, July 2005.
[10] 国际标准化组织，“信息技术-安全技术-加密算法-第3部分：分组密码”，ISO/IEC 18033-3，2005年7月。
[11] "The NESSIE project (New European Schemes for Signatures, Integrity and Encryption)", <http://www.cosic.esat.kuleuven.be/nessie/>.
[11] “尼斯工程（新的欧洲签名、完整性和加密方案）”<http://www.cosic.esat.kuleuven.be/nessie/>.
[12] Information-technology Promotion Agency (IPA), "Cryptography Research and Evaluation Committees", <http://www.ipa.go.jp/security/enc/CRYPTREC/index-e.html>.
[12] 信息技术促进局（IPA），“密码学研究和评估委员会”<http://www.ipa.go.jp/security/enc/CRYPTREC/index-e.html>.
[13] "Camellia open source software", <http://info.isl.ntt.co.jp/crypt/eng/camellia/source.html>.
[13] “Camellia开源软件”<http://info.isl.ntt.co.jp/crypt/eng/camellia/source.html>.
[14] "Camellia web site", <http://info.isl.ntt.co.jp/camellia/>.
[14] “山茶花网站”<http://info.isl.ntt.co.jp/camellia/>.
[15] Whiting, D., Housley, R., and N. Ferguson, "Counter with CBC-MAC (CCM)", RFC 3610, September 2003.
[15] Whiting，D.，Housley，R.，和N.Ferguson，“CBC-MAC（CCM）计数器”，RFC 36102003年9月。
[16] Housley, R., "Using Advanced Encryption Standard (AES) Counter Mode With IPsec Encapsulating Security Payload (ESP)", RFC 3686, January 2004.
[16] Housley，R.，“将高级加密标准（AES）计数器模式与IPsec封装安全有效载荷（ESP）结合使用”，RFC 3686，2004年1月。
Authors' Addresses
作者地址
Akihiro Kato NTT Software Corporation
加藤昭宏NTT软件公司
Phone: +81-45-212-7577
Fax: +81-45-212-9800
EMail: akato@po.ntts.co.jp
Phone: +81-45-212-7577
Fax: +81-45-212-9800
EMail: akato@po.ntts.co.jp
Masayuki Kanda NTT
神田正辅
Phone: +81-422-59-3456
Fax: +81-422-59-4015
EMail: kanda.masayuki@lab.ntt.co.jp
Phone: +81-422-59-3456
Fax: +81-422-59-4015
EMail: kanda.masayuki@lab.ntt.co.jp
Satoru Kanno NTT Software Corporation
Satoru Kanno NTT软件公司
Phone: +81-45-212-7577
Fax: +81-45-212-9800
EMail: kanno-s@po.ntts.co.jp
Phone: +81-45-212-7577
Fax: +81-45-212-9800
EMail: kanno-s@po.ntts.co.jp