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Certificate authority 27<br />

References<br />

[1] https:/ / spreadsheets. google. com/ pub?key=ttwCVzDVuWzZYaDosdU6e3w& single=true& gid=0& output=html, List of Trusted Root<br />

Certificate Authorities, 2/10/2010.<br />

[2] http:/ / www. defcon. org/ images/ defcon-17/ dc-17-presentations/ defcon-17-zusman-hacking_pki. pdf<br />

[3] Verisign, Inc. (31 January 2001). "Jan 2001 - Advisory <strong>from</strong> VeriSign, Inc." (http:/ / www. verisign. com/ support/ advisories/<br />

authenticodefraud. html). . Retrieved 2008-12-02.<br />

[4] Microsoft, Inc. (February 21, 2007). "Microsoft <strong>Security</strong> Bulletin MS01-017: Erroneous VeriSign-Issued Digital Certificates Pose Spoofing<br />

Hazard" (http:/ / support. microsoft. com/ kb/ 293818). . Retrieved 2011-Nov-09.<br />

[5] Bright, Peter (28 March 2011). "Independent Iranian hacker claims responsibility for Comodo hack" (http:/ / arstechnica. com/ security/ news/<br />

2011/ 03/ independent-iranian-hacker-claims-responsibility-for-comodo-hack. ars). Ars Technica. . Retrieved 1 September 2011.<br />

[6] Bright, Peter (30 August 2011). "Another fraudulent certificate raises the same old questions about certificate authorities" (http:/ / arstechnica.<br />

com/ security/ news/ 2011/ 08/ earlier-this-year-an-iranian. ars). Ars Technica. . Retrieved 1 September 2011.<br />

[7] Fraudulent DigiNotar Certificate Usage (http:/ / www. theregister. co. uk/ 2011/ 09/ 06/ diginotar_audit_damning_fail/ ) Retrieved 7<br />

September 2011.<br />

[8] http:/ / www. abanet. org/ rppt/ meetings_cle/ 2002/ 2002spring/ RealProperty/ Thursday/ TechnologyandtheRealEstate/<br />

OnlineTransactionManagement. pdf<br />

[9] http:/ / www. webtrust. org/<br />

[10] http:/ / news. netcraft. com/ ssl-sample-report/ CMatch/ certs<br />

[11] http:/ / pki. fedoraproject. org/ wiki/ PKI_Main_Page<br />

[12] http:/ / xca. hohnstaedt. de<br />

External links<br />

• Certificate authorities (http:/ / www. dmoz. org/ Computers/ <strong>Security</strong>/ Public_Key_Infrastructure/ PKIX/<br />

Tools_and_Services/ Third_Party_Certificate_Authorities/ / ) at the Open Directory Project<br />

• Certificate Authority Reviews (http:/ / www. sslshopper. com/ certificate-authority-reviews. html)<br />

• Certificate Authorities by Country (http:/ / www. tractis. com/ countries)<br />

CMAC<br />

In cryptography, CMAC (Cipher-based MAC) SP800-38B is a block cipher-based message authentication code<br />

algorithm. It may be used to provide assurance of the authenticity and, hence, the integrity of binary data. This mode<br />

of operation fixes security deficiencies of CBC-MAC (CBC-MAC is secure only for fixed-length messages).<br />

The core of the CMAC algorithm is a variation of CBC-MAC that Black and Rogaway proposed and analyzed under<br />

the name XCBC BR2 and submitted to NIST. BR1 The XCBC algorithm efficiently addresses the security deficiencies<br />

of CBC-MAC, but requires three keys. Iwata and Kurosawa proposed an improvement of XCBC and named the<br />

resulting algorithm One-Key CBC-MAC (OMAC) in their papers. IK2IK1 They later submitted OMAC1 IK3 , a<br />

refinement of OMAC, and additional security analysis. IK4 The OMAC algorithm reduces the amount of key material<br />

required for XCBC. CMAC is equivalent to OMAC1.<br />

To generate an ℓ-bit CMAC tag (t) of a message (m) using a b-bit block cipher (E) and a secret key (k), one first<br />

generates two b-bit sub-keys (k 1 and k 2 ) using the following algorithm (this is equivalent to multiplication by x and<br />

x 2 in a finite field GF(2 b )). Let ≪ signify a standard left-shift operator:<br />

1. Calculate a temporary value k 0 = E k (0).<br />

2. If msb(k 0 ) = 0, then k 1 = k 0 ≪ 1, else k 1 = (k 0 ≪ 1) ⊕ C; where C is a certain constant that depends only on b.<br />

(Specifically, C is the non-leading coefficients of the lexicographically first irreducible degree-b binary<br />

polynomial with the minimal number of ones.)<br />

3. If msb(k 1 ) = 0, then k 2 = k 1 ≪ 1, else k 2 = (k 1 ≪ 1) ⊕ C.<br />

As a small example, suppose b = 4, C = 0011 2 , and k 0 = E k (0) = 0101 2 . Then k 1 = 1010 2 and k 2 = 0100 ⊕ 0011 =<br />

0111 2 .

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