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| Section | Objectives |
|---|---|
| Hashing and Digital Signatures | - Digital Signature Standards - Hash Functions (MD5, SHA-1, SHA-256) - Message Authentication Codes (MAC) |
| Applied Cryptography | - VPN Security - Cryptographic Best Practices - SSL/TLS Protocols - PGP and Email Encryption |
| Asymmetric Cryptography | - RSA Algorithm - Public Key Infrastructure (PKI) - Diffie-Hellman Key Exchange - Elliptic Curve Cryptography (ECC) |
| Cryptography Fundamentals | - History and Evolution of Cryptography - Cryptographic Terminology - Symmetric vs Asymmetric Encryption |
| Symmetric Cryptography | - Stream Ciphers - Initialization Vectors (IV) - Key Management - Block Ciphers (AES, DES, 3DES) |
| Cryptanalysis and Attacks | - Social Engineering Prevention - Brute Force and Dictionary Attacks - Common Attack Vectors |
>> Introduction-to-Cryptography Fragen Beantworten <<
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31. Frage
(Which authentication method allows a customer to authenticate to a web service?)
Antwort: B
Begründung:
One-way client authentication is the method where the client (customer) proves its identity to the server (web service). In cryptographic terms, this is commonly implemented through client credentials such as client TLS certificates (mTLS from the server's perspective) or through authentication protocols layered over TLS (for example, signed tokens), but the defining direction is that the client is the party being authenticated. In a strict TLS certificate-authentication framing, client authentication occurs when the server requests a client certificate during the handshake and the client demonstrates possession of the corresponding private key (via signature in handshake messages). The server then validates the client certificate chain and authorization policy. One-way server authentication, by contrast, authenticates only the server to the client and does not identify the customer. Mutual authentication authenticates both sides simultaneously; while it includes client authentication, it is broader than what the question asks. "End-to-end authentication" describes assurance between endpoints across intermediaries, but it is not the specific "customer authenticates to service" method in certificate-based terminology. Therefore, the best answer is one-way client authentication.
32. Frage
(How does Electronic Codebook (ECB) mode encryption function?)
Antwort: D
Begründung:
ECB is the simplest block cipher mode: each plaintext block is encrypted independently using the same key and the block cipher primitive. There is no IV and no chaining, so identical plaintext blocks produce identical ciphertext blocks. This property leaks patterns and structure in the plaintext, which is why ECB is generally considered insecure for most real-world data beyond tiny, random-looking inputs. For example, images encrypted with ECB often reveal outlines because repeated pixel blocks map to repeated ciphertext blocks.
Option A describes CTR mode, option C describes CBC mode, and option B resembles feedback-based modes. ECB's independence also means it can be parallelized, but the pattern leakage is a severe weakness.
Modern practice prefers authenticated encryption modes (like GCM) or, at minimum, modes with IVs and chaining (like CBC with proper padding and MAC). Therefore, the correct statement is that ECB encrypts each block with the same key and each block is independent of the others.
33. Frage
(Which is an example of asymmetric encryption?)
Antwort: C
Begründung:
Asymmetric cryptography uses a public/private key pair where different keys are used for related operations (encryption/decryption or signature/verification). Elliptic-Curve Cryptography (ECC) is a family of asymmetric algorithms built on the mathematics of elliptic curves over finite fields. ECC supports key exchange (ECDH), digital signatures (ECDSA/EdDSA), and other primitives with smaller key sizes for comparable security to traditional discrete-log or RSA systems (e.g., a 256-bit ECC key is often comparable in security to a 3072-bit RSA key, depending on scheme and parameters). By contrast, SHA-256 is a cryptographic hash function (one-way digest), and HMAC is a keyed integrity/authentication construction built from a hash function-neither is encryption. DES is a symmetric block cipher (same key for encryption and decryption). Therefore, the example of asymmetric encryption among the options is ECC.
34. Frage
(What is an alternative to using a Certificate Revocation List (CRL) with certificates?)
Antwort: B
Begründung:
OCSP is the primary online alternative to CRLs for checking whether a certificate has been revoked.
With a CRL, a relying party periodically downloads a list of revoked certificate serial numbers published by the issuing CA (or CRL distribution point). That approach can be bandwidth-heavy, introduces latency between revocation and client awareness, and can result in clients using stale revocation data if updates are infrequent. OCSP improves this by allowing a client (or a server on the client's behalf) to query an OCSP responder in near real time about the status of a specific certificate (good, revoked, or unknown). In practice, many TLS deployments use OCSP stapling, where the server periodically fetches a signed OCSP response from the CA's responder and "staples" it to the TLS handshake, reducing client-side network calls and improving privacy (the CA doesn't learn which site the client is visiting). Thus, OCSP provides a more timely, certificate-specific revocation status mechanism than CRLs while preserving the CA's signed assurance.
35. Frage
(What is the value of 51 mod 11?)
Antwort: B
Begründung:
The value 51 mod 11 is the remainder after dividing 51 by 11. Modular arithmetic is widely used in cryptography to keep computations within a finite set of residues, such as in RSA where values are taken modulo n, or in Diffie-Hellman where exponents and group elements are reduced modulo a prime. To compute 51 mod 11, find the largest multiple of 11 less than or equal to 51. Multiples of 11 are 11, 22, 33, 44,
55. The closest without exceeding 51 is 44. Subtracting gives 51 # 44 = 7, so the remainder is 7. Therefore, 51 mod 11 = 7, matching option "07." This remainder is always in the range 0 through 10 because the modulus is
11. Such residue computations underpin the "wraparound" behavior that makes modular exponentiation and inverse computations well-defined in cryptographic groups.
36. Frage
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