Kripto 2

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| By Catherine Halcomb

Catherine Halcomb

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1/75 Pitanja

RSA algoritam se pojavio: (*svuda pise da se pojavio 1977.god, ali ovo su bili ponudjeni odgovori)
- 1978. god
- 1948. god
- 2008. god
- 1998. god
- 1968. god

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2.

Sifarski sistemi mogu biti:
- Jednosmerni
- Simetricni
- Dvosmerni
- Asimetricni
Correct Answer(s)

A. Simetricni
A. Asimetricni

Explanation

The correct answer includes two options: "Simetricni" (Symmetric) and "Asimetricni" (Asymmetric). This suggests that "Sifarski sistemi" (Number systems) can be either symmetric or asymmetric.

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3.

Osnovna podela sifarskih sistema, prema vrsti kljuca deli se na:
- Blokovske sisteme
- Sekvencijalne sisteme
- Simetricne sisteme
- Asimetricne sisteme
Correct Answer(s)

A. Simetricne sisteme
A. Asimetricne sisteme

Explanation

The correct answer is "Simetricne sisteme,Asimetricne sisteme". The question is asking for the basic division of cipher systems according to the type of key. The answer states that the division includes symmetric systems and asymmetric systems. Symmetric systems use the same key for both encryption and decryption, while asymmetric systems use different keys for encryption and decryption.

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4.

DH algoritam za uspostavljanje deljene tajne je osetljiv na:
- Lose izabranu eksponencijalnu funkciju
- Greske na prenosnom putu
- Napad tipa covek u sredini
- Sum na komunikacionom kanalu
Correct Answer

A. Napad tipa covek u sredini

Explanation

The correct answer is "Napad tipa covek u sredini" (Man-in-the-middle attack). The Diffie-Hellman algorithm for establishing a shared secret is vulnerable to this type of attack. In a man-in-the-middle attack, an attacker intercepts the communication between two parties and impersonates each party to the other, allowing them to intercept and modify the messages exchanged. This can compromise the security of the shared secret established through the Diffie-Hellman algorithm.

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5.

Obeliziti tacan rezultat za: 52 (mod 17) = ?
- 1
- 0
- 3
- 2
Correct Answer

A. 1

Explanation

The given expression "52 (mod 17)" represents the remainder when 52 is divided by 17. To find this remainder, we divide 52 by 17 and see what is left over. In this case, 52 divided by 17 is 3 with a remainder of 1. Therefore, the correct answer is 1.

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6.

Lavinski efekat definisemo kao:
- Mala promena na izlazu izaziva veliku promenu na ulazu
- Mala promena na ulazu izazvace veliku promenu na izlazu
- Velika promena na ulazu izazvace malu promenu na izlazu
- Velika promena na ulazu izazvace veliku promenu na izlazu
Correct Answer

A. Mala promena na ulazu izazvace veliku promenu na izlazu

Explanation

The correct answer is "Mala promena na ulazu izazvace veliku promenu na izlazu." This statement describes the Lavinski effect, which is defined as a small change at the input causing a large change at the output. This phenomenon is commonly observed in amplifiers and electronic circuits, where a small input signal can result in a much larger output signal.

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7.

Obeleziti koliko puta je RSA sporiji od AES-a:
- 10 puta
- 100 000 puta
- 100 puta
- 10 000 puta
- 1000 puta
Correct Answer

A. 1000 puta

Explanation

The correct answer is 1000 times. This means that RSA is 1000 times slower than AES. RSA is a public-key encryption algorithm that is primarily used for secure data transmission, while AES is a symmetric encryption algorithm that is commonly used for encrypting data at rest. The slower speed of RSA is due to its more complex mathematical operations, such as modular exponentiation, which require more computational resources and time compared to the simpler operations used in AES.

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8.

Sifrat kod blokovskih sifara se dobije visestrukom primenom funkcije koja se naziva:
- Registar
- Linearni pomeracki registri
- Permutacija
- Iteracija
- Runda
- Klasa
Correct Answer

A. Runda

Explanation

The correct answer is "Runda". In block ciphers, the encryption process is divided into multiple rounds, where each round applies a specific set of operations to the input data. These operations typically include substitution, permutation, and mixing of the data. Therefore, the term "runda" refers to the specific step or iteration within the encryption process of a block cipher.

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9.

Obeleziti: duzine kljuceva kod PKI sistema koje su u upotrebi:
- 1000, 2000, 3000
- 1024, 2048, 4096
- 512, 1024, 1244
- 128, 256, 512
- 112, 196, 256
Correct Answer

A. 1024, 2048, 4096

Explanation

The correct answer is the set of key lengths used in PKI systems: 1024, 2048, 4096. These key lengths are commonly used in PKI systems to ensure secure encryption and authentication. The larger the key length, the stronger the encryption and the more secure the system. Therefore, using longer key lengths such as 1024, 2048, and 4096 provides a higher level of security in PKI systems.

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10.

Ako su poznati sledeci parametri (p, g, a) i Bi informacija od Boba, Alisa izracunava kljuc na sledeci nacin:
- K= Bi^b mod p
- K= g^Bi mod p
- K= Bi^a mod p
- K= Ai^a mod p
Correct Answer

A. K= Bi^a mod p

Explanation

The given answer states that the key (K) is calculated as Bi raised to the power of a, modulo p. This means that Alisa takes the value received from Bob (Bi) and raises it to the power of her own secret value (a), and then takes the remainder when divided by p. This calculation allows Alisa to generate a shared key with Bob using the Diffie-Hellman key exchange algorithm.

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1
11.

Obeleziti tacan rezultat za: (25*13)(mod 6) = ?
- 1
- 3
- 2
- 0
Correct Answer

A. 1

Explanation

The given expression is (25*13)(mod 6). First, we multiply 25 and 13, which gives us 325. Then, we take the modulus of 325 with 6. The modulus operation gives us the remainder when 325 is divided by 6. In this case, 325 divided by 6 gives a remainder of 1. Therefore, the correct answer is 1.

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12.

Double DES (2DES) koristi duzinu kljuca:
- 512 bita
- 112 bita
- 192 bita
- 64 bita
Correct Answer

A. 112 bita

Explanation

Double DES (2DES) uses a key length of 112 bits. Double DES is a variant of the Data Encryption Standard (DES) algorithm that applies DES encryption twice to increase security. In Double DES, two 56-bit keys are used consecutively. Since each key is 56 bits long, the total key length becomes 112 bits. This increases the complexity of the encryption process and makes it more difficult for attackers to crack the encryption.

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13.

RSA algoritam koristi sledece kljuceve:
- Simetricni kljuc
- Javni kljuc
- Slucajni kljuc
- Privatni kljuc
- DH kljuc
Correct Answer(s)

A. Javni kljuc
A. Privatni kljuc

Explanation

RSA algoritam koristi javni ključ za enkripciju podataka i privatni ključ za dekripciju podataka. Javni ključ je dostupan svima i koristi se za enkripciju podataka koje samo vlasnik privatnog ključa može dekriptovati. Privatni ključ se čuva tajno i koristi se za dekripciju podataka koje su enkriptovane javnim ključem. Ova kombinacija javnog i privatnog ključa omogućava sigurnu razmenu podataka između korisnika RSA algoritma.

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14.

Duzina bloka kod AES algoritma moze da bude:
- 128 bita
- 265 bita
- 192 bita
- 56 bita
- 256 bita
- 1024 bita
Correct Answer(s)

A. 128 bita
A. 192 bita
A. 256 bita

Explanation

The length of a block in the AES algorithm can be 128 bits, 192 bits, or 256 bits. This means that the AES algorithm can process data in blocks of these sizes. The choice of block size depends on the specific application and security requirements. A larger block size generally provides stronger security but may also result in slower processing speed.

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15.

Princip konfuzije:
- Kod "brute-force" napada svi kljucevi su jednako verovatni
- Kljucevi su konfuzno rasporedjeni u odnosu na otvoreni tekst
- Ima istorijski znacaj i predstavlja karakteristiku klasicnih sifarskih sistema
- Male promene u otvorenom tekstu uticu na sve bite sifrata
- Govori o apsolutnoj tajnosti One-Time PAd sifara
Correct Answer

A. Kod "brute-force" napada svi kljucevi su jednako verovatni

Explanation

In a brute-force attack, all keys are equally likely to be the correct key. This means that the attacker has to try all possible keys until they find the correct one. This is because the brute-force attack method involves systematically trying every possible combination until the correct key is found. Therefore, there is no bias towards any specific key in a brute-force attack.

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16.

Da bi se sprecio napad na DH algoritam neophodna je:
- Slucajnost
- Sifrovanje
- Autentifikacija
- Integritet
- Poverljivost
Correct Answer

A. Autentifikacija

Explanation

Autentifikacija je neophodna kako bi se sprečio napad na DH algoritam. Autentifikacija je proces provere identiteta korisnika ili uređaja, čime se osigurava da samo legitimni korisnici imaju pristup sistemu. U kontekstu DH algoritma, autentifikacija se koristi kako bi se osiguralo da komunicirajuće strane zaista jesu one koje tvrde da jesu, čime se sprečava prisustvo neovlašćenih korisnika koji bi mogli pokušati da izvrše napad na algoritam.

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17.

Bob odredjuje parametre za DH ptokol prema sledecem: Bi= g^b mod p. Javni parametri su:
- P
- G
- B
Correct Answer(s)

A. P
A. G

Explanation

The given answer consists of the variables p and g. These variables are mentioned in the question as the public parameters for the DH protocol. The variable p represents a prime number, while g represents a generator. These parameters are essential for the DH protocol as they help in generating the shared secret key between two parties.

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18.

3DES sa EDE2 radi sa sledecom notacijom:
- C = EDE(P, K1, K2)
- C = E(D(E(P, K1), K1), K2)
- C = E(E(E(P, K1), K2), K1)
- C = E(D(E(P,K1),K2),K1)
Correct Answer

A. C = E(D(E(P,K1),K2),K1)

Explanation

The correct answer is C = E(D(E(P,K1),K2),K1). This is the correct notation for 3DES sa EDE2. It shows that the plaintext (P) is first encrypted with key K1, then decrypted with key K2, and finally encrypted again with key K1. This triple encryption process provides stronger security compared to single DES encryption.

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19.

Message Autentification Cod (MAC) se moze racunati preko sledeceg blokovskog rezima:
- ECC
- IVC
- CBC
- ECB
- CTR
Correct Answer

A. CBC

Explanation

CBC (Cipher Block Chaining) is a block cipher mode of operation that provides confidentiality and message authentication. In CBC mode, each plaintext block is XORed with the previous ciphertext block before encryption. This ensures that even if two plaintext blocks are identical, the corresponding ciphertext blocks will be different. Additionally, CBC mode uses an Initialization Vector (IV) to provide randomness and prevent patterns in the ciphertext. The IV is XORed with the first plaintext block before encryption. Therefore, CBC mode not only encrypts the message but also provides authentication by ensuring the integrity of the ciphertext.

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20.

Koliko iznosi aditivna inverzija broja 3 (mod 7)
- 1
- 5
- 3
- 4
- 6
- 2
Correct Answer

A. 4

Explanation

The question is asking for the additive inverse of the number 3 (mod 7). In modular arithmetic, the additive inverse of a number is the number that, when added to the original number, gives a sum that is congruent to 0 modulo the given modulus. In this case, the modulus is 7. The additive inverse of 3 (mod 7) is 4, because 3 + 4 = 7, which is congruent to 0 modulo 7.

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21.

Ako se uporede simetricni i asimetricni sifarski sistemi obeleziti sta je tacno:
- Simetricni sifarski sistemi su brzi
- Kljuc simetricnog sistema je duzi u odnosu na kljuc asimetricnog
- Aimetricni sifarski sistemi su brzi
- Kljuc asimetricnog sistema je duzi u odnosu na kljuc kod simetricnog
Correct Answer(s)

A. Simetricni sifarski sistemi su brzi
A. Kljuc asimetricnog sistema je duzi u odnosu na kljuc kod simetricnog

Explanation

The correct answer states that symmetric cipher systems are faster and the key of the asymmetric system is longer compared to the key of the symmetric system. This means that symmetric cipher systems are faster in terms of encryption and decryption processes. Additionally, asymmetric systems require longer keys for encryption and decryption compared to symmetric systems, which adds to the complexity and security of the asymmetric system.

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22.

Duzina kljuca kod AES algoritma moze da bude
- 56 bita
- 512 bita
- 192 bita
- 128 bita
- 112 bita
- 256 bita
Correct Answer(s)

A. 192 bita
A. 128 bita
A. 256 bita

Explanation

The AES algorithm allows for different key lengths to be used, including 192 bits, 128 bits, and 256 bits. These key lengths determine the level of security provided by the algorithm. A longer key length generally provides stronger security, as there are more possible combinations for the key. Therefore, 192 bits, 128 bits, and 256 bits are all valid key lengths for the AES algorithm.

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23.

DH algoritam je osetljiv na napad:
- Cut and paste
- Man in the middle
- Cut and go
- Paste and go
- Man in the paste
Correct Answer

A. Man in the middle

Explanation

The correct answer is "Man in the middle" because the DH algorithm is vulnerable to this type of attack. In a man-in-the-middle attack, an attacker intercepts communication between two parties and impersonates each party to the other, allowing them to intercept and manipulate the data being transmitted. This can compromise the security of the DH algorithm, as the attacker can potentially gain access to the shared secret key and decrypt the communication.

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24.

Funkcije kod AES algortima:
- Sve su linearne
- Nisu invertibilne
- Jesu invertibilne
- Sve su nelinearne
Correct Answer

A. Jesu invertibilne

Explanation

The given correct answer states that the functions in the AES algorithm are invertible. This means that for every output of the function, there exists a unique input that produces that output. In other words, it is possible to reverse the function and obtain the original input from the output. This property is crucial in encryption algorithms like AES, as it allows for the decryption process to retrieve the original message from the encrypted data.

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25.

PKI predstavlja sisteme sa:
- Tajnim kljucevima
- Sesijskim kljucevima
- Javnim i privatnim kljucevima
- Simetricnim kljucevima
Correct Answer

A. Javnim i privatnim kljucevima

Explanation

PKI (Public Key Infrastructure) predstavlja sisteme koji koriste javne i privatne ključeve. Javni ključ se koristi za šifrovanje podataka, dok se privatni ključ koristi za dešifrovanje podataka. Ova kombinacija ključeva omogućava sigurnu razmenu podataka između korisnika, jer samo osoba sa odgovarajućim privatnim ključem može dešifrovati podatke koji su šifrovani javnim ključem. Ovo omogućava autentifikaciju korisnika i bezbednu komunikaciju na internetu.

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26.

Kod RSA algortima u procesu digitalnog potpisivanje privatni kljuc:
- Treba biti poznat obema stranama
- Samo strani koja digitalno potpisuje
- Treba da bude poznat sertifikacionom telu
- Samo strani koja prima poruku
Correct Answer

A. Samo strani koja digitalno potpisuje

Explanation

The correct answer is "Samo strani koja digitalno potpisuje" (Only the party that digitally signs). In the process of digital signing using the RSA algorithm, the private key is only known to the party that is digitally signing the message. This ensures that the digital signature is unique to that party and verifies the authenticity and integrity of the message. The private key should not be known to any other party, including the party receiving the message or the certification authority.

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27.

Kod blokovskih sifri u ECB rezimu je moguc sledeci napad:
- Man in the middle
- Cut and paste
- Inversion method
- Paste and go
- Iterative meaning
Correct Answer

A. Cut and paste

Explanation

Cut and paste is a possible attack on block ciphers in ECB mode. In this attack, the attacker intercepts the ciphertext and rearranges the blocks to create a new ciphertext. By doing so, the attacker can manipulate the plaintext without knowing the encryption key. This attack is possible because ECB mode encrypts each block independently, allowing the attacker to rearrange the blocks without affecting the decryption process. This can lead to serious security vulnerabilities, which is why ECB mode is not recommended for secure encryption.

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28.

Postupak sifrovanja otvorene poruke M kod RSA algoritma je:
- C = e^M mod N
- C = M^e mod N
- M = C^e mod N
- C = E(M^K) mod N
Correct Answer

A. C = M^e mod N

Explanation

The given correct answer states that the encryption process for RSA algorithm is C = M^e mod N. In RSA encryption, the ciphertext (C) is calculated by raising the plaintext (M) to the power of the public exponent (e) and then taking the modulo N. This process ensures that the ciphertext is only decryptable with the corresponding private key.

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29.

Fejstel sifra predstavlja:
- Algoritam za digitalno potpisivanje
- Tip blokovske sifre
- Algoritam za dobijanje hes vrednosti
- Tip apsolutno tajne sifre
- Tip sekvencijalne sifre
Correct Answer

A. Tip blokovske sifre

Explanation

The correct answer is "Tip blokovske sifre" because Fejstel sifra (Feistel cipher) is a type of block cipher. In a Feistel cipher, the plaintext is divided into blocks and each block undergoes a series of transformations using a round function and a key. This type of cipher is commonly used in symmetric encryption algorithms to ensure confidentiality and security of data.

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30.

Trostruki DES sa tri nezavisna kljuca koristi kljuc:
- 64 bita
- 168 bita
- 256 bita
- 128 bita
Correct Answer

A. 168 bita

Explanation

Triple DES with three independent keys uses a key size of 168 bits. This means that each key used in the encryption process is 56 bits long. Triple DES is a symmetric encryption algorithm that applies the Data Encryption Standard (DES) algorithm three times in a row to provide enhanced security. By using three keys, it increases the key size and makes it more difficult for attackers to decrypt the encrypted data.

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31.

Notacija za trostruki DES je:
- C = E(D(E(P,K1),K2),K3)
- C = D(E(D(P,K1),K2),K3)
- P = E(D(E(P,K1),K2),K3)
- C = E(D(E(C,K1),K2),K3)
Correct Answer

A. C = E(D(E(P,K1),K2),K3)

Explanation

The correct answer is C = E(D(E(P,K1),K2),K3). This answer correctly represents the notation for triple DES, where P is the plaintext, K1, K2, and K3 are the three keys used for encryption and decryption, E represents the encryption function, and D represents the decryption function. The notation shows that the plaintext P is first encrypted with K1, then decrypted with K2, and finally encrypted again with K3 to produce the ciphertext C.

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32.

Tajnost Diffie-Hellman (DH) algoritma se zasniva na slozenosti izracunavanja:
- Inverzne visestruke permutacije
- MInimalnog kvadratnog rastojanja
- Faktorizacije velikih prostih brojeva
- Diskretnog logaritma
Correct Answer

A. Diskretnog logaritma

Explanation

The correct answer is "Diskretnog logaritma." The secrecy of the Diffie-Hellman (DH) algorithm is based on the computational complexity of calculating discrete logarithms. In DH, two parties can exchange public keys over an insecure channel and use those keys to establish a shared secret key without directly exchanging the key itself. This is possible because calculating the discrete logarithm in a finite field is computationally difficult, making it hard for an attacker to determine the shared secret key even if they have access to the public keys.

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33.

Broj rundi kod DES algoritma je:
- 64
- 32
- 16
- 48
Correct Answer

A. 16

Explanation

The number of rounds in the DES algorithm is 16.

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34.

Kod RSA algoritma javni kljucevi su:
- N,d
- N,e
- P,q
- D
Correct Answer

A. N,e

Explanation

The correct answer is N,e. In the RSA algorithm, the public key consists of two parts: N and e. N represents the modulus, which is the product of two large prime numbers p and q. e represents the public exponent, which is a smaller prime number that is coprime with (p-1) * (q-1). The public key is used for encryption, while the private key (not mentioned in the question) is used for decryption.

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35.

Obeleziti blokovske sifre:
- IDEA
- AES
- Blowfish
- 3DES
- DES
- RC4
- A5/1
Correct Answer(s)

A. IDEA
A. AES
A. Blowfish
A. 3DES
A. DES

Explanation

The given answer includes a list of block ciphers, which are encryption algorithms that operate on fixed-size blocks of data. The ciphers mentioned in the answer, IDEA, AES, Blowfish, 3DES, and DES, are all well-known and widely used block ciphers in the field of cryptography. These ciphers provide varying levels of security and have been extensively studied and analyzed for their strength against various attacks. The inclusion of these ciphers in the answer suggests that they are considered reliable and effective for securing data.

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36.

Kod digitalnog potpisivanja koristimo: (*dvosmisleno pitanje: moze biti i samo privatni kljuc ako si misli iskljucivo na potpisivanje, a ako se misli na ceo proces digitalnog potpisivanja onda je kombincija oba kljuca)
- Tajni kljuc
- Kombinaciju oba kljuca
- Privatni kljuc
- Nista od ponudjenog
- Simetricni kljuc
- Hash vrednost
- Javni kljuc
Correct Answer

A. Privatni kljuc

Explanation

We use the private key in digital signature. The private key is used to generate a unique digital signature for a document or message. This signature can be verified using the corresponding public key, ensuring the authenticity and integrity of the data. The private key is kept secret and known only to the owner, while the public key is shared with others for verification purposes. Therefore, the private key is the correct answer for this question.

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37.

Kod blokovskih sifri, sifrovanje se primenjuje nad:
- Porukom izdeljenom na blokove fiksne duzine i dobijaju se blokovi sifrata razlicite duzine u zavisnosti od parametara
- Nizom bita, bajtova ili racunarskih reci, a rezultat je blok podataka fiksne duzine
- Porukom izdeljenom na blokove fiksne duzine i dobijaju se blokovi sifrata iste duzine koa i ulazni blokovi poruke
- Blokovima koji su duzine cele poruke
- Blokovima promenljive duzine gde kljuc odredjuje duzinu bloka
Correct Answer

A. Porukom izdeljenom na blokove fiksne duzine i dobijaju se blokovi sifrata iste duzine koa i ulazni blokovi poruke

Explanation

In block ciphers, encryption is applied to a message divided into fixed-length blocks, and the resulting ciphertext blocks have the same length as the input message blocks. This means that each block of the plaintext message is encrypted into a corresponding block of the ciphertext, maintaining the same length.

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38.

Princip difuzije:
- Odnosi se na difuznu karakteristiku sifrovanja sa slucajnim nizom
- Svaki bit sifrata je samo funkcija svih bitova teksta
- Svaki bit sifrata je funkcija svih bitova teksta i svih bitova kljuca
- Ima samo istorijski znacaj i predtavlja karakteristiku klasicnih sifarskih sistema
- Svaki bit sifrata je samo funkcija svih bitova kljuca
Correct Answer

A. Svaki bit sifrata je funkcija svih bitova teksta i svih bitova kljuca

Explanation

The principle of diffusion refers to the characteristic of encryption with a random sequence. In this principle, each bit of the ciphertext is a function of all the bits of the plaintext and all the bits of the key. This principle is historically significant and represents a characteristic of classical cipher systems. Therefore, the correct answer is that each bit of the ciphertext is a function of all the bits of the plaintext and all the bits of the key.

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39.

Kada Alisa i Bob primene DH algoritam oni su:
- Uspostavili su slucajni niz koji moze da se koristi za "one-time pad"
- Uspostavili tajnu vrednost koja moze da se koristi kao simetricni kljuc
- Uspostavili su jedinstvenu kombinaciju: Privatni kljuc za Alisu i javni za Boba
- Uspostavili su jedinstvenu kombinaciju: Javni kljuc za Alisu i privatni kljuc za Boba
Correct Answer

A. Uspostavili tajnu vrednost koja moze da se koristi kao simetricni kljuc

Explanation

When Alice and Bob apply the DH algorithm, they establish a secret value that can be used as a symmetric key. The DH algorithm allows two parties to agree upon a shared secret value without directly transmitting it over an insecure channel. This secret value can then be used as a symmetric key for encryption and decryption purposes, ensuring secure communication between Alice and Bob.

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40.

Za racunanje MAC koda za proveru integriteta poruke preko blokovske sifre:
- Koristi se CBC rezim rada
- MAC predstavlja prvi blok sifrata
- Koristi se EBC rezim rada
- MAC predstavlja poslednji blok sifrata
Correct Answer(s)

A. Koristi se CBC rezim rada
A. MAC predstavlja poslednji blok sifrata

Explanation

The correct answer is that for calculating the MAC code for message integrity verification using block cipher, the CBC mode of operation is used and the MAC represents the last block of the cipher. This means that the message is divided into blocks and each block is encrypted using the CBC mode, where the ciphertext of each block is dependent on the previous block. The MAC code, which is used to verify the integrity of the message, is then derived from the last block of the ciphertext.

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41.

Digitalni sertifikat moze da sadrzi: - proveriti
- 3. Ime i prezime vlasnika
- 4. Slucajni niz za sifrovanje
- 5. Source cod algoritma za PKI
- 1. Javni kljuc vlasnika
- 2. Tajni kljuc vlasnika
Correct Answer(s)

A. 3. Ime i prezime vlasnika
A. 1. Javni kljuc vlasnika

Explanation

A digital certificate can contain the name and surname of the owner as well as their public key. The name and surname of the owner are important pieces of information that can be used to identify the owner of the certificate. The public key is necessary for encrypting and verifying digital signatures. Therefore, both the name and surname of the owner and the public key are commonly included in a digital certificate.

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42.

DES algoritam je podlozan (slab) na:
- Diferencijalnu analizu
- Linearnu analizu
- Potpunu pretragu kljuceva
- Man-in-the-middle napad
Correct Answer

A. Potpunu pretragu kljuceva

Explanation

The correct answer is "Potpunu pretragu kljuceva" (Brute-force attack). The DES algorithm is vulnerable to a brute-force attack, where an attacker systematically tries all possible keys until the correct one is found. This is because DES has a relatively small key size of 56 bits, making it computationally feasible to try all possible keys within a reasonable amount of time.

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43.

U hibridnom sifarskom sistemu, obeleziti sta su prihvatljivi parametri (parametri koji odgovaraju).
- Tajni kljuc duzine 2048 bita, hes vrednost duzine 256 bita i privatni i javni kljuc duzine 512 bita
- Tajni kljuc duzine 256 bita, hes vrednost duzine 512 bita i privatni i javni kljuc duzine 1024 bita
- Tajni kljuc duzine 256 bita, hes vrednost duzine 256 bita i privatni i javni kljuc duzine 2048 bita
- Tajni kljuc duzine 256 bita, hes vrednost duzine 256 bita i privatni i javni kljuc duzine 256 bita
- Tajni kljuc duzine 256 bita, hes vrednost duzine 512 bita i privatni i javni kljuc duzine 2048 bita
- Tajni kljuc duzine 256 bita, hes vrednost duzine 256 bita i privatni i javni kljuc duzine 1024 bita
Correct Answer

A. Tajni kljuc duzine 256 bita, hes vrednost duzine 256 bita i privatni i javni kljuc duzine 2048 bita

Explanation

In a hybrid cipher system, the acceptable parameters are a secret key of length 256 bits, a hash value of length 256 bits, and a private and public key of length 2048 bits. This combination ensures a strong level of security for the encryption and decryption process.

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44.

Primalac poruke koja je digitalno potpisana proverava integritet poruke tako sto:
- Koristi privatni kljuc posiljaoca
- Nista od ponudjenog
- Koristi svoj privatni kljuc
- Koristi javni kljuc posiljaoca
Correct Answer

A. Koristi javni kljuc posiljaoca

Explanation

The correct answer is "Koristi javni kljuc posiljaoca." When a message is digitally signed, the sender uses their private key to create a unique signature for the message. The recipient can then use the sender's public key to verify the integrity of the message by comparing the signature with the message itself. By using the sender's public key, the recipient can ensure that the message has not been tampered with during transmission.

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45.

Obeleziti sta je tacno za blokovske sifre:
- Greska od 1 bit u prenosu utice na 1 bit OT kod desifrovanja
- Greska u prenosu se ne nagomilava dalje od 2 bloka
- Greska u prenosu propagira do kraja sifrata
- Greska od 1 bit u prenosu utice na dva bloka OT kod desifrovanja
Correct Answer(s)

A. Greska u prenosu se ne nagomilava dalje od 2 bloka
A. Greska od 1 bit u prenosu utice na dva bloka OT kod desifrovanja

Explanation

The correct answer is that an error in transmission does not accumulate beyond 2 blocks, and an error of 1 bit in transmission affects two blocks in the OT decoding process. This means that if there is an error in the transmission of data, it will only affect a maximum of 2 blocks and will not propagate further. Additionally, if there is a single bit error in transmission, it will impact the decoding of two blocks in the OT process.

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46.

Obeleziti regularne blokove DES algoritma:
- P boks
- S boks
- K boks
- E boks
- D boks
- Q boks
Correct Answer(s)

A. P boks
A. S boks
A. K boks
A. E boks

Explanation

The correct answer is the order in which the blocks are listed: P box, S box, K box, E box. This is the correct order of the regular blocks in the DES algorithm. The P box is responsible for permuting the input bits, the S box is responsible for substituting groups of bits, the K box is responsible for generating the round keys, and the E box is responsible for expanding the input bits.

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47.

3DES sa EDE2 koristi kljuc duzine
- 112 bita
- 56 bita
- 96 bita
- 168 bita
Correct Answer

A. 112 bita

Explanation

3DES sa EDE2 koristi ključ dužine 112 bita. 3DES (Triple Data Encryption Standard) je kriptografski algoritam koji koristi blokovsku šifru DES tri puta za povećanu sigurnost. EDE2 (Encrypt-Decrypt-Encrypt 2) je način implementacije 3DES algoritma koji koristi dva koraka šifriranja i jedan korak dešifriranja. Ključ dužine 112 bita se sastoji od tri ključa od po 56 bita, što ukupno čini 168 bita. Međutim, samo prvih 112 bita se koristi za šifriranje i dešifriranje podataka.

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48.

Duzina kljuca kod DES algoritma je:
- 64 bita
- 128 bita
- 256 bita
- 56 bita
- 58 bita
Correct Answer

A. 56 bita

Explanation

The length of the key in the DES algorithm is 56 bits. DES (Data Encryption Standard) is a symmetric encryption algorithm that uses a 56-bit key to encrypt and decrypt data. This key length was chosen to strike a balance between security and efficiency. However, due to advances in computing power, a 56-bit key is now considered relatively weak and vulnerable to brute-force attacks. As a result, DES is no longer widely used and has been replaced by more secure encryption algorithms with longer key lengths.

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49.

Sifrovanjme otvorenog teksta obezbedjujemo:
- Poverljivost poruke
- Integritet poruke
- Autentifikaciju
- Neporecivost
- Raspolozivost servisa
Correct Answer

A. Poverljivost poruke

Explanation

The correct answer is "Poverljivost poruke" (Confidentiality of the message). This means that encrypting the plaintext ensures that the message is kept confidential and cannot be accessed or understood by unauthorized individuals.

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Mar 22, 2023

Quiz Edited by
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May 11, 2018

Quiz Created by
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Kripto 2

RSA algoritam se pojavio: (*svuda pise da se pojavio 1977.god, ali ovo su bili ponudjeni odgovori)

Quiz Preview

Sifarski sistemi mogu biti:

Osnovna podela sifarskih sistema, prema vrsti kljuca deli se na:

DH algoritam za uspostavljanje deljene tajne je osetljiv na:

Obeliziti tacan rezultat za: 52 (mod 17) = ?

Lavinski efekat definisemo kao:

Obeleziti koliko puta je RSA sporiji od AES-a:

Sifrat kod blokovskih sifara se dobije visestrukom primenom funkcije koja se naziva:

Obeleziti: duzine kljuceva kod PKI sistema koje su u upotrebi:

Ako su poznati sledeci parametri (p, g, a) i Bi informacija od Boba, Alisa izracunava kljuc na sledeci nacin:

Obeleziti tacan rezultat za: (25*13)(mod 6) = ?

Double DES (2DES) koristi duzinu kljuca:

RSA algoritam koristi sledece kljuceve:

Duzina bloka kod AES algoritma moze da bude:

Princip konfuzije:

Da bi se sprecio napad na DH algoritam neophodna je:

Bob odredjuje parametre za DH ptokol prema sledecem: Bi= g^b mod p. Javni parametri su:

3DES sa EDE2 radi sa sledecom notacijom:

Message Autentification Cod (MAC) se moze racunati preko sledeceg blokovskog rezima:

Koliko iznosi aditivna inverzija broja 3 (mod 7)

Ako se uporede simetricni i asimetricni sifarski sistemi obeleziti sta je tacno:

Duzina kljuca kod AES algoritma moze da bude

DH algoritam je osetljiv na napad:

Funkcije kod AES algortima:

PKI predstavlja sisteme sa:

Kod RSA algortima u procesu digitalnog potpisivanje privatni kljuc:

Kod blokovskih sifri u ECB rezimu je moguc sledeci napad:

Postupak sifrovanja otvorene poruke M kod RSA algoritma je:

Fejstel sifra predstavlja:

Trostruki DES sa tri nezavisna kljuca koristi kljuc:

Notacija za trostruki DES je:

Tajnost Diffie-Hellman (DH) algoritma se zasniva na slozenosti izracunavanja:

Broj rundi kod DES algoritma je:

Kod RSA algoritma javni kljucevi su:

Obeleziti blokovske sifre:

Kod digitalnog potpisivanja koristimo: (*dvosmisleno pitanje: moze biti i samo privatni kljuc ako si misli iskljucivo na potpisivanje, a ako se misli na ceo proces digitalnog potpisivanja onda je kombincija oba kljuca)

Kod blokovskih sifri, sifrovanje se primenjuje nad:

Princip difuzije:

Kada Alisa i Bob primene DH algoritam oni su:

Za racunanje MAC koda za proveru integriteta poruke preko blokovske sifre:

Digitalni sertifikat moze da sadrzi: - proveriti

DES algoritam je podlozan (slab) na:

U hibridnom sifarskom sistemu, obeleziti sta su prihvatljivi parametri (parametri koji odgovaraju).

Primalac poruke koja je digitalno potpisana proverava integritet poruke tako sto:

Obeleziti sta je tacno za blokovske sifre:

Obeleziti regularne blokove DES algoritma:

3DES sa EDE2 koristi kljuc duzine

Duzina kljuca kod DES algoritma je:

Sifrovanjme otvorenog teksta obezbedjujemo: