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| By Hanibani1823

Hanibani1823

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Quizzes Created: 1 | Total Attempts: 709

| Attempts: 709

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1/123 Pitanja

OTP šifra može da se koristi
- Samo jednom
- Samo dva puta
- Najvise 5 puta
- Nema pravila vezanog za koriscenje

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2.

Prilikom kreiranja korisnickih naloga, na disk se smesta par korisnicko ime-hes lozinka?
- Da
- Ne
Correct Answer

A. Da

Explanation

During the creation of user accounts, the username-password pair is stored on the disk.

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3.

Vrhovni CA je potpisan od strane
- Samog sebe
- Njemu ravnopravnog CA
- Države
- Ničeg od navedenog
Correct Answer

A. Samog sebe

Explanation

The correct answer is "Samog sebe". This means that the Supreme CA (Certification Authority) signs itself, indicating that it is its own authority and does not require validation or authorization from any other entity. This suggests that the Supreme CA has the highest level of trust and autonomy in issuing and managing digital certificates.

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4.

Koji je od sledecih formata najpogodniji za LSB supstituciju
- JPEG
- TIG
- GIF
- BMP
Correct Answer

A. BMP

Explanation

The BMP format is the most suitable for LSB (Least Significant Bit) substitution. This is because BMP files use a simple and uncompressed format, which makes it easier to manipulate individual bits without affecting the overall image quality. JPEG, TIG, and GIF formats, on the other hand, use compression techniques that can cause loss of data and introduce artifacts when modifying the least significant bits. Therefore, BMP is the preferred format for LSB substitution.

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5.

Bezuslovno sigurna sifra
- је OTP шифра
- не постоји
- је ECB
- је кодна књига
Correct Answer

A. је OTP шифра

Explanation

The correct answer is "је OTP шифра". This is because OTP (One-Time Password) is a type of encryption where a unique password is generated for each authentication attempt. It provides a high level of security as the password is only valid for a single use or a short period of time. Therefore, OTP encryption is unconditionally secure and considered one of the most secure methods of encryption available.

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6.

Jednosmerna heš funkcija generise otisak duzine 160 bitova.Koliko mogucih ulaznih poruka generise heš vrednosti 0000....0001?
- 160
- Beskonacno mnogo
- Dve
- Samo jedna
Correct Answer

A. Beskonacno mnogo

Explanation

The given correct answer is "beskonacno mnogo". A hash function generates a fixed-length output regardless of the input size. In this case, the hash function generates a 160-bit output. Since the input message can vary in length, there are infinitely many possible input messages that can produce the same 160-bit hash value. Therefore, the hash function can generate an infinite number of input messages that result in the hash value "0000....0001".

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7.

Sta se kod steganografije krije?
- Cinjenica da se prenosi poruka
- Komunikacija
- Steganografski kanal
- Steganografski medijum
Correct Answer

A. Cinjenica da se prenosi poruka

Explanation

The correct answer is "cinjenica da se prenosi poruka" which translates to "the fact that a message is being transmitted" in English. In steganography, the main purpose is to hide the existence of a message within a carrier medium. The answer suggests that what is being concealed in steganography is the fact that a message is being transmitted, rather than the actual content of the message itself. This highlights the covert nature of steganography, where the goal is to hide the communication rather than encrypting or protecting the content of the message.

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8.

Koje operacije se koriste kod OTP sifre
- Samo supstitucije
- Samo trenaspozicije
- Samo XOR
- XOR i supstitucije
Correct Answer

A. Samo XOR

Explanation

The correct answer is "samo XOR" because OTP (One-Time Pad) encryption method uses only the XOR (exclusive OR) operation. In OTP, each character of the plaintext is combined with a character from a random key using XOR, resulting in the ciphertext. XOR is a bitwise operation that produces a 1 if the corresponding bits of the operands are different, and 0 if they are the same. This operation ensures that the encryption is secure, as long as the key is truly random and used only once.

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9.

Tehnikom image downgrading-a
- Moguce je BMP sakriti u BMP
- Moguce je JPEG sakriti u JPEG
- Moguce je JPEG sakriti u BMP
- Moguce je BMP sakriti u JPEG
Correct Answer

A. Moguce je BMP sakriti u BMP

Explanation

The given answer states that it is possible to hide a BMP image within another BMP image. This means that one BMP image can be embedded or concealed within another BMP image without altering the file format. This technique is known as image downgrading, where one image is hidden within another to hide information or to achieve a certain effect.

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10.

Jednosmerna heš funkcija generiše otisak dužine 160 bitova. Koliko mogućih ulaznih poruka generiše heš vrednost 1?
- Beskonacno mnogo poruka
- Samo jedna
- Ni jedna
- Dve
Correct Answer

A. Beskonacno mnogo poruka

Explanation

The given question asks how many possible input messages can generate a hash value of 1 using a one-way hash function that produces a 160-bit output. The correct answer is "beskonacno mnogo poruka" which means "infinitely many messages" in English. This is because hash functions are designed to have a high collision resistance, meaning that it is extremely unlikely for two different input messages to produce the same hash value. Therefore, there are an infinite number of possible input messages that can generate a hash value of 1.

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11.

Koliko parametara ima funkcija HMAC
- (M,K)
- (M,K,h(M))
- (M,K,h(M),h(K))
- (M)
Correct Answer

A. (M,K)

Explanation

The correct answer is (M,K) because HMAC (Hash-based Message Authentication Code) is a cryptographic function that takes two parameters: the message (M) and the secret key (K). The function uses these parameters to generate a unique hash value that can be used to verify the integrity and authenticity of the message. The additional parameters in the other options are not necessary for the HMAC function.

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12.

Dodela prava autentifikovanom korisniku naziva se

Correct Answer
autorizacija

Explanation
The given answer is "autorizacija". This term refers to the process of granting rights or permissions to an authenticated user. It involves verifying the identity of the user and determining what actions or resources they are allowed to access. This process ensures that only authorized individuals can perform certain actions or access specific information within a system or application.

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13.

Duzina blokova kod AES algoritma moze da bude
- 64,128 ili 192
- 128,192 ili 256
- 256,512 ili 1024
- Proizvoljne duzine
Correct Answer

A. 128,192 ili 256

Explanation

The AES algorithm allows for block sizes of 128, 192, or 256 bits. These block sizes determine the amount of data that can be processed at once in the encryption or decryption process. The larger the block size, the more secure the algorithm can be, as it can handle larger amounts of data. However, larger block sizes also require more processing power and can slow down the encryption or decryption process. Therefore, the AES algorithm provides options for different block sizes to accommodate different security and performance needs.

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14.

Kod OTP šifre
- Dužina ključa mora biti jednaka dužini poruke i upotrebljava se samo jednom
- Dužina ključa je nebitna jer se skraćuje/produžava na potrebnu dužinu i upotrebljava se samo jednom
- Dužina ključa je nebitna jer se skraćuje/produžava na potrebnu dužinu i upotrebljava se samo dva puta
- Dužina ključa je jednaka dužini poruke i može da se upotrebljava nekoliko puta
Correct Answer

A. Dužina ključa mora biti jednaka dužini poruke i upotrebljava se samo jednom

Explanation

The correct answer is "Dužina ključa mora biti jednaka dužini poruke i upotrebljava se samo jednom." This means that the length of the key must be equal to the length of the message and it can only be used once. This ensures the security and confidentiality of the message by using a key that is unique to that specific message and cannot be reused.

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15.

Trostruki DES :
- Prvo šifruje sa prvim ključem, pa dešifruje sa drugim, pa šifruje sa prvim
- Prvo šifruje sa prvim ključem, pa šifruje sa drugim, pa šifruje sa trećim
- Prvo šifruje sa prvim ključem, pa šifruje sa drugim, pa dešifruje sa trećim
- Prvo dešifruje sa prvim ključem, pa šifruje sa drugim, pa dešifruje sa trećim
Correct Answer

A. Prvo šifruje sa prvim ključem, pa dešifruje sa drugim, pa šifruje sa prvim

Explanation

The correct answer is the option that states "Prvo šifruje sa prvim ključem, pa dešifruje sa drugim, pa šifruje sa prvim." This means that the encryption process starts by encrypting with the first key, then decrypting with the second key, and finally encrypting again with the first key. This sequence of encryption and decryption steps is known as triple DES (Data Encryption Standard) and provides enhanced security by applying multiple layers of encryption.

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16.

Koji algoritam ne pripada navedenoj grupi
- AES
- SHA-3
- MD5
- SHA-0
Correct Answer

A. AES

Explanation

The correct answer is AES. AES (Advanced Encryption Standard) is a symmetric encryption algorithm, while the other options (SHA-3, MD5, and SHA-0) are all cryptographic hash functions.

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17.

Ukoliko se koristi engleski alfabet prosiren ciframa za kljuc duzine 18 koliki je prostor kljuceva?
- 26 * 10 * 18
- 18 ^ (26 + 10)
- (26 + 10) ^ 18
- (26 * 10) ^ 18
Correct Answer

A. (26 + 10) ^ 18

Explanation

The question is asking for the size of the key space when using an extended English alphabet with digits for a key length of 18. The correct answer is (26 + 10) ^ 18. This is because there are 26 letters in the English alphabet and 10 digits, so the total number of characters is 26 + 10 = 36. To calculate the size of the key space, we raise this number to the power of the key length, which is 18. Therefore, the size of the key space is (26 + 10) ^ 18.

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18.

OTP je bezuslovno sigurna šifra
- Uz određene uslove
- Dokazano
- Bezuslovno sigurne šifre nema
- Pretpostavljano
Correct Answer

A. Dokazano

Explanation

The given correct answer is "Dokazano". This suggests that it has been proven or demonstrated that OTP (One-Time Password) is unconditionally secure. OTP is a type of password that is valid for only one login session or transaction, on a computer system or other digital device. It is considered highly secure because it is generated for each use and cannot be reused or intercepted by attackers. The use of OTP eliminates the risk of password theft or unauthorized access, making it a reliable and secure method of authentication.

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19.

Koja dva algoritama su po nacinu funkcionisanja slicni
- Afina sifra i Viznerova sifra
- One Time Pad i Cezarova sifra
- Jednostavni XOR i One Tme Pad
- Afina sifra i jednostavni XOR
Correct Answer

A. Jednostavni XOR i One Tme Pad

Explanation

The correct answer is "jednostavni XOR i One Time Pad." Both algorithms use the XOR operation as a fundamental component in their encryption process. XOR is a bitwise operation that combines two binary numbers and produces a result based on their differences. In both jednostavni XOR and One Time Pad, XOR is used to combine the plaintext with a randomly generated key to create the ciphertext. This makes them similar in terms of their functioning.

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20.

Frekvencijska analiza nema efekta na
- OTP
- Vizenerovu
- Hilovu
- Plejferovu
Correct Answer

A. OTP

Explanation

OTP stands for One-Time Pad, which is a cryptographic technique that uses a random key that is as long as the message being encrypted. This key is used only once and then discarded, making it impossible for anyone to decrypt the message without the key. Frequency analysis is a method used to break encryption by analyzing the frequency of letters or patterns in the encrypted message. However, since OTP uses a random key and each key is used only once, frequency analysis is ineffective against OTP encryption.

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21.

DES algoritam ima
- 16 rundi
- 14 rundi
- 10 rundi
- 12 rundi
Correct Answer

A. 16 rundi

Explanation

The correct answer is 16 rundi. The DES algorithm, which stands for Data Encryption Standard, consists of 16 rounds of encryption. Each round involves several operations, including permutation, substitution, and XOR operations, to transform the input data into the final encrypted output. These 16 rounds ensure a high level of security and make it difficult for unauthorized individuals to decrypt the encrypted data without the proper key.

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22.

Servis koji obezbedjuje proveru identiteta naziva se

Correct Answer
autentifikacija

Explanation
The correct answer is "autentifikacija" because it refers to the process of verifying the identity of a user or system. This service ensures that the claimed identity is valid and authorized, providing a secure way to access resources or information.

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23.

Jednosmerne funkcije
- Nije dokazano ni da postoje ni da ne postoje
- Dokazano je da postoje
- Dokazano je da ne postoje
- Za svaku informaciju pojedinacno se dokazuje da li je jednosmerna ili ne
Correct Answer

A. Nije dokazano ni da postoje ni da ne postoje

Explanation

The correct answer is "nije dokazano ni da postoje ni da ne postoje." This means that it has not been proven whether one-way functions exist or not.

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24.

Slucajna izmena sadrzaja poruke tokom prenosa
- Spada u narusavanje integriteta poruke
- Se detektuje drugacijim mehanizmom u odnosu na mehanizam detekcije namerne izmene
- Se sprecava ugradjom mehanizma za oporavak podataka
- Se ignorise
Correct Answer

A. Spada u narusavanje integriteta poruke

Explanation

The correct answer is "spada u narusavanje integriteta poruke" because it states that accidental modification of the message content during transmission is a violation of message integrity. This means that the message has been altered unintentionally, which can lead to the loss or corruption of data. The other options are not applicable because they either refer to intentional modifications, different detection mechanisms, or ignoring the issue altogether.

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25.

Sekvencijalni algoritmi se koriste
- Tamo gde je bitna brzina i rad u realnom vremenu
- Kao deo vecih sifrarskih sistema
- Ne koriste se vise
- Tamo gde je bitno da sifrat bude prenet bez greske
Correct Answer

A. Tamo gde je bitna brzina i rad u realnom vremenu

Explanation

Sequential algorithms are used in situations where speed and real-time processing are important, such as in larger cryptographic systems. They are not used in situations where it is important for the cipher to be transmitted without errors.

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26.

Alisa zeli da posalje Bobu poruku m sifrovanu RSA algoritmom. Alisa sifruje poruku:
- Bobovim javnim kljucem
- Svojim javnim kljucem
- Svojim privatnim kljucem
- Bobovim privatnim kljucem
Correct Answer

A. Bobovim javnim kljucem

Explanation

Alisa wants to send a message to Bob using the RSA algorithm. The RSA algorithm involves the use of public and private keys. In this case, Alisa encrypts the message using Bob's public key. This is the correct answer because when using RSA encryption, the sender encrypts the message with the recipient's public key, ensuring that only the recipient with the corresponding private key can decrypt and read the message.

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27.

Ukoliko se koristi engleski alfabet prosiren ciframa i specijalnim znacima { +, -, *, /, =, ?, !, $ } za kljuc duzine 6 koliki je prostor kljuceva?
- (26 * 10 * 8) ^ 6
- 6 ^ (26 + 10 + 12)
- (26 + 10 + 12) ^ 6
- 26 * 10 * 12 * 6
Correct Answer

A. (26 + 10 + 12) ^ 6

Explanation

The question asks for the size of the key space when using the English alphabet, extended digits, and special characters as the key. The correct answer is (26 + 10 + 12) ^ 6, which means that there are 48 possible characters (26 letters + 10 digits + 12 special characters) and the key length is 6. Thus, the key space is calculated by raising the number of possible characters to the power of the key length, resulting in a large number of possible keys.

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28.

Kod OTP sifre, duzina kljuca:
- Je iste duzine kao otvoreni tekst
- Zavisi od algoritma koji se koristi za generisanje kljuca
- Je iste ili vece duzine ali se odredjenim algoritmom skracuje na potrebnu duzinu.
- је фиксне дужине за дати алгоритам, па се из њега генерише кључ потребне величине
Correct Answer

A. Je iste duzine kao otvoreni tekst

Explanation

The correct answer states that the length of the OTP (One-Time Password) key is the same as the length of the plaintext. This means that for every character in the plaintext, there is a corresponding character in the OTP key, making them equal in length. This is important because it ensures that the OTP key is long enough to provide sufficient security and randomness for encryption purposes.

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29.

Kod blokovskih algoritama, blok sifrata se dobija visestrukom primenom odredjene funkcije na blok otvorenog teksta i medjurezultate. Jedna primena funkcije naziva se

Correct Answer
runda

Explanation
In the context of block cipher algorithms, a block cipher is obtained by repeatedly applying a specific function to a block of plaintext and intermediate results. Each application of the function is referred to as a "round".

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30.

Sigurnist DES algoritmima pociva na
- S-blokovima
- Velicini kljuca
- Broju rundi
- Generisanju podkljuca
Correct Answer

A. S-blokovima

Explanation

The security of DES algorithms is based on the use of s-boxes. S-boxes are substitution boxes that perform a nonlinear transformation on the input data, making it more resistant to cryptographic attacks. The use of s-boxes adds an extra layer of complexity and confusion to the encryption process, making it harder for an attacker to analyze and break the encryption. Therefore, the correct answer is "s-blokovima".

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31.

Prisluskivanje je napad na

Correct Answer
poverljivost

Explanation
The correct answer is "poverljivost" which translates to "confidentiality" in English. This suggests that "prisluskivanje" refers to eavesdropping or wiretapping, which is considered an attack on confidentiality. Eavesdropping involves listening in on private conversations or intercepting communication without the knowledge or consent of the parties involved. This violation of privacy breaches the confidentiality of the information being exchanged, making it the target of such an attack.

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32.

Postupak kojim se otvoreni tekst zapisuje u obliku binarnog niza naziva se

Correct Answer
kodovanje

Explanation
The correct answer is "kodovanje". This is because the question is asking for the process of converting an open text into a binary sequence, which is called "kodovanje" in Serbian.

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33.

Ukoliko se koristi engleski alfabet za kljuc duzine 26 koliki je prostor kljuceva ako slova mogu da se ponavljaju?
- 26 ^ 26
- 26 * 26
- 26!
- 62! * 5!
Correct Answer

A. 26 ^ 26

Explanation

The correct answer is 26 ^ 26. This is because there are 26 letters in the English alphabet and each letter can be used as a key. Since the letters can be repeated, there are 26 possibilities for each position in the key, resulting in a total of 26 raised to the power of 26 possible keys.

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34.

Ako je x = 01101, y = 01010 i z = 00111, koliko je (x XOR x XOR x XOR y XOR y XOR z XOR z XOR x XOR x XOR y XOR z) ?
- 00100
- 00010
- 00001
- 00000
Correct Answer

A. 00000

Explanation

The XOR operation returns 1 if the two bits being compared are different, and 0 if they are the same. In this case, x XOR x will always be 0 because comparing the same bits will always result in 0. Therefore, the entire expression (x XOR x XOR x XOR y XOR y XOR z XOR z XOR x XOR x XOR y XOR z) simplifies to y XOR z. Since y = 01010 and z = 00111, the XOR operation between them will give us 00001. Therefore, the answer is 00001.

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35.

Aktivni napad koji onemogucava funkcionisanje sistema ili pruzanje neke usluge je napad na
- Raspolozivost
- Autenticnost
- Integritet
- Poverljivost
Correct Answer

A. Raspolozivost

Explanation

The correct answer is "raspolozivost" which translates to "availability" in English. This means that the active attack that prevents the functioning of a system or the provision of a service is an attack on the availability of the system. In other words, the attacker aims to disrupt or deny access to the system or service, making it unavailable to users.

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36.

Neka je Z skup znakova nad kojim se definise kljuc. Ukoliko je kljuc duzine 5 i Z={0,1,2,3,4...a,b,c...A,B,C...}, tada je velicina prostora kljuceva ukoliko slova u kljucu ne mogu da se ponavljaju
- (62*61*60*59*58)^5
- 62*61*60*59*58
- 62! / 5!
- 62! * 5!
Correct Answer

A. 62*61*60*59*58

Explanation

The question asks for the size of the key space when the key length is 5 and the characters in the key cannot repeat. To calculate this, we need to find the number of possible choices for each character in the key.

Since there are 62 characters in the set Z, the first character of the key can be chosen in 62 ways. After choosing the first character, there are 61 characters left for the second character. Similarly, there are 60 choices for the third character, 59 choices for the fourth character, and 58 choices for the fifth character.

To find the total number of possible keys, we multiply these choices together, resulting in 62*61*60*59*58. Therefore, the correct answer is 62*61*60*59*58.

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37.

KOji od sledecih algoritama moze da se koristi za digitalno potpisivanje
- RSA
- DES
- AES
- Diffie-Hellman
Correct Answer

A. RSA

Explanation

RSA is the correct answer because it is an algorithm that can be used for digital signing. RSA is a widely used asymmetric encryption algorithm that involves the use of a public key for encryption and a private key for decryption. It can also be used for digital signatures, where the sender uses their private key to sign a message, and the receiver can verify the authenticity of the message using the sender's public key. DES and AES are symmetric encryption algorithms, while Diffie-Hellman is a key exchange algorithm, none of which are specifically designed for digital signing.

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38.

Za ECB i CBC rezime rada vazi:
- U ECB rezimu isti blok istim kljucem uvek daje isti sifrat, dok u CBC to ne mora da bude slucaj
- I u CBC i u ECB rezimu isti blok istim kljucem uvek daje razlicit sifrat
- U CBC rezimu isti blok istim kljucem uvek daje isti sifrat, dok u ECB to ne mora da bude slucaj
- I u CBC i u ECB rezimu isti blok istim kljucem uvek daje isti sifrat
Correct Answer

A. U ECB rezimu isti blok istim kljucem uvek daje isti sifrat, dok u CBC to ne mora da bude slucaj

Explanation

In ECB mode, the same block with the same key will always produce the same ciphertext. However, in CBC mode, this may not be the case. In CBC mode, each block of plaintext is XORed with the previous ciphertext block before encryption, which introduces randomness and makes the ciphertext different even if the same block and key are used. This makes CBC mode more secure than ECB mode, as it prevents patterns in the plaintext from being preserved in the ciphertext.

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39.

Generator pseudoslucajnih vrednosti na osnovu kratkog slucajnog kljuca generise

Correct Answer
radni kljuc

Explanation
The generator pseudoslucajnih vrednosti (pseudo-random value generator) uses a short random key to generate a working key. This working key is likely to be used in various cryptographic algorithms or systems. The generator takes the short random key as input and applies a pseudo-random algorithm to generate the working key. This process allows for the generation of a secure and unpredictable working key, which is essential for ensuring the confidentiality and integrity of data in cryptographic operations.

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40.

Duzina otiska (hes vrednosti ) MD5 algoritma je
- 128bita
- 512bita
- 1024bita
- 64bita
Correct Answer

A. 128bita

Explanation

The correct answer is 128bita. The MD5 algorithm generates a 128-bit hash value, which is commonly represented as a string of 32 hexadecimal characters. This hash value is used for various purposes such as data integrity checks and password storage.

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41.

Za siforvanje i digitalno potpisivanje koristi se
- Razliciti par kljuceva
- Isti par kljuceva
- Isti ili razliciti par kljuceva u zavisnosti od dogovora
- Isti li razliciti par kljuceva u zavisnosti od uslova
Correct Answer

A. Razliciti par kljuceva

Explanation

The correct answer is "razliciti par kljuceva" which means "different pairs of keys" in English. This means that for encryption and digital signing, different pairs of keys are used. This is because encryption and digital signing require separate keys for different purposes. Using the same pair of keys for both encryption and digital signing could compromise the security of the system. Therefore, it is necessary to use different pairs of keys for these operations.

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42.

Hes funkcija se ne koristi za sifrovanje poruka jer je
- Nesigurna
- Nemoguce je izvrsiti desifrovanje
- Sifrovanje je veoma sporo
- Dolazi do narusavanja integriteta poruke
Correct Answer

A. Nemoguce je izvrsiti desifrovanje

Explanation

The correct answer is "nemoguce je izvrsiti desifrovanje" which means "it is impossible to decrypt". This suggests that the function is not used for encryption because it cannot be reversed or decrypted.

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43.

Duzina kljuca koji je potreban za izracunavanje HMAC vrednosti je
- Manji od 64 bita
- Duzine poruke
- Veci od 64 bita
- Bar 128 bita
Correct Answer

A. Manji od 64 bita

Explanation

The correct answer is "Manji od 64 bita". HMAC (Hash-based Message Authentication Code) is a cryptographic hash function that requires a key for calculating the HMAC value. In this case, the length of the key needed for HMAC calculation is smaller than 64 bits.

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44.

Neporecivost je servis koji prijemnoj strani pruza neoboriv dokaz da:
- Je poruku primio od tacno odredjene osobe
- Je poruka stigla nepromenjena
- Tokom prenosa poruke nije doslo do narusivanja poverljivosti
- Je poruka proverena od strane odgovarajuceg sertifikacionog tela
Correct Answer

A. Je poruku primio od tacno odredjene osobe

Explanation

The correct answer states that "Neporecivost je servis koji prijemnoj strani pruza neoboriv dokaz da je poruku primio od tacno odredjene osobe" which means "Non-repudiation is a service that provides the receiving party with irrefutable evidence that the message was received from the exact specified person." This explanation clarifies that non-repudiation ensures that the sender cannot deny sending the message, providing a reliable proof of the sender's identity.

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45.

Simetrični algoritmi:
- Imaju problem razmene ključeva
- Koriste kraći ključ od asimetričnih algoritama
- Sporiji su od asimetričnih
- Koriste se isključivo za digitalno potpisivanje
Correct Answer

A. Imaju problem razmene ključeva

Explanation

Symmetric algorithms have a problem with key exchange. This means that both parties involved in the communication need to have the same key in order to encrypt and decrypt the data. The challenge lies in securely exchanging this key between the parties without it being intercepted by unauthorized entities. Asymmetric algorithms, on the other hand, do not have this problem as they use different keys for encryption and decryption.

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46.

Kolizija kod heš funkcija označava pojavu da
- Dve različite poruke daju istu heš vrednost
- Da postoji poruka za koju ne može da se izračuna heš vrednost
- Ista poruka može da da dve različite heš vrednosti
- Da ako je dužina heš vrednosti N, postoji niz dužine N koji nije heš vrednost ni jedne poruke
Correct Answer

A. Dve različite poruke daju istu heš vrednost

Explanation

The explanation for the given correct answer is that collision in hash functions refers to the occurrence of two different messages producing the same hash value. This means that two distinct inputs can result in the same output hash value, which can lead to potential conflicts or issues in certain applications that rely on unique hash values for data integrity and identification.

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47.

Odrediti red velicine broja operacija sifrovanja i desifrovanja na trostruki DES metodom susret u sredini
- 2^112
- 2^168
- 2^64
- 2^36
Correct Answer

A. 2^112

Explanation

The correct answer is 2^112. This is because the triple DES method uses a 112-bit key for encryption and decryption. The encryption and decryption processes each involve a series of operations, including permutation, substitution, and XOR operations. The size of the key determines the number of possible combinations, which is 2^112 in this case. Therefore, the correct answer represents the size of the key used in the triple DES method.

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48.

Jedna od nepozeljnih osobina generatora pseudoslucajnih brojeva je
- Periodicnost
- To sto ne mogu da se koriste za OTP sifru
- To sto mu je za rad potrebna pocetna vrednost
- To sto se u njihovom radu koriste komplikovane matematicke funkcije
Correct Answer

A. Periodicnost

Explanation

Periodicnost je jedna od nepoželjnih osobina generatora pseudoslučajnih brojeva. Periodicnost se odnosi na ponavljanje istih brojeva u nizu generisanih brojeva. U slučaju generatora pseudoslučajnih brojeva, periodičnost može dovesti do predvidljivosti i smanjiti sigurnost sistema koji se oslanjaju na generisane brojeve. Stoga, periodičnost je neželjena osobina generatora pseudoslučajnih brojeva.

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49.

DES algoritam je
- Blokovski algoritam Fejstelovog tipa
- Iterativni blokovski algoritam
- Algoritam koji se ne koristi za sifrovanje
- Sekvencijalni algoritam
Correct Answer

A. Blokovski algoritam Fejstelovog tipa

Explanation

The given correct answer states that DES algorithm is a "blokovski algoritam Fejstelovog tipa" which translates to "block cipher algorithm of the Feistel type" in English. This means that DES is a type of symmetric encryption algorithm that operates on fixed-size blocks of data using a Feistel network structure. It divides the input into blocks and applies a series of transformations to each block to produce the encrypted output. Therefore, the answer correctly identifies DES as a block cipher algorithm of the Feistel type.

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Quiz Review Timeline (Updated): Mar 21, 2023 +

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Current Version
Mar 21, 2023

Quiz Edited by
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Jan 25, 2017

Quiz Created by
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