Sis - Prvi Kolokvijum

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| By Mladjalug96
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Mladjalug96
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Quizzes Created: 2 | Total Attempts: 15,161
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  • 1/247 Pitanja

    Ključ je dužine 64 bita. Koliki je prostor ključeva?

    • 2^32
    • 2^64-2
    • 2^128
    • 2^64
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  • 2. 

    Ključ je dužine 56 bita. Koliki je prostor ključeva?

    • 2^56

    • 2^56-2

    • 2^112

    • 2^28

    Correct Answer
    A. 2^56
    Explanation
    The given answer, 2^56, is the correct explanation. The question states that the key is 56 bits long, and the question asks for the size of the key space. In cryptography, the key space refers to the total number of possible keys that can be generated. Since each bit can have two possible values (0 or 1), the total number of possible keys can be calculated by raising 2 to the power of the number of bits in the key. Therefore, the key space in this case is 2^56.

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  • 3. 

    Ako je heš dužine 100 bitova koliko treba napraviti poruka da bismo sigurno došlo do kolizije?

    • 2^50

    • 2^100+1

    • 2^99

    • 2^99+1

    Correct Answer
    A. 2^100+1
    Explanation
    To ensure that we definitely reach a collision, we need to create a message that is one bit longer than the hash length. In this case, the hash length is 100 bits, so we need to create a message that is 101 bits long. Therefore, the correct answer is 2^100+1.

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  • 4. 

    OTP шифра може да се користи:

    • нема правила везаног за број коришћења

    • само два пута

    • највише пет пута

    • само једном

    Correct Answer
    A. само једном
    Explanation
    The correct answer is "само једном" which translates to "only once" in English. This means that the OTP cipher can only be used one time.

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  • 5. 

    Ključ je dužine 256 bita. Koliki je prostor ključeva?

    • 2^128

    • 2^256

    • 2^512

    • 2^256-2

    Correct Answer
    A. 2^256
    Explanation
    The given question asks for the size of the key space for a key that is 256 bits long. The key space refers to the total number of possible keys that can be generated with a given key length. In this case, since the key is 256 bits long, the key space would be 2 raised to the power of 256, which is represented as 2^256. Therefore, the correct answer is 2^256.

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  • 6. 

    Neka je Z={A,B,C,...,X,Y,Z,0,1,..,9} skup nad kojim se definiše ključ. Ukoliko je ključ dužine 8 i znaci u ključu mogu da se ponavljaju, koliki je prostor ključeva:

    • (26+10)^8

    • (26*10)^8

    • 8^(26+10)

    • 26 x 10 x 8

    Correct Answer
    A. (26+10)^8
    Explanation
    The correct answer is (26+10)^8. This is because the set Z contains 26 letters (A-Z) and 10 numbers (0-9), making a total of 36 possible characters. Since the key length is 8 and the characters can be repeated, we use the formula (n^r) to calculate the total number of possible keys. In this case, n is 36 (the number of possible characters) and r is 8 (the key length), so the expression becomes (36^8), which is equivalent to (26+10)^8.

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  • 7. 

    Neka je Z={A,B,C,...,X,Y,Z,0,1,..,9} skup nad kojim se definiše ključ. Ukoliko je ključ dužine 6 i znaci u ključu mogu da se ponavljaju, koliki je prostor ključeva:

    • 26 x 10 x 6

    • 6^(26+10)

    • (26+10)^6

    • (26 x 10)^6

    Correct Answer
    A. (26+10)^6
    Explanation
    The question asks for the size of the key space given that the key is 6 characters long and the characters can repeat. The key space is determined by the number of possible choices for each character in the key. In this case, there are 26 letters in the English alphabet (A-Z) and 10 digits (0-9), making a total of 36 possible characters. Since the key can have repeated characters, the total number of possible keys is calculated by raising the total number of characters (26+10) to the power of the key length (6), resulting in (26+10)^6.

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  • 8. 

    Šifarski sistem OTP može da se koristi:

    • Dok se ne posumnja u sigurnost ključa

    • Samo jednom

    • Više od dva puta

    • Najviše dva puta

    Correct Answer
    A. Samo jednom
    Explanation
    The correct answer is "samo jednom" which means "only once" in English. This suggests that the one-time password (OTP) system can only be used for a single authentication or transaction. Once it has been used, it cannot be used again. This is a security measure to prevent unauthorized access or fraudulent activities.

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  • 9. 

    Ako je heš dužine 224 bitova koliko je potrebno napraviti poruka da bismo sigurno došli do kolizije:

    • 2^224+1

    • 2^224

    • 2^112+1

    • 2^225

    Correct Answer
    A. 2^224+1
    Explanation
    The correct answer is 2^224+1. This is because if the hash is 224 bits long, then the number of possible hash values is 2^224. To ensure that we have a collision, we need to create a message that generates a hash value that is different from all the other possible hash values. Adding 1 to 2^224 ensures that the message will generate a unique hash value and guarantees a collision.

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  • 10. 

    Ako je heš dužine 384 bitova koliko je potrebno napraviti poruka da bismo sigurno došli do kolizije:

    • 2^385

    • 2^384+1

    • 2^384

    • 2^192+1

    Correct Answer
    A. 2^384+1
    Explanation
    The correct answer is 2^384+1. This is because if the hash is 384 bits long, it means there are 2^384 possible hash values. In order to have a collision, we need to find two different messages that produce the same hash value. The birthday paradox tells us that the probability of a collision increases significantly once we have about √(2^384) = 2^192 possible hash values. Therefore, we need to add 1 to 2^384 to ensure that we have enough possible hash values to guarantee a collision.

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  • 11. 

    Neka je Z={A,B,C,...,X,Y,Z,0,1,..,9,+, ,* , /, =, ?, !, $, #, @, &, _} skup nad kojim se definiše ključ. Ukoliko je ključ dužine 6 i znaci u ključu mogu da se ponavljaju, koliki je prostor ključeva:

    • (26 x 10 x 12)^6

    • 26 x 10 x 12 x 6

    • 6^(26+10+12)

    • (26 + 10 + 12)^6

    Correct Answer
    A. (26 + 10 + 12)^6
    Explanation
    The question is asking for the size of the key space given a set of characters. The set includes 26 letters, 10 digits, and 12 special characters. The key is of length 6 and the characters in the key can be repeated. The correct answer calculates the size of the key space as (26 + 10 + 12)^6, which means that there are 48 possible characters in each position of the key and a total of 6 positions. Therefore, the total number of possible keys is (26 + 10 + 12)^6.

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  • 12. 

    Neka je Z={A,B,C,...,X,Y,Z,0,1,..,9,+, ,* , /, =, ?, !, $, #, @, &, _} skup nad kojim se definiše ključ. Ukoliko je ključ dužine 24 i znaci u ključu mogu da se ponavljaju, koliki je prostor ključeva:

    • (26 x 10 x 12)^24

    • 24^(26+10+12)

    • (26 + 10 + 12)^24

    • 26 x 10 x 12 x 24

    Correct Answer
    A. (26 + 10 + 12)^24
    Explanation
    The question is asking for the size of the key space given that the key is 24 characters long and the characters can be chosen from the set Z. The set Z contains 26 letters, 10 digits, and 12 special characters. The expression (26 + 10 + 12)^24 represents the total number of possible combinations for a 24-character key, where each character can be chosen from the set Z. Therefore, the answer is (26 + 10 + 12)^24.

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  • 13. 

    Ključ je dužine 128 bita. Koliki je prostor ključeva?

    • 2^256

    • 2^64

    • 2^128

    • 2^128-2

    Correct Answer
    A. 2^128
    Explanation
    The given question asks for the size of the key space for a key that is 128 bits long. The key space refers to the total number of possible keys that can be generated with a given key length. In this case, since the key is 128 bits long, the key space would be equal to 2 raised to the power of 128, which is 2^128. Therefore, the correct answer is 2^128.

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  • 14. 

    Neka je Z={A,B,C,...,X,Y,Z,0,1,..,9} skup nad kojim se definiše ključ. Ukoliko je ključ dužine 18 i znaci u ključu mogu da se ponavljaju, koliki je prostor ključeva:

    • (26 x 10)^18

    • 26 x 10 x 18

    • 18^(26+10)

    • (26+10)^18

    Correct Answer
    A. (26+10)^18
    Explanation
    The given question states that the set Z consists of 26 letters (A-Z) and 10 numbers (0-9). The key is defined to be of length 18 and can have repeated characters. The question asks for the size of the key space.

    The correct answer, (26+10)^18, calculates the total number of possible combinations for the key. Since there are 26 letters and 10 numbers to choose from for each position of the key, we raise the sum of 26 and 10 to the power of 18 to obtain the total number of possible keys.

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  • 15. 

    Ako je heš dužine 256 bitova koliko je potrebno napraviti poruka da bismo sigurno došli do kolizije:

    • 2^256+1

    • 2^256

    • 2^128

    • 2^(256+1)

    Correct Answer
    A. 2^256+1
    Explanation
    The correct answer is 2^256+1. This is because the question is asking how many messages need to be generated in order to ensure a collision with a hash of length 256 bits. The formula for the number of possible messages is 2^n, where n is the length of the hash in bits. Adding 1 to 2^256 ensures that at least one collision will occur.

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  • 16. 

    Реч ГОРА је шифрована различитим шифрама. Који од понуђених шифрата је добијен коришћењем шифре транспозиције:

    • АГРО

    • РАГА

    • ГАРИ

    • РОДА

    Correct Answer
    A. АГРО
    Explanation
    The word "АГРО" can be obtained by reordering the letters in the word "ГОРА". This is a characteristic of a transposition cipher, where the letters are rearranged to form a new word. The other options do not have the same letter arrangement as "АГРО", so they are not obtained using a transposition cipher.

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  • 17. 

    Ako je heš dužine 128 bitova koliko je potrebno napraviti poruka da bismo sigurno došli do kolizije:

    • 2^128

    • 2^128+1

    • 2^129

    • 2^64+1

    Correct Answer
    A. 2^128+1
    Explanation
    To ensure that we definitely reach a collision, we would need to create 2^128+1 messages. This is because a hash function with a hash length of 128 bits can produce a total of 2^128 unique hash values. Therefore, if we create 2^128+1 messages, we are guaranteed to have at least two messages that produce the same hash value, resulting in a collision.

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  • 18. 

    Ako je heš dužine 512 bitova koliko je potrebno napraviti poruka da bismo sigurno došli do kolizije:

    • 2^256+1

    • 2^(512+1)

    • 2^512+1

    • 2^512

    Correct Answer
    A. 2^512+1
    Explanation
    The correct answer is 2^512+1. This is because if the hash is 512 bits long, then there are 2^512 possible hash values. In order to have a collision, we would need to generate 2^512+1 different messages, as the +1 ensures that there is at least one duplicate hash value among the messages.

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  • 19. 

    Који је од следећих формата најпогоднији за LSB супституцију:

    • Png

    • Tif

    • Bmp

    • Jpeg

    Correct Answer
    A. Bmp
    Explanation
    BMP format is the most suitable for LSB (Least Significant Bit) substitution because it is a lossless image format that preserves all the pixel data. This means that when LSB substitution is performed, there will be no loss of information or quality in the image. Other formats like PNG, TIF, and JPEG use compression algorithms that may result in loss of data, making them less suitable for LSB substitution.

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  • 20. 

    Neka je Z={A,B,C,...,X,Y,Z,0,1,..,9,+, ,* , /, =, ?, !, $, #, @, &, _} skup nad kojim se definiše ključ. Ukoliko je ključ dužine 8 i znaci u ključu mogu da se ponavljaju, koliki je prostor ključeva:

    • 8^(26+10+12)

    • (26+10+12)^8

    • 26 x 10 x 12 x 8

    • (26 x 10 x 12)^8

    Correct Answer
    A. (26+10+12)^8
    Explanation
    The question asks for the size of the key space, given that the key is 8 characters long and can include the characters A-Z, 0-9, +, *, /, =, ?, !, $, #, @, &, and _. The key space is determined by the number of possible characters that can be used in each position of the key. In this case, there are 26 letters, 10 numbers, and 12 special characters, making a total of 48 possible characters. Since the key is 8 characters long, the size of the key space is calculated by raising the number of possible characters to the power of the key length: (26+10+12)^8.

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  • 21. 

    Ako je heš dužine 50 bitova koliko je potrebno napraviti poruka da bismo sigurno došli do kolizije:

    • 2^50

    • 2^50+1

    • 2^25+1

    • 2^25

    Correct Answer
    A. 2^50+1
    Explanation
    The correct answer is 2^50+1. This is because in order to ensure that we have a collision, we need to have at least one more message than the total number of possible hashes. In this case, the hash is 50 bits long, so there are 2^50 possible hashes. Therefore, we need 2^50+1 messages to guarantee a collision.

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  • 22. 

    Техника којом се у дигитални садржај утискују додатне информације као што су подаци о аутору, власништву, лиценцама и слично назива се:

    • LSB супституција

    • водени печат

    • Imagedowngrading

    • DCT метода

    Correct Answer
    A. водени печат
    Explanation
    The correct answer is "водени печат" (watermarking). Watermarking is a technique used to embed additional information, such as authorship, ownership, licenses, etc., into digital content. It is commonly used to protect the intellectual property rights of the content and to provide proof of ownership. Watermarks are typically invisible or difficult to remove, making it a reliable method for adding extra information to digital content.

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  • 23. 

    Речник често коришћених лозинки:

    • не постоји

    • постоји

    • у изради је

    • постоји идеја да се направи

    Correct Answer
    A. постоји
    Explanation
    The correct answer is "постоји" because it means "exists" in Serbian. The question is asking about the existence of a frequently used password dictionary, and the correct answer states that it does exist.

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  • 24. 

    Лавински ефекат је такво својство алгоритма (функције) да:

    • мале промене улаза изазивају мале промене излаза

    • мале промене улаза изазивају велике промене излаза

    • да промене улаза прате промене излаза

    • да промене на улазу немају никакав утицај на излаз

    Correct Answer
    A. мале промене улаза изазивају велике промене излаза
    Explanation
    The correct answer states that the avalanche effect is a property of an algorithm or function where small changes in the input result in large changes in the output. This means that even a slight modification in the input can cause a significant difference in the output.

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  • 25. 

    Код OTP шифре:

    • дужина кључа је небитна јер се скраћује/продужава на потребну дужину и употребљава се само два пута

    • дужина кључа мора бити једнака дужини поруке и употребљава се само једном

    • дужина кључа је једнака дужини поруке и може да се употребљава неколико пута

    • дужина кључа је небитна јер се скраћује/продужава на потребну дужину и употребљава се само једном

    Correct Answer
    A. дужина кључа мора бити једнака дужини поруке и употребљава се само једном
    Explanation
    The correct answer is that the length of the key must be equal to the length of the message and is only used once. This means that in the OTP cipher, each character of the key is used to encrypt or decrypt a corresponding character in the message, and the key is not reused for any other part of the encryption process. This ensures that the encryption is secure and that each character in the message is uniquely encrypted.

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  • 26. 

    Прислушкивање је напад на:

    • Расположивост

    • Интегритет

    • Аутентификацију

    • Поверљивост

    Correct Answer
    A. Поверљивост
    Explanation
    Listening is an attack on confidentiality.

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  • 27. 

    Безусловно сигурна шифра:

    • је кодна књига

    • је ECB

    • не постоји

    • је OTP шифра

    Correct Answer
    A. је OTP шифра
    Explanation
    OTP (One-Time Pad) is a type of encryption that is considered to be unbreakable if used correctly. It involves the use of a random key that is as long as the message being encrypted and is never reused. Each bit of the key is used only once, hence the name "one-time pad." This makes it impossible for an attacker to decrypt the message without access to the key. Therefore, OTP is a highly secure encryption method.

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  • 28. 

    Једносмерна хеш функција генерише отисак дужине 160 бита. Колико различитих порука генерише хеш облика 0000....0001:

    • 160

    • бесконачно много

    • само једна

    • две

    Correct Answer
    A. бесконачно много
    Explanation
    A one-way hash function generates a unique hash for each input message. Since the hash in this case is 160 bits long, and there are infinite combinations of 160-bit strings, the hash function can generate an infinite number of different messages. Therefore, the correct answer is "бесконачно много" (infinite).

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  • 29. 

    OTP je безусловно сигурна шифра:

    • безусловно сигурне шифре нема

    • уз одређене услове

    • претпостављано

    • доказано

    Correct Answer
    A. доказано
    Explanation
    The correct answer is "доказано" (proven). This means that OTP (One-Time Pad) is a cipher that has been proven to be secure. This implies that there is evidence or mathematical proof that demonstrates the security of OTP. It is important to note that OTP is considered to be one of the most secure encryption methods if implemented correctly, as it provides perfect secrecy.

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  • 30. 

    Предност шифровања хеш вредности асиметричним алгоритмом у односу на шифровање симетричним алгоритмом је у томе што:

    • нема потребе за "интерним" поверењем

    • може да се провери и поверљивост

    • нема потребе за употребом лиценцираног софтвера

    • је поступак аутентификације бржи

    Correct Answer
    A. нема потребе за "интерним" поверењем
    Explanation
    The advantage of encrypting hash values with an asymmetric algorithm compared to encrypting with a symmetric algorithm is that there is no need for "internal" trust. This means that the encryption can be done without relying on any internal systems or processes, making it more secure and less susceptible to potential vulnerabilities or breaches. Additionally, encrypting hash values with an asymmetric algorithm allows for the verification of confidentiality, ensuring that the data remains private and protected. There is also no requirement for the use of licensed software, making it more cost-effective and accessible. Lastly, the authentication process is faster when using an asymmetric algorithm.

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  • 31. 

    Kontrola pristupa se sastoji od:

    • Autentifikacije i raspoloživosti

    • Autentifikacije i autorizacije

    • Autorizacije i poverljivosti

    • Poverljivosti i raspoloživosti

    Correct Answer
    A. Autentifikacije i autorizacije
    Explanation
    The correct answer is "autentifikacije i autorizacije" (authentication and authorization). Access control involves two main components: authentication, which verifies the identity of a user or entity, and authorization, which determines the actions or resources that the authenticated user or entity is allowed to access. These two processes work together to ensure that only authorized individuals or entities can access specific resources or perform certain actions within a system.

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  • 32. 

    Криптосистеми који користе исти кључ за шифровање и дешифровање називају се:

    • симетрични крипто системи

    • блоковски крипто системи

    • секвенцијални крипто системи

    • тајни крипто системи

    Correct Answer
    A. симетрични крипто системи
    Explanation
    Symmetric crypto systems use the same key for both encryption and decryption. This means that the sender and receiver both have access to the same key, allowing them to encrypt and decrypt messages. This type of system is commonly used for secure communication and data protection.

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  • 33. 

    Frekvencijska analiza nema efekta na:

    • Hilovu šifru

    • Viženerovu šifru

    • Plejferovu šifru

    • OTP

    Correct Answer
    A. OTP
    Explanation
    One-Time Pad (OTP) is a type of encryption that uses a random key that is as long as the plaintext message. This key is used only once and is never reused. The security of OTP relies on the fact that the key is truly random and is kept secret. Therefore, frequency analysis, which involves analyzing the frequency of letters or patterns in the ciphertext to deduce the key or plaintext, is not effective against OTP. This is because the ciphertext produced by OTP is completely random and does not contain any patterns that can be exploited through frequency analysis.

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  • 34. 

    Код OTP шифре, дужина кључа:

    • зависи од алгоритма који се користи за генерисање кључа

    • је исте или веће дужине али се одређеним алгоритмом скраћује на потребну дужину

    • је исте дужине као отворени текст

    • је фиксне дужине за дати алгоритам, па се из њега генерише кључ потребне величине

    Correct Answer
    A. је исте дужине као отворени текст
    Explanation
    The correct answer is "је исте дужине као отворени текст" which means "is the same length as the plaintext." This means that the length of the OTP key is equal to the length of the original message that is being encrypted.

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  • 35. 

    Предност шифровања хеш вредности (приликом аутентификације и провере интегритета) асиметричним алгоритмом у односу на шифровање симетричним алгоритмом је то што:

    • нема потребе за "интерним" поверењем

    • што је поступак аутентификације бржи

    • може да се провери и поверљивост

    • нема потребе за употребом лиценцираног софтвера

    Correct Answer
    A. нема потребе за "интерним" поверењем
    Explanation
    The advantage of using asymmetric algorithm for hashing values (during authentication and integrity check) compared to using symmetric algorithm is that there is no need for "internal" trust. This means that the authentication process is faster, the confidentiality can be verified, and there is no need to use licensed software.

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  • 36. 

    Код LSB супституције колико битова максимално може да се промени а да то остане непримећено?

    • До 40%

    • До 50%

    • До 20%

    • До 30%

    Correct Answer
    A. До 50%
    Explanation
    The correct answer is "До 50%". This means that up to 50% of the bits can be changed in a LSB substitution code without being noticed. This suggests that LSB substitution is a relatively weak form of steganography, as it can only hide a limited amount of information before it becomes detectable.

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  • 37. 

    Случајна измена садржаја поруке током преноса:

    • се детектује другачијим механизмима у односу на механизме детекције намерне измене

    • се спречава уградњом механизма за опоравак података

    • се игнорише

    • спада у нарушавање интегритета поруке

    Correct Answer
    A. спада у нарушавање интегритета поруке
    Explanation
    Random changes in message content during transmission are considered a violation of message integrity. Unlike intentional changes, which can be detected using different mechanisms, random changes pose a challenge as they can go undetected. To prevent such changes, a mechanism for data recovery needs to be implemented. Ignoring random changes would compromise the integrity of the message. Therefore, the correct answer is that random changes in message content during transmission are considered a violation of message integrity.

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  • 38. 

    Шта се код стеганографије крије ?

    • комуникација

    • чињеница да се преноси порука

    • стеганографски медијум

    • стеганографски канал

    Correct Answer
    A. чињеница да се преноси порука
    Explanation
    The correct answer is "чињеница да се преноси порука" (the fact that a message is being transmitted). This answer correctly identifies the hidden information in steganography, which is the act of concealing a message within another medium. Steganography involves hiding the existence of the message itself, rather than encrypting or encoding its content.

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  • 39. 

    Кључ код Хилове шифре је матрица која:

    • не мора да буде инвертибилна уколико се користи само једном

    • не мора да буде инвертибилна

    • мора бити инвертибилна

    • не мора бити инвертибилна али све подматрице морају

    Correct Answer
    A. мора бити инвертибилна
    Explanation
    The key in Hill cipher must be invertible because in the encryption process, the key matrix is multiplied by the plaintext matrix. In order for this multiplication to be reversible (i.e., for decryption to be possible), the key matrix must be invertible. If the key matrix is not invertible, it will not be possible to accurately decrypt the ciphertext back to the original plaintext. Therefore, the key matrix in Hill cipher must be invertible.

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  • 40. 

    Aktivan napad koji onemogućava funkcionisanje sistema ili pružanje neke usluge je napad na:

    • Integritet

    • Raspoloživost

    • Poverljivost

    • Autentičnost

    Correct Answer
    A. Raspoloživost
    Explanation
    A napad na raspoloživost refers to an active attack that prevents the functioning or provision of a service. This means that the attacker aims to disrupt or disable the availability of a system or service, making it inaccessible or unusable for its intended users.

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  • 41. 

    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):

    • 00001

    • 00000

    • 00100

    • 00010

    Correct Answer
    A. 00000
    Explanation
    The given expression involves XOR operations between x, y, and z multiple times. XORing a number with itself always results in 0. Therefore, when x is XORed with itself three times, the result is 0. Similarly, when y and z are XORed with themselves once, the result is also 0. Finally, when all the resulting 0s are XORed together, the answer is 0.

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  • 42. 

    Dužina ključa kod AES algoritma može da bude:

    • 64, 128, 192 ili 256 bita

    • Proizvoljne dužine

    • 128, 192 ili 256 bita

    • 64, 128 ili 192 bita

    Correct Answer
    A. 128, 192 ili 256 bita
    Explanation
    The AES algorithm supports key lengths of 128, 192, or 256 bits. This means that the length of the key used in AES encryption can be either 128, 192, or 256 bits. These different key lengths provide varying levels of security and can be chosen based on the specific requirements of the encryption process.

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  • 43. 

    Dužina bloka kod AES algoritma može da bude:

    • 64, 128 ili 192

    • 128, 192 ili 256 bita

    • 64, 128, 192 ili 256 bita

    • Proizvoljne dužine

    Correct Answer
    A. 128, 192 ili 256 bita
    Explanation
    The length of the block in the AES algorithm can be either 128, 192, or 256 bits. This means that the block size can be any of these three options, allowing for flexibility in the encryption process. The AES algorithm is a symmetric encryption algorithm that operates on fixed-size blocks, and the block size determines the amount of data that can be processed at a time. Therefore, the correct answer is 128, 192, or 256 bits.

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  • 44. 

    Ukoliko se koriste velika slova engleskog alfabeta za ključ dužine 6, koliki je prostor klučeva ukoliko slova u ključu mogu da se ponavljaju:

    • 26^6

    • 26 x 6

    • 6^26

    • 26 x 25 x 24 x 23 x 22 x 21

    Correct Answer
    A. 26^6
    Explanation
    The question asks for the size of the key space when using uppercase English letters for a key of length 6, with the possibility of repeating letters. The correct answer is 26^6 because there are 26 possible choices for each position in the key, and since the letters can be repeated, we multiply 26 by itself 6 times.

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  • 45. 

    Који алгоритам не припада наведеној групи:

    • MD5

    • SHA - 3

    • SHA - 0

    • AES

    Correct Answer
    A. AES
    Explanation
    The given question asks to identify the algorithm that does not belong to the given group. MD5, SHA-3, and SHA-0 are all cryptographic hash functions, while AES (Advanced Encryption Standard) is a symmetric encryption algorithm. Therefore, AES is the algorithm that does not belong to the given group of hash functions.

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  • 46. 

    Нека је Z скуп знакова над којим се дефинише кључ. Уколико је кључ дужине 5 и Z={1,2,3,4,...,9,a,b,c,...,x,y,z,A,B,C,...,X,Y,Z,*,#,$}, тада је величина простора кључева уколико слова у кључу могу да се понављају:

    • 5^64

    • 5^64 - 1

    • 64^5

    • 64^5 - 1

    Correct Answer
    A. 64^5
    Explanation
    The question is asking for the size of the key space when the key length is 5 and the set of characters is Z={1,2,3,4,...,9,a,b,c,...,x,y,z,A,B,C,...,X,Y,Z,*,#,$}. The key space size is determined by the number of possible combinations of characters in the key. Since the key length is 5 and there are 64 characters in the set Z, the size of the key space is 64^5.

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  • 47. 

    Отворени текст ПЕРА шифрује се Вижнеровом шифром са кључем ПЕРА. Алфабет који се користи је српска азбука и слова су нумерисана редом бројевима од 1 до 30. Шифрат који се добија на тај начин је:

    • ЖЉИБ

    • отворени текст и кључ не смеју да буду исти

    • БИЛО

    • ОЛИБ

    Correct Answer
    A. ЖЉИБ
    Explanation
    The given answer "ЖЉИБ" is the result of encrypting the open text "БИЛО" using the Vigenere cipher with the key "ПЕРА". The Serbian alphabet is used, where each letter is assigned a number from 1 to 30. The Vigenere cipher is a polyalphabetic substitution cipher that uses a series of Caesar ciphers based on the letters of a keyword. In this case, the keyword "ПЕРА" is repeated to match the length of the open text. Each letter of the open text is shifted by the corresponding letter of the keyword, resulting in the encrypted text "ЖЉИБ".

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  • 48. 

    Аутентификација која се реализује шифровањем хеш вредности симетричним алгоритмом подразумева:

    • постојање "интерног" поверења

    • тајну комуникацију

    • претходна провера поверљивости

    • употребом MAC-a

    Correct Answer
    A. постојање "интерног" поверења
    Explanation
    The correct answer suggests that the authentication process involves the existence of "inter-node" trust. This means that there is a level of trust established between different nodes or entities involved in the authentication process. This trust allows for secure communication and verification of identities between these nodes.

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  • 49. 

    Да ли хеш функција треба да поседује својство лавинског ефекта?

    • нема става

    • да

    • пожељно је

    • не

    Correct Answer
    A. да
    Explanation
    The given correct answer is "да" (yes). This suggests that a hash function should possess the property of the avalanche effect. The avalanche effect refers to the property of a hash function where a small change in the input results in a significant change in the output. This property is desirable as it ensures that even a slight alteration in the input will produce a completely different hash value, making it difficult for attackers to predict or manipulate the output. Therefore, it is important for a hash function to have the avalanche effect to enhance its security.

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

Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.

  • Current Version
  • Mar 21, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Jun 05, 2017
    Quiz Created by
    Mladjalug96
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