📝 Text representation in multimedia (10 MCQs)

📖 From Data Communication and Networks • 28. Multimedia • 10 questions available

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What is Text representation in multimedia?

Text representation in multimedia refers to the encoding of characters and formatting information using standardized schemes like ASCII, Unicode, or markup languages such as HTML and XML, allowing textual content to be stored, transmitted, and rendered consistently across different devices and platforms within networked multimedia applications.

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📝 All Text representation in multimedia MCQs

Q1. In the LZW decoding pseudocode, what does the variable $S$ represent?

A.The next input codeword

B.The current string retrieved from the dictionary ✅

C.The previous codeword

D.A temporary counter

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The variable $S$ holds the string that is looked up in the dictionary for the current codeword. When a codeword is read, the algorithm accesses Dictionary[ $C$ ] and assigns that value to $S$ , which is then used for output and for building new entries.

Q2. If the current codeword $C$ is not found in the dictionary, which clause is executed and what immediate effect does it have on the dictionary?

A.If-clause; no new entry is added

B.If-clause; the existing entry is overwritten

C.Else-clause; the codeword is ignored

D.Else-clause; a new entry is created using

PreS + char

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: When $C$ is absent, the algorithm follows the else‑clause. This clause constructs a new dictionary entry by concatenating the previous string $PreS$ with the first character of the current string, thereby expanding the dictionary for future lookups.

Q3. Assume the dictionary initially contains entries for characters A and B. After reading codeword 2 (which corresponds to string 'BA') and then codeword 1 (string 'A'), what string is output by the algorithm for the second codeword?

A.BA

B.A ✅

C.BAA

D.AB

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The first codeword outputs 'BA' as expected. When the second codeword (1) is processed, the algorithm retrieves Dictionary[1], which is the single character 'A', and outputs that string. Thus the output associated with the second codeword is exactly 'A', independent of the previous output.

Q4. Compare the effect of the else‑clause versus the if‑clause on the size of the dictionary during decoding. Which statement is accurate?

A.Both increase size equally

B.If clause increases size, else does not

C.Else clause increases size, if clause does not ✅

D.Neither changes size

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The else‑clause is the only branch that adds a new entry to the dictionary, formed from $PreS + char$ . The if‑clause simply uses an existing entry without modification. Consequently, each execution of the else‑clause grows the dictionary, while the if‑clause leaves its size unchanged.

Q5. Considering the number of codewords processed, what is the overall time‑complexity of the LZW decoding algorithm?

O(n)

where

n

is the number of codewords ✅

O(n \log n)

O(n^{2})

O(\log n)

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Each iteration handles a single codeword with constant‑time dictionary lookups and possible entry creation. Because the work per codeword does not depend on previously processed data, the total runtime grows linearly with the number of codewords, giving a complexity of $O(n)$ .

Q6. If $PreS = 'A'$ and the next character $char = 'B'$ , what new dictionary entry is created by the else‑clause?

A.'AB' ✅

B.'BA'

C.'AA'

D.'BB'

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The else‑clause forms a new entry by concatenating the previous string $PreS$ with the newly read character. With $PreS = 'A'$ and $char = 'B'$ , the resulting string is 'AB', which is then stored in the dictionary under the next available code.

Q7. During decoding, if the algorithm encounters a codeword that is not yet present in the dictionary, what does it output?

PreS

concatenated with its first character ✅

B.The previous output string

C.An empty string

D.The codeword itself

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: When a missing codeword is read, LZW decoding uses the rule that the output is the previous string $PreS$ followed by the first character of $PreS$ . This constructs a valid string and also supplies the information needed to add a new dictionary entry.

Q8. What is the relationship between $PreS$ and $S$ in the LZW decoding process?

PreS

is always a prefix of

S

✅

PreS

equals

S

PreS

is the suffix of

S

D.No relationship

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: In each loop, $PreS$ holds the string from the previous iteration, and $S$ is the current string retrieved from the dictionary. Because $S$ is built from $PreS$ (often as $PreS + char$ ), the previous string naturally appears as a prefix of the new string.

Q9. During the decoding process, after the else‑clause creates a new entry at code 5, the subsequent if‑clause checks code 6 which is present in the dictionary. What will be the output for code 6?

A.The string stored at Dictionary[6] ✅

B.The previous string concatenated with the first character of Dictionary[6]

C.An error occurs

D.The algorithm terminates

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: When the if‑clause is taken, the algorithm simply retrieves the existing entry at Dictionary[6] and outputs that string. No new entry is added, and the decoding continues normally, producing the exact string associated with code 6.

Q10. How does LZW encoding differ from LZW decoding as illustrated in the example?

A.Encoding adds entries only when a new pattern appears, decoding adds entries on every iteration

B.Encoding never uses the else clause, decoding always does

C.Encoding builds dictionary from input, decoding reconstructs using codewords ✅

D.Both processes are identical

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Encoding constructs the dictionary by scanning the input and adding new patterns as they appear, while decoding uses the transmitted codewords to rebuild the original data, inserting new entries only when the else‑clause condition is met. This fundamental directionality distinguishes the two procedures.

🔗 Related Topics (MCQs)

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