📝 Chord protocol interface and operations (36 MCQs)

📖 From Data Communication and Networks • 29. Peer to Peer Paradigm • 36 questions available

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What is Chord protocol interface and operations?

The Chord protocol interface provides core operations like lookup, join, leave, and stabilize, which allow nodes to dynamically enter or exit the network while maintaining correct routing tables and ensuring data remains accessible.

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📝 All Chord protocol interface and operations MCQs

Q1. In arithmetic decoding, why must the original message contain a terminating symbol?

A.It marks the end of the interval

B.It allows the decoder to know when to stop ✅

C.It improves compression ratio

D.It signals the start of a new block

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The terminating symbol signals the decoder that the original message has ended, so the decoding loop can exit precisely at that point. Without it, the decoder would continue indefinitely, unable to determine where the message stops.

Q2. If the current interval is [0.685,0.690) and the code value falls at 0.688, what can be inferred about the next symbol?

A.It must be the first symbol

B.It belongs to the same sub‑interval ✅

C.Decoding must restart

D.Interval is invalid

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Since 0.688 lies within the interval [0.685,0.690), the decoder knows the next symbol will be the one whose sub‑interval contains that value, meaning it remains within the same current interval segment.

Q3. During decoding, after outputting a symbol, the interval is updated. What immediate effect does this have on the subsequent code value range?

A.It expands the range

B.It narrows the range ✅

C.It leaves the range unchanged

D.It randomizes the range

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Updating the interval after a symbol is output narrows the range because the interval is subdivided according to the probability of the decoded symbol, reducing the uncertainty for the remaining code value.

Q4. If the decoder never outputs the terminating symbol, which of the following is the most likely cause?

A.Code value outside [0,1)

B.Incorrect probability model ✅

C.Missing input bits

D.Interval never shrinks

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: An incorrect probability model leads to mismatched intervals, so the decoder cannot correctly map the code value to the terminating symbol, causing the loop to continue without ever reaching the designated end marker.

Q5. Suppose the decoder computes an interval [0.4,0.5) but the code value is 0.6. What does this indicate?

A.The code is corrupted ✅

B.The decoder should reset

C.The symbol probabilities are reversed

D.The terminating symbol was omitted

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A code value outside the computed interval suggests that the transmitted code has been altered or corrupted, because a valid arithmetic code must always lie within the current interval during decoding.

Q6. Using limited floating‑point precision can cause which of the following during arithmetic decoding?

A.Faster convergence

B.Interval underflow

C.Incorrect symbol selection ✅

D.No effect

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Finite precision can introduce rounding errors that shift interval boundaries, causing the decoder to select the wrong symbol when the code value falls near a boundary, leading to incorrect decoding results.

Q7. Given three consecutive intervals during decoding: I1 = [0.2,0.4), I2 = [0.3,0.35), I3 = [0.32,0.33). Which statement correctly describes the relationship among them?

A.I3 ⊂ I2 ⊂ I1 ✅

B.I1 ⊂ I2 ⊂ I3

C.I2 ⊂ I1 and I3 unrelated

D.All intervals are disjoint

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Each subsequent interval is a subset of the previous one because decoding continually refines the range; thus I3 is contained within I2, which is itself contained within I1.

Q8. If after processing several symbols the interval becomes [0.685,0.690) and the code value is exactly 0.690, what should the decoder do?

A.Accept as valid

B.Treat as overflow and error ✅

C.Move to next symbol

D.Adjust interval to include 0.690

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The interval is half‑open on the upper bound, so a code value equal to 0.690 lies outside the valid range; the decoder must treat this as an error or overflow condition.

Q9. In an implementation that never renormalizes the interval, which problem is most likely to appear after many symbols?

A.Interval width becomes zero

B.Interval width exceeds 1

C.Precision loss leading to wrong symbol ✅

D.Infinite loop

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Without renormalization, the interval can become extremely narrow, and limited numeric precision will cause rounding errors that shift boundaries, ultimately leading to incorrect symbol selection.

Q10. How does the loop condition differ between arithmetic encoding and decoding?

A.Encoding loops until interval size < threshold, decoding loops until terminating symbol ✅

B.Both loop until interval zero

C.Encoding loops on each symbol, decoding loops on each bit

D.No difference

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Encoding continues refining the interval until it is sufficiently small, whereas decoding proceeds symbol by symbol until the terminating symbol is emitted, making the loop conditions distinct.

Q11. Which outcome is most directly affected by an inaccurate probability model in arithmetic decoding?

A.Length of encoded message

B.Correctness of decoded symbols ✅

C.Speed of decoding

D.Memory usage

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: If the probability model does not match the encoder’s model, the interval boundaries will be wrong, causing the decoder to map the code value to incorrect symbols, directly compromising decoding correctness.

Q12. In the decoding table, cumulative frequencies are used instead of raw probabilities because...

A.They simplify interval calculations ✅

B.They reduce memory

C.They increase compression ratio

D.They avoid division

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Cumulative frequencies allow the decoder to compute sub‑interval boundaries by simple addition, avoiding division and making interval calculations more straightforward.

Q13. Which of the following best compares a naive interval‑search decoding method with a binary‑search method?

A.Naive is faster but less accurate

B.Binary‑search reduces comparisons ✅

C.Both have same complexity

D.Naive uses less memory

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Binary‑search halves the search space each step, reducing the number of comparisons needed to locate the symbol’s sub‑interval, whereas a naive linear scan examines each symbol sequentially.

Q14. Placing the terminating symbol at the end of the probability table rather than at the beginning most likely...

A.Improves compression

B.Increases decoding time ✅

C.Has no effect

D.Causes decoding errors

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: When the terminating symbol is placed later, the decoder must traverse more entries to locate it, slightly increasing the number of comparisons and thus decoding time.

Q15. If the interval after encoding a symbol is unusually wide, what can be inferred about the symbol's probability?

A.High probability ✅

B.Low probability

C.Equal probability

D.Probability unknown

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A wide interval indicates that the symbol occupies a large portion of the probability space, meaning it has a relatively high probability compared to other symbols.

Q16. Adaptive‑precision decoding differs from fixed‑precision primarily by...

A.Changing interval size dynamically

B.Using variable‑length codewords

C.Updating symbol probabilities on the fly

D.Adjusting numeric precision to prevent underflow ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Adaptive‑precision algorithms modify the numeric precision used for interval boundaries as decoding proceeds, preventing underflow and maintaining accuracy without constantly using very high precision.

Q17. Starting the decoding process with a larger initial interval, such as [0,1), versus a narrowed interval, affects...

A.Compression ratio only

B.Decoding speed and numerical stability ✅

C.Memory consumption

D.Number of symbols that can be decoded

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: A larger initial interval gives more room for refinement, which can improve numerical stability but may require more scaling operations, influencing overall decoding speed.

Q18. Which method better ensures that the terminating symbol does not bias the probability model?

A.Reserving a fixed low probability

B.Dynamically adjusting its probability after each symbol ✅

C.Excluding it from the model during encoding

D.Treating it as a regular symbol

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: By updating the terminating symbol’s probability dynamically, the model remains representative of the actual data distribution and avoids giving the end marker undue weight.

Q19. Rounding the lower bound of an interval upward by $10^{-7}$ will most likely...

A.Cause earlier termination

B.Shift symbol boundaries ✅

C.Have no effect

D.Expand the interval

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: A small upward rounding moves the lower bound, which can shift the sub‑intervals for symbols, potentially causing the decoder to select a different symbol than intended.

Q20. When decoding a symbol, the current interval is divided proportionally to symbol probabilities. If symbol A has probability 0.6, what fraction of the interval corresponds to A?

A.0.4

B.0.6 ✅

C.0.3

D.0.7

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The proportion of the interval assigned to a symbol equals its probability; thus, symbol A occupies 60 % of the interval, i.e., a fraction of 0.6.

Q21. The code value always lies...

A.At the midpoint of the interval

B.At the lower bound

C.Somewhere within the current interval ✅

D.Outside the interval

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: By definition, the code value is a real number that falls inside the current interval; it need not be at any specific bound, only within the interval limits.

Q22. Which sequence correctly describes the main steps of arithmetic decoding?

A.Input code → Find interval → Output symbol → Update interval → Repeat until terminating symbol ✅

B.Input symbol → Update interval → Output code

C.Generate random number → Map to symbol

D.None of the above

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The decoding algorithm reads the code, locates the corresponding interval, outputs the associated symbol, updates the interval, and repeats until the terminating symbol signals completion.

Q23. Suppose the cumulative probability table maps symbols X, Y, Z to intervals [0,0.2), [0.2,0.5), [0.5,1). If the code value is 0.37, which symbol is decoded?

A.X

B.Y ✅

C.Z

D.None

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: 0.37 lies between 0.2 and 0.5, which corresponds to the interval assigned to symbol Y, so Y is the decoded symbol.

Q24. Why does the interval width decrease after each decoded symbol?

A.Because probabilities sum to less than 1

B.Because the interval is subdivided according to symbol probability ✅

C.Because the decoder discards bits

D.Because of rounding errors

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Each decoded symbol narrows the interval by selecting the sub‑interval that matches the symbol’s probability, thereby reducing the remaining uncertainty.

Q25. If the terminating symbol occupies the interval [0.9,1) and the code value is 0.95, what will the decoder output next?

A.The preceding symbol

B.The terminating symbol ✅

C.An error

D.A random symbol

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: 0.95 falls within the terminating symbol’s interval, so the decoder recognises that the message has ended and outputs the terminating symbol.

Q26. During decoding, the code value falls into an interval that does not correspond to any defined symbol. Which of the following is the most appropriate response?

A.Output the last decoded symbol

B.Abort decoding and signal corruption ✅

C.Insert a placeholder symbol

D.Continue with next code value

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: If the code value cannot be mapped to any symbol, the most prudent action is to abort and indicate that the encoded data is corrupted or mismatched.

Q27. How does increasing the number of bits used to represent interval endpoints affect the decoding accuracy?

A.Improves accuracy by reducing quantization error ✅

B.Decreases accuracy

C.Has no effect

D.Causes overflow

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: More bits provide finer granularity for interval boundaries, reducing quantization errors and making the decoded symbols align more closely with the intended probabilities.

Q28. If a message contains two different terminating symbols with intervals [0.95,0.975) and [0.975,1), how should the decoder decide which one signals the end?

A.Choose the one with larger interval

B.Use the first encountered

C.Both indicate end

D.Ambiguity requires external protocol ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Having two distinct terminating symbols creates ambiguity; the decoder must rely on an external convention or protocol to determine which interval actually represents the end of the message.

Q29. What is the primary purpose of arithmetic decoding?

A.To compress data

B.To reconstruct the original message from a code value ✅

C.To generate random numbers

D.To encrypt data

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Arithmetic decoding takes the numeric code produced by arithmetic encoding and, using the probability model, reconstructs the original sequence of symbols.

Q30. In the context of arithmetic coding, a terminating symbol is...

A.A symbol with probability zero

B.A special marker indicating the end of the message ✅

C.The most frequent symbol

D.A symbol used for error correction

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The terminating symbol is a designated marker added to the source alphabet to indicate where the encoded message should stop during decoding.

Q31. According to Table 28.5, the first step after reading the input code is to...

A.Output the terminating symbol

B.Find the real number related to code ✅

C.Update the probability model

D.Reset the interval

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Table 28.5 lists the decoding algorithm, which begins by converting the received code into a real number that can be compared against interval boundaries.

Q32. The interval notation $[0.685,0.690)$ indicates that...

A.Both endpoints are inclusive

B.Lower bound inclusive, upper bound exclusive ✅

C.Both endpoints exclusive

D.Upper bound inclusive, lower bound exclusive

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Bracket notation with a parenthesis on the right means the lower bound is included in the interval while the upper bound is excluded.

Q33. When the decoder rescales the interval after it becomes too small, what is the immediate effect on the code value representation?

A.Code value is shifted left ✅

B.Code value is shifted right

C.Code value remains unchanged

D.Code value is discarded

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Rescaling typically involves left‑shifting both the interval bounds and the code value, preserving their relative positions while expanding the numeric range for further processing.

Q34. Suppose symbol probabilities are {A:0.5, B:0.3, C:0.2}. If the decoder mistakenly uses {A:0.4, B:0.4, C:0.2}, which decoding error is most likely?

A.Decoding will stop early

B.Symbols will be swapped ✅

C.Interval will overflow

D.No error

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: An incorrect probability allocation changes the sub‑interval boundaries, causing the decoder to map code values to the wrong symbols, effectively swapping symbols during reconstruction.

Q35. Given cumulative intervals: A [0,0.4), B [0.4,0.7), C [0.7,1). Code value 0.68, which symbol?

A.A

B.B ✅

C.C

D.None

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: 0.68 lies between 0.4 and 0.7, which corresponds to symbol B’s interval, so B is the decoded symbol.

Q36. Consider a message \AB\ where A has probability 0.6 and B 0.4, and terminating symbol T has probability 0.1 (renormalized). After decoding A, the interval becomes [0.0,0.6). If the code value is 0.55, what is the next decoded symbol?

A.A

B.B ✅

C.T

D.Decoding error

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: 0.55 falls within the sub‑interval for B (from 0.4 to 0.6 after renormalization),

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