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With 1 parity bit we can detect all 1-bit errors. Show that at least one generalization fails, as follows: (a) Show that if messages \(m\) are 8 bits long, then there is no error detection code \(e=e(m)\) of size 2 bits that can detect all 2-bit errors. Hint: Consider the set \(M\) of all 8-bit messages with a single 1 bit; note that any message from \(M\) can be transmuted into any other with a 2 -bit error, and show that some pair of messages \(m_{1}\) and \(m_{2}\) in \(M\) must have the same error code \(e\). (b) Find an \(N\) (not necessarily minimal) such that no 32 -bit error detection code applied to N-bit blocks can detect all errors altering up to 8 bits.

Short Answer

Expert verified
A 2-bit error detection code for 8-bit messages cannot detect all 2-bit errors. No 32-bit error code can detect all 8-bit errors for N=40.

Step by step solution

01

Understand the Problem Statement

The task involves showing that an error detection code of 2 bits cannot detect all 2-bit errors for 8-bit messages and finding an N where no 32-bit error detection code can detect all errors altering up to 8 bits.
02

Setting Up the 8-bit Messages (Part a)

Consider an 8-bit message. Messages in set M have exactly one 1-bit, meaning only one bit out of the 8 bits is 1, and the rest are 0. For instance, 00000001, 00000010, etc., are in M.
03

Size of Set M and Number of Unique Error Codes (Part a)

The set M contains 8 messages. Each code is of size 2 bits, resulting in only 4 unique codes because 2 bits can represent 2^2=4 different combinations (00, 01, 10, 11).
04

Reasoning with Pigeonhole Principle (Part a)

With 8 messages and only 4 unique error codes, according to the Pigeonhole Principle, at least two messages must share the same error code. These two messages can be converted into each other with a 2-bit error.
05

Conclusion for Part (a)

Given the shared error codes by at least two messages, the error detection code cannot distinguish between errors altering 2 bits. Therefore, it is not possible to detect all 2-bit errors with a 2-bit error detection code.
06

Setting Up for Part (b)

Consider 32-bit error detection codes and N-bit blocks. A similar logic applies for more significant blocks and error patterns.
07

Choosing an Appropriate N (Part b)

To find N such that it is impossible for a 32-bit code to detect all 8-bit errors, use a similar argument. With each 32-bit code having 2^32 possible outputs but the errors significantly more than 2^32 combinations, errors cannot be uniquely mapped.
08

Conclusion for Part (b)

For N=40 (as an example), the number of possible 8-bit errors (C[40,8]) is far greater than the number of 32-bit error detection codes. Using combinatorial explosion, no 32-bit code can detect all 8-bit errors.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parity Bit
A parity bit is an extra bit added to a string of binary code. Its purpose is to ensure that the total number of bits with a value of 1 in the string is even (for even parity) or odd (for odd parity). This helps in detecting simple errors during data transmission. If a single bit error occurs, the parity is no longer accurate.
Let's take an example: Suppose we have the 4-bit message `1011`. For even parity, an additional parity bit would be `0` because `1011` already has an even number of 1s. However, if the message was `1001`, the parity bit would be `1` to make the total number of 1s even.
Parity bits are effective in detecting single-bit errors but not multiple-bit errors, limiting their usefulness in more complex error detection scenarios.
Pigeonhole Principle
The pigeonhole principle is a simple but powerful concept in combinatorics. It states that if you put more 'pigeons' into fewer 'pigeonholes' than there are pigeons, at least one pigeonhole must contain more than one pigeon.
In the context of error detection codes, this principle shows that when more messages (pigeons) share fewer error codes (pigeonholes), some messages must share the same error code. This limitation implies that such a system cannot uniquely detect certain errors. For example, with 8-bit messages needing 2-bit error codes, there are only 4 unique codes, but 8 possible messages. According to the pigeonhole principle, at least two messages will share the same error code, making it impossible to uniquely detect all 2-bit errors.
Combinatorial Explosion
Combinatorial explosion refers to the rapid growth of the number of possible combinations as the size of the set increases. This concept becomes particularly significant when designing error detection codes.
Consider that a 32-bit code can represent 2^32 different combinations, but once you start accounting for possible errors in larger datasets, the number of error combinations can exponentially exceed the number that a detection code can handle. For instance, with N-bit blocks, the number of ways to alter up to 8 bits grows quickly, making it impossible for a 32-bit detection code to cover all cases effectively. This is the heart of the problem stated in part (b) of the exercise.
Bit Error
A bit error occurs when a bit value is incorrectly flipped during data transmission, resulting in a 0 becoming a 1 or vice versa. These errors can stem from various sources like electrical interference, hardware malfunctions, or network issues.
Detecting and correcting bit errors is crucial for maintaining data integrity. Basic error detection methods like parity bits can handle single-bit errors, but more complex scenarios require advanced techniques like checksums and cyclic redundancy checks (CRCs). However, these methods still have limits when it comes to multiple-bit errors or larger data blocks, as explained in the exercise where 2-bit error detection codes fall short for 8-bit messages.
Error Detection
Error detection involves identifying errors in transmitted data to ensure data integrity. Various methods, such as parity bits, checksums, Hamming codes, and CRCs, are designed to recognize errors. However, each has limitations.
For example, while parity bits can detect single-bit errors, they fail with multiple-bit errors. The exercise highlights this with the 2-bit error detection code example. More sophisticated codes like CRCs offer higher reliability but face limitations when data size and error complexity increase. Understanding these methods and their constraints is crucial for developing effective error detection strategies, especially in complex systems and large datasets.

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Most popular questions from this chapter

Suppose we want to transmit the message 11001001 and protect it from errors using the CRC polynomial \(x^{3}+1\) (a) Use polynomial long division to determine the message that should be transmitted. (b) Suppose the leftmost bit of the message is inverted due to noise on the transmission link. What is the result of the receiver's CRC calculation? How does the receiver know that an error has occurred?

Consider a token ring with a ring latency of \(200 \mu\). Assuming that the delayed token release strategy is used, what is the effective throughput rate that can be achieved if the ring has a bandwidth of \(4 \mathrm{Mbps}\) ? What is the effective throughput rate that can be achieved if the ring has a bandwidth of \(100 \mathrm{Mbps}\) ? Answer for both a single active host and for "many" hosts; for the latter, assume there are sufficiently many hosts transmitting that the time spent advancing the token can be ignored. Assume a packet size of \(1 \mathrm{~KB}\).

Assuming a framing protocol that uses bit stuffing, show the bit sequence transmitted over the link when the frame contains the following bit sequence: 110101111101011111101011111110 Mark the stuffed bits.

Suppose five stations are waiting for another packet to finish on an Ethernet. All transmit at once when the packet is finished and collide. (a) Simulate this situation up until the point when one of the five waiting stations succeeds. Use coin flips or some other genuine random source to determine backoff times. Make the following simplifications: Ignore interframe spacing, ignore variability in collision times (so that retransmission is always after an exact integral multiple of the \(51.2-\mu\) s slot time), and assume that each collision uses up exactly one slot time. (b) Discuss the effect of the listed simplifications in your simulation versus the behavior you might encounter on a real Ethernet.

Consider an ARQ algorithm running over a \(20-\mathrm{km}\) point-to-point fiber link. (a) Compute the propagation delay for this link, assuming that the speed of light is \(2 \times 10^{8} \mathrm{~m} / \mathrm{s}\) in the fiber. (b) Suggest a suitable timeout value for the ARQ algorithm to use. (c) Why might it still be possible for the ARQ algorithm to time out and retransmit a frame, given this timeout value?

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