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 m₁ and m2 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.

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 m₁ and m2 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.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Similar questions

Consider using both the Go-Back-N and selective-repeat strategies if the sequence number space has k bits. What is the maximum allowable sender window size?
given the maximal number of the binary sequence max seq (e.g., max seq=7 for three bits), what are the expression (in term of max seq) of the maximal sizes of sending window for go-back-n and selected-repeat protocols? you need to demonstrate the counter-examples on why window sizes larger than the maximum can fail. repeat the same question for the receiving window sizes
Problem with constructing MAC. The messages X to be authenticated consists of z independent blocks, so that: X = x1||x2|| . . . ||xz, where every xi consists of |xi| = 8 bits. The input blocks are consecutively put into the compression function ci = h(ci−1, xi) = ci−1⊕xi At the end, the MAC value: MACk(X) = cz+k mod 28 is calculated, where k is a 64-bit long shared key. Describe how exactly the (effective part of the) key k can be calculated with only one known message X.
Given the maximal number of the binary sequence MAX SEQ (e.g., MAX SEQ=7 for three bits), what are the expression (in term of MAX SEQ ONLY) of the maximal sizes of sending window for Go-back-N and Selected-Repeat Protocols? You need to demonstrate the counter-examples on why window sizes larger than the maximum can fail.
Suppose that a binary message – either 0 or 1 – must be transmitted by wire from location Ato location B. However, the data sent over the wire are subject to a channel noise disturbance,so to reduce the possibility of error, the value 2 is sent over the wire when the message is1, and the value −2 is sent when the message is 0. If X, X = {−2,2}, is the value sent atlocation A, the value received at location B, denoted as R, is given byR = X + N ,where N is the channel noise disturbance, which is independent of X. When the message isreceived at location B, the receiver decodes it according to the following rule:if R ≥.5, then conclude that message 1 was sentif R < .5, then conclude that message 0 was sentAssuming that the channel noise, N, is a unit normal random variable and that the message0 or 1 is sent with equal probability, what is the probability that we conclude that the wrongmessage was sent? This is the probability of error for this communication channel
a. Compute CBC-MAC for a message of 16 bits, “8642” (in Hexa). Assume a block size of 8 bits with an IV=F1 (in hexa). For simplicity, assume the encryption to be a simple XOR of the key with the plaintext. Let the encryption key be B4 (in Hexa). (Hint: Divide the message into blocks of 8 bits each; XOR each block with the previous cipher output; then encrypt this with the key. For the first block, XOR it with IV. Details in pages 325-326 Ch.12 of the textbook) b. Suppose Alice computes the Secret prefix MAC (page 322: secret prefix MAC(x) = h(key || x)) for the message ”AM” (in ASCII) with key “G” (in ASCII) that both Alice and Bob know. The hash function that is used is h(x1x2x3)= g(g(x1 XOR x2) XOR x3 ) where each xi is a character represented as 8 bits, and g(x) is a 8-bit string that is equal to the complement of bits in x. For example, g(10110011) = 01001100. The MAC is 8 bits. (8-bit ASCII representation of the characters is given below.) What is the Secret prefix MAC…
Consider PKCS-2.1 with n = 512 bit and SHA-1. Assuming L and PS is empty string, then describe in step-by-step that in one go maximum bytes of data can be encrypted. Now, if the modulus n is increased from 512 bit to 2048 bit then approximately how slow/fast will be the encryption, if maximum allowed bytes are encrypted.
Consider utilizing the Go-Back-N and selective-repeat methods if the sequence number space is k-bit long. What is the maximum size of the sender window we may use?
Consider the following message M=1010001101. The cyclic redundancy check (CRC) for this message using the divisor polynomial x5 + x4 + x² + 1 is?
Consider applying the Go-Back-N and selective-repeat techniques if the sequence number space is k-bit long. When sending data, what is the maximum size of the sender window that may be used?
We use DES in cipher feedback mode (CFB) to encrypt a plaintext m = m1m2 ...m100 into a ciphertext c1c2 ...c100, where each mi is 8-bit long. The ciphertext is sent to Bob. If c15 and c25 are missing and c8 and c88 are received as c8' and c88' wrongly, what mi’s can B compute correctly from the received ciphertext?
Material must be played out in a specific time intervals in order to be correctly interpreted by the recipient. Real-time protocols take varying transit times to get from source to destination. Even if packets are sent with the correct intervals, they will arrive at the other end with varying arrival times. Jitter is referred to as a fluctuation of pause time. We can buffer the packets for jitter reduction Let us say that, for the sake of argument, the noise is random, and the buffer is adequate. Armed with this knowledge, tell us what other difficulties are there?