CSC 335 Data Communications and Networking

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CSC 335 Data Communications and Networking

• Lecture 5 Error Detection, Error Correction, and
  Data Security
• Dr. Cheer-Sun Yang

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Motivation

• Error can occur during transmission.
• Signal can be distorted and cause sampling
  errors.
• Sender sends a 1101, receiver can receive 1001.
• How can a receiver be aware of that an error has
  occurred?
• Can the position of a single bit error be
  identified?
• The techniques are called error detection and
  single bit error correction.

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Transmission Errors

• External electromagnetic signals - noise
• Transmission errors - attenuation
• Lost data timing
• But, before we talk about errors, we have to
  understand how messages are actually organized
  and sent.

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Protocol Data Unit

• Messages are transmitted in a unit called
  protocol data unit (PDU).
• The data unit at data link layer is called a
  frame which is a block of data bits enclosing in
  a beginning and ending character. The beginning
  character is called the start of header (SOH) and
  the ending character is called the end of
  transmission (EOT) in our textbook.

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Data Frame
In fact, the header contains a FLAG, the address,
and some control information. There are also
several kinds of frames information,
supervisory, and others. Each information frame
is also identified by a sequence number.
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Data Frame

• Advantages If error occurs at the sender side,
  the receiver will never receive the ending
  character. When the sender sends the header
  again, the receiver can synchronize easily.
• Disadvantages
• overhead
• what if the data portion includes a character
  identical to either the header or the ending
  character?

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Bit Stuffing and Byte Stuffing

• Ideally, the frame header or the ending character
  wont occur in the data block.
• But, it could happen.
• Bit-oriented protocol uses bit stuffing to solve
  the problem.

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Bit Stuffing

• Header 01111110
• In any data character, if there are 6 continuous
  1 bits, the transmitter inserts an extra 1 bit
  after the fifth 1 bit the receiver will reverse
  the procedure.

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Byte Stuffing

• If any data block contains a SOH, EOT, or ESC
  (escape), the transmitter also does the
  following
• SOH in data becomes ESC X.
• EOT in data becomes ESC Y.
• ESC in data becomes ESC Z.

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Error Detection

• The ability to detect when a transmission error
  has occurred is called error detection.
• Additional bits should be added by the
  transmitter for detecting errors at the receiving
  side.
• Parity Bit
• Value of parity bit is such that character has
  even (even parity) or odd (odd parity) number of
  ones
• Disadvantage Even number of bit errors goes
  undetected

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Parity Checking Analysis

• Single bit error can be detected by a single
  parity bit.
• For example, the original data bits include
  11010100, the parity (even) bit is then 0.
• The data bits and the parity bit will be sent
  together as a unit to the receiver.
• If there is an error at the second bit, the
  receiver receives 100101000. The total number of
  1s is not an even number. The receiver can claim
  that the message is incorrect.

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Parity Checking (contd)

• Single bit error can be detected by a single
  parity bit.
• However, if two bits are transmitted incorrectly,
  the parity checking may not detect them.
• For example, data1101, parity bit 1
• If incorrect data 1011, parity bit 1 the
  receiver cannot tell that this message is
  incorrect.

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Any Problem?

• Parity checking will detect any single-bit error.
• Is there any glitch?
• Yes.
• What?

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Single-Bit Error

• Suppose an error occurred because of a brief
  power surge or static electricity whose duration
  is a hundredth of a second. If the data rate is
  64 Kbps, approximately 640 bits will be affected.
• When many bits are damaged, we call this a burst
  error.
• Single-bit parity checking is very likely to
  fail!!

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Single-Bit Error (contd)

• In general, if an odd number of bits change,
  parity checking will detect the error.
• If an even number of bits change, parity checking
  will not detect the error.
• The error detection rate is 50 which is not
  acceptable for a communications network.
• Does this mean that parity checking is useless?
  Well, No!
• It is the basis of more complex techniques. It is
  also useful when single-bit error could happen.

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Double-Bit Error Detection

• How about using two parity bits one checks for
  the bits in odd positions and another checks for
  bits in even positions?
• It is better but still not good enough Why?
• A_____________________________. (Exercise)

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Burst Error Detection

• Perhaps we should take a look at how data bits
  are actually transmitted.
• A block of data bits are transmitted in a unit
  called frame. We will talk about the concept of
  data frame more in next lecture.
• Lets assume that a frame looks like this for now.

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Data Frame with Parity Bits
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• Rather than sending all bits from one frame
  together, we send them separately. Transmitting
  just one bit of information is not very
  efficient, however.
• If there are more frames to be sent together, we
  can send one bit from each frame together with
  the parity bit of the new data frames. Fig 4.3
  illustrates this principle.

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• Instead of sending 10 5-bit frames with a parity
  bit, we send each column (6 10-bit frames).
• Now, suppose a burst error occurs and interferes
  with one transmission. At most, one bit from each
  original data frame is actually changed, and the
  parity check will detect them.
• How successful is this approach?

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• The only way a burst error affecting two columns
  can go undetected is when either two bits or no
  bits are actually changed in each (old) frame.
• What is the probability that this error can occur
  and will not be detected?
• Suppose column i and j were affected. Suppose
  that the bits from column i are affected
  randomly. Now consider any bit from column j, it
  is either correct or it is incorrect. Therefore
  there is a ½ chance that an error occurs to two
  bits in the same row.

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• If there are n bits in a column, there is a
  probability of (1/2)n that either two or no bits
  are changed in each row and the error goes
  undetected. If n20, the probability that the
  error will not be detected 1/1048576.

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Cyclic Redundancy Check

• A major disadvantage of previous method is the
  cost of assembling and reassembling of the
  frames. Also, the error can only be detected
  after all frames are received. Then resending all
  frames is very costly.
• Is there a way to send a frame and determine
  immediately whether there was an error?
• Yes. CRC is more efficient although more
  difficult to understand.

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Bit String Representation
The bit string 10010101110 is interpreted as
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CRC Method

• We assume all computations are done using
  modulo 2.
• Given a bit string, append several 0s to the end
  of it (we will specify how many and why later)
  and call it B. Let B(x) be the polynomial
  corresponding to B.
• Divide B(x) by some agreed-on polynomial G(x)
  (generator polynomial) and determine the
  remainder R(x).

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CRC Method (contd)

• Define T(x) B(x) R(x). Later we will show
  that T(x)/G(x) generates a 0 remainder and that
  the subtraction can be done by replacing the
  previously appended 0 bits with the bit string
  corresponding to R(x).
• Transmit T, the bit string corresponding to T(x).
• Let T represent the bit stream that receiver
  receives and T(x) the associated polynomial. The
  receiver divides T(x) by G(x). If there is a 0
  remainder the receiver concludes TT and no
  error occurred. Otherwise, the receiver concludes
  an error occurred.

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Modulo 2 Addition

• 0 0 0
• 0 1 1
• 1 0 1
• 1 1 0 (with no carry)
• Modulo 2 means to take the remainder after
  dividing by 2.
• 2 mod 2 0

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Modulo 2 Subtraction

• 0 0 0
• 0 1 1 (with no carry) (why?)
• 1 0 1
• 1 1 0

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Exclusive OR and Modulo
2 Addition and Subtraction
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How CRC works?
Suppose that we need to send the bit string
1101011, and the generator polynomial is G(x)
x4x3 1. Step 1 Append 0s to the end of the
string. The number of 0s is the same as the
degree of the the generator polynomial. Thus the
string is (B) 11010110000.
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How CRC works? (contd)
Step 2 Divide B(x) by G(x). Find the
corresponding bit representation of the remainder
polynomial R(x). In our case 1010. Step 3 Find
T(x) B(x) R(x). (Remember using modulo 2
subtraction). Transmit the string T.
11010110000 B 1010
R --------------------------------
11010111010 T (FACT)
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How CRC works? (contd)
Step 4 At the receiving side, receiver divides
the string (received) by G(x). If the remainder
is not 0, the string is incorrect.
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Summary of Cyclic Redundancy Check

• For a block of k bits transmitter generates n bit
  sequence
• Transmit kn bits which is exactly divisible by
  some number
• Receive divides frame by that number
• If no remainder, assume no error
• For math, see Section 4.3

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Hamming Code

• Retransmission takes time and is costly if
  retransmission also fails.
• Due to R. W. Hamming
• Single-bit error correction identifies the
  position of the error bit
• Multiple-bit error detection only detects that
  multiple-bit error has occurred

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General Concept

• Insert some parity checking bit at in the bit
  string instead of at the end.
• The parity bits check the parity in some specific
  location
• If there is a single-bit error, the location can
  be identified by the combination of these parity
  bits.

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A Guessing Game

• You can think of one number from 1 to 15 (Dont
  tell me what it is.)
• Is the number in one of the following?
• 1, 3, 5, 7, 9, 11, 13, 15
• Is the number in one of the following?
• 2, 3, 6, 7, 10, 11, 14, 15
• Is the number in one of the following?
• 4, 5, 6, 7, 12, 13, 14, 15
• Is the number in one of the following?
• 8, 9, 10, 11, 12, 13, 14,15

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An Example
Data to send m1 m2 m3 m4 m5 m6 m7 m8
Hamming Code
p1 p2 m1 p3 m2 m3 m4 p4 m5 m6 m7 m8
P1 even parity for positions 1,3,5,7,9,11 P2
even parity for positions 2,3,6,7,10,11 P3 even
parity for positions 4,5,6,7,12 P4 even parity
for positions 8,9,10,11,12
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Position of Extra Bits

• 20 1 the 1st position
• 21 2 the 2nd position
• 22 4 the 4th position
• 23 8 the 8th position
• 24 16 the 16th position

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Error Detection vs. Error Correction

• Which one is better?
• A Neither.
• Error correction requires more overhead time
  and extra bits
• Error detection requires retransmission.

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Error Control (next lecture)

• Detection and correction of errors
• Lost frames
• Damaged frames
• Automatic repeat request
• Error detection
• Positive acknowledgment
• Retransmission after timeout
• Negative acknowledgement and retransmission

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Data Security

• Encryption and Decryption
• Key Distribution and Protection
• Public Key Encryption
• Viruses, Worms, and Hackers

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Encryption and Decryption

• The idea is to disguise information so that
  even if it does fall into the wrong hands, it
  cannot be understood.
• Caesar Cipher
• Poly-alphabetic Cipher
• Transposition Cipher
• Bit-Level Ciphering
• Data Encryption Standard
• Clipper Chip

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Key Distribution and Protection

• Merklers Puzzles
• Shamirs Method

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Public Key Encryption

• RSA Algorithm
• Digital Signatures
• Authentication Using Hash-Bashed Schemes

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Viruses, Worms, and Hackers

• Infecting Files
• Memory-Resident Viruses
• Virus Evolution
• Virus Sources
• The Internet Worm
• Computer Hackers

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Assignments

• Read Section 4.1 - Section 4.4.
• Homework