1. Dr. Clincy Professor of CS
CS Chapter 2
Dr. Clincy
Professor of CS
Dr. Clincy
Lecture 6

2. Error Control Techniques
Approaches
Detection
Simple Parity (Double Parity) – already covered
Cyclic Redundancy Checksum – already covered
Error correction
Hamming Approach
Dr. Clincy
Lecture 6

3. Understanding Hamming Distance Concept
Dr. Clincy
Lecture 6

4. Error Correction – Hamming Code Concept
Computers make errors occasionally due to voltage spikes and etc.
Recall Encoding Concept – codes representing characters
Hamming Distance of 1: change in 1 bit creates a new code
A-000
D-001
F-110
C-011
H-101
G-111
B-010
E-100
What happens with 1 bit in error ?
Dr. Clincy
Lecture 6

5. Hamming Distance of 2 What happens with 1 bit in error ?
001
C-110
B-011
D-101
111
010
100
What happens with 1 bit in error ?
What happens with 2 bits in error ?
Dr. Clincy
Lecture 6

6. Hamming Distance of 3 What happens with 1 bit in error ?
001
110
011
101
B-111
010
100
What happens with 1 bit in error ?
What happens with 2 bits in error ?
What happens with 3 bits in error ?
Dr. Clincy
Lecture 6

7. Hamming Error Correcting Approach
1st: Determine the number of parity bits to add to the code word for checking
2nd: determine bit positions of each added parity bits
3rd: Determine what each parity bit checks
Dr. Clincy
Lecture 6

8. Hamming Code Example (Understand the “how” vs “why”)
Example: Given a 4-bit code and even parity request – recall parity
Determine number of parity bits to add: 20=1, 21=2, 22=4: also determine bit positions of parity bits – 1, 2 and 4 (with labels C1, C2 and C4)
Let the 4-bit code have labels I3, I5, I6 and I7 due to the parity bit positions
Therefore the seven bits would be transmitted in the following order: C1 C2 I3 C4 I5 I6 I7
Determine what each parity bit checks: add the parity positions to determine this (must be less than or equal to 7 in this case): C1 case: 1+2=3, 1+4=5, 1+2+4=7; C2 case: 2+1=3, 2+1+4=7, 2+4=6; C4 case: 4+1=5, 4+2=6, 4+2+1=7
For example, transmitting the 4-bit code of 0101 (I3I5I6I7) would be (red bits are parity bits)
If I3 was corrupted during transmission, C1 and C2 would detect it (1+2=3)
If I5 was corrupted during transmission, C1 and C4 would detect it (1+4=5)
If I6 was corrupted during transmission, C2 and C4 would detect it (2+4=6)
If I7 was corrupted during transmission, C1 , C2 and C4 would detect it (1+2+4=7)
Dr. Clincy
Lecture 6

9. Floating Point Numbers
Dr. Clincy
Lecture 6

10. Floating-Point Representation
The signed magnitude, one’s complement, and two’s complement representation that we have just presented deal with signed integer values only.
Without modification, these formats are not useful in scientific or business applications that deal with real number values.
Floating-point representation solves this problem.
For example, how would be stored compared to ?
How can we use the existing infrastructure of the computer to store real numbers ?
Dr. Clincy
Lecture 6

11. Scientific-Notation Vs Floating-Point Representation
Computers use a form of scientific notation for floating-point representation
Numbers written in scientific notation have three components:
Computer representation of a floating-point number consists of three fixed-size fields:
Size determines range
Size determines precision
Dr. Clincy
Lecture 6

12. Floating-Point Representation
Floating-point numbers allow an arbitrary number of decimal places to the right of the decimal point.
For example: 0.5  0.25 = 0.125
They are often expressed in scientific notation.
For example:
0.125 = 1.25  10-1
5,000,000 = 5.0  106
Dr. Clincy
Lecture 6

13. The Concept
Dr. Clincy
Lecture 6

14. The Simple Model
We introduce a hypothetical “Simple Model” to explain the concepts
In this model:
A floating-point number is 14 bits in length
The exponent field is 5 bits
The significand field is 8 bits
The significand is always preceded by an implied binary point.
Thus, the significand always contains a fractional binary value.
The exponent indicates the power of 2 by which the significand is multiplied.
Dr. Clincy
Lecture 6

15. Simple Model Illustrated
Example:
Express 3210 in the simplified 14-bit floating-point model.
We know that 32 is 25. So in (binary) scientific notation 32 = 1.0 x 25 = 0.1 x 26.
In a moment, we’ll explain why we prefer the second notation versus the first.
Using this information, we put 110 (= 610) in the exponent field and 1 in the significand as shown.
Dr. Clincy
Lecture 6

16. Problems with the Simple Model
The illustrations shown at the right are all equivalent representations for 32 using our simplified model.
Not only do these synonymous representations waste space, but they can also cause confusion.
Another problem with our system is that we have made no allowances for negative exponents. We have no way to express 0.5 (=2 -1)! (Notice that there is no sign in the exponent field.)
All of these problems can be fixed with no changes to our basic model.
Dr. Clincy
Lecture 6

17. Normalization
To resolve the problem of synonymous forms, we establish a rule that the first digit of the significand must be 1, with no ones to the left of the radix point.
This process, called normalization, results in a unique pattern for each floating-point number.
In our simple model, all significands must have the form 0.1xxxxxxxx
For example, 4.5 = x 20 = x 22 = x 23. The last expression is correctly normalized.
In our simple instructional model, we use no implied bits.
Dr. Clincy
Lecture 6

18. Negative Exponents
To provide for negative exponents, we will use a biased exponent.
A bias is a number that is approximately midway in the range of values expressible by the exponent. We subtract the bias from the value in the exponent to determine its true value.
In our case, we have a 5-bit exponent. We will use 16 for our bias. This is called excess-16 representation.
In our model, exponent values greater than 16 are positive.
an exponent of 4, would be 16+4=20
an exponent of 10, would be 16+10=26
In our model, exponent values less than 16 are negative, representing fractional numbers.
an exponent of -4, would be 16-4=12
an exponent of -10, would be 16-10=6
Dr. Clincy
Lecture 6

19. Examples Example 1: Example 2:
Express 3210 in the revised 14-bit floating-point model.
We know that 32 = 1.0 x 25 = 0.1 x 26.
To use our excess 16 biased exponent, we add 16 to 6, giving 2210 (=101102).
So we have:
Example 2:
Express in the revised 14-bit floating-point model.
We know that is So in (binary) scientific notation = 1.0 x 2-4 = 0.1 x 2 -3.
To use our excess 16 biased exponent, we add 16 to -3, giving 1310 (=011012).
Dr. Clincy
Lecture 6

20. Another Example Example:
Express in the revised 14-bit floating-point model.
We find = Normalizing, we have: = x 2 5.
To use our excess 16 biased exponent, we add 16 to 5, giving 2110 (=101012). We also need a 1 in the sign bit.
Dr. Clincy
Lecture 6