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CSE 246: Computer Arithmetic Algorithms and Hardware Design Instructor: Prof. Chung-Kuan Cheng Fall 2006 Lecture 1: Introduction and Numbers.

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Presentation on theme: "CSE 246: Computer Arithmetic Algorithms and Hardware Design Instructor: Prof. Chung-Kuan Cheng Fall 2006 Lecture 1: Introduction and Numbers."— Presentation transcript:

1 CSE 246: Computer Arithmetic Algorithms and Hardware Design Instructor: Prof. Chung-Kuan Cheng Fall 2006 Lecture 1: Introduction and Numbers

2 CSE 2462 Agenda  Administration  Motivation  Lecture 1: Numbers

3 CSE 2463 Administration  Textbook: Computer Arithmetic Algorithms and Hardware Designs, Behrooz Parhami, Oxford  Recommended: Art of Computer Programming, Volume 2, Seminumerical Algorithms (3rd Edition), Donald E. Knuth  Numerical Computing with IEEE Floating Point Arithmetic, Michael L. Overton, SIAM  Computer Arithmetic Algorithms, Israel Koren, A K Peters, Natick, Massachusetts  Digital Arithmetic, Milos D. Ercegovac and Tomas Lang, Morgan Kaufmann  CMOS VLSI Design, Neil H.E. Weste and David Harris, Addison Wesley  Principles and Practices of Interconnection Networks, William James Dally and Brian Towles, Morgan Kaufmann  In addition: set of papers to read

4 CSE 2464 Administration  No classes on the following days Tu 10/17 BIBE Tu 10/24 EPEP Tu 11/7 ICCAD

5 CSE 2465 Administration  Grading: Homework – 20% Midterm – 35% Project  Report – 25%  Presentation – 20%  Midterm: Thursday, 10/2/06

6 CSE 2466 Administration  Potential project samples: Interconnect and switch modules Data path components using FPGAs, nano technologies Low power logic styles Reconfigurable blocks Low power data path components Low power/reliable coding systems

7 CSE 2467 Agenda  Administration  Motivation  Lecture 1: Numbers

8 CSE 2468 Motivation Why do we care about arithmetic algorithms and hardware design?  Classic problems – well defined  Advancements will have a huge impact Solutions will be widely used  New paradigms Interconnect effects: clock, control, bus, signal Low power designs Wider bit width Reliability centric designs FPGAs and nano technologies

9 CSE 2469 Motivation  Should a new business focus on building market or new technology?  New technology: a market will be built around new technology

10 CSE 24610 Topics  Numbers Binary numbers, negative numbers, redundant numbers, residual numbers  Addition/Subtraction Prefix adders (zero deficiency)  Multiplication/Division  Floating point operations  Functions: (sqrt),log, exp, CORDIC  Optimization, analysis, fault tolerance

11 CSE 24611 Other Topics  Potential focus on the following topics: Power reduction Interconnect FPGAs

12 CSE 24612 Goals/Background  Why do you want to take this class? What would you like to learn? Fulfill course requirement Hardware Software Work Research Curiosity

13 CSE 24613 Agenda  Administration  Motivation  Lecture 1: Numbers

14 CSE 24614 Numbers  Special Symbols Symbols used to represent a value Roman Numerals 1 = I 100 = C 5 = V 500 = D 10 = X1000 = M 50 = L For example: 2004 = MMIV

15 CSE 24615 Numbers  Position Symbols The value depends on the position of the number  For example: 125 = 100 + 20 + 5 One 100, Two 10s, and Five 1s Another example: 1 hour, 3 minutes Positional systems includes radixes: 2, -2, 2, 2j (imaginary)

16 CSE 24616 Numbers  Summation of positional numbers Given: Value is: (where y is the base) For example:  Consider 4 -2 1 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 -2 4 5 2 3   Note that position systems provide a complete range of numbers (e.g. – 2 to 5)

17 CSE 24617 Numbers (Radix Conversion)  Repeatedly divide the integer by (R)r  Repeatedly multiply the fraction by (R)r  Example (201.31) 10 =(13001.123) 5 (x k-1, …, x 0. X -1, …, x -l ) r =(X K-1, …, X 0. X -1, …, X -L ) R

18 CSE 24618 Numbers  Avoid Division (Montgomory System)  Simplify Mod operation mod r-1, mod r+1 Example: 291 10 mod 9 = 2+9+1 mod 9 = 12 mod 9 = 3 291 10 mod 11 = 2-9+1mod 11 = -6 mod 11 = 5

19 CSE 24619 Signed Numbers  Biased numbers  Signed Bit  Complementary representation Positive number: x (mod p) Negative number:(M-x) (mod p) (Note: mod p is added implicitly) One ’ s complement Two ’ s complement Flip each bit Flip each bit + 1 Two ’ s complement can be used for subtraction 0 0 1 0 1 1 0 1 -0 0 0 1 0 1 1 0 1 -2 M=2 n -1 M=2 n

20 CSE 24620 Signed Numbers  Two ’ s complement subtraction: (M-x+M-y) mod M = M-(x+y)  Two ’ s complement conversion: Positive number: To negative:

21 CSE 24621 Signed Numbers  Two ’ s complement 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 2 3 -4 -3 -2 Proof as follows: Which leads to: Example:

22 CSE 24622 Next time  Talk about redundant numbers


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