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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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CSE 2462 Agenda Administration Motivation Lecture 1: Numbers
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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
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CSE 2464 Administration No classes on the following days Tu 10/17 BIBE Tu 10/24 EPEP Tu 11/7 ICCAD
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CSE 2465 Administration Grading: Homework – 20% Midterm – 35% Project Report – 25% Presentation – 20% Midterm: Thursday, 10/2/06
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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
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CSE 2467 Agenda Administration Motivation Lecture 1: Numbers
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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
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CSE 2469 Motivation Should a new business focus on building market or new technology? New technology: a market will be built around new technology
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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
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CSE 24611 Other Topics Potential focus on the following topics: Power reduction Interconnect FPGAs
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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
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CSE 24613 Agenda Administration Motivation Lecture 1: Numbers
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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
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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)
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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)
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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
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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
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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
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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:
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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:
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CSE 24622 Next time Talk about redundant numbers
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