CSE 246: Computer Arithmetic Algorithms and Hardware Design

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CSE 246 Computer Arithmetic Algorithms and
Hardware Design
Winter 2004 Lecture 8

• Instructor
• Prof. Chung-Kuan Cheng

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Topics

• Midterm
• Radix-4 SRT Division
• Division by a Constant
• Division by a Repeated Multiplication

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Midterm

• RNS to Decimal conversion
• (234)RNS(765) ?
• ?1 4
• ?2 5
• ?3 3

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Midterm

1. Ones Complement Adder

FA7
FA6
FA5
FA4
FA3
FA2
FA1
FA0
Loop back carryout of FA7 to the carry in of FA0

• The Delay is 8.7 because the maximum carry
  propagation is once cycle

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Midterm

• Prefix Adders
• Ripple-carry Adder

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Midterm

• Prefix Adders
• Prefix Adder

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Project Update

• Come in to speak briefly about the final project
• Status Update
• 4 or 5 630 p.m.
• Tuesday or Thursday

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Radix-4 SRT Division

• 4sj-1 qjd sj where
• qj is in -2,2 and sj-1 is in -hd,hd
• h is less than or equal to 2/3
• Therefore, sj-1 is in -2d/3, 2d/3
• And, 4sj-1 is in -8d/3, 8d/3
• s shifts to the left by 2 bits

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Radix-4 SRT Division
4sj-1
11.0
Anything about 8/3 goes against our assumption
and is therefore the infeasible region
10.1
qj2
10.0
1.1
qj1
1.0
2d/3
0.1
0.0
qj0
d
.1
.101
.110
.111
1.00
2d/3

• The overlap regions of qj denote a choice still
  allowing for recursion

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Radix-4 SRT Division

• The value of qj determines the range it governs
• For example, qj 1
• 1 2/3 5/3
• 1 2/3 1/3
• The range is 1/3 to 5/3

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Division by a Constant

• Multiplication is O(log n) but division is
  linearmuch slower
• Try to convert division to multiplication
• Property Given an odd number d m such that
  dm 2n 1
• Ex.
• d 3, m 5 35 24 1
• d 7, m 9 79 26 1
• d 11, m 93 11 93 210 - 1

E
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Division by a Constant

• 1/d m/(2n 1)
• 1/(1-r) 1rr2r3 (1r)(1r2)(1r3)(1r4)
  
• Example
• z/7 mz/(2n-1)
• log(n/6) operations

m 1
m

(12-n)(12-2n)(12-4n)
2n 1-2-n
2n
z 9
9z

(12-6)(12-12)(12-24)
26 1-2-6
26
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Division by a Repeated Multiplication

• q z/d (z/d)(x0/x0)(x1/x1)(xk-1/xk-1)
• Let xi 2-di
• d1 dxo d(2-d) 1-(1-d)2
• di1 dixi di(2-di) 1-(1-di)2
• 1-di1 (1-di)2 quadratic convergence
• For k-bit operands, we need 2m-1 multiplications
• m 2s complement
• m ceiling(log2 k) with log2 m extra bits for
  precision