1. CSE 246: Computer Arithmetic Algorithms and Hardware Design Instructor: Prof. Chung-Kuan Cheng Lecture 4

2. CSE 2462 HW 2 Due 1/27/05  11.2, 11.6, 11.7  Design a number system different from the conventional binary system. Inventing a new number system would be ideal. Show that the number system is better than the binary system in certain aspect/s. Also, design/describe the basic operations such as addition, subtraction, and comparison.

3. CSE 2463 Topics:  Adders Synchronous v.s. Asynchronous AND/OR gate v.s. Circuit Logic angle Graph angle (Prefix Adder)

4. CSE 2464 Prefix Computation  FSM example: Given:  initial state S 0 =A  A sequence of inputs: (0 0 1 1 1 0 1 0 1) Derive the sequence of outputs ABC 0/01/0 0/0 1/0 0/0 1/1 PSNextState X=0X=1 ABA BBC CBA State table BC BB BA X=0 NSPS M0M0 AC CB AA X=1 NSPS M1M1 Input Sequence: 0 1 … Compute N ’ s: N 1 =M 0 N 2 =M 0 M 0 N 3 =M 1 M 0 M 0 N 4 =M 1 M 1 M 0 M 0 … PSNS 12 AB BB CB PSNS 13 AC BC CC PSNS 14 AA BA CA

5. CSE 2465 Graph Based Approach  Consider the (g p) chain break the long paths C4C4 g3g3 g2g2 g1g1 p3p3 p2p2 p1p1 C1C1

6. CSE 2466 Graph Based Approach  Generating g 32 and p 32 C4C4 g3g3 g2g2 g1g1 p3p3 p2p2 p1p1 C1C1 g3g3 p3p3 g2g2 p2p2 g 32 p 32

7. CSE 2467 Graph Based Approach  Generating g 10 and p 10 C4C4 g3g3 g2g2 g1g1 p3p3 p2p2 p1p1 c in g1g1 p1p1 g 10 p 10

8. CSE 2468 Graph Based Approach  Generating g 30 and p 30 g 32 p 32 g 10 p 10 g 30 p 30 g3g3 p3p3 g2g2 p2p2 g 32 p 32 g1g1 p1p1 c in g 10 p 10

9. CSE 2469 Boolean Approach g 4 + p 4 ( g 3 + p 3 ( g 2 + p 2 ( g 1 + p 1 ( g 0 + p 0 c in ) ) ) ) g 4, p 4 g 3, p 3 g 2, p 2 g 1, p 1 g 0, p 0 c in g 4 +p 4 g 3, p 4 p 3 g 2 +p 2 g 1, p 2 p 1 g 0, p 0 c in g 4 +p 4 g 3 +p 4 p 3 (g 2 +p 2 g 1 ), p 4 p 3 p 2 p 1 g 0, p 0 c in g 4 +p 4 g 3 +p 4 p 3 (g 2 +p 2 g 1 )+(p 4 p 3 p 2 p 1 )g 0, (p 4 p 3 p 2 p 1 ) p 0 c in

10. CSE 24610 Prefix Adder  Given: n inputs (g i, p i ) An operation o  Compute: y i = (g i, p i ) o … o (g 1, p 1 ) ( 1 <= i <= n)  Associativity (A o B) o C = A o ( B o C)  (g ’’, p ’’ ) o (g ’, p ’ ) = (g, p)  g=g ’’ + p ’’ g ’  p=p ’’ p ’ gi=pi=gi=pi= a, i=1 a i b i, otherwise 1, i=1 a i xor b i, otherwise

11. CSE 24611 Prefix Adder: Graph Representation  Example: Ripple Carry Adder a i b i (g i, p i ) x y x o y

12. CSE 24612 Prefix Adders: Conditional Sum Adder 8 7 6 5 4 3 2 1

13. CSE 24613 Prefix Adders: Conditional Sum Adder  For output y i, there is an alphabetical tree covering inputs (x i, x i-1, …, x 1 ) 8 7 6 5 4 3 2 1  alphabetical tree:  Binary tree  Edges do not cross

14. CSE 24614 Prefix Adders: Conditional Sum Adder  From input x 1, there is a tree covering all outputs (y i, y i-1, …, y 1 ) 8 7 6 5 4 3 2 1  The nodes in this tree can be reduced to (g, p) o c = g+pc

15. CSE 24615 Prefix Adders: size and depth  Objective: Minimize # of nodes, s c(n). Minimize depth, d c(n)  Ripple Carry Adder: s c(8) = 7 d c(8) = 7 total = 14  Conditional Sum Adder: s c(8) = 12 d c(8) = 3 total = 15

16. CSE 24616 Prefix Adders: size and depth  Theorem:s c(n) +d n c(n) >= s c(n) +d n c(n) >= 2n-2 d n c(n) means the depth of the last output  Proof: Alphabetical tree of y n contains n-1 internal nodes. For each column where the prefix is not ready, at lease one extra node is needed, therefore we need at least n-(d n c(n) +1) extra nodes s c(n) >=n-1+(n – (d n c(n) +1))=2n-2-d n c(n) s c(n) + d n c(n) >= 2n-2

17. CSE 24617 Prefix Adders: Brent – Kung Adder 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 s c(16) = 26 d c(16) = 6 total = 32

18. CSE 24618 Carry Skip Adder A0A0 a 3,0 b 3,0 c in c4c4 0101 p 3,0 A1A1 a 7,4 b 7,4 c8c8 0101 p 7,4 c4c4 A2A2 a 11,8 b 11,8 c 12 0101 p 11,8 c8c8 c 12 If p 3,0 =p 3 p 2 p 1 p 0 = 1, then x = c in x

19. CSE 24619 Carry Propagation Paths  A 2 <- MUX <- MUX <- c in  A 2 <- MUX <- A 1  A 2 <- MUX <- MUX <- A 0  c 12 <- MUX <- A 2  c 12 <- MUX <- MUX <- A 1  c 12 <- MUX <- MUX <- MUX <- A 0  c 12 <- MUX <- MUX <- MUX <- MUX <- c in

20. CSE 24620 False Path  A 1 <- MUX <- A 0 <- c in is a false path If carry is from cin, then block must have p 3 p 2 p 1 p 0 = 1 Since p 3,0 = 1, g 3,0 must be 0 The carry is not generated from A 0 The carry needs not to propagate via A 0, it will go from the MUX

21. CSE 24621 Label Algorithm  Problem: Given a digraph, a set of false paths Derive the longest path of the graph  Algorithm: Color the edges on each false path a label The length of the walk of the same labels are accumulated Otherwise, change to no label