1. Instructor: Prof. Chung-Kuan Cheng
CSE246 Adder – Part II
Instructor:
Prof. Chung-Kuan Cheng

2. Zimmerman’s Heuristic Approach
Problem formulation
Given depth constraint, generate a parallel prefix adder of minimum size
Two step Heuristic
Start with a serial prefix adder
Compress to a fastest prefix structure at the cost of increasing size
LSB to MSB, low level to high level
Expand to reduce size, subject to depth constraint
MSB to LSB, high level to low level
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3. Zimmerman’s Heuristic Approach
Local compression/expansion operation
Up/down shift
The compression oper
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4. Zimmerman’s Heuristic Approach
Advantages
Simple and fast
Product depth-size optimal result in many cases
Handles non-uniform input arrival times
Disadvantage
No guarantee on optimality
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5. Prefix Adder with arbitrary input arrival time profile
Non-uniform input arrival times represented in real number
How to construct the fastest prefix adder under arbitrary input arrival time profile?
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6. Cont’ Timing model t[i:j] = max{t[i:k] , t[k-1:j] }+C
All (G,P) generators have the same delay C
Denote the output timing of generator (G,P)[i:j] as t[i:j]
Suppose in the prefix graph, (G,P)[i:j] is generated from (G,P)[j:k] and (G,P)[k-1:j], then
t[i:j] = max{t[i:k] , t[k-1:j] }+C
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7. Dynamic Programming – The idea
Image a full array of partial prefix results
All (G,P) signals of length i are on level i
Rightmost signals are wanted prefix results
Generate all the (G,P) signals row by row, from lower level to higher level
For each (G,P) signal, find the scheme that leads to best timing, i.e., find the partition point k such that
(G,P)[i:j] = (G,P)[i:k] (G,P)[k-1:j]
t[i:j] = min{max{t[i:k] , t[k-1:j] }+C}
t[n:n]
t[n-1:n-1]
t[2:2]
t[1:1]
Level 1:
Level 2:
Level n: … . k
t[n:n-1]
t[2:1]
t[n:n-2]
t[3:1]
t[n:2]
t[n-1:1]
t[n:1]
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8. Dynamic Programming A 5-bit example 2(g4p4) 4(g3p3) 3(g2p2) 1(g1p1)
Level 1 8
7(GP[3,0])
Level 4 8 7
5(GP[2,0])
Level 3 6 5
3(GP[1,0])
Level 2 7 8
8(GP[4,0])
Level 5
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9. Dynamic Programming Complexity Hints for reducing complexity
For (G,P)[i:j], search (i-j) combinations
Overall O(n3)
Hints for reducing complexity
For (G,P)[i:j], there might more than one optimal partition points, but we want just one
At least one optimal partition point of (G,P)[i:j] is bounded by the optimal partition points of (G,P)[i-1:j] and (G,P)[i:j+1]
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10. Backward Reduction I
Some of the partial prefix results are not used, hence can be removed
Level 1
Level 2
Level 3
Level 4
Level 5
(a)
(b)
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11. Backward Reduction II
Some nodes may be over tightened, and can be relaxed to reduce area
3(g4p4)
6(g3p3)
7(g2p2)
11(g1p1) 13
(G,P)[2,1] 9 8 10
(G,P)[3,1]
(G,P)[4,1]
(G,P)[4,2]
(13)
3(g4p4)
6(g3p3)
7(g2p2)
11(g1p1) 13
(G,P)[2,1]
(G,P)[3,1]
(G,P)[4,1]
(G,P)[4,2] 11
(11) 8
() 9
(9) 10
(11) 8
(9) 9 11
(11) 9
(9)
(13)
(11)
(9)
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12. A missing detail (G,P) signals allows overlap  search space increases
However, allowing overlapping does not produce better timing
(G,P)[i:j] = (G,P)[i:k] (G,P)[l:j]
l ≥k
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13. Function level optimization
Carry Skip Adder A0
a3,0 b3,0
cin c4 1
p3,0 A1
a7,4 b7,4 c8
p7,4 A2
a11,8 b11,8
c12
p11,8 x
If p3,0=p3p2p1p0 = 1, then x = cin
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14. False Path A1 <- MUX <- A0 <- cin is a false path
If carry is from cin, then block must have p3p2p1p0 = 1
Since p3,0 = 1, g3,0 must be 0
The carry is not generated from A0
The carry needs not to propagate via A0, it will go from the MUX
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15. False Path: Cycles A3,0 B3,0 Cout Cin Adder S3,0
Cycles of False Paths: Eg. 1’s complement number addition
Positive: x
Negative: (2n-1)-x
Addition
(2n-1)-x + (2n-1)-y = 2n+(2n-1)-(x+y)-1
A3,0
B3,0
Cout
Cin
Adder
S3,0
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16. Example
-3-5 = -8
110110
0+0=0
111110
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17. Multi-Operand Addition
Carry save adder: a (3,2) counter
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18. Example A (3,2) counter compresses X rows to 2/3X rows each time
Tree structure in implementation
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19. Other Counters (7,3) counter (5,3) counterCa Cb S0 S2 S1 S0
Design of (5,3) counter using full adders Ca Cb S0
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