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10/23/2017 Comp 411 - Fall 2017
Binary Multiplication
×^0 1 2 3 4 5 6 7 8 9 10000
0 0 0 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 7 8 9 2 0 2 4 6 8 10 12 14 16 18 3 0 3 6 9 12 15 18 21 24 27 4 0 4 8 12 16 20 24 28 32 36 5 0 5 10 15 20 25 30 35 40 45 6 0 6 12 18 24 30 36 42 48 54 7 0 7 14 21 28 35 42 49 56 63 8 0 8 16 24 32 40 48 56 64 72 9 0 9 18 27 36 45 54 63 72 81
× 0 1 0 0 0 1 0 1
You’ve got to be kidding… It can’t be that easy
The key to multiplication was
memorizing a digit-by-digit table…
Everything else was just adding
10/23/2017 Comp 411 - Fall 2017
Digit by digit = bit by bit
A A 2 A 1 A 0
3 B B 2 B 1 B 0 3 A A 2 B 0 A 1 B 0 A 0 B 0 3 B 0 A 3 B 1 A 2 B 1 A 1 B 1 A 0 B 1 A 3 B 2 A 2 B 2 A 1 B 2 A 0 B 2
x
+ A 3 B 3 A 2 B 3 A 1 B 3 A 0 B 3
A (^) j B (^) i is a “partial product”
Multiplying N-digit number by M-digit number gives (N+M)-digit result
Easy part: forming partial products (just an AND gate since B (^) I is either 0 or 1) Hard part: adding M, N-bit partial products
X 0 1
The “Binary” Multiplication Table
Hey, that looks like an AND gate
Binary multiplication is implemented using the same basic longhand algorithm that you learned in grade school.
10/23/2017 Comp 411 - Fall 2017
Multiplier Unit-Block
We introduce a new abstraction to aid in the construction of multipliers called the “Unsigned Multiplier Unit-block”
We did a similar thing last lecture when we converted our adder to an add/subtract unit.
A (^) k are bits of the Multiplicand and B (^) i are bits of the Multiplier.
The P (^) i,k inputs and outputs represent “partial products” which are partial results from adding together shifted instances of the Multiplicand.
The initial P (^) 0,k is zero.
A B CO CI S
FA
A i
B i
Subtract
C i
C (^) i-
S (^) i
Add/Subtract Unit Block
A B CO CI S
FA
p (^) i-1,k
A (^) k B i
C (^) k C (^) k-
p (^) i,k
Unsigned Multiply Unit Block
10/23/2017 Comp 411 - Fall 2017
Simple Combinational Multiplier
tPD = 10 * t (^) PD
not 16
NB: this circuit only works for nonnegative operands
Components N * HA N(N-1) * FA
CoB HA A S
CoB HA A S
CoB HA A S
CoB HA A S
tPD = (2*(N-1) + N) * t (^) PD
To determine the timing specification of a composite combinational circuit we find the worst-case path for every output to any input.
Is this faster than our assembly code?
10/23/2017 Comp 411 - Fall 2017
Higher-Radix Multiplication
B (^) K+1,K A = 0A = 1A = 2A Just a shift = 3*A Requires adding
A (^) N-1 A (^) N-2 … A 3 A 2 A 1 A (^0) x B^ M-1^ B^ M-2^ …^ B^3 B^2 B^1 B^0
M/2 2
Idea: If we could use, say, 2 bits of the multiplier in generating each partial product we would halve the number of rows and halve the latency of the multiplier!
B (^) K+1,K A = 0A ⇨ 0 = 1A ⇨ A = 2A ⇨ 2A or 4A – 2A = 3*A ⇨ 4A – A
Booth’s insight: rewrite 2A and 3A cases, leave 4A for next partial product to do!
10/23/2017 Comp 411 - Fall 2017
Booth Recoding of Multiplier
B 2K+
B 2K
B 2K-
action
add 0
add A
add A
add 2*A
sub 2*A
sub A
sub A
add 0
A “1” in this bit means the previous stage needed to add 4A. Since this stage is shifted by 2 bits with respect to the previous stage, adding 4A in the previous stage is like adding A in this stage!
-2*A+A
-A+A
current bit pair from previous bit pair
An encoding where each bit has the following weights:
W(B (^) 2K+1) = -2 * 2 2K W(B (^) 2K) = 1 * 2 2K W(B (^) 2K-1) = 1 * 2 2K
Hey, isn’t that a negative number?
Yep! Booth recoding works for 2-Complement integers, now we can build a signed multiplier.
10/23/2017 Comp 411 - Fall 2017
Bigger Multipliers
- Using the approaches described we can construct multipliers of arbitrary sizes, by considering every adder at the “bit” level
- We can also, build bigger multipliers using smaller ones
- Considering this problem at a higher-level leads to more “non-obvious” optimizations
×
A 4
B 4
4 P (^) H I
4 P (^) L O
10/23/2017 Comp 411 - Fall 2017
Can We Multiply With Less?
- How many operations are needed to
multiply 2, 2-digit numbers?
4 Adders
- This technique generalizes
- You can build an 8-bit multiplier using 4 4-bit multipliers and 4 8-bit adders
- O(N 2 + N) = O(N 2 )
AB
x CD
DB
DA
CB
CA
10/23/2017 Comp 411 - Fall 2017
A Trick
● The two middle partial products can be computed using a single multiplier and other partial products ● DA + CB = (C + D)(A + B) – (CA + DB) ● 3 multipliers 8 adders ● This can be applied recursively (i.e. applied within each partial product) ● Leads to O(N 1.58^ ) adders ● This trick is becoming more popular as N grows. However, it is less regular, and the overhead of the extra adders is high for small N
AB
x CD
DB
DA
CB
CA
10/23/2017 Comp 411 - Fall 2017
Let’s Try it By Hand
1) Choose 2, 2 digit numbers to multiply: ab × cd
2) Multiply digits: p1 = a x c, p2 = b x d, p3 = (c + d)(a + b)
3) Compute partial subtracted sum, SS = p3 - (p1 + p2)
4) Add as follows: p = 100 x p1 + 10 x SS + p
42 x 37 =?
42 x 37
p1 = 4 x 3 = 12, p2 = 2*7 = 14, p3 = (4+2)x(3+7) = 60
SS = 60 - (12 + 14) = 34
P = 1200 + 340 + 14 = 1554 = 42 x 37
10/23/2017 Comp 411 - Fall 2017
Next time
● We dive into floating-point arithmetic
