The Synthesis of Combinational Logic
to Generate Probabilities
Weikang Qian
Marc Riedel
Kia Bazargan
David Lilja
Electrical & Computer Engineering
University of Minnesota
A
C
B
Motivation
• Randomness and noise in nanoscale circuits
o Traditional view:
impediment
o A novel view:
resources for probabilistic computation
2
Motivation
• Random sources: Generate probabilities needed
by probabilistic algorithms
Stochastic Bit Streams: Probability 3/6
0, 1, 1, 0, 1, 0
…
Random
Sources
Deterministic
Hardware
1, 1, 0, 1, 0, 1
Probability 4/6
Issue: Costly to generate many different probabilities
directly from random sources!
Solution: Synthesize combinational logic
to generate probabilities!
3
Basic Problem
Other
Probabilities
Needed
Set S of Input
Probabilities
(|S| small)
p1
Random
Sources
p2
p3
Choose Set S ?
Logic
Circuit
?
q1
q2
q3
q4
4
Assumptions
• All signals in the circuit are random Boolean
variables.
• Elements of set S can be used many times.
• All inputs are independent.
Independent
p1
p1
p2
p2
p3
p3
p3
logic
circuit
q
5
Conventions
• “Probability”:
Probability of a signal being logical one
• “Synthesize probability q”:
Synthesize the circuit generating probability q
6
Example
0.4
0.5
P(x
P(x==1)1)==0.4
0.4
0,0,1,0,1,0,1,0,1,0
0,1,0,1,0,0,1,1,0,0
P(x = 1) = 0.4
xxx
y
y
1,0,1,1,0,1,0,0,0,0
1,0,1,1,0,0,1,0,0,1
1,0,0,1,1,0,0,1,1,0
0.2
Logic
Circuit
0.3
0.6
P(z
1)
0.2
P(z===1)
1)===0.3
0.6
P(z
zz z
0,0,0,1,0,0,1,0,0,0
0,1,0,0,1,0,1,1,1,1
0,1,0,0,0,1,0,0,0,1
AND
NOR
P(y= =1)1)==0.5
0.5
P(y
P(z = 1) = P(x = 0)
P(z==1)1)==P(x
P(x==1)0)P(y
P(y==1)0)
P(z
7
Generating Decimal Probabilities
Decimal
Probabilities
Choose Set S = {p1, p2, p3}
p1
p2
p3
Large Cost!
|S| Small!
Logic
Circuit
• Found Set S for
o |S| = 2
o |S| = 1
q1
q2
q3
q4
8
Generating Decimal Probabilities
Theorem 1:
With S = {0.4, 0.5}, we can synthesize arbitrary
decimal output probabilities.
Example: Synthesize q = 0.757
(Black dots
are inverters)
0.4
0.5
0.5
0.4
0.5
0.5
0.5
0.4
0.6
AND
0.7
AND
0.35
0.86
AND
AND
0.43
AND
0.785
AND
0.6075
AND
0.757
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Proof of Theorem 1
• Constrain design to AND gates and inverters
o AND gates: x ∙ y
o Inverters: 1 − x
• Constructive proof
o Induction on the number of digits n
• Base case
0.1 = 0.4 × 0.5 × 0.5, 0.2 = 0.4 × 0.5
0.3 = (1 − 0.4) × 0.5
0.6 / 0.7 / 0.8 / 0.9 = 1 − 0.4 / 0.3 / 0.2 / 0.1
• Inductive step
o Assume true for (n − 1). Show true for n.
10
Proof of Inductive Step
q: n digits; w: (n−1) digits; u = 10nq, numerator of q
0 ≤ q ≤ 0.2, u = 2k
conversions
w = 5q
q = 0.4 × 0.5 × w
0 ≤ q ≤ 0.1, u = 2k+1
w = 10q
q = 0.4×0.5×0.5×w
0.1 < q ≤ 0.2, u = 2k+1
w = 2 − 10q
q = (1−0.5w)×0.4×0.5
0.2 < q ≤ 0.4, u = 4k
w = 2.5q
q = 0.4 × w
cases
q
…
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Algorithm from Theorem 1
Example: Synthesize q = 0.757 from S = {0.4, 0.5}
0.757
1−
0.243 ×0.4 0.6075 1 − 0.3925 ×0.5 0.785 1 − 0.215 ×0.5 0.43
0.4
1−
0.6
×0.5
0.3
1−
0.7
×0.5
×0.4 0.14 1 − 0.86 ×0.5
0.35
Base case
cases
0 ≤ q ≤ 0.2, u = 2k
0 ≤ q ≤ 0.1, u = 2k+1
…
0.2 < q ≤ 0.3, u = 2k+1
0.3 < q ≤ 0.4, u = 2k+1
0.4 < q ≤ 0.5
0.5 < q ≤ 1
conversions
w = 5q
w = 10q
…
w = 10q − 2
w = 4 − 10q
w = 1 − 2q
w = 1− q
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Circuit Synthesized
0.4
0.757
1−
1−
0.6
0.243
(Black dots
are inverters)
×0.5
×0.4
0.3
1−
0.6075
0.4
0.5
0.5
0.4
0.5
0.5
0.5
0.4
0.7
1−
×0.5
×0.4 0.14 1 − 0.86 ×0.5
0.35
0.3925
0.6
AND
×0.5
0.785
1−
×0.5
0.43
0.7
AND
0.35
0.86
AND
AND
0.215
0.43
AND
0.785
AND
0.6075
AND
0.757
13
Generating Decimal Probabilities
Question: Does there exist a set S of size one which
can generate arbitrary decimal probabilities?
p
p
p
combinational
circuit
q
p
Yes!
14
Proof
• There exists a p (≈ 0.1295) in the unit interval such that
• Boolean function
with
, has output probability
• Boolean function
with
, has output probability
• Thus, 0.4 and 0.5 can be generated from p. (Use
Theorem 1.)
15
Implementation
• Focus on generating decimal probability form
S = {0.4, 0.5}.
• Goal: Reduce circuit depth.
0.4
0.5
0.5
0.4
0.5
0.5
0.5
0.4
0.6
AND
0.7
AND
0.35
0.86
AND
AND
0.43
AND
0.785
AND
0.6075
AND
0.757
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Balancing
• Logic Level Optimization: Balancing
Fanin
Cone
Fanin
Cone
AND
AND
a
AND
...
a
b
...
AND
b
Before Balancing
After Balancing
(a and b are primary inputs)
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Factorization of Fractions
• High Level Optimization: Factorization of Fractions
Example: Synthesize q = 0.49 from S = {0.4, 0.5}
Basic
Factor
0.4
0.5
0.5
0.5
0.4
0.5
0.4
AND
AND
0.2
0.25
AND
AND
AND
0.1
AND
AND
0.49
0.7
AND
0.4
0.5
0.5
0.98
0.49
0.7
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Factorization of Fractions
• Quality of factor pair a = al × ar
– Heuristic: Max{Digits(al), Digits (ar)}
– Digits(x) ↑  Depth of the circuit for fraction x ↑
Example: 0.1764 = 0.252×0.7 = 0.63×0.28
0.252
Sub-Ckt 1
AND
Sub-Ckt 1
0.1764
0.63
AND
0.7
Sub-Ckt 2
better!
0.1764
0.28
Sub-Ckt 2
19
Factorization of Fractions
• Consider both q and 1 − q
– Example: q = 0.37, then 1 − q = 0.63 has better factor
pair 0.7×0.9
• When both q and 1 − q cannot be factorized
– Use conversions from Theorem 1 to get w with one
less digit than q.
– Factorize w.
20
Experimental Results
• ‘Basic’: Algorithm based on Theorem 1 + Balancing
• ‘Factor’: Algorithm based on Factorization + Balancing
• Compare number of AND gates and depth of
synthesized circuits
• Present average results for different numbers of digits n
21
Experimental Results
Average number of AND gates vs. number of digits n
22
Experimental Results
Average depth vs. number of digits n
23
Summary
• Considered the problem of generating required decimal
probabilities.
• Gave a pair of probabilities / a single probability for
generating arbitrary decimal probabilities.
• Proposed implementation based on balancing and
factorization of fractions to reduce the depth of the
circuit.
• Also considered synthesis with different assumptions:
– Whether the probabilities are duplicable or not.
– Whether we are free to choose set S.
24
Future Work
• Tradeoff between circuit area and
approximation error
– Synthesize q = 0.251: an AND gate realizing 0.25 but
with error 0.001?
• Correlation on input probabilities
• Accuracy of input probabilities
– p1 = 0.4 becomes 0.4 + δ
25
Thank You!
26
Problem Flavors
• Set S: pre-determined / free to choose
• Elements: duplicable / non-duplicable
Independent
p1
p1
p2
p2
p3
p3
p3
Independent
p1
logic
circuit
q
p2
logic
circuit
q
p3
Duplicable
Non-Duplicable
27
Problem Variations
Generate Decimal
Probability
Set S
Elements
Determined? Duplicable?
Case One
No
Yes
Case Two
Yes
No
Case Three
No
No
Case Four
Yes
Yes
28
Case Two
• Set S given, elements non-duplicable
• Essentially to fill in the output column of the truth
table
• We know the probability of input combination
– e.g., r2 = (1 − p1) p2 (1 − p3)
p1 x1
p x2
2
p3 x3
logic
circuit
unknown known
z q
29
Case Two: Solution
• If z0 = z3 = z6 = 1, the output
probability is po = r0 + r3 + r6
• In general,
• Generally, a target probability
q cannot be realized exactly.
Find least error approximation
Given q, we try to find zi’s to be either 0 or 1
to minimize the objective function
--- solved by linear programming
30
Case Three
• Set S not Given and Elements non-Duplicable
p1 x1
p x2
2
p3 x3
logic
circuit
z q
Question: What is the optimal set S = {p1, p2, p3}?
An optimal set S is
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