1. SAT-Based Logic Synthesis
Alan Mishchenko
Department of EECS, UC Berkeley

2. Overview Logic synthesis, BDDs, and SAT
Canonical SAT-based computation
SAT-based formulations in logic synthesis
ISOP computation, Boolean decomposition, Boolean resubstitution, ECO
Agile SAT solving
Conclusions

3. Terminology
When a Boolean function is given in some form, logic synthesis derives a circuit for this function, while minimizing some cost (area/delay/etc)
Boolean satisfiability (SAT) is a problem of checking whether a Boolean function, represented in a non-canonical way using CNF or a circuit, is constant 0
Binary decision diagrams (BDDs) is a canonical graph-based representation of Boolean functions

4. A Recent Trend
Both BDDs and SAT are seen as computation engines used to solve practical problems
in particular, problems arising in logic synthesis
In recent years, traditional solutions based on Sums-of-Products (SOPs) and Binary Decision Diagrams (BDDs) are being gradually replaced by SAT-based ones
both quality and runtime is often improved
This presentation presents some of the latest SAT-based solutions

5. BDDs vs SAT Complementary data structures: BDDs vs SAT BDDs
trading space for time
BDDs
canonical
easy to use but hard to construct (can mem out)
SAT (CNF-based or circuit-based)
non-canonical
easy to construct but hard to use (can time out)
However, SAT is “better” in most applications
Moreover, we can make SAT computations canonical!

6. What Is Canonicity? Canonical representation Canonical computation
Representing Boolean functions in a unique way
for a given Boolean function with a given variable order, only one representation is possible
Canonical computation
Computing Boolean functions in a unique way
for a given Boolean function with a given variable order, only one result of computation is possible
In the BDD world, we did not distinguish the two
In the SAT world, we cannot have both
However, we can have canonical computation!

7. Why Canonicity Is Useful?
As in BDDs, canonicity in SAT means that a computed result (counter-example or UNSAT core) are the same
for any original circuit (if the function is given as a circuit)
for any CNF generation algorithm
for any SAT solver
for any operating system
etc
Moreover, canonicity implies minimality in some sense
the minimum number of care literals in a cube (prime cube)
the minimum support of a node (irredundant multi-level network)
A. Mishchenko, R. Brayton, A. Petkovska, M. Soeken, L. Amaru, and A. Domic, "Canonical computation without canonical representation", Proc. DAC'18.

8. Boolean Satisfiability (SAT)
Boolean variable can take value 0 or 1 (for example, a)
Literal is Boolean variable with its value (for example, a)
Clause is a disjunction of literals: a + b
CNF is a conjunction of clauses: (a + b) & (a + c)
Given CNF of a Boolean function, Boolean satisfiability checks whether it is always const 0 (‘unsat’) or not (‘sat’)
Example of a satisfiabile problem:
(a + b) & (a + c) one solution is a = 1, b = 1, c = 1
Example of an unsatisfiable problem:
(a + b) & (a + b) & (a + b) & (a + b) no solution exists

9. SAT in Practical Applications
Netlist
Answer: “SAT” or “UNSAT”
CNF
SAT solver
CNF generator
CNF
Design constraints
If SAT, a counter-example
If UNSAT, a core
User cost functions
Both counter-examples and cores are useful in SAT-based applications.
In practice, cores are often represented as subsets of assumptions that make the problem UNSAT.

10. Incremental SAT
Initial CNF
Round 1:
SAT solver
Initial assumptions
Additional CNF
Round 2:
SAT solver
New assumptions
Additional CNF
Round 3:
SAT solver
New assumptions
Assumptions are CNF clauses used only in the current round. They are handled differently from the rest of CNF clauses.

11. Background: Craig Interpolation
Given Boolean functions φ0 and φ1 such that φ0 ∧φ1 is unsatisfiable
Then, there exists Boolean function P such that
φ1 → P
P ∧ φ0 is unsatisfiable
P refers only to the common variables of φ0 and φ1 P
What is craig interpolation?
Given φA and φB are Boolean formulas, if their conjunction is unsatisfiable, then there exists another Boolean formula P with the following properties
First, φA implies P .
That means the solution space of P contain the solution space of φA.
Secondly, The conjunction of P and φB is unsatisfiable.
It means the solution space of P is disjoint from The solution space of φB.
Finally, P refers only to the common variables of φA and φB φ0 φ1 11

12. ISOP Computation
ISOP is a Sum-of-Products (SOP) that is prime and irredundant
F = ab + cd is an ISOP
G= ab + a is not an ISOP
BDD-based ISOP computation is often used
to transform functions into circuits by algebraic factoring
to improve an existing circuit by deriving a local ISOP and generating a new circuit by factoring the ISOP
S. Minato, “Fast generation of Irredundant Sum-Of-Products forms from Binary Decision Diagrams”, Proc. SASIMI’92.

13. SAT-based ISOP Computation
UNSAT proof
(finished computation)
SAT assignment
SAT solver S1 (ON-set)
SAT solver S2 (OFF-set)
(one minterm)
F(x)
F(x)
one prime cube
blocking clause
CNF
CNF
(LEXUNSAT)
ON-set
OFF-set
ISOP generator X
A. Petkovska, A. Mishchenko, D. Novo, M. Owaida, and P. Ienne, "Progressive generation of canonical sums of products using a SAT solver", "Advanced Logic Synthesis" (ed. A. Reis), Springer, 2017.

14. Boolean Decomposition
Algebraic decomposition applies factoring to an “algebraic” expression of the function, its ISOP
this does not exploit the “Boolean nature” of the function
Boolean decomposition uses functional properties to check feasibility and derive the component functions
Two popular types of Boolean decomposition
Ashenhurst decomposition
Bi-decomposition
The SAT-based formulation is proposed in:
H.-P. Lin, J.-H. R. Jiang, R.-R. Lee, “To SAT or not to SAT: Ashenhurst decomposition in a large scale”, Proc. ICCAD’08, pp
R.-R. Lee, J.-H. R. Jiang, W.-L. Hung, “Bi-decomposing large Boolean functions via interpolation and satisfiability solving”, Proc. DAC’08, pp

15. Ashenhurst Decompositionf
A-decomp g Xh Xc Xg
What is bi-decomposition?
Given a large function whose primary inputs are variable sets xa,xb,xc,
We bi-decompose f function into 2-input function h based on other functions fa, fb.
Xa is local to fa, Xb is local to fb, and Xc is the common variable of fa and fb. Xh Xc Xg
Q1: For a given {Xh, Xc, Xg}, can f be decomposed?
Q2: How to derive g and h?
Q2: What is {Xh, Xc, Xg} that minimized the cost of g and h? 15

16. f A h f f B Bi-Decomposition B-decomp XA XB XC XA XC XB
What is bi-decomposition?
Given a large function whose primary inputs are variable sets xa,xb,xc,
We bi-decompose f function into 2-input function h based on other functions fa, fb.
Xa is local to fa, Xb is local to fb, and Xc is the common variable of fa and fb. XA XC XB
Q1: For s given {XA, XC, XB}, can f be decomposed?
Q2: How to derive fA and fB?
Q2: What is {XA, XC, XB} that minimized the cost of fA and fB? 16

17. A-Decomposability Condition
When f(X) can be decomposed as f(X) = h(XH,XC,g(XG,XC))?
If and only if, in the decomposition chart of f(X) with XG on top, there are no more than two different columns, f1(XH) and f2(XH)
For example:
f(00) = XH1 & XH2
f(01) = 1
f(10) = XH1 & XH2
f(11) = XH1 & XH2
XH\XG 00 01 11 10 1

18. A-Decomposability Condition
Functions g and h exists iff
(f(XH1, XG1, XC)≠ f(XH1, XG2, XC))Λ
(f(XH2, XG2, XC)≠ f(XH2, XG3, XC))Λ
(f(XH3, XG3, XC)≠ f(XH3, XG1, XC)) is UNSAT
xg1[[XG1]]
xg2[[XG2]]
xg3[[XG3]]
xh1[[XH1]] 1
xh2[[XH2]]
xh3[[XH3]] 18

19. SAT-Based Formulation
Functions g and h exists iff
(f(XH1, XG1, XC)≠ f(XH1, XG2, XC))Λ
(f(XH2, XG2, XC)≠ f(XH2, XG3, XC))Λ
(f(XH3, XG3, XC)≠ f(XH3, XG1, XC)) is UNSAT f
XH1
XG1 XC
XG2
XH2
XG3
XH3 1 19

20. Computing Functions G and H
Apply interpolation to functions φ0 and φ1
The interpolant g = G(XG1=0, XG2, XC) is the decomposition function
Function h = H(XH, XC) is computed by resubstitution of g into f φ1 φ0 f
XH1
XG1 XC
XG2
XH2
XG3
XH3 1 20

21. SAT-Based Variable Partitioning
For each variable xi  X, add
conditional clauses for XG: ((xi2≡xi3)Λ(xi4≡xi5)Λ(xi6≡xi1)) V ai
conditional clauses for XH: ((xi1≡xi2)Λ(xi3≡xi4)Λ(xi5≡xi6)) V bi
(ai,bi)
(0,0) xi  XC
(0,1) xi  XG
(1,0) xi  XH
(1,1) xi  XH or XG f
XH1
XG1 XC
XG2
XH2
XG3
XH3 1 21

22. Practical Considerations
Positive literals of control variables ai, bi for each xi are used as assumptions in the SAT calls
To avoid trivial partitions, we assume a seed partition
Force one variable in XH and one variable in XG
Minimize assumption after the UNSAT call to find a variable partition that maximizes the number of shared variables, XC, and/or other cost functions
Can be done by LEXUNSAT (a canonical SAT computation)
Enumerate seed partitions and choose the best one
Requires incremental SAT solving 22

23. OR-B-Decomposability Example
When f(X) can be decomposed as f(X) = fA(XA)∨fB(XB) ?
If and only if, for every 1-entry in the decomposition chart, no 0-entries appear simultaneously in the same row and column
f(1101) = 0 = fA(11) +fB(01)
f(0010) = 0 = fA(00) +fB(10)
f(1110) = 1 = fA(11) +fB(10)???
XB\XA 00 01 11 10 1
fB(XB) 1
fA(XA) 1
XB\XA 00 01 11 10 1
fB(XB)
If the condition is satisfied, there exist fa,fb such that f equals fa or fb
If there is a 0-entry, we must assign 0 to this corresponding row and column
For example, if this entry (00,01) is 0, fa(00) must be 0, fb (01) must be 0.
If the column or row contains all 1’s, it must be assigned 1.
If the condition is violated, there does not exit fa,fb
fA(XA) 23

24. OR-B-Decomposability Condition
∃fA, fB such that f(X) = fA(XA) ∨ fB(XB)
⇔ For every 1-entry, no 0-entries can appear simultaneously in the corresponding row and column
⇔ f(XA,XB) ∧ ¬f(XA’,XB) ∧ ¬f(XA,XB’) is unsatisfiable
XA’ XA
XB\XA 00 01 11 10 1
fB(XB) ?
XB’
We translate the necessary and sufficient condition into Boolean formula.
By doing so, we can use SAT solver to check if the formula is unsatisfiable. XB
fA(XA) ? 24

25. SAT-Based Formulation
f(XA,XB,XC) ∧ ¬f(XA’ ,XB,XC) ∧¬f(XA,XB’ ,XC) is UNSAT 1 f f f
According to the formula, we construct the corresponding network like this.
In order to take advantage of craig interpolation to derive fa, let this part be φA formula, the other part be B formula
If we project the feasible solution of A into the common var ,the image will correspond to the onest of fa
Because their conjunction is empty, there exists an interpolant.
The interpolant refers only to the common variables Xa, Xc
It contains the onset of fa and is disjoint from the offset of fa, so it is a legal implementation of fa
3 conditions
XA’
XB’ XA XB XC 25

26. SAT-Based Computation of fA
f(XA,XB,XC) ∧ ¬f(XA’ ,XB,XC) ∧¬f(XA,XB’ ,XC) is UNSAT 1 φ0 φ1 f f f
According to the formula, we construct the corresponding network like this.
In order to take advantage of craig interpolation to derive fa, let this part be φA formula, the other part be B formula
If we project the feasible solution of A into the common var ,the image will correspond to the onest of fa
Because their conjunction is empty, there exists an interpolant.
The interpolant refers only to the common variables Xa, Xc
It contains the onset of fa and is disjoint from the offset of fa, so it is a legal implementation of fa
3 conditions
XA’
XB’ XA XB XC 26

27. SAT-Based Computation of fB
f(XA,XB,XC) ∧ ¬fA(XB,XC) ∧¬f(XA’,XB,XC) is UNSAT 1 φ0 φ1 f fA f
According to the formula, we construct the corresponding network like this.
In order to take advantage of craig interpolation to derive fa, let this part be φA formula, the other part be B formula
If we project the feasible solution of A into the common var ,the image will correspond to the onest of fa
Because their conjunction is empty, there exists an interpolant.
The interpolant refers only to the common variables Xa, Xc
It contains the onset of fa and is disjoint from the offset of fa, so it is a legal implementation of fa
3 conditions XA XB XC
XA’ 27

28. SAT-Based Variable Partitioning
For each xi  X, add
conditional clauses for XA : φ1 & [(x ≡ x1) V ai ]
conditional clauses for XB : φ0 & [(x ≡ x2) V bi ]
(ai,bi)
(0,0) xi  XC
(0,1) xi  XA
(1,0) xi  XB
(1,1) xi  XA or XB φ1 f XA XB
XA’
XB’ XC 1 φ0
We instantiate variable X into X’, X”
Xa connects first and third copy of f
Xc connects all the three copies of f
So for each var x , we use 2 control variables to represent this three conditions.
If (αx,βx) is (0,1) , it means x belongs to Xb
If (αx,βx) is (0,0) , it means x belongs to Xc
This table summaries the above condition 28

29. Boolean Resubstitution
Given function F(x) and divisors g1(x), g2(x), etc, re-express F in terms of g1, g2, etc
F(X)
F(X)
H(g) g1 g2 g3 g1 g2 g3 X X
C.-C. Lee, J.-H. R. Jiang, C.-Y. Huang, and A. Mishchenko. "Scalable exploration of functional dependency by interpolation and incremental SAT solving", Proc. ICCAD '07.

30. SAT-Based Support Computation in Boolean Resubstitution
candidate support
CNF
SAT solver
CNF
OFF-set 1 1
SAT
(invalid support)
UNSAT
(valid support) =
F(X1)
F(X2) g1 g2 g3 g1 g2 g3
Support generator
ON-set X1 X2
Alternatively, support after resubstitution can be found using LEXUNSAT

31. Engineering Change Orders (ECO)
Given networks F(x) and S(x), find target n(x) and divisors g1(x), g2(x), etc, such that n(x) can be reexpressed in terms of g1, g2, etc, so that the updated outputs of F(x) are equivalent to those of S(x) X g1 g3 g2
F(X)
n(X) X
S(X) X g1 g3 g2
n(g)
F(X) becomes S(X)

32. ECO in Terms of Primary Inputs
F(n, x)  S(x) …
Outputs
Implementation F(n, x)
Specification S(x) n
Node
Inputs x
Boolean space
Resubstitution / ECO of n in terms of x exists iff
I(x)
x n F(n, x) == S(x) is true, that is
x n F(n, x)  S(x) is false, that is
F(0, x)  S(x)
F(1, x)  S(x)
x F(0, x)  S(x) & F(1, x)  S(x) is UNSAT

33. ECO in Terms of Intermediate Divisors
F(n, x)  S(x) …
Outputs
Implementation F(n, x)
Specification S(x) n
Node
Divisors d
Inputs x 1
M2 = 1
Consider M(n, x) = [F(n, x)  S(x)] & [d == D(x)]
M1 = 1
M1(x1) =
M2(x2)
Consider M1(x1) = M(0, x1) and M1(x2) = M(1, x2)
Resubstitution / ECO of n in terms of d exists iff d1 d2
x1x2 [M1(x1) & M2(x2) & [d1 == d2]] is UNSAT x1 x2
A. Q. Dao, N.-Z. Lee, L.-C. Chen, M. P.-H. Lin, J.-H. R. Jiang, A. Mishchenko, and R. Brayton, "Efficient computation of ECO patch functions", Proc. DAC'18.

34. Handling Multiple Targets
Often ECO cannot be solved by considering only one target, leading on a multi-target problem
In this case, quantify all targets n, except one, in
x n F(n, x)  S(x)
The problem is thus reduced to 1-target
Once this target is solved, we substitute the solution into the miter, and solve the next target, until all targets are solved
Theorem: This method always finds a solution of the multi-target ECO problem, if it exists

35. Using Don’t-Cares
This formulation of resubstitution/ECO exploits internal don’t-cares of the node in the network (can also use external don’t-cares)
Moreover, the don’t-cares do not have to be computed explicitly before they are used
Instead, the circuit-based constraint is added to the SAT solver, resulting in a solution that is correct under the don’t-cares
A. Mishchenko and R. Brayton, "SAT-based complete don't-care computation for network optimization", Proc. DATE '05, pp

36. Agile SAT Solving
Use simulation and circuit structure analysis whenever possible to avoid unnecessary SAT calls
Simplify SAT-based formulations
avoid QBF if regular SAT can be used
experiment with different encodings
try to reduce the number of variables and clauses
reduce complexity of the constraints, even if it takes more variables
counter-intuitively, sometimes larger CNFs lead to shorter runtimes
Create formulations resulting in fast incremental runs
in most practical applications in the area of logic synthesis, a well-tuned formulation leads to SAT runtimes not exceeding 1 sec per call (in fact, it is a fraction of a second on average)
If constraints are complex, use lazy constraint addition
Whenever possible, give initial solution to the SAT solver

37. Conclusion Reviewed logic synthesis and Boolean satisfiability
Discussed a number of problems in logic synthesis that can be solved using SAT
ISOP computation, Boolean decomposition, resubstitution, and ECO
Shared practical advice on developing efficient SAT based solutions

38. Abstract
This presentation focuses on the use of Boolean satisfiability as a computation engine in solving typical problems arising in logic synthesis. We begin by presenting a SAT-based computation of irredundant sums-of-products (ISOPs) from the circuit structure. Next, we discuss a SAT-based formulation of several types of Boolean decomposition, resubstitution, and Engineering Change Orders (ECO). An advice is given on the efficient use of SAT solvers in a variety of practical settings.