1. Synthesis and Verification of Finite State Machines
Sungho Kang
Yonsei University

2. Outline Minimization of Incompletely Specified Machines
Binate Covering Problem
State Encoding
Decomposition and Encoding

3. Synthesis of Practical FSMs
Minimization
We have learned basic methods for minimizing, encoding, checking equivalence, and synthesizing circuits for realizing completely specified FSMs
Now we must learn to deal with the more practical case of incomplete specification
Our goal is thus to find a least cost circuit that satisfies a partial specification

4. Flow Table Compatibility Table
Incomplete specification and Compatibility
MInimization
Use don’t-cares to merge states.
Merged states must have same output sequences.
Flow Table Compatibility Table
Note each constraint represents pair
(incompatibility)

5. Strategy Derive all prime sets of compatible states
MInimization
Derive all prime sets of compatible states
Solve a covering problem to obtain minimum states.

6. Note each constraint represents pair (incompatibility)
Compatibility Constraints
MInimization
Compatibility relation: conjunction of constraints
(one for each “X”)
Note each constraint represents pair (incompatibility)

7. Computing the Maximal Compatibles
MInimization
By recursive multiplication method,
like computing the Complete Sum:
The (complete) constraint sums are multiplied out,
dropping absorbed terms when they arise.

8. Maximal compatibles are “Prime”.
Computing the Maximal Compatibles
MInimization
Maximal compatibles are “Prime”.
( No superset of these state sets are also pairwise compatible).

9. Prime Compatibles
MInimization
Unfortunately, some subsets of the maximal compatibles pairs are also prime compatibles.
Because, selection of one compatible pair may imply selection of other compatible pairs.

10. Defining Prime Compatibles
MInimization
A compatible is prime if and only if there is no
other compatible which contains it or whose class set contains class set of That is, is prime if and only if
Subsets with smaller class sets are acceptable.
(Bigger compatible,
smaller class set)

11. Class Sets and Prime Compatibles
MInimization
In minimization, we desire a minimum number of compatible sets that cover all original states. Pick from primes.
Choice of conditionally compatible set implies
choosing all implied pairs.
Set of implied compatibles pairs is called the class set, is the class set of

12. We just derived maximal compatibles that are prime
Update and Strategy
MInimization
We just derived maximal compatibles that are prime
Derive remaining prime compatibles
Solve a covering problem

13. Class Sets x1 x2 x3 x4 x5 x6 x7 a a,0 -- d,0 e,1 b,0 a,-- --
MInimization
x1 x2 x3 x4 x5 x6 x7
a a, d,0 e,1 b,0 a,
b b,0 d,1 a, a,-- a,
c b,0 d,1 a, g,0
d e, b,-- b, a,--
e b,-- e,-- a, b,-- e,-- a,1
f b,0 c, ,1 h,1 f,1 g,0 --
g c, e, g,0 f,0
h a,1 e,0 d,1 b,0 b,-- e,-- a,1
b a,d
c X ~
d b,e a,b d,e d,e a,g
e a,b a,d d,e a,b a,e X ~
f X X c,d X X
g ~ X c,d f,g X X e,h
h X X X ~ a,b a,d X X
a b c d e f g

14. Note is prime: although
Class Sets and Primes
MInimization
Note is prime: although
b a,d
c X ~
d b,e a,b d,e d,e a,g
e a,b a,d d,e a,b a,e X ~
f X X c,d X X
g ~ X c,d f,g X X e,h
h X X X ~ a,b a,d X X
a b c d e f g

15. Class Sets and Primes Note is not prime: b a,d c X ~
MInimization
Note is not prime:
b a,d
c X ~
d b,e a,b d,e d,e a,g
e a,b a,d d,e a,b a,e X ~
f X X c,d X X
g ~ X c,d f,g X X e,h
h X X X ~ a,b a,d X X
a b c d e f g

16. Note is prime: Class Sets b a,d c X ~ d b,e a,b d,e d,e a,g
MInimization
Note is prime:
b a,d
c X ~
d b,e a,b d,e d,e a,g
e a,b a,d d,e a,b a,e X ~
f X X c,d X X
g ~ X c,d f,g X X e,h
h X X X ~ a,b a,d X X
a b c d e f g

17. through are added to the list of prime compatibles
Maximal compatibles are prime
MInimization
maximal class
compatibles set
1 {a,b,d,e} {}
2 {b,c,d} {{a,b},{a,g},{d,e}}
3 {c,f,g} {{c,d}, {e,h}}
4 {d,e,h} {{a,b}, {a,d}}
11 {a,g} {}
other PCs
5 {b,c} {}
6 {c,d} {{a,g}, {d,e}}
7 {c,f} {{c,d}}
8 {c,g} {{c,d}, {f,g}}
9 {f,g} {{e,h}}
10 {d,h} {}
12 {f} {}
Note sub-compatibles
through are added to the
list of prime compatibles
before maximal compatible

18. which has an empty class set
Maximal compatibles are prime
MInimization
maximal class
compatible set
1 {a,b,d,e} {}
2 {b,c,d} {a,b},{a,g},{d,e}}
3 {c,f,g} {{c,d}, {e,h}}
4 {d,e,h} {{a,b}, {a,d}}
11 {a,g} {}
other PCs
5 {b,c} {}
6 {c,d} {{a,g}, {d,e}}
7 {c,f} {{c,d}}
8 {c,g} {{c,d}, {f,g}}
9 {f,g} {{e,h}}
10 {d,h} {}
12 {f} {}
Note that subsets and
are not prime because
they are contained in ,
which has an empty class set

19. prime because it is contained in , whose class set is the same.
Maximal compatibles are prime
MInimization
maximal class
compatible set
1 {a,b,d,e} {}
2 {b,c,d} {a,b},{a,g},{d,e}}
3 {c,f,g} {{c,d}, {e,h}}
4 {d,e,h} {{a,b}, {a,d}}
11 {a,g} {}
other PCs
5 {b,c} {}
6 {c,d} {{a,g}, {d,e}}
7 {c,f} {{c,d}}
8 {c,g} {{c,d}, {f,g}}
9 {f,g} {{e,h}}
10 {d,h} {}
12 {f} {}
Note that subset , with
class set , is not
prime because it is contained
in , whose class set
is the same.

20. are all contained in primes with empty class sets, so they
Maximal compatibles are prime
MInimization
maximal class
compatible set
1 {a,b,d,e} {}
2 {b,c,d} {a,b},{a,g},{d,e}}
3 {c,f,g} {{c,d}, {e,h}}
4 {d,e,h} {{a,b}, {a,d}}
11 {a,g} {}
other PCs
5 {b,c} {}
6 {c,d} {{a,g}, {d,e}}
7 {c,f} {{c,d}}
8 {c,g} {{c,d}, {f,g}}
9 {f,g} {{e,h}}
10 {d,h} {}
12 {f} {}
After treating subsets of size
2, we still have to check all
subsets of size 1, which have
empty class sets.
Note
are all contained in primes
with empty class sets, so they
are not prime.
But is not, so it is prime.

21. Goal: To Find Prime Compatibles
MInimization
maximal class
compatible set
1 {a,b,d,e} {}
2 {b,c,d} {a,b},{a,g},{d,e}}
3 {c,f,g} {{c,d}, {e,h}}
4 {d,e,h} {{a,b}, {a,d}}
11 {a,g} {}
other PCs
5 {b,c} {}
6 {c,d} {{a,g}, {d,e}}
7 {c,f} {{c,d}}
8 {c,g} {{c,d}, {f,g}}
9 {f,g} {{e,h}}
10 {d,h} {}
12 {f} {}
Maximal Compatibles are
prime.
Other prime compatibles are
subsets of primes such that:
is prime iff its class set
does not contain the class
set of a larger prime

22. Finding Prime Compatibles
MInimization
Enqueue known
primes
of size k
Test subcompatibles
for primality

23. For , only is enqueued For , are enqueued Finding Prime Compatibles
MInimization
For each value of k, the
for-loop of Line 1 puts
the maximal compatibles
of size k onto the queue of
primes, P.
For , only is enqueued
For , are enqueued

24. is a prime compatible if and only if
Finding Prime Compatibles
MInimization
For each enqueued
prime (of size ),
we check every subset
of size
is a prime compatible if and only if

25. Building the Reduced Machine
MInimization
x1 x2 x3 x4 x x6 x7
a a, d,0 e,1 b,0 a,
b b,0 d,1 a, a, a,
c b,0 d,1 a, g,0
d e, b,-- b, a,--
e b,-- e,-- a, b, e,-- a,1
f b,0 c, ,1 h,1 f, g,
g c, e, g,0 f,0
h a,1 e,0 d,1 b,0 b,-- e, a,1 x 1 2 3 4 5 6 7 9 , –

26. x 1 2 3 4 5 6 7 9 , x1 x2 x3 x4 x5 x6 x7 a a,0 -- d,0 e,1 b,0 a,-- --
Reduced Machine
MInimization
x1 x2 x3 x4 x x6 x7
a a, d,0 e,1 b,0 a,
b b,0 d,1 a, a, a,
c b,0 d,1 a, g,0
d e, b,-- b, a,--
e b,-- e,-- a, b, e,-- a,1
f b,0 c, ,1 h,1 f, g,
g c, e, g,0 f,0
h a,1 e,0 d,1 b,0 b,-- e, a,1
Where there is a
choice, choose 1
(as in x2-successor
of compatible 1):
{d,e} contained in
or . x 1 2 3 4 5 6 7 9 , –

27. Closed Cover : Choosing Compatibles
MInimization
Closed Cover : Choosing Compatibles
Every state of the original machine must be covered
Every implied compatible must be present in the solution

28. Coverage: Closure: Closed Cover maximal class
MInimization
maximal class
compatibles set
1 {a,b,d,e} {}
2 {b,c,d} {{a,b},{a,g},{d,e}}
3 {c,f,g} {{c,d}, {e,h}}
4 {d,e,h} {{a,b}, {a,d}}
11 {a,g} {}
other PCs
5 {b,c} {}
6 {c,d} {{a,g}, {d,e}}
7 {c,f} {{c,d}}
8 {c,g} {{c,d}, {f,g}}
9 {f,g} {{e,h}}
10 {d,h} {}
12 {f} {}
Let’s check if the following
set of compatibles forms a
closed cover:
Coverage:
Closure:

29. Covering Constraints--POS FORM
MInimization
maximal class
compatibles set
1 {a,b,d,e} {}
2 {b,c,d} {{a,b},{a,g},{d,e}}
3 {c,f,g} {{c,d}, {e,h}}
4 {d,e,h} {{a,b}, {a,d}}
11 {a,g} {}
other PCs
5 {b,c} {}
6 {c,d} {{a,g}, {d,e}}
7 {c,f} {{c,d}}
8 {c,g} {{c,d}, {f,g}}
9 {f,g} {{e,h}}
10 {d,h} {}
12 {f} {}
Every state of the original machine must be covered.

30. Covering Constraints--POS FORM
MInimization
maximal class
compatibles set
1 {a,b,d,e} {}
2 {b,c,d} {{a,b},{a,g},{d,e}}
3 {c,f,g} {{c,d}, {e,h}}
4 {d,e,h} {{a,b}, {a,d}}
11 {a,g} {}
other PCs
5 {b,c} {}
6 {c,d} {{a,g}, {d,e}}
7 {c,f} {{c,d}}
8 {c,g} {{c,d}, {f,g}}
9 {f,g} {{e,h}}
10 {d,h} {}
12 {f} {}
Every state of the original machine must be covered. c a b d e f g h

31. Finding a Minimum Closed Cover
MInimization
Associate a variable to the prime compatible
For each , form the coverage constraint
1 {a,b,d,e} {}
2 {b,c,d} {{a,b},{a,g},{d,e}}
3 {c,f,g} {{c,d}, {e,h}}
4 {d,e,h} {{a,b}, {a,d}}
This cover is not closed, since is excluded

32. is the set of prime compatibles with non-empty class sets
Closure Constraints
MInimization
is the set of prime compatibles
with non-empty class sets
Note
class sets
1 {a,b,d,e} {}
2 {b,c,d} {{a,b},{a,g},{d,e}}
3 {c,f,g} {{c,d}, {e,h}}
4 {d,e,h} {{a,b}, {a,d}}
11 {a,g} {}
5 {b,c} {}
6 {c,d} {{a,g}, {d,e}}
7 {c,f} {{c,d}}

33. Covering and Closure Constraints--POS FORM
MInimization

34. Covering Constraints--Matrix FORM
MInimization
Row Dominance
Col Dominance?
(see below)

35. Closure Constraints--Matrix FORM
MInimization
class sets
1 {a,b,d,e} {}
2 {b,c,d} {{a,b},{a,g},{d,e}}
3 {c,f,g} {{c,d}, {e,h}}
4 {d,e,h} {{a,b}, {a,d}}
11 {a,g} {}
5 {b,c} {}
6 {c,d} {{a,g}, {d,e}}
7 {c,f} {{c,d}}
For each pair in the class
set of each compatible ,
form the clause
where ranges over the
indices of compatibles that
contain .

36. Closure Constraints--Matrix FORM
MInimization
Cover rows by including a 1-col OR excluding a 0-col

37. Covering Constraints Closure Constraints Closed Covering Problem
Minimization
Find a minimum set
of columns which
cover all rows:
{1,4,5,9}
Covering
Constraints
A row is covered by
either including a
1-col or excluding a
0-col.
Closure
Constraints

38. Binate Covering Problem
Similar to unate covering
Matrix
Variables on columns
Sum expressions on the rows
Solution may not exist when product is 0

39. The Binate Covering Problem
Note: replaced by to emphasize POS semantics
Also there is one addition (for empty solution space)

40. Unacceptable (anti-essential)Variables
Binate Covering
When is essential we say that is unacceptable
When is essential, we may delete all rows of the
matrix which has a zero in the column

41. Formally: Row dominates row if is satisfied,
Row Dominance
Binate Covering
Row 1 ( ) dominates row
2 ( ) since row 2 matches
row 1 at all care entries.
Row 1 may be deleted.
Formally: Row dominates row if is satisfied,
in a Boolean sense, whenever is satisfied, that is,

42. Let and be two columns of . We say that
Column Dominance
Binate Covering
Let and be two columns of We say that
dominates if, for each row of , one of
the following conditions hold:
Example: reduced
column dominates

43. Maximal Independent Set
Binate Covering
Two rows are independent if it is not possible to satisfy both clauses by assigning one variable to 1
Thus in finding the MIS, we ignore rows (clauses) that contain 0s, since these are satisfied by assigning variables to 0

44. Infeasible Subproblems
Binate Covering
cannot occur in original problem (first call to
the recursive procedure). But it can happen after
one or more recursions:
This is detected by REDUCTION, which
discovers that both and are essential

45. dominates dominates is essential dominates Solution: x x x x x x x x x
Reduction
Binate Covering
dominates
dominates x x x x x x x x x x x x 1 2 3 4 1 2 3 4 1 2 3 4 1 - f 1 - f 1 - f 1 1 1 - 1 - f - 1 - f - 1 - f 2 2 2 1 - - 1 f 1 - - 1 f 1 - - 1 f 3 3 3 1 1 f 1 1 f 1 1 f 4 4 4
is essential
dominates x x x x x x x x 1 2 3 4 1 2 3 4
Solution: 1 - f 1 - f 1 1 - 1 - f - 1 - f 2 2 1 - - 1 f 1 - - 1 f 3 3 1 1 f 1 1 f 4 4

46. State Encoding The number of possible assignments is very high
If one uses k bits to encode p states, there are (2k)!/(2k - p)! possible assignments
If one considers two assignments obtained by permutation or complementation of some of the bits as essentially the same assignment, then there are (2k -1)! / (2k - p)! k! distinct assignments

47. Practical Encoding Algorithms
State Encoding
Mustang tries to identify pairs of states by receiving adjacent pairs
Two codes are adjacent if they only differ in one bit
The first objective is to build a graph representing the attraction between each pair of states
Two states that have a strong attraction should be given adjacent codes
How to build attraction graph
In the fanout-oriented algorithm, whenever two states, si and sj have a common fanout state, the weight of the edge (si, sj) of the attraction graph is increased
In the fanin-oriented algorithm, if si and sj have a common fanin state, the weight of the edge (si, sj) of the attraction graph is increased
Once the graph of the attractions is found, we try to assign codes to pairs of states that have strong attractions

48. Fanout Oriented Algorithm
State Encoding
Build two matrices
The first with one row for each present state and one column for each next state
The second with one row for each present state and one column for each output

49. Embedding Algorithm Assign codes to states
State Encoding
Assign codes to states
Select first the node for which the sum of the weights of the Nb heaviest incident edges is maximum

50. Fanin Oriented Algorithm
State Encoding
Build two matrices
The first with one row for each next state and one column for each present state
The second with one row for each next state and two columns for each output
One column is for the true input and the other is for the complement

51. Decomposition and Encoding
Rather than aiming directly at minimizing the number of literals in the next-state functions, one may actually try to minimize the support of the functions
Reduction of the number of literals and simplification of the interconnections

52. Partitions
Decomposition
A partition  is on a set S is a collection of disjoint subsets of S whose set union is S, i.e.  = { Ba } such that Ba  Bb =  for a  b and  { Ba } = S
Each subset is called a block of the partition
If 1 and 2 are partitions on S, then 1 2 is the partition on S such that s  t(1 2 ) if and only if s  t(1 ) and s  (2 ), whereas, 1 + 2 is the partition on S such that s  t(1 + 2 ) if and only if there exists a sequence in S s=s0 s1 s2… sn = t for which either si  si+1 (1) or si  si+1 (2), 0  i  n-1

53. Partitions with Substitution Property
Decomposition
A partition  on the set of states of the machine is said to have the substitution property if and only if s  t() implies that (s,a)  (t,a) () a  I
A sequential machine M has a non-trivial parallel decomposition of its state behavior if and only if there exist two nontrivial S.P. partitions 1 and 2 on M such that 1 2 = 0
Independent component
Dependent component

54. Computation of SP Partitions
Decomposition
First generate the minimal SP partitions and then sum them until considering all possible sums
The minimal partitions are those obtained by requiring that two states only are included in a block

55. General Decomposition and Encoding
Need to resort to something more general than SP partitions, namely, partition pairs
A partition pair (, ’) on the machine is an ordered pair of partitions on S such that s  t() implies that (s,a)  (t,a) (’) a  I
The knowledge of the block of  containing the present state and of the current input allows one to compute the block ’ of that will contain the next state.
It is evident that if (, ) is a partition pair, then  has substitution property
Partition pairs generalize SP partitions