Sequential System Synthesis
-- State Encoding
The State Encoding Problem
 Goal: Given n states, assign a unique code (of
length of at least log n) to each state such that
the cost of binary logic level implementation is
minimized. (state assignment problem)
 Cost of the implementation:
 Number of literals
 Speed
 Testability
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Example: Why State Encoding Matters?
 Consider a fragment of the cube
table for an FSM:
 Let y1y2 and Y1Y2 be the current
state and next state. We want to
represent Y1,Y2, and output O as a
function of y1,y2, and input x.
x
0
0
CS NS
A C
B C
O
1
1
x
0
0
y1y2 Y1Y2
O
1
1
01
10
11
11
 Assignment 1: A=01, B=10, C=11
Y1=…+x’y1’y2+…+x’y1y2’+…
x y1y2 Y1Y2 O
Y2=…+x’y1’y2+…+x’y1y2’+…
0 00 11 1
O=…+x’y1’y2+…+x’y1y2’+…
0 10 11 1
 Assignment 2: A=00, B=10, C=11
Y1=…+x’y1’y2’+…+x’y1y2’+…=…+x’y2’+…
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Lessons We Have Learned
 Two codes are adjacent if they only differ in one
bit. (e.g. 00 and 01 are adjacent, 01 and 10 are not.)
 Any k-bit code has k adjacent codes.
 If two states are given adjacent codes, we will be
able to extract common cubes from the next state
and output functions.
 Common fanout (next) state
 Common fanin (current) state
x
a
b
CS NS
A
B
A C
O
1
1
 Identify pairs that should receive adjacent codes.
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Attraction Graph
 The attraction graph of a FSM is a weighted,
undirected, complete graph that represents the
attraction between each pair of states.
 Node: state
 Edge: the attraction between the two states
 If two states share a common fanout/fanin state,
their attraction should increase.
 Two states that have a strong attraction should
be assigned adjacent codes.
 How to measure the attraction quantitatively?
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Fanout-Oriented Algorithm
 State Transition Matrix S|s|x|s|:
0/0
 Row: one per (current) state
 Column: one per (next) state
 Entry: (non-negative) number of arcs going from state si
A
1/1
1/1
(row) to state sj (column)
C
 Output Matrix Z|s|x|O|:
1/0
1 1 0 


1 0 1 
1 0 1 


(row) with output zj (column)
 Let Nb be the number of encoding bits, then the
 W(1,2) = 2(1 1 0)(1 0 1)T+(1)(0)T=2
 W(1,3) = 2(1 1 0)(1 0 1)T+(1)(1)T=3
 W(2,3) = 2(1 0 1)(1 0 1)T+(0)(1)T=4
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2
3
C
B
0/0
 Row: one per (current) state
 Column: one per output
 Entry: (non-negative) number of arcs going out of state si
attraction between states si and sj is given
by Nb•Si•SjT+Zi•ZjT
0/0
4
B
1
 
 0
1
 
6
Fanin-Oriented Algorithm
^
 State Transition Matrix S
|s|x|s|:
0/0
 Row: one per (next) state
 Column: one per (current) state
 Entry: (non-negative) number of arcs going from state sj
A
1/1
1/1
(column) to state si (row)
C
 Input Matrix X|s|x|I|:
1/0
1 1 1


 1 0 0
 0 1 1


(row) with input xj (column)
 Let Nb be the number of encoding bits, then the
attraction between states si and sj is given
^ •S
^ T+X •X T
by Nb•S
i
j
i
j
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0/0
 Row: one per (current) state
 Column: |I| per input
 Entry: (non-negative) number of arcs entering state si
 W(1,2) = 2(1 1 1)(1 0 0)T+(2 1)(0 1)T=3
 W(1,3) = 2(1 1 1)(0 1 1)T+(2 1)(1 1)T=7
 W(2,3) = 2(1 0 0)(0 1 1)T+(0 1)(1 1)T=1
0/0
A
3
7
C
1
B
 2 1


 0 1
 1 1


7
The Encoding Algorithm
 Given the attraction graph (matrix W(i,j)), assign
a unique Nb-bit code to each state.
 For each node si, find the Nb largest attractions W(i,j)
and sum them up;
 Pick the one with the largest sum and assign it code
0…0;
 Assign the Nb adjacent codes of 0…0 to the Nb
neighbor states that have the Nb largest attractions;
 Remove the Nb+1 node and repeat until all nodes get
a Nb-bit code;
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Example: The Encoding Algorithm
0/0
A
1/1
1/1
C
Fanout-Oriented Algorithm:
0/0
1/0
B
0/0
Fanin-Oriented Algorithm:
01
A
A
2
3
C
4
00
B
2
3
C
00
A
4
3
7
B
C
10
Y1=y1’y2x
Y2=y1’y2x + y1y2’x’ + y1’y2’x
Z = y1’x
(13 literals)
A
1
B
3
7
C
10
1
B
01
Y1=y1’y2x’ + y1’y2x
Y2=y1’y2’x
Z = y2’x
(11 literals)
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Decomposition and Encoding: Motivation
 There are many different ways to encode a FSM
with four states: {A,B,C,D}.
 If we know that pairs (A,B),(C,D); (A,C),(B,D)
should receive adjacent codes, then there are
only 8 different assignments:








A=00, B=01, C=10, D=11
A=00, B=10, C=01, D=11
A=01, B=00, C=11, D=10
A=01, B=11, C=00, D=10
A=10, B=11, C=00, D=01
A=10, B=00, C=11, D=01
A=11, B=01, C=10, D=00
A=11, B=10, C=01, D=00
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Notations on Partitions
 A partition  on a set S is a collection of disjoint subsets
(called blocks) of S whose union is S.
 ={(1,2),(3,4,6),(5,8),(7)} is a partition of S={1,2,3,4,5,6,7,8}
 0={(1),(2),(3)} and 1={(1,2,3)} are two trivial partitions of S={1,2,3}
 For s,t S, st () means that they belong to the same block of
partition 
 For two partitions 1 and 2, their meet 1•2 is also a partition
where st (1•2) iff st (1) and st (2); their join 1+2 is the
partition where st (1+2) iff there exists a sequence in S:
s=s0,s1,…,sn=t, such that for all i=0,…n-1, either si si+1(1) or si
si+1(2).
 Given 1={(1,2),(3,4,6),(5,8),(7)}, 2={(1,6),(2,5,8),(3,4,7)},
1•2={(1),(2),(3,4),(5,8),(6),(7)}, 1+2={(1,2,3,4,5,6,7,8)}
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Substitution Property
 A partition  on set S of machine M=<I,S,,S0,O,>
has the substitution property (S.P.) iff st()
implies that (s,a)(t,a)() for all aI.
 Theorem. A sequential machine M has a nontrivial parallel decomposition of its states iff there
exist two non-trivial S.P. partitions 1 and 2 on M
such that meet 1•2=0.
 Theorem. IF a sequential machine has one nontrivial S.P. partition, then it has a non-trivial serial
decomposition.
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Parallel Decomposition
 1 = {(0,1,2),(3,4,5)} and
2={(0,5),(1,4),(2,3)} are two
S.P. partitions and 1•2=0.
 Let A=(0,1,2), B=(3,4,5), and
X=(0,5),Y=(1,4),Z=(2,3), we
can build two machines.
 The output, in this case, is
the product of the outputs of
the two newly built machines.
0
3
5
4
1
0
2
0
1
2
3
4
5
0
1
A
B
A
0
B
A
B
1
1
2
2
1
4
3
3
X
Y
Z
0
0
0
1
0
0
0
Z
X
Y
1
Z
Z
Y
0
0
1
G1
G2
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Serial Decomposition
 1 = {(1,2,3),(4,5)} is the only
S.P. partition.
 We choose another partition
2={(1,4),(2,5),(3)} s.t. 1•2=0.
 Let A=(1,2,3), B=(4,5), and
X=(1,4),Y=(2,5),Z=(3), we can
rebuild this machine as
A
0
A
1
B
0
B
1
A
0
A
1
B
0
B
1
X X
X
Z
X
1
0
1
1
Y Z
Y
Y
Y
1
0
0
1
Z Y
X
-
-
0
0
-
-
1
2
3
4
5
A,X
A,Y
A,Z
B,X
B,Y
0
1
3
2
3
2
1
4
5
4
1
2
0
1
1
0
1
0
1
0
0
0
1
1
0
X
Z
Y
Z
Y
1
X
Y
X
X
Y
0
1
1
0
1
0
1
0
0
0
1
1
G1
G2
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Computation of the S.P. Partition
 Recall that “A partition  on set S of machine
M=<I,S,,S0,O,> has the substitution property (S.P.) iff
st() implies that (s,a)(t,a)() for all aI.”
 So S.P. partition is independent of output.
0
1
1
1
2
2
4
3
3
5
2
4
1
2
5
1
3




If (1,2)(1,4),(2,3)(1,2,3,4) (transitivity){(1,2,3,4,5)} trivial.
If (1,3)(1,5)(1,3,5)(2,3)(1,2,3,5) (1,2,3,4,5) trivial.
If (1,4), form a block.
 If (2,3) (4,5) (1,4,5) {(1,4,5),(2,3)}
 If (2,5) (1,4) {(1,4),(2,5),(3)}
 If (2)
 if (3,5) (1,5),(2,3), contradiction.
 If (3) {(1,4),(2),(3),(5)}
Another partition: {(1),((2,3),(4,5)}
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Decomposition and State Encoding
 We need two bits xy to encode a sequential state





machine M with four states: A,B,C,D.
Suppose 1={(A,B), (C,D)} and 2={(A,C),(B,D)} are two
non-trivial S.P. partitions.
Clearly 1•2=0 so we have a non-trivial parallel
decomposition, M1={ab,cd} and M2={ac,bd}, both with
only two states.
If we use bit x to encode M1 and bit y to encode M2 as
ab=0, cd=1; and ac=0, bd=1.
Then M can be encoded as A=00,B=01,C=10,D=11.
SE problem becomes finding S.P. partitions and applying
parallel/serial decompositions.
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