1. State Assignment of synchronous FSM based on partitions
Some Training
State Assignment of synchronous FSM based on partitions

2. We generate closed partitions
Example 1 x s 1 Z A H B F C G D E
To encode machine M we need 3 two-block partitions such that:
We generate closed partitions

3. Graph of successor pairs:
Example continued. x s 1 Z A H B F C G D E
Graph of successor pairs:
A,B
F,H
C,D
E,F
A,C
G,E
G,H
B,D

4. Example 1 continued x s 1 Z A H B F C G D E
A,D
D,H
+ =2
B,F

5. Example 1 continued
Unfortunately:
Thus we need one more partition :

6. Example 1 continued t
Encoding wrt 1 2 A B C D E F G H 1 1 1

7. Example 1 continued
With such encoding two excitation functions Q1’ i Q2’ of this machine will depend on one internal variable, and the third one Q3’ (in the worst case) on three variables, i.e.:
Q1’ = f(x,Q1)
Q2’ = f(x,Q2)
Q3’ = f(x,Q1,Q2,Q3)
If you do not trust, please encode, calculate cost functions Q1’, Q2’, Q3’ and check.

8. If I am wrong
I will buy beer
For everybody

9. Comment
Every other encoding will lead to more complex excitation functions.
In particular for binary encoding: A B C D E F G H
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Q1’ = f(x,Q1)
Q2’ = f(x,Q1,Q2,Q3)
Q3’ = f(x,Q1,Q2,Q3)

10. Partition compatible with inputs:
Example 2 x s 1 A D C B E F
Partition compatible with inputs:
I is closed

11. Example 2 continued Q1’ = f(Q1,Q2) Q2’ = f(Q1,Q2) Q3’ = ???
Encoding wrt I
wrt O A B C D E F 1 1
Q2’ = f(Q1,Q2)
Q3’ = ???
y = f(x,Q3)

12. Intuitive Encoding Example of a counter with input Enable E S 1 S0 S11 S0 S1 S2 S3 S4  S7
counter E
clock Q

13. Modulo 8 counter Transition table encoded with binary natural code1
Q2Q1Q0 S0 S1
000
001 S2
010 S3
011 S4
100 S5
101 S6
110 S7
111 s0 S’
Q2’Q1’Q0’

14. Encoded table E
Q2Q1Q0 1
000
001
010
011
100
101
110
111
Q2’Q1’Q0’

15. Excitation function table for D flip-flops
Q2Q1Q0 1
000
001
010
011
100
110
111
101
Q2’Q1’Q0’ D2 D1 D0
D2 =
D1 =
D0 =

16. Encoded excitation table for T flip-flops
Q2Q1Q0 1
000
001
010
011
100
110
111
101
Q2’Q1’Q0’ T2 T1 T0
T2 =
T1 =
T0 =

17. Q T EQ E = Diagram of the counter1 2 Q T EQ E = T Q
Clock
Enable
Discuss this design, how flip flops are selected, how to generalize to any number of bits

18. Is not sufficient for encoding
Example 3 x s 1 A F B E C D
Encoding wrt 1  A B C D E F 1
0 0
0 1
1 0
Is not sufficient for encoding

19. Example 3 continued Q1’ = D1 = f(x,Q1) Q2’ = D2 = f(x,Q1,Q2,Q3)
Let us then use closed partition: x s 1 A F B E C D
A 0 0 0
B 0 0 1
C 1 0 1
D 1 0 0
E 0 1 0
F 1 1 0
Q1’ = D1 = f(x,Q1)
And what with remaining?
Unfortunately only one variable we were able to encode according to the closed partition, therefore:
We do not have to calculate excitation functions to know that the first one, D1 will…
Q2’ = D2 = f(x,Q1,Q2,Q3)
Q3’ = D3 = f(x,Q1,Q2,Q3)

20. Example 3… May be there are more closed partitions: x s 1 A F B E C D1 A F B E C D
We will show that besides 1
is 2
Encoding wrt 1 2 A B C D E F 1
0 0
0 1
1 0
This is unique encoding

21. Example 3 cont
With this encoding the first excitation function Q1’ of this machine will depend on one internal variable, and the second and third together (Q2’, Q3’) on two internal variables, like this:
Q1’ = f(x,Q1)
Q2’ = f(x,Q2,Q3)
Q3’ = f(x,Q2,Q3)
If you do not trust, encode and calculate excitation functions
Q1’, Q2’, Q3’ and check.
Do this!