Grammar – some common problems A quick training session from.
Some Training
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1 Some Training State Assignment of synchronous FSM based on partitions
description
Some Training. State Assignment of synchronous FSM based on partitions. To encode machine M we need 3 two-block partitions such that:. Example 1. We generate closed partitions. G,H. C,D. G,E. A,C. A,B. F,H. E,F. B,D. Example continued . Gra ph of successor pairs :. - PowerPoint PPT Presentation
Transcript of Some Training
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Some Training
State Assignment of synchronous FSM based on partitions
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ExampleExample 1 1x
s 0 1 Z
A H B 0B F A 0C G D 0D E C 1E A C 0F C D 0G B A 0H D B 0
We generate closed partitions
To encode machine M we need 3 two-block partitions such that:
(0) cba
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Example continued.
xs 0 1 Z
A H B 0B F A 0C G D 0D E C 1E A C 0F C D 0G B A 0H D B 0
Graph of successor pairs:
HG,F,E,;DC,B,A,1
A,B F,H
C,D
E,F A,CG,E
G,HB,D
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Example 1 continued
xs 0 1 Z
A H B 0B F A 0C G D 0D E C 1E A C 0F C D 0G B A 0H D B 0
GF,C,B,;HE,D,A,
GF,;HE,CB,;DA, ;
GF,CB,;HE,D,A, ;
HE,DA,;GF,C,B, ;
A,D
D,H
B,F+ =2
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Example 1 continued
HG,F,E,;DC,B,A,1
GF,C,B,;HE,D,A,2
Unfortunately:
Thus we need one more partition :
)0( GF,;HE,;CB,;DA,21
)(0τ 21
FE,D,C,;HG,B,A,τ
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Example 1 continued
HG,F,E,;DC,B,A,1
GF,C,B,;HE,D,A,2 00001111
01100110
Encoding wrt 1 2
ABCDEFGH
FE,D,C,;HG,B,A,τ
00111100
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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.
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For everybody
If I am wrong
I will buy beer
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Comment
Q1’ = f(x,Q1)
Q2’ = f(x,Q1,Q2,Q3)
Q3’ = f(x,Q1,Q2,Q3)
0 0 0 0 0 10 1 0
0 1 11 0 01 0 11 1 01 1 1
ABCDEFGH
Every other encoding will lead to more complex excitation functions.
In particular for binary encoding:
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ExampleExample 2 2x
s 0 1 0 1
A D C 0 1B C D 0 0C E F 0 1D F E 0 0E B A 0 1F A B 0 0 FD,B,;EC,A,O
FE,;DC,;BA,I
Partition compatible with inputs:
Π(0)ΠΠOI
I is closed
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Example 2 continued
FE,;DC,;BA,Π I FD,B,;EC,A,ΠO
0 00 00 10 11 01 0
010101
Encoding wrt I wrt O
ABCDEF
Q1’ = f(Q1,Q2)
Q2’ = f(Q1,Q2)
Q3’ = ???
y = f(x,Q3)
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ES
0 1
S0 S0 S1
S1 S1 S2
S2 S2 S3
S3 S3 S4
S7 S7 S0
Intuitive Encoding
counterE
clock
Q
Example of a counter with input Enable
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E
S0 1 E
Q2Q1Q0 0 1
S0 S0 S1 000 000 001
S1 S1 S2 001 001 010
S2 S2 S3 010 010 011
S3 S3 S4 011 011 100
S4 S4 S5 100 100 101
S5 S5 S6 101 101 110
S6 S6 S7 110 110 111
S7 S7 s0 111 111 000
S’ Q2Q1Q0 Q2’Q1’Q0’
Modulo 8 counter
Transition TableTransition table encoded with binary natural code
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EQ2Q1Q0 0 1 E
Q2Q1Q0 0 1
000 000 001 000 000 001
001 001 010 001 001 010
010 010 011 011 011 100
011 011 100 010 010 011
100 100 101 110 110 111
101 101 110 111 111 000
110 110 111 101 101 110
111 111 000 100 100 101
Q2Q1Q0 Q2’Q1’Q0’ Q2Q1Q0 Q2’Q1’Q0’
Encoded table
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EQ2Q1Q0 0 1 E
Q2Q1Q0 0 1 0 1 0 1
000 000 001 000 0 0 0 0 0 1
001 001 010 001 0 0 0 1 1 0
011 011 100 011 0 1 1 0 1 0
010 010 011 010 0 0 1 1 0 1
110 110 111 110 1 1 1 1 0 1
111 111 000 111 1 0 1 0 1 0
101 101 110 101 1 1 0 1 1 0
100 100 101 100 1 1 0 0 0 1
Q2Q1Q0 Q2’Q1’Q0’ Q2Q1Q0 D2 D1 D0
Excitation function table for D flip-flops
D2 = 1QQ20QQ2
D1 = EQ1 D0 = EQ0EQ22Q1Q0EQ 1Q0EQ
0QQ1 0EQ
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EQ2Q1Q0 0 1 E
Q2Q1Q0 0 1 0 1 0 1
000 000 001 000 0 0 0 0 0 1
001 001 010 001 0 0 0 1 0 1
011 011 100 011 0 1 0 1 0 1
010 010 011 010 0 0 0 0 0 1
110 110 111 110 0 0 0 0 0 1
111 111 000 111 0 1 0 1 0 1
101 101 110 101 0 0 0 1 0 1
100 100 101 100 0 0 0 0 0 1
Q2Q1Q0 Q2’Q1’Q0’ Q2Q1Q0 T2 T1 T0
Encoded excitation table for T flip-flops
T2 = EQ1Q0 T1 = T0 = EQ0 E
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Diagram of the counter
T Q
Q Clock
T Q
Q
Enable T Q
Q
11102
01
0
QTQEQTEQTET
Discuss this design, how flip flops are selected, how to generalize to any number of bits
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Example 3Example 3x
s 0 1 0 1
A A F 0 0B E C 0 1C C E 0 1D F A 1 0E B F 1 1F D E 0 0
FDCEBA ,,;,,1
0 0 0 10 10 01 01 0
Encoding wrt 1
ABCDEF
001101
FECBDAτ ,;,;,
Is not sufficient for encoding
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Example 3 continued
xs 0 1 0 1
A A F 0 0B E C 0 1C C E 0 1D F A 1 0E B F 1 1F D E 0 0
A 0 0 0 B 0 0 1C 1 0 1D 1 0 0E 0 1 0F 1 1 0
Let us then use closed partition:
Q1’ = D1 = f(x,Q1)
We do not have to calculate excitation functions to know that the first one, D1
will…
Unfortunately only one variable we were able to encode according to the closed partition, therefore:
And what with remaining?
Q2’ = D2 = f(x,Q1,Q2,Q3)
Q3’ = D3 = f(x,Q1,Q2,Q3)
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Example 3…
xs 0 1 0 1
A A F 0 0B E C 0 1C C E 0 1D F A 1 0E B F 1 1F D E 0 0
FEDBCA ,;,;,2 π
FDCEBA ,,;,,1 π
Encoding wrt 1
0 0 0 10 00 11 01 0
ABCDEF
001101
2
)(0 21
This is unique encoding
May be there are more closed partitions:
We will show that besides 1
is 2
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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 functionsQ1’, Q2’, Q3’ and check.
Do this!
