presentation on row reduction techniques

1

Presentation on Row Reduction Techniques

Presented By :Name: Amit GroverEmail : [email protected]

mailto:[email protected]





2

Topics covered in this presentation

• What is Row Reduction.

• What are the reasons for doing Row Reduction

• What are state variables and external inputs.• Algorithms used for row reduction

– For completely specified state machines.

• Pairchart

– For incompletely specified state machines

• Pairchart

• Skill algorithm

• Bargain hunter

• MEU Method

• Applications

3

What is Row Reduction ??

4

Definition

• Given a sequential machine, our aim is to find the finite state machine which have same behavior as the given machine but has reduced states.

5

Advantages of Row Reduction

• Cost Reduction– We know Flip Flops used as memory elements are costly and each

time a new state is added to the machine we are adding new memory element to it.

– Addition of this new element needs space in the logic circuits and thus increases the size of the logic circuit

– Adds cost the logic circuit

– These additions adds significant amount of cost to the Logic circuit and we don’t want that, so our aim is to reduce the cost as much as possible without effecting the functionality of the logic circuit.

6

State Variables and External Inputs.

External Inputs

00 01 11 10

1

2

3

4

Sta

te

Va r

i ab l

e s

7

Completely Specified Machines.

• A machine which has no don’t cares in its table is called a completely specified state machine (CSSM).

• Let’s see an example to find out what do we mean by having no don’t cares.

8

Completely Specified State Machine(CSSM)

2,0 3,1

4,1 1,0

2,0 1,1

3,1 6,0

6,0 5,1

4,1 5,0

1

2

3

4

5

6

Sta

t e

Va r

i ab

l es

External InputsA B

No don’t cares

9

Incompletely Specified State Machines (ISSM)

• Those machines whose next state or output has don’t care is called as ISSM.

• Let’s see with diagram what is ISSM.

10

Incompletely Specified State Machine (ISSM)

-,0 3,1

4,1 -,0

2,0 1,-

3,1 -,-

6,0 5,1

-,- -,0

1

2

3

4

5

6

Sta

te

Va r

i ab l

e sExternal InputsA B

11

Algorithms for Row Reductions

12

Completely Specified Machine

• Pairchart Algorithm.- Things given to you.

- Flow table.

- Aim is to reduce the number of rows as much as possible.Steps Involvedi. Draw Pairchart.ii. Check for output Incompatibility and cross out all those which are

incompatible.iii. Check for next state incompatibilities and cross out all those which are

incompatibles.iv. Repeat step iii till you get no next state incompatible state.!!These things will be more clear with a example which follows..

13

Given CSSM

2,0 3,1

4,1 1,0

2,0 1,1

3,1 6,0

6,0 5,1

4,1 5,0

1

2

3

4

5

6

A B

Figure 1

14

How to read CSSM

2,0 3,1

4,1 1,0

2,0 1,1

3,1 6,0

6,0 5,1

4,1 5,0

A B

• If you are in State 3 and input is A you go to State 2 with output 0.

• And so on……

• If you are in State 1 and input is A you go to State 2 with output 0.• If you are in State 1 and input is B you go to State 3 with output 1.

1

2

3

4

5

6

15

Step 1 for Pairchart Algorithm

• Look at the Maximum Number of States you have.– In our case it is 6. (Refer previous given CSSM).

– Now write the States from 1->6 horizontally in increasing order and from Top to Bottom in increasing order.

• This Step is shown in next slide….

16

Implementation of Step 1 of Pairchart Algorithm

1 2 3 4 5 6

1

2

3

4

5

6

Inc

reas

ing

Ord

er

Increasing Order

17

Step 2 of Pair Chart Algorithm

• We can remove half of the pairchart (diagonally) because it is symmetric along the diagonal.

• So we are effectively duplicating along the diagonal.

• So after removing half of the portion from diagonal, we get something like this.(shown in next slide).

18

Implementation of Step 2 of Pairchart Algorithm

1 2 3 4 5 6

1

2

3

4

5

6

Inc

reas

ing

Ord

er

Increasing Order

Cutting from diagonal

19

1 2 3 4 5 6

1

2

3

4

5

6

Inc

reas

ing

Ord

er

Increasing Order

Figure2 2,0 3,1

4,1 1,0

2,0 1,1

3,1 6,0

6,0 5,1

4,1 5,0

1

A B

2

3

4

5

6

20

Step 3 of Pairchart Algorithm

• Let’s start filling the Table in Figure 2.– Output Compatibility

• We will take one column at a time and will check if it is output compatible with each and every other state.

• Question comes what is output compatibility??– State 1 is Output Compatible with State 2 iff for same

input both gives same Output.

• Refer to next slide for better explanation..

21

1 2 3 4 5 6

1

2

3

4

5

6

2,0 3,1

4,1 1,0

2,0 1,1

3,1 6,0

6,0 5,1

4,1 5,0

1

A B

2

3

4

5

6

Column 1- Check if 1 is output compatible with 2 for both input A and B

Answer: No. For same Input A or B State 1 and State 2 has different output.

-So we put a cross in block at the inter section of 1 and 2.

X

22

1 2 3 4 5 6

1

2

3

4

5

6

2,0 3,1

4,1 1,0

2,0 1,1

3,1 6,0

6,0 5,1

4,1 5,0

1

A B

2

3

4

5

6

Column 1-check if 1 is output compatible with 3 for both input A and B

Answer: Yes. For same Input A and B State 1 and State 3 has same output.

-So we don’t do anything with the block at the inter section of 1 and 3.

X

23

1 2 3 4 5 6

1

2

3

4

5

6

2,0 3,1

4,1 1,0

2,0 1,1

3,1 6,0

6,0 5,1

4,1 5,0

1

A B

2

3

4

5

6

X

X

X

X

X

X

X

X

X

Similarly we will fill for all the blocks

24

Pairchart Algorithm continued…

• Refer to given flow table and take one column at

a time in the pairchart and fill each block in that column (excluding those blocks which are having ‘X’ sign) with, where the state representing that column will go when input A and B is given..

Refer to next slide for better explanation.

25

1 2 3 4 5 6

1

2

3

4

5

6

2,0 3,1

4,1 1,0

2,0 1,1

3,1 6,0

6,0 5,1

4,1 5,0

1

A B

2

3

4

5

6

X

X

X

X

X

X

X

X

X

Taking Column represented by state 1 - From pairchart, state 1 goes to state 2 with input A and to state 3 with input B.

2 3

2 3

- So we will write 2 and 3 in each block in column 1 and Row 1 as shown..

26

Pairchart algorithm continued..

• We will continue with all the rows and column and what we get is shown in next slide…

• Let’s see how that will look like.

27

1 2 3 4 5 6

1

2

3

4

5

6

2,0 3,1

4,1 1,0

2,0 1,1

3,1 6,0

6,0 5,1

4,1 5,0

1

A B

2

3

4

5

6

X

X

X

X

X

X

X

X

X

Taking next Column represented by state 2 - From pairchart, state 2 goes to state 4 with input A and to state 1 with input B.

2 3

2 3

4 1

4 1

- So we will write 4 and 1 in each block in column 2 and Row 2 (if no ‘X’) as shown..

28

1 2 3 4 5 6

1

2

3

4

5

6

2,0 3,1

4,1 1,0

2,0 1,1

3,1 6,0

6,0 5,1

4,1 5,0

1

A B

2

3

4

5

6

X

X

X

X

X

X

X

X

X

2 3

2 3

4 1

4 1

2 1

2 1

Taking next Column represented by state 3

- State 3 goes to 2 and 1 so we write it in column 3 and also in row 3 -> 2 and 1

- So we will write 2 and 1 in each block in column 2 and Row 2 (if no ‘X’) as shown..

29

1 2 3 4 5 6

1

2

3

4

5

6

2,0 3,1

4,1 1,0

2,0 1,1

3,1 6,0

6,0 5,1

4,1 5,0

1

A B

2

3

4

5

6

X

X

X

X

X

X

X

X

X

2 3

2 3

4 1

4 1

2 1

2 1

3 6

3 6

6 5 6 5

4 5 4 5

Lets fill the whole table accordingly

30

Pairchart Algorithm continued..

• So after filling out all the blocks in the pairchart what next…

• We need to check each and every block and see if that block is going to some other block which has a ‘X’. (Next State Compatibility)– If Yes then we cross this block.

– Else keep moving and check for other states.

• Let’s see how it looks like.

31

1 2 3 4 5 6

1

2

3

4

5

6

2,0 3,1

4,1 1,0

2,0 1,1

3,1 6,0

6,0 5,1

4,1 5,0

1

A B

2

3

4

5

6

X

X

X

X

X

X

X

X

X

2 3

2 3

4 1

4 1

2 1

2 1

3 6

3 6

6 5 6 5

4 5 4 5

Column 1 - Check each block in column 1 for e.g block at the intersection of 1 and 3.

Check Row 2 Column 2 and vice versa..

2 = 2, so it is okay.Reason is on Slide 17

Check Row 1 and Column 3 or vice versa.

In this case, (3,1) references itself, and is not yet X’d,

so it is okay.

32

1 2 3 4 5 6

1

2

3

4

5

6

2,0 3,1

4,1 1,0

2,0 1,1

3,1 6,0

6,0 5,1

4,1 5,0

1

A B

2

3

4

5

6

X

X

X

X

X

X

X

X

X

2 3

2 3

4 1

4 1

2 1

2 1

3 6

3 6

6 5 6 5

4 5 4 5

Column 1 continue…. - Checking next block in column 1 i.e block at the intersection of 1 and 5.

Check Row 2 Column 6.It does not have a ‘X’.

So Okay

Check Row 3 and Column 5 It does not have a ‘X’.

So Okay

33

1 2 3 4 5 6

1

2

3

4

5

6

2,0 3,1

4,1 1,0

2,0 1,1

3,1 6,0

6,0 5,1

4,1 5,0

1

A B

2

3

4

5

6

X

X

X

X

X

X

X

X

X

2 3

2 3

4 1

4 1

2 1

2 1

3 6

3 6

6 5 6 5

4 5 4 5

Column 2 - Check each block in column 2 for e g block at the intersection of 2 and 4.

Check Row 4 Column 3 and vice versa

It has a Cross So Mark This block with ‘X’Column 3 Row 4.

No Need to check this block further.

X

34

1 2 3 4 5 6

1

2

3

4

5

6

2,0 3,1

4,1 1,0

2,0 1,1

3,1 6,0

6,0 5,1

4,1 5,0

1

A B

2

3

4

5

6

X

X

X

X

X

X

X

X

X

2 3

2 3

4 1

4 1

2 1

2 1

3 6

3 6

6 5 6 5

4 5 4 5

Let’s fill the Whole Pairchart

X

x

35


• If you understand everything till this point then assume that you understand most of this algorithm.

• There is one last step still left for this algorithm.

• Guess what it can be…– Remember what was our aim!!!

• To reduce the number of states required to represent the same machine.

36


• It ‘s very easy..

– Draw the graph and link all those states which are compatible to each other with the help of pairchart.

– Let’s see how it looks like…

37

13

6

25

->Look at column 1 6 is compatible with none

Refer Pairchart on slide 35

-> Look at column 1 1 is compatible with 3 and 5-> Look at column 2 2 is compatible with 6

-> Look at column 3 3 is compatible with 5

-> Look at column 4 4 is compatible with none

-> Look at column 5 5 is compatible with none

4

38


• Now we need to build up the largest possible circle which covers as much states as possible.– It shows that 135 , 24 and 3 are the states

required to represent the same machine..– So let’s see how the new state machine looks

like..

39


13

6

25

4

40


26,0 135,1

4,1 135,0

135,1 26,0

135

26

4

A BNew State Assignment

-Refer to our original table and see where State (1 3 5) go for Input A . eg in our case it goes to 2 and 6 with output 0. and for input B it goes to 135.

2,0 3,1

4,1 1,0

2,0 1,1

3,1 6,0

6,0 5,1

4,1 5,0

1

A B

3

4

5

6

2

- Similarly we will do for 26 and 4.Note : - For 4 we have 3 but there is no state

as 3 but 3 is covered by 135 and thus we will write 135,1. Repeat for Input B.

41

Review of what we did till now

• Till Now we talked about – What State Reduction is..– We talked about Algorithm which can help us in

reducing the rows to reduce the cost of the machine for a given CSSM..

• What next ??? • Now we will talk about ISSM machines and

algorithms used for it…• If you really understood till this point it won’t be

difficult to understand the rest of stuff..

42

ISSM Algorithm

• ISSM Algorithm is continuation of CSSM pairchart algorithm..

• But there are some more additional algorithms..• Steps

– Pairchart (we already discussed)

– Skill algorithm( To get MC’s (Maximum Compatibility Sets)

– Bargain Hunter Algorithm.

– MEU

43

ISSM - Pairchart

• Since we had talked about Pairchart, I will skip the details and will proceed further.. – But one thing to keep in mind is that in ISSM

we have don’t cares and in the pairchart where ever we encounter don’t cares we write don’t cares directly..

– Example is shown in the next slide.

44

ISSM: Pairchart Algorithm

1 - 2 a - B

4 a 3 b - -

- - 3 b 8 -

1 - 2 - 5 -

3 - 7 - 5 -

- b - b - b

- - 6 - - a

2 - 4 - - b

1

2

3

4

5

6

7

8

A B CGiven:

451 2 3 4 5 6

1

2

3

4

5

6

1 - 2 a - B

4 a 3 b - -

- - 3 b 8 -

1 - 2 - 5 -

3 - 7 - 5 -

- b - b - b

- - 6 - - a

2 - 4 - - b

1

2

3

4

5

6

7

8

8

7

7 8

X

After Multiple Passes (Already discussed already)We come to final Table

X

X

X

X

X

X

X

X

X

X

X

X

X X X X

Shows that 2&3 state is

compatible to each other

46

Skill Algorithm

• This algorithms helps us to get minimal covering machine for ISSM.

• This algorithms gives the MC’s (Maximum Compatibility.

• Let’s see how it works…

47

Skill Algorithm Continued…

• It consists of different columns– Column K will consists of all the states we have in decreasing

order.

– Column Sk will contain all those states which are compatible with state in K (We have to refer or pairchart for that).

– We compare column Sk and L’ and concatenate state in column K with states which are common to Sk and L’.

– Compare Column L’ and states in previous row and its L column and find out all those states which are not covered by each other.

– Next slide will make these points much clear..

48

Skill Algorithm Continue..

K Sk L’ L

8 NULL NULL 8

7

6

5

4

3

2

1

MC’s

Refer Pairchart for State 8 in Column 8

8 is not compatible with any one so right NULL in Sk and L’ and 8 in L.

49

Skill Algorithm Continued..

K Sk L’ L

8 NULL NULL 8

7 NULL 7 7,8

6

5

4

3

2

1

Look forstates which

intersect

Concatenate 7 with states which

Intersects.In this case nothing intersect so

Put just 7 in L’

Nothing covers each other so

7 is not compatible with any one so..Sk will have NULL.

MC’s


50


K Sk L’ L

8 NULL NULL 8

7 NULL 7 7,8

6 7 67,6 67,8

5

4

3

2

1

Look for states which intersect

Concatenate 6 with states which intersects.In this case 7 intersects so It would be 67 in L’But 7 doesnot intersect with 8 so we will write

6 also in L’

67 covers 6 and 7 But not 8 so67,8 in L

6 is compatible with 7 so..Sk will have 7.

MC’s


51

Skill Algorithm Continue..

K Sk L’ L

8 NULL NULL 8

7 NULL 7 7,8

6 7 67,6 67,8

5 67 567,5 567,8

4

3

2

1


Intersects.

Concatenate 5 with states which intersects.In this case 67 intersects so L’ will have 567, 5

Remember we have to write 5 in L’ although it is covered by 567 but it will be taken care in

Column L.

567 covers 5,67But not 8 so567,8 in L


MC’s


52


K Sk L’ L

8 NULL NULL 8

7 NULL 7 7,8

6 7 67,6 67,8

5 67 567,5 567,8

4 6 46,4 567,46,8

3

2

1


Intersects.

Concatenate 4 with states which intersects.In this case 6 intersects so L’ will have 46,4 in L’.

46 covers 4so 567,8, 46in L


MC’s


53


K Sk L’ L

8 NULL NULL 8

7 NULL 7 7,8

6 7 67,6 67,8

5 67 567,5 567,8

4 6 46,4 567,46,8

3 567 3567,3 46,3567,8

2

1


Intersects.

Concatenate 3 with states which intersects.In this case 567 intersects so L’ will have 367,3

3567 covers 5,367 but not 8,46 so 3567,8, 46in L


MC’s


54


K Sk L’ L

8 NULL NULL 8

7 NULL 7 7,8

6 7 67,6 67,8

5 67 567,5 567,8

4 6 46,4 567,46,8

3 567 3567,3 46,3567,8

2 347 24,237,2 46,24,3567,8,237

1

Look forstates whichIntersects.

Concatenate 2 with states which intersects.In this case 4,37 is common so

It would be 24,23,2 in L’

24,237 covers 2so 3567,8, 46,237,24in L


MC’s


55


K Sk L’ L

8 NULL NULL 8

7 NULL 7 7,8

6 7 67,6 67,8

5 67 567 567,8

4 6 46 567,46,8

3 567 3567 46,3567,8

2 347 24,237 46,24,3567,8,237

1 4 14,1 46,8,3567,24,237,14


Intersects.

Concatenate 1 with states which intersects.In this case 4 intersect so L’ will have 14,1

14 covers 1 so 3567,8,46,237,24,14in L


MC’s


56

Skill Algorithm continued..

– We are done with Skill Algorithm.– But still have Bargain Hunter algorithm.– Let’s take another ISSM and recalculate Skill

algorithm and then proceed with Bargain Hunter.

– Reason for taking another ISSM is because we will get better view of these algorithms.

57

1 2 3 4 5 6

1

2

3

4

5

6

-,0 - , - 4,-

-,0 -,0 3,-

1,1 2,- -,-

2,- 4,- -,-

-,- 4,1 -,1

5,- -,0 2,0

1

0

2

3

4

5

6

43

X

X

1 22 4

X

24

1 2

X

42 X

25

Pairchart Algorithm already seen

X

58

Skill Algorithm

K Sk L’ L

6 NULL 6 6

5 NULL 5 5,6

4 5 45 45,6

3 456 345,36 345,36

2 4 24 24,36,345

1 2456 16,145,124 124,16,145,345,36

MCs

Till this point we knew how to draw it..

59

Bargain Hunter Algorithm

• This Algorithms uses the MCs found from Skill algorithm as inputs and calculate CCs (Compatibility classes and CS (Compatible sets) we get PCs(Prime compatibles).

• This how flow of algorithms are correlated.• MCs(Skill Algorithm) ->Bargain Hunter

(PCs) ->MEU->Minimal Covering Machine.

60


• Steps 1– Create CCs from MCs.

• Take each MC from highest to lowest and take all possible combinations of those MCs to get CC.

• Next slide will explain this in a better way.

61


MCs ->124,16,145,345,36 from SKILL AlgorithmCC

345

124

145

36

16

34

35

45

12

14

24

15

3

4

5

1

2

6

MCs from Skill Algorithm

All Possible Subsets of MCs

62

Bargain Hunter continued..

• Step 2.– Now we will take each CC one at a time and

will look at the table (given to us) and check where each CC goes if we give input as 0 , 1 ,2.

– Let’s see how it looks like.

63


CC Inputs

0 1 2

CS

345 12 24 -

124 2 4 34

145 2 4 4

36 15 2 2

16 5 - 16

34 12 24 -

35 1 24 -

45 2 4 -

12 - - 34

14 2 4 4

24 2 4 3

15 - 4 4

3 1 2 -

4 2 4 -

5 - 4 -

1 - 4 -

2 - - 3

6 5 - 2

-,0 - , - 4,-

-,0 -,0 3,-

1,1 2,- -,-

2,- 4,- -,-

-,- 4,1 -,1

5,- -,0 2,0

1

2

3

4

5

6

0 1 2

64


CC Inputs

0 1 2

CS

345 12 24 -

124 2 4 34

145 2 4 4

36 15 2 2

16 5 - 16

34 12 24 -

35 1 24 -

45 2 4 -

12 - - 34

14 2 4 4

24 2 4 3

15 - 4 4

3 1 2 -

4 2 4 -

5 - 4 -

1 - 4 -

2 - - 3

6 5 - 2

-,0 - , - 4 ,-

-,0 -,0 3,-

1,1 2,- -,-

2,- 4,- -,-

-,- 4,1 -,1

5,- -,0 2,0

1

2

3

4

5

6

0 1 2

Similarly we can do for all CCs

65


• Step 3– Time to calculate CS

• Way we calculate CS is– Ignore all the singletons like 2,3,1,4,5,6 etc.

– For each Row see corresponding block when Input is

0 ,1 ,2 and copy all those values which does not cover each other.

– Let’s see diagram to find out what is it all about.

66


CC Inputs

0 1 2

CS

345 12 24 - 12,24

124 2 4 34 34

145 2 4 4 NULL

36 15 2 2 15

16 5 - 16 24

34 12 24 - 12,24

35 1 24 - 24

45 2 4 - NULL

12 - - 34 34

14 2 4 4

NULL

24 2 4 3 NULL

15 - 4 4 NULL

3 1 2 - NULL

4 2 4 - NULL

5 - 4 - NULL

1 - 4 - NULL

2 - - 3 NULL

6 5 - 2 NULL

-,0 - , - 4,-

-,0 -,0 3,-

1,1 2,- -,-

2,- 4,- -,-

-,- 4,1 -,1

5,- -,0 2,0

12

3

4

5

6

0 1 2

67


CC Inputs

0 1 2

CS

345 12 24 - 12,24

124 2 4 34 34

145 2 4 4 NULL

36 15 2 2 15

16 5 - 16 24

34 12 24 - 12,24

35 1 24 - 24

45 2 4 - NULL

12 - - 34 34

14 2 4 4

NULL

24 2 4 3 NULL

15 - 4 4 NULL

3 1 2 - NULL

4 2 4 - NULL

5 - 4 - NULL

1 - 4 - NULL

2 - - 3 NULL

6 5 - 2 NULL

-,0 - , - 4,-

-,0 -,0 3,-

1,1 2,- -,-

2,- 4,- -,-

-,- 4,1 -,1

5,- -,0 2,0

12

3

4

5

6

0 1 2

IgnoreSingleton

Similarly for all other Rows we can find CS

68


• Step 4– Now it’s time to apply bargain method– Copy only CC and CS Column because you

don’t need any other column.• The way it works is that we try to get the best deal.

– What I mean is that if some one is giving you apples for $2 and also for $4 then you go for the one with $2 and delete $4 entry..

– The same concept works for us as shown in next slide.

69


CC CS

345 12,24

124 34

145 NULL

36 15

16 24

34 12,24

35 24

45 NULL

12 34

14 NULL

24 NULL

15 NULL

3 NULL

4 NULL

5 NULL

1 NULL

2 NULL

6 NULL

CC 145 is better than CC 45


CC 145 is better than CC 14 CC 145 is better than CC 15





You can’t remove 6 because of 16 because 6 gives NULL but 16 is coming with some cost (CS 24)

Better deal because CC 345 gives us CS 12, 24 but CC 34 gives 12 , 24 so of course 345 is better than 34 because it is larger than 34

70


CC CS PC

345 12,24 345

124 34 124

145 NULL 145

36 15 36

16 24 16

35 24 35

24 NULL 24

3 NULL 3

6 NULL 6

Prime Compatibles

Those CCs which are left are our PCs

71

Now it’s time for MEU method

– Steps:• We start with NULL set.• Uresin class will be all state singletons (i.e., {q}, q is a state)

– We will count the number of occurrences of each state in the PCs

» Then we choose the state with least number of occurrences.» We check what all PCs cover this state » We take each PC and calculate its Uresin class» Uresin class consists of all states and (class set) sets that

were in the previous Uresin class but not a subset of the current PC, together with any set in the class set of the current PC that is not a subset of (possibly equal to) an earlier PC.

» This will be more clear with the diagram in next slide.

72

MEU Method PCs : 345, 124, 145, 36 ,16 ,35, 24, 3,6

NULL

(1,2,3,4,5,6)

3 2 4 4 3 3

Starting Point

Uresin class

Number of occurrencesin PCs

PCs : 345,124,145,36,16,35,24,3,6

State 2 is occurring the least.choose 2 and write all those PCs which covers 2 i.e PC 24,124

24

124

Number of times 1 is coming is 3

73

MEU MethodPCs : 345, 124, 145, 36 ,16 ,35, 24, 3,6

NULL(1,2,3,4,5,6)

24(1,3,5,6)

3 4 3 3

(34,5,6)

36

16

6

345

Uresin class

PC 24 has CS NULL and

includes 2 and 4 but not 1,3,5,6

so Uresin of 24 is (1,3,5,6)

PC 124 has CS 34 and thus

includes 1,2 3,and 4

but not 5,6

so Uresinof 124 is (34,5,6)

Uresin class

124

Number of occurrencesin PCs

1, 3, 3

We will choose 1/5/6 I am choosing right mostYou can choose whatever you want

We will choose 34Left most

74

MEU Method

NULL(1,2,3,4,5,6)

24(1,3,5,6)

(34,5,6)

36

16

6

345

PCs : 345,124,145,36,16,35,24,3,6

PC 36 has CS 15 and thuswhen compared with previous PC24’s Uresin class

It includes everything

So it’s Uresin is 15

124

(15)

PC 16 has CS 24, but 24 does not appear

in the Uresin class of 16 because 24 is contained

in earlier node 24. That’s why this node is NOT eliminated by P1 - see Pruning Principle

slides below).

PC 6 has CS NULL and thus

when compared with previous PC24’s Uresin class

It includes 6 but not 1,3,5

thus it’s Uresin is 1,3,5

(1,3,5)PC 345 has

CS 12,24 and Observe that it covers previous

PC 124 completely.It shows that Uresin will

be 6 only(6)

PC CS

345 12,24

124 34

145 NULL

36 15

16 24

35 24

24 NULL

3 NULL

6 NULL

(3,5)

75

MEU MethodPCs : 345, 124, 145, 36 ,16 ,35, 24, 3,6

NULL(1,2,3,4,5,6)

24(1,3,5,6)

(34,5,6)

36

16

6

345

PCs : 345,124,145,36,16,35,24,3,6

124

(15)

(1,3,5)

(12,24,6)

145(NULL)

PC 145 has CS NULL and it

Covers his previous PC36’s Uresin class

completelySo it’s Uresin is NULL

Since there is nothing else So no need to check other

PC’s, we will stop here and will go

ahead to create Minimum

Covering machine

15 is covered by145

Minimum covering machine

(3,5)

76

Minimum Covering Machine

- We saw that PC’s 24,36,145 will form the minimal covering machine.- To create minimum covering machine we will draw a table with states 24, 36145 and we will see that when we give input 0,1,2 then where will these states go.- Let’s see how they look like.

77

Minimal covering machine-,0 - , - 4,-

-,0 -,0 3,-

1,1 2,- -,-

2,- 4,- -,-

-,- 4,1 -,1

5,- -,0 2,0

1

2

3

4

5

6

145

24

36

0 1 2Input

States

0 1 2

State 1 4 and 5 goes to state 2. when Input is 0 and State 2 is

covered by state 24 in our minimal cover machine so we will write 24 and output is 0

24,0

78


-,0 -,0 3,-

1,1 2,- -,-

2,- 4,- -,-

-,- 4,1 -,1

5,- -,0 2,0

1

2

3

4

5

6

145

24

36

0 1 2Input

States

0 1 2

State 1 4 and 5 goes to state 2 when Input is 1 and state 4 is

covered by state 24 in our minimal cover machine so we will write 24 and output is 1

24,0 24,1

79


-,0 -,0 3,-

1,1 2,- -,-

2,- 4,- -,-

-,- 4,1 -,1

5,- -,0 2,0

1

2

3

4

5

6

145

24

36

0 1 2Input

States

0 1 2

State 145 goes to state 4 when Input is 0 and state 2 is

covered by state 24 in our minimal covering machine so

we will write 24 and output is 1

24,0 24,1 24,1

80


-,0 -,0 3,-

1,1 2,- -,-

2,- 4,- -,-

-,- 4,1 -,1

5,- -,0 2,0

1

2

3

4

5

6

145

24

36

0 1 2Input

States

0 1 2

24,0 24,1 24,1

24,0 36,124,0

24,024,0145,1Similarly we will fill for all the states

81

Pruning Principles for the MEU Method

82

Rules to prune the MEU Branches• I am going to discuss only 2 rules here

– First Pruning Principle (P1):• We can prune any node in MEU whose latest CC

covers an earlier CC in its label. Why? Because:– The covered CC will be a redundant member of any

closed cover this node eventually leads to. No such closed cover can therefore be minimal.

– Second Pruning Principle (P2):• We can remove any node whose Uresin class

contains a PC in its label– Because each branch from that node eventually becomes

pruned by the first pruning principle.

83

Pruning Principles continued...

• These pruning principles need an example that clearly demonstrates how they work.

• The example we discussed before does not display pruning principles, so I am taking and example from Prof Ellison’s book to show it with animation as to how to apply these pruning principles.

84

Things given to us..

• Now in this example I am going to skip the basic algorithm because we had already seen two examples previously as to how to make MEU diagram.

• In this example I assume that we are given CCs and their corresponding CSs.

85

MEU Pruning ExampleCC CS

1345 124,26,56

156 145,246

124 13,26

126 14,15,24

345 24

145 NULL

134 12,26,56

56 45

16 15,24

15 14

24 13

12 26

26 14

34 NULL

2 NULL

6 NULL

Assuming this Table is given to us.

Next Slide only demonstrate How to apply Pruning Principles

Details discussed before are not repeated.

86

MEU Example illustrating Pruning Principles

NULL(1,2,3,4,5,6)

1345

(124,26,56)

345

(24,1,6)

134 ( 12,26,56)

34( 1,2,5,6)

124(26,56)

124(13,26)

24(13,6)

126(15,24,56)

26(12,56)

Omitted (see book for complete exampleHad to be completed

in accord with breadth-first principle.

IncompleteCan be finished as shown

in previous examples.

1345 X P1

134(12,26,56)

124(15,56)

24(15,56)

156 X P2 (145,246,12)

56(45,12)

Uresin

Pruned by P1because CC

1345 contains earlier node,

345

Pruned by P2because the Uresin

class of CC 156 has member 246

that covers earlier node, 26

87

References

• Techniques in Advanced Logic Switching Theory book by Prof. Ellison

• Examples are taken from Discussion sessions and class lectures and above mentioned book for easy understanding for future students.

88

Declaration

• I alone prepared and wrote this project. I received no help from any one else. This material is not copied or paraphrased from any other source except where specifically indicated. I grant my permission for this project to be placed on the course homepage during future semesters. I understand that I could receive an F for the course retroactively, even after graduation, if this work us later found to be plagiarized.

Amit Grover

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