1. 1 Presentation on Row Reduction Techniques Presented By : Name: Amit Grover Email : grover@usc.edugrover@usc.edu

2. 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. 3 What is Row Reduction ??

4. 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. 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. 6 State Variables and External Inputs. External Inputs 00 01 11 10 12341234 State Variables

7. 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. 8 Completely Specified State Machine (CSSM) 2,03,1 4,11,0 2,01,1 3,16,0 5,1 4,15,0 123456123456 State Variables External Inputs AB No don’t cares

9. 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. 10 Incompletely Specified State Machine (ISSM) -,03,1 4,1-,0 2,01,- 3,1-,- 6,05,1 -,--,0 123456123456 State Variables External Inputs AB

11. 11 Algorithms for Row Reductions

12. 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 Involved i.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. 13 Given CSSM 2,03,1 4,11,0 2,01,1 3,16,0 5,1 4,15,0 123456123456 AB Figure 1

14. 14 How to read CSSM 2,03,1 4,11,0 2,01,1 3,16,0 5,1 4,15,0 AB 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. 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. 16 Implementation of Step 1 of Pairchart Algorithm 1 2 3 4 5 6 123456123456 Increasing Order

17. 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. 18 Implementation of Step 2 of Pairchart Algorithm 1 2 3 4 5 6 123456123456 Increasing Order Cutting from diagonal

19. 19 123456123456 123456123456 Increasing Order Figure2 2,03,1 4,11,0 2,01,1 3,16,0 5,1 4,15,0 1 AB 2 3 4 5 6

20. 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. 21 123456123456 123456123456 2,03,1 4,11,0 2,01,1 3,16,0 5,1 4,15,0 1 AB 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. 22 123456123456 123456123456 2,03,1 4,11,0 2,01,1 3,16,0 5,1 4,15,0 1 AB 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. 23 123456123456 123456123456 2,03,1 4,11,0 2,01,1 3,16,0 5,1 4,15,0 1 AB 2 3 4 5 6 X X X X X X X X X Similarly we will fill for all the blocks

24. 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. 25 123456123456 123456123456 2,03,1 4,11,0 2,01,1 3,16,0 5,1 4,15,0 1 AB 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. 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. 27 123456123456 123456123456 2,03,1 4,11,0 2,01,1 3,16,0 5,1 4,15,0 1 AB 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 41 41 - So we will write 4 and 1 in each block in column 2 and Row 2 (if no ‘X’) as shown..

28. 28 123456123456 123456123456 2,03,1 4,11,0 2,01,1 3,16,0 5,1 4,15,0 1 AB 2 3 4 5 6 X X X X X X X X X 2 3 2 3 41 41 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. 29 123456123456 123456123456 2,03,1 4,11,0 2,01,1 3,16,0 5,1 4,15,0 1 AB 2 3 4 5 6 X X X X X X X X X 2 3 2 3 41 41 2 1 3 6 6 5 4 5 Lets fill the whole table accordingly

30. 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. 31 123456123456 123456123456 2,03,1 4,11,0 2,01,1 3,16,0 5,1 4,15,0 1 AB 2 3 4 5 6 X X X X X X X X X 2 3 2 3 41 41 2 1 3 6 6 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. 32 123456123456 123456123456 2,03,1 4,11,0 2,01,1 3,16,0 5,1 4,15,0 1 AB 2 3 4 5 6 X X X X X X X X X 2 3 2 3 41 4 1 2 1 3 6 6 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. 33 123456123456 123456123456 2,03,1 4,11,0 2,01,1 3,16,0 5,1 4,15,0 1 AB 2 3 4 5 6 X X X X X X X X X 2 3 2 3 41 41 2 1 3 6 6 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. 34 123456123456 123456123456 2,03,1 4,11,0 2,01,1 3,16,0 5,1 4,15,0 1 AB 2 3 4 5 6 X X X X X X X X X 2 3 2 3 41 41 2 1 3 6 6 5 4 5 Let’s fill the Whole Pairchart X x

35. 35 Pairchart Algorithm continued.. 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. 36 Pairchart Algorithm continued.. 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. 37 1 3 6 2 5 ->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. 38 Pairchart Algorithm continued.. 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. 39 Pairchart Algorithm continued.. 1 3 6 2 5 4

40. 40 Pairchart Algorithm continued.. 26,0135,1 4,1135,0 135,126,0 135 26 4 AB New 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. 2,03,1 4,11,0 2,01,1 3,16,0 5,1 4,15,0 1 AB 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. 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. 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. 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. 44 ISSM: Pairchart Algorithm 1-2a-B 4a3b-- --3b8- 1-2-5- 3-7-5- -b-b-b --6--a 2-4--b 1 2 3 4 5 6 7 8 A B C Given:

45. 45 1 2 3 45 6 123456123456 1-2a-B 4a3b-- --3b8- 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 XXXX Shows that 2&3 state is compatible to each other

46. 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. 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. 48 Skill Algorithm Continue.. KSkSk L’L 8NULL 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. 49 Skill Algorithm Continued.. KSkSk L’L 8NULL 8 7 77,8 6 5 4 3 2 1 Look for states 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 Refer Pairchart for State 7 in Column 7

50. 50 Skill Algorithm Continued.. KSkSk L’L 8NULL 8 7 77,8 6767,667,8 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 so 67,8 in L 6 is compatible with 7 so.. Sk will have 7. MC’s Refer Pairchart for State 6 in Column 6

51. 51 Skill Algorithm Continue.. KSkSk L’L 8NULL 8 7 77,8 6767,667,8 5 67567,5567,8 4 3 2 1 Look for states which 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 5,67 But not 8 so 567,8 in L 5 is compatible with 67 so.. Sk will have 67. MC’s Refer Pairchart for State 5 in Column 5

52. 52 Skill Algorithm Continued.. KSkSk L’L 8NULL 8 7 77,8 6767,667,8 5 67567,5567,8 4 646,4567,46,8 3 2 1 Look for states which Intersects. Concatenate 4 with states which intersects. In this case 6 intersects so L’ will have 46,4 in L’. 46 covers 4 so 567,8, 46 in L 4 is compatible with 6 so.. Sk will have 6. MC’s Refer Pairchart for State 4 in Column 4

53. 53 Skill Algorithm Continued.. KSkSk L’L 8NULL 8 7 77,8 6767,667,8 5 67567,5567,8 4 646,4567,46,8 35673567,346,3567,8 2 1 Look for states which 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, 46 in L 3 is compatible with 567 so.. Sk will have 567. MC’s Refer Pairchart for State 3 in Column 3

54. 54 Skill Algorithm Continued.. KSkSk L’L 8NULL 8 7 77,8 6767,667,8 567567,5567,8 4646,4567,46,8 35673567,346,3567,8 2347 24,237,2 46,24,3567,8,237 1 Look for states which Intersects. 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 2 so 3567,8, 46,237,24 in L 2 is compatible with 347 so.. Sk will have 347. MC’s Refer Pairchart for State 2 in Column 2

55. 55 Skill Algorithm Continued.. KSkSk L’L 8NULL 8 7 77,8 6767,667,8 567567567,8 4646567,46,8 3567356746,3567,8 234724,23746,24,3567,8,237 1414,146,8,3567, 24,237,14 Look for states which Intersects. Concatenate 1 with states which intersects. In this case 4 intersect so L’ will have 14,1 14 covers 1 so 3567,8,46,237,24,14 in L 1 is compatible with 4 so.. Sk will have 4. MC’s Refer Pairchart for State 1 in Column 1

56. 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. 57 123456123456 123456123456 -,0-, -4,- -,0 3,- 1,12,--,- 2,-4,--,- 4,1-,1 5,--,02,0 1 0 2 3 4 5 6 4343 X X 1 2 2 4 X 2424 1 2 X 4242 X 2525 Pairchart Algorithm already seen X

58. 58 Skill Algorithm KSkSk L’L 6NULL66 5 55,6 454545,6 3456345,36 242424,36,345 1245616,145,124124,16,145, 345,36 MCs Till this point we knew how to draw it..

59. 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. 60 Bargain Hunter Algorithm 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. 61 Bargain Hunter Algorithm MCs ->124,16,145,345,36 from SKILL Algorithm CC 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. 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. 63 Bargain Hunter Algorithm CC Inputs 0 1 2 CS 34512 24 - 1242 4 34 1452 4 4 3615 2 2 165 - 16 3412 24 - 351 24 - 452 4 - 12- - 34 142 4 4 242 4 3 15- 4 4 31 2 - 42 4 - 5- 4 - 1 2- - 3 65 - 2 -,0-, -4,- -,0 3,- 1,12,--,- 2,-4,--,- 4,1-,1 5,--,02,0 1 2 3 4 5 6 012012

64. 64 Bargain Hunter Algorithm CC Inputs 0 1 2 CS 34512 24 - 1242 4 34 1452 4 4 3615 2 2 165 - 16 3412 24 - 351 24 - 452 4 - 12- - 34 142 4 4 242 4 3 15- 4 4 31 2 - 42 4 - 5- 4 - 1 2- - 3 65 - 2 -,0-, -4,- -,0 3,- 1,12,--,- 2,-4,--,- 4,1-,1 5,--,02,0 1 2 3 4 5 6 012012 Similarly we can do for all CCs

65. 65 Bargain Hunter Algorithm 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. 66 Bargain Hunter Algorithm CC Inputs 0 1 2 CS 34512 24 -12,24 1242 4 3434 1452 4 4NULL 3615 2 215 165 - 1624 3412 24 -12,24 351 24 -24 452 4 -NULL 12- - 3434 142 4 4NULL 242 4 3NULL 15- 4 4NULL 31 2 -NULL 42 4 -NULL 5- 4 -NULL 1- 4 -NULL 2- - 3NULL 65 - 2NULL -,0-, -4,- -,0 3,- 1,12,--,- 2,-4,--,- 4,1-,1 5,--,02,0 1 2 3 4 5 6 012012

67. 67 Bargain Hunter Algorithm CC Inputs 0 1 2 CS 34512 24 -12,24 1242 4 3434 1452 4 4NULL 3615 2 215 165 - 1624 3412 24 -12,24 351 24 -24 452 4 -NULL 12- - 3434 142 4 4NULL 242 4 3NULL 15- 4 4NULL 31 2 -NULL 42 4 -NULL 5- 4 -NULL 1- 4 -NULL 2- - 3NULL 65 - 2NULL -,0-, -4,- -,0 3,- 1,12,--,- 2,-4,--,- 4,1-,1 5,--,02,0 1 2 3 4 5 6 012012 Ignore Singleton Similarly for all other Rows we can find CS

68. 68 Bargain Hunter Algorithm 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. 69 Bargain Hunter Algorithm CCCS 34512,24 12434 145NULL 3615 1624 3412,24 3524 45NULL 1234 14NULL 24NULL 15NULL 3 4 5 1 2 6 CC 145 is better than CC 45 CC 124 is better than CC 12 CC 145 is better than CC 14 CC 145 is better than CC 15 CC 145 is better than CC 4 CC 145 is better than CC 5 CC 145 is better than CC 1 CC 24 is better than CC 2 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. 70 Bargain Hunter Algorithm CCCSPC 34512,24345 12434124 145NULL145 361536 162416 352435 24NULL24 3NULL3 6 6 Prime Compatibles Those CCs which are left are our PCs

71. 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. 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 occurrences in 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. 73 MEU Method PCs : 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 occurrences in PCs 1, 3, 3 We will choose 1/5/6 I am choosing right most You can choose whatever you want We will choose 34 Left most

74. 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 thus when compared with previous PC 24’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 PC 24’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) PCCS 34512,24 12434 145NULL 3615 1624 3524 NULL 3 6 (3,5)

75. 75 MEU Method PCs : 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 PC 36’s Uresin class completely So 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. 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. 77 Minimal covering machine -,0-, -4,- -,0 3,- 1,12,--,- 2,-4,--,- 4,1-,1 5,--,02,0 1 2 3 4 5 6 145 24 36 0 1 2 Input States 01 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. 78 Minimal covering machine -,0-, -4,- -,0 3,- 1,12,--,- 2,-4,--,- 4,1-,1 5,--,02,0 1 2 3 4 5 6 145 24 36 0 1 2 Input States 01 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. 79 Minimal covering machine -,0-, -4,- -,0 3,- 1,12,--,- 2,-4,--,- 4,1-,1 5,--,02,0 1 2 3 4 5 6 145 24 36 0 1 2 Input States 01 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

80. 80 Minimal covering machine -,0-, -4,- -,0 3,- 1,12,--,- 2,-4,--,- 4,1-,1 5,--,02,0 1 2 3 4 5 6 145 24 36 0 1 2 Input States 01 2 24,0 24,1 24,0 36,1 24,0 145,1 Similarly we will fill for all the states

81. 81 Pruning Principles for the MEU Method

82. 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. 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. 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. 85 MEU Pruning Example CCCS 1345124,26,56 156145,246 12413,26 12614,15,24 34524 145NULL 13412,26,56 5645 1615,24 1514 2413 1226 14 34NULL 2 6 Assuming this Table is given to us. Next Slide only demonstrate How to apply Pruning Principles Details discussed before are not repeated.

86. 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 example Had to be completed in accord with breadth- first principle. Incomplete Can 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 P1 because CC 1345 contains earlier node, 345 Pruned by P2 because the Uresin class of CC 156 has member 246 that covers earlier node, 26

87. 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. 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