Synthesis of Two Level Circuits

Transcription

1 Synthesis of Two Level Circuits Sungho Kang Yonsei University

2 Outline Design Otimality Sums of Products and Products of Sums Imlicants and Prime Imlicants Iterated Consensus Recursive Comutation of Prime Imlicants Selecting a Subset of Primes Unate Covering Problem Branch-and-Bound Algorithm Multile Outut Functions 2

3 Design Otimality Secifications Constraints Cost Function 3

4 Two Level Logic Imlementation in two levels Seed Simlicity The number of gates and the number of gate inuts are the measure of the cost of a two level imlementation 4

5 Sums of Products A letter is a constant or a variable A literal is a letter or its comlement A roduct term is a formula of one of the following forms a non-constant literal a conjunction of non-constant literals where no letter aears more than once A sum term is a formula of one of the following forms 0 a non-constant literal a disjunction of non-constant literals where no letter aears more than once 5

6 Sums of Products An SOP (sum of roducts) is a join of meets, henceforth called OR of ANDs: f=xy z+z y+wxyz Conjunctions (ANDs) of literals are functions. They are also called cubes, terms, and imlicants An SOP is otimum, if no other SOP reresenting the same function has fewer cubes or literals Disjunctive Normal Form(DNF) - SOP Conjunctive Normal Form(CNF) - POS 6

7 Imlicants with Don t Cares x y wyz Discriminant (ON-set) Discriminant don t care Discriminant 0 (OFF-set) f = xy z + x y + wxyz, d = w xyz Cubes x y and wyz are imlicants of f : ( c= x y = ) ( x = 0, y = ) f = = 7

8 Imlicants with Don t Cares xyz Discriminant (ON-set) Discriminant don t care Discriminant 0 (OFF-set) f = xy z + x y + wxyz, d = w xyz Cube c=xyz is an imlicant of f+d 8

9 K-mas and Boolean 4-cubes y z Note these minterms are Distance wxy z w x 0 0 wxyz B 4 = {0,} 4 Note: Each vertex is a minterm of F 4 ({0,}) f=wx(y z+yz )=wxy z+wxyz 9

10 Prime Imlicants An imlicant of f is a rime imlicant if and only if no other imlicant of f contains. If so, the cube formed from by deleting any literal of must intersect the OFF-set An otimal SOP can be formed from only rime imlicants. 0

11 Cube xy is a rime imlicant Cube xy is not an imlicant. It contains offset minterm wx y z, and similarly for cube y. x xy xz Cube xz is also a rime imlicant. wx y z Fmin = xz + xy

12 Prime Imlicants Can Be Redundant yz xy xz f = xy z + x y + wxyz, d = w xyz xz, xy yz are all rime imlicants yz is redundant 2

13 The Comlete Sum Another canonical reresentation of a function f is the sum of all its rime imlicants, called the comlete sum For f = ab + bc the comlete sum is f = ab +bc + ac. Note that the consensus term ac is also a rime imlicant of f The comlete sum is unique, but usually subotimal 3

14 The Comlete Sum The comlete sum is O(3 n /n), where n is the number of inut variables The minterm canonical form is only O(2 n ) Nevertheless the comlete sum is imortant because of the role it lays in logic minimization 4

15 Comlete Sum with Don t Cares yz xz xy f = xy z + x y + wxyz, d = w xyz F = CS( f + d) = x y + xz + yz CS(f+d) denotes the comlete sum of f+d 5

16 Comlete Sum: 4-Cube Vs. K-Ma yz xy xz = xy 2= xz 3= yz Note Cube (xy ) has x=0, and y= 6

17 Comlete Sum: 4-Cube Vs. K-Ma K-Ma 4-Cube 7

18 Essential Prime Imlicants Cube is an essential rime imlicant of f if it contains a minterm not contained by any other rime imlicant of f All essential rime imlicants are resent in every otimal SOP. 8

19 Quine s Prime Imlicant Theorem A minimal SOP must always consist of a sum of rime imlicants if any definition of cost is used in which the addition of a single literal to any formula increases the cost of the formula 9

20 Essential Prime Imlicants yz xy xz Only xz, xy are essential ( xz exclusively contains minterms wx yz and wxyz of f ) 20

21 Essential Prime Imlicants 2

22 Essential Primes: K-Ma The selected function/file is: a'bc'de'+a'b'c'de'f'+a'b'c'de'f+b'c'de'+acdf+ac'de+bcdf'+ab'cde'f' essential rimes = + 2 = a'c'de'+bcdf' 22

23 Tabular Method Exress the function in minterm canonical form Method Comare adjacent airs If only one letter is different, a new term is added where the letter is relaced by a don t care All terms that are used to form new terms are marked These are reeated until no more new terms are added All unmarked terms are rime imlicants 23

24 Iterated Consensus Xy+Xy = X : distance- merging A comlete sum as a SOP formula comosed of all the rime imlicants of the function it reresents A SOP formula is a comlete sum if and only if No term includes any other term The consensus of any two terms of the formula either does not exist or is contained in some term of the formula 24

25 Comuting the Prime Imlicants Theorem If F and F 2 are comlete sums, then ABS(F F 2 ) is the comlete sum for the function F F 2 The function ABS converts the roduct to a sum and removes cubes contained in other cubes ABS((wy+xy)(y+z)) = ABS(wy+wyz+xy+xyz) = wy+xy CS(f) = ABS([x +CS(f(0,x 2,,x n ))][x +CS(f(,x 2,,x n ))]) 25

26 Comuting the Comlete Sum Procedure CS(F){ ~ F = SIMPLIFY(F) ~ ~ if ( F is unate) return ABS( F) ~ x = SELECT(SUPPORT( F)) ~ CS = CS( F ) x ~ CS = CS( F ) 0 x return ABS((x + CS ) (x + CS } 0 )) Notation: Small cas =>rocedure call Bold uright => keywords 26

27 Data Trace for Proc. CS F(v,w,x,y,z) = v xyz+v w x+v x z +v wxz+w yz +vw z+vwx z F(0,w,x,y,z) = xyz+w x+x z +wxz+w yz F(0,w,x,y,0) = w x+x +w y F(0,0,x,y,0) = x+x +y= F(0,,x,y,0) = x CS(F(0,w,x,y,0)) = ABS((w+)(w +x )) = w +x F(0,w,x,y,) = xy+w x+wx F(0,w,0,y,) = 0 F(0,w,,y,) = y+w +w= CS(F(0,w,x,y,)) = ABS((x+0)(x +)) = x CS(F(0,w,x,y,z))=ABS((z+w +x )(z +x))=w x+w z +x z +xz F(,w,x,y,z) = w yz +w z+wx z F(,0,x,y,z) = yz +z = y+z F(,,x,y,z) = x z CS(F(,w,x,y,z)) = ABS((w+y+z)(w +x z)) = ABS(wx z+w y+x yz+w z+x z) = w y+w z+x z CS(F(v,w,x,y,z)) = ABS((v+w x+w z +x z +xz)(v +w y+w z+x z)) = vw y+vw z+vx z+v w x+w xy+w xz+v w z +w yz +v x z +v xz 27

28 ABS(F) Given an SOP F, ABS(F) is the SOP derived from F by deleting any cubes of F whose minterms are contained in those of another cube of F. C = abc, Lits( C) = { abc,, }, C2 = bc, Lits( C2) = { b, c } ( C C2) ( Lits( C2) Lits( C)) ( Minterms( C) Minterms( C2)) ABS( xx 3+ xxx 34+ x23 x + x234 xx ) = xx 3+ x23 x 28

29 Data Trace for Proc. CS ~ F = ~ SIMPLIFY( F) ~ 2 if( F is unate) return ABS( ~ F) 3 x= SELECT(SUPPORT( ~ F)) 4 CS= CS( F ~ x ) 5 CS0= CS( Fx ) 6 return ABS([ x + CS ] [ x+ CS 0]) F = xyz + w x + x z + wxz + w yz ~ CS- F = w x + x z + xz + w yz ~ CS-2 F is NOT unate Consensus term xz CS-3 select variable x ~ CS-4CS = CS( F = w + z + w yz x ) ~ CS - F = w + z Absorbed ~ CS -2 F is unate:return (w + z) ~ CS- 5CS CS( F = z + w yz 0= x ) ~ CS - F = z ~ CS -2 F is unate:return (z ) No single-cube CS-6 ABS([x + (w + z)][x + (z )]) containment CS-6 return (x z + w x + w z + zx) 29

30 Picking a Subset of the Primes F = yz + xy + yz + xyz + x z MCF( F) = xyz + xyz + xyz + xyz+ xyz CS( F) = x y+ x + z y + z yz FMin = 34( + 2) xyz xyz xyz xy z xyz

31 Picking a Subset of the Primes F = yz + xy + yz + xy z + x z MCF( F) = xyz + xyz + xyz + xyz + xyz + xyz CS( F) = x y+ x + z y + z yz+ xy + xz xyz xyz xyz xy z xy z xyz

32 The Trouble with Enumeration POS to SOP conversion, like many enumeration algorithms, is in class NP (u to 2 n minterms) Thus either it has no efficient (sub-exonential) imlementation, or a host of other exhaustively studied roblems, like the traveling salesman roblem, also have efficient solutions So if you find a linear or quadratic algorithm for any of these roblems, you ll be rich and famous 32

33 Picking a Subset of the Primes F = yz + xy + yz + xy z + x z MCF( F) = xyz + xyz + xyz + xyz + xyz + xyz CS( F) = x y+ x + z y + z yz+ xy + xz Cyclic! No essentials, all rimes redundant xyz xyz xyz xy z xy z xyz

34 Generalization Logic Minimization is just one of many imortant roblems that can be formulated as a matrix covering roblem Others include Technology Maing, FSM State Minimization with DCs, and many others from various fields So we study this roblem intensively 34

35 Unate SOPs An SOP in which each variable aears only as ositive literals or only as negative literals is called Unate Unate: wy+wyz+xy+xyz = wy+xy Non-Unate: xz+xy (Note ossible consensus) If F is a unate SOP, ABS(F) is the comlete sum of the function reresented by F, and all the rimes are essential 35

36 Unate SOPs A non-unate SOP binate The roblem that is obtained by relaxing the assumtion that the constraint equation is unate is called binate covering roblem 36

37 The Astronaut and the Cookies Otimal Nutrition: A good diet should contain adequate amounts of roteins (P), vitamins (V), fats (F), and cookies (C). An astronaut, who must travel light, can choose from five different rearations: Prearation (logic variable ) contains: V and P; Prearation 2 (logic variable 2 ) contains: V and F; Prearation 3 (logic variable 3 ) contains: P and F; Prearation 4 (logic variable 4 ) contains: V; Prearation 5 (logic variable 5 ) contains: C. Can the astronaut have a balanced diet with only two rearations? Proteins: ( + 3 ), Vitamins: ( ),.. 37

38 The Astronaut and the Cookies Brute Force Enumeration Aroach: Encode candidates with binary variables i (like rimes) Exress constraints in POS form (like minterm coverage) Convert POS to SOP (enumerate all ossibilities) Pick smallest cube (fewest literals) = (Proteins)( Vitamins)(Fats)(Cookies) =( + 3) ( )( 2 + 3)( 5) =( + 3( 2 + 4))( 2 + 3)( 5) =( )( 2 + 3)( 5) + 3( 2 + 4)( 5) = All ossible solutions! 38

39 Reductions: Essential Columns Variables with singleton constraints, like P 5 in the cookies covering roblem, are essential and are resent in every otimum solution If Row i of M contains a single nonzero in Col j, then add j to the artial solution and delete all rows of M with a nonzero in Col j Col j, lus the otimum solution to the reduced matrix, is an otimum solution of the original roblem 39

40 40 Essential Columns: Examle =( ( ) )( )( ) 0 0 R 0 R 0 0 R M R R 0 0 R R M = = S={ 5 } S 5, lus an otimum solution (say 2 ) to the reduced matrix, is an otimum solution of the original roblem

41 Reductions: Row Dominance Row i of M dominates Row k if every nonzero of Row k is matched by a nonzero of Row i in the same column, that is, Mkj = Mij =, j Any set of columns that covers Row k will also cover Row i Hence dominating rows may be deleted without affecting the size of the otimum solution Such deletion may lead to column dominance (later today) 4

42 Row Dominance: Examle M = 2 0 0, M = S = S Note: Row 4 dominates Row 42

43 Reductions: Column Dominance Col j of M dominates Col k if every nonzero of Col k is matched by a nonzero of Col j in the same row, that is Mik = Mij =, i Any set of columns that contains Col j will also cover all rows covered by Col k. Hence dominated columns may be deleted without affecting the size of the otimum solution Such deletion may lead to row dominance or even reveal new essential columns 43

44 Column Dominance: Examle M = 2 0 0, M = S = S Col dominates Cols 4 and 5 Row 4 co-dominates Row Note cyclic core 44

45 Matrix Reduction Procedure REDUCE( M ) EC = COLS_ OF_ SINGLETON_ ROWS( M ) 2 " delete cols in EC and rows with cols in EC" 3 " Add cols in EC to otimum solution" 4 " delete rows which dominate other rows" 5 " delete cols which are dominated by other cols" 6 if( M ) reeat - 6 Essential Column reduction Row Dominance reduction Column Dominance reduction Iterate: Result is unique Cyclic Core 45

46 ) )( ( ) )( )( ( : ) )( ( ) ( ) ))( )( ( ( ) )( )( )( ( : M M + + = + + = = = = Note: Col dominance kills alternative solutions M M = = , Note: Row dominance kills constraints Matrix Reduction

47 Exloring Search Sace When the constraint matrix cannot be further reduced If the matrix has no rows left, roblem solved Otherwise, cyclic To solve large enumeration roblems, use imlicit enumeration Branch and Bound 47

48 Lower Bounds M = Note rows,2,3 are column disjoint, that is, the nonzero column sets {,3},{2,4},{5,6} are airwise disjoint Thus at least 3 columns are required to cover the rows of M Thus if we somehow find a solution of size 3, we can take it and quit --- it is an otimum solution 48

49 Lower Bounds Row i of M is disjoint from Row k if no nonzero of Row i is matched by a nonzero of Row k in the same column, and conversely. That is, M = M = 0, M = M = 0, j ij kj kj ij m distinct columns are required to cover m air-wise disjoint rows Hence the cost of covering a matrix that contains m air-wise disjoint rows is at least m Lower bounds can rune vast regions of the search 49

50 Lower Bounds An indeendent set of rows is called maximal if it intersects every other row of the covering matrix Unless some of the decisions already made in building the set are reversed, the set cannot grow larger while retaining its indeendence 50

51 Quick Lower Bound Algorithm Procedure MIS_ QUICK( M ) MIS = 2 do { 3 i = 4 MIS = MIS {} i 5 M = 6 } while( M > 0 ) continue 7 return row set MIS CHOOSE_SHORTEST_ ROW ( M ) DELETE_ INTERSECTING_ ROWS ( M) This is a chea heuristic: Finding best lower bound can be harder than solving the original covering roblem 5

52 MIS_QUICK Examle Procedure MIS_ QUICK( M ) MIS = 2 do { 3 i = 4 MIS = MIS {} i 5 M = 6 } while( M > 0) continue 7 return row set MIS MIS = {} CHOOSE_SHORTEST_ ROW DELETE_ INTERSECTING_ ROWS ( M ) ( M) MIS = {,} 3 MIS = {, 35, } 52

53 Examle M 0000 w =2 w 2 =5 000 w 2 =3 w2 2 = w 3 =2 w 2 3 =6 000 w 4 =3 w 2 4 = w 5 =2 w 2 5 = w 6 =2 w 2 6 = w 7 =2 w 2 7 =6 MIS={} on the first ass MIS={,3} on the second ass 53

54 Examle For second heuristic, MIS={} on the first ass M 0000 w 3 =2 w2 3 = w 5 =2 w2 5 = w 6 =2 w2 6 =3 MIS={,5} on the second ass MIS={,5,6} on the third ass 54

55 Effective : Enumeration Accet the inevitability of exonential worst case erformance -- Imlicitly Enumerate the search sace Reduce the roblem into simlest equivalent roblem--at least one otimum solution remains Prune the search sace by comuting lower bounds 55

56 Branch and Bound Algorithm Procedure BCP( M, U, currentsol){ ( M, currentsol) = REDUCE( M, currentsol) if ( terminalcase( M)){ \\ M = 0 if ( COST( currentsol) < U){ U = COST( currentsol) return ( currentsol) } else return(" no (better) solution (in this subsace) ") } L = LOWER_ BOUND( M, currentsol) if ( L U) return(" no (better) solution (in this subsace) ") xi = CHOOSE_ VAR( M) \\ longest column S = BCP( M, U, currentsol { x }) xi if ( COST( S ) = L) return( S ) 0 S = BCP( M, U, currentsol) return } x i BEST_ SOLUTION 0 i ( S, S ) 56

57 Choice of Slitting Variable A column that covers many rows is tyically a better candidate than a column that covers few row A ossible refinement of this strategy consists of favoring columns that cover many rows This criterion is based on the assumtion that shorter rows have a lower chance of being covered 57

58 Solved Problem 4-4: Reduction Initial uer bound: U=5+= Cols 2,4 Dominate,3 Resectively Row 4 (Co-) dominates Row Cyclic Core Note: No essential columns 58

59 Solved Problem 4-4: Slitting Row counts MIS={}, U=6 L=0+ MIS =0+= (L U) (So recur) Column counts Slitting Variable Heuristic: Pick long columns which cover many short rows 59

60 Solved Problem 4: Binary Recursion = currentsol={2} EC={4} U=+= = currentsol={} EC={4,5} U=0+2=2 60

61 Examle Showing Lower Bounding Initial uer bound: U=6+= Cyclic Core: no essential columns, Row dominance, or Column dominance Row Intersects rows 2-4 MIS={}, L=0+= Slit on Column 6

62 Examle Showing Lower Bounding Recursive Call for = = currentsol={ }, EC={ 2 }, U=+= {P, P 2 } is a best solution which includes P (THE?) 62

63 Examle Showing Lower Bounding Recursive Call for =0 = 0 EC={}, MIS={,2}, L=0+2=2 U Cyclic core 63

64 3x Examle M x Killed by row MIS={,3,5,7}, L=0+4=4 Ignore Heuristic of icking longest column! Pick Col M 0 Dro Col M Dro Col, and Rows,4,2 64

65 3x Examle x x M ded x cyclic Cols 2,4 are dominated, so Col 3 becomes essential MIS={7,9}, L=2+2 = 4 Pick longest column, Col 5 M 2 00 Dro Col 5 M 2 0 Dro Col 5, and Rows 5,9-3 65

66 Examle M ddd U=5 Cyclic (takes 2 Cols), Cost = 3+2=5=U, 3, 5 6, 7 L=4 7 6 L=4 3 U= ( M, currentsol) = REDUCE( M, currentsol) if ( terminalcase( M)) { if ( COST( currentsol) < U) { U = COST( currentsol) return ( currentsol) } else return(" no (better) solution") } L = LOWER_ BOUND ( M, currentsol) if ( L U) return(" no (better) solution") xi = CHOOSE_ VAR( M) S = BCP( Mx, U, currentsol { xi}) 544 i if ( COST( S ) = L) return( S ) no 0 S = BCP( Mx, U, currentsol) i 0 return BEST_ SOLUTION ( S, S ) 66

67 Examle L=4 L=4 3 U=5 M U= MIS={5,0,}, L=2+3=5 xi ( M, currentsol) = REDUCE( M, currentsol) if ( terminalcase( M)) { if ( COST( currentsol) < U) { U = COST( currentsol) return ( currentsol) } else return(" no (better) solution") } L = LOWER_ BOUND ( M, currentsol) if ( L U) return(" no (better) solution") xi = CHOOSE_ VAR( M) S = BCP( M, U, currentsol { x }) if ( COST( S ) = L) return( S ) 0 S BCP( M, U, currentsol) = x return i BEST_ SOLUTION 0 i ( S, S ) 67

68 Examle x x x M 0 0 x x x x d EdE E L=4 U= d L= MIS={5,7}, L=2+3 = 5 5, 7 2, 4, 6 7 L=U=5 U=5 L=U=5 68

69 Examle Summary L=4 U=5 7 L= L=U=5 U=5 L=U=5 5 Recursive Calls 5 Reductions 5 Lower Bounds 3 Slitting Choices 69

70 Examle M Columns 2, 4, are essential After the rows they cover are eliminated, Column 0 is dominated by 5 Also row 3 is dominated by For the artial solution 2=4==, the reduced matrix has MIS={,4}, so the lower bound is 3+ MIS =3+2=5 70

71 Examle If we slit on 5, for 5=, 7 becomes essential So we can get a cost of 5, with a solution =4==5=7= =8=9=0 7= For 5=0, 5 and 8 become essential, leading to the solution =4==6=8= =9=0 6=8= 7

72 Minimizing Weighted Cost To make cost sensitive to transistor count, set the the weight of a rime imlicant to its literal count COST(x y)=2, COST(uvwx yz )=6 Thus a dominated column is only removed if the dominating column has lower (or the same) cost The rest of BCP is unaffected 72