STGs may contain redundant states, i.e. states whose. State minimization is the transformation of a given

Transcription

1 Completely Specied Machines STGs may contain redundant states, i.e. states whose function can be accomplished by other states. State minimization is the transformation of a given machine into an equivalent machine with no redundant states.

2 Completely Specied Machines Two states, s i and s j of machine M are distinguishable if and only if there exists a nite input sequence which when applied to M causes dierent output sequences depending on whether M started in s i or s j. Such a sequence is called a distinguishing sequence for (s i s j ). If there exists a distinguishing sequence of length k for (s i s j ), they are said to be k- distinguishable. Example: PS NS, z x =0 x = A E, 0 D, B F, 0 D, 0 C E, 0 B, D F, 0 B, 0 E C, 0 F, F B, 0 C, 0 states A and B are -distinguishable, since a input applied to A yields an output, versus an output 0 from B. states A and E are 3-distinguishable, since input sequence applied to A yields output 00, versus an output 0 from B. 2

3 Completely Specied Machines States s i and s j are said to be equivalent i no distinguishing sequence exists for (s i s j ). If s i is equivalent to s j and s j is equivalent to s k, then s i is equivalent to s k. So state equivalence is an equivalence relation (i.e. it is a reexive, symmetric and transitive relation). An equivalence relation partitions the elements of a set into equivalence classes. Property. If s i and s j are equivalent states, their corresponding X-successors, for all inputs X, are also equivalent. Procedure: Partition states of M so that two states are in the same equivalence class i they are equivalent. 3

4 Completely Specied Machines PS NS, z x =0 x = A E, 0 D, B F, 0 D, 0 C E, 0 B, D F, 0 B, 0 E C, 0 F, F B, 0 C, 0 P i : partition using distinguishing sequences of length i. Partition Distinguishing Sequence P 0 =(ABCDEF) P =(ACE)(BDF) x = P 2 =(ACE)(BD)(F) x = x = P 3 = (A C) (E) (B D) (F) x = x = x = P 4 = (A C) (E) (B D) (F) Algorithm terminates when P k = P k+. 4

5 Completely Specied Machines Outline of state minimization procedure: All states equivalent to each other form an equivalence class. All the states in an equivalence class may be combined into one state in the reduced (quotient) machine. These equivalence classes form a partition of the setofstates. Start with all states in a partition of a single block. Iteratively rene this partition by separating the - distinguishable states, 2-distinguishable states and so on. In general, when obtaining P k+ from P k, place in the same block of P k+ the states that are (k+)- equivalent, and in dierent blocks states that are (k + )-distinguishable. 5

6 Completely Specied Machines Theorem. The equivalence partition is unique. Theorem. If two states, s i and s j, of machine M are distinguishable, then they are distinguishable by a sequence of length n; or less, where n is the number of states in M. Denition: Two machines, M and M 2,are said to be equivalent i, for every state in M there is a corresponding equivalent state in M 2 and vice versa. Theorem. For every machine M there is a minimal machine M red that is equivalent to M and is unique up to isomorphism. Reduced machine obtained from previous example: PS NS, z x =0 x =, 0,, 0,, 0, 0, 0, 0 6

7 State Minimization of CSMs: Complexity Algorithm DFA DFA min Input: A nite automaton M =(Q q 0 F) with no unreachable states. Output: A minimal nite automaton M 0 =(Q 0 0 q 0 0 F0 ). Method:. t := 2 Q 0 := f undened g Q := F Q 2 := Q n F. 2. while there is 0 <i t, a 2 with(q i a) 6 Q j, for allj t do (a) Choose such an i, a 2, and j t with (Q i a) \ Q j 6=. (b) Q t+ := fq 2 Q i j (q a) 2 Q j g Q i := Q i n Q t+ t := t +. do. 3. (* Let [q] denote the equivalence class the state q is in and fq i g denote the set of all equivalence classes. *) Q 0 := fq Q 2 ::: Q t g: q 0 0 := [q 0]: F 0 := f[q] 2 Q 0 jq 2 F g: 0 ([q] a):=[(q a)] for allq 2 Q, a 2. 7

8 State Minimization of CSMs: Complexity Standard Implementation: O(kn 2 ), where n = jqj and k = jj Modication of the body of the while loop:. Choose such an i, a 2, and choose j j2 t with j 6= j 2, (Q i a) \ Q j 6=, and(q i a) \ Q j2 6=. 2. If jfq 2 Q i j (q a) 2 Q j gj jfq 2 Q i j (q a) 2 Q j2 gj then Q t+ := fq 2 Q i j (q a) 2 Q j g else Q t+ := fq 2 Q i j (q a) 2 Q j2 g Q i := Q i n Q t+ t := t +. Note: jq t+ j=2jq i j. Therefore, for all q 2 Q the name of the class which q contains changes at most logn times. Goal: Develop an implementation such that all work can be assigned to transitions containing a state for whcih the name of the corresponding class is changed. Suitable data structures achieve an O(kn log n) implementation Details in N. Blum IPL '96 [Original O(kn log n) algorithm in Hopcroft 97] 8

9 BDD Im- State Minimization of CSMs: plementation E 0 (x y) = YjSj i= (x i y i ) E j+ (x y) = E j (x y) ^ 8i9(o z w)[t (x i z o) ^ T (y i w o) ^ E j (z w)] 9

10 Incompletely Speci- ed Machines Statement of the problem: given an incompletely specied machine M, nd a machine M 0 such that: on any input sequence, M 0 produces the same outputs as M, whenever M is specied. no machine M 00 with fewer states than M 0 does the job. 0

11 Incompletely Speci- ed Machines Machine M: PS NS, z x =0 x = s s3, 0 s2, 0 s2 s2, - s3, 0 s3 s3, s2, 0 Attempt to reduce this case to usual state minimization of completely specied machines. Force the don't cares to all their possible values and choose the smallest of the completely speci- ed machines so obtained. In our case it means to state minimize two completely specied machines obtained from M, by setting the don't care to either 0 or.

12 Incompletely Speci- ed Machines Suppose that the ; is set to be a 0. Machine M 0 : PS NS, z x =0 x = s s3, 0 s2, 0 s2 s2, 0 s3, 0 s3 s3, s2, 0 States s and s2 are equivalent if s3 and s2 are equivalent, but s3 and s2 assert dierent outputs under input 0, so s and s2 are not equivalent. States s and s3 are not equivalent either. So this completely specied machine cannot be reduced further. 2

13 Incompletely Speci- ed Machines Suppose that the ; is set to be a. Machine M 00 : PS NS, z x =0 x = s s3, 0 s2, 0 s2 s2, s3, 0 s3 s3, s2, 0 States s is incompatible with both s2 and s3. States s3 and s2 are equivalent. So number of states is reduced from 3 to 2. Machine M 00 red : PS NS, z x =0 x = A A, A, 0 B A, 0 A, 0 3

14 Can this always be done? Machine M: PS NS, z x =0 x = s s3, 0 s2, 0 s2 s2, - s, 0 s3 s, s2, 0 4

15 Machine M 2 : Machine M 3 : PS NS, z x =0 x = s s3, 0 s2, 0 s2 s2, 0 s, 0 s3 s, s2, 0 PS NS, z x =0 x = s s3, 0 s2, 0 s2 s2, s, 0 s3 s, s2, 0 Machines M 2 and M 3 are formed by lling in the unspecied entry in M with 0 and, respectively. Both machines M 2 and M 3 cannot be reduced. Conclusion: M cannot be minimized further! But is it a correct conclusion? 5

16 Notice that we want to 'merge' two states when, for any input sequence, they generate the same output sequence, but only where both outputs are specied. This suggests the notion of a compatible set of states: a set of states that agree on the outputs where they are all specied. Machine M: PS NS, z x =0 x = s s3, 0 s2, 0 s2 s2, - s, 0 s3 s, s2, 0 In this case we have two compatible sets: A = (s,s2) and B = (s3,s2). A reduced machine M red can be built as follows. Machine M red : PS NS, z x =0 x = A B, 0 A, 0 B A, A, 0 6

17 Can we simply look for a set of compatibles of minimum cardinality, such that any original state is in at least one compatible? (This would be nice since it would lead to a simple unate covering problem.) No. To build a reduced machine we must be able to send compatibles into compatibles. So choosing a given compatible may imply that some other compatibles must be chosen too. PS NS, z s I s3,0 I 2 s,- I 3 - I 4 - s2 s6, - s2, 0 s, - - s3 -, -, - s4, 0 - s4 s,0 -, - - s5, s5 -, - s5, - s2, s, s6 -, - s2, s6, - s4, A set of compatibles that cover all states is: (s3s6), (s4s6), (ss6), (s4s5),(s2s5). But (s3s6) requires (s4s6), (s4s6) requires (s4s5), (s4s5) requires (ss5), (ss6) requires (ss2), (ss2) requires (s3s6), (s2s5) requires (ss2). So, this selection of compatibles requires too many other compatibles... 7

18 PS NS, z s I s3,0 I 2 s,- I 3 - I 4 - s2 s6, - s2, 0 s, - - s3 -, -, - s4, 0 - s4 s,0 -, - - s5, s5 -, - s5, - s2, s, s6 -, - s2, s6, - s4, Another set of compatibles that covers all states is (ss2s5), (s3s6), (s4s5). But compatible (ss2s5) requires (s3s6), (s3s6) requires (s4s6) (which requires (s4s5)) and (s4s5) requires (ss5). So must select also compatible (s4s6). Selection of minimum set of compatibles closed with respect to next state implication is a binate covering problem!!! 8

19 More formally: When a next state in unspecied, the future behaviour of the machine is unpredictable. This suggests the denition of admissible input sequence. Denition. An input sequence is admissible, or applicable, for a starting state of a machine if no unspecied next state is encountered, except possibly at the nal step. Denition. State s i of machine M is said to cover, or contain, state s j of M 2 provided every input sequence applicable to s j is also applicable to s i, and its application to both M and M 2 when they are initially ins i and s j, results in identical output sequences whenever the outputs of M 2 are specied. Denition. Machine M is said to cover machine M 2 i, for every state s j in M 2, there is a corresponding state s i in M such that s i covers s j. The problem of state minimization for an incompletely specied machine M is to nd a machine M 0 which covers M such that for any other machine M 00 which covers M, the number of states of M 0 does not exceed the number of states of M 00. 9

20 Machine M: Machine M 0 : PS NS, z x =0 x = s s3, 0 s2, 0 s2 s2, - s, 0 s3 s, s2, 0 PS NS, z x =0 x = A B, 0 A, 0 B A, A, 0 State A of M 0 covers states s and s2 ofm and state B of M 0 covers states s2 and s3 ofm. Therefore M 0 covers M. Note that M started in s under input sequence 0 0 generates 0 - -, while M 0 started in A under input sequence 0 0 generates 0 0. The output generated by M 0 corresponding to the don't care entry in M is not always the same!!! 20

21 Machine M 2 : Machine M 3 : PS NS, z x =0 x = s s3, 0 s2, 0 s2 s2, 0 s, 0 s3 s, s2, 0 PS NS, z x =0 x = s s3, 0 s2, 0 s2 s2, s, 0 s3 s, s2, 0 M 2 and M 3 are formed by lling in the unspecied entry in M with 0 and, respectively. Neither state A nor B of M 0 covers state s2 ofm 2 or M 3 and hence M 0 would not cover M 2 or M 3. 2

22 If machine M can be covered by M 0 containing fewer states than M, then some state of M 0 must cover more than one state of M. If a set of states of M can be covered by the same state of M 0, this set is called a compatible set. Intuitively: the states in a compatible set can be combined into a single state in the reduced machine. Denition: States s i and s j are compatible i they never generate dierent specied output sequences for any admissible input sequence. Example: PS NS, z A I - I 2 - I 3 E, I 4 - B C, 0 A, B, 0 - C C, 0 D, - A, 0 D E - B, 0 E, - B, - C, - - B, 0 (AC) is a compatible pair, (AD) is a compatible pair if and only if (BE) is a compatible pair. (ACD) is a compatible set. 22

23 A set of states is compatible if and only if every pair of states in that set is compatible. (BC) is a compatible pair, (AC) is a compatible pair, but (AB) is not compatible pair, so (ABC) is not a compatible set. The compatibility relation is not an equivalence relation! 23

24 Compatible sets are computed as those sets of states that do not contain any incompatible pair of states. States s i and s j are incompatible i they are not compatible. Denition: States s i and s j are output incompatible i 9i k such that (i k s i ) 6= (i k s j ), if both are specied. The set of all pairs of incompatible states can be computed as follows:. Compute output incompatible pairs. 2. Add any pair of states (s i s j ) if 9i k such that ((i k s i ) (i k s j )) is a previously determined incompatible pair of states. 3. Repeat 2. until no new pairs can be added to the incompatible state pairs set. 24

25 Example: PS Compatibles of example: C = (BE), C 2 = (AD), C 3 = (CD), C 4 = (BC), C 5 =(ACD), C 6 = (DE), C 7 =(AC), C 8 = (A), C 9 = (B), C 0 = (C), C = (D), C 2 =(E). NS, z I I 2 I 3 I 4 A - - E, - B C, 0 A, B, 0 - C C, 0 D, - A, 0 D - E, B, - - E B, 0 - C, - B, 0 25

26 A class of compatibles is of special interest: maximal compatibles. Sets of compatible states which are not subsets of any other compatible set of states are called maximal compatibles. In the case of completely specied machines, each equivalence class is a maximal compatible. Maximal compatibles of previous example: (BE), (BC), (ACD), (DE). If machine M is to be reduced to M 0, the states of M 0 must correspond to compatible sets of states of M. If a state of M 0 corresponds to a compatible set C i, then the next state of M 0 under input I j must correspond to some compatible set C m such that the next state entries in M under I j of all states in C j are contained in C m. (Why is this?) 26

27 Denition: If C i is a set of compatible states and C ij = fs k js k = (I j s i ),ands i 2 C i g i.e. C ij is the set of next states of the states in C i for input I j, then C ij is said to be implied by the set C i for inputi j. Denition: Let C i be a compatible set of states and C ij be the set of next states implied by C i for inputi j. C ij = fs k j s k = (I j s l ) s l 2 C i g The sets C ij implied by C i for all inputs I j are the implied classes of C i. Denition: A set of compatible sets C = fc C 2 :::g is closed if for everyc i 2 C all the implied sets C ij are contained in some element of C for all inputs I j. 27

28 The problem of minimizing the number of states reduces to nding a closed set C of compatible states, of minimum cardinality, which covers every state of the original machine, i.e. a minimum closed cover. Note that:. The set of all maximal compatibles of a completely specied FSM is the unique minimum closed cover. 2. For an incompletely specied FSM, a closed cover consisting of maximal compatibles only, may be a larger cover than a closed cover in which some or all of the compatibles are not maximal. This is because each compatible has a dierent set of implied compatibles, and removing a state from a compatible may result in an implied compatible that is already there, rather than a new one. This means that we may have to search over the set of all compatibles to nd the minimum cover. 28

29 But the set of all compatibles is a very large set. Is there a smaller set of compatibles (larger than the set of all maximal compatibles) that guarantees nding a minimum closed cover? Yes. Prime compatibles (Grasselli and Luccio, 965) Denition: A compatible set of states that is not \dominated" by any other compatible set is called a prime compatible set. Denition similar tothatforprime implicants of logic functions. Of course here one must specify what is meant by dominance of a compatible. 29

30 Let C i be a compatible set of states, and C ij be the set of next states implied by C i for inputi j. Denition: The class set P i implied by C i is the set of all C ij implied by C i for all inputs I j such that:. C ij has more than one element 2. C ij 6 C i 3. C ij 6 C ik if C ik 2 P i Denition: A compatible C i dominates a compatible C j if. C i C j, and 2. P i P j i.e. C i dominates C j if C i covers all states covered by C j and the conditions on the closure of C i are a subset of the conditions on the closure of C j. Denition: A compatible set of states that is not dominated by any other compatible set is called a prime compatible set. 30

31 Example: PS NS, z I I 2 I 3 I 4 A - - E, - B C, 0 A, B, 0 - C C, 0 D, - A, 0 D - E, B, - - E B, 0 - C, - B, 0 Prime compatibles and respective class sets: p =(BE) p 2 =(AD) p 3 =(CD) p 4 =(BC) p 5 =(ACD) p 6 =(DE) p 7 =(AC) p 9 =(B) p =(D) p 2 =(E) f(cb)g f(be)g f(ed)g f(da)g f(ed)(be)g f(bc)g fg fg fg fg: 3

32 The following procedure computes all prime compatibles. At the beginning, the set of prime compatibles is empty.. Order the maximal compatibles by decreasing size, say n is the size of the largest. 2. Add to the set of prime compatibles the maximal compatibles of size n. 3. For i =ton ; : (a) Generate all compatibles of size n;i and compute their implied classes. The compatibles of size n ; i are generated starting from the maximal compatibles of size n to n ; i + (only those that do not have a void class set). (b) Add to the set of primes the compatibles of size n ; i not dominated by any prime already in the set. (c) Add to the set of primes all maximal compatibles of size n ; i. 32

33 The following facts are true: A compatible already added to the set of primes, cannot be excluded by a newly generated compatible. In the previous algorithm, the same compatible can be generated more than once by dierent maximal compatibles. The question arises of nding the most ecient algorithm to generate the compatibles. Only the compatibles generated from maximal compatibles with non-empty class set need be considered, because a maximal compatible with an empty class set dominates any compatible that it generates. A single state s i can be a prime compatible if every compatible set C i with more than one state and containing s i implies a set with more than one state. Theorem For any FSM M there is a minimum equivalent FSM M red whose states all correspond to prime compatible sets of M. 33

34 A minimum closed cover can be determined by rst constructing a conjunctive form expression and then nding a satisfying assignment to it which has the fewest variables assigned TRUE. This is a binate covering problem. The table of the problem has columns that correspond to prime compatibles and rows that correspond to covering and closure conditions. It can be solved by branch and bound. At each step the table is reduced, by eliminating rows and columns, according to dominance criteria. These generalize the row and column dominance used for solving unate covering problems. 34

35 p =(BE) p 2 =(AD) p 3 =(CD) p 4 =(BC) p 5 =(ACD) p 6 =(DE) p 7 =(AC) p 9 =(B) p =(D) p 2 =(E) f(cb)g f(be)g f(ed)g f(da)g f(ed)(be)g f(bc)g fg fg fg fg: Computation of clauses in previous example. Clauses due to coverage: A: (p 5 + p 2 + p 7 ) B: (p + p 4 + p 9 ) C: (p 3 + p 4 + p 5 + p 7 ) D: (p 2 + p 3 + p 5 + p 6 + p ) E: (p + p 6 + p 2 ) Clauses due to closure: p ) p 4 : (p + p 4 ) p 2 ) p : (p 2 + p ) p 3 ) p 6 : (p 3 + p 6 ) p 4 ) (p 2 + p 5 ): (p 4 + p 2 + p 5 ) p 5 ) (p p 6 ): (p 5 + p )( p 5 + p 6 ) p 6 ) p 4 : (p 6 + p 4 ) 35

36 Need to nd a satisfying assignment for the following expression with the fewest TRUE assignments. C = (p 5 + p 2 + p 7 )(p + p 4 + p 9 )(p 3 + p 4 + p 5 + p 7 ) (p 2 + p 3 + p 5 + p 6 + p )(p + p 6 + p 2 ) (p + p 4 )( p 2 + p )( p 3 + p 6 )( p 4 + p 2 + p 5 ) (p 5 + p )( p 5 + p 6 )( p 6 + p 4 ): The assignment p = p 2 = p 4 = TRUE and all other p j = FALSE is a satisfying assignment with the fewest TRUE assignments. 36

37 This corresponds to M red = f(b E), (A D), (B C)g giving the following reduced machine. Note: PS NS, z I I 2 I 3 I 4 (A D)! -,, - (B C)!, 0,, 0, 0 (B E)!, 0,, 0, 0. Minimum form not unique. 2. A state in the original machine may be split between two states in the reduced machine. 3. resulting machine may still be incompletely speci- ed. 37

38 A A B B C C D D E E C E C E B A D E D A B C D E C, D E, A C, D B, D A, E A B C D (a) (b) An incompletely-specied machine and its implication table (A, B) (A, B) C C D D E E C E (A, B) (A, B) D E D (A, B) (A, E) (A, E) (B, C, D) (B, C, D) (B, C, D) (A, E) (B, C, D) (A, E) 0 0 (a) (b) Twominimal realizations of an incompletely-specied machine 38

39 Problem: An implied class can be contained by more than one compatible. Choose as next state in the reduced machine the compatible that optimizes some cost function, for instance the number of rows of the state transition table of the reduced machine. In general, dene a mapping problem: given a closed set of compatibles which covers all the states of the original machine, nd a mapping of the implied classes into the compatibles, so as to minimize the cost of the resulting machine. The problem can be modeled as the one of determining the set of unique representatives which minimizes row count after minimization of a symbolic relation (analogous to a boolean relation with symbolic input and output elds). 39

40 Complexity oftheproblem: Problem is NP-hard [Peeger 973] Exact solution requires computing prime compatibles (in the worst-case, order of 3 n=3 ) and then a binate covering step. Existing tools perform well on examples with few prime compatibles. Are inadequate for examples with many prime compatibles. Newly developed tools use BDD's and implicitmethods (Kam and Villa). They have been shown to be able to handle some problems with trillions of prime compatibles. [Kam and Villa have developed a package called SILK] 40