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ECE465. FSM State Minimization for Completely Specified Machines Shantanu Dutt. Acknowledgement: Slides prepared by Huan Ren from Prof. Dutt’s Lecture Notes (some modifications made by Prof. Dutt). A. C. Reset. 0 (even). 1 (odd). B. D. State Minimization of Seq. Ckts .

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Ece465

ECE465

FSM State Minimization for Completely Specified Machines

Shantanu Dutt

Acknowledgement: Slides prepared by Huan Ren from Prof. Dutt’s Lecture Notes

(some modifications made by Prof. Dutt)


State minimization of seq ckts

A

C

Reset

0 (even)

1 (odd)

B

D

State Minimization of Seq. Ckts.

  • Removal of redundant states

  • Important because:

    • Cost(a): # of FFs # of states

    • Cost(b): logic complexity # of states

    • Easier to diagnose faults if there are no redundant states

  • Example: Odd-parity detection

  • 0/1

    Reset

    1/1

    1/1

    Can this sub-optimal design be corrected by systematic techniques?

    1/0

    0/0

    0/0

    1/0

    1/0

    0/0

    0/1

    1/1

    0/1

    Minimal FSM

    Non-minimal FSM


    Definitions

    2-equiv.

    Definitions

    • Two states Si and Sj are 1-equivalent, if any input sequence (can be multiple bits) of length 1 produce identical output responses.

    • Two states Si and Sj are k-equivalent, if any input sequence of length k produces identical output sequences.

    • Example 1


    Examples contd

    3-equiv.

    Examples (contd.)

    • Example1 (contd.)


    Examples contd1

    0/1

    1/1

    A

    C

    1/0

    2-equiv.

    0/0

    0/0

    B

    D

    1/0

    1/1

    0/1

    Examples (contd.)

    • Example 2: Parity detection

    R


    Definitions contd

    1/0

    0/1

    Sk

    Sp

    Si

    Implied pairs

    of Si, Sj

    Implied pairs

    of Si, Sj

    0/1

    1/0

    Sl

    Sq

    Sj

    Definitions (Contd.)

    • States Si,……,Sj of a seq. ckt. are said to be equivalent if and only if for every possible i/p seq. of any length, the o/p’s produced by the ckt will be identical irrespective of whether the ckt is in states Si,……,Sj when the i/p seq. starts.

       Si,……,Sj are k-equivalent for all k

    • Alternative (and more practical!) definition for equivalence:

      • Si and Sjare equiv. if and only if for every possible i/p Ip of length 1

        • (i) The o/p’s produced by states Si= o/p’s produced by Sj.

        • (ii) The next states Skfor Si and Slfor Sj are equivalent.

    • Definition: Implied pairs (or implied next-state [NS] pairs) of a pair of states Si, Sj is a pair of states Sk,Sl which are N.S.’s of Si and Sj for the same 1-length I/P Ip

    This is a recursive definition that helps in detecting more equiv. states once some sets of equiv. states are detected.


    Equivalent state detection using the 2 nd state equiv defn

    0/1

    E

    1/0

    State pair

    C

    B

    1/0

    0/1

    Equivalent State Detection Using the 2nd State Equiv Defn

    • Idea 1:

      • Using defn. 2’s negation—Si, Sj are non-equiv. if any of their implied pairs are not equiv. (i.e., not k-equiv. for some k), iteratively find non-equiv state pairs

      • Stop when no more non-equiv state pairs cannot be found

      • The remaining state pairs are equivalent (a la Sherlock Holmes!—after all alternatives are eliminated the remaining possibility has to be correct, however, implausible it may seem to be)

      • This is the partitioning method

    • Idea 2:

      • Bootstrap the equiv state-pair finding process by finding a base & easy-to-detect-&-verify pattern that implies equivalency, e.g., see the foll. pattern which implies equiv. of states B and C:

      • Use these initial pair of equiv states to

        determine other equiv state pairs

        using defn. 2.

      • This is the implication table method


    Definitions contd1

    Equiv.

    Ip1/Op1

    Si

    Sj

    Ip1/Op1

    Equiv.

    Sk1

    Sk2

    Sl1

    Sl2

    Equiv.

    O/P O1

    O2

    O1

    O2

    Same

    Definitions (Contd.)

    • The second definition is the same as the first one

      • Necessity for (i) is clear

      • Necessity for (ii), let Sk, Sl not be equivalent

        => There is an input seq. I1,……,It that produces different o/p’s starting from states Sk and Sl

        => The seq. Ip, I1,……,It produces diff. o/p starting from Si and Sj

        => Si, Sjare not equiv.

    • Consider any i/p seq Ip, I1,……,It

    • After recv’ing Ip Si & Sj produce the same o/p and fo to equiv states (by defn) Skp & Slp

    • Since Skp & Slp are equiv states, seq I1,……,Itwill produce identical o/ps from either state

    • Thus Si & Sj produce identical o/ps for any arbitrary seq, and are thus equivalent

    • Sufficiency


    Definitions contd2

    Definitions (Contd.)

    • For a set S of elements, a binary relation R is a set of ordered pairs (Si, Sj) s.t. Si, Sj belong to S, and SiΔSj, where Δ is the actual relation represented by R

    • A relation R on elements of S is reflexive if xΔx, e.g., Δ is ≤

    • A relation R on elements of S is symmetricif xΔy yΔx,

      e.g., Δ is ≤

    • A relation R on elements of S is transitiveif xΔy and yΔz  xΔz e.g., ≤

    • A relation R on elements of S is equivalentif it is reflexive, symmetric and transitive. Example: sibling relationship, divides

    • A relation R is a compatibility relation if it is reflexive and symmetric. Example: friendship, gcd (greatest common divisor) > 1


    Definitions contd3

    Definitions (Contd.)

    • Note 1: If there is an equivalence relation R on a set S, the the elements of S can be partitioned into disjoint subsets called equivalence classes, where elements in each subset are related to each other by relation R:

      Important concept for completely specified FSMs—each equiv. class of equiv. state does not conflict w/ any other class, and thus can replace each equiv. class in a well-defined way by a single state

    • Note 2: If there is a compatibility relation R on a set S, then R defines subsets of S referred to as compatibility classes (the elements in a subset Si are related to each other by R). These subsets are not disjoint, in general.

      Important concept for incompletely specified FSMs—need to find maximal set of mutually compatible states for maximum reduction in the # of states

    Compatibiity classes may intersect

    Equivalent classes are disjoint


    Minimization method for completely specified fsms

    Minimization Method for Completely Specified FSMs

    • Two methods. Partitioning and Implication Table

    • The Partitioning Method:

    Use condition (i) of the alternative equivalency defn to determine initial partition P1=subsets/blocks of 1-equiv. states;

    /* k-equivalency & equivalency are equivalence classes */

    i=1;

    Repeat

    For each subset/block Cj in Pi do

    Begin

    Two or more states in Cj are placed in the same block of Pi+1 iff for each I/P value of length 1 their next states lie in the same block of Pi. Otherwise the relevant states are separated into diff. blocks based on where their next states lie.

    End

    Pi+1=set of all new blocks created. /* each bl. contains (i+1)-equiv states */

    i = i+1;

    Until (Pi=Pi-1)/* Note: this uses condition (ii) of the alternative equivalency defn to construct equivalent state sets */


    Partitioning method contd

    Partitioning Method (contd)

    • Let Pk be the final partition

      • Pk is an equivalence relation.

      • All blocks in Pk contain equivalent sets

      • Eliminate all but one state in each block Ci denoted repr(Ci) of Pk and reduces the FSM as follows

        • A) For arc(s) from states in Ci to a state in Cj, draw an arc (w/ the same input/output values) with same labels from repr(Ci) to repr(Cj). Do not duplicate arcs.

        • b) Resulting FSM is minimized.


    Partitioning method example 1

    P1=(A,B) (C,D)

    A

    C

    Reset

    A

    D

    B

    D

    Partitioning Method: Example 1

    0/1

    P1=(A,B) (C,D)

    1/1

    X=0: NS: B A C D

    X=1: NS: C D A B

    1/0

    0/0

    0/0

    P2=(A,B) (C,D)

    No change from P1; stop

    1/0

    1/1

    0/1

    Non-minimal STD

    1/1

    1/0

    0/0

    0/1

    Minimal STD


    Partitioning method example 2

    P1=(A,B,C) (D,E)

    Need to check for each I/P of length 1

    X=0 C C B D E

    X=1 B E E B A

    N.S.’s

    P2=(A) (B,C) (D,E)

    N.S. will not change, but whether N.S.’s belong to same blocks will

    X=0 C B D E

    X=1 E E B A

    0

    1

    P3=(A) (B,C) (D) (E)

    X=0 C B

    X=1 E E

    A

    B/1

    B/0

    B

    B/1

    E/0

    D

    D/0

    B/1

    E

    E/0

    A/1

    Partitioning Method: Example 2

    0

    1

    A

    C/1

    B/0

    B

    C/1

    E/0

    C

    B/1

    E/0

    D

    D/0

    B/1

    E

    E/0

    A/1

    P4=(A) (B,C) (D) (E) = P3


    Partitioning method why it works basic idea

    k-equiv, but

    not (k+1)-equiv.

    A1

    A2

    B1

    B2

    (k-1)-equiv, but

    not k-equiv.

    Partitioning Method—Why It works (basic idea)?

    • Lemma 1: If there are two states A1, B1 that are k-equiv. but not (k+1)-equiv then the algorithm will detect this in iteration k

      Proof: By definition of (k+1)-equivalency, A1, B1 will have at least one implied next-state pair (A2, B2) that are (k-1)-equiv but not k-equiv (otherwise A1, B1 will be (k+1)-equiv.)

       In iteration k, A2, B2 will be in diff. blocks (assuming lemma true for smaller values of k—proof by induction)

       this will be detected in iter. k, and A1, B1 will thus be separated in diff. blocks at the end of iter. k.

    • Theorem 1: All blocks contain equivalent states after the algorithm terminates.

      Proof: Suppose not for 2 states X, Y that are in the same block when algo terminates, and let algo terminate in iter. m.

       By Lemma 1, X, Y are at least (m+1)-equiv.

       By our assumption, then there must be a j > m s.t.. X, Y are j-equiv but not (j+1)-equiv.

       By Lemma 1, the algo will catch this in iter. j.

       By Lemma 1, if we run the algo up to iter. j, the algo will separate X, Y in separate blocks in this iter. However, algo terminated in iter. m means Pm = Pm+1.

       If we run the algo for more iterations, its partition input remains unchanged compared to iter. m, and it will thus producing the same result as iter. m

       We will get:: Pm = Pm+1 = Pm+2 = ……. = Pj+1.

       Thus X,Y remains in the same block in iter. j

       X, Y are (j+1)-equiv and we reach a contradiction w/ our

      assumption.

       Theorem 1 has to be true


    Partitioning method why it works detailed proofs

    A1

    A2

    B1

    B2

    Partitioning Method—Why It works? (detailed proofs)

    • Lemma 1: If there are two states that are k-equiv. but not (k+1)-equiv then the algorithm will detect this in iteration k

    • Proof by induction

      • Basis (k=0): Not 1-equiv. is caught at iteration 0 (before iteration 1, i.e., when forming P1)

      • Hypothesis: True for k=m

      • To prove for k=m+1: Use property: If two states A1 and B1 are (m+1)-equiv but not (m+2)-equiv., at least one of their N.S. pairs (A2, B2) has to be m-equiv. but not (m+1)-equiv. Can be easily proved by contradiction—do as an exercise.

    • Hence, by ind. hypothesis, A2 and B2 are in diff. blocks of Pm+1 (i.e., their non-(m+1)-equiv. is detected in iter m) according to the hypothesis

    • Thus, by the procedure, A1 and B1 will be in diff. blocks of Pm+2 at the end of iteration m+1, i.e., their non-(m+2)-equiv is detected in iter m+1

    m equiv.

    m

    1 equiv.

    Am+1

    Bm+1

    not 1 equiv.

    Am+2

    Bm+2


    Partitioning method contd1

    Partitioning Method (contd)

    • Theorem 1: All blocks contain equivalent states after the algorithm terminates.

      • Proof: Assume the program terminates at iteration m

      • Following lemma 1, if two states are not i-equiv. for I ≤ m, they are put into different blocks.

      • Since Pm+1=Pm, for any future iteration j (j>m), i.e., if we were to run the procedure up to iteration j, the partition will not change: we will get

        Pm= Pm+1= Pm+2= …… = Pj

      • If any two states in Pk are not equiv  they are not j-equiv. for j > m, they will be put in two blocks in iteration j > m  procedure will continue till at least iteration j >m  Pm+1 != Pm, (otherwise the partitions will never change after Pm) and we reach a contradiction. Thus all states in a block are equivalent.

    • Theorem 2: All equivalent states are in the same blocks, i.e., two equivalent states will not be in different blocks.

      • Proof Outline: Follows from the procedure: If two states are in different blocks, then they were detected as not k-equivalent for some k, and thus are not equivalent. Thus no 2 equivalent states can be in different blocks after the procedure terminates. Hence each block will contain the maximal set of equivalence states, i.e., each block is an equivalence class


    State minimization implication table method

    State Minimization—Implication Table Method

    • Definition

      • Implied pairs of a pairs of states Si, Sj is a pair of states Sk,Sl which are N.S.’s of Si and Sj for the same 1-length I/P Ip

    • Implication Table Procedure (good manual method for small problems)

      • Step 1: Form an implication table (a diagonaled square) by vertically listing all states except the first, and horizontally listing all states except the last.

    0

    1

    • Step 2: Put X’s for state pairs in the table that are not 1- equiv.

    A

    C/1

    B/0

    B

    C/1

    E/0

    C

    B/1

    E/0

    D

    D/0

    B/1

    E

    E/0

    A/1


    Implication table procedure

    0/1

    E

    1/0

    BE

    State pair

    C

    B

    BCBE

    1/0

    0/1

    AB

    Implication Table Procedure

    • Step 3:

    • a) Put a ‘\/’ mark if the implied pairs in a cell:

      • either contains only the states that define the cell

      • or is the same state (e.g., (E,E))—singleton states

    • Step 3: Use cond. (ii) of defn. 2 for equiv. states for other squares.

    • b) For the remaining cells, write all implied pairs (incl. the same state pair and singleton states) of the states defining the cell that not meet the above two conditions

    0

    1

    B

    A

    C/1

    B/0

    C

    B

    C/1

    E/0

    C

    B/1

    E/0

    D

    D

    D/0

    B/1

    E

    E

    E/0

    A/1

    A

    B

    C

    D


    Implication table procedure contd

    A1B1

    Equiv. states

    AB

    C,D

    Equiv. states

    A2B2

    A1,B1

    A3B3

    (a)

    E,E

    U2V2

    U1V1

    X

    UV

    Singleton state

    Non-equiv

    states

    Equiv. states

    (b)

    Implication

    arc

    Closed System

    (c)

    Implication Table Procedure (contd)

    • Step 4:

      • Repeat

        • Put a mark ‘\/’ if all implied pairs have the mark ‘\/’ in their cell

        • Put a X if any implied pair has an X in its cell

      • Until no change.

    • Step 5: Remaining cells (w/ only implied states or ‘\/’) are equiv.

    • Note: Cells w/ only implied pairs form closed systems (see defn in next slide) of implied pairs w/ neither ‘\/’ nor X marks in them. I/Ps to any state pairs in this system will either:

      • remain in the closed system producing identical o/ps (since these state pairs are at least 1-equivalent)

      • or go out of the system to either:

        • (a) singleton states o

        • (b) equivalent states,

      • and from thereon also produce identical o/ps. Thus these state pairs are equivalent.

    A2,B2

    A4,B4

    • Form the largest equivalent sets by transitivity: (Si, Sj) and (Sj, Sk)=>(Si, Sj, Sk)

      • Note: (Si, Sk) will also be equivalent from the implication table, since “state equivalence” is an equivalence relation

    X,Y

    Equiv. states

    A3,B3


    Implication table procedure1

    BE

    BCBE

    AB

    Implication Table Procedure

    Only B, C are equivalent.

    Choose B as the repr. of (B,C)

    B

    X

    C

    X

    0

    1

    0

    1

    D

    A

    B/1

    B/0

    A

    C/1

    B/0

    E

    B

    B/1

    E/0

    B

    C/1

    E/0

    X

    D

    D/0

    B/1

    C

    B/1

    E/0

    A

    B

    C

    D

    E

    E/0

    A/1

    D

    D/0

    B/1

    E

    E/0

    A/1


    Implication table example contd

    AD

    BE

    CF

    A closed system of state pairs w/ implication arcs

    • An implication arc

    • AD  BE means BE is the implied state pair of AD

    • A closed system of state pairs is a group of state pairs s.t.:

    • (a) there is a directed path betw. any 2 state pairs, and

    • (b) there are no outgoing arcs from any state pair in the group to a state pair outside this group that is not the same state or equiv states

    Implication Table Example (contd)

    B

    C

    D

    BE

    E

    ADCF

    F

    AD

    G

    Equiv. Classes:

    (AD)

    (BE)

    (CF)

    (G)

    (H)

    H

    CHBG

    X

    A

    B

    C

    D

    E

    F

    G


    Ece465

    AF

    BDAF

    DFAF

    DF

    BD

    DGAF

    BGAF

    DGAF

    BCAF

    BC

    BCDF

    B

    C

    D

    E

    F

    G

    Equivalence classes:

    (AF)

    (BC)(BH)=>(CH) => (BCH)

    (D)

    (E)

    (G)

    H

    A

    B

    C

    D

    E

    F

    G


    Ece465

    Choose B as the repr. of (BCH) & A as repr. of (A F)

    Equivalence classes:

    (AF)

    (BC)(BH)=>(CH) => (BCH)

    (D)

    (E)

    (G)


    Backup

    Backup


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