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Synthesis For Finite State Machines

Synthesis For Finite State Machines. FSM (Finite State Machine) Optimization. State tables. identify and remove equivalent states. State minimization. assign unique binary code to each state. State assignment. use unassigned state-codes as don’t care. Combinational logic optimization.

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Synthesis For Finite State Machines

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  1. Synthesis For Finite State Machines

  2. FSM (Finite State Machine) Optimization State tables identify and remove equivalent states State minimization assign unique binary code to each state State assignment use unassigned state-codes as don’t care Combinational logic optimization net-list

  3. FSM Optimization 00 01 -0 S3 S2 0- 10 -1 11 01 -0 1- S4 S1 11 PI PO Combinational Logic v1 u1 NS PS v2 u2

  4. State Minimization Goal : identify and remove redundant states (states which can not be observed from the FSM I/O behavior) Why : 1. Reduce number of latches • assign minimum-length encoding • only as the logarithm of the number of states 2. Increase the number of unassigned states codes • heuristic to improve state-assignment and logic-optimization

  5. State Minimization Definition • Completely-specified state machine • two states are equivalent if outputs are identical for all input combinations Next states are equivalent for all input combinations • equivalence of states is an equivalence relation which partitions the states into disjoint equivalence classes • Incompletely specified state machines

  6. Classical State Minimization 1. Partition states based on input output values asserted in the state 2. Define the partitions so that all states in a partition transition into the same next-state partition (under corresponding inputs)

  7. Example Ex : 0 A B 0 1 A C 0 0 B D 0 (A,B,C,D,E,F,H)(G) 1 B E 0 0 C F 0 (A,B,C,E,F,H)(G)(D) 1 C A 0 0 D H 0 (A,C,E,H)(G)(D)(B,F) 1 D G 0 0 E B 0 (A,C,E)(G)(D)(B,F)(H) 1 E C 0 0 F D 0 1 F E 0 0 G F 1 1 G A 0 0 H H 0 1 H A 0

  8. State Assignment • Assign unique code to each state to produce logic-level description • utilize unassigned codes effectively as don’t cares • Choice for S state machine • minimum-bit encoding log S • maximum-bit encoding • one-hot encoding • using one bit per state • something in between • Modern techniques • hypercube embedding of face constraint derived for collections of states (Kiss,Nova) • adjacency embedding guided by weights derived between state pairs (Mustang)

  9. Hypercube Embedding Technique • Observation : one -hot encoding is the easiest to decode Am I in state 2,5,12 or 17? binary : x4’x3’x2’x1x0’(00010) + x4’x3’x2x1’x0 (00101) + x4’x3x2x1’x0’(01100) + x4x3’x2’x1’x0 (10001) one hot : x2+x5+x12+x17 But one hot uses too many flip flops. • Exploit this observation 1. two-level minimization after one hot encoding identifies useful state group for decoding 2. assigning the states in each group to a single face of the hypercube allows a single product term to decode the group to states.

  10. State Group Identification Ex: state machine input current-state next state output 0 start S6 00 0 S2 S5 00 0 S3 S5 00 0 S4 S6 00 0 S5 start 10 0 S6 start 01 0 S7 S5 00 1 start S4 01 1 S2 S3 10 1 S3 S7 10 1 S4 S6 10 1 S5 S2 00 1 S6 S2 00 1 S7 S6 00 Symbolic Implicant : represent a transition from one or more state to a next state under some input condition.

  11. Representation of Symbolic Implicant Symbolic cover representation is related to a multiple-valued logic. Positional cube notation : a p multiple-valued logic is represented as P bits (V1,V2,...,Vp) Ex: V = 4 for 5-value logic (00010) represent a set of values by one string V = 2 or V = 4 (01010)

  12. Minimization of Multi-valued Logic Find a minimum multiple-valued-input cover - espresso Ex: A minimal multiple-valued-input cover 0 0110001 0000100 00 0 1001000 0000010 00 1 0001001 0000010 10

  13. State Group Consider the first symbolic implicant 0 0110001 0000100 00 • This implicant shows that input “0” maps “state-2” or “state-3” or “state-7” into “state-5” and assert output “00” • This example shows the effect of symbolic logic minimization is to group together the states that are mapped by some input into the same next-state and assert the same output. • We call it “state group” if we give encodings to the states in the state group in adjacent binary logic and no other states in the group face, then the states group can be implemented as a cube.

  14. Group Face • group face : the minimal dimension subspace containing the encoding assigned to that group. Ex: 0 010 0**0 group face 0100 0110

  15. Hyper-cube Embedding c state groups : {2,5,12,17} {2,6,17} b a 12 17 6 5 2 2 17 wrong! 6 5 12

  16. Hyper-cube Embedding c state groups : {2, 6, 17} {2, 4, 5} b a 6 17 2 5 4 17 6 2 4 wrong! 5

  17. How to Check if a State Assignment Satisfies the Constraint Matrix? Step1: Step2: Find the group face of the encoding For all states, check if a state that does not belong to a state group intersects that group face

  18. 1 * * 0 * 0 0 0 * Example Constraint matrix A, state encoding S and group-face matrix F 010 110 101 000 001 011 100 0110001 1001000 0001001 S = A = Step1: Group face F = A˙S = Step2: Check encoding of state-6 = [011] Since it does not belong to group 1, 2 and 3, Encoding of state-6 satisfies the constraint check [0 1 1] ∩ [1 * *] = [0 1 1] ∩ [0 * 0] = [0 1 1] ∩ [* 0 0] =

  19. Other State Encoding If encoding of state-6 = [111], check Do not satisfy the constraint. [1 1 1] ∩ [1 * *] = 111 [1 1 1] ∩ [0 * 0] = [1 1 1] ∩ [* 0 0] =

  20. Algorithm for State Assignment Step 1: Step 2: Step 3: Step 4: Step 5: Select an uncoded state (or a state subset). Determine the encodings for that state (states) satisfying the constraint relation. If no encoding exists, increase the state code dimension and go to Step 2. Assign an encoding to the selected state (states). If all states have been encoded, stop. Else go to Step 1.

  21. Step 3 • Can always increase the coding length by one bit • New state assignment: • For states already assigned, append 0 at the end • For the new state, ns, ns does not belong to any state group, encoding of ns = [c | 1]  c is any vector ns belongs to some state group, encoding of ns = [c | 1] c is the encoding of any state that belongs to the state group case1: case2:

  22. 00 10 01 11 0101 1010 1100 A = S = 000 100 010 110 01011 10100 11000 A’ = S’ = Example Ex: To encode a new state (state-5), we have a new constraint matrix, For the states already assigned, we have a new encoding, For the new state (state-5), we have encodings ns = [1 0 1] or [1 1 1]

  23. Hyper-cube Embedding Method • Advantage : • use two-level logic minimizer to identify good state group • almost all of the advantage of one-hot encoding, but fewer state-bit

  24. Adjacency-Based State Assignment Basic algorithm: (1) Assign weight w(s,t) to each pair of states • weight reflects desire of placing states adjacent on the hypercube (2) Define cost function for assignment of codes to the states • penalize weights for the distance between the state codes eg. w(s,t) * distance(enc(s),enc(t)) (3) Find assignment of codes which minimize this cost function summed over all pairs of states. • heuristic to find an initial solution • pair-wise interchange (simulated annealing) to improve solution

  25. Adjacency-Based State Assignment • Mustang : weight assignment technique based on loosely maximizing common cube factors

  26. How to Assign Weight to State Pair • Assign weights to state pairs based on ability to extract a common-cube factor if these two states are adjacent on the hyper-cube.

  27. Fan-Out-Oriented (examine present-state pairs) • Present state pair transition to the same next state S1 S3 S2 $$$ S1 S2 $$$$ $$$ S3 S2 $$$$ Add n to w(S1,S3) because of S2

  28. Fan-Out-Oriented S3 S1 • present states pair asserts the same output $/j $/j S4 S2 $$$ S1 S2 $$$1$ $$$ S3 S4 $$$1$ Add 1 to w(S1 ,S3) because of output j

  29. Fanin-Oriented (exam next state pair) • The same present state causes transition to next state pair. $$$ S1 S2 $$$$ $$$ S1 S4 $$$$ Add n/2 to w(S2,S4) because of S1 S1 S4 S2

  30. Fanin-Oriented (exam next state pair) S1 S3 • The same input causes transition to next state pair. $0$ S1 S2 $$$$ $0$ S3 S4 $$$$ Add 1 to w(S2,S4) because of input i i i S2 S4

  31. Which Method Is Better? • Which is better? FSMs have no useful two-level face constraints => adjacency-embedding FSMs have many two-level face constraints => face-embedding

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