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Sequential Logic Optimization

Sequential Logic Optimization. State Minimization Algorithms for State Minimization State, Input, and Output Encodings Minimize the Next State and Output logic. Optimization in Context. Understand the word specification Draw a picture Derive a state diagram and Symbolic State Table

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Sequential Logic Optimization

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  1. Sequential Logic Optimization • State Minimization • Algorithms for State Minimization • State, Input, and Output Encodings • Minimize the Next State and Output logic CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 1

  2. Optimization in Context • Understand the word specification • Draw a picture • Derive a state diagram and Symbolic State Table • Determine an implementation approach (e.g., gate logic, ROM, FPGA, etc.) • Perform STATE MINIMIZATION • Perform STATE ASSIGNMENT • Map Symbolic State Table to Encoded State Tables for implementation (INPUT and OUTPUT encodings) CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 2

  3. Finite State Machine Optimization • State Minimization • Fewer states require fewer state bits • Fewer bits require fewer logic equations • In general, one equation per output bit BUT • What counts are literals, not numbers of equations • Encodings: State, Inputs, Outputs • State encoding with fewer bits has fewer equations to implement • However, each may be more complex • State encoding with more bits (e.g., one-hot) has simpler equations • Complexity directly related to complexity of state diagram • Input/output encoding may or may not be under designer control CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 3

  4. Input Next State Output Sequence Present State X=0 X=1 X=0 X=1 Reset S0 S1 S2 0 0 0 S1 S3 S4 0 0 1 S2 S5 S6 0 0 00 S3 S0 S0 0 0 01 S4 S0 S0 1 0 10 S5 S0 S0 0 0 11 S6 S0 S0 1 0 S0 0/0 1/0 S1 S2 These two states must be identical – why? These two states must be identical – why? 0/0 1/0 0/0 1/0 S3 S4 S5 S6 1/0 1/0 1/0 1/0 0/0 0/1 0/0 0/1 State Minimization Example • Sequence Detector for 010 or 110 CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 4

  5. Input Next State Output Sequence Present State X=0 X=1 X=0 X=1 Reset S0 S1 S2 0 0 0 S1 S3 S4 0 0 1 S2 S5 S6 0 0 00 S3 S0 S0 0 0 01 S4 S0 S0 1 0 10 S5 S0 S0 0 0 11 S6 S0 S0 1 0 S0 0/0 1/0 S1 S2 0/0 1/0 0/0 1/0 S3 S4 S5 S6 1/0 1/0 1/0 1/0 0/0 0/1 0/0 0/1 State Minimization Example • Sequence Detector for 010 or 110 States S3, S5 are identical because can’t detect from here, next step is always S0 States S4, S6 are identical because they detect on 0, next step is always S0 CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 5

  6. Input Next State Output Sequence Present State X=0 X=1 X=0 X=1 Reset S0 S1 S2 0 0 0 S1 S3/5 S4/6 0 0 1 S2 S3/5 S4/6 0 0 00/11 S3/5 S0 S0 0 0 01/10 S4/6 S0 S0 1 0 These states are identical – why? State Minimization Example • Sequence Detector for 010 or 110 S0 0/0 1/0 S1 S2 0/0 1/0 0/0 1/0 S3/5 S4/6 0/1 1/0 1/0 0/0 CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 6

  7. Input Next State Output Sequence Present State X=0 X=1 X=0 X=1 Reset S0 S1/2 S1/2 0 0 0/1 S1/2 S3/5 S4/6 0 0 00/11 S3/5 S0 S0 0 0 01/10 S4/6 S0 S0 1 0 State Minimization Example • Sequence Detector for 010 or 110 S0 0,1/0 S1/2 1/0 0/0 S3/5 S4/6 0/1 1/0 1/0 0/0 CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 7

  8. Algorithmic Approach to State Minimization • Goal – identify and combine states that have equivalent behavior • Equivalent States: • Same output • For all input combinations, states transition to same or equivalent states • Algorithm Sketch • 1. Place all states in one set • 2. Initially partition set based on output behavior • 3. Successively partition resulting subsets based on next state transitions • 4. Repeat (3) until no further partitioning is required • states left in the same set are equivalent • Polynomial time procedure (in size of state set) CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 8

  9. Input Next State Output Sequence Present State X=0 X=1 X=0 X=1 Reset S0 S1 S2 0 0 0 S1 S3 S4 0 0 1 S2 S5 S6 0 0 00 S3 S0 S0 0 0 01 S4 S0 S0 1 0 10 S5 S0 S0 0 0 11 S6 S0 S0 1 0 S0 0/0 1/0 S1 S2 0/0 1/0 0/0 1/0 S3 S4 S5 S6 1/0 1/0 1/0 1/0 0/0 0/1 0/0 0/1 State Minimization Example • Sequence Detector for 010 or 110 CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 9

  10. Input Next State Output Sequence Present State X=0 X=1 X=0 X=1 Reset S0 S1 S2 0 0 0 S1 S3 S4 0 0 1 S2 S5 S6 0 0 00 S3 S0 S0 0 0 01 S4 S0 S0 1 0 10 S5 S0 S0 0 0 11 S6 S0 S0 1 0 Method of Successive Partitions ( S0 S1 S2 S3 S4 S5 S6 ) ( S0 S1 S2 S3 S5 ) ( S4 S6 ) ( S0 S3 S5 ) ( S1 S2 ) ( S4 S6 ) ( S0 ) ( S1 S2 ) ( S3 S5 ) ( S4 S6 ) S1 is equivalent to S2 S3 is equivalent to S5 S4 is equivalent to S6 CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 10

  11. Input Next State Output Sequence Present State X=0 X=1 X=0 X=1 Reset S0 S1' S1' 0 0 0 + 1 S1' S3' S4' 0 0 X0 S3' S0 S0 0 0 X1 S4' S0 S0 1 0 S0 X/0 S1’ 0/0 1/0 S4’ S3’ X/0 0/1 1/0 Minimized FSM • State minimized sequence detector for 010 or 110 7 States reduced to 4 States 3 bit encoding replaced by 2 bit encoding CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 11

  12. Another Example • 4-Bit Sequence Detector: output 1 after each 4-bit input sequence consisting of the binary strings 0110 or 1010 CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 12

  13. State Transition Table • Group states with same next state and same outputs S’10 CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 13

  14. Iterate the Row Matching Algorithm S’7 CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 14

  15. Iterate One Last Time S’3 S’4 CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 15

  16. Final Reduced State Machine 15 states (min 4 FFs) reduced to 7 states (min 3 FFs) CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 16

  17. 00 10 S0[1] S1 [0] 00 01 10 11 present next state output state 00 01 10 11 S0 S0 S1 S2 S3 1 S1 S0 S3 S1 S4 0 S2 S1 S3 S2 S4 1 S3 S1 S0 S4 S5 0 S4 S0 S1 S2 S5 1 S5 S1 S4 S0 S5 0 11 01 00 01 00 S2[1] S3 [0] 01 10 10 11 11 01 10 10 S4[1] S5 [0] 11 00 00 01 11 More Complex State Minimization • Multiple input example inputs here symbolic state transition table CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 17

  18. present next state output state 00 01 10 11 S0' S0' S1 S2 S3' 1 S1 S0' S3' S1 S3' 0 S2 S1 S3' S2 S0' 1 S3' S1 S0' S0' S3' 0 S1 present next state output state 00 01 10 11 S0 S0 S1 S2 S3 1 S1 S0 S3 S1 S4 0 S2 S1 S3 S2 S4 1 S3 S1 S0 S4 S5 0 S4 S0 S1 S2 S5 1 S5 S1 S4 S0 S5 0 S0-S1 S1-S3 S2 S2-S2 S3-S4 S0-S1 S3-S0 S3 S1-S4 minimized state table (S0==S4) (S3==S5) S4-S5 S0-S0 S1-S0 S1-S1 S3-S1 S4 S2-S2 S2-S2 S3-S5 S4-S5 S0-S1 S1-S1 S3-S4 S0-S4 S5 S1-S0 S4-S0 S4-S5 S5-S5 S0 S1 S2 S3 S4 Minimized FSM • Implication Chart Method • Cross out incompatible states based on outputs • Then cross out more cells if indexed chart entries are already crossed out CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 18

  19. “Polynomial” Algorithm • n (n+1)/2 squares for n states • At each iteration, must cross off at least one square, or done • ð ≤ n(n+1)/2 iterations at most • Filling out a square takes O(num input combinations) • ðTotal Running time is O(n2 * num input combinations) • “Polynomial”?...well • Representation of finite state machine is often log number of states x log number of input combinations • Consider a counter • Algorithm is exponential in this representation CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 19

  20. Minimizing Incompletely Specified FSMs • Equivalence of states is transitive when machine is fully specified • But its not transitive when don't cares are presente.g., state outputS0 – 0 S1 is compatible with both S0 and S2 S1 1 – but S0 and S2 are incompatible S2 – 1 disagree in output 2 • No polynomial time algorithm exists for determining best grouping of states into equivalent sets that will yield the smallest number of final states CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 20

  21. X Q1 Q0 Q1+ Q0+0 0 0 0 00 0 1 0 00 1 1 0 01 0 0 0 11 0 1 1 11 1 1 1 1– 1 0 0 0X Q1 Q0 Q1+ Q0+0 0 0 0 00 0 1 0 00 1 1 0 01 0 0 0 11 0 1 1 11 1 1 1 1– 1 0 0 0 X’ 00[0] X’ 01[1] X X’ 11 [0] X X Minimizing States May Not Yield Best Circuit • Example: edge detector - outputs 1 when last two input changes from 0 to 1 Q1+ = X (Q1 xor Q0) Q0+ = X Q1’ Q0’ CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 21

  22. X Q1 Q0 Q1+ Q0+0 0 0 0 00 0 1 0 00 1 1 0 01 0 0 0 11 0 1 1 11 1 1 1 1– 1 0 0 0X Q1 Q0 Q1+ Q0+0 0 0 0 00 0 1 0 00 1 1 0 01 0 0 0 11 0 1 1 11 1 1 1 1– 1 0 0 0 X’ 00[0] X’ 01[1] X X’ 11 [0] X X Edge Detector Implementation Q1+ = X (Q1 xor Q0) Q0+ = X Q1’ Q0’ CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 22

  23. X’ 00[0] X’ X X 10 [0] 01[1] X’ X’ 11[0] X X Another Implementation of Edge Detector • "Ad hoc" solution - not minimal but cheap and fast CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 23

  24. State Assignment • Choose bit vectors to assign to each “symbolic” state • With n state bits for m states there are 2n! / (2n – m)! [log n <= m <= 2n] • 2n codes possible for 1st state, 2n–1 for 2nd, 2n–2 for 3rd, … • Huge number even for small values of n and m • Intractable for state machines of any size • Heuristics are necessary for practical solutions • Optimize some metric for the combinational logic • Size (amount of logic and number of FFs) • Speed (depth of logic and fanout) • Dependencies (decomposition) CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 24

  25. Key Idea Behind State Assignment • Goal is to minimize the logic realizing a state machine • But we can’t generate logic and optimize (takes too long) • Need to get a good proxy for logic complexity • Key insight: If state bits don’t change, don’t need logic • How many bit changes in one-hot? • One-hot encoding • One AND gate per incoming edge to state • One OR gate per state • Logic is cheap, but far too many bits (one per state!) • Therefore: minimize bit changes subject to number-of-state-bits constraint CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 25

  26. Note minimize-changes heuristic doesn’t always work! • Consider a counter • Can get n bit changes (consider counting from -1 to 0) • Logic is very small • But is generally a good proxy • Key: minimizing number of bit changes (usually) minimizes Karnaugh maps CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 26

  27. State Assignment Strategies • Possible Strategies • Sequential – just number states as they appear in the state table • Random – pick random codes • One-hot – use as many state bits as there are states (bit=1 –> state) • Output – use outputs to help encode states • Heuristic – rules of thumb that seem to work in most cases • No guarantee of optimality – another intractable problem CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 27

  28. One-hot State Assignment • Simple • Easy to encode, debug • Small Logic Functions • Each state function requires only predecessor state bits as input • Good for Programmable Devices • Lots of flip-flops readily available • Simple functions with small support (signals its dependent upon) • Impractical for Large Machines • Too many states require too many flip-flops • Decompose FSMs into smaller pieces that can be one-hot encoded • Many Slight Variations to One-hot • One-hot + all-0 CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 28

  29. State Maps and Counting Bit Changes Bit Change Heuristic S0 0 1 S1 S2 S3 S4 S0 -> S1: 2 1S0 -> S2: 3 1S1 -> S3: 3 1S2 -> S3: 2 1S3 -> S4: 1 1S4 -> S1: 2 2Total: 13 7 CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 29

  30. Equations for Each Assignment • State Equations: • S1+ = S0 X’ + S4 • S2+ = S0 X, • S3 +=S1 + S2, • S4+ = S3 S0 Assignment One: 0 1 S1 S2 • Q2 = S1 + S2 • Q1 = S4 + S2 + S3 • Q0 = S4 + S2 + S1 S3 S4 • Q2 = Q2’Q1’Q0’ + Q2’Q1Q0 • Q1 = Q2’Q1Q0’ + Q2Q1Q0 + Q2Q1’Q0 + Q2’Q1’Q0’ X • Q0 = Q2’Q1Q0’ + Q2’Q1’Q0’ + Q2’Q1Q0 • Q2 = S0 X’ + S4 + S0 X • Q1 = S3 + S2 + S1 + S0 X • Q0 = S3 + S0 X’ + S4 + S0 X CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 30

  31. Karnaugh Maps Q1Q0 • Q2 = Q2’Q1’Q0’ + Q2’Q1Q0 • Q1 = Q2’Q1Q0’ + Q2Q1Q0 + Q2Q1’Q0 + Q2’Q1’Q0’ X • Q0 = Q2’Q1Q0’ + Q2’Q1’Q0’ + Q2’Q1Q0 00 01 11 10 XQ2 00 01 11 Q1Q0 10 00 01 11 10 Q2 Q1 0 Q1Q0 1 00 01 11 10 Q2 Q0 0 1 Q2 CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 31

  32. Equations for Each Assignment • State Equations: • S1+ = S0 X’ + S4 • S2+ = S0 X, • S3 +=S1 + S2, • S4+ = S3 S0 Assignment Two: 0 1 S1 S2 • Q2 = S4 • Q1 = S4 + S2 + S3 • Q0 = S4 + S3 + S1 S3 S4 • Q2 = Q2’Q1Q0 • Q1 = Q2’Q1Q0 + Q2‘Q1Q0’ + Q2’Q1’Q0 + Q2’Q1’Q0’ X • Q0 = Q2’Q1Q0 + Q2’Q1Q0’ + Q2’Q1’Q0 + Q2Q1Q0+ Q2’Q1’Q0’ X • Q2 = S3 • Q1 = S3 + S2 + S1 + S0 X • Q0 = S3 + S2 + S1 + S4 + S0 X CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 32

  33. Karnaugh Maps Q1Q0 00 01 11 10 • Q2 = Q2’Q1Q0 • Q1 = Q2’Q1Q0 + Q2‘Q1Q0’ + Q2’Q1’Q0 + Q2’Q1’Q0’ X • Q0 = Q2’Q1Q0 + Q2’Q1Q0’ + Q2’Q1’Q0 + Q2Q1Q0+ Q2’Q1’Q0’ X XQ2 00 01 11 Q1Q0 10 00 01 11 10 XQ2 00 Q1 01 Q1Q0 00 01 11 10 11 Q2 0 10 1 Q0 Q2 CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 33

  34. a b i / j i / k c a i / j k / l b c a c i / j i / j b d Adjacency Heuristics for State Assignment • Adjacent codes to states that share a common next state • Group 1's in next state map • Adjacent codes to states that share a common ancestor state • Group 1's in next state map • Adjacent codes to states that have a common output behavior • Group 1's in output map I Q Q+ Oi a c ji b c k c = i * a+ i * b I Q Q+ Oi a b jk a c l b = i * ac = k * a I Q Q+ Oi a b ji c d j j = i * a + i * cb = i * ad = i * c CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 34

  35. Successor/Predecessor Heuristics High Priority: S’3 and S’4 share common successor state (S0) Medium Priority: S’3 and S’4 share common predecessor state (S’1) Low Priority: 0/0: S0, S’1, S’3 1/0: S0, S’1, S’3, S’4 Heuristics for State Assignment CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 35

  36. Heuristics for State Assignment CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 36

  37. Another Example High Priority: S’3, S’4 S’7, S’10 Medium Priority: S1, S2 2 x S’3, S’4 S’7, S’10 Low Priority: 0/0: S0, S1, S2, S’3, S’4, S’7 1/0: S0, S1, S2, S’3, S’4, S’7, S10 CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 37

  38. Choose assignment for S0 = 000 Place the high priority adjacency state pairs into the State Map Repeat for the medium adjacency pairs Repeat for any left over states, using the low priority scheme Two alternativeassignments at the left Example Continued CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 38

  39. Why Do These Heuristics Work? • Attempt to maximize adjacent groupings of 1’s in the next state and output functions CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 39

  40. General Approach to Heuristic State Assignment • All current methods are variants of this • 1) Determine which states “attract” each other (weighted pairs) • 2) Generate constraints on codes (which should be in same cube) • 3) Place codes on Boolean cube so as to maximize constraints satisfied (weighted sum) • Different weights make sense depending on whether we are optimizing for two-level or multi-level forms • Can't consider all possible embeddings of state clusters in Boolean cube • Heuristics for ordering embedding • To prune search for best embedding • Expand cube (more state bits) to satisfy more constraints CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 40

  41. Inputs Present State Next State OutputsC TL TS ST H F0 – – HG HG 0 00 10 – 0 – HG HG 0 00 10 1 1 – HG HY 1 00 10 – – 0 HY HY 0 01 10 – – 1 HY FG 1 01 101 0 – FG FG 0 10 000 – – FG FY 1 10 00– 1 – FG FY 1 10 00– – 0 FY FY 0 10 01– – 1 FY HG 1 10 01 Output-Based Encoding • Reuse outputs as state bits - use outputs to help distinguish states • Why create new functions for state bits when output can serve as well • Fits in nicely with synchronous Mealy implementations HG = ST’ H1’ H0’ F1 F0’ + ST H1 H0’ F1’ F0HY = ST H1’ H0’ F1 F0’ + ST’ H1’ H0 F1 F0’ FG = ST H1’ H0 F1 F0’ + ST’ H1 H0’ F1’ F0’ HY = ST H1 H0’ F1’ F0’ + ST’ H1 H0’ F1’ F0 Output patterns are unique to states, we do notneed ANY state bits – implement 5 functions(one for each output) instead of 7 (outputs plus2 state bits) CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 41

  42. Current State Assignment Approaches • For tight encodings using close to the minimum number of state bits • Best of 10 random seems to be adequate (averages as well as heuristics) • Heuristic approaches are not even close to optimality • Used in custom chip design • One-hot encoding • Easy for small state machines • Generates small equations with easy to estimate complexity • Common in FPGAs and other programmable logic • Output-based encoding • Ad hoc - no tools • Most common approach taken by human designers • Yields very small circuits for most FSMs CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 42

  43. Sequential Logic Implementation Summary • Implementation of sequential logic • State minimization • State assignment • Implications for programmable logic devices • When logic is expensive and FFs are scarce, optimization is highly desirable (e.g., gate logic, PLAs, etc.) • In Xilinx devices, logic is bountiful (4 and 5 variable TTs) and FFs are many (2 per CLB), so optimization is not so crucial an issue as in other forms of programmable logic • This makes sparse encodings like One-Hot worth considering CS 150 - Spring 2008 – Lec #24: Seq Logic Optimization - 43

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