Synthesis of sequential circuits Steps in synthesis of a sequential circuit Specify the FSM (state table or state diagram) Minimize the states State assignment/encoding Specification: PS x NS z S1 S3 1 S5 S2 S4 0/1 0/0 S3 S1 S4 1/1 1/1 0/0 0/1 0/1 1/0 1/1 S2 S5 1/1

State minimization (Completely specified FSM) Completely specified  no don’t cares Progressively refine a set of equivalence classes, j For the state machine on the previous slide 0 = all states in one class = {{S1, S2, S3, S4, S5}} 1: distinguished on the basis of output value 1 = {{S1,S2},{S3,S4},{S5}} 2: distinguished if NS for two states in a partition are in different partitions. Example: For x=1, NS for S3  {S1,S2}; NS for S4 {S5} 2 = {{S1,S2},{S3}{S4},{S5}} Continue finding j+1 given j in a similar way Stop when j+1 = j since this implies that k = j for all k > j In this example, 3 = 2 Final set of equivalence classes = {{S1,S2},{S3}{S4},{S5}}

State minimization (Completely specified FSM) – Contd. Final set of equivalence classes = {{S1,S2},{S3}{S4},{S5}} State table shown at right Complexity: Let ns = # states At most O(ns) iterations Each iteration is O(ns) Total complexity is O(ns2) A more clever implementation can achieve O(ns log ns) PS x NS z S1 S3 1 S5 S4

State minimization (Incompletely specified FSM) PS x NS z S1 S3 1 S5 d S2 S4 Idea of equivalence does not hold any more Equivalence involves Reflexivity (si <comp> si) Symmetry (si <comp> sj)  (sj <comp> si) Transitivity (si <comp> sj) and (sj <comp> sk)  (si <comp> sk) For the state table here, looking at compatibility only in terms of the output value (similar to constructing 1 for a completely specified FSM) s1 <comp> s2, s2 <comp> s3 but s1 is not compatible with s3 Need to work with compatibilities instead of equivalences

State minimization (Incompletely specified FSM) – Contd. Example: state table shown on previous slide List (possibly) compatible states and clearly incompatible states List conditions under which states could be compatible Compatible Incompatible {S1,S2} {S1,S3} {S1,S5}  {S3,S4} {S1,S4} {S2,S3}  {S1,S5} {S2,S5} {S2,S4}  {S3,S4} {S3,S5} {S3,S4}  {S2,S4}, {S1,S5} {S4,S5} {S1,S5} are compatible if {S3,S4} are compatible Now use list of incompatible states to iteratively strike out states from compatible set and move them to the incompatible set No such update possible here, although in general it may be possible Combine now as {S2,S3,S4}  {S1,S5} and {S1,S5}  {S3,S4}: consistent Result: combine {S2,S3,S4} into one state and {S1,S5} into another

State encoding Assign Boolean codes to states Two extremes One-hot encoding: n bits for n states, only one bit set to 1 Example: S1 = 1000, S2 = 0100, S3 = 0010, S4 = 0001 Large # of state bits (and hence FF’s), simple combinational logic due to extensive don’t care space Minimum-bit encoding: log2 n bits for n states Example: S1 = 00, S2 = 01, S3 = 10, S4 = 11 Small # of state bits, possibly more complex combinational logic More commonly used Can also be in between the extremes

State assignment heuristics “Fanout oriented” Same NS from two PS under a given input should be assigned neighboring codes Example: two inputs X1,X2 and one output Z Rationale: state table will include which implies that 0-01  fon(NS2), fon(NS1), fon(Z) 01/1 01 00 11 PS2 PS1 X1 X2 NS2 NS1 Z 1

State assignment heuristics – contd. “Fanin oriented” If a PS goes to two different NS’s, they should be assigned neighboring codes Example: two inputs X1,X2 and one output Z Rationale: state table will include which implies that 011-  fon(NS1), fon(Z) 01 10/1 11/1 10 11 PS2 PS1 X1 X2 NS2 NS1 Z 1

A simple state-assignment algorithm Considers fanin oriented case only and defines “attractions” Construct matrices ns x ns between states Entry (i,j) = # of transitions from i to j ns x noutputs between states and outputs Entry (i,j) = # output transitions with value 1 0/0 S1 1/1 1/1 0/0 1/0 0/0 S3 S2 NS1 NS2 NS3 PS1 1 PS2 PS3 Z PS1 1 PS2 PS3

State assignment algorithm Attraction metric between states Si and Sj = Nb PSi PSjT + zi zjT where Nb = # state bits, PSk = row k of state matrix zk = row k of output matrix Attraction12 = 2[110][101]T + [1][0] = 2 Attraction23 = 2[101][101]T + [0][1] = 4 Attraction13 = 2[110][101]T + [1][1] = 3 Build attraction graph Start with node with max sum of edge attractions and assign it an encoding (here, S3 is arbitrarily assigned 00) Assign encodings in order of attraction to neighbors to reduce Hamming distance (here, S2 = 01, S1 = 10) Repeat until all states are assigned codes (here, we are done) 10 S1 3 2 4 S3 S2 00 01