Sequential System Synthesis -- State Encoding

ENEE 6442 The State Encoding Problem > Goal: Given n states, assign a unique code (of length of at least  log n  ) to each state such that the cost of binary logic level implementation is minimized. (state assignment problem) > Cost of the implementation: =Number of literals =Speed =Testability

ENEE 6443 Example: Why State Encoding Matters? > Consider a fragment of the cube table for an FSM: =Let y 1 y 2 and Y 1 Y 2 be the current state and next state. We want to represent Y 1,Y 2, and output O as a function of y 1,y 2, and input x. 1CB0 1CA0ONSCSx =Assignment 1: A=01, B=10, C=11 Y 1 =…+x’y 1 ’y 2 +…+x’y 1 y 2 ’+… Y 2 =…+x’y 1 ’y 2 +…+x’y 1 y 2 ’+… O=…+x’y 1 ’y 2 +…+x’y 1 y 2 ’+… O Y1Y2Y1Y2Y1Y2Y1Y2 y1y2y1y2y1y2y1y2x O Y1Y2Y1Y2Y1Y2Y1Y2 y1y2y1y2y1y2y1y2x =Assignment 2: A=00, B=10, C=11 Y 1 =…+x’y 1 ’y 2 ’+…+x’y 1 y 2 ’+…=…+x’y 2 ’+…

ENEE 6444 Lessons We Have Learned > Two codes are adjacent if they only differ in one bit. (e.g. 00 and 01 are adjacent, 01 and 10 are not.) > Any k-bit code has k adjacent codes. > If two states are given adjacent codes, we will be able to extract common cubes from the next state and output functions. =Common fanout (next) state =Common fanin (current) state > Identify pairs that should receive adjacent codes. 1CAb 1BAaONSCSx

ENEE 6445 Attraction Graph > The attraction graph of a FSM is a weighted, undirected, complete graph that represents the attraction between each pair of states. =Node: state =Edge: the attraction between the two states > If two states share a common fanout/fanin state, their attraction should increase. > Two states that have a strong attraction should be assigned adjacent codes. > How to measure the attraction quantitatively?

ENEE 6446 Fanout-Oriented Algorithm > State Transition Matrix S |s|x|s| : =Row: one per (current) state =Column: one per (next) state =Entry: (non-negative) number of arcs going from state s i (row) to state s j (column) > Output Matrix Z |s|x|O| : =Row: one per (current) state =Column: one per output =Entry: (non-negative) number of arcs going out of state s i (row) with output z j (column) > Let N b be the number of encoding bits, then the attraction between states s i and s j is given by N b S i S j T +Z i Z j T =W(1,2) = 2(1 1 0)(1 0 1) T +(1)(0) T =2 =W(1,3) = 2(1 1 0)(1 0 1) T +(1)(1) T =3 =W(2,3) = 2(1 0 1)(1 0 1) T +(0)(1) T =4 A C B 1/1 1/0 0/0 1/1 0/0 0/0 A C B 3 4 2

ENEE 6447 Fanin-Oriented Algorithm > State Transition Matrix S |s|x|s| : =Row: one per (next) state =Column: one per (current) state =Entry: (non-negative) number of arcs going from state s j (column) to state s i (row) > Input Matrix X |s|x|I| : =Row: one per (current) state =Column: |I| per input =Entry: (non-negative) number of arcs entering state s i (row) with input x j (column) > Let N b be the number of encoding bits, then the attraction between states s i and s j is given by N b S i S j T +X i X j T =W(1,2) = 2(1 1 1)(1 0 0) T +(2 1)(0 1) T =3 =W(1,3) = 2(1 1 1)(0 1 1) T +(2 1)(1 1) T =7 =W(2,3) = 2(1 0 0)(0 1 1) T +(0 1)(1 1) T =1 A C B 1/1 1/0 0/0 1/1 0/0 0/0 A C B ^^^

ENEE 6448 The Encoding Algorithm > Given the attraction graph (matrix W(i,j)), assign a unique N b -bit code to each state. =For each node s i, find the N b largest attractions W(i,j) and sum them up; =Pick the one with the largest sum and assign it code 0…0; =Assign the N b adjacent codes of 0…0 to the N b neighbor states that have the N b largest attractions; =Remove the N b +1 node and repeat until all nodes get a N b -bit code;

ENEE 6449 Example: The Encoding Algorithm A C B A C B Fanout-Oriented Algorithm: A C B A C B Fanin-Oriented Algorithm: Y 1 =y 1 ’y 2 x Y 2 =y 1 ’y 2 x + y 1 y 2 ’x’ + y 1 ’y 2 ’x Z = y 1 ’x (13 literals) A C B 1/1 1/0 0/0 1/1 0/0 0/0 Y 1 =y 1 ’y 2 x’ + y 1 ’y 2 x Y 2 =y 1 ’y 2 ’x Z = y 2 ’x (11 literals)

ENEE Decomposition and Encoding: Motivation > There are many different ways to encode a FSM with four states: {A,B,C,D}. > If we know that pairs (A,B),(C,D); (A,C),(B,D) should receive adjacent codes, then there are only 8 different assignments: =A=00, B=01, C=10, D=11 =A=00, B=10, C=01, D=11 =A=01, B=00, C=11, D=10 =A=01, B=11, C=00, D=10 =A=10, B=11, C=00, D=01 =A=10, B=00, C=11, D=01 =A=11, B=01, C=10, D=00 =A=11, B=10, C=01, D=00

ENEE Notations on Partitions > A partition  on a set S is a collection of disjoint subsets (called blocks) of S whose union is S. =  ={(1,2),(3,4,6),(5,8),(7)} is a partition of S={1,2,3,4,5,6,7,8} =0={(1),(2),(3)} and 1={(1,2,3)} are two trivial partitions of S={1,2,3} =For s,t  S, s  t (  ) means that they belong to the same block of partition  =For two partitions  1 and  2, their meet  1  2 is also a partition where s  t (  1  2 ) iff s  t (  1 ) and s  t (  2 ); their join  1 +  2 is the partition where s  t (  1 +  2 ) iff there exists a sequence in S: s=s 0,s 1,…,s n =t, such that for all i=0,…n-1, either s i  s i+1 (  1 ) or s i  s i+1 (  2 ). =Given  1 ={(1,2),(3,4,6),(5,8),(7)},  2 ={(1,6),(2,5,8),(3,4,7)},  1  2 ={(1),(2),(3,4),(5,8),(6),(7)},  1 +  2 ={(1,2,3,4,5,6,7,8)}

ENEE Substitution Property > A partition  on set S of machine M= has the substitution property (S.P.) iff s  t(  ) implies that  (s,a)  (t,a)(  ) for all a  I. > Theorem. A sequential machine M has a non- trivial parallel decomposition of its states iff there exist two non-trivial S.P. partitions  1 and  2 on M such that meet  1  2 =0. > Theorem. IF a sequential machine has one non- trivial S.P. partition, then it has a non-trivial serial decomposition.

ENEE Parallel Decomposition >  1 = {(0,1,2),(3,4,5)} and  2 ={(0,5),(1,4),(2,3)} are two S.P. partitions and  1  2 =0. > Let A=(0,1,2), B=(3,4,5), and X=(0,5),Y=(1,4),Z=(2,3), we can build two machines. > The output, in this case, is the product of the outputs of the two newly built machines BAB 0ABA101YYZ 0ZXY 0ZZX10 G1G1G1G1 G2G2G2G2

ENEE Serial Decomposition >  1 = {(1,2,3),(4,5)} is the only S.P. partition. > We choose another partition  2 ={(1,4),(2,5),(3)} s.t.  1  2 =0. > Let A=(1,2,3), B=(4,5), and X=(1,4),Y=(2,5),Z=(3), we can rebuild this machine as G1G1G1G1 G2G2G2G2 10YYB,Y 11XZB,X 00XYA,Z 01YZA,Y 01XXA,X XYZ 1001YYYZY 1101XZXXX B1B1B1B1 B0B0B0B0 A1A1A1A1 A0A0A0A0 B1B1B1B1 B0B0B0B0 A1A1A1A1 A0A0A0A0

ENEE Computation of the S.P. Partition > Recall that “ A partition  on set S of machine M= has the substitution property (S.P.) iff s  t(  ) implies that  (s,a)  (t,a)(  ) for all a  I.” > So S.P. partition is independent of output > If (1,2)  (1,4),(2,3)  (1,2,3,4) (transitivity)  {(1,2,3,4,5)} trivial. > If (1,3)  (1,5)  (1,3,5)  (2,3)  (1,2,3,5)  (1,2,3,4,5) trivial. > If (1,4), form a block. > If (2,3)  (4,5)  (1,4,5)  {(1,4,5),(2,3)} > If (2,5)  (1,4)  {(1,4),(2,5),(3)} > If (2) > if (3,5)  (1,5),(2,3), contradiction. > If (3)  {(1,4),(2),(3),(5)} > Another partition: {(1),((2,3),(4,5)}

ENEE Decomposition and State Encoding > We need two bits xy to encode a sequential state machine M with four states: A,B,C,D. > Suppose  1 ={(A,B), (C,D)} and  2 ={(A,C),(B,D)} are two non-trivial S.P. partitions. > Clearly  1  2 =0 so we have a non-trivial parallel decomposition, M 1 ={ab,cd} and M 2 ={ac,bd}, both with only two states. > If we use bit x to encode M 1 and bit y to encode M 2 as ab=0, cd=1; and ac=0, bd=1. > Then M can be encoded as A=00,B=01,C=10,D=11. > SE problem becomes finding S.P. partitions and applying parallel/serial decompositions.