Flip-Flops Matching of Sequential Circuit

Flip-Flops Matching of Sequential Circuit Speaker : Adonis Lin Advisor : Chun-Yao Wang 2007.7.31 Department of Computer Science National Tsing Hua University, Taiwan

Outline • Problem Formulation • Research Application • Limits of previous work • Introduction of our work • Conclusion • Future Work

Problem Formulation • Given • A product machine of two sequential circuits • Two separate circuit space (only same PI) • A sequential circuit • One circuit space (overlap fan in cone of FFs) • Initial state • 0 initial sate , X initial state , specific initial state • Objective • Find flip-flop corresponding or constant flip-flop based on initial state

Research Application • Simplify product machine & sequential circuit • Use equivalence、complement、constant relation of FFs • It can modify delay of circuit (reduce clock skew) • Reduce size of BDD (reachability analysis) • Use Flip-Flop corresponding prevent BDD size from explosion • Map sequential equivalence checking to combinational equivalence checking

R2 Fan out cone Fan out cone R1 R1 R3 Research Application (cont’d) • How to simplify? • Reduce # of FFs in sequential circuit • Reduce FFs => reduce space size • Timing change (increase or reduce) • EX : R1 = R2 * R3

Previous Work (1/2) • Guess Flip-Flop corresponding • Equivalent & Complement relation • Prove Flip-Flop corresponding you guess • Use mathematics induction (M.I.) • If FFC can be found, this FFC is for initial states which matching the rule you guess

Previous Work (2/2) • Found FFC • Three groups • Vg1, Vg2, Vg3 • 8 (23) initial states matching this FFC Initial states must hold this FFC, and this FFC could be proven later !!

Limits of Previous Work • It can not focus on one initial state because of its proving method (M.I.) • It can not find functional relationship • It can only find general FFC which some initial states match • For matching some initial state, only a few FFCs can be found • Practically, we want special FFC for one initial state

M1 S1 S2 S3 X Y S4 S5 M2 Special Example • Example (Five state bits in PM) • Flip-flop matching (Underinitial state : S1=1,S2=S3=S4=S5=0) • S1 = S4 * S5 = S2 * S3 • S2 = S4 • S3 = S5 －－－－

M1 S1 S2 S3 X -X*V + (-V) X -X* V3 + V2 X*V Y X -X* V3 -X*V -X*V + (-V) 1 X* V3 V 0 0 V3 - V - V V V3 V2 - V V2 -V V1 S4 X*V V V3 S5 X*(- (V1+V2)) X*V M2 V2 - V 0 0 - V - V V2 V1 -V V X*V V1 - (V1+V2) - V Use Previous Work With signal V3 Without signal V3 Previous work can not find any FFC, even though begin with proper initial state

Weakness of Previous Work • Because it can not find functional relationship, S1 should be set as V3 instead of (-V1 * -V2) • Even though it can find functional relationship, it still can not find FFC in this case. • Prove (S2 = S4), (S3 = S5), (S1 = S4 * S5 = S2 * S3 ) • Initial state : S1=1, S2=S3=S4=S5=0 (O)……..<1> • Initial state : S1=0, S2=S3=S4=S5=1 (X)……..<2> • Only initial state <1> hold this FFC, but proving this FFC need initial state <1> & initial state <2> hold this FFC －－－－

Run Flow • Step.1 • Set all Flip-Flops to specific initial state • Step.2 • Do < pi – time frame > propagation, it stop when no new FFs can be set by pi • Step.3 • Mark FFs (with value 0 or 1) constant FFs under this initial state • Step.4 • Base on FF = f( pi-time frames ) to find relationship between FFs

M1 S1 S2 S3 X Y S4 S5 M2 Example (Time Frame 0) • Find FFC under initial state S1=1, S2=S3=S4=S5=0 1 0 1 0 0 1 0 0 0 1 0

M1 S1 S2 S3 X Y S4 S5 M2 Example (Time Frame 1) • After time frame 0, 3 new FFs set by <PI – time frame> 0 0 0 0

M1 S1 S2 S3 X Y S4 S5 M2 Example (Time Frame 2) • After time frame 1, 2 new FFs set by <PI – time frame>

M1 S1 S2 S3 X Y S4 S5 M2 Example (Time Frame 3) • After time frame 2, no new FFs set by <PI – time frame> • Stop !!

Equivalent ! Equivalent ! FFC Checking (1/2) • Check FFC(1) < Using time frame 1 -- PI > • S1 = • S2 = • S3 = • S4 = • S5 = • FFC(1) • S1 = S4 * S5 = S2 * S3 • S2 = S4 • S3 = S5 －－－－

Equivalent ! Equivalent ! FFC Checking (2/2) • Check FFC(2) < Using time frame 2 -- PI > • S1 = • S2 = • S3 = • S4 = • S5 = • FFC(2) • S1 = S4 * S5 = S2 * S3 • S2 = S4 • S3 = S5 －－－－

Conclusion • FFC(2) = FFC(1), this FFC is proved and it will exist for any time frame by M.I. • We can find special FFC under one initial state • We can also find functional relationship • How can we find exact & unique FFC every time frame for checking ?

Future Work • Find exact & unique FFC every time frame • Study SAT Solver • Think whether it can apply to our work or not • Study latest paper about our research

Flip-Flops Matching of Sequential Circuit