Example using bridge over trouble wrappers : automated

interface synthesis

Notation introduction

• When the protocol is in a state and a clock tick occurs, the transition whose guard evaluates to true is taken.

• Guards check : an action is blocking if it has a guard. – Presence : c? (c must be 1)– Absence : c# (c must be 0)

• Req! : a control signal is issued• Address!y : a data signal is issued with value y• They didn’t tell what is the symble “:” means

Pipeline and handshake (1)

0 1 2 4 5

0 1 2 3

0 1 2 3

Pipeline and handshake (2)

0 1 2 3 1 2 4 5

0 1 2 3 4 2 3 5

0 1 2 3 1 2 3

Pipeline definition

P = (Q,S,D,V,A,->,q0,qf)

Q : 0,1,2,3,4,5

S : Req, Ack, Rdy

D : Address, Data

V : internal variable

A : actions

-> : Q x A x Q state transition relation e.x. (0,A1,3)

q0 : 0 qf : 5

Handshake definition

P = (Q,S,D,V,A,->,q0,qf)

Q : 0,1,2,3

S : SEL,READ,ENABLE

D : ADDR,RData

V : internal variable

A : actions

-> : Q x A x Q

q0 : 0 qf : 3

Interface definition• Given 2 protocols

– P1 = (Q1,S1,D1,V1,A1,->,r0,rf)– P2 = (Q2,S2,D2,V2,A2,->,t0,tf)

• A specification f : D1 -> D2 relating their data channels must be provided.• I = (Q,S,D,V,A,->,qo,qf)• Q bt (Qt,Qr) where Qt bt P1, Qr bt P2• S = S1 union S2• D = D1 union D2• V : data buffers there is one buffer for each di and f(di)• A : set containing one action complementary to each of the actions in A1

and A2• -> : relate the different states in Q as per the algorithm• Q0 = <{r0},{t0}> is the initial control state and qf = <{rf},{tf}> is the final sta

te• In addition, I will have a set of counters X = {xi} with one counter for each

pair of data channels.

step1

• A complete description of the control state of the interface and its counters in that state is written as [<R,T>,X]– R is a set of states from P1– T is a set of states from P2– X is the status of the counters

• valid(S,X) is true if, for every data write operation in a candidate action S, all required data has been read in S.

Algorithm 1 interface synthesisQ = empty setQ is the state space of IPendingStates = {[{r0},{t0},X]}Let [R,T,X] be some state in the set pendingStatesWhile pendingState != empty do for all r bt R, S1 : r –S1->, t bt T, S2 : t-S2->do S1`:=computecomplement(S1) S2`:=computecomplement(S2) if valid(S1’US2’,X) then R`:=computetarget(R,S1) T`:=computetarget(T,S2) X`:=modifycounter(x,S1`US2`) addtransition : [R,T,X] -> [R’,T’,X’] if[R’,T`,X`] nbt QUpendingStates then Add [R`,T`,X`] to pending States end if end if end for Add [R,T,X] to Q and remove from pending Statesend whilePrune()

Problems about the algorithm

1. ComputeComplement(S1)input : a transitionoutput : another transitionis the function return an opposite arrow with the same variable on it?

2. Valid(S,X)input : S, transition; X, counteroutput : true if every data required in S is available

3. Computetarget(source,S) return the set that state in S can reach via X (transition)

4. ModifyCounters(X,S)if a data is required at the transition, the counter is increase. If a data is sent, the counter is decresed.