Combinational Circuits Verification. 2. Verification by Equivalence Checking. Combinational Equivalence Checking (con t)

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1 Comintionl Ciruits Verifition 2. Verifition y Equivlene Cheking Pge Comintionl Ciruits Verifition 2.2 Propositionl Logi (Clulus) 2.5 Propositionl Resolution 2. Stålmrk s Proedure 2.9 Redued Ordered Binry Deision Digrms (ROBDDs) 2.23 Sequentil Ciruits Verifition 2.55 Reltionl Representtion of FSMs 2.56 Reltionl Produt of FSMs 2.6 Rehility Anlysis on FSMs 2.62 Equivlene Cheking Tools 2.7 Referenes 2.73 Consist of n interonnetion of logi gtes AND, OR, NOT, NAND, NOR, XOR, XNOR, nd loks implementing more omple logi (Boolen) funtions. No logil loops, i.e., topologilly there my e loops, ut they re not sensitile under ny (vlid) input omintion, even suh loops my e prohiited / not produed y utomted nlysis / synthesis tools Gol Given two Boolen netlists, hek if the orresponding outputs of the two iruits re equl for ll possile inputs Two iruits re equivlent iff the Boolen funtion representing the outputs of the networks re logilly equivlent Identify equivlene points nd implitions etween the two iruits to simplify equivlene heking Sine typil design proeeds y series of lol hnges, in most ses there re mny implitions / equivlent suiruits in the two iruits to e ompred Vrious tutology/stisfiility heking lgorithms sed on heuristis (prolem is NPomplete, ut mny work well on rel pplitions...) In this ourse we onsider three min omintionl equivlene heking methods: - Propositionl resolution method (tutology/stisfiility heking) - Stålmrk s method (reent ptented lgorithm, very effiient nd populr) - ROBDD-sed method (Boolen funtion onverted into ROBDD s representtion) 2. (of 73) 2.2 (of 73) Comintionl Equivlene Cheking Comintionl Equivlene Cheking (on t) Epliit Proof Impliit Proof f = flg: T/F Tutology Chek d f (,,) out f2 (f = f2) = T Propositionl resolution Stålmrk s proedure y q p f2(,y,) out 2 y hek! ROBDDs ROBDDs 2.3 (of 73) 2.4 (of 73)

2 Propositionl Logi (Clulus) Propositionl Logi (ont d) Synt P, Q, R,... propositionl symols (tomi propositions) t: true; f: flse onstnts P: not P P Q: P nd Q P Q: P or Q; P Q: if P then Q (proposition equivlent to P Q) P Q: P if nd only if Q, i.e., P equivlent to Q (proposition equivlent to (P Q) P Q) ) Semntis Given through the Truth Tle: Formul F is stisfile (onsistent) iff it is true under t lest one interprettion Formul F is unstisfile (inonsistent) iff it is flse under ll interprettions Formul F is vlid iff it is true (onsistent) under ll interprettions Interprettion I stisfies formul F (I is model of F) iff F is true under I. Nottion: I F Theorem: A formul F is vlid ( tutology) iff F is unstisfile. Nottion: The reltionship etween F to F n e visulied y mirror priniple : All formuls in propositionl logi F P Q P P Q P Q P Q P Q t t f t t t t t f f f t f f f t t f t t f f f t f f t t An interprettion is funtion from the propositionl symols to {t, f} Vlid formuls G Stisfile, ut non-vlid formuls F F Unstisfile formuls To determine if F is stisfile or vlid, test finite numer (2 n ) of interprettions of the n tomi propositions ourring in F... ut it is n eponentil method... stisfiility is n NP-omplete prolem G 2.5 (of 73) 2.6 (of 73) Propositionl Logi (ont d) Proofs A proof of proposition is derived using ioms, theorems, nd inferene rules (n inferene rule permits deduing onlusions sed on the truth of ertin premises) A logi formul F is deduile from the set S of sttements if there is finite proof of F strting from elements of S. Nottion: S F Emple: A simple proof system Aioms: K: A (B A) S: (A (B C)) ((A B) (A C)) DN: A A Inferene rule (Modus Ponens): {A B, A} B A proof of A A () (2) (3) (4) (5) (A ((D A) A)) ((A (D A)) (A A)) A ((D A) A) (A (D A)) (A A) A (D A) A A y S ([B\D A], [C\A]) y K ([B\D A]) y MP, (), (2) y K y MP, (3), (4). Propositionl Logi (ont d) Reltion etween synt nd semntis Truth tles provide mens of deiding truth Propositionl logi is: - omplete: everything tht is true my e proven, i.e., if S A then S A - onsistent (sound): nothing tht is flse my e proven. i.e., if S A then S A - deidle: there is n lgorithm for deiding the truth of ny proposition, i.e., test finite (eponentil) numer of truth ssignments 2.7 (of 73) 2.8 (of 73) 2

3 Flse Negtive & Flse Positive Comintionl Equivlene Cheking Let P e proposition ( property) nd A verifition method (lgorithm). Flse Negtive: (similr to inompleteness) A(P) reports true interprettion, (P) = true A(P) reports flse ( interprettion, (P) = true)! Flse Positive: (similr to inonsisteny, unsoundness) A(P) reports flse interprettion, (P) = flse A(P) reports true ( interprettion, (P) = flse)! (, (P) = flse) (, (P) = true) Determine if two epressions f nd f2 denote the sme truth tle Applition: Determine if two omintionl logi iruit designs C nd C2 implement the sme truth tle (logi (Boolen) funtion) - Etrt representtion of logi epressions f nd f2 - Verify if (f f2) is vlid formul, i.e., f f2 is unstisfile using stisfiility lgorithms (Propositionl Resolution methods), or (f f2) nd (f2 f) hold (where f nd f2 re trnsformed to implition form using Stålmrk s proedure), or f nd f2 hve the sme nonil form using, e.g., Redued Binry Deision Digrms 2.9 (of 73) 2. (of 73) Propositionl Resolution A Literl L is n tomi proposition A or its negtion A A Cluse C is finite set of disjuntive literls (C = L L 2 L 3 C is true iff one of its elements is true. The empty luse is lwys flse. Let A, A 2,... e tomi propositions nd L i, j literls Conjuntive Norml Form (CNF): onjuntion of disjuntions of literls n m i F=( ( L i, j )), where L i, j {A,A 2,...} { A, A 2,...} i= j= Disjuntive Norml Form (DNF): disjuntion of onjuntions of literls n m i F=( ( L i, j )), where L i, j {A,A 2,...} { A, A 2,...} i= j= Eh L i, j {A,A2,...} { A, A2,...} ppers in eh disjunt (onjunt) t most one! Theorem: For every logi formul F, there is n equivlent CNF nd n equivlent DNF Cnonil Conjuntive Form (CCF): CNF in whih eh L ppers etly one Cnonil Disjuntive Form (DCF): DNF in whih eh L ppers etly one Propositionl Resolution (ont d) Resolution is proof method underlying some utomti theorem provers sed on simple syntti trnsformtion nd refuttion. Refuttion is proedure to show tht given formul is unstisfile Resolution proedure: - To prove F, we trnslte F into set of luses, eh disjuntion of tomi formule or their negtions. - Eh resolution step tkes two luses nd yields new one. - The method sueeds if it produes the empty luse ( ontrdition), thus refuting F. 2. (of 73) 2.2 (of 73) 3

4 Propositionl Resolution (ont d) Propositionl Resolution (ont d) Let F=(L,... L,n )... (L k,... L k,nk ) where literls L i, j {A,A 2,...} { A, A 2,...} F n e viewed s set of luses: F={{L,,..., L,n },..., {L k,..., L k,nk }}, where - Comm seprting two literls within luse orresponds to - Comm seprting two luses orresponds to Let L e literl in luse C (L C ) nd its omplement L in luse C 2 (L C 2 ), Cluse R is resolvent of C nd C 2 if: R = (C {L}) (C 2 {L}) Emple: F = {{p, r}, {q, r}, { q}, { p, v}, { s}, {s, v}}. {p, r} {q, r p, v {s, v {p, q} { q} { p, s { s {p} { } { p} A (resolution) dedution of C from F is finite sequene C, C 2,..., C n of luses suh tht eh C i is either in F or resolvent of C j, C k, (j, k < i) Res(F) = F R where R is resolvent of two luses in F Lemm. F nd F R re equivlent Define Res (F) = F, Res n+ (F) = Res( Res n (F) ), n Let Res * (F) = Res n (F) n Theorem. F is unstisfile iff Res * (F) Algorithm: to deide stisfiility of formul F in CNF (luse set): repet G:=F; F:=Res(F) until (( F) or (F = G); if F then F is unstisfile else F is stisfile. 2.3 (of 73) 2.4 (of 73) Summry of si ide: Propositionl Resolution (ont d) Two iruits C nd C2 Propositinl Resolution - Emple Gol: G in DNF is vlid? G is unstisfile F = G out out2 C C2 =? in CNF: F = (L,.. L,n )... (L k,... L k,nk ) F = {{L,,..., L,n },..., {L k,..., L k,nk }} Resolution Refuttion proedure F={ } (ontrdition) Propositionl Resolution C: out = C2: out2 = ( ) ( ) (Mu: out2 = ( s ) (s ) ) G = (out out2) 2.5 (of 73) 2.6 (of 73) 4

5 G = (out out 2) ( out out 2) (DNF) Theorem Proving =true? F = G = out out 2) ( out out 2)) = Flse? (unstifile!) out out2 CNF F = out out 2) (out out 2) = ( ) ) )]) ( ) ) )]) =... = ( ) ) ) out = ( s ) (s ) = ( ) ) s out2 = Literls: {{ }, { }, {, }} = ( ) = ( ) ) = ) = = { } {, {} { } derive empty luse { } { } out2 = out 2.7 (of 73) 2.8 (of 73) Stålmrk s Proedure Trnsform propositionl formul G (in liner time) in nested implition form, e.g.: G = (p (q r)) s G is now represented using set of triplets { i,, y}, mening i ( y), e.g.: (p (q r)) s eomes {(, q, r), ( 2, p, ), ( 3, 2, s)}; G = 3 To prove formul vlid, ssume tht it is flse nd try to find ontrdition (use for flse nd for true, s in swithing (Boolen) lger) Derivtion rules: (/ mens reple y ) r (, y, ) y/, / mening flse (y ) implies y = true nd = flse r2 (, y, ) / mening (y true) implies = true r3 (,, ) / mening (flse ) implies = true r4 (,, ) / mening (true ) implies = r5 (, y, ) / y mening (y ) implies = y r6 (,, ) /, / mening ( ) implies = true nd = true r7 (, y, y) / mening (y y) implies = true Emple: G = (p (q p)) : {(, q, p), ( 2, p, )}, ssume G = 2 =, i.e., (, p, ) By r : p = nd =, sustitute for nd get (, q, ) (whih is terminl triplet) Agin y r this is ontrdition sine / is derived for in r, hene 2 = G = (true) Stålmrk s Proedure (ont d) Not ll formuls n e proved with these rules, need form of rnhing: Dilemm rule T = set of triplets, D i, i =, 2, re derivtions, results U[S ] nd V[S 2 ], onlusion T[S] T T[/] D U[S ] T[S] T[/] D 2 V[S 2 ] Assume = derive result, then ssume = nd lso derive result. - If either derivtion gives ontrdition, the result is the other derivtion - If oth re ontrditions, then T ontins ontrdition - Otherwise return the intersetion of the result of the two derivtions, sine ny informtion gined from = nd = must e independent of tht vlue Emple: T = { (, p, p), (, p, p) } nnot e resolved using r - r7 T[p/] = {(,, ), (,, )} where (,, ) is ontrdition T[p/] = {(,, ), (,, )} where (,, ) is gin ontrdition Hene T[S] results in ontrdition. 2.9 (of 73) 2.2 (of 73) 5

6 Stålmrk s Proedure (ont d) Trnsformtion from nd-or-not logi to implition form: not: G = A A (, A, ), G = or: G= A B A B, y, B (y, A, ), G = nd: G= A B A B, y, (y, A, ), (, B, ), G = Emple of equivlene heking: y. C = {(y, e, ), (e,, ), (, f, ), (f,, g), (g,, )} Chek y t nd t y y t: Form C C2 {(, y, t)} whih y r yields [y/, t/] nd fter sustitution {(, e, ), (e,, ), (, f, ), (f,, g), (g,, ), (, h, s), (h,, ), (s, u, ), (u, r, v), (v,, ), (r, w, ), (w,, p), (p,, )} giving y r gin [h/, s/] nd.... r C2 = {(t, h, s), (h,, ), (s, u, ), (u, r, v), (v,, ), (r, w, ), (w,, p), (p,, )} s t Stålmrk s Proedure (ont d) Emple of equivlene heking (ont d): {(, e, ), (e,, ), (, f, ), (f,, g), (g,, ), (,, ), (, u, ), (u, r, v), (v,, ), (r, w, ), (w,, p), (p,, )} pply r nd r5 nd get [u/, e/, / f, g/, v/, r/ w, p/ ] whih yields {(,, f), (f,, ), (,, ), (, w, ), (w,, )} Applition of Dilemm rule to, sy, yields: = : {(,, f), (f,, ), (,, ), (, w, ), (w,, )} yields [f/, w/] y r2, thus {(,, ), (,, ), (,, ), (,, )} i.e., (,, ) is ontrdition = : {(,, f), (f,, ), (,, ), (, w, ), (w,, )} ontrdition gin Conlusion: y t holds. Similrly for t y The two iruits re equivlent. 2.2 (of 73) 2.22 (of 73) Binry Deision Digrms (BDDs) Clssil representtion of logi funtions: Truth Tle, Krnugh Mps, Sum-of-Produts, ritil omplees, et. Critil drwks: - My not e nonil form or is too lrge (eponentil) for useful funtions, Equivlene nd tutology heking is hrd - Opertions like omplementtion my yield representtion of eponentil sie Redued Ordered Binry Deision Digrms (ROBDDs) A nonil form for Boolen funtions Often sustntilly more ompt thn trditionl norml forms Cn e effiiently mnipulted Introdued minly y R. E. Brynt (986). Vrious etensions eist tht n e dpted to the sitution t hnd (e.g., the type of iruit to e verified) Binry Deision Trees A Binry deision Tree (BDT) is rooted, direted grph with terminl nd nonterminl verties Eh nonterminl verte v is leled y vrile vr(v) nd hs two suessors: - low(v) orresponds to the se where the vrile v is ssigned - high(v) orresponds to the se where the vrile v is ssigned Eh terminl verte v is leled y vlue(v) {, } Emple: BDT for two-it omprtor, f(, 2,, 2 ) = ( ) ( 2 2 ) Unordered 2 vriles (of 73) 2.24 (of 73) 6

7 Binry Deision Trees (ont d) We n deide if truth ssignment = (,..., n ) stisfies formul in BDT in liner time in the numer of vriles y trversing the tree from the root to terminl verte: -If vr(v) is, the net verte on the pth is low(v) -If vr(v) is, the net verte on the pth is high(v) -If v is terminl verte then f() = f v (,..., n ) = vlue(v) -If v is nonterminl verte with vr(v)= i, then the struture of the tree is otined y Shnon s epnsion f v (,..., n ) = i f low(v) (,..., n )] [ i f high(v) (,..., n )] For the omprtor, (, 2,, 2 ) leds to terminl verte leled y, i.e., f(,,, ) = Binry deision trees re redundnt: - In the omprtor, there re 6 sutrees with roots leled y 2, ut not ll re distint Merge isomorphi sutrees: - Results in direted yli grph (DAG), inry deision digrm (BDD) Cnonil Form property Redued Ordered BDD A nonil representtion for Boolen funtions is desirle: two Boolen funtions re logilly equivlent iff they hve isomorphi representtions This simplifies heking equivlene of two formuls nd deiding if formul is stisfile Two BDDs re isomorphi if there eists ijetion h etween the grphs suh tht - Terminls re mpped to terminls nd nonterminls re mpped to nonterminls - For every terminl verte v, vlue(v) = vlue(h(v)), nd - For every nonterminl verte v: vr(v) = vr(h(v)), h(low(v)) = low(h(v)), nd h(high(v)) = high(h(v)) Brynt (986) showed tht BDDs re nonil representtion for Boolen funtions under two restritions: () the vriles pper in the sme order long eh pth from the root to terminl (2) there re no isomorphi sutrees or redundnt verties Redued Ordered Binry Deision Digrms (ROBDDs) 2.25 (of 73) 2.26 (of 73) Cnonil Form Property Requirement (): Impose totl order < on the vriles in the formul: if verte u hs nonterminl suessor v, then vr(u) < vr(v) Requirement (2): repetedly pply three trnsformtion rules (or impliitly in opertions suh s disjuntion or onjuntion) Cnonil Form Property (ont d) 2. Remove duplite nonterminls: if nonterminls u nd v hve vr(u) = vr(v), low(u) = low(v) nd high(u) = high(v), eliminte one of the two verties nd rediret ll inoming rs to the other verte shre.remove duplite terminls: eliminte ll ut one terminl verte with given lel nd rediret ll rs to the eliminted verties to the remining one 3. Remove redundnt tests: if nonterminl verte v hs low(v) = high(v), eliminte v nd rediret ll inoming rs to low(v) 2.27 (of 73) 2.28 (of 73) 7

8 Creting the ROBDD for y Cnonil Form Property (ont d) () y y A nonil form is otined y pplying the trnsformtion rules until no further pplition is possile Brynt showed how this n e done y proedure lled Redue in liner time Applitions: - heking equivlene: verify isomorphism etween ROBDDs - non-stisfiility: verify if ROBDD hs only one terminl node, leled y - tutology: verify if ROBDD hs only one terminl node, leled y () eomes eomes Emple: ROBDD of 2-it Comprtor with vrile order < < 2 < 2 : () y y (of 73) 2.3 (of 73) ROBDD Emples ROBDD Emples (on t) OR out = f (,) = AND out = f (,) = out BDD ROBDD BDD ROBDD 2.3 (of 73) 2.32 (of 73) 8

9 ROBDD Emples (on t) ROBDD Emples (on t) XOR out = f(,) = NAND s out = f(,) = ( ) BDD ROBDD BDD out red T.T. ROBDD out (of 73) 2.34 (of 73) Vrile Ordering Prolem Vrile Ordering Prolem - Emple The sie of n ROBDD depends ritilly on the vrile order For order < 2 < < 2, the omprtor ROBDDeomes: y 2 y2 3 y y y y y y y y y y2 y 2 y2 y 2 For n n-it omprtor: < <... < n < n gives 3n+2 verties (linerompleity) <... < n <... < n, gives 32 n verties (eponentilompleity!) 2 y2 y2 y2 y2 y3 y3 y3 3 y (of 73) 2.36 (of 73) 9

10 Vrile Ordering Prolem (ont d) Logi Opertions on ROBDDs The prolem of finding the optiml vrile order is NP-omplete Some Boolen funtions hve eponentil sie ROBDDs for ny order (e.g., multiplier) Heuristis for Vrile Ordering Heuristis developed for finding good vrile order (if it eists) Intuition for these heuristis omes from the oservtion tht ROBDDs tend to e smller when relted vriles re lose together in the order (e.g., ripple-rry dder) Vriles ppering in suiruit re relted: they determine the suiruit s output should usully e lose together in the order Dynmi Vrile Ordering Useful if no ovious stti ordering heuristi pplies During verifition opertions (e.g., rehility nlysis) funtions hnge, hene initil order is not good lter on Good ROBDD pkges periodilly internlly reorder vriles to redue ROBDD sie Bsi pproh sed on neighoring vrile ehnge... < < <......< < <... Among numer of trils the est is tken, nd the ehnge is repeted Residul funtion (oftor): {, } f i (,..., n ) = f(,..., i-,, i+,..., n ) ROBDD of f i omputed y depth-first trversl of the ROBDD off: For ny verte v whih hs pointer to verte w suh tht vr(w) = i, reple the pointer y low(w) if is nd y high(w) if is. If not in nonil form, pply Redue to otin ROBDD of f i. All 6 two-rgument logi opertions on Boolen funtion implemented effiiently on ROBDDs in liner time in the sie of the rgument ROBDDs (of 73) 2.38 (of 73) Logi Opertions on ROBDDs (ont d) Logi Opertions on ROBDDs (ont d) Bsed on Shnnon s epnsion f = [ f ] [ f ] Brynt (986) gve uniform lgorithm, Apply, for omputing ll 6 opertions: f * f : n ritrry logi opertion on Boolen funtions f nd f v nd v : the roots of the ROBDDs for f nd f, = vr(v) nd = vr(v ) Consider severl ses depending on v nd v () v nd v re oth terminl verties: f * f = vlue(v) * vlue(v ) (2) = : use Shnnon s epnsion f * f = [ (f * f )] (f * f )] to rek the prolem into two suprolems, eh is solved reursively The root is v with vr(v) = Low(v) is (f * f ) High(v) is (f * f ) (3) < : f =f = f sine f does not depend on In this se the Shnnon s epnsion simplifies to f * f = [ (f * f )] (f * f )], similr to (2) nd ompute suprolems reursively, (4) < : similr to the se ove Improvement using the if-then-else (ITE) opertor: ITE(F, G, H) = F. G + F. H where F, G nd H re funtions Reursive lgorithm sed on the following, v is the top vrile (lowest inde): ITE(F, G, H) = = = v.(f.g + F.H) v + v.(f.g + F.H) v v.(f v.g v + F v.h v ) + v.(f v.g v +F v.h v ) (v, ITE(F v, G v, H v ), ITE(F v, G v,h v )) With terminl ses eing: F = ITE(, F, G) = ITE(, G, F) = ITE(F,, ) = ITE(G, F, F) we define NOT(F) = ITE(F,, ) OR(F, G) = ITE(F,, G) AND(F, G) = ITE(F, G, ) XOR(F, G) = ITE(F, G, G) LEQ(F, G) = ITE(F, G, ) et (of 73) 2.4 (of 73)

11 Logi Opertions on ROBDDs (ont d) By using dynmi progrmming, it is possile to mke the ITE lgorithm polynomil: () The result must e redued to ensure tht it is in nonil form; - reord onstruted nodes (unique tle); - efore reting new node, hek if it lredy eists in this unique hsh tle (2) Reord ll previously omputed funtions in hsh tle (omputed tle); - must e implemented effiiently s it my grow very quikly in sie; - efore omputing ny funtion, hek tle for solution lredy otined Complement edges n redue the sie of n ROBDD y ftor of 2 - Only one terminl node is leled - Edges hve n ttriute (dot) to indite if they re inverting or not - To mintin noniity, dot n pper only on low(v) edges - Complementtion hieved in O() Comprtor: F time y pling dot on the funtion edge - F nd F n shre entry in omputed tle 2 - Adpttion of ITE esy 2 Test for F G n e omputed y speilied ITE_CONSTANT lgorithm Tsk: ompute ROBDD for f (,) BDD Opertors - Emple y out = f (,) ) f = y = ( ) ( ) order,. out out (of 73) 2.42 (of 73) BDD Opertors - Emples (on t) 2) f = y BDD f = BDD BDD y = Conj (BDD,BDD y ) BDD BDD y = : = = : = = : = = : = = = Other Deision Digrms Multiterminl BDD (MTBDD): Pseudo-Boolen funtions B n N, terminl nodes re integers Binry Moment Digrms (BMD): for representing nd verifying rithmeti opertions, word-level representtion Ordered Kroneker Funtionl BDDs (OKFBDD): Bsed on XOR opertions nd OBDD Free BDDs (FBDD): Different vrile order long different pths in the grph Zero suppressed BDDs (ZBDD) Comintion of vrious forms of DDs integrted in DD softwre pkges: Drehsler et l (U. Freiurg, Germny), Clrke et l (Crnegie Mellon U., USA) Etension to represent systems of liner nd Boolen onstrints (DTU) Multiwy Deision Digrms (MDG): Representtion for suset of equtionl firstorder logi for modeling stte mhines with strt nd onrete dt (U. of Montrel) Well known ROBDD pkges: CMU (s used in SMV from Crnegie Mellon U.) CUDD, U. of Colordo t Boulder (s used in VIS from UC t Berkeley) Industril pkges: Intel, Luent, Cdene, Synopsys, Bull Systems, et (of 73) 2.44 (of 73)

12 Applitions of ROBDDs Comintionl Equivlene Chequing - Emple ROBDD: Constrution DD from iruit desription: - Depth-first vs. redth-first onstrution (keep only few levels in memory, rest on disk; prolem with dynmi reordering) - Prtitioning of Boolen spe, eh prtition represented y seprte grph - Bottom-up vs. top-down, introduing deomposition points Internl orrespondenes in the two iruits equivlent funtions, or omple reltions Two iruits C nd C2 out out 2 C C2 =? ATPG-sed: Comine iruits with n XOR gte on the outputs, show ineistene of test for fult s- - on the output (i.e., the output would hve to e driven to mening tht there is differene in the two iruits) Use ATPG nd lerning to determine equivlent iruit nodes Fst rndom simultion: Detet quikly esy differenes Rel tools: Use omintion of tehniques, fst nd less powerful first, slow ut et lter C: C2: if then else MUX: if then else isomorph 2.45 (of 73) 2.46 (of 73) Comintionl Equivlene Cheking Multipleor Emple Mutipleor Emple (on t) Build ROBDD for Imp: Speifition:if = then out = else out = Build ROBBD for Spe: out Implementtion: out = ( ) ( ) ROBDD: order:,, out out 2.47 (of 73) : y: y disjuntion (or) out ROBDD2 order:,, isomorph to ROBDD! 2.48 (of 73) 2

13 Multipleor Emple (on t) Comprtor Emple Alterntive wy to uild ROBDD2: out = order:,, out 2 ˆ ˆ 2 f ˆ ˆ Spe: f ˆ = ˆ if ˆ = ˆ BBD ROBBD isomorph to ROBDD Refinement: f (, 2,, 2 ) = if ( = ) 2 = 2 ) Impliit: ˆ = 2 ˆ = = = f (...) 2.49 (of 73) 2.5 (of 73) Comprtor Emple (ont d) Comprtor Emple (ont d) y y = : = y y T: T2: disj: T, T2, T2 = = 2 2 f= T T2 f ( ) T3: 2 2 T4: 2 2 disj: T3, T4, T T4 T3 2.5 (of 73) 2.52 (of 73) 3

14 Comprtor Emple (ont d) Equivlene Cheking in Prtie Usully, omintionl iruits implement rithmeti nd logi opertions, nd net-stte nd output funtions of finite-stte mhines (sequentil iruits) onj. T 2, T 34,, 2, 2 order:,, 2, 2 ) independent Isomorph to the spe Verifying the ehvior of the gte-level implementtion ginst the RTL design of digitl systems n often e redued to verifying the omintionl iruits - Equivlene omprison etween the net-stte nd output funtions (omintionl iruits) - Requires tht oth hve the sme stte spe (nd of ourse inputs nd outputs), knowing the mpping etween sttes helps... - Cn lso e used to verify gte-level implementtion ginst gte-level model etrted from lyout - This kind of verifition is useful for onfirming the orretness of mnul hnges or synthesis tools If the stte spe is not the sme, sequentil (ehviorl equivlene) of FSM must e onsidered (of 73) 2.54 (of 73) Sequentil Ciruits nd Finite Stte Mhines Reltionl Representtion of FSM Com. Logi f(, y) Com. Logi g(, y) r y y r s Clok To verify the ehvior of suh iruits we need effiient representtion for the mnipultion of net-stte nd output funtions nd sets of sttes Using hrteristi funtions of reltions nd sets r = (r,..., r s ) vetor of memory its stte vriles, memorie enoded sttes y = (y,..., y s ) vetor of present stte vlues y = (y,..., y s ) vetor of net stte vlues = (,..., m ) vetor of input its enode input symols = (,..., n ) vetor of output its enode output symols f = output funtion, f(, y) = Mely, f(y) = Moore g = net-stte funtion Here we onsider FSM synhronied on lok trnsitions synhronous sequentil iruits Representtion of Reltions nd Sets If R is n-ry reltion over {,} then R n e represented y (the ROBDD of) its hrteristi funtion: f R (v,...,v n ) = iff (v,...,v n ) R - Sme tehnique n e used to represent sets of sttes Trnsition reltion N of sequentil iruit is represented y its Boolen hrteristi funtion over inputs nd stte vriles: N(, y,..., y s,y,..., y s ) Emple: synhronous modulo 8 ounter, N(y, y ) = N (y, y ) N (y, y ) N 2 (y, y 2 ) y 2 y 2 y y Register (3 its) y y Net stte y : y = y y = y y y 2 = (y y y 2 Trnsition reltion N(y,y ): N (y,y ) = (y y ) N (y,y ) = (y y y ) N 2 (y,y ) = (y 2 (y y y 2 ) 2.55 (of 73) 2.56 (of 73) 4

15 Reltionl Representtion of FSM (ont d) Reltionl Representtion of FSM (ont d) Quntified Boolen Formuls (QBF) Needed to onstrut omple reltions nd mnipulte FSMs V={v,v 2,..., v n } = set of Boolen (propositionl) vriles QBF(V) is the smllest set of formuls suh tht - every vrile in V is formul - if f nd g re formuls, then f, f g, f g re formuls - if f is formul nd v V, then v.f nd v.f re formuls A truth ssignment for QBF(V) is funtion : V {,} If {,}, then [v ] represents [v ](w) = if v = w [v ](w) = (w) if v w f is formul in QBF(V) nd is truth ssignment: f if f is true under. Quntified Boolen Formuls (ont d) QBF formuls hve the sme epressive power s ordinry propositionl formuls; however, they my e more onise QBF Semntis: reltion is defined reursively: v iff (v)=; f iff f; f g iff f or g; f g iff f nd g; v.f iff [v ] for [v ] f; v.f iff [v ] f nd [v ] f. Every QBF formul n represent n n-ry Boolen reltion onsisting of those truth ssignments for the vriles in V tht mkes the formul true: Boolen hrteristi funtion of the reltion. f = f f,. f = f f In prtie, speil lgorithms needed to hndle quntifiers effiiently (e.g., on ROBDD) 2.57 (of 73) 2.58 (of 73) Sequentil Equivlene Cheking Reltionl Produt of FSMs Bsi Ide: To prove the equivlene of two FSMs M nd M 2 (with the sme input nd output lphet), produt mhine is formed whih tests the equlity of outputs of the two individul mhines in every stte M M 2 Produt Mhine Flg M nd M 2 re equivlent iff the produt mhine produes Flg = true output in every stte rehle from the initil stte Coudert et l. were first to reognie the dvntge of representing set of sttes with ROBDD s: Symoli Bredth-First Serh of the trnsition grph of the produtmhine Their tehnique ws initilly pplied to heking mhine equivlene nd lter etended y MMilln, et l. to symoli model heking of temporl logi formuls (in CTL) Reltionl Produts implementtion using ROBDD A typil tsk in verifition: ompute reltionl produts with strtion of vriles: v.[f(v) g(v)] Algorithm RelProd omputes it in one pss over ROBDDs f(v) nd g(v), insted of onstruting f(v) g(v) RelProd uses omputed tle (result he), nd is sed on Shnnon s epnsion Entries in the he hve the form (f, g, E, h), where E is set of vriles tht re eistentilly qulified out nd f, g nd h re (pointers to) ROBDDs If n entry indeed y f, g nd E is in the he, then previous ll to RelProd (f, g, E) hs returned h, it is not reomputed Algorithm works well in prtie, even if it hs theoretil eponentil ompleity 2.59 (of 73) 2.6 (of 73) 5

16 Reltionl Representtion of FSMs (ont d) Reltionl Produt Algorithm RelProd (f, g: ROBDD, E: set of vriles) if f=flse g=flse then return flse else if f=true g=true then return true else if (f, g, E, h) is hed then return h else let nd y e the top vriles of f nd g, respetively let e the topmost of nd y, h:=relprod(f =, g =,E) h:=relprod(f =, g =,E) if E then h:=or(h, h) {ROBDD: h h} else h:=ifthenelse(, h, h) endif insert (f, g, E, h) in he return h endif Rehility Anlysis on FSMs Computing Set of Rehle Sttes Rehle stte omputtion (stte enumertion) is needed for FSM equivlene nd model heking S = set of sttes, represented y the ROBDD S (V) Find those sttes S rehle in t most one trnsition from S : ROBDD s S (y) nd N(y, y ), ompute n ROBDD representing S: S =S { s s [s S (s, s ) N]} S (y )=S (y ) y i [S (y) N(y,y )] yi y S S S 2 S 2 = S {s s [s S (s, s ) N]} S 2 (y )=S (y ) y i [S (y) N(y,y )] yi y 2.6 (of 73) 2.62 (of 73) Rehility Anlysis on FSMs (ont d) BDD Enoding Rehility Anlysis (ont d) In generl, the sttes rehle in t most k+ steps re represented y: S k+ (y ) = S (y ) y i [ S k (y) N(y.y )] yi y As eh set of sttes is superset of the previous one, nd the totl numer of sttes is finite, t some point, we must hve S k+ = S k, k 2 s the numer of sttes Rehility omputtion n e viewed s finding lest fipoint Wht out inputs? Eistentilly quntify them out in the reltionl produt (equivlent to losing the system with non-deterministi soure of vlues for ) M M2 Bsi ide: = Flg Produt Mhine ) onnet oth mhines to equlity hek of outputs 2) ompute set of rehle sttes 2) representing set of sttes using ROBDD 2) omputing imges of BDDs of ll net sttes (using trnsition reltions) 2) rehility itertion (using imges strting from one initil stte until sequene emerges) R = initil BDD.. R i + = R i Imge R i onvergene; 2.63 (of 73) 2.64 (of 73) 6

17 ROBDD Enoding (ont d) Sequentil Equivlene Cheking Emple Representing set of sttes using ROBDDs ) Connet oth mhines to equlity hek of outputs Set {} {} {,,,,, } i i Formul ROBDD lk out i Q D lk y out mu Q D lk 2 mu y Q D lk y2 flg 2.65 (of 73) 2.66 (of 73) Sequentil Equivlene Cheking Emple (on t) Sequentil Equivlene Cheking Emple (ont d) 2) Representing set of sttes using ROBDD 2) Compute imge of set { } Initil Stte: = = 2 = {,,} set set = { } formul ROBDD 2 imge: trnsition reltion: 2 2 i = i = y= i y= i i 2 y2= i 2 i BDD BDD BDD 2.67 (of 73) 2.68 (of 73) 7

18 Sequentil Equivlene Cheking Emple (ont d) Emple (ont d) 2) Compute imges of set { } { y = i Tnsition Reltion y = i i 3 y2 = i 2 i : : 2: i= i= BDD = (i = ) i = (i = ) = (i = ) i = i = ) i = i = ) = i = ) {{,,}, {,,}} 2 2) Rehility itertion R = 2 R = 2 2 R R 2 = R In terms of sets: R = {} R = {, } R2 = {,} R2 =R BDD2 2 ROBDD disj. BDD BDD2 Converged ll sttes rehed! 2.69 (of 73) 2.7 (of 73) Equivlene Cheking Tools Equivlene Cheking Tools (ont d) Commeril tools: Chryslis: Design Verifier Synopsys: Formlity Cdene: Conforml Verysys: Torndo AHL: ChekOff-E Applition: Used to prove equivlene of two sequentil iruits tht hve the sme stte vriles (or t lest the sme stte spe nd known mpping etween sttes) y verifying tht they hve the sme net-stte nd output funtions Used in ple of gte vs. RTL verifition y simultion Reommendtions: Use modulr design, reltively smll modules, k - 2k gtes Mintin hierrhy during synthesis (not flttening) nd efore lyout: equivlene n e proven hierrhilly muh fster, espeilly for rithmeti iruits ChekOff-E Commeril produt y Astrt Hrdwre Ltd. (UK) nd Siemens AG(Germny) Performs ehviorl omprison of two Finite Stte Mhines Input EDIF netlist + lirry or VHDL VHDL suset (superset of synthesile synhronous VHDL) - no rel time luses (fter, wit for), no onditionl loop sttements Interprets VHDL simultion semntis to uild Miro FSM Converts to Mro FSM y merging trnsition until stilition t eh time t Mro FSM is strting point for ny verifition; representtion in ROBDD Produt disontinued! 2.7 (of 73) 2.72 (of 73) 8

19 Referenes. V. Spershneider, G. Antoniou. Logi: A Foundtion for Computer Siene. Addison- Wesley, S. Reeves, M. Clrke. Logi for Computer Siene. Addison-Wesley, Aln J. Hu, Forml Hrdwre Verifition with BDDs: An Introdution, IEEE Pifi Rim Conferene on Communitions, Computers, nd Signl Proessing, pp , J. Jin, A. Nryn, M. Fujit, nd A. Sngiovnni-Vinentelli, Forml Verifition of Comintionl Ciruits, VLSI Design, R.E. Brynt. Grph-Bsed Algorithms for Boolen Funtion Mnipultion. IEEE Trnstions on Computers, C-35(8), pp , August R.E. Brynt. Symoli Boolen Mnipultion with Ordered Binry Deision Digrms. ACM Computing Surveys, 24(3), 992, pp R.E. Brynt. Binry Deision Digrms nd Beyond: Enling Tehnologies for Forml Verifition. Interntionl Conferene on Computer-Aided Design, pp , S. Minto. Binry Deision Digrms nd Applitions for VLSI CAD. Kluwer Ademi Pulishers, M. Sheern, G. Stålmrk. A tutoril on Stålmrk s proof proedure for propositionl logi. Forml Methods in Systems Design, Kluwer, 999..O. Coudert nd J.C. Mdre, A Unified Frmework for the Forml Verifition of Sequentil Ciruits, Int. Conferene on Computer-Aided Design, pp , 99..H. Touti, H. Svoj, B. Lin, R.K. Bryton, nd A. Sngiovnni-Vinentelli, Impliit Stte Enumertion of Finite Stte Mhines Using BDD s, Int. Conferene on Computer-Aided Design, pp. 3-33, (of 73) 9