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A Context-free Relation NS <s, e> <(, e > NP VP <), e > NP <np, e> <(, e > DET N <), e > VP <vp, e > <(, e > V NP <), e > DET <det, the> N <n, dog> N <n, cat> V <v, chased>       / Regular Relations LThe empty set and {a} for all a in Se Se ... are regular relations If R1, R2, and R are regular n-relations, then so are R1R2 = {xy | x R1 and y R2} (concatenation) R1 R2 (union) " R* = Ri (n-way Kleene closure) i=0 There are no other regular relations| %    -   % '0n-way Regular Expressions `a:b:c* e:f:g denotes {<aie, bif, cig> | i 0}^1  11All Correspondences Every n-way regular expression describes a regular n-relation. Every regular n-relation is described by an n-way regular expression. Every n-tape finite-state transducer accepts a regular n-relation. Every regular n-relation is accepted by an n-tape finite-state transducer.,0 2Regular  HIf L1, L2, and L are regular languages and R1, R2, and R are regular relations, the following relations are regular: R1 R2 R1R2 R* R-1 R1 R2 Id(L) L1 L2 Rev(R)fu0  #E   3 Not Regular R1 R2 RH  44n-way Automata Transducers "An n-way automaton is defined by a quintuple similar to the ones that define ordinary finite-state machines (S, Q, q, F, d) Where S is a finite alphabet, Q is a finite set of states q Q is the initial state F in Q is the set of final states d maps Q Se ... Se to 2Qm   n 5;   5Union and Iteration  6 Concatenation  7Range and Domain .Dom(R) = R/S* Range(R) = S*/R Accepting FSMs derived from T(R) by replacing all transition labels a:b by a (domain) or b (range) Regular languages.< z %   8R-1  9Id(L)  :Extending d  vTo state sets To strings The machine accepts a string x just in case is not empty.&w <: w;Cartesian Product  <"T accepts L1 L2@  Proof: by induction. Thus, T enters a final state on iff M(L1) enters a final state on x and M(L2) enters a final state on y. $ =R1 R2B   >Images xxRy <x,y> R x/R = {y | <x,y> R} R/y = {x | <x,y> R} <  =?Images are Regular  @Rev(R)  A Not Regular R1 R2 RH  B  Pumping Lemma It is possible to delete a part of any sufficiently long substring of a regular language and leave a string the is a member of the language  C!Intersection is not Regular  P  ` ̙33` ` ff3333f` 333MMM` f` f` 3>?" dZ@$|? " KZ@ d " @@K`@d n?" dZ(: ::XXuuPR    @ ` ` p>  >   (  )  Z$M$wawa1 ?.    ;Body Text Second Level Third Level Fourth Level Fifth Level     <   `M$wawa1 ?   ? Slide Title    ZH$wawa1?%  [*^B  6>?^  ZdL$wawa1?%g  YBasic Mathematics - 6  S  ? GGGggg untitled 2 UM (  )  Z$J$gg1 ? @   ;Body Text Second Level Third Level Fourth Level Fifth Level     <  ZJ$UnUn1?p@  Page *Z ,,;;p  01 ?    6  S  ? a( 0 N(     ZdO$UnUn1?p@  Page *Z ,,;;6  S  ? a('d F>@(    # lP$wawa1 ?p     # lDQ$wawa1 ? `     "H  0޽h ? GGGggg 'w ldP(    # lR$wawa1 ?     # lDK$wawa1 ?0p   4   `["wawa1))?  "Se = S { e }T0Z  H  0޽h ? GGGggg '{ `( pJ]`' p  HA1? cP   # ldR$wawa1 ?`     # lZ"wawa1 ?0     NA 1?p( X $  NA 1?  X $  NA 1? 8(X $  NA 1? xH X $p   HA1?hp   HA1?      ZX wawa1? Ea    ZX wawa1?   cM(L1)0     ZDW wawa1? '   cM(L2)0    Z^ wawa1?   Fe   Z^ wawa1?2   Fe p  HA1?|   ZD] wawa1?G Fe   Z\ wawa1?Go Fe   Z[ wawa1?w cM(L1)0    Z[ wawa1?a cM(L2)0    ZDZ wawa1?j QO  Fe   ZY wawa1?Z '? Fe   Za wawa1?^ gC  HM(L) H  0޽h ? GGGggg $ C;p(    # ldZ"wawa1 ?     # lY wawa1 ?z     h  Td[ wawa1?   s-... can all characterize any regular language.. .H  0޽h ? GGGgggC #٠  k(     # l$\ wawa1 ?     # l\ wawa1 ?. 0    p  HA 1?6pp  HA 1?  8  ZV wawa1? d: Q S 2Qf  H  Z$b wawa1?  (d: Q S Q Q =2Qh    T$_ wawa1? g L x2A State is final if any member of the set is final33 3  T$wawa1?R g  USearch time is linear for a deterministic, but exponential for a nondeterministic FSMVV VH  0޽h ? GGGggg4 # $\(  $ $ # l_ wawa1 ?   p $ HA 1?cV $  `_ wawa1))?B ?  Pruned: Smaller d is a partial function. No dead states Lookup is fasterDA8 Ij $@ BG/*1?T <  $ Nwawa1? 7/|  ZDead (failure) state H $ 0޽h ? GGGggg  #P 8 0 ( (  ),  ( ( # ld^ wawa1 ?   p ( HA 1?@ ^Fp ( HA 1? @ ^  ( Twawa1? !  y5Minimized: No two states have congruent suffix graphs66 6p (@ HZG/*1?4L| d ( <1?!F  ( Zwawa1?' E Ca   ( Zwawa1? y Cb   ( Zwawa1? 5  Ca   ( ZDwawa1?g G   Cb   ( Zwawa1?5 Da   ( Zwawa1?7 U  Da  ( Zwawa1? Da  (  `$wawa1))?GO 9 EMinimal machine is unique (up to renaming of states and arc ordering)FF FH ( 0޽h ? GGGggg $` h`,(  , , # lwawa1 ?    , # lwawa1 ?J     , Zwawa1?   j"L(M) is the language accepted by M#0Z# #& ,  fwawa1))?   NCompute with machine  They are finite!(( (H , 0޽h ? GGGggg #y\ {s0 (  0 0 # ldwawa1 ?   d2 0 <1?td2 0 <1?T4 d2 0 <1?$| d2 0 <1?T <d2 0 <1?  \ < d2 0 <1? 4l ^B  0 61? d 0< ^B  0 61?L ^B  0 61?$,   0 ZDwawa1?A E1   0 Zwawa1?' g E2  0 Zwawa1?9 Ea  0 ZTwawa1?G  Eb  0 Z4wawa1?2    F1a  0 ZԈwawa1?b G  F2b  0 Z4wawa1? q  }2b is final if 2 and b is final0    0 Zwawa1?7V  NZ$  0 ZTwawa1? - 2  G=$  0 Z4wawa1?B' Ex  0 ZԜwawa1?  Ex  0 Zwawa1?r W  Ex H 0 0޽h ? GGGgggZ  #t 4(  4 4 # ldwawa1 ?    4 # lwawa1 ?     4 Ttwawa1?E?   <Suppose x not in L x takes M(L) into rQ-F. \ x takes M into rF \ xL(M).xL& LXB 4 08c?PPXB 4 08c? 4 Twawa1?_  / 4Suppose x in L x takes M(L) into rQ. \ x takes M into rQ-F \ xL(M).xH" Hr 4 Twawa1?A o  If M(L) were not deterministic, a string x could take M(L) into rF and sF. It would therefore take L(M) into rF and sF }A$ }H 4 0޽h ? GGGggg !@ 0(8( ? 8 8 # lwawa1 ?    8 # lwawa1 ?b-    H 8 0޽h ? GGGggg "l 4,<(  < < # ltwawa1 ?    < # lwawa1 ?FE    < Zԙwawa1?  XY =        B <  `D wawa1))?b g?n  d One can construct families of string relations that parallel the usual classes of formal languages ed e < TE wawa1?  N 2 L < TTF wawa1?g  The n-way concatenation of two string-tuples is the tuple of strings formed by string concatenation of the corresponding elements. H < 0޽h ? GGGggg #Z _W@(  @ @ # lF wawa1 ?    @ # lG wawa1 ?.    ' @ ZG wawa1? \  Ys ( np ( det n ) vp ( v np ( det n ))) the dog chased the catZZ ZH @ 0޽h ? GGGggg !\ 0(D(  D D # l4H wawa1 ?    D # lH wawa1 ?. P    H D 0޽h ? GGGggg   0( H( H wa H H # lTI wawa1 ?    H # lJ wawa1 ?obq    H H 0޽h ? GGGggg "F\ 0(0L(  L L # ltJ wawa1 ?    L # l4K wawa1 ?gpy    H L 0޽h ? GGGggg ٘ kc@P(  P P # lK wawa1 ?    P # lTL wawa1 ?.    3 P ZM wawa1?' Ga ?... and the following languages Dom(R) Range(R) L/R R/L x/R R/x<@  @H P 0޽h ? GGGggg^ D PT(  T T # ltM wawa1 ?    T # lM wawa1 ?    XB T 08c?^ T 6(8c?HH T 0޽h ? GGGggg a RJ`X( 1? X X # lN wawa1 ?    X # lTO wawa1 ?.     X  `P wawa1))?R D HFrom now on, we limit our discussion to binary relations and transducersII IH X 0޽h ? GGGggg`  k   p\ (  \p \ HA1?  \ # lO wawa1 ?    \ Zwawa1?w cT(R1)0   \ Zwawa1?gg Je:e  \ Zwawa1?qV  Je:e  \ Zwawa1?BW' Je:e  \ ZDwawa1?G Fe   \ Zwawa1?r WW  Je:e   \ Zwawa1?B G '  cT(R2)0    \ NA  1?@ 8X  $  \ NA  1?@ pX  $d  \ <1?Ld \ <1?   \ Zwawa1?q'V  HT(R) p \ HA1?(x h H \ 0޽h ? GGGggg4   `\(  ` ` # lwawa1 ?   p ` HA1?Hh  ` Zwawa1?W cT(R1)0   ` Z$wawa1?' cT(R2)0   ` NA 1?8xX $d ` <1?  ` ZDwawa1?w  Je:e   ` Zwawa1?"   Je:e H ` 0޽h ? GGGggg  >6d( 1 d d # ldwawa1 ?    d # lwawa1 ?P1-     H d 0޽h ? GGGggg  h(  h h # ldwawa1 ?    h0 6A 1 ?  h $ @ h NA 1?\$X $H h 0޽h ? GGGgggj < l(  l l # ldwawa1 ?    l NA 1?D<X $ l NA 1?X X $ l NA 1? X $H l 0޽h ? GGGggg " e] p( ZlU&$> p p # lwawa1 ?    p # lDwawa1 ?}(    p NA 1?h ZX $ p NA 1?X rX $ p NA 1?R X $ p NA  1? ( X  $ p NA "1?  j X "$  p NA $1?  X $$  p  `wawa1))?  i#Parallel extensions for transducers$$ $H p 0޽h ? GGGggg   t(  t t # lwawa1 ?    t NA &1?P0X &$ t NA (1?P H8 X ($ t Zwawa1? $  Gwhere $ t ZDwawa1?R? &N.B. L L ID(L)F  t NA *1?@ X *$ t Zwawa1?H ELet 2  t ZHwawa1?   0Claim: T accepts L1 L2J   H t 0޽h ? GGGgggv X< x(  x x # lIwawa1 ?    x # lDJwawa1 ?.     x NA! ,1?0X ,$H x 0޽h ? GGGgggH ^ |p(  lU&| | | # lJwawa1 ?    | NA" .1?X .$ | NA /1?@ X /$ | NA# 11? P X 1$ | NA$ 31? ph X 3$H | 0޽h ? GGGggg w 6.(    # lIwawa1 ?     # ldKwawa1 ?@     NA% 51?* P X 5$   Z$Lwawa1?7 C  HR/intractable = {intractable, iNtractable} iNtractable/R = {intractable} IH IH  0޽h ? GGGggg ; $( F4   # lLwawa1 ?     NA& 71? 0@pX 7$  NA' 91?pP X 9$H  0޽h ? GGGggg p +# (    # lKwawa1 ?   p  HA(1?H p  HG/*1?    TMwawa1?g gK  GStart H  0޽h ? GGGggg^ 0 0(    # lNwawa1 ?     # lNwawa1 ?    XB  08c?^  6(8c?HH  0޽h ? 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Every regular n-relation is described by an n-way regular expression. Every n-tape finite-state transducer accepts a regular n-relation. Every regular n-relation is accepted by an n-tape finite-state transducer.,0 2Regular  HIf L1, L2, and L are regular languages and R1, R2, and R are regular relations, the following relations are regular: R1 R2 R1R2 R* R-1 R1 R2 Id(L) L1 L2 Rev(R)fu0  #E   3 Not Regular R1 R2 RH  C!Intersection is not Regular  E"R is not regular because ... 44n-way Automata Transducers "An n-way automaton is defined by a quintuple similar to the ones that define ordinary finite-state machines (S, Q, q, F, d) Where S is a finite alphabet, Q is a finite set of states q Q is the initial state F in Q is the set of final states d maps Q Se ... 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Ln) W          n*$L1 L2 is Regular$  +  L is Regular Get deterministic, complete M(L) = <S, Q, q, F, d > M = < S , Q, q, F , d> where F =Q-F L(M)=L?Z`$    `, Properties of Regular Languages ~Closure Intersection Union Complementation Concatenation Iteration Reversal Decidable Predicates Emptiness Equality FinitenessHDD - String Relations ;n-way concatenation X = Y =  (  <. A Context-free Relation NS <s, e> <(, e > NP VP <), e > NP <np, e> <(, e > DET N <), e > VP <vp, e > <(, e > V NP <), e > DET <det, the> N <n, dog> N <n, cat> V <v, chased>       / Regular Relations LThe empty set and {a} for all a in Se Se ... are regular relations If R1, R2, and R are regular n-relations, then so are R1R2 = {xy | x R1 and y R2} (concatenation) R1 R2 (union) " R* = Ri (n-way Kleene closure) i=0 There are no other regular relations| %    -   % '0n-way Regular Expressions `a:b:c* e:f:g denotes {<aie, bif, cig> | i 0}^1  11All Correspondences Every n-way regular expression describes a regular n-relation. Every regular n-relation is described by an n-way regular expression. Every n-tape finite-state transducer accepts a regular n-relation. Every regular n-relation is accepted by an n-tape finite-state transducer.,0 2Regular  HIf L1, L2, and L are regular languages and R1, R2, and R are regular relations, the following relations are regular: R1 R2 R1R2 R* R-1 R1 R2 Id(L) L1 L2 Rev(R)fu0  #E   3 Not Regular R1 R2 RH  C!Intersection is not Regular  E"R is not regular because ... 44n-way Automata Transducers "An n-way automaton is defined by a quintuple similar to the ones that define ordinary finite-state machines (S, Q, q, F, d) Where S is a finite alphabet, Q is a finite set of states q Q is the initial state F in Q is the set of final states d maps Q Se ... Se to 2Qm   n 5;   5Union and Iteration  6 Concatenation  7Range and Domain .Dom(R) = R/S* Range(R) = S*/R Accepting FSMs derived from T(R) by replacing all transition labels a:b by a (domain) or b (range) Regular languages.< z %   8R-1  9Id(L)  :Extending d  vTo state sets To strings The machine accepts a string x just in case is not empty.&w <: w;Cartesian Product  <"T accepts L1 L2@  Proof: by induction. Thus, T enters a final state on iff M(L1) enters a final state on x and M(L2) enters a final state on y. $ =R1 R2B   >Images xxRy <x,y> R x/R = {y | <x,y> R} R/y = {x | <x,y> R} <  =?Images are Regular  @Rev(R)  B  Pumping Lemma It is possible to delete a part of any sufficiently long substring of a regular language and leave a string the is a member of the language  /GHIJKLMNOP Q R S T UVWXYZ[\]^_`abcdefg h!P4r` F> (    # lwawa1 ?p P   # lwawa1 ? `   P "H  0޽h ? GGGggg  ldP(    # lwawa1 ?  P   # lwawa1 ?0p P 4   `wawa1))?  "Se = S { e }T0Z  H  0޽h ? GGGggg P  5-`( pJ]`' p  HA1? cP   # lwawa1 ?`  P   # lTwawa1 ?0 P   TA ?1?p( X $  TA ?1?  X $  TA ?1? 8(X $  TA ?1? xH X $p   HA1?hp   HA1?      Zwawa1? Ea    Zwawa1?   cM(L1)0     Ztwawa1? '   cM(L2)0    ZԢwawa1?   Fe   Z4wawa1?2   Fe p  HA1?|   Zwawa1?G Fe   Zwawa1?Go Fe   ZTwawa1?w cM(L1)0    Zwawa1?a cM(L2)0    Zwawa1?j QO  Fe   Ztwawa1?Z '? Fe   Zԥwawa1?^ gC  HM(L) H  0޽h ? GGGggg V C;p(    # l4wawa1 ?  P   # lwawa1 ?z  P  h  Twawa1?   s-... can all characterize any regular language.. .H  0޽h ? GGGgggC   k(     # lTwawa1 ?  P   # lwawa1 ?. 0  P p  HA 1?6pp  HA 1?  8  Ztwawa1? d: Q S 2Qf  H  Z4wawa1?  (d: Q S Q Q =2Qh    Twawa1? g L x2A State is final if any member of the set is final33 3  Twawa1?R g  USearch time is linear for a deterministic, but exponential for a nondeterministic FSMVV VH  0޽h ? GGGggg4 j0 $\(  $ $ # lTwawa1 ?  P p $ HA 1?cV $  `wawa1))?B ?  Pruned: Smaller d is a partial function. No dead states Lookup is fasterDA8 Ij $@ BG/*1?T <  $ Ntwawa1? 7/|  ZDead (failure) state H $ 0޽h ? GGGggg  ؐ 8 0 ( (  ),  ( ( # lԫwawa1 ?  P p ( HA 1?@ ^Fp ( HA 1? @ ^  ( T4wawa1? !  y5Minimized: No two states have congruent suffix graphs66 6p (@ HZG/*1?4L| d ( <1?!F  ( ZPwawa1?' E Ca   ( ZtPwawa1? y Cb   ( ZPwawa1? 5  Ca   ( Z4Qwawa1?g G   Cb   ( ZQwawa1?5 Da   ( ZQwawa1?7 U  Da  ( ZTRwawa1? Da  (  `Rwawa1))?GO 9 EMinimal machine is unique (up to renaming of states and arc ordering)FF FH ( 0޽h ? GGGggg L` h`,(  , , # lSwawa1 ?  P  , # ltSwawa1 ?J  P  , Z4Twawa1?   j"L(M) is the language accepted by M#0Z# #& ,  fTwawa1))?   NCompute with machine  They are finite!(( (H , 0޽h ? GGGggg qT {s0 (  0 0 # lTUwawa1 ?  P d2 0 <1?td2 0 <1?T4 d2 0 <1?$| d2 0 <1?T <d2 0 <1?  \ < d2 0 <1? 4l ^B  0 61? d 0< ^B  0 61?L ^B  0 61?$,   0 ZUwawa1?A E1   0 ZVwawa1?' g E2  0 ZtVwawa1?9 Ea  0 ZVwawa1?G  Eb  0 Z4Wwawa1?2    F1a  0 ZWwawa1?b G  F2b  0 ZTXwawa1? q  }2b is final if 2 and b is final0    0 ZXwawa1?7V  NZ$  0 ZYwawa1? - 2  G=$  0 ZtYwawa1?B' Ex  0 ZYwawa1?  Ex  0 Z4Zwawa1?r W  Ex H 0 0޽h ? GGGgggZ    4(  4 4 # lZwawa1 ?  P  4 # lZwawa1 ?  P  4 TT[wawa1?E?   <Suppose x not in L x takes M(L) into rQ-F. \ x takes M into rF \ xL(M).xL& LXB 4 08c?PPXB 4 08c? 4 T[wawa1?_  / 4Suppose x in L x takes M(L) into rQ. \ x takes M into rQ-F \ xL(M).xH" Hr 4 Twawa1?A o  If M(L) were not deterministic, a string x could take M(L) into rF and sF. It would therefore take L(M) into rF and sF }A$ }H 4 0޽h ? GGGggg   0(8( ? 8 8 # lwawa1 ?  P  8 # lDwawa1 ?b-  P H 8 0޽h ? GGGggg  x 4,<(  < < # lwawa1 ?  P  < # lwawa1 ?FE P  < Zwawa1?  XY =        B <  `$wawa1))?b g?n  d One can construct families of string relations that parallel the usual classes of formal languages ed e < Twawa1?  N 2 L < Twawa1?g  The n-way concatenation of two string-tuples is the tuple of strings formed by string concatenation of the corresponding elements. H < 0޽h ? GGGggg   _W@(  @ @ # lDwawa1 ?  P  @ # lwawa1 ?.  P ' @ Zwawa1? \  Ys ( np ( det n ) vp ( v np ( det n ))) the dog chased the catZZ ZH @ 0޽h ? GGGggg   0(D(  D D # lwawa1 ?  P  D # l$wawa1 ?. P  P H D 0޽h ? GGGggg  0( H( H wa H H # lwawa1 ?  P  H # lwawa1 ?obq  P H H 0޽h ? GGGggg  0(0L(  L L # lDwawa1 ?  P  L # lwawa1 ?gpy  P H L 0޽h ? GGGggg  kc@P(  P P # lwawa1 ?  P  P # ldwawa1 ?.  P 3 P Zwawa1?' Ga ?... and the following languages Dom(R) Range(R) L/R R/L x/R R/x<@  @H P 0޽h ? GGGggg^  PT(  T T # l$wawa1 ?  P  T # lwawa1 ?  P XB T 08c?^ T 6(8c?HH T 0޽h ? GGGggg 0T P .( ?   # lwawa1 ?  P p  HA)1?p p  HA)1?8    Zwawa1?W Ex:a   ZDwawa1?^ Ex:b   Zwawa1?g/  He:e   Zwawa1?  Re:a   Zdwawa1?/  He:e H  0޽h ? GGGgggF  p( <  l  C D  P pB  HDjJ??P@P  <1?P Lif it were, you could use it, together with union to construct intersection!MM M`  c $A+ X??PX  XH  0޽h ? GGGggg kl RJ`X( 1? X X # lwawa1 ?  P  X # lDwawa1 ?.  P  X  `wawa1))?R D HFrom now on, we limit our discussion to binary relations and transducersII IH X 0޽h ? GGGgggl  3   p\ (  \p \ HA1?  \ # lwawa1 ?  P  \ Zdwawa1?w cT(R1)0   \ Zwawa1?gg Je:e  \ Z$wawa1?qV  Je:e  \ Z4wawa1?BW' Je:e  \ Zwawa1?G Fe   \ Zwawa1?r WW  Je:e   \ ZTwawa1?B G '  cT(R2)0    \ TA  ?1?@ 8X  $  \ TA  ?1?@ pX  $d  \ <1?Ld \ <1?   \ Zwawa1?q'V  HT(R) p \ HA1?(x h H \ 0޽h ? GGGggg: E  `b(  ` ` # lwawa1 ?  P p ` HA1?Hh  ` Ztwawa1?W cT(R1)0   ` ZԊwawa1?' cT(R2)0   ` TA ?1?8xX $d ` <1?  ` Z4wawa1?w  Je:e   ` Zwawa1?"   Je:e H ` 0޽h ? GGGggg @ >6d( 1 d d # lwawa1 ?  P  d # lTwawa1 ?P1-  P  H d 0޽h ? GGGggg  ( h(  h h # lwawa1 ?  P  h0 <A ?1 ?  h $ t  h TA ?1?\$X $H h 0޽h ? GGGggg|   l(  l l # ltwawa1 ?  P  l TA ?1?D<X $ l TA ?1?X X $ l TA ?1? X $H l 0޽h ? GGGggg   p( ZlU&$> p p # lԍwawa1 ?  P  p # l4wawa1 ?}( P  p TA ?1?h ZX $ p TA ?1?X rX $ p TA ?1?R X $ p TA  ?1? ( X  $ p TA "?1?  j X "$  p TA $?1?  X $$  p  `wawa1))?  i#Parallel extensions for transducers$$ $H p 0޽h ? GGGggg  *" t(  t t # lwawa1 ?  P  t TA &?1?P0X &$ t TA (?1?P H8 X ($ t Ztwawa1? $  Gwhere $ t Z4wawa1?R? &N.B. L L ID(L)F  t TA *?1?@ X *$ t Zwawa1?H ELet 2  t ZTwawa1?   0Claim: T accepts L1 L2J   H t 0޽h ? GGGggg| Ԅ  x(  x x # lwawa1 ?  P  x # lwawa1 ?.  P  x TA! ,?1?0X ,$H x 0޽h ? GGGggg` @ |(  lU&| | | # ltwawa1 ?  P  | TA" .?1?X .$ | TA /?1?@ X /$ | TA# 1?1? P X 1$ | TA$ 3?1? ph X 3$H | 0޽h ? GGGggg  <4(    # lԓwawa1 ?  P   # ld wawa1 ?@ P   TA% 5?1?* P X 5$   Z wawa1?7 C  HR/intractable = {intractable, iNtractable} iNtractable/R = {intractable} IH IH  0޽h ? GGGggg \ 0(( F4   # l$ wawa1 ?  P   TA& 7?1? 0@pX 7$  TA' 9?1?pP X 9$H  0޽h ? GGGggg   +# (    # l wawa1 ?  P p  HA(1?H p  HG/*1?    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B$2n4TaK* B$}/̪MxB$2)籅;|]sS |>B$Ks^Sp hC!_B$(8ѣ<Mn8tXԂTB$B~baHT PޔArB$ 3bS?yѐRQB$\̜>yB$ M = < S , Q, q, F , d> where F =Q-F L(M)=L?Z`$   , Properties of Regular Languages~Closure Intersection Union Complementation Concatenation Iteration Reversal Decidable Predicates Emptiness Equality FinitenessLDD- String Relations;n-way concatenation X = Y =  ( ,#. A Context-free RelationNS <s, e> <(, e > NP VP <), e > NP <np, e> <(, e > DET N <), e > VP <vp, e > <(, e > V NP <), e > DET <det, the> N <n, dog> N <n, cat> V <v, chased>      >(%#1/ Regular RelationsLThe empty set and {a} for all a in Se Se ... are regular relations If R1, R2, and R are regular n-relations, then so are R1R2 = {xy | x R1 and y R2} (concatenation) R1 R2 (union) " R* = Ri (n-way Kleene closure) i=0 There are no other regular relations| %    -   %>K B0n-way Regular Expressions`a:b:c* e:f:g denotes {<aie, bif, cig> | i 0}^1 ,1All CorrespondencesEvery n-way regular expression describes a regular n-relation. Every regular n-relation is described by an n-way regular expression. Every n-tape finite-state transducer accepts a regular n-relation. Every regular n-relation is accepted by an n-tape finite-state transducer.,02Regular HIf L1, L2, and L are regular languages and R1, R2, and R are regular relations, the following relations are regular: R1 R2 R1R2 R* R-1 R1 R2 Id(L) L1 L2 Rev(R)fu0  #E  3 Not RegularR1 R2 RH C!Intersection is not Regular E"R is not regular because ...44n-way Automata Transducers"An n-way automaton is defined by a quintuple similar to the ones that define ordinary finite-state machines (S, Q, q, F, d) Where S is a finite alphabet, Q is a finite set of states q Q is the initial state F in Q is the set of final states d maps Q Se ... 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A Context-free RelationNS <s, e> <(, e > NP VP <), e > NP <np, e> <(, e > DET N <), e > VP <vp, e > <(, e > V NP <), e > DET <det, the> N <n, dog> N <n, cat> V <v, chased>      >(%#1/ Regular RelationsLThe empty set and {a} for all a in Se Se ... are regular relations If R1, R2, and R are regular n-relations, then so are R1R2 = {xy | x R1 and y R2} (concatenation) R1 R2 (union) " R* = Ri (n-way Kleene closure) i=0 There are no other regular relations| %    -   %>K B0n-way Regular Expressions`a:b:c* e:f:g denotes {<aie, bif, cig> | i 0}^1 ,1All CorrespondencesEvery n-way regular expression describes a regular n-relation. Every regular n-relation is described by an n-way regular expression. Every n-tape finite-state transducer accepts a regular n-relation. Every regular n-relation is accepted by an n-tape finite-state transducer.,02Regular HIf L1, L2, and L are regular languages and R1, R2, and R are regular relations, the following relations are regular: R1 R2 R1R2 R* R-1 R1 R2 Id(L) L1 L2 Rev(R)fu0  #E  3 Not RegularR1 R2 RH C!Intersection is not Regular E"R is not regular because ...44n-way Automata Transducers"An n-way automaton is defined by a quintuple similar to the ones that define ordinary finite-state machines (S, Q, q, F, d) Where S is a finite alphabet, Q is a finite set of states q Q is the initial state F in Q is the set of final states d maps Q Se ... 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Kaplan and Martin Kay#Regular Languages,The empty set and {a} for all a in Se are regular languages If L1, L2, and L are regular languages, then so are L1L2 = {xy | x L1 and y L2} (concatenation) L1 L2 (union) " L* = Li (Kleene closure) i=0 There are no other regular languagespG  U0 %# +   %  ,yTBi#Correspondence Theorem (Kleene)% Types of FSMsn+ - Deterministic + - e-free + - Minimal + - Complete&8 !& Determinizing "Nondeterministic Deterministic( #'Complete vs. Pruned ( Minimization)&Proof Strategy for Language PropertiesTo prove f(L1, ... Ln) Get machines M(L1), ... M(Ln) Transform M(L1), ... M(Ln) M Show L(M) = f(L1, ... Ln)4 W         P*$L1 L2 is Regular$ +  L is RegularGet deterministic, complete M(L) = <S, Q, q, F, d > M = < S , Q, q, F , d> where F =Q-F L(M)=L?Z`$   , Properties of Regular Languages~Closure Intersection Union Complementation Concatenation Iteration Reversal Decidable Predicates Emptiness Equality FinitenessLDD- String Relations;n-way concatenation X = Y =  ( ,#. A Context-free RelationNS <s, e> <(, e > NP VP <), e > NP <np, e> <(, e > DET N <), e > VP <vp, e > <(, e > V NP <), e > DET <det, the> N <n, dog> N <n, cat> V <v, chased>      >(%#1/ Regular RelationsLThe empty set and {a} for all a in Se Se ... are regular relations If R1, R2, and R are regular n-relations, then so are R1R2 = {xy | x R1 and y R2} (concatenation) R1 R2 (union) " R* = Ri (n-way Kleene closure) i=0 There are no other regular relations| %    -   %>K B0n-way Regular Expressions`a:b:c* e:f:g denotes {<aie, bif, cig> | i 0}^1 ,1All CorrespondencesEvery n-way regular expression describes a regular n-relation. Every regular n-relation is described by an n-way regular expression. Every n-tape finite-state transducer accepts a regular n-relation. Every regular n-relation is accepted by an n-tape finite-state transducer.,02Regular HIf L1, L2, and L are regular languages and R1, R2, and R are regular relations, the following relations are regular: R1 R2 R1R2 R* R-1 R1 R2 Id(L) L1 L2 Rev(R)fu0  #E  3 Not RegularR1 R2 RH C!Intersection is not Regular E"R is not regular because ...44n-way Automata Transducers"An n-way automaton is defined by a quintuple similar to the ones that define ordinary finite-state machines (S, Q, q, F, d) Where S is a finite alphabet, Q is a finite set of states q Q is the initial state F in Q is the set of final states d maps Q Se ... Se to 2Qm   n 5;  5Union and Iteration 6 Concatenation 7Range and Domain.Dom(R) = R/S* Range(R) = S*/R Accepting FSMs derived from T(R) by replacing all transition labels a:b by a (domain) or b (range) Regular languages.N z %  'l8R-1 9Id(L) :Extending d vTo state sets To strings The machine accepts a string x just in case is not empty.&w <:;Cartesian Product <"T accepts L1 L2@ Proof: by induction. Thus, T enters a final state on iff M(L1) enters a final state on x and M(L2) enters a final state on y. $CF=R1 R2B  >ImagesxxRy <x,y> R x/R = {y | <x,y> R} R/y = {x | <x,y> R} < :?Images are Regular @Rev(R) B  Pumping LemmaIt is possible to delete a part of any sufficiently long substring of a regular language and leave a string the is a member of the language /GHJKLMNOP Q R S T UVWXYZ[\]^_`abcdefg h!P@sx,, ` e(HH(d"h    c [  (     # l lwawa1 ?     # llwawa1 ?. 0   &  Zlwawa1? d: Q S 2Qf 6  Zlwawa1?  (d: Q S Q Q =2Qh   T`lwawa1? g L,$D, f2A State is final if any member of the set is final33A  T@lwawa1?R g ,$D, USearch time is linear for a deterministic, but exponential for a nondeterministic FSMVV$  C Adeterminize1.tiff 002C4ABB Macintosh HD ABA78158:P 8p$  C Adeterminize2.tiff 002C4ABB Macintosh HD ABA78158: X H  0޽h ? GGGggg  rj$(  $$ $ C A Roman.tiff 002C4ABB Macintosh HD ABA78158:L  $ # l@wawa1 ?    $  `bwawa1))?B ? ,$D, Pruned: Smaller d is a partial function. No dead states Lookup is fasterJA8l  P   $P  ,$D, $B NҺGB:H1? p  <  $ Ncwawa1?P   HDead (failure) stateH $ 0޽h ? GGGggg  (&(  ),  ( ( # l0)mwawa1 ?    ( T)mwawa1? !  g5Minimized: No two states have congruent suffix graphs66 (  `0/mwawa1))?GO 9 wEMinimal machine is unique (up to renaming of states and arc ordering)FF$ ( C A Minimization.tiff 002C4ABB Macintosh HD ABA78158:P @8 H ( 0޽h ? GGGggg   $( ?   # lkwawa1 ?   $  C A Transducer.tiff 002C4ABB Macintosh HD ABA78158:  H  0޽h ? GGGggg  & \(  \ \ # liwawa1 ?    \ Z@iwawa1?G 4e  \ TA  ?1?@ 8X  $X  \ TA  ?1?@ pX  $Xd  \ <1?Ld \ <1?  $ \ C A0Titeration.tiff 002C4ABB Macintosh HD ABA78158: $ \ C A  Tunion.tiff 002C4ABB Macintosh HD ABA78158: 0 H \ 0޽h ? GGGgggL   `t(  ` ` # l0piwawa1 ?    ` TA ?1?8xX $Xd ` <1? $  ` C A  Tconcat.tiff 002C4ABB Macintosh HD ABA78158:( H ` 0޽h ? GGGggg9 xp^RЀ3ÿ lHbP  LPÄ!?$< F`B&Vx `D Ymɯ?GlFXϐÐJ\@H ߙ!!$,b`by'b] JJ2cP`ʈ 2r/o J!8ـb$,/JgEi% %y )%E Fz T 3"~ x  (a 4D0JH`+K" xp^RЀ3ÿ lHbP  LPÄ!?$< F`B&Vx `D Ymɯ?Ga gaH%~. PC$a̐ϐPtGC 101GBٮ % @w1(0AyeDU?2#'0S倄ofrQ~q~ZkaibIf~kJfI~8չrUH_! ZHr@w11r125U 5'25R |. xp^RЀ3ÿ lHbP  LPÄ!?$< F`B&Vx `D EI}022ş!!l.@@f3C>C.CI Y$/ OR e222$20dݑǠ* dV(4@4 #0Bq#00IYy_Hf&秕(&d)d)S_+7P%L^ψe Ѐi$9}: F@m)r<&0V`g5 o(wClj0|{|}l U(   |   | Equation Equation.20Equation|  Equation Equation.20Equation| Equation Equation.20Equation| Equation Equation.20Equation| Equation Equation.20Equation| Equation Equation.20Equation|Equation Equation.20Equation|EquationT     P !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNO QRSTUVWXYZ[\]^_`abcdefghiklmnopqrstuvwxyz{|}~j Equation.20Equation|Equation Equation.20Equation|Equation Equation.20Equation|Equation Equation.20Equation| Equation Equation.20Equation|"Equation Equation.20Equation|$Equation Equation.20Equation|&Equation Equation.20Equation|(Equation Equation.20Equation|*Equation Equation.20Equation|,Equation Equation.20Equation|.Equation Equation.20Equation|oma Ps  l&DTimes New Roman Ps  l& DSymbolew Roman Ps  l&0DCourierw Roman Ps  l&@DGreekrw Roman Ps  l&PDMT Extra Roman Ps  l& a .@  @@``  @n?" dd@  @@`` h ` `,KDF(      !"#$%   &'()*+,-./0123456789:;<=>?@ABCDEF@$0B$&ܦ¦H(:)WH]dC7 0B$PF ËFQYA0B$G[z0b$obR`Se5U1'N ,0b$TRop#/ȸsKX0b$`&Xu~.bto0b$AU`TBck0b$IU xwN<s'0b$/x3GUQq&`/0b$%+Jh<y6nN0b${]<(,sv%0$0$0B$E -}7 6vN90B$݌8+E<0$0B$q6G"NS4_0B$APq0(}aca0B$ u. 0'=d0B$x8Sg%x'X+h0B$n =]Ue5*k0B$7fcձ.h9_n0B$9Hh[Αq0B$Sa$|FDXt0B$ ޞǙˍ]kw0B$tSsBӧS{0B$o%stT;Ů<P~0B$X hjD79UL@<0B$57"q΀n.}:qO0B$5nz6 UG!0B$p|M-i0B$  `"$0B$fF ,=R0B$ 8yq #0B$¾֋Ou rN00B$g^ Ҝll@l;p0B$dNzݺ9]p0B$Ɉ`K09Ar0B$EP<˟^H.70$0b$EKj^OmbmKBP#?0B$Oy KH"tq.x0$0b$ U2'ds1e.0$0b$FoNf=׼nE0b$VK6n΅ zA!$]0C x1?@(v|@|! g4dddds{@Pm hppp@ <4BdBd!`gʚ;huʚ;2Nʚ;<4!d!d!`uʚ;<4dddd!`uʚ;ph___PPT2001D<4X? %0"FFinite-State Methods in Natural-Language Processing: Basic MathematicsRonald M. Kaplan and Martin Kay#Regular Languages,The empty set and {a} for all a in Se are regular languages If L1, L2, and L are regular languages, then so are L1L2 = {xy | x L1 and y L2} (concatenation) L1 L2 (union) " L* = Li (Kleene closure) i=0 There are no other regular languagespG  U0 %# +   %  ,yTBi#Correspondence Theorem (Kleene)% Types of FSMsn+ - Deterministic + - e-free + - Minimal + - Complete&8 !& Determinizing "Nondeterministic Deterministic( #'Complete vs. Pruned ( Minimization)&Proof Strategy for Language PropertiesTo prove f(L1, ... Ln) Get machines M(L1), ... M(Ln) Transform M(L1), ... M(Ln) M Show L(M) = f(L1, ... Ln)4 W         P*$L1 L2 is Regular$ +  L is RegularGet deterministic, complete M(L) = <S, Q, q, F, d > M = < S , Q, q, F , d> where F =Q-F L(M)=L?Z`$   , Properties of Regular Languages~Closure Intersection Union Complementation Concatenation Iteration Reversal Decidable Predicates Emptiness Equality FinitenessLDD- String Relations;n-way concatenation X = Y =  ( ,#. A Context-free RelationNS <s, e> <(, e > NP VP <), e > NP <np, e> <(, e > DET N <), e > VP <vp, e > <(, e > V NP <), e > DET <det, the> N <n, dog> N <n, cat> V <v, chased>      >(%#1/ Regular RelationsLThe empty set and {a} for all a in Se Se ... are regular relations If R1, R2, and R are regular n-relations, then so are R1R2 = {xy | x R1 and y R2} (concatenation) R1 R2 (union) " R* = Ri (n-way Kleene closure) i=0 There are no other regular relations| %    -   %>K B0n-way Regular Expressions`a:b:c* e:f:g denotes {<aie, bif, cig> | i 0}^1 ,1All CorrespondencesEvery n-way regular expression describes a regular n-relation. Every regular n-relation is described by an n-way regular expression. Every n-tape finite-state transducer accepts a regular n-relation. Every regular n-relation is accepted by an n-tape finite-state transducer.,02Regular HIf L1, L2, and L are regular languages and R1, R2, and R are regular relations, the following relations are regular: R1 R2 R1R2 R* R-1 R1 R2 Id(L) L1 L2 Rev(R)fu0  #E  3 Not RegularR1 R2 RH C!Intersection is not Regular E"R is not regular because ...44n-way Automata Transducers"An n-way automaton is defined by a quintuple similar to the ones that define ordinary finite-state machines (S, Q, q, F, d) Where S is a finite alphabet, Q is a finite set of states q Q is the initial state F in Q is the set of final states d maps Q Se ... Se to 2Qm   n 5;  5Union and Iteration 6 Concatenation 7Range and Domain.Dom(R) = R/S* Range(R) = S*/R Accepting FSMs derived from T(R) by replacing all transition labels a:b by a (domain) or b (range) Regular languages.N z %  'l8R-1 9Id(L) :Extending d vTo state sets To strings The machine accepts a string x just in case is not empty.&w <:;Cartesian Product <"T accepts L1 L2@ Proof: by induction. Thus, T enters a final state on iff M(L1) enters a final state on x and M(L2) enters a final state on y. $CF=R1 R2B  >ImagesxxRy <x,y> R x/R = {y | <x,y> R} R/y = {x | <x,y> R} < :?Images are Regular @Rev(R) B  Pumping LemmaIt is possible to delete a part of any sufficiently long substring of a regular language and leave a string the is a member of the language /GHJKLMNOP Q R S T UVWXYZ[\]^_`abcdefg h!PGPsx,, ` e(HH(d"h   P((( : (l ( C  z`   $ ( C A* char.tiff 002C4ABB Macintosh HD ABA78158:0 $ ( C A- concat.tiff 002C4ABB Macintosh HD ABA78158:   $ ( C A/iteration.tiff 002C4ABB Macintosh HD ABA78158:`  )8    ( ` [  ( 6z $   R "   ( 6`z ~ d  ?L"  ( 6zw F  ?*"  ( 6z g l  A=$  ( 6`z v < o  ?U"  ( 6 z ;  Ai$ ( 6z C   C=& ( 6z ; d  ?0" ( 6`z 4  D& ( 6 ~ d  ?L" ( 6` F   Ai$$ ( C A union.tiff 002C4ABB Macintosh HD ABA78158:@P  ( TA w?1? 8X w$X ( TA x?1?hPX x$X ( TA y?1? @X X y$X ( Z wawa1 ?` e5Every regular language L is accepted by some FSM M(L)6 6H ( 0޽h ? 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%6"FFinite-State Methods in Natural-Language Processing: Basic MathematicsGGRonald M. Kaplan and Martin Kay #Regular Languages,The empty set and {a} for all a in Se are regular languages If L1, L2, and L are regular languages, then so are L1L2 = {xy | x L1 and y L2} (concatenation) L1 L2 (union) " L* = Li (Kleene closure) i=0 There are no other regular languagespG  U0 %#   +            %  ,ySAi#Correspondence Theorem (Kleene) % Types of FSMsn+ - Deterministic + - e-free + - Minimal + - Complete.8 P!& Determinizing "Nondeterministic Deterministic, #'Complete vs. Pruned ( Minimization )&Proof Strategy for Language Properties''To prove f(L1, ... Ln) Get machines M(L1), ... M(Ln) Transform M(L1), ... M(Ln) M Show L(M) = f(L1, ... Ln)P W           P*$L1 L2 is Regular* +  L is Regular Get deterministic, complete M(L) = <S, Q, q, F, d > M = < S , Q, q, F , d> where F =Q-F L(M)=L?~`$   , Properties of Regular Languages ~Closure Intersection Union Complementation Concatenation Iteration Reversal Decidable Predicates Emptiness Equality Finiteness`PDPPPD- String Relations;n-way concatenation X = Y =  P( P      ,#. A Context-free RelationNS <s, e> <(, e > NP VP <), e > NP <np, e> <(, e > DET N <), e > VP <vp, e > <(, e > V NP <), e > DET <det, the> N <n, dog> N <n, cat> V <v, chased>      >(%#1/ Regular RelationsLThe empty set and {a} for all a in Se Se ... are regular relations If R1, R2, and R are regular n-relations, then so are R1R2 = {xy | x R1 and y R2} (concatenation) R1 R2 (union) " R* = Ri (n-way Kleene closure) i=0 There are no other regular relationsr| %      -            %>JA0n-way Regular Expressions`a:b:c* e:f:g denotes {<aie, bif, cig> | i 0}1    ,1All CorrespondencesEvery n-way regular expression describes a regular n-relation. Every regular n-relation is described by an n-way regular expression. Every n-tape finite-state transducer accepts a regular n-relation. Every regular n-relation is accepted by an n-tape finite-state transducer.P,02Regular  HIf L1, L2, and L are regular languages and R1, R2, and R are regular relations, the following relations are regular: R1 R2 R1R2 R* R-1 R1 R2 Id(L) L1 L2 Rev(R)u0   #  E         3 Not Regular R1 R2 RV C!Intersection is not Regular E"R is not regular because ...44n-way Automata Transducers"An n-way automaton is defined by a quintuple similar to the ones that define ordinary finite-state machines (S, Q, q, F, d) Where S is a finite alphabet, Q is a finite set of states q Q is the initial state F in Q is the set of final states d maps Q Se ... Se to 2Qm   n 5;  5Union and Iteration 6 Concatenation 7Range and Domain.Dom(R) = R/S* Range(R) = S*/R Accepting FSMs derived from T(R) by replacing all transition labels a:b by a (domain) or b (range) Regular languages.z z %  'k8R-1  9Id(L) :Extending d lTo state sets To strings The machine accepts a string x just in case is not empty..m P<0;Cartesian Product <"T accepts L1 L2L Proof: by induction. Thus, T enters a final state on iff M(L1) enters a final state on x and M(L2) enters a final state on y. $CE=R1 R2N  >ImagesxxRy <x,y> R x/R = {y | <x,y> R} R/y = {x | <x,y> R} < PP9?Images are Regular @Rev(R) B  Pumping LemmaIt is possible to delete a part of any sufficiently long substring of a regular language and leave a string the is a member of the language /GHJKLMNOP Q R S T UVWXYZ[\]^_`abcdefg h!PGPsx,, ` e(HH(d"h 0  ` ̙33` ` ff3333f` 333MMM` f` f` 3>?" dZ@$|? " KZ@ d " @@K`@d n?" dZ(: ::XXuuPR    @ ` ` p>  >  D% p,(  ,N , Z wawa1 ?".   RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  ,  `@wawa1 ?"  T Click to edit Master title style! !  , <`1?"0 <Martin Kay Chart Translation * 2  B , s *޽h ? GGGggg0(___PPT2001seanwhatdslvsrcsHpcnttwntdatazvarzdstxcntycnt  ChartTranslation F>0(    # lGwawa1 ?,p    # l@Gwawa1 ?, `    "H  0޽h ? GGGggg  ZR@(    # l`zwawa1 ?,     # lzwawa1 ?,0p  "   `zwawa1))?  "Se = S { e }T0Z H  0޽h ? GGGggg  P(,( : (l ( C  z,`   $ ( C A* char.tiff 002C4ABB Macintosh HD ABA78158:0 $ ( C A- concat.tiff 002C4ABB Macintosh HD ABA78158:   $ ( C A/iteration.tiff 002C4ABB Macintosh HD ABA78158:`  )8    ( ` [  ( 6z $   R "   ( 6`z ~ d  ?L"  ( 6zw F  ?*"  ( 6z g l  A=$  ( 6`z v < o  ?U"  ( 6 z ;  Ai$ ( 6z C   C=& ( 6z ; d  ?0" ( 6`z 4  D& ( 6 ~ d  ?L" ( 6` F   Ai$$ ( C A union.tiff 002C4ABB Macintosh HD ABA78158:@P  ( TA w?1? 8X w$X ( TA x?1?hPX x$X ( TA y?1? @X X y$X ( Z wawa1 ?` i5Every regular language L is accepted by some FSM M(L)6 6H ( 0޽h ? GGGggg 7Equation Equation.20Equation|9 Equation Equation.20EquationXFEquation Equation.30,Microsoft Equation 3.0|wjEquation Equation 1)`(    # l wawa1 ?,     # lwawa1 ?,z    h  Twawa1?   a-... can all characterize any regular language..H  0޽h ? GGGggg   c [  (     # lwawa1 ?,     # lwawa1 ?,. 0   &  Zwawa1? d: Q S 2Qf 6  Zwawa1?  (d: Q S Q Q =2Qh   T wawa1? g ,$D, f2A State is final if any member of the set is final33A  Twawa1?` @ h,$D, USearch time is linear for a deterministic, but exponential for a nondeterministic FSMVV$  C Adeterminize1.tiff 002C4ABB Macintosh HD ABA78158:P 8p$  C Adeterminize2.tiff 002C4ABB Macintosh HD ABA78158: X H  0޽h ? GGGggg  | $(  $$  $ C A Roman.tiff 002C4ABB Macintosh HD ABA78158:L  $ # lzwawa1 ?,    $  `zwawa1))?B o,$D, Pruned: Smaller d is a partial function. No dead states Lookup is fasterLA8l  P   $P  ,$D, $B NPGB:H1? p  <  $ Nwawa1?P   HDead (failure) state  $ <91? W HFollow a dashed line iff there is no solid one for the current characterII0H $ 0޽h ? GGGggg  (&(  ),  ( ( # lнwawa1 ?,    ( T0wawa1? ! * g5Minimized: No two states have congruent suffix graphs66 (  `wawa1))?GO  wEMinimal machine is unique (up to renaming of states and arc ordering)FF$ ( C A Minimization.tiff 002C4ABB Macintosh HD ABA78158:P @8 H ( 0޽h ? GGGggg  D<,(  , , # lwawa1 ?,    , # lwawa1 ?,J    , Zpwawa1?   X"L(M) is the language accepted by M#0Z# ,  f0wawa1))? q ~NCompute with machine  They are finite!((H , 0޽h ? GGGggg   04(  0 0 # lP…wawa1 ?,   d2 0 <1?td2 0 <1?T4 d2 0 <1?$| d2 0 <1?T <d2 0 <1?  \ < d2 0 <1? 4l ^B  0 61? d 0< ^B  0 61?L ^B  0 61?$,   0 ZÅwawa1?A 31  0 ZpÅwawa1?' g 32 0 Z0ąwawa1?9 3a 0 Ząwawa1?G  3b 0 ZŅwawa1?2    41a 0 Zpƅwawa1?b G  42b 0 Zwawa1?   l 2b is final if 2 and b are final0!  0 Z0wawa1?7  <Z$ 0 Zwawa1? - 2  5=$ 0 ZPwawa1?B' 3x 0 Zwawa1?  3x 0 Zwawa1?r W  3xH 0 0޽h ? GGGggg    z r 4 (  4 4 # lwawa1 ?,    4 # lPwawa1 ?,    4 Twawa1?E? ,$D, XSuppose x not in L x takes M(L) into rQ-F. \ x takes M into r F \ x L(M).Q&XB 4 08c?P`PXB 4 08c?0  4 Twawa1?P  ,$D, TSuppose x in L x takes M(L) into r Q. \ x takes M into r Q-F \ x L(M).O#V 4 Twawa1? < ,$D, If M(L) were not deterministic, a string x could take M(L) into r F and sF. It would therefore take L(M) into r F and s F ~B' H 4 0޽h ? GGGggg   0(8( ? 8 8 # lPwawa1 ?,    8 # lwawa1 ?,b-   H 8 0޽h ? GGGggg   <4<(  < < # lp wawa1 ?,    < # l wawa1 ?,FE   < Z0!wawa1?  XY =       ( <  `!wawa1))?b g?4  d One can construct families of string relations that parallel the usual classes of formal languagesee < T"wawa1?  < 2p < Twawa1?gu  The n-way concatenation of two string-tuples is the tuple of strings formed by string concatenation of the corresponding elements.,&JH < 0޽h ? GGGggg   @G(  @ @ # l`wawa1 ?,    @ # lwawa1 ?,.    @ Z wawa1? \  Ys ( np ( det n ) vp ( v np ( det n ))) the dog chased the catZZb4H @ 0޽h ? GGGggg   0( D(  D D # lwawa1 ?,    D # l`wawa1 ?,. P   H D 0޽h ? GGGggg  0(0H( H wa H H # lwawa1 ?,    H # lwawa1 ?,obq   H H 0޽h ? GGGggg  0(@L(  L L # l@wawa1 ?,    L # lwawa1 ?,gpy   H L 0޽h ? GGGggg  c[PP(  P P # lwawa1 ?,    P # l`wawa1 ?,.   + P Zwawa1?' k ?... and the following languages Dom(R) Range(R) L/R R/L x/R R/xF@ H P 0޽h ? GGGggg^  `T(  T T # l wawa1 ?,    T # lwawa1 ?,   XB T 08c?^ T 6(8c?HH T 0޽h ? GGGggg  p $( ?   # l@wawa1 ?,   $  C A Transducer.tiff 002C4ABB Macintosh HD ABA78158:  H  0޽h ? GGGggg4  t( <  l  C ,   pB  HDjJ??PPP  <1?P |Lif it were, you could use it, together with union to construct intersection!MM`  c $A+ X??PX  XH  0޽h ? GGGggg  tlX( 1? X X # l`wawa1 ?,    X # l wawa1 ?,.   < X  `wawa1))?p @ b,$D, xHFrom now on, we limit our discussion to binary relations and transducersIIH X 0޽h ? GGGggg  & \(  \ \ # l0Džwawa1 ?,    \ Zwawa1?G 4e  \ TA  ?1?@ 8X  $X  \ TA  ?1?@ pX  $Xd  \ <1?Ld \ <1?  $ \ C A0Titeration.tiff 002C4ABB Macintosh HD ABA78158: $ \ C A  Tunion.tiff 002C4ABB Macintosh HD ABA78158: 0 H \ 0޽h ? GGGgggL   `t(  ` ` # lwawa1 ?,    ` TA ?1?8xX $Xd ` <1? $  ` C A  Tconcat.tiff 002C4ABB Macintosh HD ABA78158:( H ` 0޽h ? GGGggg  >6d( 1 d d # lwawa1 ?,    d # l0wawa1 ?,P1-    H d 0޽h ? GGGggg  ( h(  h h # lPwawa1 ?,    h0 <A ?1 ?,  h $h p h TA ?1?\$X $XH h 0޽h ? GGGggg|   l(  l l # lwawa1 ?,    l TA ?1?D<X $X l TA ?1?X X $X l TA ?1? X $XH l 0޽h ? GGGggg  wo p( ZlU&$> p p # lwawa1 ?,    p # lpwawa1 ?,}(   p TA ?1?h ZX $X p TA ?1?X rX $X p TA ?1?: X $X p TA  ?1?F ( X  $X p TA "?1?`  X "$X  p TA $?1?8 P X $$X  p  `wawa1))?  W#Parallel extensions for transducers$$H p 0޽h ? GGGgggJ   tr(  t t # lwawa1 ?,    t TA &?1?P0X &$X t TA (?1?P H8 X ($X t ZPwawa1? $  5where t Zwawa1?R? &N.B. L L ID(L)F t TA *?1?@ X *$X t Zwawa1?H 3Let   t Zwawa1?   0Claim: T accepts L1 L2J  H t 0޽h ? GGGggg|   x(  x x # l wawa1 ?,    x # lp!wawa1 ?,.    x TA! ,?1?0X ,$XH x 0޽h ? GGGggg`   |(  lU&| | | # l!wawa1 ?,    | TA" .?1?X .$X | TA /?1?@ X /$X | TA# 1?1? P X 1$X | TA$ 3?1? ph X 3$XH | 0޽h ? GGGggg  ^V0(    # l0"wawa1 ?,     # l"wawa1 ?,@    TA% 5?1?* P X 5$XB  Z"wawa1?7 C  HR/intractable = {intractable, iNtractable} iNtractable/R = {intractable} IH,  H  0޽h ? GGGggg  0(@( F4   # lP#wawa1 ?,     TA& 7?1? 0@pX 7$X  TA' 9?1?pP X 9$XH  0޽h ? GGGggg    (    # lGwawa1 ?,   p  HA(1?H p  HG/*1?    T`Gwawa1?g gK  5StartH  0޽h ? GGGgggx ! P ( WlU&l   # lGwawa1 ?,     # l Gwawa1 ?,. 0  d  <1?t lp  HG/*1?L|   T0:wawa1?G m  VStatei"  p  HG/*1? |   T:wawa1?G I f m  VStatei"  XB  0p?XB  0p?` ` H  0޽h ? GGGgggr" ]MP%eriEs9{?=dD79U<zHB q}EiSm%D wm 6X( RdO)& O@Pictures 9PowerPoint Document(;bSummaryInformation(      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHLMQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuP     !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKMQRSUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~DocumentSummaryInformation8Current User*"_b Martin KayData FoxPro Help66  RegCountKey1.Help>>UserDataMicrosoft HelpV6  RegCountKey4xPro Driver HelpHelpAlias mkpowerbooke;BD Q+E ODBC FoxPro Driver Help Y.20Equation|xkEquation Equation.20Equation|ylEquation Equation.20Equation0/0(  0;[0 0 000$([\{b00 000000000  0=] 0 0 0000 2 3 !A0C0E0G0I0c00000000000000000!%),.:;?]}acdeghijklmnop`DTimesica Neue raPsgh &DTimes New Roman raPsgh & DSymbolew Roman raPsgh &0DCourierw Roman raPsgh &@DGreekrw Roman raPsgh &PDMT Extra Roman raPsgh &`DComic Sans MSn raPsgh &pDHelvetica Neue UltraLight& a .@  @@``  @n?" dd@  @@`` p h h0LC%$%(         !"#   &'()*+,-./0123456789:;<=>?@ABCDEF@b$N93wglPmiH0B$&ܦ¦H(:)WH]dC7 0B$PF ËFQYA0B$G[z0b$obR`Se5U1'N ,0b$TRop#/ȸsKX0b$`&Xu~.bto0$0b$IU xwN<s'0b$/x3GUQq&`/0b$%+Jh<y6nN0b${]<(,sv%0$0$0B$E -}7 6vN90B$݌8+E<0$0B$q6G"NS4_0B$APq0(}aca0B$ u. 0'=d0B$x8Sg%x'X+h0B$n =]Ue5*k0B$7fcձ.h9_n0B$9Hh[Αq0B$Sa$|FDXt0B$ ޞǙˍ]kw0B$tSsBӧS{0B$o%stT;Ů<P~0B$X hjD79UL@<0B$57"q΀n.}:qO0B$5nz6 UG!0B$p|M-i0B$  `"$0B$fF ,=R0B$ 8yq #0B$¾֋Ou rN00B$g^ Ҝll@l;p0B$dNzݺ9]p0B$Ɉ`K09Ar0B$EP<˟^H.70$0b$EKj^OmbmKBP#?0B$Oy KH"tq.x0$0b$ U2'ds1e.0$0b$FoNf=׼nE0b$VK6n΅ zA!$]0C x1?@(v|@|m g4dddds{@Pm ppp@ <4BdBdm`gʚ;huʚ;2Nʚ;<4!d!dm`uʚ;<4ddddm`uʚ;ph___PPT2001D<4X? %s2"FFinite-State Methods in Natural-Language Processing: Basic MathematicsRonald M. Kaplan and Martin Kay#Regular Languages,The empty set and {a} for all a in Se are regular languages If L1, L2, and L are regular languages, then so are L1L2 = {xy | x L1 and y L2} (concatenation) L1 L2 (union) " L* = Li (Kleene closure) i=0 There are no other regular languagespG  U0 %#   +            %  ,ySAi#Correspondence Theorem (Kleene)% Types of FSMsn+ - Deterministic + - e-free + - Minimal + - Complete(8 P!& Determinizing "Nondeterministic Deterministic* #'Complete vs. Pruned ( Minimization)&Proof Strategy for Language PropertiesTo prove f(L1, ... Ln) Get machines M(L1), ... M(Ln) Transform M(L1), ... M(Ln) M Show L(M) = f(L1, ... Ln)B W           P*$L1 L2 is Regular$ +  L is RegularGet deterministic, complete M(L) = <S, Q, q, F, d > M = < S , Q, q, F , d> where F =Q-F L(M)=L?l`$   , Properties of Regular Languages~Closure Intersection Union Complementation Concatenation Iteration Reversal Decidable Predicates Emptiness Equality Finiteness\PDPPPD- String Relations;n-way concatenation X = Y =  P( P      ,#. A Context-free RelationNS <s, e> <(, e > NP VP <), e > NP <np, e> <(, e > DET N <), e > VP <vp, e > <(, e > V NP <), e > DET <det, the> N <n, dog> N <n, cat> V <v, chased>      >(%#1/ Regular RelationsLThe empty set and {a} for all a in Se Se ... are regular relations If R1, R2, and R are regular n-relations, then so are R1R2 = {xy | x R1 and y R2} (concatenation) R1 R2 (union) " R* = Ri (n-way Kleene closure) i=0 There are no other regular relations| %      -            %>JA0n-way Regular Expressions`a:b:c* e:f:g denotes {<aie, bif, cig> | i 0}p1    ,1All CorrespondencesEvery n-way regular expression describes a regular n-relation. Every regular n-relation is described by an n-way regular expression. Every n-tape finite-state transducer accepts a regular n-relation. Every regular n-relation is accepted by an n-tape finite-state transducer.P,02Regular HIf L1, L2, and L are regular languages and R1, R2, and R are regular relations, the following relations are regular: R1 R2 R1R2 R* R-1 R1 R2 Id(L) L1 L2 Rev(R)u0   #  E         3 Not RegularR1 R2 RH C!Intersection is not Regular E"R is not regular because ...44n-way Automata Transducers"An n-way automaton is defined by a quintuple similar to the ones that define ordinary finite-state machines (S, Q, q, F, d) Where S is a finite alphabet, Q is a finite set of states q Q is the initial state F in Q is the set of final states d maps Q Se ... Se to 2Qm   n 5;  5Union and Iteration 6 Concatenation 7Range and Domain.Dom(R) = R/S* Range(R) = S*/R Accepting FSMs derived from T(R) by replacing all transition labels a:b by a (domain) or b (range) Regular languages.j z %  'k8R-1 9Id(L) :Extending d lTo state sets To strings The machine accepts a string x just in case is not empty.*m P<0;Cartesian Product <"T accepts L1 L2@ Proof: by induction. Thus, T enters a final state on iff M(L1) enters a final state on x and M(L2) enters a final state on y. $CE=R1 R2B  >ImagesxxRy <x,y> R x/R = {y | <x,y> R} R/y = {x | <x,y> R} < PP9?Images are Regular @Rev(R) B  Pumping LemmaIt is possible to delete a part of any sufficiently long substring of a regular language and leave a string the is a member of the language /GHJKLMNOP Q R S T UVWXYZ[\]^_`abcdefg h!PGPsx,, ` e(HH(d"h r  bm RdO)4>Ez@Pictures 9PowerPoint Document(;bSummaryInformation(      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHLQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuxDocumentSummaryInformation8Current User*"_b Martin KayData FoxPro Help66  RegCountKey1.Help>>UserDataMicrosoft HelpV6  RegCountKey4xPro Driver HelpHelpAlias mkpowerbooke;BD Q+E ODBC FoxPro Driver Help Y