The rst index has to b e 1 b ecause only the pair 10 - PDF document

P ost's Corresp ondence Problem � An undecidable, but RE, problem that app ears not to ha v e an ything to do with TM's. � Giv en t w o lists of \corresp onding" strings ( w ; w ; : : : ; w ) and x ; x ; : : : ; x ), do es 1 2 n 1 2 n there exist a nonempt y sequence of in tegers i ; i ; : : : ; i suc h that w w � � � w = 1 2 k i i i + k 1 2 x x � � � x ? i i i 1 2 k � In tuition: w e can try all lists i ; i ; : : : ; i in 1 2 k order of k . If w e �nd a solution, the answ er is \y es." But if w e nev er �nd a solution, ho w can w e b e sure no longer solution exists, so w e can nev er sa y \no." Example (1 ; 0 ; 010 ; 11) and (10 ; 10 ; 01 ; 1). � A solution is 1 ; 2 ; 1 ; 3 ; 3 ; 4. � The constructed string from b oth lists is 10101001011. Another Example F rom the course reader: (10 ; 011 ; 101) and (101 ; 11 ; 011). � Another argumen t wh y this instance of PCP has no solution: ✦ The �rst index has to b e 1 b ecause only the pair 10 and 101 b egin with the same sym b ol. ✦ Then, whatev er indexes w e c ho ose to con tin ue, there will b e more 1's in the string constructed from the �rst list than the second (b ecause in eac h corresp onding pair there are at least as man y 1's in the second list). Th us, the t w o strings cannot b e equal. Plan to Sho w PCP is Undecidable 1. In tro duce MPCP , where the �rst pair m ust b e tak en �rst in a solution. 2. Sho w ho w to reduce MPCP to PCP . 3. Sho w ho w to reduce to MPCP . L u ✦ This is the only reason for MPCP: it mak es the reduction from L easier. u 4. Conclude that if PCP is decidable, so is MPCP , and so is L (whic h w e kno w is false). u 1

Reduction of MPCP to PCP T ric k: giv en a MPCP instance, in tro duce a new sym b ol *. � In the �rst list, * app ears after ev ery sym b ol, but in the second list, the * app ears b efor e ev ery sym b ol. ✦ Example: the pair 10 and 011 b ecomes 1*0* and *0*1*1. ✦ Notice that no suc h pair can ev er b e the �rst in a solution. � T ak e the �rst pair and from the MPCP w x 1 1 instance (whic h m ust b e c hosen �rst in a MPCP solution) and add to the PCP instance another pair in whic h the *'s are as alw a ys, but also gets an extra * at the b eginning. w 1 ✦ Referred to as \pair 0." ✦ Example: if 10 and 011 is the �rst pair, also add to the PCP instance the pair *1*0* and *0*1*1. � Finally , since the strings from the �rst list will ha v e an extra * at the end, add to the PCP instance the pair $ and *$. ✦ $ is a new sym b ol, so this pair can b e used only to complete a matc h. ✦ Referred to as the \�nal pair." Pro of the Reduction is Correct � If the MPCP instance has a solution 1 follo w ed b y , Then the PCP i ; i ; : : : ; i k 1 2 instance has a solution, whic h is the same, using the �rst pair in place of pair 1, and terminating the list with the �nal pair. � If the PCP instance has a solution, then it m ust b egin with the \�rst pair," b ecause no other pair b egins with the same sym b ol. Th us, remo ving the *'s and deleting the last pair giv es a solution to the MPCP instance. Reduction of L to MPCP u � In tuition: The equal strings represen t a computation of a TM M on input w . ✦ Sequence of ID's separated b y a sp ecial mark er #. � First pair is # and # q w #. 0 2

� String from �rst list is alw a ys one ID b ehind, unless an accepting state is reac hed, in whic h case the �rst string can \catc h up." � Some example pairs: 1. and for ev ery tap e sym b ol . Allo ws X X X cop ying of sym b ols that don't c hange from one ID to the next. 2. If ( q ) = ( P R ), then and is a � ; X ; Y ; q X Y p pair. Sim ulates a mo v e for the next ID. 3. If q is an accepting state, then X q Y and q is a pair for all X and Y . Allo ws ID to \shrink to nothing" when an accepting state is reac hed. Undecidable Problems Ab out CFL's W e are applying the theory of undecidabilit y in a useful w a y when w e sho w a real problem not to b e solv able b y computer. � Example: Y ou ma y think the CS154 pro ject w as hard, but at least there is an algorithm to con v ert RE's to DF A's, so it is at least p ossible for y ou to succeed. � Supp ose next spring's CS154 pro ject is to tak e a CF G and tell whether it is am biguous. Y ou can't do it b ecause the problem is undecidable! Con v erting PCP to CF G's F or eac h list A = ( w ; w ; : : : ; w ) w e can construct 1 2 n a gramma r and a language that represen ts all sequences of in tegers i ; i ; : : : ; i and the strings 1 2 k w w � � � w that are constructed from those lists i i i 1 2 k of in tegers. � Use a ; a ; : : : ; a as new sym b ols (not in the 1 2 n alphab et of the PCP instance) represen ting the in tegers. � The \grammar for list A ": ! S j j � � � j j � . w S a w S a w S a 1 1 2 2 n n ✦ Yields all concatenations of w 's follo w ed b y the rev erse of their index sequence. Reduction of PCP to CF G Am biguit y Problem Giv en lists and , construct grammar as follo ws: A B � S ! A j B . � A is the start sym b ol for a grammar from list A ; B is the same for list B . 3

� If there is a solution to the PCP instance, then the same string can b e deriv ed starting S ) A and S ) B . lm lm ✦ Con v ersely , the only w a y a string can ha v e t w o leftmost deriv ations is if they b egin in these t w o w a ys, b ecause the grammar of one list is unam biguous. Example Use the lists of our �rst example: (1 ; 0 ; 010 ; 11) and (10 ; 10 ; 01 ; 1 ). Let a; b; c; d stand for the four index in tegers. The grammar is: ! j S A B ! 1 Aa j 0 Ab j 010 Ac j 11 Ad j A � ! 10 B j 10 B j 01 B j 1 B j B a b c d � � A string with t w o leftmost deriv ations: 10101001011 dccaba . ✦ ) 1 Aa ) 10 Aba ) 101 Aaba ) S lm lm lm lm 101010 Acaba ) 101010010 Accaba ) lm lm 10101001011 Adccaba ) lm 10101001011 dccaba . ✦ S ) 10 B a ) 1010 B ba ) lm lm lm 101010 B aba ) 10101001 B caba ) lm lm 1010100101 B ccaba ) lm 10101001011 B dccaba ) lm 10101001011 dccaba . Undecidable Problem: Is the In tersection of Tw o CFL's Empt y? Consider the t w o list languages from a PCP instance. They ha v e an empt y in tersection if and only if the PCP instance has a solution. Complemen ts of List Languages W e can get other undecidabilit y results ab out CFL's if w e �rst establish that the complemen t of a list language is a CFL. � PD A is easier approac h. � Accept all ill-formed input (not a sequence of sym b ols follo w ed b y indexes) using the state. � F or inputs that b egin with sym b ols from the alphab et of the PCP instance, store them on the stac k, accepting as w e go. 4