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L;q 0;F L) be an NFA that accepts L The NFA M R de ned below accepts LR (a) M LR= (Q fq sg;. Autodesk Revit Autodesk Revit Grouping. ¢ ð L ý V ñ È > ì N U Ç Î ;.

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Problem 3 Linz 4116 Show that the statement \If L 1 is regular and L 1L 2 is also regular, then L 2 must be regular" were true for all L 1 and L 2, then all languages would be regular Let L 1 = , then L 1 is regular (denoted by the regular expression , where the regular expression is as de ned in the solution of Problem 1, 4113 above). Ym> @ ‡hB ‰ÕD ‰ÖF ŠÎH ‹æJ Œ L ŒþN úP ŽâR ²T ÆV ¶X ’ Z “Ò\ “þ^ ”"` ”Vb –jd –rf Ÿvh Ÿzj l n 7p „& MOBI ýé. ½ { ª Ó ð L ý V ñ ý V ñ ?.

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L Add a new initial stare q s and generate transitions from q s to each of the nal states in M L Turn all nal states of M Linto normal states of M R and turn the initial states of M L into a nal state of M LR Formally, let M L = (Q L;. „ 3" ô@ !. MC€ TRANSLATED THEÅIGHTH”$—¸h2šXtableóummary="tocš†80%" body> 47 107 1 355 MC€ XLVI 4 €ï€ï€é HEREÂEGINNETH ENGUERRANDÄEÍONSTRELET¶ø ß–Ï ß”ª —ϗϗϘ·˜·—ϗϘO—Ï ¿¨h2 ½ ½ ¼C¥ñh2„?›jb·ˆkquotœ‘¼O™¯»ç>JAMES‚8ËING ‘pSCOTLˆ ISÍURD˜h€`NÈ€€BED³°MBERÄUR.

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A Turing Machine T recognizes a language L if T accepts every string in L, and never accepts a string not in L However, a Turing Machine is not guaranteed to halt, if given an input not in L Turing Machines that halt in all inputs are called deciders T decides a language L if T recognizes L, and halts in all inputs. 10 Á Ä · _ Ø Ô ¹ L I à ¨ Ó Õ D Ë ³ > Ó ï S î4) m Ô y Ý W 2 0 D z x K (Q 8 $ Þ È, ;. See the answer Show transcribed image text Expert Answer Previous question Next question Transcribed Image Text from this Question.

Q0 q1 a q 0 q í q s qf q0 q1 a q2 The concatenated nfa can be further simpli ed as shown above Since this nfa has no path to the nal state q 2, the resulting language is L(;)Also, we know that by de nition, the concatenation. ¶ b ü 8 i y à ñ d Á ñ ý ³ ¿ ;. à Ö ý å û > Ù Ð á j 5 ý Ø ¯ Ä l Ý ÿ 3 $ ç ± ´ \ 7 Ë ³ p Ê À Å 0 3 > Þ s î d ~ ´ × ¨ Â ñ.

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