Sipser 1.54 - 1.54) Consider the language F = {aibjck...

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Unformatted text preview: 1.54) Consider the language F = {aibjck |i,j,k 0 and if i = 1 then j = k} (A) Show that F is not regular. Assume that F is regular and conforms appropriately. If F is regular, then the reverse of F is regular; Use the pumping lemma, and let p be its hypothetical pumping number. Consider w = cpbpa , w FR. Attempting to pump w will yield some cp+ibpa, which is not in the reverse of F, as p+i > p. Thus FR is not regular as it violates the pumping lemma, so F is not regular. (B) Show that F acts like a regular language in the pumping lemma For any string w in F, there are four possible forms: aaa*b*c* (two or more as, followed by any number of as, followed by any number of bs, followed by any number of cs), abici (exactly one a, followed by some number of bs, possibly zero, followed by an equal number of cs), bb*c* (no as, followed by at least one b, followed by any number of cs, possibly zero), or c*(no as or bs, followed by any number of cs, possibly zero). We must now show that for each of these cases, w is suitable for pumping (let us simply assume p = 1, although it would be easy to expand this). Case 1: aaa*b*c* Let us parse w as xyz, with x = aa, y = b, and z = b*c* (i.e. whatever is left over from w after the removal of the two as and first b). Here, for p = 1, we can pump w successfully, since for all i 0, xyiz = aab*c* F, |y| > 0, and |xy| p. Case 2: abici Let us parse w as xyz, with x = , y = a, and z = bici (i.e. whatever is left over from w after the removal of the first a). Here, for p = 1, we can pump w successfully, since for all i 0, xyiz = a*bici F, |y| > 0, and |xy| p. Case 3: bb*c* Let us parse w as xyz, with x = , y = b, and z = b*c* (i.e. whatever is left over from w after the removal of the first b). Here, for p = 1, we can pump w successfully, since for all i 0, xyiz = b*c* F, |y| > 0, and |xy| p. Case 4: c* Let us parse w as xyz, with x = , y = c, and z = c (i.e. whatever is left over from w after the removal of the first c). Here, for p = 1, we can pump w successfully, since for all i 0, xyiz = c* F, |y| > 0, and |xy| p. Thus, F acts like a regular language in the pumping lemma for p = 1. (C) There is no contradiction of the pumping lemma here; all the lemma states is that if a language doesn't conform, then it cannot be regular, but a nonregular language may or may not conform. ...
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This note was uploaded on 04/06/2008 for the course EECS 376 taught by Professor Stout during the Spring '08 term at University of Michigan.

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