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slega [8]
2 years ago
15

Prove the following languages are nonregular, once using the pumping lemma and once using the Myhill-Nerode theorem. When using

the pumping lemma, try to use the most general (least restrictive) case:i) The language where the longest string of consecutive a's is longer than the longest string of consecutive b'sii) aibjajiii) aibjck, where k ≤ i+jiv) aibjck, where i is neither the least nor greatest out of i,j, and k.
Engineering
1 answer:
VashaNatasha [74]2 years ago
8 0

Answer:

For any string, we use s = xyz

Explanation:

The pumping lemma says that for any string s in the language, with length greater than the pumping length p, we can write s = xyz with |xy| ≤ p, such that xyi z is also in the language for every i ≥ 0. For the given language, we can take p = 2.

Here are the cases:

  • Consider any string a i b j c k in the language. If i = 1 or i > 2, we take x = \epsilon   and y = a. If i = 1, we must have j = k and adding any number of a’s still preserves the membership in the language. For i > 2, all strings obtained by pumping y as defined above, have two or more a’s and hence are always in the language.
  • For i = 2, we can take    and y = aa. Since the strings obtained by pumping in this case always have an even number of a’s, they are all in the language.
  • Finally, for the case i = 0, we take x = \epsilon  , and y = b if j > 0 and y = c otherwise. Since strings of the form b j c k are always in the language, we satisfy the conditions of the pumping lemma in this case as well.
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Explanation:

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Once the forces were established, we can write components of the Second Newtons Law on vertical and horizontal axes, considering that horizontal axis is directed backwards- in the same direction with the acceleration:

For car A on the vertical axis the equation is: -mAg+NA=0

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For car B, on the vertical axis the equation is: -mBg+NB=0

For car B, on the horizontal axis, the equation is: T=mBa

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After the transformations, the equations for acceleration and force in the coupling will be:

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Let <em>f(z)</em> = (4<em>z </em>² + 2<em>z</em>) / (2<em>z </em>² - 3<em>z</em> + 1).

First, carry out the division:

<em>f(z)</em> = 2 + (8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1)

Observe that

2<em>z </em>² - 3<em>z</em> + 1 = (2<em>z</em> - 1) (<em>z</em> - 1)

so you can separate the rational part of <em>f(z)</em> into partial fractions. We have

(8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1) = <em>a</em> / (2<em>z</em> - 1) + <em>b</em> / (<em>z</em> - 1)

8<em>z</em> - 2 = <em>a</em> (<em>z</em> - 1) + <em>b</em> (2<em>z</em> - 1)

8<em>z</em> - 2 = (<em>a</em> + 2<em>b</em>) <em>z</em> - (<em>a</em> + <em>b</em>)

so that <em>a</em> + 2<em>b</em> = 8 and <em>a</em> + <em>b</em> = 2, yielding <em>a</em> = -4 and <em>b</em> = 6.

So we have

<em>f(z)</em> = 2 - 4 / (2<em>z</em> - 1) + 6 / (<em>z</em> - 1)

or

<em>f(z)</em> = 2 - (2/<em>z</em>) (1 / (1 - 1/(2<em>z</em>))) + (6/<em>z</em>) (1 / (1 - 1/<em>z</em>))

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\displaystyle\frac1{1-z}=\sum_{n=0}^\infty z^n

Replace <em>z</em> with 1/<em>z</em> to get

\displaystyle\frac1{1-\frac1z}=\sum_{n=0}^\infty z^{-n}

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Now condense <em>f(z)</em> into one series:

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So, the inverse <em>Z</em> transform of <em>f(z)</em> is \boxed{6+2^{2-n}}.

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