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slega [8]
3 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]3 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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Check the explanation

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3 years ago
A vector AP is rotated about the Z-axis by 60 degrees and is subsequently rotated about X-axis by 30 degrees. Give the rotation
Musya8 [376]

Answer:

R = \left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]

Explanation:

The mappings always involve a translation and a rotation of the matrix. Therefore, the rotation matrix will be given by:

Let \theta and \alpha be the the angles 60⁰ and 30⁰ respectively

that is \theta = 60⁰ and

\alpha = 30⁰

The matrix is given by the following expression:

\left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]

The angles can be evaluated and left in the surd form.

4 0
2 years ago
From an aerial photograph, one observes that on a level section of a (multilane) highway, 25% of the vehicles are trucks, 75% ar
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Answer:

(a) Flow rate of vehicles = No of vehicles per mile * Speed

=No of cars per mile * Speed +No of trucks per mile * Speed

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(b) Let Density of vehicles on grade = x

Density on flat * Speed =Density on grade * Speed

So,( 0.75*50) * 60 + (0.25*50) * 40 = (0.75* x) * 55 + (0.25* x) * 25

So, x= 57.89

So, Density is around 58 Vehicles per Mile.

(c) Percentage of truck by aerial photo = 25%

(d)Percentage of truck bystationary observer on the grade= 25*30/60 * 25/55 =22.73 %

4 0
3 years ago
A drilling operation is performed on a steel part using a 12.7 mm diameter twist drill with a point angle of 118 degrees. The ho
Masteriza [31]

Answer:

a. Rotational speed of the drill  = 375.96 rev/min

b. Feed rate  = 75 mm/min

c. Approach allowance  = 3.815 mm

d. Cutting time  = 0.67 minutes

e. Metal removal rate after the drill bit reaches full diameter. = 9525 mm³/min

Explanation:

Here we have

a. N = v/(πD) = 15/(0.0127·π) = 375.96 rev/min

b. Feed rate = fr = Nf = 375.96 × 0.2 = 75 mm/min

c. Approach allowance = tan 118/2 = (12.7/2)/tan 118/2 = 3.815 mm

d. Approach allowance T∞ =L/fr = 50/75 = 0.67 minutes

e. R = 0.25πD²fr = 9525 mm³/min.

7 0
3 years ago
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