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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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Answer:

Without Recursives

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Using Recursive

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using namespace std;

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3 0
3 years ago
A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °
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Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30  (Btu/hr)/(ft ⋅ °F)= \frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3}  (Btu/s)/(ft.F)

Thermal Conductivity is SI units:

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Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

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T_2=50 C= 122 F

Part A:

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A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J

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A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu

PArt B:

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Q=\frac{k*A*(T_1-T_2)}{L}

On rearranging:

T_2=T_1-\frac{Q*L}{KA}

T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C

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