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Oksanka [162]
2 years ago
9

What is the purpose for this experiment

Engineering
1 answer:
Natali5045456 [20]2 years ago
6 0

Answer:

I dont know

Explanation:

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This question allows you to practice proving a language is non-regular via the Pumping Lemma. Using the Pumping Lemma (Theorem 1
Ulleksa [173]

Answer:

<em>L is not a regular language with formal proofs  </em>

Explanation:

<em>(a) To prove that L is not a regular language, we will use a proof by contradiction. the assumption entails  that L is a regular language. Then by the Pumping Lemma for Regular Languages, </em>

<em>there exists a pumping length p for L such that for any string s ∈ L where |s| ≥ p, </em>

<em>s = xyz subject to the following conditions: </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L</em>

<em />

<em>(b) To determine that L is not a regular language, we mke use of proof by contradiction.  lets assume, that L is regular. Then by the Pumping Lemma for Regular Languages, it states also,</em>

<em>The pumping length, p for L such that for any string s ∈ L where |s| ≥ p, s = xyz subject  to the condtions as follows : </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L. </em>

<em>Choose s = 0p10p </em>

<em>. Clearly, |s| ≥ p and s ∈ L. By condition (b) above, it follows is shown. by the first condition x and y are zeros.</em>

<em>for some  k > 0. Per (c), we can take i = 0 and the resulting string will still be in L. Thus,  xy0 </em>

<em>z should be in L. xy0 </em>

<em>z = xz = 0(p−k)10p </em>

<em>It is shown that is is  not in L. This is a  contraption with the pumping lemma.  our assumption that L is regular is  incorrect, and L is not a regular language</em>

6 0
3 years ago
The quantity of bricks required increases with the surface area of the wall, but the thickness of a masonry wall does not affect
Yakvenalex [24]

Answer:

false

Explanation:

5 0
3 years ago
Read 2 more answers
If the contact surface between the 20-kg block and the ground is smooth, determine the power of force F when t = 4 s. Initially,
grigory [225]

Answer:

The power of force F is 115.2 W

Explanation:

Use following formula

Power  = F x V

F_{H} = F cos0

F_{H} = (30) x 4/5

F_{H} = 24N

Now Calculate V using following formula

V = V_{0} + at

V_{0} = 0

a = F_{H} / m

a = 24N / 20 kg

a = 1.2m / S^{2}

no place value in the formula of V

V = 0 + (1.2)(4)

V = 4.8 m/s

So,

Power = F_{H} x V

Power = 24 x 4.8

Power = 115.2 W

3 0
3 years ago
Disconnecting a circuit while in operation can create a voltage blank
zlopas [31]

Answer:

what is the question

Explanation:

confused

5 0
3 years ago
2. Write a Java program that generates a new string by concatenating the reversed substrings of even indexes and odd indexes sep
Nana76 [90]

Answer:

  1. public class Main {
  2.    public static void main(String[] args) {
  3.        String testString = "abscacd";
  4.        String evenStr = "";
  5.        String oddStr = "";
  6.        for(int i=testString.length() - 1; i >= 0; i--){
  7.            if(i % 2 == 0){
  8.                evenStr += testString.charAt(i);
  9.            }
  10.            else{
  11.                oddStr += testString.charAt(i);
  12.            }
  13.        }
  14.        System.out.println(evenStr + oddStr);
  15.    }
  16. }

Explanation:

Firstly, let declare a variable testString to hold an input string "abscacd" (Line 1).

Next create another two String variable, evenStr and oddStr and initialize them with empty string (Line 5-6). These two variables will be used to hold the string at even index and odd index, respectively.

Next, we create a for loop that traverse the characters of the input string from the back by setting initial position index i to  testString.length() - 1  (Line 8). Within the for-loop, create if and else block to check if the current index, i is divisible by 2, (i % 2 == 0), use the current i to get the character of the testString and join it with evenStr. Otherwise, join it with oddStr (Line 10 -14).

At last, we print the concatenated evenStr and oddStr (Line 18).  

4 0
3 years ago
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