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Andreyy89
2 months ago
12

PDAs with two stacks are strictly more powerful than PDAs with one stack. Prove that 2-stack PDAs are not a valid model for CFLs

by giving an example of a language that is not context-free and yet accepted by a 2-stack PDA. Describe (no math necessary) how a 2-stack PDA would accept that language.
Engineering
1 answer:
iogann1982 [368]2 months ago
6 0

Answer:

?pooooooooooooooooooooooop

Explanation:

?pooooooooooooooooooooooop

You might be interested in
At what forward voltage does a diode conduct a current equal to 10,000 Is ? In terms of Is , what current flows in the same diod
pantera1 [306]

Answer:a) The forward voltage is 0.23 V

b) The current that flows  

I_{d} = (1.45*10^{12}I_{s})A

Explanation:

The forward voltage refers to the minimum voltage required for a diode to start conducting. The formula used is given by:

a) At which forward voltage does a diode allow a current equal to 10,000 Is? In terms of Is

I_{d} = I_{s}(e^{\frac{v_{f} }{0.025} }-1)

Where:

Id denotes the diode current = 10000Is,

Vd stands for the forward voltage at which conduction begins,

Is indicates the saturation current.

I_{d} = I_{s}(e^{\frac{v_{f} }{0.025} }-1)

10000I_{s} = I_{s}(e^{\frac{v_{f} }{0.025} }-1)

By dividing through by Is,

10000 = (e^{\frac{v_{f} }{0.025} }-1)

10000 +1= e^{\frac{v_{f} }{0.025} }

10001= e^{\frac{v_{f} }{0.025} }

Taking the natural log of both sides,

ln(10001)= {\frac{v_{f} }{0.025} }

9.21= {\frac{v_{f} }{0.025} }

Multiplying through by 0.025 yields  

= 0.23 V

This indicates the forward voltage at which the diode conducts a current equal to 10,000 Is is 0.23 V

b) What current flows in the same diode when the forward voltage is 0.7 V?

I_{d} = I_{s}(e^{\frac{v_{f} }{0.025} }-1)

I_{d} = I_{s}(e^{\frac{0.7}{0.025} }-1)

I_{d} = I_{s}(1.45*10^{12} -1)

I_{d} = (1.45*10^{12}I_{s})A

6 0
1 month ago
Number pattern Write a recursive method called print Pattern() to output the following number pattern. Given a positive integer
alex41 [359]

Response:

Refer to the explanation

Clarification:

Code:

import java.util.Scanner;

public class NumberPattern {

public static int x, count;

public static void displayNumPattern(int num1, int num2) {

if (num1 > 0 && x == 0) {

System.out.print(num1 + " ");

count++;

displayNumPattern(num1 - num2, num2);

} else {

x = 1;

if (count >= 0) {

System.out.print(num1 + " ");

count--;

if (count < 0) {

System.exit(0);

}

displayNumPattern(num1 + num2, num2);

}

}

}

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

displayNumPattern(num1, num2);

}

}

See attached example output

3 0
2 months ago
Read 2 more answers
Given num_rows and num_cols, print a list of all seats in a theater. Rows are numbered, columns lettered, as in 1A or 3E. Print
Viktor [391]

Answer:

This is the solution code in Python:

  1. alphabets = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J']
  2. user_input = input("Enter number of rows and columns: ")
  3. myArr = user_input.split(" ")
  4. num_rows = int(myArr[0])
  5. num_cols = int(myArr[1])
  6. seats = []
  7. for i in range(num_rows):
  8. row = []
  9. for j in range(num_cols):
  10. row.append(alphabets[j])
  11. seats.append(row)
  12. output = ""
  13. for i in range(len(seats)):
  14. for j in range(len(seats[i])):
  15. output += str(i + 1) + seats[i][j] + " "
  16. print(output)

Explanation:

Initially, we create a small list of alphabets from A to J (Line 1).

We then request the user to enter the number of rows and columns (Line 3). Given that the input comes as a string (e.g., "2 3"), we utilize the split() method to separate the numbers into individual items in a list (Line 4). The first item (row number) is assigned to variable num_rows, while the second item (column number) goes to num_cols.

Subsequently, we construct the seats list with a nested for-loop (Lines 10-15). Once the seats list is formed, another nested for-loop generates the required output string as per the question (Lines 19-21).

Finally, the output is printed (Line 23). For example, an input of 2 3 results in the output:

1A 1B 1C 2A 2B 2C

8 0
3 months ago
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