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Charra [1.4K]
3 years ago
5

Problem 3. Consider the following recurrence, defined for n a power of 4 (for the time of some algorithm): T(n) = 3 if n = 1 2T(

n/4) + 4n + 1 otherwise (a) Calculate T(16) by hand. Show your work. (b) Use the tree method to solve the recurrence exactly, assuming n is a power of 4. (c) Use the formula to calculate T(1). Show your work. (d) Use the formula to calculate T(4). Show your work. (e) Use the formula to calculate T(16). Show your work. (f) Use Mathematical Induction to prove that your formula is correct in general.

Computers and Technology
1 answer:
yan [13]3 years ago
5 0

Answer:

See attached pictures.

Explanation:

See attached pictures for explanation.

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kvv77 [185]

Answer:

static int checkSymbol(char ch)

{

switch (ch)

{

case '+':

case '-':

return 1;

case '*':

case '/':

return 2;

case '^':

return 3;

}

return -1;

}

static String convertInfixToPostfix(String expression)

{

String calculation = new String("");

Stack<Character> operands = new Stack<>();

Stack<Character> operators = new Stack<>();

 

for (int i = 0; i<expression.length(); ++i)

{

char c = expression.charAt(i);

if (Character.isLetterOrDigit(c))

operands.push(c);

else if (c == '(')

operators.push(c);

 

else if (c == ')')

{

while (!operators.isEmpty() && operators.peek() != '(')

operands.push(operators.pop());

 

if (!operators.isEmpty() && operators.peek() != '(')

return NULL;    

else

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}

else

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while (!operators.isEmpty() && checkSymbol(c) <= checkSymbol(operators.peek()))

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}

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operands.push(operators.pop());

while (!operands.isEmpty())

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return calculation;

}

Explanation:

  • Create the checkSymbol function to see what symbol is being passed to the stack.
  • Create the convertInfixToPostfix function that keeps track of the operands and the operators stack.
  • Use conditional statements to check whether the character being passed is a letter, digit, symbol or a bracket.
  • While the operators is  not empty, keep pushing the character to the operators stack.
  • At last reverse and return the calculation which has all the results.
4 0
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krek1111 [17]

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4 0
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Firdavs [7]
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Answer:

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Explanation:

From the given information:

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We can conclude that the longest codeword that could possibly be for "n" symbol is n-1 bits.

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