Question

In: Computer Science

C++ Use BinaryNodeTree to solve the following questions in a program: 1) What is the value...

C++ Use BinaryNodeTree to solve the following questions in a program:

1) What is the value of the prefix expression +−∗ 235/↑ 2 3 4?

2) What is the postfix form of the expression ((x + y) ↑ 2) + ((x − 4)/3)?

3) What is the value of the postfix expression723 ∗ − 4 ↑ 9 3/+?

Solutions

Expert Solution

QUE : 1) What is the value of the prefix expression +−∗ 235/↑ 2 3 4?

Symbol Opnd 1 Opnd 2 Value Opndstack
4 4
3 4,3
2 4,3,2
↑ 2 3 8 4
4,8
/ 8 4 2
2
5 2,5
3 2,5,3
2 2,5,3,2
* 2 3 6 2,5
2,5,6
- 6 5 1 2
2,1
+ 1 2 3
VALUE = 3

VALUE For given prefix expression  +−∗ 235/↑ 2 3 4 is 3.

_______________________________________________________________________

QUE: 2) What is the postfix form of the expression ((x + y) ↑ 2) + ((x − 4)/3)?

ANS:

Input String Output Stack Operator Stack
((x + y) ↑ 2) + ((x − 4)/3) (
((x + y) ↑ 2) + ((x − 4)/3) ((
((x + y) ↑ 2) + ((x − 4)/3) x ((
((x + y) ↑ 2) + ((x − 4)/3) x ((
((x + y) ↑ 2) + ((x − 4)/3) x ((+
((x + y) ↑ 2) + ((x − 4)/3) x ((+
((x + y) ↑ 2) + ((x − 4)/3) x y ((+
((x + y) ↑ 2) + ((x − 4)/3) x y+ (
((x + y) ↑ 2) + ((x − 4)/3) x y+ (
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ (
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ (
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ 2 (
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ 2
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ 2
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ 2 +
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ 2 +
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ 2 +(
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ 2 +((
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ 2 x +((
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ 2 x +((
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ 2 x − +((
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ 2 x − +((
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ 2 x − 4 +((
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ 2 x − 4 +(
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ 2 x − 4 +(/
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ 2 x − 43 +(/
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ 2 x − 43/ +
((x + y) ↑ 2) + ((x − 4)/3) x y+ ↑ 2 x − 43/+

Que : 3) What is the value of the postfix expression723 ∗ − 4 ↑ 9 3/+?

ANS :

723 ∗ − 4 ↑ 9 3/+

The first character scanned is "723", which is an operand, so push it to the stack.

723

  
Stack Expression

___________________________________________________________________________________________

723 ∗ − 4 ↑ 9 3/+

The next character scanned is "∗", which is an operator, so pop its two operands from the stack. Pop 723 from the stack for the right operand and then pop undefined from the stack to make the left operand.

undefined ∗ 723 = 0
Stack Expression

Next, push the result of undefined ∗ 723 (0) to the stack.

0

  
Stack Expression

__________________________________________________________________________________________

723 ∗ − 4 ↑ 9 3/+

The next character scanned is "−", which is an operator, so pop its two operands from the stack. Pop 0 from the stack for the right operand and then pop undefined from the stack to make the left operand.

undefined − 0 = 0
Stack Expression

Next, push the result of undefined − 0 (0) to the stack.

0

  
Stack Expression

_______________________________________________________________________________________

723 ∗ − 4 ↑ 9 3/+

The next character scanned is "4", which is an operand, so push it to the stack.

4
0

  
Stack Expression

_______________________________________________________________________________________

723 ∗ − 4 ↑ 9 3/+

The next character scanned is "↑", which is an operator, so pop its two operands from the stack. Pop 4 from the stack for the right operand and then pop 0 from the stack to make the left operand.

0 ↑ 4 = 0
Stack Expression

Next, push the result of 0 ↑ 4 (0) to the stack.

0

  
Stack Expression

_________________________________________________________________________________________

723 ∗ − 4 ↑ 9 3/+

The next character scanned is "9", which is an operand, so push it to the stack.

9
0

  
Stack Expression

_________________________________________________________________________________________

723 ∗ − 4 ↑ 9 3/+

The next character scanned is "3/+", which is an operator, so pop its two operands from the stack. Pop 9 from the stack for the right operand and then pop 0 from the stack to make the left operand.

0 3/+ 9 = 0
Stack Expression

Next, push the result of 0 3/+ 9 (0) to the stack.

0

  
Stack Expression

Since we are done scanning characters, the remaining element in the stack (0) becomes the result of the postfix evaluation.

Postfix notation: 723 ∗ − 4 ↑ 9 3/+
Result: 0

venereology answered 1 year ago
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