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Assume that the operators + , -, x are left associative and ^ is right associative. The order of precedence (from highest to lowest) is ^, x, +, The postfix expression corresponding to the infix expression a + b x c - d ^ e ^ f is
- a)abc x + def ^ ^ -
- b)abc x + de ^ f ^
- c)ab + c x d - e ^ f ^
- d)- + a x be ^ ^ def
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Assume that the operators + , -, x are left associative and ^ is right...
According to conditions given:
⇒ [(a + (b x c)) - (d ^ (e ^ f))]
⇒ [(a + ( b c x)) - (d ^ (e f ^))]
⇒ [(abc x +) - (def^^)]
⇒ abc x + def^^ -
⇒ [(a + (b x c)) - (d ^ (e ^ f))]
⇒ [(a + ( b c x)) - (d ^ (e f ^))]
⇒ [(abc x +) - (def^^)]
⇒ abc x + def^^ -
Most Upvoted Answer
Assume that the operators + , -, x are left associative and ^ is right...
Explanation:
To convert an infix expression to a postfix expression, we need to follow certain rules based on the precedence and associativity of the operators. Let's break down the given infix expression and convert it to postfix expression step by step.
Given infix expression: a b x c - d ^ e ^ f
Step 1: Start with an empty stack and an empty output string.
Step 2: Scan the infix expression from left to right.
- When an operand (a variable or constant) is encountered, append it to the output string.
- a -> output string: a
- b -> output string: ab
- c -> output string: abc
- d -> output string: abcd
- e -> output string: abcde
- f -> output string: abcdef
Step 3: When an operator is encountered, do the following:
- If the stack is empty, push the operator onto the stack.
- If the stack is not empty, compare the precedence of the operator with the top of the stack.
- If the precedence of the operator is higher, push the operator onto the stack.
- If the precedence of the operator is lower or equal, pop operators from the stack and append them to the output string until an operator with lower precedence is encountered or the stack becomes empty. Then push the current operator onto the stack.
Operator precedence:
- ^ (highest precedence)
- x
- (lowest precedence)
Step 4: Repeat steps 2 and 3 until all the tokens in the infix expression have been scanned.
- x -> stack: x, output string: abcdef
- -> stack: x, output string: abcdefx
- c -> stack: xc, output string: abcdefx
- - -> stack: xc-, output string: abcdefx
- d -> stack: xc-d, output string: abcdefx
- ^ -> stack: xc-d^, output string: abcdefx
- e -> stack: xc-d^e, output string: abcdefx
- ^ -> stack: xc-d^e^, output string: abcdefx
- f -> stack: xc-d^e^f, output string: abcdefx
Step 5: After scanning all the tokens, pop any remaining operators from the stack and append them to the output string.
- stack: xc-d^e^f, output string: abcdefx^
Step 6: The resulting postfix expression is "abcdefx^", which is the correct answer.
Therefore, the postfix expression corresponding to the given infix expression "a b x c - d ^ e ^ f" is "abcdefx^".
To convert an infix expression to a postfix expression, we need to follow certain rules based on the precedence and associativity of the operators. Let's break down the given infix expression and convert it to postfix expression step by step.
Given infix expression: a b x c - d ^ e ^ f
Step 1: Start with an empty stack and an empty output string.
Step 2: Scan the infix expression from left to right.
- When an operand (a variable or constant) is encountered, append it to the output string.
- a -> output string: a
- b -> output string: ab
- c -> output string: abc
- d -> output string: abcd
- e -> output string: abcde
- f -> output string: abcdef
Step 3: When an operator is encountered, do the following:
- If the stack is empty, push the operator onto the stack.
- If the stack is not empty, compare the precedence of the operator with the top of the stack.
- If the precedence of the operator is higher, push the operator onto the stack.
- If the precedence of the operator is lower or equal, pop operators from the stack and append them to the output string until an operator with lower precedence is encountered or the stack becomes empty. Then push the current operator onto the stack.
Operator precedence:
- ^ (highest precedence)
- x
- (lowest precedence)
Step 4: Repeat steps 2 and 3 until all the tokens in the infix expression have been scanned.
- x -> stack: x, output string: abcdef
- -> stack: x, output string: abcdefx
- c -> stack: xc, output string: abcdefx
- - -> stack: xc-, output string: abcdefx
- d -> stack: xc-d, output string: abcdefx
- ^ -> stack: xc-d^, output string: abcdefx
- e -> stack: xc-d^e, output string: abcdefx
- ^ -> stack: xc-d^e^, output string: abcdefx
- f -> stack: xc-d^e^f, output string: abcdefx
Step 5: After scanning all the tokens, pop any remaining operators from the stack and append them to the output string.
- stack: xc-d^e^f, output string: abcdefx^
Step 6: The resulting postfix expression is "abcdefx^", which is the correct answer.
Therefore, the postfix expression corresponding to the given infix expression "a b x c - d ^ e ^ f" is "abcdefx^".
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Assume that the operators + , -, x are left associative and ^ is right associative. The order of precedence (from highest to lowest) is ^, x, +, The postfix expression corresponding to the infix expression a + b x c - d ^ e ^ f isa)abc x + def ^ ^-b)abc x + de ^ f ^c)ab + c x d - e ^ f ^d)- + a x be ^^ defCorrect answer is option 'A'. Can you explain this answer?
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