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Assume that the operators $+, -, \times$ are left associative and $^\hat{}$ is right associative. The order of precedence (from highest to lowest) is $^\hat{}, \times, +, -$. The postfix expression corresponding to the infix expression $a+ b \times c-d^\hat{}e^\hat{}f$ is

  1. $abc\times+def^\hat{}{} ^\hat{}-$

  2. $abc\times+de^\hat{}f^\hat{}-$

  3. $ab+c\times d-e^\hat{}f^\hat{}$

  4. $-+a\times bc^\hat{}{}^\hat{}def$

asked in DS by Veteran (59.5k points)
edited by | 2.9k views

3 Answers

+32 votes
Best answer

Answer is (A).

Here is the procedure first :
Scan Infix Expression from left to right whenever you see operand just print it.
But, in case of operator
if(stack is empty) then push it.
if(precedence(tos) $<$ precedence(current operator) ) push it. where, tos means top of stack.
else if (precedence(tos) $>$ precedence(current operator) ) pop and print.
else if (precedence(tos) $==$  precedence(current operator) ) then check for associativity.In case Operators are Left to right then pop and print it otherwise push the current operator (Right to Left Associativity)
And once you have scanned infix expression completely. Make sure pop all the element and print it in same order.

Here, the infix expression is $a+b \times c−d ^\hat{} e ^\hat{}f$
$a$ : print it
$+$ : push into the Operator Stack
$b$ : print it
$*$ : its having higher precedence than $+$ then push into Operator Stack
$c$ : print it
$'-'$ : $'-'$ is having less precedence than $'*'$ so pop from operator stack and print $'*'$.after this stack will be having $'+'$ on top.which is having same precedence as $'-'$ but both are left to right associative then just pop $+$ and print it. Now stack is empty. Push $'-'$ to it.
$d$ : print it
'$^\hat{}$' : top of the stack is having $'-'$.'$^\hat{}$' has higher precedence than $'-'$.so simply push '$^\hat{}$' into stack
$e$ : print it.
'$^\hat{}$' : Now top of the stack is '$^\hat{}$' and has same precedence so associativity will come to picture. Since '$^\hat{}$' is right associative as given in question. So '$^\hat{}$' will be pushed.
$f$ : print it.

Now, we have scanned entire infix expression.Now pop the stack until it becomes empty.This way you will get $a b c * + d e f ^\hat{} \ ^\hat{}-$.

answered by Junior (505 points)
edited by
+5 votes

Use Bracket according to precedance 

^  - Precedance from Right to Left and others left to right

((a + (b * c )) - (d ^ (e ^ f))) then solve it 

answered by Loyal (6.7k points)
0 votes
answer - A
answered by Loyal (9k points)
+4
One easy way is:

1. Add brackets as per precedence and order of evaluation

2. Move operators to the right of immediate paranthesis for postfix.

 

Here, after adding brackets,

((a+(b*c))-(d^(e^f)))

Now after moving operators to immediate right paranthesis, we obtain,

abc*+def^^-

Sorry for typos.


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