GATE CSE 1998 | Question: 19b

Question

Compute the post fix equivalent of the following expression $3^*\log(x+1)-\frac{a}{2}$

Comments

@msrsaditya

In mathematics operators are special kind of functions. 

In programming, "operator" and "function" have special meaning, simillar to mathematical ones, but not exactly the same. 

For eg. C considers sizeof, +, - as operator and not as function; printf is an example of a function.

 

But in this case, log is an operator/function, just like any other operators like *, +, -, /. 

But it is an unary operator like the ++, -- operations.

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so log will be treated same as the exponentiation operator?

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log is just an unary operator, nothing special. Note: in my previous comment I did a mistake of taking ^ as unary operator.

Answer 1

According to http://faculty.washington.edu/jstraub/dsa/aexp/

The priority of the operators follows the usual conventions:

• The highest priority is assigned to unary operators (note that, in this context, a function such as sin  is considered a unary operator). All unary operators have the same priority.
• Exponentiation has the second highest priority.
• The third highest priority is assigned to the multiplication and division operators.
• The lowest priority is given to the addition and subtraction operators. 

Example:->>
Infix expression:   $3  * \log(  10  )$
Postfix expression:
$=3  *   (10 \log)$   //(Priority of unary operator log forces $\log( 10 )$ to evaluate first.)
$=3  \ 10  \log  *$

Now for our case $3*\log(x + 1) - a / 2$

First content inside parenthesis will be evaluated

So, $x+1$ will become $x\;1+$

Now among $(*,/,\log,+,-)$ operators, $\log$ has highest priority as it is the only unary operator

So, $\log(x\;1+)$ will become ${x\;1+\log}$

Now suppose ${z= x1+\log}$ and we get  $3* z - a / 2$

$\implies 3z* a\;2/ -$

Now, substitute $z= x\;1+\log$ and we get

$\mathbf{3x\;1+\log*  a\;2/  -}$ as answer.

Comments

thanx a lot

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But in C lang, there is no exponentiation operator!

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great answer

 

Answer 2

We can do directly(manually) =>

3*log(x + 1) - a / 2

= ((3*(log(x + 1))) - (a / 2))

= ((3*log(x1+))) - (a2/))

= ((3*x1+log) - (a2/))

= ((3x1+log*) - (a2/))

= 3x1+log*a2/- 

Answer 3

$3 x 1 + \log * a 2 / -$

Comments

i m not getting it ..... can you please explain me ...

 

by the expanding it .... fully

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brother how to write the postfix of these two??
1. a++ ,  do i have to expand like a=a+1 and then proceed?  Please explain in detail.
2 ++a + b++

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Suhail96

i think the first one i,e a++ is already a postfix expression.

and the 2^nd one  i.e ++a+b++  can be written as   a++b+++ .

pls verify once!

Answer 4

Given infix expression : $3*log(x+1)- \frac{a}{2}$

Expression tree :

Do a postorder traversal of the above tree to get postfix expression: $3x1+log*a2/-$

Comments

@Radha mohan

how you are so sure the x+1 is on the right child of log in expression tree.

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ASNR1010

but it doesn’t matter in postorder traversal.

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wow, I don't know we can even use this method, thanks

Answer 5

Using Stack (Standard method)

Answer 6

Answer attached

Comments

great..