What would be the Prefix notation for the given equation?
A^B^C^D
a) ^^^ABCD
b) ^A^B^CD
c) ABCD^^^
d) AB^C^D
Explanation: Reverse the equation or scan the equation from right to left. Apply the infix-prefix algorithm. Here we have to remember that the exponentiation has order of associativity from right to left. Therefore the stack goes on pushing ^. Therefore resulting in ^^^ABCD.
What would be the Prefix notation for the given equation?
a+b-c/d&e|f
a) |&-+ab/cdef
b) &|-+ab/cdef
c) |&-ab+/cdef
d) |&-+/abcdef
Explanation: Reverse the equation or scan the equation from right to left. Apply the infix-prefix algorithm. The preference order in ascending order are as follows |&+*/.
What would be the Prefix notation for the given equation?
(a+(b/c)*(d^e)-f)
a) -+a*/^bcdef
b) -+a*/bc^def
c) -+a*b/c^def
d) -a+*/bc^def
Explanation: Reverse the equation or scan the equation from right to left. Apply the infix-prefix algorithm. The preference order in ascending order are as follows +*/^. Brackets have the highest priority. The equations inside the brackets are solved first.
What would be the Prefix notation and Postfix notation for the given equation?
A+B+C
a) ++ABC and AB+C+
b) AB+C+ and ++ABC
c) ABC++ and AB+C+
d) ABC+ and ABC+
Explanation: For prefix notation there is a need of reversing the giving equation and solving it as a normal infix-postfix question. We see that it doesn’t result as same as normal infix-postfix conversion.
What would be the Prefix notation for the given equation? a|b&c
a) a|&bc
b) &|abc
c) |a&bc
d) ab&|c
Explanation: The order of preference of operators is as follows (descending): & |.
The equation a|b&c will be parenthesized as (a|(b&c)) for evaluation.
Therefore the equation for prefix notation evaluates to |a&bc.
When an operand is read, which of the following is done?
a) It is placed on to the output
b) It is placed in operator stack
c) It is ignored
d) Operator stack is emptied
Explanation: While converting an infix expression to a postfix expression, when an operand is read, it is placed on to the output. When an operator is read, it is placed in the operator stack.
