conversion from infix to prefix

Question

I am preparing for an exam where i couldn\'t understand the convertion of infix notation to polish notation for the below expression:

(a–b)/c*(d + e – f / g)         


        

Answer 1

(a–b)/c*(d + e – f / g)

remember scanning the expression from leftmost to right most start on parenthesized terms follow the WHICH COMES FIRST rule... *, /, % are on the same level and higher than + and -.... so (a-b) = -bc prefix (a-b) = bc- for postfix another parenthesized term: (d + e - f / g) = do move the / first then plus '+' comes first before minus sigh '-' (remember they are on the same level..) (d + e - f / g) move / first (d + e - (/fg)) = prefix (d + e - (fg/)) = postfix followed by + then - ((+de) - (/fg)) = prefix ((de+) - (fg/)) = postfix

(-(+de)(/fg)) = prefix so the new expression is now -+de/fg (1) ((de+)(fg/)-) = postfix so the new expression is now de+fg/- (2)

(a–b)/c* hence

1. (a-b)/c*(d + e – f / g) = -bc prefix [-ab]/c*[-+de/fg] ---> taken from (1) / c * do not move yet so '/' comes first before '*' because they on the same level, move '/' to the rightmost : /[-ab]c * [-+de/fg] then move '*' to the rightmost

   • / [-ab]c[-+de/fg] = remove the grouping symbols = */-abc-+de/fg --> Prefix

2. (a-b)/c*(d + e – f / g) = bc- for postfix [ab-]/c*[de+fg/-]---> taken from (2) so '/' comes first before '' because they on the same level, move '/' to the leftmost: [ab-]c[de+fg/-]/ then move '' to the leftmost [ab-] c [de+fg/-]/ = remove the grouping symbols= a b - c d e + f g / - / * --> Postfix