09 STACK APPLICATION DATA STRUCTURES AND ALGORITHMS REVERSE POLISH NOTATION


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1 DATA STRUCTURES AND ALGORITHMS 09 STACK APPLICATION REVERSE POLISH NOTATION IMRAN IHSAN ASSISTANT PROFESSOR, AIR UNIVERSITY, ISLAMABAD LECTURES ADAPTED FROM: DANIEL KANE, NEIL RHODES DEPARTMENT OF CS & ENGINEERING UNIVERSITY OF CALIFORNIA, SAN DIEGO
2 ALGEBRAIC EXPRESSION X + Y * Z An algebraic expression is a legal combination of operands and the operators. Operand is the quantity (unit of data) on which a mathematical operation is performed. Operand may be a variable like x, y, z or a constant like 5, 4,0,9,1 etc. Operator is a symbol which signifies a mathematical or logical operation between the operands. Example of familiar operators include +,,*, /, ^ Considering these definitions of operands and operators now we can write an example of expression as x+y*z. Infix, Postfix and Prefix notations are three different but equivalent ways of writing expressions. It is easiest to demonstrate the differences by looking at examples of operators that take two operands. 2
3 INFIX NOTATION X + Y Operators are written inbetween their operands. This is the usual way we write expressions. An expression such as A * ( B + C ) / D is usually taken to mean something like: "First add B and C together, then multiply the result by A, then divide by D to give the final answer. Infix notation needs extra information to make the order of evaluation of the operators clear: rules built into the language about operator precedence and associativity, and brackets ( ) to allow users to override these rules. For example, the usual rules for associativity say that we perform operations from left to right, so the multiplication by A is assumed to come before the division by D. Similarly, the usual rules for precedence say that we perform multiplication and division before we perform addition and subtraction. 3
4 POSTFIX NOTATION X Y + Also known as "Reverse Polish notation Operators are written after their operands. The infix expression given above is equivalent to A B C + * D / The order of evaluation of operators is always lefttoright, and brackets cannot be used to change this order. Because the "+" is to the left of the "*" in the example above, the addition must be performed before the multiplication. Operators act on values immediately to the left of them. For example, the "+" above uses the "B" and "C". We can add (totally unnecessary) brackets to make this explicit: ( (A (B C +) *) D /) Thus, the "*" uses the two values immediately preceding: "A", and the result of the addition. Similarly, the "/" uses the result of the multiplication and the "D". 4
5 PREFIX NOTATION + X Y Also known as "Polish notation Operators are written before their operands. The expressions given above are equivalent to / * A + B C D to make this clear: (/ (* A (+ B C) ) D) 5
6 OPERATOR PRIORITIES TIE BREAKER, DELIMITERS How do you figure out the operands of an operator? a + b * c a * b + c / d This is done by assigning operator priorities. priority(*) = priority(/) > priority(+) = priority() When an operand lies between two operators, the operand associates with the operator that has higher priority. Tie Breaker When an operand lies between two operators that have the same priority, the operand associates with the operator on the left. a + b  c a * b / c / d Delimiters Subexpression within delimiters is treated as a single operand, independent from the remainder of the expression. (a + b) * (c d) / (e f) 6
7 WHY POSTFIX? INFIX EXPRESSION IS HARD TO PARSE Why to use these weird looking PREFIX and POSTFIX notations when we have simple INFIX notation? To our surprise INFIX notations are not as simple as they seem specially while evaluating them. To evaluate an infix expression we need to consider Operators Priority and Associative property For example expression 3+5*4 evaluate to 32 i.e. (3+5)*4 or to 23 i.e. 3+(5*4). To solve this problem Precedence or Priority of the operators were defined. Operator precedence governs evaluation order. An operator with higher precedence is applied before an operator with lower precedence. Need operator priorities, tie breaker, and delimiters. This makes computer evaluation more difficult than is necessary. Postfix and prefix expression forms do not rely on operator priorities, a tie breaker, or delimiters. So it is easier to evaluate expressions that are in these forms. 7
8 INFIX TO POSTFIX (RPN) ALGORITHM 1. Examine the next element in the input. 2. If it is operand, output it. 3. If it is opening parenthesis, push it on stack. 4. If it is an operator, then 1. If stack is empty, push operator on stack. 2. If the top of stack is opening parenthesis, push operator on stack 3. If it has higher priority than the top of stack, push operator on stack. 4. Else pop the operator from the stack and output it, repeat step 4 5. If it is a closing parenthesis, pop operators from stack and output them until an opening parenthesis is encountered. pop and discard the opening parenthesis. 6. If there is more input go to step 1 7. If there is no more input, pop the remaining operators to output. 8
9 EXAMPLE INFIX TO POSTFIX Let the incoming the Infix expression be: A * (B + C) D / E Stage 1: Stack is empty and we only have the Infix Expression. Infix Notation A * (B + C) D / E Postfix Notation STACK NULL 9
10 EXAMPLE INFIX TO POSTFIX Let the incoming the Infix expression be: A * (B + C) D / E Stage 2: First token is Operand A and is Appended to Output as it is. Infix Notation * (B + C) D / E NULL Postfix Notation A STACK 10
11 EXAMPLE INFIX TO POSTFIX Let the incoming the Infix expression be: A * (B + C) D / E Stage 3: Next token is * and it is pushed into the Stack Infix Notation (B + C) D / E NULL * Postfix Notation A STACK 11
12 EXAMPLE INFIX TO POSTFIX Let the incoming the Infix expression be: A * (B + C) D / E Stage 4: Next token is ( and it is pushed into the Stack as well Infix Notation B + C) D / E NULL ( * Postfix Notation A STACK 12
13 EXAMPLE INFIX TO POSTFIX Let the incoming the Infix expression be: A * (B + C) D / E Stage 5: Next token is B and will go to output expression as it is Infix Notation + C) D / E NULL ( * Postfix Notation A B STACK 13
14 EXAMPLE INFIX TO POSTFIX Let the incoming the Infix expression be: A * (B + C) D / E Stage 6: Next token is + and is pushed in stack. NULL + ( * Infix Notation C ) D / E Postfix Notation A B STACK 14
15 EXAMPLE INFIX TO POSTFIX Let the incoming the Infix expression be: A * (B + C) D / E Stage 7: Next token is C and is appended to output NULL + ( * Infix Notation ) D / E Postfix Notation A B C STACK 15
16 EXAMPLE INFIX TO POSTFIX Let the incoming the Infix expression be: A * (B + C) D / E Stage 8: Next token is ) pop all elements and append to output Infix Notation D / E NULL * Postfix Notation A B C + STACK 16
17 EXAMPLE INFIX TO POSTFIX Let the incoming the Infix expression be: A * (B + C) D / E Stage 9: Next token is . Precedence of * is high. So Pop * and Push  Infix Notation D / E NULL  Postfix Notation A B C + * STACK 17
18 EXAMPLE INFIX TO POSTFIX Let the incoming the Infix expression be: A * (B + C) D / E Stage 10: Next token is D. Append to Output Infix Notation / E NULL  Postfix Notation A B C+ * D STACK 18
19 EXAMPLE INFIX TO POSTFIX Let the incoming the Infix expression be: A * (B + C) D / E Stage 11: Next token is /. Precedence of / is high, so Push / in stack. Infix Notation E NULL /  Postfix Notation A B C + * D STACK 19
20 EXAMPLE INFIX TO POSTFIX Let the incoming the Infix expression be: A * (B + C) D / E Stage 12: Last token is E. Append to Output Infix Notation NULL /  Postfix Notation A B C + * D E STACK 20
21 EXAMPLE INFIX TO POSTFIX Let the incoming the Infix expression be: A * (B + C) D / E Stage 13: Input Expression Complete. Pop all to complete Postfix Infix Notation NULL Postfix Notation A B C + * D E /  STACK 21
22 RPN USING STACK REVERSEPOLISH NOTATION The easiest way to parse reversepolish notation is to use an operand stack: operands are processed by pushing them onto the stack when processing an operator: pop the last two items off the operand stack, perform the operation, and push the result back onto the stack 22
23 RPN EXAMPLE REVERSEPOLISH NOTATION Evaluate the following reversepolish expression using a stack:
24 RPN EXAMPLE REVERSEPOLISH NOTATION Push 1 onto the stack
25 RPN EXAMPLE REVERSEPOLISH NOTATION Push 2 onto the stack
26 RPN EXAMPLE REVERSEPOLISH NOTATION Push 3 onto the stack
27 RPN EXAMPLE REVERSEPOLISH NOTATION Pop 3 and 2 and push onto the stack
28 RPN EXAMPLE REVERSEPOLISH NOTATION Push 4 onto the stack
29 RPN EXAMPLE REVERSEPOLISH NOTATION Push 5 onto the stack
30 RPN EXAMPLE REVERSEPOLISH NOTATION Push 6 onto the stack
31 RPN EXAMPLE REVERSEPOLISH NOTATION Pop 6 and 5 and Push 5 x 6 onto the stack
32 RPN EXAMPLE REVERSEPOLISH NOTATION Pop 30 and 4 and Push 430 onto the stack
33 RPN EXAMPLE REVERSEPOLISH NOTATION Push 7 onto the stack
34 RPN EXAMPLE REVERSEPOLISH NOTATION Pop 7 and 26 and Push 26 x 7 onto the stack
35 RPN EXAMPLE REVERSEPOLISH NOTATION Pop 182 and 5 and Push onto the stack
36 RPN EXAMPLE REVERSEPOLISH NOTATION Pop 117 and 1 and Push 1 ( 177 ) onto the stack
37 RPN EXAMPLE REVERSEPOLISH NOTATION Push 8 onto the stack
38 RPN EXAMPLE REVERSEPOLISH NOTATION Push 9 onto the stack
39 RPN EXAMPLE REVERSEPOLISH NOTATION Pop 9 and 8 and Push 8 x 9 onto the stack
40 RPN EXAMPLE REVERSEPOLISH NOTATION Pop 72 and 178 and Push onto the stack
41 RPN REVERSEPOLISH NOTATION Thus evaluates to the value on the top: 250 The equivalent infix notation is ((1 ((2 + 3) + ((4 (5 6)) 7))) + (8 9)) We reduce the parentheses using orderofoperations: 1 ( (4 5 6) 7) Incidentally, = 132 which has the reversepolish notation of
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