Stacks & QueuesInfix Calculator

CSC 172

SPRING 2004

LECTURE 13

Announcements

CIF – cool living space Social Engineering Quiz Project – issued today, due Friday 3/26

Read Weiss Chapter 11 Read about Trees

Weiss Chap 18

WORKSHOP LEADERSHIP

Consider being a CSC 171 workshop leader?Good grade in CSC 171?Belief in the workshop program?Seeking to develop your leadership skills?Pay and credit?

Interest ? “SW Eng & Advanced Java Programming”

The “game course” Multithreading, Graphics, UIs, SWeng

CSC 290 4 credits – normal grading – ok for CSC ULFall term 2004CSC 172 pre-req

Interest?

“Recreational Graphics”The “Maya” course – animation & ray tracingCSC 3901 or 2 credits – pass/fail – CSC credit, but not ULThursdays 4:50-6:05?No pre-reqs

A useful stack algorithm Postfix evaluation

We can rewrite the infix expression 1+2As the postfix expression 1 2 +

“Think” like a computer “load value ‘1’ into accumulator “load value ‘2’ into register A Add value in register A to value in accumulator

How about 1+2+3+4 ?How about 2*3+4?How about 2+3*4?

How to implement?

Can you write method that evaluates postfix expressions?

double postfixeval(Object[] items) Where objects in items[] are either

DoubleCharacter

Postfix evaluation using a stack

1. Make an empty stack

2. Read tokens until EOFa. If operand push onto stack

b. If operator i. Pop two stack values

ii. Perform binary operation

iii. Push result

3. At EOF, pop final result

Trace by hand

1 2 3 4 + + + 2 3 * 4 + 2 3 4 * +

Infix to postfix

1 + 2 * 3

== 7 (because multiplication has higher precedence)

10 – 4 – 3

== 3 (because subtraction proceeds left to right)

Infix to postfix

4 ^ 3 ^ 2== 262144!= 4096

234

234

Generally,

Rather than:

Precidence

A few simple rules:

( ) > ^ > * / > + -

Subtraction associates left-to-right

Exponentiation associates right to left

Infix Evaluation

1 – 2 – 4 ^ 5 * 3 * 6 / 7 ^ 2 ^ 2

== -8

(1 – 2) – ( ( ( ( 4 ^ 5) * 3) * 6) / (7 ^ ( 2 ^ 2 ) ) )

Could you write a program to evaluate stuff like this?

Postfix

If we expressed

(1 – 2) – ( ( ( ( 4 ^ 5) * 3) * 6) / (7 ^ ( 2 ^ 2 ) ) )

As

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

Then, we could use the postfix stack evaluator

Postfix evaluation using a stack

1. Make an empty stack

2. Read tokens until EOFa. If operand push onto stack

b. If operator i. Pop two stack values

ii. Perform binary operation

iii. Push result

3. At EOF, pop final result

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

1

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

21

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

-1

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

4-1

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

5 4-1

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

1024-1

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

31024

-1

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

3072-1

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

63072

-1

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

18432-1

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

718432

-1

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

27

18432-1

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

227

18432-1

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

47

18432-1

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

204118432

-1

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

7-1

1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

-8

But how to go from infix to postfix? Could you write a program to do it? What data structures would you use

Stack Queue

How about a simple case just using “+” 1+ 2 + 7 + 4 1 2 7 4 + + +

Operands send on to output? Operator push on stack? Pop ‘em all at the end?

More complex

2 ^ 5 – 1 == 2 5 ^ 1 –

Modify the simple rule?

If you are an operator, pop first, then push yourself?

1 + 2 + 7 + 4

1 2 + 7 + 4 + ok

Even more complex

3 * 2 ^ 5 - 1

3 2 5 ^ * 1 –

If you are an operator:

Pop if the top of the stack is higher precedence than

Infix to postfix Stack AlgorithmOperands : Immediately output

Close parenthesis: Pop stack until open parenthesis

Operators: 1. Pop all stack symbols until a symbol of lower

precedence (or a right-associative symbol of equal precedence) appears.

2. Push operator

EOF: pop all remaining stack symbols

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

1

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

-

1

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

-

1 2

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

^-

1 2

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

^-

1 2 3

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

^^-

1 2 3

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

^^-

1 2 3 3

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

-

1 2 3 3 ^ ^ -

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

(-

1 2 3 3 ^ ^ -

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

(-

1 2 3 3 ^ ^ - 4

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

+(-

1 2 3 3 ^ ^ - 4

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

+(-

1 2 3 3 ^ ^ - 4 5

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

*+(-

1 2 3 3 ^ ^ - 4 5

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

*+(-

1 2 3 3 ^ ^ - 4 5 6

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

-

1 2 3 3 ^ ^ - 4 5 6 * +

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

*-

1 2 3 3 ^ ^ - 4 5 6 * +

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

*-

1 2 3 3 ^ ^ - 4 5 6 * + 7

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

1 2 3 3 ^ ^ - 4 5 6 * + 7 * -

1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

1 2 3 3 ^ ^ - 4 5 6 * + 7 * -

((1 – (2 ^ (3 ^ 3))) – (( 4 + (5 * 6)) * 7))

InputTo evaluationstack