long story short my lecturer is crap, and was showing us infix to prefix stacks via an overhead projector and his bigass shadow was blocking everything so i missed the important stuff
he was referring to push and pop, push = 0 pop = x
he gave an example but i cant see how he gets his answer at all,
2*3/(2-1)+5*(4-1)
step 1 Reverse : )1-4(*5+)1-2(/3*2 ok i can see that
he then went on writing x's and o's operations and i got totally lost
answer 14-5*12-32*/+ then reversed again to get +/*23-21*5-41
if some one could explain to me the push pop so i could understand i would be very greatful, i have looked online but alot stuff im finding seems to be a step above this, so i really need to get an understanding here first
This question is related to
stack
terminology
infix-notation
after all these good examples adam shankman still can't make sense of it. I think you should open up some code and try it. The second you try a myStack.Push(1) and myStack.Pop(1) you really should get the picture. But by the looks of it, even that will be a challenge for you!
Hopefully this will help you visualize a Stack, and how it works.
Empty Stack:
| |
| |
| |
-------
After Pushing A, you get:
| |
| |
| A |
-------
After Pushing B, you get:
| |
| B |
| A |
-------
After Popping, you get:
| |
| |
| A |
-------
After Pushing C, you get:
| |
| C |
| A |
-------
After Popping, you get:
| |
| |
| A |
-------
After Popping, you get:
| |
| |
| |
-------
A Stack is a LIFO (Last In First Out) data structure. The push and pop operations are simple. Push puts something on the stack, pop takes something off. You put onto the top, and take off the top, to preserve the LIFO order.
edit -- corrected from FIFO, to LIFO. Facepalm!
to illustrate, you start with a blank stack
|
then you push 'x'
| 'x'
then you push 'y'
| 'x' 'y'
then you pop
| 'x'
A stack in principle is quite simple: imagine a rifle's clip - You can only access the topmost bullet - taking it out is called "pop", inserting a new one is called "push".
A very useful example for that is for applications that allow you to "undo".
Imagine you save each state of the application in a stack. e.g. the state of the application after every type the user makes.
Now when the user presses "undo" you just "pop" the previous state from the stack. For every action the user does - you "push" the new state to the stack (that's of course simplified).
About what your lecturer specifically was doing - in order to explain it some more information would be helpful..
The rifle clip analogy posted by Oren A is pretty good, but I'll try another one and try to anticipate what the instructor was trying to get across.
A stack, as it's name suggests is an arrangement of "things" that has:
(think of it as a literal stack of books on your desk and you can only take something from the top)
Pushing something on the stack means "placing it on top". Popping something from the stack means "taking the top 'thing'" off the stack.
A simple usage is for reversing the order of words. Say I want to reverse the word: "popcorn". I push each letter from left to right (all 7 letters), and then pop 7 letters and they'll end up in reverse order. It looks like this was what he was doing with those expressions.
push(p) push(o) push(p) push(c) push(o) push(r) push(n)
after pushing the entire word, the stack looks like:
| n | <- top
| r |
| o |
| c |
| p |
| o |
| p | <- bottom (first "thing" pushed on an empty stack)
======
when I pop() seven times, I get the letters in this order:
n,r,o,c,p,o,p
conversion of infix/postfix/prefix is a pathological example in computer science when teaching stacks:
Post fix conversion to an infix expression is pretty straight forward:
(scan expression from left to right)
So if we have 53+2* we can convert that to infix in the following steps:
*When you reach the end of the expression, if it was formed correctly you stack should only contain one item.
By introducing 'x' and 'o' he may have been using them as temporary holders for the left and right operands of an infix expression: x + o, x - o, etc. (or order of x,o reversed).
There's a nice write up on wikipedia as well. I've left my answer as a wiki incase I've botched up any ordering of expressions.
Ok. As the other answerers explained, a stack is a last-in, first-out data structure. You add an element to the top of the stack with a Push operation. You take an element off the top with a Pop operation. The elements are removed in reverse order to the order they were put inserted (hence Last In, First Out). For example, if you push the elments 1,2,3 in that order, the number 3 will be at the top of the stack. A Pop operation will remove it (it was the last in) and leave 2 at the top of the stack.
Regarding the rest of the lecture, the lecturer tried to describe a stack-based machine that evaluates arithmetic expressions. The machine operates by continuously popping 3 elements from the top of the stack. The first two elements are operands and the third is an operator (+, -, *, /). It then applies this operator on the operands, and pushes the result onto the stack. The process continues until there is only one element on the stack, which is the value of the expression.
So, suppose we begin by pushing the values "+/*23-21*5-41" in left-to-right order onto the stack. We then pop 3 elements from the top. The last in is first out, which means the first 3 element are "1", "4", and "-" in that order. We push the number 3 (the result of 4-1) onto the stack, then pop the three topmost elements: 3, 5, *. Push the result, 15, onto the stack, and so on.
The algorithm to go from infix to prefix expressions is:
-reverse input
TOS = top of stack
If next symbol is:
- an operand -> output it
- an operator ->
while TOS is an operator of higher priority -> pop and output TOS
push symbol
- a closing parenthesis -> push it
- an opening parenthesis -> pop and output TOS until TOS is matching
parenthesis, then pop and discard TOS.
-reverse output
So your example goes something like (x PUSH, o POP):
2*3/(2-1)+5*(4-1)
)1-4(*5+)1-2(/3*2
Next
Symbol Stack Output
) x )
1 ) 1
- x )- 1
4 )- 14
( o ) 14-
o 14-
* x * 14-
5 * 14-5
+ o 14-5*
x + 14-5*
) x +) 14-5*
1 +) 14-5*1
- x +)- 14-5*1
2 +)- 14-5*12
( o +) 14-5*12-
o + 14-5*12-
/ x +/ 14-5*12-
3 +/ 14-5*12-3
* x +/* 14-5*12-3
2 +/* 14-5*12-32
o +/ 14-5*12-32*
o + 14-5*12-32*/
o 14-5*12-32*/+
+/*23-21*5-41
Simply:
pop: returns the item at the top then remove it from the stack
push: add an item onto the top of the stack.
Source: Stackoverflow.com