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Need code that adds brackets to multi-leveled expressions (group by indentation levels)

Posted on 2004-04-18
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Last Modified: 2010-04-17
I need a simple and elegant algorithm that will add properly
positioned brackets between 2 or more multi-leveled expressions.
The diagram below shows exactly what I am looking for:

!! NOTE: VIEW IN FIXED FONT SUCH AS COURIER NEW!!

=============================================================
8 POSSIBLE SCENARIOS FOR 3 EXPRESSIONS USING UP TO 2 LEVELS
=============================================================
               (1)   (2)   (3)   (4)   (5)   (6)   (7)   (8)
              +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+
GROUP LEVEL:  |1|2| |1|2| |1|2| |1|2| |1|2| |1|2| |1|2| |1|2|
              +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+
EXPRESION A:  |A| | |A| | |A| | |A| | | |A| | |A| | |A| | |A|
              +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+
EXPRESION B:  |B| | |B| | | |B| | |B| |B| | |B| | | |B| | |B|
              +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+
EXPRESION C:  |C| | | |C| |C| | | |C| |C| | | |C| |C| | | |C|
              +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+
=============================================================
BRACKET SOLUTIONS FOR THE 8 SCENARIOS SHOWN ABOVE
=============================================================
(1)       A  +  B  +  C
(2)       A  +  B  + (C)
(3)       A  + (B) +  C
(4)       A  + (B  +  C)
(5)      (A) +  B  +  C
(6)      (A) +  B  + (C)
(7)      (A  +  B) +  C
(8)      (A  +  B  +  C)

Things become more complicated as we add more levels and
expressions. Is there an elegant approach?.



Thanks


0
Comment
Question by:const71
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6 Comments
 
LVL 15

Expert Comment

by:Tommy Hui
ID: 10856313
This can be thought of as a mapping between the parentheses and binary numbers.

In your sample with 8 possibilities, the possible combinations are equal to 2^3 (two levels of three expressions).

You can also think of your expressions as binary numbers:

1. 000
2. 001
3. 010
4. 011
5. 100
6. 101
7. 110
8. 111

The pattern is everywhere there are sequences of 1, you'll need to put parentheses around things.
0
 

Author Comment

by:const71
ID: 10856318
I know that, but i need code that does this for me. I have some Visual Basic code as a starting point if that will help...
0
 

Author Comment

by:const71
ID: 10856360
Heres the code that I have so far

http://www3.sympatico.ca/cnterekas/EXPRESSION.ZIP


Thanks
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LVL 2

Accepted Solution

by:
sitbon earned 500 total points
ID: 10856928
Hi,

I didn't look at your code yet, but you need tp write a simple expression parser, just for the reason that adding even one more little thing will ruin the code if you don't. What you probably should do is start with a string that has a line of code in it. Strip everything but the characters (ie remove spaces and newlines). The easiest solution is a recursive one:

function evaluate( left, right, operator )
{
    if left has a " ( " then
        find first " ) " from the end (of hte left part), split in between
        left = evaluate( left->left, left->right, left->op )

    return operator->do_something(left,right)
}

This probably isn't the speediest, but you're probably not worried about performance (after all, you *are* using VB lol) so this is the simplest solution for you.

So for the expression

        A +  ( B + C )

the logic would be:

      evaluate( "A", "(B+C)","+") ->

          return add(  "A", add("B","C") )
                                    ^---- evaluate("B+C") -> return add->do_something("B","C")

and so on.

good luck! look at some scripting engine source code... should help.
0
 
LVL 2

Expert Comment

by:sitbon
ID: 10856964
>>    if left has a " ( " then
>>        find first " ) " from the end (of hte left part), split in between
>>        left = evaluate( left->left, left->right, left->op )

sorry, I should indicate that you need to do the same for the right side :)  This forms an expression tree:

X = A + ( B + (C + D) )

 Gives
 
        X
       / \
     /     \
   /         \
  A          +
             /  \
           /      \
         /          \
        B           +
                    /  \
                  /      \
                /          \
               C           D


as you can see, each node either has 2 children or none - nodes with two children are operators and nodes with none are operands. This binary tree allows for some efficient login operations, so you don't have to limit yourself to recursion. If it were me, I'd use a stack.
0
 

Author Comment

by:const71
ID: 10858646
great!  thanks
0

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