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Traversing all possible combinations

 
Vladas Razas
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We have multiple points in 2D space described by their X and Y coords. I have to program an algorythm which would display all possible squares defined by 4 points combination. As I understand it comes to making all possible 4 member combinations of out of number of entities.

Anybody could tell me such algorythm?
Thank you
 
Vladas Razas
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I forgot to mention that all points are stored in DB... Pershaps I will have to load them all into memory (otherwise I would need professional mathematician help).
 
Henry Wong
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The easiest way in "making all possible 4 member combinations" is probably to do it recursively.

Henry
 
Jayesh Lalwani
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Do you want squares or do you want any 4 sided figures?

Possibly, not the most efficient way of doing this, but have you thought about doing a self-join.. 4 times. If the table is small enough, the database could probably do it in memory. Don't flame me.. just a suggestion
 
Vladas Razas
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3.1.Business requirements
Initial data set consists of random set of distinct points on a plane. Each point is defined using integer x, y coordinates from the range [0..100]. Candidate task is to find all squares that these points form on the plane. Result should be presented as list of squares (as coordinates of vertex points) as well as graphically. Example of graphical presentation is shown in picture below.

http://putfile.com/pic.php?pic=10/27803104929.jpg&s=x7

User can perform following actions from the web page.
User can see a list of points (as x, y coordinates).
User can add new point.
User can delete selected point.
User can execute task that finds all squares and displays results.


I guess that means squares - rectangles with all sides equal? At least webster defines "square" this way. Sorry english not my native language. Having that in mind this must not be just all 4-point combinations
 
Vladas Razas
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But still that might be useful to have algorythm that would allow to iterate 4 point combinations and then check if they are rectangles
 
Ryan McGuire
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[ October 06, 2005: Message edited by: Ryan McGuire ]
 
Stan James
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For any two points you get a line and for any line there are only two possible squares. Would it be faster to take each pair and see if there are squares around them than to take each quad and see if it's a square?
 
Ryan McGuire
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Originally posted by Stan James:
For any two points you get a line and for any line there are only two possible squares. Would it be faster to take each pair and see if there are squares around them than to take each quad and see if it's a square?


Myabe but...
The number of points in the example given was only 18. Even if that were doubled to 36, the brute force algorithm with four nested loops will still execute pretty darn quick.

Besides, taking each pair of points, determining the possible squares, and then searching the array of points to see if the other two points are present still involves at least three nested loops.

You might get O(n^3), while four nested loops achieves O(n^4), but the simpler overhead of the n^4 algorithm may be faster for small values of n. It depends on what executed in your loops.
 
Stan James
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With enough points it might get to be worth while to index or sort them and look for squares, but for this number I agree, just try em all.
 
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