# Traversing all possible combinations

Vladas Razas

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posted 11 years ago

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

Anybody could tell me such algorythm?

Thank you

Vladas Razas

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Posts: 385

Jayesh Lalwani

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Vladas Razas

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Posts: 385

posted 11 years ago

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

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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Ryan McGuire

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Stan James

(instanceof Sidekick)

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posted 11 years ago

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?

A good question is never answered. It is not a bolt to be tightened into place but a seed to be planted and to bear more seed toward the hope of greening the landscape of the idea. John Ciardi

Ryan McGuire

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posted 11 years ago

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.

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

(instanceof Sidekick)

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posted 11 years ago
A good question is never answered. It is not a bolt to be tightened into place but a seed to be planted and to bear more seed toward the hope of greening the landscape of the idea. John Ciardi

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.