iOS find my point if inside a circle.

chhoda

macrumors 6502
Original poster
Oct 25, 2008
285
1
I have two cgpoints say pt1 pt2

pt2 is center of a circle.

how can i find if pt1 is within the perifery of a 50px radius circle centered at pt2 ?

thanks in advance
ch
 

jonnymo5

macrumors 6502
Jan 21, 2008
279
0
Texas
Sounds like you need the distance formula.

Go to www.google.com and type this into the search bar. "what is the distance between two points"

If your result is greater than your radius, your point is not in your circle.
 

ArtOfWarfare

macrumors G3
Nov 26, 2007
8,579
4,019
Are you familiar with

A^2 = B^2 + C^2?

A is 50
B is the difference in X's between your two points.
C is the difference in Y's between your two points.

The formula defines the boundary of your point. So you'd actually want to use
A^2 > B^2 + C^2
to find out if your two points were within 50 of each other.
 

mobilehaathi

macrumors G3
Aug 19, 2008
9,351
6,219
The Anthropocene
I'd think something like:
Code:
if (sqrt(b*b + c*c) <= 50.0)
would be more clear.
More clear, yes, and hopefully the OP understands now. However, wouldn't a sqrt function be rather expensive compared to an extra multiplication? Of course that only matters if the check is happening frequently.
 

ArtOfWarfare

macrumors G3
Nov 26, 2007
8,579
4,019
This is probably one of the first times i see this formula being applied and used in real life, i always thought its just another pointless topic that the school forces you to learn.
Math is meaningless without programming.

(Ok, fine, the other sciences give meaning to math as well...)

Edit: And math team. Somehow math team gives math purpose.
 

ArtOfWarfare

macrumors G3
Nov 26, 2007
8,579
4,019
I made a quick app that tests which is faster, this is the output:
Code:
2012-01-30 13:13:53.915 SquareRootCost3[22875:f803] Beginning Test with squares.
2012-01-30 13:13:53.941 SquareRootCost3[22875:f803] Finished test with squares - found 369212
2012-01-30 13:13:53.941 SquareRootCost3[22875:f803] Beggining test with roots.
2012-01-30 13:13:53.954 SquareRootCost3[22875:f803] Finished test with roots - found 369212
So the squares took 0.026 seconds while the roots took 0.013 seconds... it appears to take half as long to use squares. Odd.

My code, for anyone interested, is this:
Code:
    srand(time(NULL));
    
    int i = 0;
    float a[500000], b[500000], c[500000];
    for (i = 0; i < 500000; i++)
    {
        a[i] = rand();
        b[i] = rand();
        c[i] = rand();
    }
    
    long int squareans = 0;
    long int rootans = 0;
    
    NSLog(@"Beginning Test with squares.");
    for (int i = 0; i < 500000; i++)
    {
        if (a[i]*a[i] < (b[i]*b[i] + c[i]*c[i]))
        {
            squareans++;
        }
    }
    
    NSLog(@"Finished test with squares - found %ld", squareans);
    NSLog(@"Beggining test with roots.");
    for (int i = 0; i < 500000; i++)
    {
        if (a[i] < sqrtf(b[i]*b[i] + c[i]*c[i]))
        {
            rootans++;
        }
    }
    
    NSLog(@"Finished test with roots - found %ld", rootans);
 

mobilehaathi

macrumors G3
Aug 19, 2008
9,351
6,219
The Anthropocene
I made a quick app that tests which is faster, this is the output:
Code:
2012-01-30 13:13:53.915 SquareRootCost3[22875:f803] Beginning Test with squares.
2012-01-30 13:13:53.941 SquareRootCost3[22875:f803] Finished test with squares - found 369212
2012-01-30 13:13:53.941 SquareRootCost3[22875:f803] Beggining test with roots.
2012-01-30 13:13:53.954 SquareRootCost3[22875:f803] Finished test with roots - found 369212
So the squares took 0.026 seconds while the roots took 0.013 seconds... it appears to take half as long to use squares. Odd.
Cool, that was my intuition from taking a numerical analysis course long ago, but thanks for demonstrating it. :D

----------

sqrt() can probably take advantage of bitwise operators.
Hmm, it will probably change based on implementation. What is the difference between sqrt() and sqrtf()?

Edit: Oh it is for floats. It'll probably be slower than one optimized for ints.

Edit2: Ok, not quite. sqrtf() explicitly takes floats, where as sqrt() is overloaded and can take floats or doubles (or other stuff in C++).
 
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