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iOS find my point if inside a circle.
- Thread starter chhoda
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Sounds like you need the distance formula.
Go to http://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.
Go to http://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.
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.
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.
No. But I am familiar with A^2 + B^2 = C^2.
I'd think something like:
Code:
if (sqrt(b*b + c*c) <= 50.0)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.
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.
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.
I made a quick app that tests which is faster, this is the output:
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:
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 369212So 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);Cool, that was my intuition from taking a numerical analysis course long ago, but thanks for demonstrating it.
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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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