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- Thread starter chhoda
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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.

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.Are you familiar with

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

I'd think something like: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.

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.A^2 + B^2 = C^2[/URL].

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.I'd think something like:

would be more clear.Code:`if (sqrt(b*b + c*c) <= 50.0)`

Math is meaningless without programming.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.

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

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

I don't know. Do you?However, wouldn't a sqrt function be rather expensive compared to an extra multiplication?

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
```

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.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.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`

----------

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

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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