Powers vs Multiplying
BlitzMax Forums/BlitzMax Programming/Powers vs Multiplying
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I just ran some test..seems Powers are a bit on the slow side. They seem to remain constant in execution regardless of the power size. I would think powers would be more optimized than this.Vari1 = 2 Local Counter1, Counter2 Local PH Local EndTime StartMilli = MilliSecs() EndTime = StartMilli + 1000 Repeat Counter2:+1 PH = Vari1^10 Until EndTime < MilliSecs() Print "^" + Counter2 StartMilli = MilliSecs() EndTime = StartMilli + 1000 Repeat Counter1:+1 PH = Vari1 * Vari1 * Vari1 * Vari1 * Vari1 * Vari1 * Vari1 * Vari1 * Vari1 * Vari1 Until EndTime < MilliSecs() Print "*" + Counter1 Delay 100 Print Counter1/Float(Counter2) End |
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| Problem is: Did you ever try to create a power of 2.72343 in your way? ;-) It ends up constant becase a^b = a * ln(b) on which ln is the worse and calculation intense part. If you defined your variables correctly (with type not this sluggish B3D style), you could have used :float which would be faster, because yours uses int which ends with double calculation. Gives more accurate results but takes its time. I always use * if I know the power and if the power is not > 4 (got used to that in B3D where the multiplication is faster for a^x with x < 4) |
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| Well, for one thing... a^b = a * ln(b) I don't think that's correct. For another thing, Pow takes into account fractional numbers and not just integers, so it's more complex than simply multiplying a number by itself n times, hence why it takes longer. Just thought I'd add that in case you didn't understand Dreamora's gobbledegook. |
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| What do you think is not correct? The equation itself is. The actual implementation might differ, since ln is the slowest mathematical operation with non-matrices. but still the implementation bases on this and uses IEEE optimations which use the restriction correctness of the types to gain in speed (as SQRT does as well) |
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| So you're saying that 5^2.5 = 5*ln(2.5), right? If so, my calculator disagrees. |
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| 5^2.5 = e^(ln(5^2.5)) = e^(2,5 * ln(5)) = 5 * e^2.5 oops yeah you are right, was wrong, mixed it vice versa :-( |
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| . There is no reason why the compiler can't replace x^c with ( x * x * ... x ) for sufficiently small values of c. Using multiplication is O( n ) and calculating with exp and ln is constant, but a very large one. |
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| Um, no, it's still wrong. You can't jump from exp( 2.5 * ln( 5 ) ) -which is right to 5 * exp( 2.5 ) -which is wrong |
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| narv yeah its monday morning and boring I stop telling strange stuff thats crap. |
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