Re^9: Number functions.. primes (ugly code runs faster)

Here are some of the problems.

Some made assumptions I knew the regex was slowing down my code. I had no idea the regex was the problem.
Some made assumptions of my prowess in math. If it isn't addition, subtraction, multiplication, or division; I won't think to use it. So, using sqrt is confusing.
Only aaron_baugher commented directly on my code and showed me where I overdid it with the check for divisibility by 9. Everyone else wrote their own versions without commenting on the code I provided first.
No one commented their code! I'm still stuck back at choroba's first prime example at the square root test. Adding the use of square roots still bothers me for some reason.

Like with many car shows my hubby watches where the show glosses over a lot, it is like you took the logo off the car, trashed the car, and put the logo on another car. I still don't know what is so wrong with the original car (function) that the only thing worth saving was the logo (function name). So take one part out at a time, explain why it is broken, then install the new part telling the audience why the new part is better. Pretty please?

No matter how hysterical I get, my problems are not time sensitive. So, relax, have a cookie, and a very nice day!

Lady Aleena

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Re^10: Number functions.. primes (ugly code runs faster) by trippledubs (Deacon) on Apr 08, 2015 at 20:09 UTC
I also struggle with Math!! A number cannot be prime if it is a perfect square. 16 is perfect square of 4. 16 not prime. A number is only prime if two other integers cannot be multiplied together to equal that number. Take 37, sqrt = 6.08 So as you iterate over the numbers.. 37 % 2 == 0, nope. Another way to say is that 2 * something = 37 with no remainder 37 % 3 == 0, nope. Another way to say is that 3 * something = 37 with no remainder 37 % 4 == 0, nope. Another way to say is that 4 * something = 37 with no remainder 37 % 5 == 0, nope. Another way to say is that 5 * something = 37 with no remainder 37 % 6 == 0, nope. Another way to say is that 6 * something = 37 with no remainder Our next number would be 7. But we know that 6.08 times any number smaller than 6.08 is < 37 and 6.08 times any number larger than 6.08 is > 37 So, in our next iteration, 7 * something = 37 with no remainder, for that something to multiply 7 and make 37, would have to be smaller than 6.08, and we already tested all those. Does that sound right?	[reply]
Re^10: Number functions.. primes (ugly code runs faster) by choroba (Cardinal) on Apr 09, 2015 at 08:47 UTC
I tried to explain the usage of sqrt in Re^3: Number functions I have lying around. لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ	[reply]