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Thread: Big numbers and speed
 01172014, 03:17 AM #1Señor Member
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Big numbers and speed
Java Code:class Hmwk1 { public static void main ( String[] args ) { long counter = 0L; long max = 1000000000000L; for(long y = 1L; y <= max; y++) { for(long x = 1L; x < y; x++) { if((x * y) % (x + y) == 0) { counter++; if(counter % 10000 == 0) { System.out.println("Counter: " + counter + ". y = " + y + "."); } } } } System.out.println("Counter: " + counter); } }
Thanks
EDIT: I'll clarify, for anyone who isn't clear on what the problem is. I've been running this program for nearly 45 minutes now, and y = 550,000. y will need to go all the way up to 10^12...and since x has to count up all the way to y, the bigger the y value, the more xs need to be generated  the speed is going to get exponentially slower.Last edited by AlexGraal; 01172014 at 03:43 AM.
 01172014, 03:20 AM #2Senior Member
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Re: Big numbers and speed
you could try using a timer and when ever it ticks add to x until it reaches a certain number then add to the y and then back to the x
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 01172014, 03:29 AM #3Señor Member
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Re: Big numbers and speed
Ok maybe this is just my lack of experience with java, but I fail to see how that is going to speed anything up.
Does a forloop run slower than a timer can tick? I don't really see how that helps me program to run faster...
 01172014, 03:32 AM #4Senior Member
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 01172014, 03:42 AM #5Señor Member
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Re: Big numbers and speed
Explaining the purpose fully takes an unnecessary amount of typing for me to explain. Long story short, I need to check that x + y will divide evenly into x*y, where x < y <= max, which for this problem, happens to be 10^12, or 1000000000000. That is the problem. I see no other way to approach this...
 01172014, 04:40 AM #6Senior Member
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Re: Big numbers and speed
You will never see your program run to completion. The inner loop goes 1 time, then 2 times, then 3 times, up to 10^12 times. So it will run n(n+1)/2 times where n is 10^12. So that's about (10^24)/2 or 5 x 10^23. Assuming a trillion iterations per second (to simplify the math cause it ain't gonna happen), that would be .5 trillion seconds or about 15,000 years. I suggest you get a book on algebra and perhaps number theory and start reading. In the meantime, I will think about this. Of course, my above analysis could be off.
Note: Since an odd number is not divisible by an even number you may be able to speed up your loops by changing the increment. But that alone is not sufficient. You simply need a better algorithm (it's possible the solutions could be derived too). And one more tidbit while I'm thinking of it. A long holds a 63 bit positive number. Multiplying by .30103 (log 2 base 10) you get 18.9 or 19 digits. So your counter will increment into oblivion long before your loops finish.
Regards,
JimLast edited by jim829; 01172014 at 10:31 PM.
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 01172014, 04:57 AM #7Señor Member
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