Will Any Current 100M Digit LL Tests Finish? - Page 6

CRGreathouse · 2008-12-22, 18:13

Quote:

Originally Posted by jinydu

Well I've come up with a more elegant "coordinate system". See posts 50 and 52.

Yes, but I don't see how you get t2 = 0.

jinydu · 2008-12-22, 20:30

Quote:

Originally Posted by CRGreathouse

Yes, but I don't see how you get t2 = 0.

Well we want to choose $t_2$ so that $t_2+e^{-t_2}$ is minimized. Differentiating both sides gives $1-e^{-t_2}$ , which is negative when $t_2<0$ and positive when $t_2>0$ . Thus the minimum occurs at $t_2=0$ .

lpmurray · 2008-12-22, 21:09

when I started working on my 100million digit number it was running on my duel 2.4ghz server. The software only allowed me to use 1 processor and it ran at 2.2sec per itt. Then when I could use both processors it went down to 1.1 sec per itt. and now that its on my duel core laptop its running about .38 per itt when nothing else is running. When I started this thing it had a completion time sometime around 2020 now its down to 2012, next year when I get my quad-core I figure it will drop down to late 2010. So upgrading does work I can't wait to start my next number a billion digit if prime 95 will handle it. the sooner it gets started the sooner it gets done

CRGreathouse · 2008-12-22, 22:17

Quote:

Originally Posted by jinydu

Well we want to choose $t_2$ so that $t_2+e^{-t_2}$ is minimized. Differentiating both sides gives $1-e^{-t_2}$ , which is negative when $t_2<0$ and positive when $t_2>0$ . Thus the minimum occurs at $t_2=0$ .

But you need to minimize subject to both variables, not just to t_2.

jinydu · 2008-12-23, 01:04

Quote:

Originally Posted by CRGreathouse

But you need to minimize subject to both variables, not just to t_2.

The expression to be minimized is $t_2+e^{-t_2}-e^{-u}u$ , where $t_2$ can be any real number and $u$ can be any non-negative number. The minimum is attained when $t_2+e^{-t_2}$ and $-e^{-u}u$ are minimized independently.

More generally, if x and y can be chosen independently, then any expression of the form f(x)+g(y) can be minimized by choosing x so as to minimize f(x) and choosing y so as to minimize g(y).

CRGreathouse · 2008-12-23, 01:22

Fine, whatever, but your answer is still wrong. You gave (t1, t2) = (-1, 0), or converting back to normal time with Moore's Law, about 26 months before now and now, respectively, for the computer and the upgrade. But for a calculation taking 80 years with present technology, that upgrade path will take ~78 years to finish. Waiting 4 and 6 years for the initial order and upgrade would take ~10 years to finish.

fivemack · 2008-12-23, 03:16

CRGreathouse: You're not converting back to conventional units of time correctly; t=0 is not the present moment.

Quote:

Taking a hint from the N = 1 case, let us use the time needed for computing speed to increase by a factor of e as our unit of time, and set our clock so that t = 0 a computer bought at t = 0 will finish the test when t = 1.

Under the Moore's-law assumption, the unit of time is 26 months (18/ln(2)), and jindyu's 't=0' for a calculation that takes eighty years today is quite a long time in the future (roughly 94 months).

So his answer translates as 'do nothing for 66 months, then buy a computer and run it for 26 months, then buy another computer and run that for 26 months to finish the job', which doesn't seem completely unreasonable.

starrynte · 2008-12-23, 03:54

the magical number 26 again!
so the generalized problem is:

Quote:

Given y computers you can upgrade to after your first purchase, wait x months until you can run an assignment such that, upgrading the computer every 26 months, the assignment completes in 26 months after the final upgrade.

?
(If x is negative, then you should start now, though I don't know when to upgrade in that case)

CRGreathouse · 2008-12-23, 04:04

Quote:

Originally Posted by fivemack

CRGreathouse: You're not converting back to conventional units of time correctly; t=0 is not the present moment.

So jinydu's post #57 was wrong, and the answer should have been "t2 = 0 is arbitrary"?

Quote:

Originally Posted by fivemack

Under the Moore's-law assumption, the unit of time is 26 months (18/ln(2)), and jindyu's 't=0' for a calculation that takes eighty years today is quite a long time in the future (roughly 94 months).

How do you do this conversion? (0 -> 94)

jinydu · 2008-12-23, 04:16

CRGreathouse, I think you still haven't understood my time-measurement system. Let's start from the beginning.

First, we get to choose what unit we use to measure time. Of course, we could always choose seconds, minutes, hours, days or years; but all of these are rather "unnatural" choices. Instead, look at what the problem gives us. We know that the speed of market computers grows exponentially. This means that the time required for the speed of a market computer to double (or triple, or quadruple, or grow by any constant factor) is constant. So a more natural choice of time unit is computer speed doubling time. But this is still not the best choice; as you may have learned in a math class, the most elegant base to use for exponentials is base e = 2.718281828... So let us define 1 time unit to be the time needed for the speed of market computers to rise by a factor of e.

Next, we choose a unit for computing work. Clearly, the most natural choice is to define 1 unit of computing work to be the amount of computation needed to do the given LL test. Thus, the speed of a computer in the natural units is just 1/(number of time units needed to do LL test).

Now every instant of time is naturally associated with a computing speed. We must choose an instant to call time 0. Evidently, the most natural choice is to define t = 0 to be the unique time at which a market computer has speed 1.

---------------------------------------------------------------------------------------------------------

To starrynte

I'm assuming that you can choose over all time when to buy the computers. So either you're at $t=-\infty$ (whatever that means), making plans, or you able to send a message back in time, instructing your past self on when to buy a computer. Unrealistic perhaps, but mathematically elegant.

Also, I haven't solved the general problem yet. I only claim to have done it in the case of 1 and 2 computers.

CRGreathouse · 2008-12-23, 04:44

Quote:

Originally Posted by jinydu

Let's start from the beginning.

First, we get to choose what unit we use to measure time. Of course, we could always choose seconds, minutes, hours, days or years; but all of these are rather "unnatural" choices. Instead, look at what the problem gives us. We know that the speed of market computers grows exponentially. This means that the time required for the speed of a market computer to double (or triple, or quadruple, or grow by any constant factor) is constant. So a more natural choice of time unit is computer speed doubling time. But this is still not the best choice; as you may have learned in a math class, the most elegant base to use for exponentials is base e = 2.718281828... So let us define 1 time unit to be the time needed for the speed of market computers to rise by a factor of e.

That's insulting.

Quote:

Originally Posted by jinydu

Clearly, the most natural choice is to define 1 unit of computing work to be the amount of computation needed to do the given LL test. Thus, the speed of a computer in the natural units is just 1/(number of time units needed to do LL test).

Thanks for that; I didn't recall seeing that earlier.

View Poll Results: Will Any Current 100M Digit LL Tests Finish?
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2008-12-23, 01:22	#61
CRGreathouse Aug 2006 3×1,993 Posts	Fine, whatever, but your answer is still wrong. You gave (t1, t2) = (-1, 0), or converting back to normal time with Moore's Law, about 26 months before now and now, respectively, for the computer and the upgrade. But for a calculation taking 80 years with present technology, that upgrade path will take ~78 years to finish. Waiting 4 and 6 years for the initial order and upgrade would take ~10 years to finish. Last fiddled with by CRGreathouse on 2008-12-23 at 01:22

2008-12-23, 04:16	#65
jinydu Dec 2003 Hopefully Near M48 11011011110₂ Posts	CRGreathouse, I think you still haven't understood my time-measurement system. Let's start from the beginning. First, we get to choose what unit we use to measure time. Of course, we could always choose seconds, minutes, hours, days or years; but all of these are rather "unnatural" choices. Instead, look at what the problem gives us. We know that the speed of market computers grows exponentially. This means that the time required for the speed of a market computer to double (or triple, or quadruple, or grow by any constant factor) is constant. So a more natural choice of time unit is computer speed doubling time. But this is still not the best choice; as you may have learned in a math class, the most elegant base to use for exponentials is base e = 2.718281828... So let us define 1 time unit to be the time needed for the speed of market computers to rise by a factor of e. Next, we choose a unit for computing work. Clearly, the most natural choice is to define 1 unit of computing work to be the amount of computation needed to do the given LL test. Thus, the speed of a computer in the natural units is just 1/(number of time units needed to do LL test). Now every instant of time is naturally associated with a computing speed. We must choose an instant to call time 0. Evidently, the most natural choice is to define t = 0 to be the unique time at which a market computer has speed 1. --------------------------------------------------------------------------------------------------------- To starrynte I'm assuming that you can choose over all time when to buy the computers. So either you're at $t=-\infty$ (whatever that means), making plans, or you able to send a message back in time, instructing your past self on when to buy a computer. Unrealistic perhaps, but mathematically elegant. Also, I haven't solved the general problem yet. I only claim to have done it in the case of 1 and 2 computers. Last fiddled with by jinydu on 2008-12-23 at 04:24

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2008-12-22, 21:09	#58
lpmurray Sep 2002 89 Posts	when I started working on my 100million digit number it was running on my duel 2.4ghz server. The software only allowed me to use 1 processor and it ran at 2.2sec per itt. Then when I could use both processors it went down to 1.1 sec per itt. and now that its on my duel core laptop its running about .38 per itt when nothing else is running. When I started this thing it had a completion time sometime around 2020 now its down to 2012, next year when I get my quad-core I figure it will drop down to late 2010. So upgrading does work I can't wait to start my next number a billion digit if prime 95 will handle it. the sooner it gets started the sooner it gets done