Site admin (and creator) Kestas can check the server logs. He may catch this thread or you can email him.

It's possible you're right; but every single other time someone has made a similar statement it turns out they screwed up their orders.

Have a look in the error log yourself.
There's a link to it at the bottom of the board.php page

Sorry, I meant the order logs. That will tell you what orders you input

I think what he's saying is that he's sure he didn't actually enter those orders. The orders do match up with what he describes as the error. He just doesn't really think he would have done that.

Personally - I think that user error makes a lot more sense than the PHP script that takes the values out of those dropdowns and translates them into code and put them into the database just happened to screw up this one time out of - what did Kestas say the other day - 2 BILLION orders.

yeah, there haven't been two billion orders input - they're just using identifiers that are in the billions I think

Actually, there have been 2 billion jacob. The old ones have been removed, and the data stored through a different method, but there have apparently been over 2000000000! (that's an exclamation mark, not factorial before anyone points it out)

haha 2,000,000,000 factorial would be HUGE and I mean HUGE. You just won't believe how vastly hugely mindbogglingly big it is. I mean you may think it's a long way down the road to the chemist, but that's just peanuts to 2B factorial

Yes. Since the main issue of the thread is solved, do you know of any way to work out the scale of said number? I can't think of a simple way to work out just what sort of field we're thinking!
10^10^10^10 is my first completely fact-less guess

It would have about 20,000,000,000 digits

Talking about big numbers, I read this a few months ago:
How big is a googol?
Imagine that the entire observable universe was filled completely with marbles. Now - take all of the marbles away and fill it again with new marbles. Do this every single second for a billion billion years and you'll reach a googol marbles.

Wow, thats impressive.
Hmm, Ghost I know you're a mathematical genius, but how?
I soon realised that estimate was vastly off, but was just thinking how much...

I still don't believe there have been 2 billion orders, but I don't want to take over this thread anymore. See my other thread for my reasoning =)

below 25! it has fewer digits than the number
around 25! it has has 25 digits - 1.55x10^25
around 84! it has 1.5(84) digits = 3.3x10^126
around 170! is where Excel reaches it's limit, but it's not quite to 2xnumber or digits, but you can see it's getting there.
170! = 7.3x10^308

so to get a rough estimate of sccale - you could plot x versus log(x!) to get an idea of the erlationship between x and the number of digits in x!

using a quadradic polynomial fit, I get:
y = .0031x^2 + 1.3421x - 8.67

Granted this is only for the first 170 data points, who knows how much a little deviation is going to be off at the 20Bth, but at least there's some rigor to this method

so throwing in 20B for x you get ... 1.24x10^18 digits.

so you'd have 10^10^18 or a 1 followed by 1,000,000,000,000,000,000 zeros. That is a freaking Quintillion zeros

now lets think about the number of orders - now that our math brains have woken up a bit.

7949 is the highest game ID, let's call it 8000 games knowing that it will be a slight over estimate. Each year has 2 diplomacy phases with 7 people and 34 units to order, that 476 orders per year for diplomacy. Say SCs are trading a lot and they average 2 builds / destroys per person per build phase, and maybe 10 units have to retreat each retreat phase.

That gives an additional 48 orders per year, or about 525 total orders per year. If the average game end date is 1914, then that means at most 7350 orders per game (people will have died out prior to this, complicating slightly the estimates, but I'm just seeing if we're in the ball park of 2B)
let's just round that up to 8000 too to make the math real simple - relalizing we going for an upper limit here.

that 64M. Maybe Jacob is right. Unless someone thinks I underestimated anything (I was purposely trying to over estimate)

476 orders per year? You're multiplying things you shouldn't be.
There are never more than 34 orders per main phase as there are only 34 total units available. For simplicity sake we'll assume the lower # of units at start balance out the build and retreats for a game. That's probably a bit low but it gives us 68 orders per year. If the average game is 15 years then we'd have 15x68x8000=8.16 million orders. If we estimate generously it's still 20x75x8000= 12 million.

you're right danger - the 34 will be split between the 7 ppl - good catch, but that makes the number even lower.

It would have about 20,000,000,000 digits

Should read

20,000,000,000,000,000,000 digits

Which is roughly what you said philcore.

How did you arrive at your figure?

I wasn't going to resurect this thread, but since you did GM, I was wondering about DM's claim about a googol marbles, so I did some rough calcs to see where I'd end up with my set of assumptions.

a googol is 10^100

The universe is approx 15 billion years old, so expanding at the speed of light for 15B years gives it a radius of 15x10^9 Light years.

Light travels at 3x10^8 m/s so multiply that by 3600 s/h and then by 24 h/day then by 365.25 days/year to get 9.5x10^15 m/year.

Mult by 15x10^9 to get the radius of the universe in meters: 1.4x10^26

The area of a sphere (assuming a spherical universe, of course) is 4/3Pi*Radius cubed. This gives 1.2x10^79 cubic meters in the universe. If you assume a marble has a radius of about 1 cm, then it's volume is about 4 cubic centimeters, then there are 25^3 marbles in a cubic meter, so you could fit:

2x10^83 marbles in the universe at one time so you'd have to fill the universe up with marbles 5x10^16 times to get to a google. If you did it once per second, that would be 31Million times per year and you'd have to do that every second for 1.6 Billion years. Not quite the Billion Billion Dangermouse read a couple of months ago. Maybe they were using larger marbles or a younger universe. but still, it's good to be able to come up with your own math for these kinds of comparisons. Especially when your waiting for your hard drive to defrag ...

Enrico Fermi was famous for these types of calculations, in fact in Physics they're called Fermi calculations.

I got my figure from the very rough and ready estimate for the number of digits that is n*k, where the number is n and has k digits.

or maybe nx10^k ?

cause then 20B times 10^11 = 20x10^18 which is what you wrote the second time.

pretty cool and not too far off from my calculation taking only the bottom 0.0000009% of the data set and hoping it continued like that for a LOOONG while.

Hang on... Yeah, I screwed that up.

It is nk, not n x 10^k

right, it should be about 220Billion.

I think

1,000,000!= 8.2639316883... × 10^5,565,708 (wiki)

so 1,000,000*7=7,000,000 isn't a bad estimate for the digits.

Also, despite the size of the marble, it is not fluidic so there are gaps in between them, so you can't fit that number in the universe. It will be some smaller number that fits.

2billion orders?
Bullshit!

< 34*2 normal orders per game year
< 11*2 retreats per game year
< 22 builds per winter
=>
< 112 orders per year
< 5000 games
< 35 years per game

5000*35*112 = 19600000 < 20000000 (20 million)

if I'm not missing anything then there should be less that 20 million orders in the system.

Draugnar - I thought about that too but that would just be an adjustment to the radius - make the radius slightly larger than it really is to compensate for the empty space. I didn't want to look up the calculation for hexoganal close packing. But good point, just the same

Ghost - after seeing how far off my original calculation was based on your nk, I checked a boundry condition. You know that for any number greater than 1, n! < n^n, so I did 20B^20B just to see where the digits would be:

(2x10^10)^(2x10^10)
= 2^(2x10^10) x (10^10)^(2x10^10) // (xy)^z = x^z*y^z
= 2^20,000,000,000 x 10^(20x10^10) // (x^y)^z = x^(yz)
= 2^20,000,000,000 x 10^200,000,000,000

now to get the bases the same - we use the old power of 2 approx that 2^10 = 10^3 (where 1kB = 1024 bytes comes from)

= (2^10)^2,000,000,000 x 10^200,000,000,000
= (10^3)^2,000,000,000 x 10^200,000,000,000
= 10^6,000,000,000 x 10^200,000,000,000
= 10^206,000,000,000 //(a^x)(a^y) = a^(x+y)

and violla! 20B^20B is greater than 20B! and yet it has digits in the 200Billion range, much less than my original value and much closer to yours.

Congrats, you win the digits contest!! (But I get extra credit for showing my work!)

@TheGhostmaker & philcore:
have you heard of the stirling approximation
lim (n to inf) n! /( (n/e)^n*sqrt(2 pi n)) = 1
=>n!~ (n/e)^n * sqrt(2 pi n)

in our case 2000000000!= (2*10^9)! ~ (2/e *10^9)^(2*10^9) * sqrt(2* pi * 2 *10^9)
1/10 < 2/e < 1 so
2000000000! > (10^8)^(2*10^9) * 10^5 = 10^(8*2*10^9) * 10^5 = 10^(21*10^9)
which means that 2000000000! has more than 2.1*10^10 digits
and
2000000000! < (10^9)^(2*10^9) * 10^5 = 10^(18*10^9) * 10^5 = 10^(23*10^9)
which means that 2000000000! has less than 2.3*10^10 digits

in conclusion:
2.1*10^10 = 21.000.000.000 < digits in 2.000.000.000! < 23.000.000.000 = 2.3*10^10

sceptic - I've never heard of that, but it is interesting, I'll have to check it out. It also agrees pretty well with both Ghost's and my calculations ...

... provided you start of with 20B as we did and not 2B ;-)

still pretty cool stuff.

New question for those of you how love math puzzels.
how many trailing zeros does 2.000.000.000! have?
in other words
2.000.000.000! = a * 10^t with both a and t being natural numbers what is the highest possible value for t.

e.g for 12000 the answer would be 3 because
12000 = 1200 * 10^1 = 120 * 10^2 = 12 * 10^3

here's the math for 20Billion:
in our case 20000000000!= (2*10^10)! ~ (2/e *10^10)^(2*10^10) * sqrt(2* pi * 2 *10^10)
1/10 < 2/e < 1 so
20000000000! > (10^9)^(2*10^10) * 10^5 = 10^(9*2*10^10) * 10^5 = 10^(23*10^10)
which means that 20000000000! has more than 2.3*10^11 digits
and
20000000000! < (10^10)^(2*10^10) * 10^6 = 10^(20*10^10) * 10^6 = 10^(26*10^10)
which means that 20000000000! has less than 2.6*10^11 digits

in conclusion:
2.3*10^11 = 210.000.000.000 < digits in 20.000.000.000! < 230.000.000.000 = 2.6*10^11

oops:
in conclusion should be:
2.3*10^11 = 230.000.000.000 < digits in 20.000.000.000! < 260.000.000.000 = 2.6*10^11

I'm thinking any time you hit a multiple of 5 it will add a zero - so 20 times per order of magnitude trailing zeros

not quite some numbers cause two or more trailing zeros.
200! = 200 * ... * 150 * ... * 120 * .. * 100 *...

ahh right. That complicates things a little. My brain is tired from all of the math today though. But I did at least give it some thought - mad props to me for that!!!

philcore + (1x10^0)!