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Post by ~Memzak~ on Aug 3, 2010 18:35:43 GMT
But will it be in a new category?
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Post by Fringe Pioneer on Aug 3, 2010 18:37:18 GMT
It has already been implemented by QwertyuiopThePie as a sub-board in the General Talk board.
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Post by Artifact123 on Aug 8, 2010 19:31:55 GMT
Here is a suggestion...
If you still didn,t post after 2 months since you created your account, it will be deleted but you may still create a new account.
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Post by QwertyuiopThePie on Aug 25, 2010 8:20:21 GMT
A code cannot be made for that, sorry. Wouldn't work.
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Post by ~Memzak~ on Aug 25, 2010 8:55:11 GMT
Besides, there are many long lost members who eventually go inactive and then come back....
This would give them quite an unwelcome receiving....
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Post by Artifact123 on Sept 11, 2010 10:57:34 GMT
Maybe the price for a custom board could be lowered to the original 5000 ¥? But 7500 ¥ would be good too.
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Post by ganondorfchampin on Sept 11, 2010 19:33:23 GMT
I know proboards is an english-language website, but maybe there can be one board for other languages.
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Post by Fringe Pioneer on Sept 11, 2010 19:46:14 GMT
With that comes one problem: how do staff properly moderate in such boards, in the event that the language used is unknown to the moderators?
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Post by ganondorfchampin on Sept 11, 2010 20:05:46 GMT
Good point. What languages do the mods know?
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Post by QwertyuiopThePie on Sept 12, 2010 1:38:38 GMT
English, and... English. People are free to use other languages in their posts, so long as they provide a google translation or something.
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Post by Artifact123 on Sept 12, 2010 6:18:19 GMT
Well, GV knows Spanish and i could always check that Board and report it to the mods when i notice they are flaming, trolling or insulting in Dutch and Spanish. Those are the languages i know.
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Post by Fringe Pioneer on Sept 12, 2010 7:06:21 GMT
No one in the forum uses those languages for standard communication, and it seems that at long last we have two native Japanese speakers, a language which no staff member knows. Not only would "foreign" language boards be hard to moderate, but on account of the paucity of alternate language speakers, the boards would die off...
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Post by QwertyuiopThePie on Sept 12, 2010 10:41:24 GMT
We always have Google Translate.
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Post by sparkpowder on Sept 15, 2010 11:20:04 GMT
How about a updated ranking system. You would be able to choose what stars you wanted [when you purchase the rights i.e. the custom stars]. You would be able to switch between them at any time.
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Post by QwertyuiopThePie on Sept 15, 2010 22:53:27 GMT
You can't choose what stars you want, except for the custom stars. Proboards doesn't work that way. An updated ranking system would be good, though. Here's our current: Would you say we should make it go in a steady curve up to ten stars? I think a logarithmic approach may be the best one, though we would need to make it slower at the beginning. No more than one star per jump, but when it gets high enough it changes to two, etc... Maybe the delta/derivative/something of a logarithmic graph? Or something like that? Veers, any help here?
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Post by ganondorfchampin on Sept 17, 2010 1:52:55 GMT
The derivative of the logarithmic graph is 1/x. Definitely not.
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Post by Fringe Pioneer on Sept 17, 2010 3:26:57 GMT
Well, assuming the logarithm is base e, anyways.
You want a function that reaches a finite limit as the independent variable (P for "posts") approaches infinity. The function should also intersect the origin, and it should always increase, although the amount it increases should itself decrease (i.e. the derivative should be positive but decrease to 0).
The natural log (ln x) approaches infinity as it goes to infinity, which is not what we want. It's asymptote is vertical, which is not what we want, either. The natural base (ex) approaches infinity as it goes to infinity, which is not what we want. It's asymptote is horizontal, so some simple transformations to the natural base function should prove useful.
Take the negative of its independent variable to reflect the function over the dependent variable axis (e-x). Now the function approaches a finite limit (specifically, 0) as the independent variable goes to infinity. The function is always decreasing, so we need to reflect the function over the independent variable axis.
Take the negative of the entire function to reflect the function over the independent variable axis (-e-x). Now the function approaches the same finite limit as the independent variable goes to infinity AND it is always increasing. Now the function is always negative, which makes no sense when counting how many stars we need for how many posts someone has. Let's translate the function upwards.
Add one to the function to translate the function upward (1 - e-x). Now the limit goes to one as the independent variable approaches infinity, people with no posts get no stars (instead of -1 stars), and the function is still always increasing. Surely, we want posters to attain more than one star, so let's scale the function.
Multiply the entire function by the maximum number of stars one should attain (M - Me-x). Now the limit goes to the maximum we specified as the independent variable approaches infinity, users with no posts still get no stars, and the function is always increasing.
Change the base to the factor by which posts will increase in order to increase the dependent variable by one with each factor, for example 10, so that the dependent variable will always be an integer (M - MB-x). If the base is the same as the maximum number of stars, you can take the factor of the maximum number of posts off the base and add 1 to the exponent (M - B1-x).
One problem persists - at S(1), the dependent variable will become M - (MB-1) = M - M/B = M(B - 1)/B, which can become very close to M at moderate values of B. To correct this, change the rate of change by multiplying the independent variable with the reciprocal of a number D (M - MB-X/D). Now the exponent will smaller than before, and not result in as large a number at small values of X. To find a good D, we need to determine another point on the function that ought to be reasonable, so that X posts will result in M/2 (half the maximum) stars.
M/2 = M - MB-X/D M/2 - M = -MB-X/D -M/2 = -MB-X/D M/2 = MB-X/D M/2 * 1/M = B-X/D 1/2 = B-X/D logB(1/2) = -X/D (logB(1/2))/-X = 1/D
I could go on to say that D = -X/(logB(1/2)), but because D appears as the denominator of -P, it would save time to leave the solution at 1/D = (logB(1/2))/-X so that all we have to do is multiply that value to -P when we plug it in.
So, a nice function would go along the lines of S(P) = M - MB-P*(logB(1/2))/-X. Remembering the properties of logarithms, the equation can be simplified to S(P) = M - M(1/2)P/X
I'm assuming we want no more than 10 stars, so M = 10. Since S(X) = M/2, you should probably determine at what value you want the posts to result in the maximum number of stars and halve that X value.
Voila, a fitting function!
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Post by QwertyuiopThePie on Sept 17, 2010 14:22:46 GMT
Yay!
I'm guessing just all 10s from above Forum Legend for now since very few people are above that rank anyway, so about 21 'posts'. I'll just round it to 20, we can have Posting Legend have 10 as well.
I'm afraid I'll need to implement it later: I have something to which I need to go.
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Post by Fringe Pioneer on Sept 17, 2010 16:55:45 GMT
S(0) = 10 - 10*(1/2)0/20 = 10 - 10 = 0
S(10) = 10 - 10*(1/2)10/20 = 10 - 10(1/2)1/2 = 10 - 5*21/2 ~ 10 - 7.07 ~ 2.93
S(20) = 10 - 10*(1/2)20/20 = 10 - 10(1/2)1 = 10 - 5 = 5
S(40) = 10 - 10*(1/2)40/20 = 10 - 10(1/2)2 = 10 - 5/2 = 7.5
S(60) = 10 - 10*(1/2)60/20 = 10 - 10(1/2)3 = 10 - 5/4 = 8.875
S(80) = 10 - 10*(1/2)80/20 = 10 - 10(1/2)4 = 10 - 5/8 = 9.375
S(100) = 10 - 10*(1/2)100/20 = 10 - 10(1/2)5 = 10 - 5/16 = 9.6875
0 = 10 - 10(1/2)P/20 -10 = -10(1/2)P/20 1 = (1/2)P/20 log1/21 = P/20 0 = P/20 P = 0
1 = 10 - 10(1/2)P/20 9/10 = (1/2)P/20 20*log1/20.9 = P = 20*ln(0.9)/ln(0.5) = ~3.040
4 posts to get 1 star
2 = 10 - 10(1/2)P/20 4/5 = (1/2)P/20 20*ln(0.8)/ln(0.5) = ~6.439
7 posts to get 2 stars
3 = 10 - 10(1/2)P/20 20*ln(0.7)/ln(0.5) = ~10.291
11 posts to get 3 stars
4 = 10 - 10(1/2)P/20 20*ln(0.6)/ln(0.5) = ~14.739
15 posts to get 4 stars
5 = 10 - 10(1/2)P/20 20*ln(0.5)/ln(0.5) = 20
20 posts exactly to get 5 stars
6 = 10 - 10(1/2)P/20 20*ln(0.4)/ln(0.5) = ~26.439
27 posts to get 6 stars
7 = 10 - 10(1/2)P/20 20*ln(0.3)/ln(0.5) = ~34.739
35 posts to get 7 stars
8 = 10 - 10(1/2)P/20 20*ln(0.2)/ln(0.5) = ~46.439
47 posts to get 8 stars
9 = 10 - 10(1/2)P/20 20*ln(0.1)/ln(0.5) = ~66.439
67 posts to get 9 stars
9.5 = 10 - 10(1/2)P/20 20*ln(0.05)/ln(0.5) = ~86.439
87 posts to get the lowest number that can be rounded up to ten stars, otherwise you can never get 10 stars.
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Post by QwertyuiopThePie on Sept 18, 2010 1:13:39 GMT
Oh, to arrange it by posts? I thought you meant by ranks. My bad. Hmm, a max of 5,000 posts, so that would be... Ehm... Brb, doing math. I'll take the 20 vital ranks and multiply them by 5 so it goes to 100, should be close enough for this. That gives us: 0 2 3 4 5 6 7 8 8 8 9 9 9 9 9 9 9 10 10 10 And all 10s from there. Seem good to everyone? We can split the stars evenly, in order from grey, red, to yellow, or any other order if people want. Here's a graph: www.wolframalpha.com/input/?i=10-10%281%2F2%29^%285x%2F20%29Thanks Veers!
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Post by Canopy on Sept 27, 2010 1:35:39 GMT
I think there should be a powder game Special Board.
That is my idea.
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Post by QwertyuiopThePie on Sept 27, 2010 5:33:07 GMT
Is there really enough to do in PG that would be considering hacking? Anything that would give you a particular advantage, I mean? Since there really isn't enough, it's not worth making one.
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Post by Artifact123 on Oct 31, 2010 8:09:48 GMT
A Music Board as a Sub-Board in general Talk? Or maybe merge it with Non Dan-Ball games? Also, if it gets approved, Ideas Badge!
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Post by ~Memzak~ on Oct 31, 2010 10:15:41 GMT
Um, now that I look over this, perhaps from the rank Dan-Ball Master to Supreme the stars shouldn't go back to one and go 8, 9, 10 and then stay at 10.
Just a suggestion. (since this thread is being revived now)
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Post by QwertyuiopThePie on Oct 31, 2010 20:35:33 GMT
Memzak, did you read our conversation? We were talking the whole time about how they don't go back to one. The updated list is in my very last post.
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Post by ~Memzak~ on Nov 3, 2010 17:55:29 GMT
Are you sure they don't go back to one? Because the reason I ignored your list was because it wasn't in effect. The old list was still in effect.
As instead of having 10 stars I had 7. And now I have 1, which is clearly the old list.
Just saying...
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Post by Fringe Pioneer on Nov 3, 2010 19:40:48 GMT
I should probably point out something I overlooked when I did this. Concerning finding a good value for D; just because you find a value of D that will make half the maximum number of points return half the maximum number of stars doesn't necessarily mean that the maximum number of posts will return a close approximation for the maximum number of stars. Instead, it would be better to define the function so the maximum number of posts will result in 95% of the maximum number of stars (i.e. 9.5 stars, which is the smallest number that rounds to 10) (since there is an asymptote at the actual maximum number of stars and is ergo impossible to achieve). A better function (relative to my last one) would be S(P) = 10 - 10(0.05) P/X, where X is the maximum number of posts, which intersects at (0 posts, 0 stars) and (X posts, 9.5 stars). In order to get to that function, I had to make sure that the maximum number of posts would result in nearly the maximum number of stars. Since it is impossible to get exactly the maximum number of stars, I had to get it "really close." Where L is a limiting value that is very close to M, the maximum number of stars, I got rid of M/2 = M - MB -X/Dand used L = M - MB -X/DL - M = -MB -X/D(L - M)/-M = B -X/Dlog B((L - M)/-M) = -X/D log B((L - M)/-M)/-X = D -1For reasons stated in my other post, I will keep this solved for the reciprocal of D rather than going all the way to solve for D. A good function, better than the one I suggested last time, would go along the lines of S(P) = M - MB -P*logB((L - M)/-M)/-X. The statement can be rewritten as follows. S(P) = M - MB -P*logB((L - M)/-M)/-XS(P) = M - MB P*logB((M - L)/M)/XS(P) = M - M((M - L)/M) P/XS(P) = M - M(M/M - L/M) P/XS(P) = M - M(1 - L/M) P/XNow, keeping in mind that M is the maximum number of stars, L is a number very close to that, and X is the maximum number of posts, the equation will be rather useless until you determine how close L is to M. A substitution in terms of M will be necessary. We could say that L is 95% of M, so that S(P) = M - M(1 - (0.95M)/M) P/XS(P) = M - M(1 - 0.95) P/XS(P) = M - M(0.05) P/XThis will return S(0) = 0 and S(X) = 0.95M. If M = 10, then S(X) = 9.5. Naturally, 9.5 rounds up to 10, which will allow us to get to 10 by rounding despite that there is an asymptote at S(P) = M.
0 = 10 - 10(0.05)P/5000 -10 = -10(0.05)P/5000 1 = 0.05P/5000 log0.051 = 0 = P/5000 P = 0
1 = 10 - 10(0.05)P/5000 9/10 = (0.05)P/5000 5000*log0.050.9 = P = 5000*ln(0.9)/ln(0.05) = ~175.851
176 posts to get 1 star
2 = 10 - 10(0.05)P/5000 4/5 = (0.05)P/5000 5000*ln(0.8)/ln(0.05) = ~372.436
373 posts to get 2 stars
3 = 10 - 10(0.05)P/5000 5000*ln(0.7)/ln(0.05) = ~595.305
596 posts to get 3 stars
4 = 10 - 10(0.05)P/5000 5000*ln(0.6)/ln(0.05) = ~852.589
853 posts to get 4 stars
5 = 10 - 10(0.05)P/5000 5000*ln(0.5)/ln(0.05) = ~1156.891
1157 posts to get 5 stars
6 = 10 - 10(0.05)P/5000 5000*ln(0.4)/ln(0.05) = ~1529.327
1530 posts to get 6 stars
7 = 10 - 10(0.05)P/5000 5000*ln(0.3)/ln(0.05) = ~2009.480
2010 posts to get 7 stars
8 = 10 - 10(0.05)P/5000 5000*ln(0.2)/ln(0.05) = ~2686.218
2687 posts to get 8 stars
9 = 10 - 10(0.05)P/5000 5000*ln(0.1)/ln(0.05) = ~3843.109
3844 posts to get 9 stars
9.5 = 10 - 10(1/2)P/20 5000*ln(0.05)/ln(0.05) = 5000
5000 posts to get exactly 9.5, the lowest number that can be rounded up to ten stars, otherwise you can never get 10 stars.
We can go with that, or we can always go with QwertyuiopThePie's alternative function, which seems to work well, too...
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Post by ~Memzak~ on Nov 4, 2010 6:26:35 GMT
Going with your version would be mathematically correct... (sorta) ...but we'd have to change posting ranks to 176, 373, 596, etc...
Perhaps we should just use qwerty's method to avoid having to edit the post numbers needed for a rank.
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Post by QwertyuiopThePie on Nov 4, 2010 13:38:20 GMT
Yeah, I prefer a system arranged by rank instead of by post count.
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Post by ~Memzak~ on Nov 5, 2010 6:51:59 GMT
So then can we agree on a system arranged by rank? Veers? Would you mind if it was arrange by rank?
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