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> Powerlevel Estimate Calculation, Help Us Help You :)
Rokugo
post Aug 14 2008, 12:44 AM
Post #41


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Okay, so I did a few fits based on the data Aceo gave. I found that a exponential decay model doesn't work very well, so I made up my own that seemed reasonable, A/x^(.1) + B, where x is the powerrank and A and B are constants fit to the following:
A = 19296684
B = -9194589

This leads to the following fitted values for the list given above.
CODE

Rank    name    vita    mana    powerlevel    Estimated from fit
1    winroute    0    0    0    10103095
2    aubrey    0    0    0    8810853.79926045
3    holangi    0    0    0    8095438.27667388
4    santa    0    0    0    7605150.1259473
5    SubChess    0    0    0    7234448.46146807
6    chinchin    0    0    0    6937643.84071649
7    trunk    0    0    0    6690886.56618751
8    suwan    0    0    0    6480188.82072903
9    Cristiana    0    0    0    6296661.25095365
10    xtroubsx    0    0    0    6134311.93775229
11    Sindella    0    0    0    5988915.42734338
12    Ceez    2219665    1806667    5832999    5857383.4345505
13    kEum    0    0    0    5737392.23752356
14    CheBakY    0    0    0    5627150.75651326
15    Blason    0    0    0    5525250.14218327
16    Conro    3266817    1007284    5281385    5430562.80875807
17    ColdPrincess    0    0    0    5342172.64321688
18    ssjgoku    0    0    0    5259325.53130107
19    goldenflames    2173120    1523634    5220388    5181393.50675192
20    pady    0    0    0    5107848.26593085
21    ShindanMuSan    0    0    0    5038241.26365958
22    KOFighter    0    0    0    4972188.52486502
23    KAKin    0    0    0    4909358.89479556
24    raikou    0    0    0    4849464.83614668
25    fRIGHT    2019995    1398161    4816317    4792255.13961477
26    MoonWater    2000000    1380400    4760800    4737509.09062655
27    aNGeLiCiouS    0    0    0    4685031.75739242
28    Momentum    2702282    997984    4698250    4634650.15180806
29    Neuro    1136437    1767502    4671441    4586210.07657397
30    Valandil    0    0    0    4539573.51678031
31    LadyBug    2523256    1021134    4565524    4494616.4671729
32    Zells    0    0    0    4451227.11081395
33    LordMoses    1932332    1266105    4464542    4409304.28324813
34    guitar    0    0    0    4368756.17023663
35    Eunice    0    0    0    4329499.19780137
36    Najun    0    0    0    4291457.08156563
37    SHeiLLa    0    0    0    4254560.00879373
38    Werrty    0    0    0    4218743.93156402
39    yfandes    0    0    0    4183949.95348292
40    dethsgift    0    0    0    4150123.79550744
41    taleen    0    0    0    4117215.32897146
42    silex    0    0    0    4085178.16594637
43    SpellCasterX    0    0    0    4053969.29871385
44    TyKo    2301179    827808    3956795    4023548.78146968
45    Emitt    0    0    0    3993879.44847445
46    DemonHealer    0    0    0    3964926.66376882
47    Cashel    1678100    1116179    3910458    3936658.09831606
48    FeiWulong    0    0    0    3909043.53105237
49    juangary    0    0    0    3882054.67084045
50    Soyeong    0    0    0    3855664.99675229
51    SoliTa    1504182    1131451    3767084    3829849.61446902
52    Bowtiesan    0    0    0    3804585.12688999
53    Syncopate    1434800    1150000    3734800    3779849.51730159


Also, it looks like this on the data available:
IPB Image

Amazingly enough, this fit works pretty decent even down to the 1000 level. I looked at a couple of values here and there and it appears that the estimate is correct to within 10% at the 500 rank and within 20% at the 1000 rank. Not bad at all for only 13 data points. Anyway, this is all useless for the majority of what you're doing, but if you are really motivated to do an accurate estimate of the top few, this sort of fitting might be one way to generate such an estimate. I would also be very interested in what this sort of estimate would look like if I applied the remainder of the top 1000 data.

One last nerdy aside: Now that we have a (somewhat) accurate function, we can easily estimate what the sum of the powerranks of the top 1000 players are. All we have to do is integrate this function from 1 to 1000 (well, the exact limits of integration to use here are a little trickier, but we'll use 1 to 1000 as a first estimate). The indefinite integral is A * x^0.9/0.9 + Bx. When we apply the limits to this we get 1560440780. Interestingly enough, if you look at only the top 25 players, you get 167825534, so about 10% of the powerrank in the top 1000 is held by only the top 25 people. Wow!



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AceoStar
post Aug 14 2008, 01:19 PM
Post #42


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Im the most interested in whatever distribution will work best across the board. We can manually add weight to the tops if we have to.

Working with the actual rank is something I hadn't planned on, so I need to rework a bit of my code to make that available. Results tonight, for real this time :-p


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Rokugo
post Aug 14 2008, 06:07 PM
Post #43


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Sure, so the idea that's been bouncing around in my head is that if you are unhappy with the results of the linear interpolation, we can try a nonlinear interpolation based off a function form that we know can be fitted accurately to all of the powerrank stats data. But if the linear interpolation gives you satisfactory results, then we don't have to worry about any of that.

If I understand what you're going for correctly, I want to comment that for the last few tops (the ones above the highest known top), linear interpolation can't really be used with much success without manually putting in "weighting" fudge factors, but these weighting fudge factors would have to be adjusted manually very often (if you are planning on doing what I think you are, but I may be wrong), and in order to calculate meaningful fudge factors, you'd have to know something about the hidden top stats, anyway.

I think the best simple way to calculate the powerlevels of those last few would be just to make yourself a line whose slope you know is reasonable simiiar, such as this: let p0 be the powerlevel of the person at the highest known rank. Let that rank be n. Then a good first approximation for those last few dudes might be something like:

p(rank) = p0*(1+.05*(n-rank)). Currently, this would give results like this:

CODE

Rank    Name    Vita    Mana    Powerrank    Estimated powerrank currently    Estimated powerrank if Ceez stopped showing stats
Rank    Name    Vita    Mana    Powerrank    Estimated powerrank currently    Estimated powerrank if Ceez stopped showing stats
1    winroute    0    0    0    9041148.45    9242423.75
2    aubrey    0    0    0    8749498.5    8978354.5
3    holangi    0    0    0    8457848.55    8714285.25
4    santa    0    0    0    8166198.6    8450216
5    SubChess    0    0    0    7874548.65    8186146.75
6    chinchin    0    0    0    7582898.7    7922077.5
7    trunk    0    0    0    7291248.75    7658008.25
8    suwan    0    0    0    6999598.8    7393939
9    Cristiana    0    0    0    6707948.85    7129869.75
10    xtroubsx    0    0    0    6416298.9    6865800.5
11    Sindella    0    0    0    6124648.95    6601731.25
12    Ceez    2219665    1806667    5832999    5832999    6337662
13    kEum    0    0    0    0    6073592.75
14    CheBakY    0    0    0    0    5809523.5
15    Blason    0    0    0    0    5545454.25
16    Conro    3266817    1007284    5281385    0    5281385

I estimated what it would be if Ceez stopped showing to make sure that it still gave sane estimates with a different dataset, which it seems to do.

Maybe every year or so the shape of the top part of the curve will change sufficiently to merit a different slope, but this shouldn't occur too often as it would by creating a line with the two lowest data points.

In summary:
Regular linear interpolation: Using two points, construct a line and use that to estimate points near the two you know (works best for points between the two known points). This will probably work wonderfully for everyone below the highest known top, and if it doesn't we can discuss some nonlinear interpolation methods.

"Ghetto" linear interpolation: Constructing a line with the last two known tops and using it to estimate everyone above them is begging for large errors (Consider if the two highest known tops are just about to pass each other, suddenly winroute's estimate is 5m. One month later when one has gained many stats and one has been on vacation, the estimate may bounce up to 10m).

Instead the "Ghetto" method I describe above assumes a reasonable slope and uses that and the highest known point to make a line. This allows it to grow accurately with the power chart, assuming that the average slope of the tops doesn't change much (which it doesn't really, and if it does, it is very slowly, on the order of several months to a year). This gives us just one fudge factor that we only have to manually adjust every few months at the very most.


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Adam
post Aug 15 2008, 10:49 AM
Post #44


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... Or we can just get a Chongun or something to check out Winroute's stats, tell us the PowerLevel only, and use linear interpolation to estimate the stats for the players inbetween him and Ceez.

Now all we need is an easily susceptible Chongun...


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Rokugo
post Aug 15 2008, 08:05 PM
Post #45


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Yeah, Aceo pointed out that that would be a lot easier. tongue.gif We pretty much know what win's stats are, so it definitely works a lot better than what I was talking about.


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Hi, I'm Rokugo. You probably don't know me.
There is a 95% chance the above post is sarcasm and is not to be taken seriously.
I play Real Life to help escape NexusTK.
"Baby, for you I'd divide by zero" <--- My # 1 pickup line
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SilentS
post Aug 15 2008, 08:35 PM
Post #46


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QUOTE(KaN @ Aug 15 2008, 09:49 AM) [snapback]56893[/snapback]

... Or we can just get a Chongun or something to check out Winroute's stats, tell us the PowerLevel only, and use linear interpolation to estimate the stats for the players inbetween him and Ceez.

Now all we need is an easily susceptible Chongun...
*shrugs* And risk not only being kicked from the path, but for the path to lose the spell again..........


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HaKSaW
post Aug 15 2008, 08:52 PM
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Why have it if you don't use it? Rather pointless.


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Adam
post Aug 16 2008, 09:24 AM
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You could just give a general estimate of the Power level only.

No one needs to know, SilentS... wink.gif


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