| Author | Durability formula of memorial artifact for 10th year |
|---|
24 + (battles_fought) ^ 0.38, rounded down.
If it is below 30. round it to 30. |
Mmmm not sure where you got this formula.
According to your formula, theoretically I should have 42.18 or so and rounded down is 42 but I was given 38.
Either the base is closer is 20 or the exponent is .35 or maybe a different combination altogether.
I've checked it on people with more combats and your formula still overshoots the total. |
| this formula works for me. Maybe a coincidence and not sure about the source :) |
| Maybe a coincidence, also not sure about the source :) |
You and her are within what I would say is the middle range (11000-16000) and she probably made this formula based on her own artifact so it should work for you since you are near the same area.
This formula overshoots in the low ranges that I checked (1600-2000) and overshoots in high range (~59000) by a bit.
If it is also overshooting at the high range and the low range you are probably looking at a lower exponent and a higher base. Just my guess, someone can check my reasoning.
The two other people I used to check besides myself and their corresponding artifacts:
~59480
https://www.lordswm.com/pl_info.php?id=578344
https://www.lordswm.com/art_info.php?id=10scroll&uid=322972519&crc=i191dad42e
~1640
https://www.lordswm.com/pl_info.php?id=6473573
https://www.lordswm.com/art_info.php?id=10scroll&uid=323184707&crc=ib424348cd |
I took 45 data points from 920-99073 combats fought and plotted them.
The best equation I got is:
Durability = 5.8362 (Battles fought) ^ 0.2464
Anything below 30 becomes 30. Also above 100 rounds down to 100.
with a 99.85% accuracy |
Manuscript of history has price ~ 300 gold/1durability.
Manuscript of history 30/30 ~ 9,000 gold.
Manuscript of history 90/90 ~ 27,000 gold. |
for darkelf84:
Ahahah, funny little dude! |
for darkelf84:
I would gladly buy a 90/90 from you for double the price :P |
To Slust and PLaY LikE a PRO
Manuscript of history 0/90 ~ 0 gold.
:D |
for KnightofDusk:
Thanks for correction.
You can try if this formula works:
5.6 * Sqrt(Sqrt(battles_fought))
and round it to nearest integer.
This worked for me, for a few friends I checked, for virtual_vitrea, and for the two cases you listed. |
for SwiftGirl:
your formula is practically identical to knightofdusk's post 6 since you are using 5.6*(x^0.25) and he used 5.836(x^0.246). The only place i would expect that his multiplier would not be able to compensate for the greater exponential reduction is for very high value of x beyond currently testable upper limit.
well the point is that your formula will give a slightly bigger value of y at higher x. His will do the opposite. |
| How do you guys come up with such formula ? |
Don't worry she understands this. It is more of a reverse engineering approach, a question of finding the original equation.
She believes that the admin would not use an ugly decimal such as mine and I completely agree and have had the same thought.
I simply stated the equation from them Excel data analysis.
Where did the formula come from?
There are several ways to do it.
You can simply do as I did and collect data and find the best fit power function line in Excel.
Another way you can do it is with some log math and maybe some statistical analysis.
y=a*x^b
therefore if you log both sides of the equation:
log(y) = log(a) + b*log(x)
This should give you a linear function and you can then find the slope of the equation and use it for b and then solve for a.
To do this in the first place, it is key that you notice that is a power function.
However this method may get a little messy simply because of the fact that partial durability is not given.
Essentially you would need to find a best fit line. For this you can look up Linear Regression. |
Don't worry she understands this. It is more of a reverse engineering approach, a question of finding the original equation.
i am aware that you need data points for both x and y to figure out the formula. I don't know if something simplistic like that can be called reverse engineering.
Also, the anniversary fsp formula is dependent on battles fought and it has a similar 0.34 decimal :p
I honestly cannot understand why you think its ugly to use three of four decimal places. It only adds precision. |
Yes it is simplistic and I did not call it reverse engineering, but I'm saying the idea is comparable to reverse engineering approach in that you are left with a product and need to find out how it was made to begin with.
Reverse engineering is obviously much more complicated and I don't dare claim the ability to operate in this field.
I'm not sure what .34 decimal you are pointing to unless you mean the ~.25 decimal here.
You could say that long decimals are more precise but only in that we are finding a equation that best represents our presented data.
However, if you were the developer, you would not use a random decimal to program a durability output. Why would you pick something such as 0.2453545?
Yes that is precise but if you were creating a program you would see no reason to pick such a random amount of decimal places.
You would definitely not pick such a precise number when durability has no fractional output and therefore all outputs would be rounded to a whole number anyways.
As such precision as an argument is completely negligible because precision was not a desired aspect to begin with.
You may still thinking in terms of data analyst, but that is exactly what I mean by a "more of reverse engineering approach".
What I mean is to think in terms of the person creating the program.
Perhaps you are in the field and don't appreciate my comparison to reverse engineering. If so I apologize; it was not my intent. |
Btw how did you figure out the equation with the data? Some kind of a program to make an exponential curve of best fit?
And how did you even collect and transfer the data? Some kind of a software? |
I'm not sure what .34 decimal you are pointing to unless you mean the ~.25 decimal here.
i think the free fsp you receive if you pick that for your anniversary reward is based on that formula.
and yes, i did understand what you meant by a developer not wanting to pick that sort of a number but the thing is.. the idea of anniversary artifact as a reward not a new concept. I mean we have had different arts each year and i think you would be convinced if i mentioned just as a possibility that the formula is merely adjusted each year based on something like player base in the server and overall range of battles fought such that durability is always between 30 and 100 without having to round down to 100 for ~100% of the player base. So this is just an example why maybe the formula is not picked at random by a developer meaning its value is something relative and therefore rounding off .24xxx to .25 is not necessary. Of course this is just a possibility that i am mentioning because initially i would want to think that the formula is created such that nobody on the server is actually capable of receiving more than 100 durability.
P.S I used your formula to calculate the number of battles required to get more than 100 durability, and i found that we need more than 10^5 battles. Currently i am sure nobody has more than 100k battles although the top player is very very close to 100k mark :) Based on this observation, my idea is still valid :) |
In fact by removing the outliers of 30 durability and adding more data points you get an equation where the limit of the exponent is approaching 0.25. So I am quite confident she is right that 0.25 is the correct original exponent. As the resulting best fit line has a R^2 of 0.9998.
As for the scalar/multiplier, it seems to be approaching either 45/8 (or 5.625) as SwiftGirl suggested to me or possibly 10^0.75 (~5.6234) from the observation of the log formula.
If you wish to do the best fit line by hand you are best off using this formula:
log(y) = log(a) + b*log(x)
Then you can treat a power function linearly and apply the same concepts with linear best fit creation.
If you aren't familiar with linear regression one method you can use is the least squares regression.
Where Y = a + bx
Slope(b) = (N∑XY - (∑X)(∑Y)) / (N∑X^2 - (∑X)^2)
a = Mean of Y - (b)(Mean of X)
N = number of data points
∑ = sum
R^2 you can look up yourself.
If you don't want to do it by hand just do it on Excel like me.
I collected data by hand by looking for the artifact and then subtracting battles fought since start of anniversary.
On Excel it is pretty simple to do best fit lines a video tutorial would help you in no time better than my block of text.
I uploaded a picture in my album since I don't believe external links are allowed.
https://www.lordswm.com/photo_pl_photos.php?aid=281746&pl_id=4809520
You are free to message me if you would like a copy of the spreadsheet.
You make an excellent point that there is a possibility that the formula is adjusted every year since the artifacts have never passed 100. However, since the power is so clearly 0.25 perhaps the scalar is the point of adjustment since it is the more ambiguous selection. They would simply need to solve 2 = 0.25(max combat) log^a. This way maximum can be adjusted for without adjusting the spread. |
However, since the power is so clearly 0.25 perhaps the scalar is the point of adjustment since it is the more ambiguous selection.
Yes, i did not know that we were sure about 0.25 :) Based on that alone, I can give a lot more credibility to your idea that it is a chosen number without any particular logic behind it. I still haven't verified that information however i am glad you shared your excel spreadsheet for more assurance. |
