 |
|
Level I 82 / 82 solved |
Level II 110 / 123 solved |
Level III 16 / 59 solved |
Level X 9 / 18 solved |
Overall Hall-of-Fame (All time)
The Overall Hall-of-Fame contains the sum of all achieved points of all solved challenges for all users.
You will get at least 100 points for a level I challenge, 1,000 points for a level II challenge, and 10,000 points for a level III challenge (minimum points per challenge). As closer to the date it was published you solve it as more points you'll get: The maximum is the double of the minimum points when you send in the correct solution within a day after the publishing date. If you solve a challenge some weeks after it was published you will only get about 110 % of the minimum points. The points will be fewer every day, but will never fall below 100 % of the minimum points.
If you want to know more on how the points are calculated, take a look at the formula shown at the end of the Overall Hall-of-Fame table.
Using the drop-down list at the right side on top of the following table you can select the displayed time frame of the Overall Hall-of-Fame.
| Rank |
User (#4791) |
#Level I
(#22250) |
#Level II
(#4906) |
#Level III
(#89) |
#Level X
(#68) |
Points (9,490,717) |
|---|---|---|---|---|---|---|
 |
George Lasry (george4096) | 42 | 70 | 14 | 3 | 299,488 |
 |
D3d4lu5 (D3d4lu5) | 82 | 99 | 11 | 1 | 278,422 |
 |
Alain (Integral) | 75 | 78 | 7 | 1 | 207,924 |
 |
Armin Krauss (argh) | 82 | 83 | 2 | 3 | 185,115 |
| 5. | Michael Möller (michaelm) | 65 | 35 | 8 | 0 | 173,129 |
| 6. | Kurt Gebauer (Yokozuna) | 79 | 77 | 2 | 2 | 159,388 |
| 7. | Viktor (Veselovský) | 75 | 72 | 2 | 4 | 151,602 |
| 8. | Nicolas (nicosPavlov) | 67 | 66 | 5 | 0 | 151,166 |
| 9. | Bart den Hartog (Bart13) | 80 | 74 | 2 | 2 | 132,400 |
| 10. | Em Doulgerakis (Greko) | 72 | 61 | 3 | 1 | 124,033 |
Formula
To calculate the sum of all points, we add the
dynamically calculated points for the level 1, 2 and 3 to the manually awarded
points for level X.
The sum of the dynamically calculated points arises from this equation:
$$
Points = (\sum_{i=1}^L (\sum_{j=1}^{N_i}\frac{10^i * 10}{f(d)}))
$$
\(L=3\) because we currently have three dynamically calculated levels.
\(N_i\) denotes the number of challenges solved in level \(i\). The function
\(f(d)\) is used to calculate the point ratio depending on the number of days
between the challenge start and the date, the user sent in a valid solution:
$$
f(d) = 1-\frac{1}{2^{1-c}*(d+1)^c}
$$
Here, the coefficient \(c\) depends on the level \(i\) and is defined as
follows:
$$
c = \begin{cases}
1, & \text{if } i = 1, \\
0.25, & \text{if } i = 2, \\
0.1, & \text{if } i = 3.
\end{cases}
$$
This means that the points for higher levels decline slower than for lower
levels.
For level X, the points are not calculated dynamically but rather decided upon manually.
The MTC3 team may adapt this formula according to new requirements [including a change of the coefficient \(c\)
or the base scales (\(100, 1000, 10000 = (10*10^i),\ i={1,2,3}\))].
This is version 2 of the formula since the start of MTC3.
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