简单的任务难度评估的算法（ELO）
简单的任务难度评估的算法（ELO）
适用于简单的任务如题目，象棋谜题等。不保证合适
\[P = \frac{1}{1 + 10 ^ {\frac{R_0 - R_p}{400}}}\\
R_n = R_0 + K(O - P)
\]
其中 \(R_n\) 为新 Rating，\(R_0\) 为旧 Rating, \(R_p\) 为玩家旧 Rating.
若玩家回答正确，\(O = 1\)，否则 \(O = 0\).
\(K_0\) 为常数，推荐为 \(32\).
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对于任务的 Rating 更新，\(K = -\frac{(K_0 + \frac{1}{n})|R_0 -R_p|}{200}\)，其中 \(n\) 为尝试过该题的人数.
-
对于玩家的 Rating 更新，\(K = K_0\).
C++ 实现
#include <iostream>
#include <cmath>
double puzzle_elo = 1500;
const double K0 = 32.0;
int n = 0;
int main()
{
while (true)
{
// One attempt
double player_elo, O;
// Correct -> O = 1
// Wrong -> O = 0
std::cin >> player_elo >> O;
n++;
const double P = 1.0 / (1 + std::pow(10.0, (puzzle_elo - player_elo) / 400.0));
// Rating update
puzzle_elo -= ((K0 + 1 / n) * std::fabs(((puzzle_elo - player_elo) / 200.0))) * (O - P);
player_elo += K0 * (O - P);
std::cout << "Puzzle: " << puzzle_elo << "\n" << "Player: " << player_elo << "\n";
}
return 0;
}
Python 实现
puzzle_elo = 1500
n = 0
K0 = 32
while True:
player_elo = float(input('Player Rating: '))
O = float(input('Correct?: '))
n += 1
P = 1 / (1 + 10 ** ((puzzle_elo - player_elo) / 400))
puzzle_elo -= (K0 + 1 / n) * abs((puzzle_elo - player_elo) / 200) * (O - P)
player_elo += K0 * (O - P)
print('Puzzle: {}\nPlayer: {}'.format(puzzle_elo, player_elo))
