8 // integration step size
9 static const double step_size = 10.0;
13 double prob_score(double a, double rd);
14 double prob_score_real(double a, double prodai, double rd_norm);
15 double prodai(double a);
17 // probability of match ending 10-a when winnerR - loserR = RD
22 // | Poisson[lambda1, t](a) * Erlang[lambda2, 10](t) dt
27 // where lambda1 = 1.0, lambda2 = 2^(rd/455)
29 // The constant of 455 is chosen carefully so to match with the
30 // Glicko/Bradley-Terry assumption that a player rated 400 points over
31 // his/her opponent will win with a probability of 10/11 =~ 0.90909.
33 double prob_score(double a, double rd)
35 return prob_score_real(a, prodai(a), rd/455.0);
38 // Same, but takes in Product(a+i, i=1..9) as an argument in addition to a. Faster
39 // if you already have that precomputed, and assumes rd is already divided by 455.
40 double prob_score_real(double a, double prodai, double rd_norm)
43 pow(2.0, -a*rd_norm) * pow(2.0, 10.0*rd_norm) * pow(pow(2.0, -rd_norm) + 1.0, -a)
45 double denom = 362880 * pow(1.0 + pow(2.0, rd_norm), 10.0);
49 // Calculates Product(a+i, i=1..9) (see above).
50 double prodai(double a)
52 return (a+1)*(a+2)*(a+3)*(a+4)*(a+5)*(a+6)*(a+7)*(a+8)*(a+9);
56 // Computes the integral
61 // | ProbScore[a] (r2-r1) Gaussian[mu2, sigma2] (dr2) dr2
66 // For practical reasons, -inf and +inf are replaced by 0 and 3000, which
67 // is reasonable in the this context.
69 // The Gaussian is not normalized.
71 // Set the last parameter to 1.0 if player 1 won, or -1.0 if player 2 won.
72 // In the latter case, ProbScore will be given (r1-r2) instead of (r2-r1).
74 double opponent_rating_pdf(double a, double r1, double mu2, double sigma2, double winfac)
77 double prodai_precompute = prodai(a);
79 for (double r2 = 0.0; r2 < 3000.0; r2 += step_size) {
80 double x = r2 + step_size*0.5;
81 double probscore = prob_score_real(a, prodai_precompute, (r1 - x)*winfac);
82 double z = (x - mu2)/sigma2;
83 double gaussian = exp(-(z*z/2.0));
84 sum += step_size * probscore * gaussian;
89 // normalize the curve so we know that A ~= 1
90 void normalize(vector<pair<double, double> > &curve)
93 for (vector<pair<double, double> >::const_iterator i = curve.begin(); i != curve.end(); ++i) {
94 peak = max(peak, i->second);
97 double invpeak = 1.0 / peak;
98 for (vector<pair<double, double> >::iterator i = curve.begin(); i != curve.end(); ++i) {
103 // computes matA * matB
104 void mat_mul(double *matA, unsigned ah, unsigned aw,
105 double *matB, unsigned bh, unsigned bw,
109 for (unsigned y = 0; y < bw; ++y) {
110 for (unsigned x = 0; x < ah; ++x) {
112 for (unsigned c = 0; c < aw; ++c) {
113 sum += matA[c*ah + x] * matB[y*bh + c];
115 result[y*bw + x] = sum;
120 // computes matA^T * matB
121 void mat_mul_trans(double *matA, unsigned ah, unsigned aw,
122 double *matB, unsigned bh, unsigned bw,
126 for (unsigned y = 0; y < bw; ++y) {
127 for (unsigned x = 0; x < aw; ++x) {
129 for (unsigned c = 0; c < ah; ++c) {
130 sum += matA[x*ah + c] * matB[y*bh + c];
132 result[y*bw + x] = sum;
137 void print3x3(double *M)
139 printf("%f %f %f\n", M[0], M[3], M[6]);
140 printf("%f %f %f\n", M[1], M[4], M[7]);
141 printf("%f %f %f\n", M[2], M[5], M[8]);
144 void print3x1(double *M)
146 printf("%f\n", M[0]);
147 printf("%f\n", M[1]);
148 printf("%f\n", M[2]);
151 // solves Ax = B by Gauss-Jordan elimination, where A is a 3x3 matrix,
152 // x is a column vector of length 3 and B is a row vector of length 3.
153 // Destroys its input in the process.
154 void solve3x3(double *A, double *x, double *B)
156 // row 1 -= row 0 * (a1/a0)
158 double f = A[1] / A[0];
166 // row 2 -= row 0 * (a2/a0)
168 double f = A[2] / A[0];
176 // row 2 -= row 1 * (a5/a4)
178 double f = A[5] / A[4];
187 // row 1 -= row 2 * (a7/a8)
189 double f = A[7] / A[8];
195 // row 0 -= row 2 * (a6/a8)
197 double f = A[6] / A[8];
203 // row 0 -= row 1 * (a3/a4)
205 double f = A[3] / A[4];
217 // Give an OK starting estimate for the least squares, by numerical integration
218 // of x*f(x) and x^2 * f(x). Somehow seems to underestimate sigma, though.
219 void estimate_musigma(vector<pair<double, double> > &curve, double &mu_result, double &sigma_result)
223 double sum_area = 0.0;
225 for (unsigned i = 1; i < curve.size(); ++i) {
226 double x1 = curve[i].first;
227 double x0 = curve[i-1].first;
228 double y1 = curve[i].second;
229 double y0 = curve[i-1].second;
230 double xm = 0.5 * (x0 + x1);
231 double ym = 0.5 * (y0 + y1);
232 sum_area += (x1-x0) * ym;
233 mu += (x1-x0) * xm * ym;
234 sigma += (x1-x0) * xm * xm * ym;
237 mu_result = mu / sum_area;
238 sigma_result = sqrt(sigma) / sum_area;
241 // Find best fit of the data in curves to a Gaussian pdf, based on the
242 // given initial estimates. Works by nonlinear least squares, iterating
243 // until we're below a certain threshold.
244 void least_squares(vector<pair<double, double> > &curve, double mu1, double sigma1, double &mu_result, double &sigma_result)
248 double sigma = sigma1;
251 double matA[curve.size() * 3]; // N x 3
252 double dbeta[curve.size()]; // N x 1
254 // A^T * A: 3xN * Nx3 = 3x3
257 // A^T * dβ: 3xN * Nx1 = 3x1
263 //printf("A=%f mu=%f sigma=%f\n", A, mu, sigma);
265 // fill in A (depends only on x_i, A, mu, sigma -- not y_i)
266 for (unsigned i = 0; i < curve.size(); ++i) {
267 double x = curve[i].first;
270 matA[i + 0 * curve.size()] =
271 exp(-(x-mu)*(x-mu)/(2.0*sigma*sigma));
274 matA[i + 1 * curve.size()] =
275 A * (x-mu)/(sigma*sigma) * matA[i + 0 * curve.size()];
278 matA[i + 2 * curve.size()] =
279 matA[i + 1 * curve.size()] * (x-mu)/sigma;
283 for (unsigned i = 0; i < curve.size(); ++i) {
284 double x = curve[i].first;
285 double y = curve[i].second;
287 dbeta[i] = y - A * exp(- (x-mu)*(x-mu)/(2.0*sigma*sigma));
291 mat_mul_trans(matA, curve.size(), 3, matA, curve.size(), 3, matATA);
292 mat_mul_trans(matA, curve.size(), 3, dbeta, curve.size(), 1, matATdb);
295 solve3x3(matATA, dlambda, matATdb);
301 // terminate when we're down to three digits
302 if (fabs(dlambda[0]) <= 1e-3 && fabs(dlambda[1]) <= 1e-3 && fabs(dlambda[2]) <= 1e-3)
307 sigma_result = sigma;
310 int main(int argc, char **argv)
312 double mu1 = atof(argv[1]);
313 double sigma1 = atof(argv[2]);
314 double mu2 = atof(argv[3]);
315 double sigma2 = atof(argv[4]);
316 int score1 = atoi(argv[5]);
317 int score2 = atoi(argv[6]);
318 vector<pair<double, double> > curve;
321 for (double r1 = 0.0; r1 < 3000.0; r1 += step_size) {
322 double z = (r1 - mu1) / sigma1;
323 double gaussian = exp(-(z*z/2.0));
324 curve.push_back(make_pair(r1, gaussian * opponent_rating_pdf(score2, r1, mu2, sigma2, 1.0)));
327 for (double r1 = 0.0; r1 < 3000.0; r1 += step_size) {
328 double z = (r1 - mu1) / sigma1;
329 double gaussian = exp(-(z*z/2.0));
330 curve.push_back(make_pair(r1, gaussian * opponent_rating_pdf(score1, r1, mu2, sigma2, -1.0)));
334 double mu_est, sigma_est, mu, sigma;
336 estimate_musigma(curve, mu_est, sigma_est);
337 least_squares(curve, mu_est, sigma_est, mu, sigma);
338 printf("%f %f\n", mu, sigma);