blob: 842f7ed804a26e2331b08aae91b067ab36f51e5f [file] [log] [blame]
/*
* Copyright 2018 Google Inc.
*
* Use of this source code is governed by a BSD-style license that can be
* found in the LICENSE file.
*/
#include "../skcms.h"
#include "GaussNewton.h"
#include "LinearAlgebra.h"
#include "TransferFunction.h"
#include <assert.h>
#include <string.h>
bool skcms_gauss_newton_step(float (* t)(float x, const void*),
const void* t_ctx,
float (* f)(float x, const void*, const float P[4]),
const void* f_ctx,
void (*grad_f)(float x, const void*, const float P[4], float dfdP[4]),
const void* g_ctx,
float P[4],
float x0, float x1, int N) {
// We'll sample x from the range [x0,x1] (both inclusive) N times with even spacing.
//
// We want to do P' = P + (Jf^T Jf)^-1 Jf^T r(P),
// where r(P) is the residual vector t(x) - f(x,P)
// and Jf is the Jacobian matrix of f(), ∂r/∂P.
//
// Let's review the shape of each of these expressions:
// r(P) is [N x 1], a column vector with one entry per value of x tested
// Jf is [N x 4], a matrix with an entry for each (x,P) pair
// Jf^T is [4 x N], the transpose of Jf
//
// Jf^T Jf is [4 x N] * [N x 4] == [4 x 4], a 4x4 matrix,
// and so is its inverse (Jf^T Jf)^-1
// Jf^T r(P) is [4 x N] * [N x 1] == [4 x 1], a column vector with the same shape as P
//
// Our implementation strategy to get to the final ∆P is
// 1) evaluate Jf^T Jf, call that lhs
// 2) evaluate Jf^T r(P), call that rhs
// 3) invert lhs
// 4) multiply inverse lhs by rhs
//
// This is a friendly implementation strategy because we don't have to have any
// buffers that scale with N, and equally nice don't have to perform any matrix
// operations that are variable size.
//
// Other implementation strategies could trade this off, e.g. evaluating the
// pseudoinverse of Jf ( (Jf^T Jf)^-1 Jf^T ) directly, then multiplying that by
// the residuals. That would probably require implementing singular value
// decomposition, and would create a [4 x N] matrix to be multiplied by the
// [N x 1] residual vector, but on the upside I think that'd eliminate the
// possibility of this skcms_gauss_newton_step() function ever failing.
// 0) start off with lhs and rhs safely zeroed.
skcms_Matrix4x4 lhs = {{ {0,0,0,0}, {0,0,0,0}, {0,0,0,0}, {0,0,0,0} }};
skcms_Vector4 rhs = { {0,0,0,0} };
// 1,2) evaluate lhs and evaluate rhs
// We want to evaluate Jf only once, but both lhs and rhs involve Jf^T,
// so we'll have to update lhs and rhs at the same time.
float dx = (x1-x0)/(N-1);
for (int i = 0; i < N; i++) {
float x = x0 + i*dx;
float resid = t(x,t_ctx) - f(x,f_ctx,P);
float dfdP[4] = {0,0,0,0};
grad_f(x,g_ctx,P, dfdP);
for (int r = 0; r < 4; r++) {
for (int c = 0; c < 4; c++) {
lhs.vals[r][c] += dfdP[r] * dfdP[c];
}
rhs.vals[r] += dfdP[r] * resid;
}
}
// If any of the 4 P parameters are unused, this matrix will be singular.
// Detect those cases and fix them up to indentity instead, so we can invert.
for (int k = 0; k < 4; k++) {
if (lhs.vals[0][k]==0 && lhs.vals[1][k]==0 && lhs.vals[2][k]==0 && lhs.vals[3][k]==0 &&
lhs.vals[k][0]==0 && lhs.vals[k][1]==0 && lhs.vals[k][2]==0 && lhs.vals[k][3]==0) {
lhs.vals[k][k] = 1;
}
}
// 3) invert lhs
skcms_Matrix4x4 lhs_inv;
if (!skcms_Matrix4x4_invert(&lhs, &lhs_inv)) {
return false;
}
// 4) multiply inverse lhs by rhs
skcms_Vector4 dP = skcms_Matrix4x4_Vector4_mul(&lhs_inv, &rhs);
P[0] += dP.vals[0];
P[1] += dP.vals[1];
P[2] += dP.vals[2];
P[3] += dP.vals[3];
return true;
}
float skcms_eval_curve(float x, const void* vctx) {
const skcms_Curve* curve = (const skcms_Curve*)vctx;
if (curve->table_entries == 0) {
return skcms_TransferFunction_eval(&curve->parametric, x);
}
// TODO: today we should always hit an entry exactly, but if that changes, lerp?
// (We add half to account for slight int -> float -> int round tripping issues.)
int ix = (int)( x*(curve->table_entries - 1) + 0.5f );
if (curve->table_8) {
return curve->table_8[ix] * (1/255.0f);
} else {
uint16_t be;
memcpy(&be, curve->table_16 + 2*ix, 2);
uint16_t le = ((be << 8) | (be >> 8)) & 0xffff;
return le * (1/65535.0f);
}
}