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Roll skia/third_party/skcms 8b9d1f9db8a8..d2f489e79366 (1 commits)

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https://skia.googlesource.com/skcms.git/+log/8b9d1f9db8a8..d2f489e79366

2019-01-07 mtklein@google.com rewrite skcms_TransferFunction_invert


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CQ_INCLUDE_TRYBOTS=luci.chromium.try:linux-blink-rel
TBR=bsalomon@google.com

Change-Id: If6cbe7d36b2076eb5742eb1fb21747724b0d1822
Reviewed-on: https://skia-review.googlesource.com/c/181860
Reviewed-by: skia-autoroll <skia-autoroll@skia-public.iam.gserviceaccount.com>
Commit-Queue: skia-autoroll <skia-autoroll@skia-public.iam.gserviceaccount.com>
This commit is contained in:
skia-autoroll 2019-01-07 19:46:57 +00:00 committed by Skia Commit-Bot
parent 0b3b2e774a
commit d0b577fd2b
2 changed files with 44 additions and 62 deletions
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104
third_party/skcms/skcms.cc vendored
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@ -1403,80 +1403,62 @@ float skcms_TransferFunction_eval(const skcms_TransferFunction* tf, float x) {
: powf_(tf->a * x + tf->b, tf->g) + tf->e);
}
// TODO: Adjust logic here? This still assumes that purely linear inputs will have D > 1, which
// we never generate. It also emits inverted linear using the same formulation. Standardize on
// G == 1 here, too?
bool skcms_TransferFunction_invert(const skcms_TransferFunction* src, skcms_TransferFunction* dst) {
// Original equation is: y = (ax + b)^g + e for x >= d
// y = cx + f otherwise
//
// so 1st inverse is: (y - e)^(1/g) = ax + b
// x = ((y - e)^(1/g) - b) / a
//
// which can be re-written as: x = (1/a)(y - e)^(1/g) - b/a
// x = ((1/a)^g)^(1/g) * (y - e)^(1/g) - b/a
// x = ([(1/a)^g]y + [-((1/a)^g)e]) ^ [1/g] + [-b/a]
//
// and 2nd inverse is: x = (y - f) / c
// which can be re-written as: x = [1/c]y + [-f/c]
//
// and now both can be expressed in terms of the same parametric form as the
// original - parameters are enclosed in square brackets.
skcms_TransferFunction tf_inv = { 0, 0, 0, 0, 0, 0, 0 };
// This rejects obviously malformed inputs, as well as decreasing functions
if (!tf_is_valid(src)) {
return false;
}
// There are additional constraints to be invertible
bool has_nonlinear = (src->d <= 1);
bool has_linear = (src->d > 0);
// We're inverting this function, solving for x in terms of y.
// y = (cx + f) x < d
// (ax + b)^g + e x ≥ d
// The inverse of this function can be expressed in the same piecewise form.
skcms_TransferFunction inv = {0,0,0,0,0,0,0};
// Is the linear section not invertible?
if (has_linear && src->c == 0) {
// We'll start by finding the new threshold inv.d.
// In principle we should be able to find that by solving for y at x=d from either side.
// (If those two d values aren't the same, it's a discontinuous transfer function.)
float d_l = src->c * src->d + src->f,
d_r = powf_(src->a * src->d + src->b, src->g) + src->e;
if (fabsf_(d_l - d_r) > 1/512.0f) {
return false;
}
inv.d = d_l; // TODO(mtklein): better in practice to choose d_r?
// Is the nonlinear section not invertible?
if (has_nonlinear && (src->a == 0 || src->g == 0)) {
return false;
// When d=0, the linear section collapses to a point. We leave c,d,f all zero in that case.
if (inv.d > 0) {
// Inverting the linear section is pretty straightfoward:
// y = cx + f
// y - f = cx
// (1/c)y - f/c = x
inv.c = 1.0f/src->c;
inv.f = -src->f/src->c;
}
// If both segments are present, they need to line up
if (has_linear && has_nonlinear) {
float l_at_d = src->c * src->d + src->f;
float n_at_d = powf_(src->a * src->d + src->b, src->g) + src->e;
if (fabsf_(l_at_d - n_at_d) > (1 / 512.0f)) {
return false;
}
}
// The interesting part is inverting the nonlinear section:
// y = (ax + b)^g + e.
// y - e = (ax + b)^g
// (y - e)^1/g = ax + b
// (y - e)^1/g - b = ax
// (1/a)(y - e)^1/g - b/a = x
//
// To make that fit our form, we need to move the (1/a) term inside the exponentiation:
// let k = (1/a)^g
// (1/a)( y - e)^1/g - b/a = x
// (ky - ke)^1/g - b/a = x
// Invert linear segment
if (has_linear) {
tf_inv.c = 1.0f / src->c;
tf_inv.f = -src->f / src->c;
}
float k = powf_(1.0f / src->a, src->g); // TODO(mtklein): evaluate as 1 / powf(src->a, src->g)?
inv.g = 1.0f / src->g;
inv.a = k;
inv.b = -k * src->e;
inv.e = -src->b / src->a;
// Invert nonlinear segment
if (has_nonlinear) {
tf_inv.g = 1.0f / src->g;
tf_inv.a = powf_(1.0f / src->a, src->g);
tf_inv.b = -tf_inv.a * src->e;
tf_inv.e = -src->b / src->a;
}
// TODO(mtklein): we'd like to guarantee the edge cases more strongly:
// inv(src(0)) = 0
// inv(src(d)) = d
// inv(src(1)) = 1
if (!has_linear) {
tf_inv.d = 0;
} else if (!has_nonlinear) {
// Any value larger than 1 works
tf_inv.d = 2.0f;
} else {
tf_inv.d = src->c * src->d + src->f;
}
*dst = tf_inv;
return true;
*dst = inv;
return tf_is_valid(dst);
}
// ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ //
@ -1722,7 +1704,7 @@ bool skcms_ApproximateCurve(const skcms_Curve* curve,
int mid = (L + N) / 2;
float mid_x = mid / (N - 1.0f);
float mid_y = eval_curve(curve, mid_x);
tf.g = log2f_(mid_y) / log2f_(mid_x);
tf.g = log2f_(mid_y) / log2f_(mid_x);;
tf.a = 1;
tf.b = 0;
tf.e = tf.c*tf.d + tf.f

2
third_party/skcms/version.sha1 vendored
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@ -1 +1 @@
8b9d1f9db8a814a730e0b72850bc0e757213a920
d2f489e793667c594102183b841cbeb84fb010a4