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/*
* Copyright (C) 2015 The Android Open Source Project
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
*
* http://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*/
package com.android.messaging.util;
import android.view.animation.Interpolator;
/**
* Class that acts as an interpolator to match the cubic-bezier css timing function where p0 is
* fixed at 0,0 and p3 is fixed at 1,1
*/
public class CubicBezierInterpolator implements Interpolator {
private final float mX1;
private final float mY1;
private final float mX2;
private final float mY2;
public CubicBezierInterpolator(final float x1, final float y1, final float x2, final float y2) {
mX1 = x1;
mY1 = y1;
mX2 = x2;
mY2 = y2;
}
@Override
public float getInterpolation(float v) {
return getY(getTForXValue(v));
}
private float getX(final float t) {
return getCoordinate(t, mX1, mX2);
}
private float getY(final float t) {
return getCoordinate(t, mY1, mY2);
}
private float getCoordinate(float t, float p1, float p2) {
// Special case start and end.
if (t == 0.0f || t == 1.0f) {
return t;
}
// Step one - from 4 points to 3.
float ip0 = linearInterpolate(0, p1, t);
float ip1 = linearInterpolate(p1, p2, t);
float ip2 = linearInterpolate(p2, 1, t);
// Step two - from 3 points to 2.
ip0 = linearInterpolate(ip0, ip1, t);
ip1 = linearInterpolate(ip1, ip2, t);
// Final step - last point.
return linearInterpolate(ip0, ip1, t);
}
private float linearInterpolate(float a, float b, float progress) {
return a + (b - a) * progress;
}
private float getTForXValue(final float x) {
final float epsilon = 1e-6f;
final int iterations = 8;
if (x <= 0.0f) {
return 0.0f;
} else if (x >= 1.0f) {
return 1.0f;
}
// Try gradient descent to solve for t. If it works, it is very fast.
float t = x;
float minT = 0.0f;
float maxT = 1.0f;
float value = 0.0f;
for (int i = 0; i < iterations; i++) {
value = getX(t);
double derivative = (getX(t + epsilon) - value) / epsilon;
if (Math.abs(value - x) < epsilon) {
return t;
} else if (Math.abs(derivative) < epsilon) {
break;
} else {
if (value < x) {
minT = t;
} else {
maxT = t;
}
t -= (value - x) / derivative;
}
}
// If the gradient descent got stuck in a local minimum, e.g. because the
// derivative was close to 0, use an interval bisection instead.
for (int i = 0; Math.abs(value - x) > epsilon && i < iterations; i++) {
if (value < x) {
minT = t;
t = (t + maxT) / 2.0f;
} else {
maxT = t;
t = (t + minT) / 2.0f;
}
value = getX(t);
}
return t;
}
}
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