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Given the following text description, write Python code to implement the functionality described below step by step
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Description:
 Copyright 2020 Google LLC.
Licensed under the Apache License, Version 2.0 (the "License");
Step1:  Goal
We want to build a model $h_\theta(s) \rightarrow a^$ which predicts the mode $a^$ of some target distribution, for which we have unnormalized log-probabilities, $y$.
Stated another way, we want to find an appropriate loss L s.t. it is tractable to solve
$$
\text{argmin}\theta \sum\limits{i=1}^N L(\pi(a_i \mid h_\theta(s_i)), y_i)
$$
where $\theta$ are model parameters, $h$ is a model that depends on context $s$ and outputs the mode of $\pi$, and $\pi(a \mid \text{mode})$ is the predicted probability of action $a$. $y$ is information about the target distribution. We want the mode of $\pi$ to correspond to areas where $y$ is maximized.
Candidate
$$
L(a, h_\theta(s), y) = \left\lvert -\lvert h_\theta(s) - a\rvert^p - y \right\rvert^{1/p}
$$
where $p$ is close to 0.
Step2:  Generalize the loss above to
$$
L(a, h_\theta(s), y) = \left\lvert -\lvert h_\theta(s) - a\rvert^p - y \right\rvert^{q}
$$
where $q = 1/p$. Bringing $p$ arbitrarily close to 0 converges to the non-differentiable $L^0$ loss. Choosing $p=0.1$ seems to be small enough for practable purposes.
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Python Code:
#@title Default title text
# 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
#
# https://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.
Explanation: Copyright 2020 Google LLC.
Licensed under the Apache License, Version 2.0 (the "License");
End of explanation
import numpy as np
import tensorflow.compat.v2 as tf
import matplotlib.pyplot as plt
tf.enable_v2_behavior()
n = 20
lower, upper = -2, 2
modes_x, modes_y = zip(*[  # (x, y)
    (-1., -20),
    (0.0, -20),
    (1., -20),
])
x = np.concatenate((modes_x, np.random.uniform(lower, upper, size=n-len(modes_x))))
y = np.random.uniform(-1000, -300, size=n)
y[:len(modes_x)] = modes_y
y /= np.max(np.abs(y))
plt.scatter(x, y)
plt.show()
Explanation: Goal
We want to build a model $h_\theta(s) \rightarrow a^$ which predicts the mode $a^$ of some target distribution, for which we have unnormalized log-probabilities, $y$.
Stated another way, we want to find an appropriate loss L s.t. it is tractable to solve
$$
\text{argmin}\theta \sum\limits{i=1}^N L(\pi(a_i \mid h_\theta(s_i)), y_i)
$$
where $\theta$ are model parameters, $h$ is a model that depends on context $s$ and outputs the mode of $\pi$, and $\pi(a \mid \text{mode})$ is the predicted probability of action $a$. $y$ is information about the target distribution. We want the mode of $\pi$ to correspond to areas where $y$ is maximized.
Candidate
$$
L(a, h_\theta(s), y) = \left\lvert -\lvert h_\theta(s) - a\rvert^p - y \right\rvert^{1/p}
$$
where $p$ is close to 0.
End of explanation
# Mode
p=0.1
q=1/p
t = np.linspace(-2,2,200).reshape(-1, 1)
L = np.mean(np.power(np.abs(-np.power(np.abs(t - x[np.newaxis,:]), p) - y[np.newaxis,:]), q), axis=1)
plt.plot(t, L)
plt.show()
# Median
p=1.0
q=2.0
t = np.linspace(-2,2,200).reshape(-1, 1)
L = np.mean(np.power(np.abs(-np.power(np.abs(t - x[np.newaxis,:]), p) - y[np.newaxis,:]), q), axis=1)
#plt.ylim(0.5, 1.)
plt.plot(t, L)
plt.show()
# Mean
p=2.0
q=2.0
t = np.linspace(-2,2,200).reshape(-1, 1)
L = np.mean(np.power(np.abs(-np.power(np.abs(t - x[np.newaxis,:]), p) - y[np.newaxis,:]), q), axis=1)
#plt.ylim(0.5, 1.)
plt.plot(t, L)
plt.show()
Explanation: Generalize the loss above to
$$
L(a, h_\theta(s), y) = \left\lvert -\lvert h_\theta(s) - a\rvert^p - y \right\rvert^{q}
$$
where $q = 1/p$. Bringing $p$ arbitrarily close to 0 converges to the non-differentiable $L^0$ loss. Choosing $p=0.1$ seems to be small enough for practable purposes.
End of explanation
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