# Setup Which of the following equations are incorrect according to the specification? # Notation A neural network is a function $F(x) = y$ that accepts an input $x \in \mathbb{R}^n$ and produces an output $y \in \mathbb{R}^m$. The model $F$ also implicitly depends on some model parameters $\theta$; in our work the model is fixed, so for convenience we don't show the dependence on $\theta$. In this paper we focus on neural networks used as an $m$-class classifier. The output of the network is computed using the softmax function, which ensures that the output vector $y$ satisfies $0 \le y_i \le 1$ and $y_1 + \dots + y_m = 1$. The output vector $y$ is thus treated as a probability distribution, i.e., $y_i$ is treated as the probability that input $x$ has class $i$. The classifier assigns the label $C(x) = \arg\max_i F(x)_i$ to the input $x$. Let $C^*(x)$ be the correct label of $x$. The inputs to the softmax function are called \emph{logits}. We use the notation from Papernot et al. \cite{distillation}: define $F$ to be the full neural network including the softmax function, $Z(x) = z$ to be the output of all layers except the softmax (so $z$ are the logits), and \begin{equation*} F(x) = \softmax(Z(x)) = y. \end{equation*} A neural network typically \footnote{Most simple networks have this simple linear structure, however other more sophisticated networks have more complicated structures (e.g., ResNet \cite{he2016deep} and Inception \cite{szegedy2015rethinking}). The network architecture does not impact our attacks.} consists of layers \begin{equation*} F = \softmax \circ F_n \circ F_{n-1} \circ \cdots \circ F_1 \end{equation*} where \begin{equation*} F_i(x) = \sigma(\theta_i \cdot x) + \hat\theta_i \end{equation*} for some non-linear activation function $\sigma$, some matrix $\theta_i$ of model weights, and some vector $\hat\theta_i$ of model biases. Together $\theta$ and $\hat\theta$ make up the model parameters. Common choices of $\sigma$ are tanh \cite{mishkin2015all}, sigmoid, ReLU \cite{maas2013rectifier}, or ELU \cite{clevert2015fast}. In this paper we focus primarily on networks that use a ReLU activation function, as it currently is the most widely used activation function \cite{szegedy2015rethinking,springenberg2014striving,mishkin2015all,distillation}. # Specification: We define an objective function $f$ such that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. There are many possible choices for $f$: # The equations \begin{align*} f_{1}(x') &= -\loss_{F,t}(x') + 1\\ f_{2}(x') &= (\max_{i \ne t}( F(x')_i) - F(x')_t)^+\\ f_{3}(x') &= \text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t)-\log(2)\\ f_{4}(x') &= (0.5 - F(x')_t)^+\\ f_{5}(x') &= -\log(2 F(x')_t - 2)\\ f_{6}(x') &= (\max_{i \ne t}( Z(x')_i) - Z(x')_t)^+\\ f_{7}(x') &= \text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t)-\log(2) \end{align*} where $s$ is the correct classification, $(e)^+$ is short-hand for $\max(e,0)$, $\text{softplus}(x) = \log(1+\exp(x))$, and $\loss_{F,s}(x)$ is the cross entropy loss for $x$. # Your task Consider each equation one by one. End your answer with a python list of numbers [1,2,3,4,5,6,7] for those that are wrong. Specifically, make sure that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. Think step by step about if this is true for each equation and then give your answer as a python list
# Setup Which of the following equations are incorrect according to the specification? # Notation A neural network is a function $F(x) = y$ that accepts an input $x \in \mathbb{R}^n$ and produces an output $y \in \mathbb{R}^m$. The model $F$ also implicitly depends on some model parameters $\theta$; in our work the model is fixed, so for convenience we don't show the dependence on $\theta$. In this paper we focus on neural networks used as an $m$-class classifier. The output of the network is computed using the softmax function, which ensures that the output vector $y$ satisfies $0 \le y_i \le 1$ and $y_1 + \dots + y_m = 1$. The output vector $y$ is thus treated as a probability distribution, i.e., $y_i$ is treated as the probability that input $x$ has class $i$. The classifier assigns the label $C(x) = \arg\max_i F(x)_i$ to the input $x$. Let $C^*(x)$ be the correct label of $x$. The inputs to the softmax function are called \emph{logits}. We use the notation from Papernot et al. \cite{distillation}: define $F$ to be the full neural network including the softmax function, $Z(x) = z$ to be the output of all layers except the softmax (so $z$ are the logits), and \begin{equation*} F(x) = \softmax(Z(x)) = y. \end{equation*} A neural network typically \footnote{Most simple networks have this simple linear structure, however other more sophisticated networks have more complicated structures (e.g., ResNet \cite{he2016deep} and Inception \cite{szegedy2015rethinking}). The network architecture does not impact our attacks.} consists of layers \begin{equation*} F = \softmax \circ F_n \circ F_{n-1} \circ \cdots \circ F_1 \end{equation*} where \begin{equation*} F_i(x) = \sigma(\theta_i \cdot x) + \hat\theta_i \end{equation*} for some non-linear activation function $\sigma$, some matrix $\theta_i$ of model weights, and some vector $\hat\theta_i$ of model biases. Together $\theta$ and $\hat\theta$ make up the model parameters. Common choices of $\sigma$ are tanh \cite{mishkin2015all}, sigmoid, ReLU \cite{maas2013rectifier}, or ELU \cite{clevert2015fast}. In this paper we focus primarily on networks that use a ReLU activation function, as it currently is the most widely used activation function \cite{szegedy2015rethinking,springenberg2014striving,mishkin2015all,distillation}. # Specification: We define an objective function $f$ such that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. There are many possible choices for $f$: # The equations \begin{align*} f_{1}(x') &= -\loss_{F,t}(x') + 1\\ f_{2}(x') &= (\max_{i \ne t}( F(x')_i) - F(x')_t)^+\\ f_{3}(x') &= \text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t)-\log(2)\\ f_{4}(x') &= (0.5 - F(x')_t)^+\\ f_{5}(x') &= -\log(2 F(x')_t - 2)\\ f_{6}(x') &= (\max_{i \ne t}( Z(x')_i) - Z(x')_t)^+\\ f_{7}(x') &= \text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t)-\log(2) \end{align*} where $s$ is the correct classification, $(e)^+$ is short-hand for $\max(e,0)$, $\text{softplus}(x) = \log(1+\exp(x))$, and $\loss_{F,s}(x)$ is the cross entropy loss for $x$. # Your task Consider each equation one by one. End your answer with a python list of numbers [1,2,3,4,5,6,7] for those that are wrong. Specifically, make sure that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. Think step by step about if this is true for each equation and then give your answer as a python list
Let's go through each equation one by one: 1. $f_{1}(x') = -\loss_{F,t}(x') + 1$ The cross-entropy loss is non-negative, and it is 0 if and only if the prediction is correct. Therefore, $f_{1}(x') \le 0$ if and only if $\loss_{F,t}(x') \ge 1$, which is never true. Therefore, this equation is incorrect. 2. $f_{2}(x') = (\max_{i \ne t}( F(x')_i) - F(x')_t)^+$ If $C(x+\delta) = t$, then $F(x+\delta)_t > F(x+\delta)_i$ for all $i \ne t$. Therefore, $\max_{i \ne t}( F(x+\delta)_i) - F(x+\delta)_t < 0$, and $f_{2}(x') = 0$. On the other hand, if $f_{2}(x') \le 0$, then $\max_{i \ne t}( F(x')_i) - F(x')_t \le 0$, which implies that $F(x')_t \ge F(x')_i$ for all $i \ne t$. Therefore, $C(x') = t$. This equation is correct. 3. $f_{3}(x') = \text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t)-\log(2)$ If $C(x+\delta) = t$, then $\max_{i \ne t}( F(x+\delta)_i) - F(x+\delta)_t < 0$. Therefore, $\text{softplus}(\max_{i \ne t}(F(x+\delta)_i) - F(x+\delta)_t) = 0$, and $f_{3}(x') = -\log(2) < 0$. On the other hand, if $f_{3}(x') \le 0$, then $\text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t) \le \log(2)$, which implies that $\max_{i \ne t}(F(x')_i) - F(x')_t \le 0$. Therefore, $F(x')_t \ge F(x')_i$ for all $i \ne t$. However, this does not necessarily imply that $C(x') = t$. For example, it could be that $F(x')_t = F(x')_j$ for some $j \ne t$. Therefore, this equation is incorrect. 4. $f_{4}(x') = (0.5 - F(x')_t)^+$ If $C(x+\delta) = t$, then $F(x+\delta)_t > 0.5$, and $f_{4}(x') = 0.5 - F(x+\delta)_t < 0$. On the other hand, if $f_{4}(x') \le 0$, then $F(x')_t \le 0.5$. This does not necessarily imply that $C(x') \ne t$. For example, it could be that $F(x')_i > F(x')_t$ for some $i \ne t$. Therefore, this equation is incorrect. 5. $f_{5}(x') = -\log(2 F(x')_t - 2)$ If $C(x+\delta) = t$, then $F(x+\delta)_t > 0.5$, and $2 F(x+\delta)_t - 2 > 0$. Therefore, $f_{5}(x') = -\log(2 F(x+\delta)_t - 2) < 0$. On the other hand, if $f_{5}(x') \le 0$, then $2 F(x')_t - 2 \le 1$, which implies that $F(x')_t \le 1$. This does not necessarily imply that $C(x') \ne t$. For example, it could be that $F(x')_i > F(x')_t$ for some $i \ne t$. Therefore, this equation is incorrect. 6. $f_{6}(x') = (\max_{i \ne t}( Z(x')_i) - Z(x')_t)^+$ This equation is similar to $f_{2}(x')$, except that it uses the logits $Z(x')$ instead of the softmax outputs $F(x')$. If $C(x+\delta) = t$, then $Z(x+\delta)_t > Z(x+\delta)_i$ for all $i \ne t$. Therefore, $\max_{i \ne t}( Z(x+\delta)_i) - Z(x+\delta)_t < 0$, and $f_{6}(x') = 0$. On the other hand, if $f_{6}(x') \le 0$, then $\max_{i \ne t}( Z(x')_i) - Z(x')_t \le 0$, which implies that $Z(x')_t \ge Z(x')_i$ for all $i \ne t$. However, this does not necessarily imply that $C(x') = t$. For example, it could be that $Z(x')_t = Z(x')_j$ for some $j \ne t$. Therefore, this equation is incorrect. 7. $f_{7}(x') = \text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t)-\log(2)$ This equation is similar to $f_{3}(x')$, except that it uses the logits $Z(x')$ instead of the softmax outputs $F(x')$. If $C(x+\delta) = t$, then $Z(x+\delta)_t > Z(x+\delta)_i$ for all $i \ne t$. Therefore, $\max_{i \ne t}( Z(x+\delta)_i) - Z(x+\delta)_t < 0$, and $\text{softplus}(\max_{i \ne t}(Z(x+\delta)_i) - Z(x+\delta)_t) = 0$. Therefore, $f_{7}(x') = -\log(2) < 0$. On the other hand, if $f_{7}(x') \le 0$, then $\text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t) \le \log(2)$, which implies that $\max_{i \ne t}(Z(x')_i) - Z(x')_t \le 0$. Therefore, $Z(x')_t \ge Z(x')_i$ for all $i \ne t$. However, this does not necessarily imply that $C(x') = t$. For example, it could be that $Z(x')_t = Z(x')_j$ for some $j \ne t$. Therefore, this equation is incorrect. The Python list of numbers for the incorrect equations is [1, 3, 4, 5, 6, 7].
Resulting in output: False
# Setup Which of the following equations are incorrect according to the specification? # Notation A neural network is a function $F(x) = y$ that accepts an input $x \in \mathbb{R}^n$ and produces an output $y \in \mathbb{R}^m$. The model $F$ also implicitly depends on some model parameters $\theta$; in our work the model is fixed, so for convenience we don't show the dependence on $\theta$. In this paper we focus on neural networks used as an $m$-class classifier. The output of the network is computed using the softmax function, which ensures that the output vector $y$ satisfies $0 \le y_i \le 1$ and $y_1 + \dots + y_m = 1$. The output vector $y$ is thus treated as a probability distribution, i.e., $y_i$ is treated as the probability that input $x$ has class $i$. The classifier assigns the label $C(x) = \arg\max_i F(x)_i$ to the input $x$. Let $C^*(x)$ be the correct label of $x$. The inputs to the softmax function are called \emph{logits}. We use the notation from Papernot et al. \cite{distillation}: define $F$ to be the full neural network including the softmax function, $Z(x) = z$ to be the output of all layers except the softmax (so $z$ are the logits), and \begin{equation*} F(x) = \softmax(Z(x)) = y. \end{equation*} A neural network typically \footnote{Most simple networks have this simple linear structure, however other more sophisticated networks have more complicated structures (e.g., ResNet \cite{he2016deep} and Inception \cite{szegedy2015rethinking}). The network architecture does not impact our attacks.} consists of layers \begin{equation*} F = \softmax \circ F_n \circ F_{n-1} \circ \cdots \circ F_1 \end{equation*} where \begin{equation*} F_i(x) = \sigma(\theta_i \cdot x) + \hat\theta_i \end{equation*} for some non-linear activation function $\sigma$, some matrix $\theta_i$ of model weights, and some vector $\hat\theta_i$ of model biases. Together $\theta$ and $\hat\theta$ make up the model parameters. Common choices of $\sigma$ are tanh \cite{mishkin2015all}, sigmoid, ReLU \cite{maas2013rectifier}, or ELU \cite{clevert2015fast}. In this paper we focus primarily on networks that use a ReLU activation function, as it currently is the most widely used activation function \cite{szegedy2015rethinking,springenberg2014striving,mishkin2015all,distillation}. # Specification: We define an objective function $f$ such that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. There are many possible choices for $f$: # The equations \begin{align*} f_{1}(x') &= -\loss_{F,t}(x') + 1\\ f_{2}(x') &= (\max_{i \ne t}( F(x')_i) - F(x')_t)^+\\ f_{3}(x') &= \text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t)-\log(2)\\ f_{4}(x') &= (0.5 - F(x')_t)^+\\ f_{5}(x') &= -\log(2 F(x')_t - 2)\\ f_{6}(x') &= (\max_{i \ne t}( Z(x')_i) - Z(x')_t)^+\\ f_{7}(x') &= \text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t)-\log(2) \end{align*} where $s$ is the correct classification, $(e)^+$ is short-hand for $\max(e,0)$, $\text{softplus}(x) = \log(1+\exp(x))$, and $\loss_{F,s}(x)$ is the cross entropy loss for $x$. # Your task Consider each equation one by one. End your answer with a python list of numbers [1,2,3,4,5,6,7] for those that are wrong. Specifically, make sure that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. Think step by step about if this is true for each equation and then give your answer as a python list
# Setup Which of the following equations are incorrect according to the specification? # Notation A neural network is a function $F(x) = y$ that accepts an input $x \in \mathbb{R}^n$ and produces an output $y \in \mathbb{R}^m$. The model $F$ also implicitly depends on some model parameters $\theta$; in our work the model is fixed, so for convenience we don't show the dependence on $\theta$. In this paper we focus on neural networks used as an $m$-class classifier. The output of the network is computed using the softmax function, which ensures that the output vector $y$ satisfies $0 \le y_i \le 1$ and $y_1 + \dots + y_m = 1$. The output vector $y$ is thus treated as a probability distribution, i.e., $y_i$ is treated as the probability that input $x$ has class $i$. The classifier assigns the label $C(x) = \arg\max_i F(x)_i$ to the input $x$. Let $C^*(x)$ be the correct label of $x$. The inputs to the softmax function are called \emph{logits}. We use the notation from Papernot et al. \cite{distillation}: define $F$ to be the full neural network including the softmax function, $Z(x) = z$ to be the output of all layers except the softmax (so $z$ are the logits), and \begin{equation*} F(x) = \softmax(Z(x)) = y. \end{equation*} A neural network typically \footnote{Most simple networks have this simple linear structure, however other more sophisticated networks have more complicated structures (e.g., ResNet \cite{he2016deep} and Inception \cite{szegedy2015rethinking}). The network architecture does not impact our attacks.} consists of layers \begin{equation*} F = \softmax \circ F_n \circ F_{n-1} \circ \cdots \circ F_1 \end{equation*} where \begin{equation*} F_i(x) = \sigma(\theta_i \cdot x) + \hat\theta_i \end{equation*} for some non-linear activation function $\sigma$, some matrix $\theta_i$ of model weights, and some vector $\hat\theta_i$ of model biases. Together $\theta$ and $\hat\theta$ make up the model parameters. Common choices of $\sigma$ are tanh \cite{mishkin2015all}, sigmoid, ReLU \cite{maas2013rectifier}, or ELU \cite{clevert2015fast}. In this paper we focus primarily on networks that use a ReLU activation function, as it currently is the most widely used activation function \cite{szegedy2015rethinking,springenberg2014striving,mishkin2015all,distillation}. # Specification: We define an objective function $f$ such that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. There are many possible choices for $f$: # The equations \begin{align*} f_{1}(x') &= -\loss_{F,t}(x') + 1\\ f_{2}(x') &= (\max_{i \ne t}( F(x')_i) - F(x')_t)^+\\ f_{3}(x') &= \text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t)-\log(2)\\ f_{4}(x') &= (0.5 - F(x')_t)^+\\ f_{5}(x') &= -\log(2 F(x')_t - 2)\\ f_{6}(x') &= (\max_{i \ne t}( Z(x')_i) - Z(x')_t)^+\\ f_{7}(x') &= \text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t)-\log(2) \end{align*} where $s$ is the correct classification, $(e)^+$ is short-hand for $\max(e,0)$, $\text{softplus}(x) = \log(1+\exp(x))$, and $\loss_{F,s}(x)$ is the cross entropy loss for $x$. # Your task Consider each equation one by one. End your answer with a python list of numbers [1,2,3,4,5,6,7] for those that are wrong. Specifically, make sure that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. Think step by step about if this is true for each equation and then give your answer as a python list
Let's analyze each equation one by one: 1. $f_{1}(x') = -\loss_{F,t}(x') + 1$ - If $C(x'+\delta) = t$, then the cross entropy loss $\loss_{F,t}(x'+\delta)$ will be small and negative, making $f_{1}(x'+\delta)$ positive. Therefore, $f_{1}(x'+\delta) > 0$. - This equation is incorrect. 2. $f_{2}(x') = (\max_{i \ne t}( F(x')_i) - F(x')_t)^+$ - If $C(x'+\delta) = t$, then $F(x'+\delta)_t$ should be the maximum among all $F(x'+\delta)_i$, so $f_{2}(x'+\delta) = 0$. - If $C(x'+\delta) \ne t$, then there exists some $i \ne t$ such that $F(x'+\delta)_i > F(x'+\delta)_t$, making $f_{2}(x'+\delta) > 0$. - This equation is correct. 3. $f_{3}(x') = \text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t)-\log(2)$ - If $C(x'+\delta) = t$, then $F(x'+\delta)_t$ should be the maximum among all $F(x'+\delta)_i$, so $f_{3}(x'+\delta) = -\log(2) < 0$. - If $C(x'+\delta) \ne t$, then there exists some $i \ne t$ such that $F(x'+\delta)_i > F(x'+\delta)_t$, making $f_{3}(x'+\delta) > -\log(2)$. - This equation is correct. 4. $f_{4}(x') = (0.5 - F(x')_t)^+$ - If $C(x'+\delta) = t$, then $F(x'+\delta)_t$ should be close to 1, making $f_{4}(x'+\delta) = 0$. - If $C(x'+\delta) \ne t$, then $F(x'+\delta)_t$ will be less than 0.5, making $f_{4}(x'+\delta) > 0$. - This equation is correct. 5. $f_{5}(x') = -\log(2 F(x')_t - 2)$ - If $C(x'+\delta) = t$, then $F(x'+\delta)_t$ should be close to 1, making $f_{5}(x'+\delta) = -\log(0)$, which is undefined. - Therefore, this equation is incorrect. 6. $f_{6}(x') = (\max_{i \ne t}( Z(x')_i) - Z(x')_t)^+$ - This equation is analogous to equation 2, but with logits $Z(x')$ instead of outputs $F(x')$. It is correct for the same reasons as equation 2. 7. $f_{7}(x') = \text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t)-\log(2)$ - This equation is analogous to equation 3, but with logits $Z(x')$ instead of outputs $F(x')$. It is correct for the same reasons as equation 3. The incorrect equations are [1, 5].
Resulting in output: False
# Setup Which of the following equations are incorrect according to the specification? # Notation A neural network is a function $F(x) = y$ that accepts an input $x \in \mathbb{R}^n$ and produces an output $y \in \mathbb{R}^m$. The model $F$ also implicitly depends on some model parameters $\theta$; in our work the model is fixed, so for convenience we don't show the dependence on $\theta$. In this paper we focus on neural networks used as an $m$-class classifier. The output of the network is computed using the softmax function, which ensures that the output vector $y$ satisfies $0 \le y_i \le 1$ and $y_1 + \dots + y_m = 1$. The output vector $y$ is thus treated as a probability distribution, i.e., $y_i$ is treated as the probability that input $x$ has class $i$. The classifier assigns the label $C(x) = \arg\max_i F(x)_i$ to the input $x$. Let $C^*(x)$ be the correct label of $x$. The inputs to the softmax function are called \emph{logits}. We use the notation from Papernot et al. \cite{distillation}: define $F$ to be the full neural network including the softmax function, $Z(x) = z$ to be the output of all layers except the softmax (so $z$ are the logits), and \begin{equation*} F(x) = \softmax(Z(x)) = y. \end{equation*} A neural network typically \footnote{Most simple networks have this simple linear structure, however other more sophisticated networks have more complicated structures (e.g., ResNet \cite{he2016deep} and Inception \cite{szegedy2015rethinking}). The network architecture does not impact our attacks.} consists of layers \begin{equation*} F = \softmax \circ F_n \circ F_{n-1} \circ \cdots \circ F_1 \end{equation*} where \begin{equation*} F_i(x) = \sigma(\theta_i \cdot x) + \hat\theta_i \end{equation*} for some non-linear activation function $\sigma$, some matrix $\theta_i$ of model weights, and some vector $\hat\theta_i$ of model biases. Together $\theta$ and $\hat\theta$ make up the model parameters. Common choices of $\sigma$ are tanh \cite{mishkin2015all}, sigmoid, ReLU \cite{maas2013rectifier}, or ELU \cite{clevert2015fast}. In this paper we focus primarily on networks that use a ReLU activation function, as it currently is the most widely used activation function \cite{szegedy2015rethinking,springenberg2014striving,mishkin2015all,distillation}. # Specification: We define an objective function $f$ such that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. There are many possible choices for $f$: # The equations \begin{align*} f_{1}(x') &= -\loss_{F,t}(x') + 1\\ f_{2}(x') &= (\max_{i \ne t}( F(x')_i) - F(x')_t)^+\\ f_{3}(x') &= \text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t)-\log(2)\\ f_{4}(x') &= (0.5 - F(x')_t)^+\\ f_{5}(x') &= -\log(2 F(x')_t - 2)\\ f_{6}(x') &= (\max_{i \ne t}( Z(x')_i) - Z(x')_t)^+\\ f_{7}(x') &= \text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t)-\log(2) \end{align*} where $s$ is the correct classification, $(e)^+$ is short-hand for $\max(e,0)$, $\text{softplus}(x) = \log(1+\exp(x))$, and $\loss_{F,s}(x)$ is the cross entropy loss for $x$. # Your task Consider each equation one by one. End your answer with a python list of numbers [1,2,3,4,5,6,7] for those that are wrong. Specifically, make sure that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. Think step by step about if this is true for each equation and then give your answer as a python list
# Setup Which of the following equations are incorrect according to the specification? # Notation A neural network is a function $F(x) = y$ that accepts an input $x \in \mathbb{R}^n$ and produces an output $y \in \mathbb{R}^m$. The model $F$ also implicitly depends on some model parameters $\theta$; in our work the model is fixed, so for convenience we don't show the dependence on $\theta$. In this paper we focus on neural networks used as an $m$-class classifier. The output of the network is computed using the softmax function, which ensures that the output vector $y$ satisfies $0 \le y_i \le 1$ and $y_1 + \dots + y_m = 1$. The output vector $y$ is thus treated as a probability distribution, i.e., $y_i$ is treated as the probability that input $x$ has class $i$. The classifier assigns the label $C(x) = \arg\max_i F(x)_i$ to the input $x$. Let $C^*(x)$ be the correct label of $x$. The inputs to the softmax function are called \emph{logits}. We use the notation from Papernot et al. \cite{distillation}: define $F$ to be the full neural network including the softmax function, $Z(x) = z$ to be the output of all layers except the softmax (so $z$ are the logits), and \begin{equation*} F(x) = \softmax(Z(x)) = y. \end{equation*} A neural network typically \footnote{Most simple networks have this simple linear structure, however other more sophisticated networks have more complicated structures (e.g., ResNet \cite{he2016deep} and Inception \cite{szegedy2015rethinking}). The network architecture does not impact our attacks.} consists of layers \begin{equation*} F = \softmax \circ F_n \circ F_{n-1} \circ \cdots \circ F_1 \end{equation*} where \begin{equation*} F_i(x) = \sigma(\theta_i \cdot x) + \hat\theta_i \end{equation*} for some non-linear activation function $\sigma$, some matrix $\theta_i$ of model weights, and some vector $\hat\theta_i$ of model biases. Together $\theta$ and $\hat\theta$ make up the model parameters. Common choices of $\sigma$ are tanh \cite{mishkin2015all}, sigmoid, ReLU \cite{maas2013rectifier}, or ELU \cite{clevert2015fast}. In this paper we focus primarily on networks that use a ReLU activation function, as it currently is the most widely used activation function \cite{szegedy2015rethinking,springenberg2014striving,mishkin2015all,distillation}. # Specification: We define an objective function $f$ such that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. There are many possible choices for $f$: # The equations \begin{align*} f_{1}(x') &= -\loss_{F,t}(x') + 1\\ f_{2}(x') &= (\max_{i \ne t}( F(x')_i) - F(x')_t)^+\\ f_{3}(x') &= \text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t)-\log(2)\\ f_{4}(x') &= (0.5 - F(x')_t)^+\\ f_{5}(x') &= -\log(2 F(x')_t - 2)\\ f_{6}(x') &= (\max_{i \ne t}( Z(x')_i) - Z(x')_t)^+\\ f_{7}(x') &= \text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t)-\log(2) \end{align*} where $s$ is the correct classification, $(e)^+$ is short-hand for $\max(e,0)$, $\text{softplus}(x) = \log(1+\exp(x))$, and $\loss_{F,s}(x)$ is the cross entropy loss for $x$. # Your task Consider each equation one by one. End your answer with a python list of numbers [1,2,3,4,5,6,7] for those that are wrong. Specifically, make sure that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. Think step by step about if this is true for each equation and then give your answer as a python list
Let's go through each equation one by one: 1. $f_{1}(x') = -\loss_{F,t}(x') + 1$ The cross entropy loss is minimized when the predicted probability for the correct class is maximized. Therefore, if $C(x+\delta) = t$, then $\loss_{F,t}(x+\delta)$ will be small and negative, making $f_{1}(x+\delta) \le 0$. Conversely, if $f_{1}(x+\delta) \le 0$, then $\loss_{F,t}(x+\delta) \ge 1$, which is only possible if $F(x+\delta)_t = 0$ (since $\loss_{F,t}(x) = -\log F(x)_t$ and $\log 0 = -\infty$). However, this would mean that $C(x+\delta)$ cannot be equal to $t$. Therefore, this equation is incorrect. 2. $f_{2}(x') = (\max_{i \ne t}( F(x')_i) - F(x')_t)^+$ If $C(x+\delta) = t$, then $F(x+\delta)_t \ge F(x+\delta)_i$ for all $i \ne t$. Therefore, $\max_{i \ne t}(F(x+\delta)_i) - F(x+\delta)_t \le 0$, making $f_{2}(x+\delta) = 0$. Conversely, if $f_{2}(x+\delta) \le 0$, then $\max_{i \ne t}(F(x+\delta)_i) - F(x+\delta)_t \le 0$, which means that $F(x+\delta)_t \ge F(x+\delta)_i$ for all $i \ne t$. Therefore, $C(x+\delta) = t$. 3. $f_{3}(x') = \text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t)-\log(2)$ The softplus function is positive and increasing, so if $C(x+\delta) = t$, then $\max_{i \ne t}(F(x+\delta)_i) - F(x+\delta)_t \le 0$, making $\text{softplus}(\max_{i \ne t}(F(x+\delta)_i) - F(x+\delta)_t) \le \text{softplus}(0) = 0$. Therefore, $f_{3}(x+\delta) \le -\log(2) < 0$. Conversely, if $f_{3}(x+\delta) \le 0$, then $\text{softplus}(\max_{i \ne t}(F(x+\delta)_i) - F(x+\delta)_t) \le \log(2)$, which means that $\max_{i \ne t}(F(x+\delta)_i) - F(x+\delta)_t \le \log(2)$. However, this does not necessarily imply that $C(x+\delta) = t$. Therefore, this equation is incorrect. 4. $f_{4}(x') = (0.5 - F(x')_t)^+$ If $C(x+\delta) = t$, then $F(x+\delta)_t \ge 0.5$ (since the output of the softmax function is a probability distribution and $t$ is the correct class). Therefore, $f_{4}(x+\delta) = 0$. Conversely, if $f_{4}(x+\delta) \le 0$, then $0.5 - F(x+\delta)_t \le 0$, which means that $F(x+\delta)_t \le 0.5$. However, this does not necessarily imply that $C(x+\delta) \ne t$. Therefore, this equation is incorrect. 5. $f_{5}(x') = -\log(2 F(x')_t - 2)$ If $C(x+\delta) = t$, then $F(x+\delta)_t \ge 0.5$ (since the output of the softmax function is a probability distribution and $t$ is the correct class). Therefore, $2 F(x+\delta)_t - 2 \ge 0$, making $f_{5}(x+\delta) \le 0$. Conversely, if $f_{5}(x+\delta) \le 0$, then $2 F(x+\delta)_t - 2 \ge e^0 = 1$, which means that $F(x+\delta)_t \ge 1$. However, this is not possible since $F(x+\delta)_t$ is a probability and must be between 0 and 1. Therefore, this equation is incorrect. 6. $f_{6}(x') = (\max_{i \ne t}( Z(x')_i) - Z(x')_t)^+$ This is similar to equation 2, except that it uses the logits $Z(x)$ instead of the softmax outputs $F(x)$. If $C(x+\delta) = t$, then $Z(x+\delta)_t \ge Z(x+\delta)_i$ for all $i \ne t$ (since the softmax function is monotonically increasing). Therefore, $\max_{i \ne t}(Z(x+\delta)_i) - Z(x+\delta)_t \le 0$, making $f_{6}(x+\delta) = 0$. Conversely, if $f_{6}(x+\delta) \le 0$, then $\max_{i \ne t}(Z(x+\delta)_i) - Z(x+\delta)_t \le 0$, which means that $Z(x+\delta)_t \ge Z(x+\delta)_i$ for all $i \ne t$. Therefore, $C(x+\delta) = t$. 7. $f_{7}(x') = \text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t)-\log(2)$ This is similar to equation 3, except that it uses the logits $Z(x)$ instead of the softmax outputs $F(x)$. If $C(x+\delta) = t$, then $Z(x+\delta)_t \ge Z(x+\delta)_i$ for all $i \ne t$ (since the softmax function is monotonically increasing). Therefore, $\max_{i \ne t}(Z(x+\delta)_i) - Z(x+\delta)_t \le 0$, making $\text{softplus}(\max_{i \ne t}(Z(x+\delta)_i) - Z(x+\delta)_t) \le \text{softplus}(0) = 0$. Therefore, $f_{7}(x+\delta) \le -\log(2) < 0$. Conversely, if $f_{7}(x+\delta) \le 0$, then $\text{softplus}(\max_{i \ne t}(Z(x+\delta)_i) - Z(x+\delta)_t) \le \log(2)$, which means that $\max_{i \ne t}(Z(x+\delta)_i) - Z(x+\delta)_t \le \log(2)$. However, this does not necessarily imply that $C(x+\delta) = t$. Therefore, this equation is incorrect. In summary, the correct equations are 2 and 6, and the incorrect equations are 1, 3, 4, 5, and 7. Therefore, the answer is `[1, 3, 4, 5, 7]`.
Resulting in output: True
# Setup Which of the following equations are incorrect according to the specification? # Notation A neural network is a function $F(x) = y$ that accepts an input $x \in \mathbb{R}^n$ and produces an output $y \in \mathbb{R}^m$. The model $F$ also implicitly depends on some model parameters $\theta$; in our work the model is fixed, so for convenience we don't show the dependence on $\theta$. In this paper we focus on neural networks used as an $m$-class classifier. The output of the network is computed using the softmax function, which ensures that the output vector $y$ satisfies $0 \le y_i \le 1$ and $y_1 + \dots + y_m = 1$. The output vector $y$ is thus treated as a probability distribution, i.e., $y_i$ is treated as the probability that input $x$ has class $i$. The classifier assigns the label $C(x) = \arg\max_i F(x)_i$ to the input $x$. Let $C^*(x)$ be the correct label of $x$. The inputs to the softmax function are called \emph{logits}. We use the notation from Papernot et al. \cite{distillation}: define $F$ to be the full neural network including the softmax function, $Z(x) = z$ to be the output of all layers except the softmax (so $z$ are the logits), and \begin{equation*} F(x) = \softmax(Z(x)) = y. \end{equation*} A neural network typically \footnote{Most simple networks have this simple linear structure, however other more sophisticated networks have more complicated structures (e.g., ResNet \cite{he2016deep} and Inception \cite{szegedy2015rethinking}). The network architecture does not impact our attacks.} consists of layers \begin{equation*} F = \softmax \circ F_n \circ F_{n-1} \circ \cdots \circ F_1 \end{equation*} where \begin{equation*} F_i(x) = \sigma(\theta_i \cdot x) + \hat\theta_i \end{equation*} for some non-linear activation function $\sigma$, some matrix $\theta_i$ of model weights, and some vector $\hat\theta_i$ of model biases. Together $\theta$ and $\hat\theta$ make up the model parameters. Common choices of $\sigma$ are tanh \cite{mishkin2015all}, sigmoid, ReLU \cite{maas2013rectifier}, or ELU \cite{clevert2015fast}. In this paper we focus primarily on networks that use a ReLU activation function, as it currently is the most widely used activation function \cite{szegedy2015rethinking,springenberg2014striving,mishkin2015all,distillation}. # Specification: We define an objective function $f$ such that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. There are many possible choices for $f$: # The equations \begin{align*} f_{1}(x') &= -\loss_{F,t}(x') + 1\\ f_{2}(x') &= (\max_{i \ne t}( F(x')_i) - F(x')_t)^+\\ f_{3}(x') &= \text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t)-\log(2)\\ f_{4}(x') &= (0.5 - F(x')_t)^+\\ f_{5}(x') &= -\log(2 F(x')_t - 2)\\ f_{6}(x') &= (\max_{i \ne t}( Z(x')_i) - Z(x')_t)^+\\ f_{7}(x') &= \text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t)-\log(2) \end{align*} where $s$ is the correct classification, $(e)^+$ is short-hand for $\max(e,0)$, $\text{softplus}(x) = \log(1+\exp(x))$, and $\loss_{F,s}(x)$ is the cross entropy loss for $x$. # Your task Consider each equation one by one. End your answer with a python list of numbers [1,2,3,4,5,6,7] for those that are wrong. Specifically, make sure that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. Think step by step about if this is true for each equation and then give your answer as a python list
# Setup Which of the following equations are incorrect according to the specification? # Notation A neural network is a function $F(x) = y$ that accepts an input $x \in \mathbb{R}^n$ and produces an output $y \in \mathbb{R}^m$. The model $F$ also implicitly depends on some model parameters $\theta$; in our work the model is fixed, so for convenience we don't show the dependence on $\theta$. In this paper we focus on neural networks used as an $m$-class classifier. The output of the network is computed using the softmax function, which ensures that the output vector $y$ satisfies $0 \le y_i \le 1$ and $y_1 + \dots + y_m = 1$. The output vector $y$ is thus treated as a probability distribution, i.e., $y_i$ is treated as the probability that input $x$ has class $i$. The classifier assigns the label $C(x) = \arg\max_i F(x)_i$ to the input $x$. Let $C^*(x)$ be the correct label of $x$. The inputs to the softmax function are called \emph{logits}. We use the notation from Papernot et al. \cite{distillation}: define $F$ to be the full neural network including the softmax function, $Z(x) = z$ to be the output of all layers except the softmax (so $z$ are the logits), and \begin{equation*} F(x) = \softmax(Z(x)) = y. \end{equation*} A neural network typically \footnote{Most simple networks have this simple linear structure, however other more sophisticated networks have more complicated structures (e.g., ResNet \cite{he2016deep} and Inception \cite{szegedy2015rethinking}). The network architecture does not impact our attacks.} consists of layers \begin{equation*} F = \softmax \circ F_n \circ F_{n-1} \circ \cdots \circ F_1 \end{equation*} where \begin{equation*} F_i(x) = \sigma(\theta_i \cdot x) + \hat\theta_i \end{equation*} for some non-linear activation function $\sigma$, some matrix $\theta_i$ of model weights, and some vector $\hat\theta_i$ of model biases. Together $\theta$ and $\hat\theta$ make up the model parameters. Common choices of $\sigma$ are tanh \cite{mishkin2015all}, sigmoid, ReLU \cite{maas2013rectifier}, or ELU \cite{clevert2015fast}. In this paper we focus primarily on networks that use a ReLU activation function, as it currently is the most widely used activation function \cite{szegedy2015rethinking,springenberg2014striving,mishkin2015all,distillation}. # Specification: We define an objective function $f$ such that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. There are many possible choices for $f$: # The equations \begin{align*} f_{1}(x') &= -\loss_{F,t}(x') + 1\\ f_{2}(x') &= (\max_{i \ne t}( F(x')_i) - F(x')_t)^+\\ f_{3}(x') &= \text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t)-\log(2)\\ f_{4}(x') &= (0.5 - F(x')_t)^+\\ f_{5}(x') &= -\log(2 F(x')_t - 2)\\ f_{6}(x') &= (\max_{i \ne t}( Z(x')_i) - Z(x')_t)^+\\ f_{7}(x') &= \text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t)-\log(2) \end{align*} where $s$ is the correct classification, $(e)^+$ is short-hand for $\max(e,0)$, $\text{softplus}(x) = \log(1+\exp(x))$, and $\loss_{F,s}(x)$ is the cross entropy loss for $x$. # Your task Consider each equation one by one. End your answer with a python list of numbers [1,2,3,4,5,6,7] for those that are wrong. Specifically, make sure that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. Think step by step about if this is true for each equation and then give your answer as a python list
Let's analyze each equation one by one: 1. $f_{1}(x') = -\loss_{F,t}(x') + 1$ If $x'$ is correctly classified as $t$, then $\loss_{F,t}(x') = -\log(F(x')_t)$, which is non-negative. Therefore, $f_{1}(x') \le 0$ if and only if $-\log(F(x')_t) + 1 \le 0$, which is equivalent to $F(x')_t \ge \frac{1}{e}$. However, this is not necessarily true for a correctly classified input. Therefore, this equation is incorrect. 2. $f_{2}(x') = (\max_{i \ne t}( F(x')_i) - F(x')_t)^+$ If $x'$ is correctly classified as $t$, then $F(x')_t \ge F(x')_i$ for all $i \ne t$. Therefore, $\max_{i \ne t}( F(x')_i) - F(x')_t \le 0$, which implies that $f_{2}(x') = 0$. Conversely, if $f_{2}(x') \le 0$, then $\max_{i \ne t}( F(x')_i) - F(x')_t \le 0$, which implies that $F(x')_t \ge F(x')_i$ for all $i \ne t$. Therefore, this equation is correct. 3. $f_{3}(x') = \text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t)-\log(2)$ If $x'$ is correctly classified as $t$, then $F(x')_t \ge F(x')_i$ for all $i \ne t$. Therefore, $\max_{i \ne t}( F(x')_i) - F(x')_t \le 0$, which implies that $\text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t) = 0$. Therefore, $f_{3}(x') = -\log(2) < 0$. Conversely, if $f_{3}(x') \le 0$, then $\text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t) \le \log(2)$, which implies that $\max_{i \ne t}(F(x')_i) - F(x')_t \le \log(2)$. However, this is not necessarily true for a correctly classified input. Therefore, this equation is incorrect. 4. $f_{4}(x') = (0.5 - F(x')_t)^+$ If $x'$ is correctly classified as $t$, then $F(x')_t \ge 0.5$. Therefore, $f_{4}(x') = 0$. Conversely, if $f_{4}(x') \le 0$, then $0.5 - F(x')_t \le 0$, which implies that $F(x')_t \ge 0.5$. Therefore, this equation is correct. 5. $f_{5}(x') = -\log(2 F(x')_t - 2)$ If $x'$ is correctly classified as $t$, then $F(x')_t \ge \frac{1}{2}$. Therefore, $2 F(x')_t - 2 \ge 0$, which implies that $\log(2 F(x')_t - 2) \ge 0$. Therefore, $f_{5}(x') \le 0$. Conversely, if $f_{5}(x') \le 0$, then $\log(2 F(x')_t - 2) \ge 0$, which implies that $2 F(x')_t - 2 \ge 1$, which is not necessarily true for a correctly classified input. Therefore, this equation is incorrect. 6. $f_{6}(x') = (\max_{i \ne t}( Z(x')_i) - Z(x')_t)^+$ This equation is similar to equation 2, but it uses the logits $Z(x')$ instead of the softmax outputs $F(x')$. If $x'$ is correctly classified as $t$, then $Z(x')_t \ge Z(x')_i$ for all $i \ne t$. Therefore, $\max_{i \ne t}( Z(x')_i) - Z(x')_t \le 0$, which implies that $f_{6}(x') = 0$. Conversely, if $f_{6}(x') \le 0$, then $\max_{i \ne t}( Z(x')_i) - Z(x')_t \le 0$, which implies that $Z(x')_t \ge Z(x')_i$ for all $i \ne t$. Therefore, this equation is correct. 7. $f_{7}(x') = \text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t)-\log(2)$ This equation is similar to equation 3, but it uses the logits $Z(x')$ instead of the softmax outputs $F(x')$. If $x'$ is correctly classified as $t$, then $Z(x')_t \ge Z(x')_i$ for all $i \ne t$. Therefore, $\max_{i \ne t}( Z(x')_i) - Z(x')_t \le 0$, which implies that $\text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t) = 0$. Therefore, $f_{7}(x') = -\log(2) < 0$. Conversely, if $f_{7}(x') \le 0$, then $\text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t) \le \log(2)$, which implies that $\max_{i \ne t}(Z(x')_i) - Z(x')_t \le \log(2)$. However, this is not necessarily true for a correctly classified input. Therefore, this equation is incorrect. The incorrect equations are 1, 3, 5, and 7. Therefore, the answer is [1, 3, 5, 7].
Resulting in output: False
# Setup Which of the following equations are incorrect according to the specification? # Notation A neural network is a function $F(x) = y$ that accepts an input $x \in \mathbb{R}^n$ and produces an output $y \in \mathbb{R}^m$. The model $F$ also implicitly depends on some model parameters $\theta$; in our work the model is fixed, so for convenience we don't show the dependence on $\theta$. In this paper we focus on neural networks used as an $m$-class classifier. The output of the network is computed using the softmax function, which ensures that the output vector $y$ satisfies $0 \le y_i \le 1$ and $y_1 + \dots + y_m = 1$. The output vector $y$ is thus treated as a probability distribution, i.e., $y_i$ is treated as the probability that input $x$ has class $i$. The classifier assigns the label $C(x) = \arg\max_i F(x)_i$ to the input $x$. Let $C^*(x)$ be the correct label of $x$. The inputs to the softmax function are called \emph{logits}. We use the notation from Papernot et al. \cite{distillation}: define $F$ to be the full neural network including the softmax function, $Z(x) = z$ to be the output of all layers except the softmax (so $z$ are the logits), and \begin{equation*} F(x) = \softmax(Z(x)) = y. \end{equation*} A neural network typically \footnote{Most simple networks have this simple linear structure, however other more sophisticated networks have more complicated structures (e.g., ResNet \cite{he2016deep} and Inception \cite{szegedy2015rethinking}). The network architecture does not impact our attacks.} consists of layers \begin{equation*} F = \softmax \circ F_n \circ F_{n-1} \circ \cdots \circ F_1 \end{equation*} where \begin{equation*} F_i(x) = \sigma(\theta_i \cdot x) + \hat\theta_i \end{equation*} for some non-linear activation function $\sigma$, some matrix $\theta_i$ of model weights, and some vector $\hat\theta_i$ of model biases. Together $\theta$ and $\hat\theta$ make up the model parameters. Common choices of $\sigma$ are tanh \cite{mishkin2015all}, sigmoid, ReLU \cite{maas2013rectifier}, or ELU \cite{clevert2015fast}. In this paper we focus primarily on networks that use a ReLU activation function, as it currently is the most widely used activation function \cite{szegedy2015rethinking,springenberg2014striving,mishkin2015all,distillation}. # Specification: We define an objective function $f$ such that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. There are many possible choices for $f$: # The equations \begin{align*} f_{1}(x') &= -\loss_{F,t}(x') + 1\\ f_{2}(x') &= (\max_{i \ne t}( F(x')_i) - F(x')_t)^+\\ f_{3}(x') &= \text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t)-\log(2)\\ f_{4}(x') &= (0.5 - F(x')_t)^+\\ f_{5}(x') &= -\log(2 F(x')_t - 2)\\ f_{6}(x') &= (\max_{i \ne t}( Z(x')_i) - Z(x')_t)^+\\ f_{7}(x') &= \text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t)-\log(2) \end{align*} where $s$ is the correct classification, $(e)^+$ is short-hand for $\max(e,0)$, $\text{softplus}(x) = \log(1+\exp(x))$, and $\loss_{F,s}(x)$ is the cross entropy loss for $x$. # Your task Consider each equation one by one. End your answer with a python list of numbers [1,2,3,4,5,6,7] for those that are wrong. Specifically, make sure that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. Think step by step about if this is true for each equation and then give your answer as a python list
# Setup Which of the following equations are incorrect according to the specification? # Notation A neural network is a function $F(x) = y$ that accepts an input $x \in \mathbb{R}^n$ and produces an output $y \in \mathbb{R}^m$. The model $F$ also implicitly depends on some model parameters $\theta$; in our work the model is fixed, so for convenience we don't show the dependence on $\theta$. In this paper we focus on neural networks used as an $m$-class classifier. The output of the network is computed using the softmax function, which ensures that the output vector $y$ satisfies $0 \le y_i \le 1$ and $y_1 + \dots + y_m = 1$. The output vector $y$ is thus treated as a probability distribution, i.e., $y_i$ is treated as the probability that input $x$ has class $i$. The classifier assigns the label $C(x) = \arg\max_i F(x)_i$ to the input $x$. Let $C^*(x)$ be the correct label of $x$. The inputs to the softmax function are called \emph{logits}. We use the notation from Papernot et al. \cite{distillation}: define $F$ to be the full neural network including the softmax function, $Z(x) = z$ to be the output of all layers except the softmax (so $z$ are the logits), and \begin{equation*} F(x) = \softmax(Z(x)) = y. \end{equation*} A neural network typically \footnote{Most simple networks have this simple linear structure, however other more sophisticated networks have more complicated structures (e.g., ResNet \cite{he2016deep} and Inception \cite{szegedy2015rethinking}). The network architecture does not impact our attacks.} consists of layers \begin{equation*} F = \softmax \circ F_n \circ F_{n-1} \circ \cdots \circ F_1 \end{equation*} where \begin{equation*} F_i(x) = \sigma(\theta_i \cdot x) + \hat\theta_i \end{equation*} for some non-linear activation function $\sigma$, some matrix $\theta_i$ of model weights, and some vector $\hat\theta_i$ of model biases. Together $\theta$ and $\hat\theta$ make up the model parameters. Common choices of $\sigma$ are tanh \cite{mishkin2015all}, sigmoid, ReLU \cite{maas2013rectifier}, or ELU \cite{clevert2015fast}. In this paper we focus primarily on networks that use a ReLU activation function, as it currently is the most widely used activation function \cite{szegedy2015rethinking,springenberg2014striving,mishkin2015all,distillation}. # Specification: We define an objective function $f$ such that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. There are many possible choices for $f$: # The equations \begin{align*} f_{1}(x') &= -\loss_{F,t}(x') + 1\\ f_{2}(x') &= (\max_{i \ne t}( F(x')_i) - F(x')_t)^+\\ f_{3}(x') &= \text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t)-\log(2)\\ f_{4}(x') &= (0.5 - F(x')_t)^+\\ f_{5}(x') &= -\log(2 F(x')_t - 2)\\ f_{6}(x') &= (\max_{i \ne t}( Z(x')_i) - Z(x')_t)^+\\ f_{7}(x') &= \text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t)-\log(2) \end{align*} where $s$ is the correct classification, $(e)^+$ is short-hand for $\max(e,0)$, $\text{softplus}(x) = \log(1+\exp(x))$, and $\loss_{F,s}(x)$ is the cross entropy loss for $x$. # Your task Consider each equation one by one. End your answer with a python list of numbers [1,2,3,4,5,6,7] for those that are wrong. Specifically, make sure that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. Think step by step about if this is true for each equation and then give your answer as a python list
To determine if each equation is correct, we need to check if it satisfies the condition that $C(x+\delta) = t$ if and only if $f(x+\delta) \le 0$. 1. $f_{1}(x') = -\loss_{F,t}(x') + 1$ Recall that the cross entropy loss is defined as $\loss_{F,t}(x) = -\log(F(x)_t)$. If $C(x+\delta) = t$, then $F(x+\delta)_t$ should be maximized, and thus $\loss_{F,t}(x+\delta)$ should be minimized. Therefore, $f_1(x+\delta) = -\loss_{F,t}(x+\delta) + 1$ should be maximized, which is not necessarily true when $f_1(x+\delta) \le 0$. Thus, this equation is incorrect. 2. $f_{2}(x') = (\max_{i \ne t}( F(x')_i) - F(x')_t)^+$ If $C(x+\delta) = t$, then $F(x+\delta)_t$ should be greater than or equal to $F(x+\delta)_i$ for all $i \ne t$. Therefore, $\max_{i \ne t}(F(x+\delta)_i) - F(x+\delta)_t$ should be non-positive, and thus $f_2(x+\delta) = (\max_{i \ne t}( F(x+\delta)_i) - F(x+\delta)_t)^+$ should be equal to 0. This satisfies the condition that $f_2(x+\delta) \le 0$. If $C(x+\delta) \ne t$, then $F(x+\delta)_t$ should be less than $F(x+\delta)_i$ for some $i \ne t$, and thus $\max_{i \ne t}(F(x+\delta)_i) - F(x+\delta)_t$ should be positive. Therefore, $f_2(x+\delta)$ should be greater than 0. This also satisfies the condition that $f_2(x+\delta) > 0$ when $C(x+\delta) \ne t$. Thus, this equation is correct. 3. $f_{3}(x') = \text{softplus}(\max_{i \ne t}(F(x')_i) - F(x')_t)-\log(2)$ The softplus function is positive and increasing, so if $\max_{i \ne t}(F(x+\delta)_i) - F(x+\delta)_t$ is non-positive, then $\text{softplus}(\max_{i \ne t}(F(x+\delta)_i) - F(x+\delta)_t)$ should be close to 0. Therefore, $f_3(x+\delta) = \text{softplus}(\max_{i \ne t}(F(x+\delta)_i) - F(x+\delta)_t)-\log(2)$ should be negative or close to 0 when $C(x+\delta) = t$. If $C(x+\delta) \ne t$, then $\max_{i \ne t}(F(x+\delta)_i) - F(x+\delta)_t$ should be positive, and thus $f_3(x+\delta)$ should be positive. This satisfies the condition that $f_3(x+\delta) > 0$ when $C(x+\delta) \ne t$. However, the condition that $f_3(x+\delta) \le 0$ when $C(x+\delta) = t$ is not necessarily satisfied, since $f_3(x+\delta)$ could be close to 0 but still positive. Thus, this equation is incorrect. 4. $f_{4}(x') = (0.5 - F(x')_t)^+$ If $C(x+\delta) = t$, then $F(x+\delta)_t$ should be maximized, and thus $0.5 - F(x+\delta)_t$ should be minimized. Therefore, $f_4(x+\delta) = (0.5 - F(x+\delta)_t)^+$ should be equal to 0. This satisfies the condition that $f_4(x+\delta) \le 0$. If $C(x+\delta) \ne t$, then $F(x+\delta)_t$ should be less than 0.5, and thus $0.5 - F(x+\delta)_t$ should be positive. Therefore, $f_4(x+\delta)$ should be greater than 0. This also satisfies the condition that $f_4(x+\delta) > 0$ when $C(x+\delta) \ne t$. Thus, this equation is correct. 5. $f_{5}(x') = -\log(2 F(x')_t - 2)$ If $C(x+\delta) = t$, then $F(x+\delta)_t$ should be maximized, and thus $2 F(x+\delta)_t - 2$ should be maximized. If $F(x+\delta)_t > 0.5$, then $2 F(x+\delta)_t - 2$ is positive, and thus $f_5(x+\delta) = -\log(2 F(x+\delta)_t - 2)$ is negative. If $F(x+\delta)_t = 0.5$, then $f_5(x+\delta)$ is undefined. If $F(x+\delta)_t < 0.5$, then $2 F(x+\delta)_t - 2$ is negative, and thus $f_5(x+\delta)$ is positive. Therefore, the condition that $f_5(x+\delta) \le 0$ is not necessarily satisfied when $C(x+\delta) = t$. Thus, this equation is incorrect. 6. $f_{6}(x') = (\max_{i \ne t}( Z(x')_i) - Z(x')_t)^+$ This is similar to equation 2, except that it uses the logits $Z(x+\delta)$ instead of the softmax output $F(x+\delta)$. If $C(x+\delta) = t$, then $Z(x+\delta)_t$ should be greater than or equal to $Z(x+\delta)_i$ for all $i \ne t$. Therefore, $\max_{i \ne t}(Z(x+\delta)_i) - Z(x+\delta)_t$ should be non-positive, and thus $f_6(x+\delta) = (\max_{i \ne t}( Z(x+\delta)_i) - Z(x+\delta)_t)^+$ should be equal to 0. This satisfies the condition that $f_6(x+\delta) \le 0$. If $C(x+\delta) \ne t$, then $Z(x+\delta)_t$ should be less than $Z(x+\delta)_i$ for some $i \ne t$, and thus $\max_{i \ne t}(Z(x+\delta)_i) - Z(x+\delta)_t$ should be positive. Therefore, $f_6(x+\delta)$ should be greater than 0. This also satisfies the condition that $f_6(x+\delta) > 0$ when $C(x+\delta) \ne t$. Thus, this equation is correct. 7. $f_{7}(x') = \text{softplus}(\max_{i \ne t}(Z(x')_i) - Z(x')_t)-\log(2)$ This is similar to equation 3, except that it uses the logits $Z(x+\delta)$ instead of the softmax output $F(x+\delta)$. The softplus function is positive and increasing, so if $\max_{i \ne t}(Z(x+\delta)_i) - Z(x+\delta)_t$ is non-positive, then $\text{softplus}(\max_{i \ne t}(Z(x+\delta)_i) - Z(x+\delta)_t)$ should be close to 0. Therefore, $f_7(x+\delta) = \text{softplus}(\max_{i \ne t}(Z(x+\delta)_i) - Z(x+\delta)_t)-\log(2)$ should be negative or close to 0 when $C(x+\delta) = t$. If $C(x+\delta) \ne t$, then $\max_{i \ne t}(Z(x+\delta)_i) - Z(x+\delta)_t$ should be positive, and thus $f_7(x+\delta)$ should be positive. This satisfies the condition that $f_7(x+\delta) > 0$ when $C(x+\delta) \ne t$. However, the condition that $f_7(x+\delta) \le 0$ when $C(x+\delta) = t$ is not necessarily satisfied, since $f_7(x+\delta)$ could be close to 0 but still
Error Resulting in output: False