SimCLR论文中损失函数求导梯度

题记

关于《A Simple Framework for Contrastive Learning of Visual Representations》这篇文章已经有很多大神整理过了，具体可见
【深度学习】详解 SimCLR
和AI的未来：自监督，谷歌SimCLR-A Simple Framework for Contrastive Learning of Visual Representations。我在这里就不详细阐述，在看论文时对文中loss函数求导有疑问。在网上没有查阅到相关资料，在请教师兄后才得以解决，特作此笔记，希望能帮助到有同样疑惑的小伙伴。

NT-Xent求导梯度

在算法中损失函数是这样的，
$$\ell (i,j)=-\log \frac{\exp \left(s_{i,j}/\tau \right)}{{\sum }_{k=1}^{2N}{1}_{[k\ne i]}\exp \left(s_{i,k}/\tau \right)}$$
其中$s_{i,j}$为相似性度量，是 $\mathrm{sim}(u,v)=u^{\top }v/\Vert u\Vert \Vert v\Vert$的缩写，$1$是判别函数，保证$k\ne i$.
在表格中是这样的，
u T v + / τ − log ⁡ ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) \boldsymbol{u}^{T} \boldsymbol{v}^{+} / \tau-\log \sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right) uTv+/τ−logv∈{v+,v−}∑​exp(uTv/τ)
这里实际上没加负号（因为可以是梯度下降法也可以是梯度上升法优化），并且将$log$乘进去了，我们默认其是以$e$为底。
接下来将其看作$\mathbf{u}$的函数，对其进行求导
f ( u ) ′ = v + / τ − ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) v / τ ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) f(\boldsymbol{u})^{'}=\boldsymbol{v}^{+} / \tau-\frac{{\sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right)} \boldsymbol{v}/ \tau}{ {\sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right)}} f(u)′=v+/τ−∑v∈{v+,v−}​exp(uTv/τ)∑v∈{v+,v−}​exp(uTv/τ)v/τ​
最后一项涉及三层复合函数求导，公式为

$$f[g(h(x)){]}^{\prime }=f^{\prime }[g(h(x))]g^{\prime }[h(x)]h^{\prime }(x)$$
这里要做的是把第一项给通分
f ( u ) ′ = ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) v + / τ − ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) v / τ ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) f(\boldsymbol{u})^{'}=\frac{ {\sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right)}{\boldsymbol{v}^{+} / \tau-\sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right)} \boldsymbol{v}/ \tau}{ {\sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right)}} f(u)′=∑v∈{v+,v−}​exp(uTv/τ)∑v∈{v+,v−}​exp(uTv/τ)v+/τ−∑v∈{v+,v−}​exp(uTv/τ)v/τ​
下一步是把分子求和公式展开，可得

f ( u ) ′ = exp ⁡ ( u T v + / τ ) v + / τ + ∑ v − exp ⁡ ( u T v − / τ ) v + / τ − exp ⁡ ( u T v + / τ ) v + / τ − ∑ v − exp ⁡ ( u T v − / τ ) v − / τ ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) f(\boldsymbol{u})^{'}=\frac{ {\exp \left(\boldsymbol{u}^{T} \boldsymbol{v} ^{+}/ \tau\right)} \boldsymbol{v} ^{+}/ \tau+{\sum_{\boldsymbol{v} ^{-}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} ^{-}/ \tau\right)}{\boldsymbol{v}^{+} / \tau- \exp \left(\boldsymbol{u}^{T} \boldsymbol{v}^{+} / \tau\right) \boldsymbol{v} ^{+}/ \tau-\sum_{\boldsymbol{v} ^{-}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v}^{-} / \tau\right)} \boldsymbol{v}^{-}/ \tau}{ {\sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right)}} f(u)′=∑v∈{v+,v−}​exp(uTv/τ)exp(uTv+/τ)v+/τ+∑v−​exp(uTv−/τ)v+/τ−exp(uTv+/τ)v+/τ−∑v−​exp(uTv−/τ)v−/τ​
可以看到分子的第一项和第三项一样，将其消掉可得
f ( u ) ′ = ∑ v − exp ⁡ ( u T v − / τ ) v + / τ − ∑ v − exp ⁡ ( u T v − / τ ) v − / τ ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) = ∑ v − exp ⁡ ( u T v − / τ ) v + / τ ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) − ∑ v − exp ⁡ ( u T v − / τ ) v − / τ ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) f(\boldsymbol{u})^{'}=\frac{{\sum_{\boldsymbol{v} ^{-}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} ^{-}/ \tau\right)}{\boldsymbol{v}^{+} / \tau-\sum_{\boldsymbol{v} ^{-}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v}^{-} / \tau\right)} \boldsymbol{v}^{-}/ \tau}{ {\sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right)}}=\frac{{\sum_{\boldsymbol{v} ^{-}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} ^{-}/ \tau\right)}{\boldsymbol{v}^{+} / \tau } }{ {\sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right)}} -\frac{{\sum_{\boldsymbol{v} ^{-}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v}^{-} / \tau\right)} \boldsymbol{v}^{-}/ \tau}{ {\sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right)}} f(u)′=∑v∈{v+,v−}​exp(uTv/τ)∑v−​exp(uTv−/τ)v+/τ−∑v−​exp(uTv−/τ)v−/τ​=∑v∈{v+,v−}​exp(uTv/τ)∑v−​exp(uTv−/τ)v+/τ​−∑v∈{v+,v−}​exp(uTv/τ)∑v−​exp(uTv−/τ)v−/τ​
到这一步已经和论文给出的非常相似了，将第一项分子再加上正样本减去正样本可得
f ( u ) ′ = ∑ v − exp ⁡ ( u T v − / τ ) v + / τ + exp ⁡ ( u T v + / τ ) v + / τ − exp ⁡ ( u T v + / τ ) v + / τ ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) − ∑ v − exp ⁡ ( u T v − / τ ) v − / τ ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) f(\boldsymbol{u})^{'}=\frac{{\sum_{\boldsymbol{v} ^{-}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} ^{-}/ \tau\right)}{\boldsymbol{v}^{+} / \tau }+ \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} ^{+}/ \tau\right)\boldsymbol{v}^{+} / \tau-\exp \left(\boldsymbol{u}^{T} \boldsymbol{v} ^{+}/ \tau\right)\boldsymbol{v}^{+} / \tau}{ {\sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right)}} -\frac{{\sum_{\boldsymbol{v} ^{-}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v}^{-} / \tau\right)} \boldsymbol{v}^{-}/ \tau}{ {\sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right)}} f(u)′=∑v∈{v+,v−}​exp(uTv/τ)∑v−​exp(uTv−/τ)v+/τ+exp(uTv+/τ)v+/τ−exp(uTv+/τ)v+/τ​−∑v∈{v+,v−}​exp(uTv/τ)∑v−​exp(uTv−/τ)v−/τ​
将第一项的分子中正负样本合并可得
f ( u ) ′ = ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) v + / τ − exp ⁡ ( u T v + / τ ) v + / τ ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) − ∑ v − exp ⁡ ( u T v − / τ ) v − / τ ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) f(\boldsymbol{u})^{'}=\frac{{\sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right)}\boldsymbol{v}^{+} / \tau-\exp \left(\boldsymbol{u}^{T} \boldsymbol{v} ^{+}/ \tau\right)\boldsymbol{v}^{+} / \tau}{ {\sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right)}} -\frac{{\sum_{\boldsymbol{v} ^{-}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v}^{-} / \tau\right)} \boldsymbol{v}^{-}/ \tau}{ {\sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right)}} f(u)′=∑v∈{v+,v−}​exp(uTv/τ)∑v∈{v+,v−}​exp(uTv/τ)v+/τ−exp(uTv+/τ)v+/τ​−∑v∈{v+,v−}​exp(uTv/τ)∑v−​exp(uTv−/τ)v−/τ​
提取公因式后化简得
f ( u ) ′ = ( v + τ ) ( 1 − exp ⁡ ( u T v + / τ ) ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) ) − ( ∑ v − exp ⁡ ( u T v − / τ ) ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) ) ( v − τ ) f(\boldsymbol{u})^{'}=(\frac{\boldsymbol{v}^{+}}{\tau} )(1- \frac{\exp \left(\boldsymbol{u}^{T} \boldsymbol{v} ^{+}/ \tau\right)}{ {\sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right)}}) -(\frac{{\sum_{\boldsymbol{v} ^{-}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v}^{-} / \tau\right)} }{ {\sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right)}})(\frac{\boldsymbol{v}^{-}}{\tau} ) f(u)′=(τv+​)(1−∑v∈{v+,v−}​exp(uTv/τ)exp(uTv+/τ)​)−(∑v∈{v+,v−}​exp(uTv/τ)∑v−​exp(uTv−/τ)​)(τv−​)
令 Z ( u ) = ∑ v ∈ { v + , v − } exp ⁡ ( u T v / τ ) {Z}\left(\mathcal{u}\right)={\sum_{\boldsymbol{v} \in\left\{\boldsymbol{v}^{+}, \boldsymbol{v}^{-}\right\}} \exp \left(\boldsymbol{u}^{T} \boldsymbol{v} / \tau\right)} Z(u)=∑v∈{v+,v−}​exp(uTv/τ),可得
$$f(\mathbf{u}{)}^{{}^{\prime }}=(\frac{{\mathbf{v}}^{+}}{\tau })(1-\frac{\exp \left({\mathbf{u}}^T{\mathbf{v}}^{+}/\tau \right)}{Z\left(\mathbf{u}\right)})-(\frac{{\sum }_{{\mathbf{v}}^{-}}\exp \left({\mathbf{u}}^T{\mathbf{v}}^{-}/\tau \right)}{Z\left(\mathbf{u}\right)})(\frac{{\mathbf{v}}^{-}}{\tau })$$
再稍微一整理就得到论文中最后的结果

$$f(\mathbf{u}{)}^{{}^{\prime }}=(1-\frac{\exp ({\mathbf{u}}^T{\mathbf{v}}^{+}/\tau )}{Z(\mathbf{u})})/\tau {\mathbf{v}}^{+}-\sum _{{\mathbf{v}}^{-}}\frac{\exp ({\mathbf{u}}^T{\mathbf{v}}^{-}/\tau )}{Z(\mathbf{u})}/\tau {\mathbf{v}}^{-}$$

需要注意的是，论文最后结果中的${\mathbf{v}}^{-}$和${\mathbf{v}}^{+}$都应该是看做后面乘上去的（如果作为分母的话，是要加括号的，推出来的公式也确实如此）；
此外，$Z\left(\mathbf{u}\right)$的定义论文也没有给出，不过通过推导应该是上文所描述的。

NT-Logistic求导梯度

表格中的损失函数是这样的
$$log\sigma ({\mathbf{u}}^T{\mathbf{v}}^{+}/\tau )+log\sigma (-{\mathbf{u}}^T{\mathbf{v}}^{-}/\tau )$$
这里实际上也没加负号（因为可以是梯度下降法也可以是梯度上升法优化）。接下来将其看作$\mathbf{u}$的函数，对其进行求导,我们默认其是以$e$为底,激活函数$\sigma (\cdot )$为 sigmoid 函数
$$f(\mathbf{u}{)}^{{}^{\prime }}=\frac{1}{\sigma ({\mathbf{u}}^T{\mathbf{v}}^{+}/\tau )}\sigma ({\mathbf{u}}^T{\mathbf{v}}^{+}/\tau )(1-\sigma ({\mathbf{u}}^T{\mathbf{v}}^{+}/\tau ))(\frac{{\mathbf{v}}^{+}}{\tau })+\frac{1}{\sigma (-{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau )}\sigma (-{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau )(1-\sigma (-{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau ))(-\frac{{\mathbf{v}}^{-}}{\tau })$$
化简得
$$f(\mathbf{u}{)}^{{}^{\prime }}=(1-\sigma ({\mathbf{u}}^T{\mathbf{v}}^{+}/\tau ))(\frac{{\mathbf{v}}^{+}}{\tau })-(1-\sigma (-{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau ))(\frac{{\mathbf{v}}^{-}}{\tau })$$
将sigmoid 函数带入并通分
$$f(\mathbf{u}{)}^{{}^{\prime }}=\frac{1+e^{-{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau }-1}{1+e^{-{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau }}(\frac{{\mathbf{v}}^{+}}{\tau })-\frac{1+e^{{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau }-1}{1+e^{{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau }}(\frac{{\mathbf{v}}^{-}}{\tau })$$
整理得
$$f(\mathbf{u}{)}^{{}^{\prime }}=\frac{e^{-{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau }}{1+e^{-{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau }}(\frac{{\mathbf{v}}^{+}}{\tau })-\frac{e^{{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau }}{1+e^{{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau }}(\frac{{\mathbf{v}}^{-}}{\tau })$$
第一项分子分母同乘$e^{{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau }$；第二项分子分母同乘$e^{-{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau }$得
$$f(\mathbf{u}{)}^{{}^{\prime }}=\frac{1}{e^{{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau }+1}(\frac{{\mathbf{v}}^{+}}{\tau })-\frac{1}{e^{-{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau }+1}(\frac{{\mathbf{v}}^{-}}{\tau })$$
最后可得
$$f(\mathbf{u}{)}^{{}^{\prime }}=\sigma (-{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau )(\frac{{\mathbf{v}}^{+}}{\tau })-\sigma ({\mathbf{u}}^T{\mathbf{v}}^{+}/\tau )(\frac{{\mathbf{v}}^{-}}{\tau })$$
可以发现和论文给出结果一致，论文给出结果为
$$\sigma (-{\mathbf{u}}^T{\mathbf{v}}^{+}/\tau )/\tau {\mathbf{v}}^{+}-\sigma ({\mathbf{u}}^T{\mathbf{v}}^{+}/\tau )/\tau {\mathbf{v}}^{-}$$

参考博文及感谢

部分内容参考以下链接，这里表示感谢 Thanks♪(･ω･)ﾉ
参考博文1 【深度学习】详解 SimCLR
https://blog.youkuaiyun.com/qq_39478403/article/details/128358529
参考博文2 AI的未来：自监督，谷歌SimCLR-A Simple Framework for Contrastive Learning of Visual Representations
https://zhuanlan.zhihu.com/p/372073905
参考博文3 神经网络中的常用激活函数和导数
https://blog.youkuaiyun.com/lw_power/article/details/90291928
参考博文4 矩阵求导公式的数学推导（矩阵求导——基础篇）
https://zhuanlan.zhihu.com/p/273729929
参考博文5 向量对向量求导
https://zhuanlan.zhihu.com/p/449988999