Transformer-Evolution-Paper
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On this page
  • 整体思路以及计算方式
  • 训练以及loss
  • 代码
  • 实验以及适用场景
  • 细节
  • 简评
  1. Normalize_And_Residual

Improving Deep Transformer with Depth-Scaled Initialization and Merged Attention

PreviousBatch Normalization Biases Residual Blocks Towards the Identity Function in Deep NetworksNextRealFormer Transformer Likes Residual Attention

Last updated 2 years ago

论文地址：

整体思路以及计算方式

论文工作量两点：

  1. 解释了深层Transformer为什么难以训练，针对这点，提出了一种新的初始化方式；

  2. 结合作者之前的AAN，提出加速解码的方案；

第一点：

作者的观点是，深层Transformer难以训练是因为反传的原因，而主要的方面是因为残差和Layer Norm。

回顾Transformer的计算公式：

r=RC⁡(z,f(z))o=LN⁡(r)\begin{aligned} \mathrm{r}&=\operatorname{RC}(\mathrm{z}, f(\mathrm{z})) \\ \mathrm{o}&=\operatorname{LN}(\mathrm{r}) \end{aligned}ro​=RC(z,f(z))=LN(r)​

其中fff表示MHA\mathrm{MHA}MHA或FFN\mathrm{FFN}FFN，RC\mathrm{RC}RC表示残差连接，LN\mathrm{LN}LN表示layer norm。

计算反传可得：

δr=∂o∂rδo=diag⁡(gσr)(I−1−r‾r‾⊤d)δoδz=∂r∂zδr=(1+∂f∂z)δr,\begin{aligned} &\delta_{r}=\frac{\partial \mathrm{o}}{\partial \mathrm{r}} \delta_{o}=\operatorname{diag}\left(\frac{\mathrm{g}}{\sigma_{r}}\right)\left(\mathrm{I}-\frac{1-\overline{\mathrm{r}} \overline{\mathrm{r}}^{\top}}{d}\right) \delta_{o} \\ &\delta_{z}=\frac{\partial \mathrm{r}}{\partial \mathrm{z}} \delta_{r}=\left(1+\frac{\partial f}{\partial \mathrm{z}}\right) \delta_{r}, \end{aligned}​δr​=∂r∂o​δo​=diag(σr​g​)(I−d1−rr⊤​)δo​δz​=∂z∂r​δr​=(1+∂z∂f​)δr​,​

其中r‾\overline {\mathrm r}r表示归一化的输入，所以r‾r‾⊤\overline{\mathrm{r}} \overline{\mathrm{r}}^{\top}rr⊤表示方差，注意到δr\delta_rδr​和方差正相关。

接着作者考虑了如下量：

β=βLN⋅βRC=∥δz∥2∥δr∥2⋅∥δr∥2∥δo∥2\beta=\beta_{\mathrm{LN}} \cdot \beta_{\mathrm{RC}}=\frac{\left\|\delta_{z}\right\|_{2}}{\left\|\delta_{r}\right\|_{2}} \cdot \frac{\left\|\delta_{r}\right\|_{2}}{\left\|\delta_{o}\right\|_{2}}β=βLN​⋅βRC​=∥δr​∥2​∥δz​∥2​​⋅∥δo​∥2​∥δr​∥2​​

从直觉上看，我们希望β≈1\beta\approx 1β≈1，因为太大会造成梯度消失，太小会造成梯度爆炸。那么β,βLN,βRC\beta,\beta_{\mathrm{LN}} , \beta_{\mathrm{RC}}β,βLN​,βRC​的值大概是多少？作者经过分析，给出如下结论：

  • FFN\mathrm{FFN}FFN的β<1\beta<1β<1，MHA\mathrm{MHA}MHA的β>1\beta>1β>1，包含Self和Cross Attention；

  • 无论是FFN\mathrm{FFN}FFN还是MHA\mathrm{MHA}MHA，都有βLN<1\beta_{\mathrm{LN}}<1βLN​<1 , βRC>1\beta_{\mathrm{RC}}>1βRC​>1；

    • 在Transformer中，βLN∈[0.82,0.86],βRC∈[1.10,1.22]\beta_{\mathrm{LN}}\in [0.82, 0.86], \beta_{\mathrm{RC}}\in [1.10,1.22]βLN​∈[0.82,0.86],βRC​∈[1.10,1.22]

由于βRC\beta_{\mathrm{RC}}βRC​和方差正相关，所以减少方差可以使得βRC\beta_{\mathrm{RC}}βRC​更接近1，作者给出的方案是修改初始化方案：

W∈Rdi×do∼U(−γαl,γαl)\mathrm{W} \in \mathbb{R}^{d_{i} \times d_{o}} \sim \mathcal{U}\left(-\gamma \frac{\alpha}{\sqrt{l}}, \gamma \frac{\alpha}{\sqrt{l}}\right)W∈Rdi​×do​∼U(−γl​α​,γl​α​)

其中lll表示层数。

第二点：

把NMT中Decoder的Self Attention换成AAN，从而提高解码效率：

MATT⁡(Sl−1)=SAAN⁡(Sl−1)+ATT⁡(Sl−1,HL)S‾l=LN(RC⁡(Sl−1,MATT⁡(Sl−1)))\begin{aligned} &\operatorname{MATT}\left(\mathrm{S}^{l-1}\right)=\operatorname{SAAN}\left(\mathrm{S}^{l-1}\right)+\operatorname{ATT}\left(\mathrm{S}^{l-1}, \mathrm{H}^{L}\right) \\ &\overline{\mathrm{S}}^{l}=\mathrm{L N}\left(\operatorname{RC}\left(\mathrm{S}^{l-1}, \operatorname{MATT}\left(\mathrm{S}^{l-1}\right)\right)\right) \end{aligned}​MATT(Sl−1)=SAAN(Sl−1)+ATT(Sl−1,HL)Sl=LN(RC(Sl−1,MATT(Sl−1)))​

训练以及loss

没有变化。

代码

实验以及适用场景

初始化部分适用于所有场景，解码部分适用于Decoder；实验测试了NMT，带来了一定的效果。

细节

暂无。

简评

第一点提供了一种分析的思路，值得复现。

https://arxiv.org/abs/1908.11365
https://github.com/bzhangGo/zero/tree/master/docs/depth_scale_init_and_merged_attention