bert,transformer架构图及面试题

atten之后经过一个全连接层+残差+层归一化

python 复制代码

`class BertSelfOutput(nn.Module):
    def __init__(self, config):
        super().__init__()
        self.dense = nn.Linear(config.hidden_size, config.hidden_size)
        self.LayerNorm = nn.LayerNorm(config.hidden_size, eps=config.layer_norm_eps)
        self.dropout = nn.Dropout(config.hidden_dropout_prob)

    def forward(self, hidden_states: torch.Tensor, input_tensor: torch.Tensor) -> torch.Tensor:
        hidden_states = self.dense(hidden_states)  # 全连接 768->768
        hidden_states = self.dropout(hidden_states)
        hidden_states = self.LayerNorm(hidden_states + input_tensor) # 残差和层归一化
        return hidden_states`

残差的作用：避免梯度消失

归一化的作用：避免梯度消失和爆炸，加速收敛

然后再送入一个两层的前馈神经网络

python 复制代码

`class BertIntermediate(nn.Module):
    def __init__(self, config):
        super().__init__()
        self.dense = nn.Linear(config.hidden_size, config.intermediate_size)
        if isinstance(config.hidden_act, str):
            self.intermediate_act_fn = ACT2FN[config.hidden_act]
        else:
            self.intermediate_act_fn = config.hidden_act

    def forward(self, hidden_states: torch.Tensor) -> torch.Tensor:
        hidden_states = self.dense(hidden_states)  # [1, 16, 3072] 映射到高维空间：768 -> 3072
        hidden_states = self.intermediate_act_fn(hidden_states)
        return hidden_states`

python 复制代码

`class BertOutput(nn.Module):
    def __init__(self, config):
        super().__init__()
        self.dense = nn.Linear(config.intermediate_size, config.hidden_size)
        self.LayerNorm = nn.LayerNorm(config.hidden_size, eps=config.layer_norm_eps)
        self.dropout = nn.Dropout(config.hidden_dropout_prob)

    def forward(self, hidden_states: torch.Tensor, input_tensor: torch.Tensor) -> torch.Tensor:
        hidden_states = self.dense(hidden_states)  # 3072 -> 768
        hidden_states = self.dropout(hidden_states)
        hidden_states = self.LayerNorm(hidden_states + input_tensor)  # 残差和层归一化
        return hidden_states`

面试题：为什么注意力机制中要除以根号dk

答：因为q和k做点积后值会很大，会导致反向传播时softmax函数的梯度很小。除以根号dk是为了保持点积后的值均值为0，方差为1.(q和k都是向量)

证明：已知q和k相互独立，且是均值为0，方差为1。

则D(qi*ki)=D(qi)*D(ki)=1

除以dk则D((qi*ki)/根号dk)=1/dk，每一项是这个值，但是根据上面红框的公式，一共有dk项求和，值为1

所以(q*k)/dk的方差就等1

（背景知识）方差性质：

D(CX)=C^2D(X) ,其中C是常量