% teach:
% gap: 读者会用 autograd 但不知道手写 KDA backward 的具体展开
% takeaway: backward = 逆序遍历时间步, 每步求 dq/dk/dv/dg/dbeta + 累积 dS; GVA 反传 = view+sum
% jump: 为什么 dS_dec 要加 dS_acc 和 -k⊗dr 两项
% omit: Triton backward 优化

\section{反向传播推导}

\subsection{Forward 回顾}

逐步写下 forward（省略 batch/head 下标）：

\begin{align}
  S_{\mathrm{dec}} &= \exp(g_t) \odot S_{t-1}  \tag{F1} \\
  p_t &= k_t^T S_{\mathrm{dec}}  \tag{F2} \\
  r_t &= v_t - p_t  \tag{F3} \\
  a_t &= \beta_t \cdot k_t  \tag{F4} \\
  S_t &= S_{\mathrm{dec}} + a_t \otimes r_t  \tag{F5} \\
  o_t &= q_t^T S_t  \tag{F6}
\end{align}

\subsection{反传公式（BPTT，$T \to 0$）}

设 $dS_{\mathrm{acc}}$ 是从时间步 $t$ 开始累积到 $S_t$ 上的梯度。逆序遍历：

\paragraph{Step 1: $o_t = q_t^T S_t$}

\begin{align}
  dS_{\mathrm{acc}} &\mathrel{+}= q_t \otimes do_t
  & \xrightarrow{\texttt{einsum('bhk,bhv->bhkv')}}
  & \quad \shape{B, HV, K, V} \\
  dq_t &= do_t^T S_t
  & \xrightarrow{\texttt{einsum('bhv,bhkv->bhk')}}
  & \quad \shape{B, HV, K}
\end{align}

\paragraph{Step 2: $S_t = S_{\mathrm{dec}} + a_t \otimes r_t$}

外积的反传：$d(a \otimes r) = (\cdot)$，分解为：

\begin{align}
  da_t &= \sum_v r_{t,v} \cdot dS_{\mathrm{acc},\cdot,v}
  & \xrightarrow{\texttt{einsum('bhv,bhkv->bhk')}}
  & \quad \shape{B, HV, K} \\
  dr_t &= \sum_k a_{t,k} \cdot dS_{\mathrm{acc},k,\cdot}
  & \xrightarrow{\texttt{einsum('bhk,bhkv->bhv')}}
  & \quad \shape{B, HV, V}
\end{align}

\paragraph{Step 3: $a_t = \beta_t \cdot k_t$}

\begin{align}
  d\beta_t &= \sum_k k_{t,k} \cdot da_{t,k}
  & \xrightarrow{\texttt{einsum('bhk,bhk->bh')}}
  & \quad \shape{B, HV} \\
  dk_t^{(a)} &= \beta_t \cdot da_t
  & & \quad \shape{B, HV, K}
\end{align}

\paragraph{Step 4: $r_t = v_t - p_t = v_t - k_t^T S_{\mathrm{dec}}$}

\begin{align}
  dv_t &= dr_t & & \shape{B, HV, V} \\
  dk_t^{(r)} &= -S_{\mathrm{dec}}^T \cdot dr_t
  & \xrightarrow{\texttt{einsum('bhv,bhkv->bhk')}}
  & \quad \shape{B, HV, K} \\
  dS_{\mathrm{dec}}^{(r)} &= -k_t \otimes dr_t
  & \xrightarrow{\texttt{einsum('bhv,bhk->bhkv')}}
  & \quad \shape{B, HV, K, V}
\end{align}

\paragraph{Step 5: 合并 $dS_{\mathrm{dec}}$ 并传递 $dg_t$, $dS_{t-1}$}

\[
  dS_{\mathrm{dec}}^{\mathrm{total}} = dS_{\mathrm{acc}} + dS_{\mathrm{dec}}^{(r)}
  = dS_{\mathrm{acc}} - k_t \otimes dr_t
\]

因为 $S_{\mathrm{dec}} = \exp(g_t) \odot S_{t-1}$：

\begin{align}
  dg_t &= S_{\mathrm{dec}} \odot dS_{\mathrm{dec}}^{\mathrm{total}}
  & \xrightarrow{\texttt{einsum('bhkv,bhkv->bhk')}}
  & \quad \shape{B, HV, K} \\
  dS_{t-1} &= \exp(g_t) \odot dS_{\mathrm{dec}}^{\mathrm{total}}
  & & \quad \shape{B, HV, K, V}
\end{align}

\paragraph{Step 6: 合并 $dk_t$ 和 GVA 归约}

\[
  dk_t = dk_t^{(a)} + dk_t^{(r)}
  = \beta_t \cdot da_t - S_{\mathrm{dec}}^T \cdot dr_t
\]

GVA 反传（$H_V \to H$）：
\[
  dq_H = dq_{H_V}.\texttt{view}(B, T, H, G, K).\texttt{sum}(\text{dim}=3) \cdot \mathrm{scale}
\]
\[
  dk_H = dk_{H_V}.\texttt{view}(B, T, H, G, K).\texttt{sum}(\text{dim}=3)
\]

\subsection{代码对照}

\begin{codemathtop}{ops/reference/recurrent.py — KDAFunction.backward}
\begin{lstlisting}
for t in range(T - 1, -1, -1):
    q_t, k_t, b_t = q_ts[:,t], k_ts[:,t], b_ts[:,t]
    S_dec, r_t, a_t = S_decs[:,t], r_ts[:,t], a_ts[:,t]
    exp_g_t, do_t = exp_g_ts[:,t], do[:,t]

    # Step 1: o_t = q_t . S_t
    S_t = S_dec + einsum('bhk,bhv->bhkv', a_t, r_t)
    dS_acc += einsum('bhk,bhv->bhkv', q_t, do_t)
    dq_e[:,t] = einsum('bhv,bhkv->bhk', do_t, S_t)

    # Step 2: outer product grads
    da_t = einsum('bhv,bhkv->bhk', r_t, dS_acc)
    dr_t = einsum('bhk,bhkv->bhv', a_t, dS_acc)

    # Step 3: a_t = beta_t * k_t
    dbeta[:,t] = einsum('bhk,bhk->bh', k_t, da_t)
    dk_t_a = b_t.unsqueeze(-1) * da_t

    # Step 4: r_t = v_t - k_t . S_dec
    dv[:,t] = dr_t
    dS_dec_from_r = -einsum('bhv,bhk->bhkv', dr_t, k_t)
    dk_t_r = -einsum('bhv,bhkv->bhk', dr_t, S_dec)

    # Step 5: S_dec = exp(g) * S_{t-1}
    dS_dec_total = dS_acc + dS_dec_from_r
    dk_e[:,t] = dk_t_a + dk_t_r
    dg[:,t] = einsum('bhkv,bhkv->bhk', S_dec, dS_dec_total)
    dS_acc = exp_g_t.unsqueeze(-1) * dS_dec_total

# Step 6: GVA reduce
dq_H = dq_e.view(B,T,H,G,K).sum(dim=3) * scale
dk_H = dk_e.view(B,T,H,G,K).sum(dim=3)
\end{lstlisting}
\end{codemathtop}

\begin{warningbox}{$dS_{\mathrm{dec}}^{\mathrm{total}}$ 为什么包含两项？}
$S_t = S_{\mathrm{dec}} + a_t \otimes r_t$，$S_{\mathrm{dec}}$ 同时参与了：
\begin{enumerate}[nosep]
\item 直接传递到 $dS_{\mathrm{acc}}$（作为 $S_t$ 的一部分被读出）
\item 通过 $r_t = v_t - k_t \cdot S_{\mathrm{dec}}$ 参与 delta rule
\end{enumerate}
所以 $dS_{\mathrm{dec}}^{\mathrm{total}} = dS_{\mathrm{acc}} + dS_{\mathrm{dec}}^{(r)}$，
两条路径的梯度要\textbf{加}起来（chain rule 分叉处求和）。
\end{warningbox}

\subsection{本章小结}

KDA backward 是 BPTT 展开：逆序遍历时间步，每步 6 个 einsum + 一次 $dS$ 累积更新。
GVA 反传在最后做 \texttt{view+sum}。手写 backward 的关键是正确处理
$dS_{\mathrm{dec}}$ 的两条梯度路径（直接传递 + 通过 $r_t$ 的 delta rule 路径）。