多路径链式法则与反向传播Multi-Path Chain Rule & Backpropagation
同一个输入通过多条路径影响损失时，怎样合并这些影响？How do we combine effects when one input influences a loss along multiple paths?
前置知识：Prerequisites: B08：一元链式法则；B09：梯度；B10：损失与更新。B08: single-variable chain rule; B09: gradients; B10: loss and updates.
复习前置：Review prerequisites: B08 · 一元链式法则The Single-Variable Chain Rule · B09 · 偏导数与梯度Partial Derivatives & Gradients · B10 · 损失函数与梯度下降Loss Functions & Gradient Descent
学完后你能：After this lesson you can: 计算共享节点两条路径的导数贡献，并正确累加。Calculate and correctly accumulate derivatives from two paths to a shared node.
动手观察Explore
单步执行 x→u、x→v、u+v 的前向与反向计算，观察回到 x 的两条贡献如何累加。Step through the forward and backward computation of x→u, x→v and u+v; observe the two contributions accumulating at x.
从观察到解释From observation to explanation
先做前向计算Compute forward first
令 u=x²、v=3x、L=u+v。x=2 时，u=4、v=6、L=10。计算图中 x 分别连接到 u 和 v，再在加法节点汇合。Let u=x², v=3x and L=u+v. At x=2, u=4, v=6 and L=10. The graph branches from x to u and v, then merges at addition.
每条路径相乘Multiply along each path
加法节点对 u 和 v 的局部导数都为 1。经 u 返回的贡献是 1×2x；经 v 返回的贡献是 1×3。x=2 时分别是 4 和 3。The addition node has local derivative 1 with respect to each input. The contribution through u is 1×2x; through v it is 1×3. At x=2 these are 4 and 3.
共享节点相加Add at shared nodes
两条路径同时影响 x，所以总导数为 4+3=7。反向传播把局部导数沿边相乘，在共享节点累加；重新运行前要把旧累计值清零。Both paths affect x, so the total derivative is 4+3=7. Backpropagation multiplies local derivatives along edges and accumulates them at shared nodes; clear old accumulated values before another pass.
公式与符号Formula and symbols
x 是共享输入，u 与 v 是中间结果，L 是标量输出；每项表示一条路径的局部导数乘积，两个贡献在 x 汇合时相加。x is a shared input, u and v are intermediate values, and L is a scalar output. Each term is a product of local derivatives along one path; contributions add where they meet at x.
留意这个区别Watch for this
分支贡献要相加，不能覆盖先到的值；如果不清零就重复反传，会错误地把两次结果叠加。Add branch contributions rather than overwriting the first. Repeating a backward pass without clearing accumulated gradients incorrectly adds the two runs.
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