09.5 One weight used several times

Content and key concepts

On the previous page, from one error to all weights.
Each way makes its contribution, it needs to be folded up, not rewritten by the final result.
And we're going to save the behavior and we'll go back to the error, and the deposits of the different paths to the total weight add up. It's an automatic differentiation, not a new way to guess the direction.
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One node used twice
Let u=x2,L=u+u,x=2,Forward gives u=4,L=8, Add transmits two contributions,1 to total2, then the square transmits rad(x)=2x2x=8.
+=
It's going to be four: forward stays the right one, backward breaks down.
d
L
d
x
=
(
1
+
1
)
d
u
d
x
=
2
⋅
2
x
.
\frac{dL}{dx}=(1+1)\frac{du}{dx}=2\cdot2x.
d
x
d
L
​
=
(
1
+
1
)
d
x
d
u
​
=
2
⋅
2
x
.
Topological sorts visit each
knot
One time, and edges keep the chord. It is not possible to remove repeats. In the batch MLP, the general parameter also affects a few examples; its gradiant sums up all these ways with the averaging of loss.
Research.
Create L=(x+x)*(x + x). Compare the option with the general u=x+X and the version with the different u,v containing the same expressions. The correct derivative8x will match, although the number of transactions varies.
Backward again.
Some of the playworks accumulate gradiants between the batches. Our engine before a full backward wears all the achievable knots: the second call repeats the result, does not double it. This is an explicit implementation agreement, not a universal PyTorch contract.
x=2
u=x*x
grad_u=1+1
grad_x=grad_u*x+grad_u*x
assert u+u==8 and grad_x==8
The code repeats the operations just taken care of.
Completely checked Python-example Chapter
Available with a separate file; the short piece shown does not replace its implementation.
Nodes is a one-time thing, but the frequency of edges is kept.
The program calculates derivatives. How does this question continue?
Python examples of Chapter 09
Academic base