¿Cuál es la complejidad del tiempo para entrenar una red neuronal usando la retropropagación?

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Supongamos que un NN contiene $$nnn$$ capas ocultas, $$mmm$$ ejemplos de entrenamiento, $$xxx$$ características $$yn inin_i$$ nodos en cada capa. ¿Cuál es la complejidad del tiempo para entrenar a este NN utilizando la propagación inversa?

Tengo una idea básica sobre cómo encuentran la complejidad temporal de los algoritmos, pero aquí hay 4 factores diferentes a considerar aquí, es decir, iteraciones, capas, nodos en cada capa, ejemplos de entrenamiento y quizás más factores. Encontré una respuesta aquí, pero no estaba lo suficientemente clara.

¿Existen otros factores, además de los que mencioné anteriormente, que influyen en la complejidad temporal del algoritmo de entrenamiento de un NN?

Ver también https://qr.ae/TWttzq .
nbro

Respuestas:

11

No he visto una respuesta de una fuente confiable, pero trataré de responderla yo mismo, con un ejemplo simple (con mi conocimiento actual).

En general, tenga en cuenta que el entrenamiento de un MLP utilizando la propagación inversa generalmente se implementa con matrices.

Complejidad temporal de la multiplicación de matrices

La complejidad temporal de la multiplicación de matrices para $$M i j ∗ M j kMij∗MjkM_{ij} * M_{jk}$$ es simplemente $$O ( i ∗ j ∗ k )O(i∗j∗k)\mathcal{O}(i*j*k)$$ .

Tenga en cuenta que aquí estamos asumiendo el algoritmo de multiplicación más simple: existen algunos otros algoritmos con una complejidad temporal algo mejor.

Algoritmo de avance

El algoritmo de propagación de avance es el siguiente.

Primero, para pasar de la capa $$iii$$ a la $$jjj$$ , debes

$$S j = W j i ∗ Z i$$

Sj=WjiZi

Luego aplicas la función de activación

$$Z j = f ( S j )$$

Zj=f(Sj)

Si tenemos $$NNN$$ capas (incluidas las capas de entrada y salida), esto se ejecutará $$N - 1N−1N-1$$ veces.

Ejemplo

Como ejemplo, calculemos la complejidad del tiempo para el algoritmo de avance de un MLP con $$444$$ capas, donde $$iii$$ denota el número de nodos de la capa de entrada, $$jjj$$ el número de nodos en la segunda capa, $$kkk$$ el número de nodos en el tercera capa $$ylll$$ el número de nodos en la capa de salida.

Como hay $$444$$ capas, necesita $$333$$ matrices para representar los pesos entre estas capas. Denotémoslos por $$W j iWjiW_{ji}$$ , $$W k jWkjW_{kj}$$ y $$W l kWlkW_{lk}$$ , donde $$W j iWjiW_{ji}$$ es una matriz con $$jjj$$ filas y $$iii$$ columnas ( $$W j iWjiW_{ji}$$ contiene los pesos que van de la capa$$iii$$a la capa$$jjj$$ ).

Supongamos que tienes $$ttt$$ ejemplos de entrenamiento. Para propagar de la capa $$iii$$ a $$jjj$$ , tenemos primero

$$S j t =Wji∗Zit$$

Sjt=WjiZit

and this operation (i.e. matrix multiplcation) has $$O(j∗i∗t)O(j∗i∗t)\mathcal{O}(j*i*t)$$ time complexity. Then we apply the activation function

$$Zjt=f(Sjt)$$

Zjt=f(Sjt)

and this has $$O(j∗t)O(j∗t)\mathcal{O}(j*t)$$ time complexity, because it is an element-wise operation.

So, in total, we have

$$O(j∗i∗t+j∗t)=O(j∗t∗(t+1))=O(j∗i∗t)$$

O(jit+jt)=O(jt(t+1))=O(jit)

Using same logic, for going $$j→kj→kj \to k$$, we have $$O(k∗j∗t)O(k∗j∗t)\mathcal{O}(k*j*t)$$, and, for $$k→lk→lk \to l$$, we have $$O(l∗k∗t)O(l∗k∗t)\mathcal{O}(l*k*t)$$.

In total, the time complexity for feedforward propagation will be

$$O(j∗i∗t+k∗j∗t+l∗k∗t)=O(t∗(ij+jk+kl))$$

O(jit+kjt+lkt)=O(t(ij+jk+kl))

I'm not sure if this can be simplified further or not. Maybe it's just $$O(t∗i∗j∗k∗l)O(t∗i∗j∗k∗l)\mathcal{O}(t*i*j*k*l)$$, but I'm not sure.

Back-propagation algorithm

The back-propagation algorithm proceeds as follows. Starting from the output layer $$l→kl→kl \to k$$, we compute the error signal, $$EltEltE_{lt}$$, a matrix containing the error signals for nodes at layer $$lll$$

$$Elt=f′(Slt)⊙(Zlt−Olt)$$

Elt=f(Slt)(ZltOlt)

where $$⊙⊙\odot$$ means element-wise multiplication. Note that $$EltEltE_{lt}$$ has $$lll$$ rows and $$ttt$$ columns: it simply means each column is the error signal for training example $$ttt$$.

We then compute the "delta weights", $$Dlk∈Rl×kDlk∈Rl×kD_{lk} \in \mathbb{R}^{l \times k}$$ (between layer $$lll$$ and layer $$kkk$$)

$$Dlk=Elt∗Ztk$$

Dlk=EltZtk

where $$ZtkZtkZ_{tk}$$ is the transpose of $$ZktZktZ_{kt}$$.

$$Wlk=Wlk−Dlk$$

Wlk=WlkDlk

For $$l→kl→kl \to k$$, we thus have the time complexity $$O(lt+lt+ltk+lk)=O(l∗t∗k)O(lt+lt+ltk+lk)=O(l∗t∗k)\mathcal{O}(lt + lt + ltk + lk) = \mathcal{O}(l*t*k)$$.

Now, going back from $$k→jk→jk \to j$$. We first have

$$Ekt=f′(Skt)⊙(Wkl∗Elt)$$

Ekt=f(Skt)(WklElt)

Then

$$Dkj=Ekt∗Ztj$$

Dkj=EktZtj

And then

$$Wkj=Wkj−Dkj$$

Wkj=WkjDkj

where $$WklWklW_{kl}$$ is the transpose of $$WlkWlkW_{lk}$$. For $$k→jk→jk \to j$$, we have the time complexity $$O(kt+klt+ktj+kj)=O(k∗t(l+j))O(kt+klt+ktj+kj)=O(k∗t(l+j))\mathcal{O}(kt + klt + ktj + kj) = \mathcal{O}(k*t(l+j))$$.

And finally, for $$j→ij→ij \to i$$, we have $$O(j∗t(k+i))O(j∗t(k+i))\mathcal{O}(j*t(k+i))$$. In total, we have

$$O(ltk+tk(l+j)+tj(k+i))=O(t∗(lk+kj+ji))$$

O(ltk+tk(l+j)+tj(k+i))=O(t(lk+kj+ji))

which is same as feedforward pass algorithm. Since they are same, the total time complexity for one epoch will be $$O(t∗(ij+jk+kl)).$$

O(t(ij+jk+kl)).

This time complexity is then multiplied by number of iterations (epochs). So, we have $$O(n∗t∗(ij+jk+kl)),$$

O(nt(ij+jk+kl)),
where $$nnn$$ is number of iterations.

Notes

Note that these matrix operations can greatly be paralelized by GPUs.

Conclusion

We tried to find the time complexity for training a neural network that has 4 layers with respectively $$iii$$, $$jjj$$, $$kkk$$ and $$lll$$ nodes, with $$ttt$$ training examples and $$nnn$$ epochs. The result was $$O(nt∗(ij+jk+kl))O(nt∗(ij+jk+kl))\mathcal{O}(nt*(ij + jk + kl))$$.

We assumed the simplest form of matrix multiplication that has cubic time complexity. We used batch gradient descent algorithm. The results for stochastic and mini-batch gradient descent should be same. (Let me know if you think the otherwise: note that batch gradient descent is the general form, with little modification, it becomes stochastic or mini-batch)

Also, if you use momentum optimization, you will have same time complexity, because the extra matrix operations required are all element-wise operations, hence they will not affect the time complexity of the algorithm.

I'm not sure what the results would be using other optimizers such as RMSprop.

Sources

The following article http://briandolhansky.com/blog/2014/10/30/artificial-neural-networks-matrix-form-part-5 describes an implementation using matrices. Although this implementation is using "row major", the time complexity is not affected by this.

http://briandolhansky.com/blog/2013/9/27/artificial-neural-networks-backpropagation-part-4

Your answer is great..I could not find any ambiguity till now, but you forgot the no. of iterations part, just add it...and if no one answers in 5 days i'll surely accept your answer
DuttaA

@DuttaA I tried to put every thing I knew. it may not be 100% correct so feel free to leave this unaccepted :) I'm also waiting for other answers to see what other points I missed.
M.kazem Akhgary

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For the evaluation of a single pattern, you need to process all weights and all neurons. Given that every neuron has at least one weight, we can ignore them, and have $$O(w)O(w)\mathcal{O}(w)$$ where $$www$$ is the number of weights, i.e., $$n∗nin∗nin * n_i$$, assuming full connectivity between your layers.

The back-propagation has the same complexity as the forward evaluation (just look at the formula).

So, the complexity for learning $$mmm$$ examples, where each gets repeated $$eee$$ times, is $$O(w∗m∗e)O(w∗m∗e)\mathcal{O}(w*m*e)$$.

The bad news is that there's no formula telling you what number of epochs $$eee$$ you need.

From the above answer don't you think itdepends on more factors?
DuttaA

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@DuttaA No. There's a constant amount of work per weight, which gets repeated e times for each of m examples. I didn't bother to compute the number of weights, I guess, that's the difference.
maaartinus

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I think the answers are same. in my answer I can assume number of weights w = ij + jk + kl. basically sum of n * n_i between layers as you noted.
M.kazem Akhgary

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A potential disadvantage of gradient-based methods is that they head for the nearest minimum, which is usually not the global minimum.

This means that the only difference between these search methods is the speed with which solutions are obtained, and not the nature of those solutions.

An important consideration is time complexity, which is the rate at which the time required to find a solution increases with the number of parameters (weights). In short, the time complexities of a range of different gradient-based methods (including second-order methods) seem to be similar.

Six different error functions exhibit a median run-time order of approximately O(N to the power 4) on the N-2-N encoder in this paper:

Lister, R and Stone J "An Empirical Study of the Time Complexity of Various Error Functions with Conjugate Gradient Back Propagation" , IEEE International Conference on Artificial Neural Networks (ICNN95), Perth, Australia, Nov 27-Dec 1, 1995.

Summarised from my book: Artificial Intelligence Engines: A Tutorial Introduction to the Mathematics of Deep Learning.

Hi J. Stone. Thanks for trying to contribute to the site. However, please, note that this is not a place for advertising yourself. Anyway, you can surely provide a link to your own books if they are useful for answering the questions and provided you're not just trying to advertise yourself.
nbro

@nbro If James Stone can provide an insightful answer - and it seems so - then i'm fine with him also mentioning some of his work. Having experts on this network is a solid contribution to the quality and level.