पीठ-प्रसार का उपयोग करके तंत्रिका नेटवर्क के प्रशिक्षण के लिए समय जटिलता क्या है?

18

मान लीजिए कि एक एनएन में $n$ छिपी हुई परतें हैं, $m$ प्रशिक्षण के उदाहरण, $x$ विशेषताएं, और प्रत्येक परत में $n_i$ नोड्स हैं। बैक-प्रचार का उपयोग करके इस एनएन को प्रशिक्षित करने की समय जटिलता क्या है?

मुझे इस बारे में एक बुनियादी विचार है कि वे एल्गोरिदम के समय की जटिलता को कैसे देखते हैं, लेकिन यहाँ पर विचार करने के लिए 4 अलग-अलग कारक हैं जैसे कि पुनरावृत्तियाँ, परतें, प्रत्येक परत में नोड्स, प्रशिक्षण उदाहरण और शायद अधिक कारक। मुझे यहाँ एक उत्तर मिला लेकिन यह पर्याप्त स्पष्ट नहीं था।

क्या अन्य कारक हैं, जिनके अलावा मैंने ऊपर उल्लेख किया है, जो एक एनएन के प्रशिक्षण एल्गोरिथ्म की समय जटिलता को प्रभावित करते हैं?

— DuttaA
स्रोत

Https://qr.ae/TWttzq भी देखें ।

— नबर

12

मैंने एक विश्वसनीय स्रोत से उत्तर नहीं देखा है, लेकिन मैं इसका उत्तर स्वयं देने की कोशिश करूँगा, एक साधारण उदाहरण के साथ (मेरे वर्तमान ज्ञान के साथ)।

सामान्य तौर पर, ध्यान दें कि पीठ-प्रसार का उपयोग करते हुए एक एमएलपी को प्रशिक्षण आमतौर पर मैट्रिसेस के साथ लागू किया जाता है।

मैट्रिक्स गुणन का समय जटिलता

के लिए आव्यूह गुणन के समय जटिलता $M_{ij} * M_{jk}$ बस है $\mathcal{O}(i*j*k)$ ।

ध्यान दें कि हम यहां सरलतम गुणा एल्गोरिथ्म मान रहे हैं: कुछ बेहतर समय जटिलता के साथ कुछ अन्य एल्गोरिदम मौजूद हैं।

Feedforward पास एल्गोरिथ्म

फीडफ़ॉर्वर्ड प्रचार एल्गोरिथ्म इस प्रकार है।

सबसे पहले, लेयर $i$ से $j$ तक जाने के लिए , आप करते हैं

S_{j} = W_{j i} * Z_{i}

$S_j = W_{ji}*Z_i$

फिर आप सक्रियण फ़ंक्शन लागू करते हैं

Z_{j} = f (S_{j})

$Z_j = f(S_j)$

अगर हमारे पास $N$ लेयर्स हैं (इनपुट और आउटपुट लेयर सहित), तो यह चलेगा $N-1$ बार ।

उदाहरण

एक उदाहरण के रूप में, चलो $4$ परतों के साथ एक एमएलपी के लिए आगे पास एल्गोरिथ्म की समय जटिलता की गणना करते हैं, जहां $i$ इनपुट परत के नोड्स की संख्या को दर्शाता है, दूसरी परत में नोड की संख्या को $j$ , नोड की संख्या को $k$ । तीसरी परत और आउटपुट परत में नोड्स की संख्या $l$ ।

$4$ $3$ $W_{ji}$ $W_{kj}$ $W_{lk}$ , where $W_{ji}$ is a matrix with $j$ rows and $i$ columns ( $W_{ji}$ thus contains the weights going from layer $i$ to layer $j$ ).

Assume you have $t$ training examples. For propagating from layer $i$ to $j$ , we have first

S_{j t} = W_{j i} * Z_{i t}

$S_{jt} = W_{ji} * Z_{it}$

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

Z_{j t} = f (S_{j t})

$Z_{jt} = f(S_{jt})$

and this has $\mathcal{O}(j*t)$ 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)

$\mathcal{O}(j*i*t + j*t) = \mathcal{O}(j*t*(t + 1)) = \mathcal{O}(j*i*t)$

$j \to k$ $\mathcal{O}(k*j*t)$ , and, for $k \to l$ , we have $\mathcal{O}(l*k*t)$ .

In total, the time complexity for feedforward propagation will be

O (j * i * t + k * j * t + l * k * t) = O (t * (i j + j k + k l))

$\mathcal{O}(j*i*t + k*j*t + l*k*t) = \mathcal{O}(t*(ij + jk + kl))$

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

Back-propagation algorithm

$l \to k$ $E_{lt}$ $l$

E_{l t} = f^{'} (S_{l t}) ⊙ (Z_{l t} - O_{l t})

$E_{lt} = f'(S_{lt}) \odot {(Z_{lt} - O_{lt})}$

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

We then compute the "delta weights", $D_{lk} \in \mathbb{R}^{l \times k}$ (between layer $l$ and layer $k$ )

D_{l k} = E_{l t} * Z_{t k}

$D_{lk} = E_{lt} * Z_{tk}$

where $Z_{tk}$ is the transpose of $Z_{kt}$ .

We then adjust the weights

W_{l k} = W_{l k} - D_{l k}

$W_{lk} = W_{lk} - D_{lk}$

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

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

E_{k t} = f^{'} (S_{k t}) ⊙ (W_{k l} * E_{l t})

$E_{kt} = f'(S_{kt}) \odot (W_{kl} * E_{lt})$

Then

D_{k j} = E_{k t} * Z_{t j}

$D_{kj} = E_{kt} * Z_{tj}$

And then

W_{k j} = W_{k j} - D_{k j}

$W_{kj} = W_{kj} - D_{kj}$

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

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

O (l t k + t k (l + j) + t j (k + i)) = O (t * (l k + k j + j i))

$\mathcal{O}(ltk + tk(l + j) + tj (k + i)) = \mathcal{O}(t*(lk + kj + ji))$

which is same as feedforward pass algorithm. Since they are same, the total time complexity for one epoch will be

O (t * (i j + j k + k l)) .

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

This time complexity is then multiplied by number of iterations (epochs). So, we have

O (n * t * (i j + j k + k l)),

$O(n*t*(ij + jk + kl)),$ where

n

$n$ 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 $i$ , $j$ , $k$ and $l$ nodes, with $t$ training examples and $n$ epochs. The result was $\mathcal{O}(nt*(ij + jk + kl))$ .

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.

If you're not familiar with back-propagation, check this article:

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

— M.kazem Akhgary
स्रोत

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

4

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 $\mathcal{O}(w)$ where $w$ is the number of weights, i.e., $n * n_i$ , 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 $m$ examples, where each gets repeated $e$ times, is $\mathcal{O}(w*m*e)$ .

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

— maaartinus
स्रोत

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

— DuttaA

1

@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

1

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

1

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.

— James V Stone
स्रोत

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.

— javadba

Dear nbro, That is a fair comment. I dislike adverts too. But it is possible for a book and/or paper to be relevant to a question, as I believe it is in this case. regards, Jim Stone

— James V Stone