A Step by Step Backpropagation Example

March 17, 2015February 23, 2024Mazur

Background

Backpropagation is a common method for training a neural network. There is no shortage of papers online that attempt to explain how backpropagation works, but few that include an example with actual numbers. This post is my attempt to explain how it works with a concrete example that folks can compare their own calculations to in order to ensure they understand backpropagation correctly.

Backpropagation in Python

You can play around with a Python script that I wrote that implements the backpropagation algorithm in this Github repo.

Continue learning with Emergent Mind

If you find this tutorial useful and want to continue learning about AI/ML, I encourage you to check out Emergent Mind, a new website I’m working on that uses GPT-4 to surface and explain cutting-edge AI/ML papers:

In time, I hope to use AI to explain complex AI/ML topics on Emergent Mind in a style similar to what you’ll find in the tutorial below.

Now, on with the backpropagation tutorial…

Overview

For this tutorial, we’re going to use a neural network with two inputs, two hidden neurons, two output neurons. Additionally, the hidden and output neurons will include a bias.

Here’s the basic structure:

In order to have some numbers to work with, here are the initial weights, the biases, and training inputs/outputs:

The goal of backpropagation is to optimize the weights so that the neural network can learn how to correctly map arbitrary inputs to outputs.

For the rest of this tutorial we’re going to work with a single training set: given inputs 0.05 and 0.10, we want the neural network to output 0.01 and 0.99.

The Forward Pass

To begin, lets see what the neural network currently predicts given the weights and biases above and inputs of 0.05 and 0.10. To do this we’ll feed those inputs forward though the network.

We figure out the total net input to each hidden layer neuron, squash the total net input using an activation function (here we use the logistic function), then repeat the process with the output layer neurons.

Total net input is also referred to as just net input by some sources.

Here’s how we calculate the total net input for $h_1$ :

$net_{h1} = w_1 * i_1 + w_2 * i_2 + b_1 * 1$

$net_{h1} = 0.15 * 0.05 + 0.2 * 0.1 + 0.35 * 1 = 0.3775$

We then squash it using the logistic function to get the output of $h_1$ :

$out_{h1} = \frac{1}{1+e^{-net_{h1}}} = \frac{1}{1+e^{-0.3775}} = 0.593269992$

Carrying out the same process for $h_2$ we get:

$out_{h2} = 0.596884378$

We repeat this process for the output layer neurons, using the output from the hidden layer neurons as inputs.

Here’s the output for $o_1$ :

$net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1$

$net_{o1} = 0.4 * 0.593269992 + 0.45 * 0.596884378 + 0.6 * 1 = 1.105905967$

$out_{o1} = \frac{1}{1+e^{-net_{o1}}} = \frac{1}{1+e^{-1.105905967}} = 0.75136507$

And carrying out the same process for $o_2$ we get:

$out_{o2} = 0.772928465$

Calculating the Total Error

We can now calculate the error for each output neuron using the squared error function and sum them to get the total error:

$E_{total} = \sum \frac{1}{2}(target - output)^{2}$

Some sources refer to the target as the ideal and the output as the actual.

The $\frac{1}{2}$ is included so that exponent is cancelled when we differentiate later on. The result is eventually multiplied by a learning rate anyway so it doesn’t matter that we introduce a constant here [1].

For example, the target output for $o_1$ is 0.01 but the neural network output 0.75136507, therefore its error is:

$E_{o1} = \frac{1}{2}(target_{o1} - out_{o1})^{2} = \frac{1}{2}(0.01 - 0.75136507)^{2} = 0.274811083$

Repeating this process for $o_2$ (remembering that the target is 0.99) we get:

$E_{o2} = 0.023560026$

The total error for the neural network is the sum of these errors:

$E_{total} = E_{o1} + E_{o2} = 0.274811083 + 0.023560026 = 0.298371109$

The Backwards Pass

Our goal with backpropagation is to update each of the weights in the network so that they cause the actual output to be closer the target output, thereby minimizing the error for each output neuron and the network as a whole.

Output Layer

Consider $w_5$ . We want to know how much a change in $w_5$ affects the total error, aka $\frac{\partial E_{total}}{\partial w_{5}}$ .

$\frac{\partial E_{total}}{\partial w_{5}}$ is read as “the partial derivative of $E_{total}$ with respect to $w_{5}$ “. You can also say “the gradient with respect to $w_{5}$ ”.

By applying the chain rule we know that:

$\frac{\partial E_{total}}{\partial w_{5}} = \frac{\partial E_{total}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial w_{5}}$

Visually, here’s what we’re doing:

We need to figure out each piece in this equation.

First, how much does the total error change with respect to the output?

$E_{total} = \frac{1}{2}(target_{o1} - out_{o1})^{2} + \frac{1}{2}(target_{o2} - out_{o2})^{2}$

$\frac{\partial E_{total}}{\partial out_{o1}} = 2 * \frac{1}{2}(target_{o1} - out_{o1})^{2 - 1} * -1 + 0$

$\frac{\partial E_{total}}{\partial out_{o1}} = -(target_{o1} - out_{o1}) = -(0.01 - 0.75136507) = 0.74136507$

$-(target - out)$ is sometimes expressed as $out - target$

When we take the partial derivative of the total error with respect to $out_{o1}$ , the quantity $\frac{1}{2}(target_{o2} - out_{o2})^{2}$ becomes zero because $out_{o1}$ does not affect it which means we’re taking the derivative of a constant which is zero.

Next, how much does the output of $o_1$ change with respect to its total net input?

The partial derivative of the logistic function is the output multiplied by 1 minus the output:

$out_{o1} = \frac{1}{1+e^{-net_{o1}}}$

$\frac{\partial out_{o1}}{\partial net_{o1}} = out_{o1}(1 - out_{o1}) = 0.75136507(1 - 0.75136507) = 0.186815602$

Finally, how much does the total net input of $o1$ change with respect to $w_5$ ?

$net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1$

$\frac{\partial net_{o1}}{\partial w_{5}} = 1 * out_{h1} * w_5^{(1 - 1)} + 0 + 0 = out_{h1} = 0.593269992$

Putting it all together:

$\frac{\partial E_{total}}{\partial w_{5}} = \frac{\partial E_{total}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial w_{5}}$

$\frac{\partial E_{total}}{\partial w_{5}} = 0.74136507 * 0.186815602 * 0.593269992 = 0.082167041$

You’ll often see this calculation combined in the form of the delta rule:

$\frac{\partial E_{total}}{\partial w_{5}} = -(target_{o1} - out_{o1}) * out_{o1}(1 - out_{o1}) * out_{h1}$

Alternatively, we have $\frac{\partial E_{total}}{\partial out_{o1}}$ and $\frac{\partial out_{o1}}{\partial net_{o1}}$ which can be written as $\frac{\partial E_{total}}{\partial net_{o1}}$ , aka $\delta_{o1}$ (the Greek letter delta) aka the node delta. We can use this to rewrite the calculation above:

$\delta_{o1} = \frac{\partial E_{total}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} = \frac{\partial E_{total}}{\partial net_{o1}}$

$\delta_{o1} = -(target_{o1} - out_{o1}) * out_{o1}(1 - out_{o1})$

Therefore:

$\frac{\partial E_{total}}{\partial w_{5}} = \delta_{o1} out_{h1}$

Some sources extract the negative sign from $\delta$ so it would be written as:

$\frac{\partial E_{total}}{\partial w_{5}} = -\delta_{o1} out_{h1}$

To decrease the error, we then subtract this value from the current weight (optionally multiplied by some learning rate, eta, which we’ll set to 0.5):

$w_5^{+} = w_5 - \eta * \frac{\partial E_{total}}{\partial w_{5}} = 0.4 - 0.5 * 0.082167041 = 0.35891648$

Some sources use $\alpha$ (alpha) to represent the learning rate, others use $\eta$ (eta), and others even use $\epsilon$ (epsilon).

We can repeat this process to get the new weights $w_6$ , $w_7$ , and $w_8$ :

$w_6^{+} = 0.408666186$

$w_7^{+} = 0.511301270$

$w_8^{+} = 0.561370121$

We perform the actual updates in the neural network after we have the new weights leading into the hidden layer neurons (ie, we use the original weights, not the updated weights, when we continue the backpropagation algorithm below).

Hidden Layer

Next, we’ll continue the backwards pass by calculating new values for $w_1$ , $w_2$ , $w_3$ , and $w_4$ .

Big picture, here’s what we need to figure out:

$\frac{\partial E_{total}}{\partial w_{1}} = \frac{\partial E_{total}}{\partial out_{h1}} * \frac{\partial out_{h1}}{\partial net_{h1}} * \frac{\partial net_{h1}}{\partial w_{1}}$

Visually:

We’re going to use a similar process as we did for the output layer, but slightly different to account for the fact that the output of each hidden layer neuron contributes to the output (and therefore error) of multiple output neurons. We know that $out_{h1}$ affects both $out_{o1}$ and $out_{o2}$ therefore the $\frac{\partial E_{total}}{\partial out_{h1}}$ needs to take into consideration its effect on the both output neurons:

$\frac{\partial E_{total}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial out_{h1}} + \frac{\partial E_{o2}}{\partial out_{h1}}$

Starting with $\frac{\partial E_{o1}}{\partial out_{h1}}$ :

$\frac{\partial E_{o1}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial out_{h1}}$

We can calculate $\frac{\partial E_{o1}}{\partial net_{o1}}$ using values we calculated earlier:

$\frac{\partial E_{o1}}{\partial net_{o1}} = \frac{\partial E_{o1}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} = 0.74136507 * 0.186815602 = 0.138498562$

And $\frac{\partial net_{o1}}{\partial out_{h1}}$ is equal to $w_5$ :

$net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1$

$\frac{\partial net_{o1}}{\partial out_{h1}} = w_5 = 0.40$

Plugging them in:

$\frac{\partial E_{o1}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial out_{h1}} = 0.138498562 * 0.40 = 0.055399425$

Following the same process for $\frac{\partial E_{o2}}{\partial out_{h1}}$ , we get:

$\frac{\partial E_{o2}}{\partial out_{h1}} = -0.019049119$

Therefore:

$\frac{\partial E_{total}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial out_{h1}} + \frac{\partial E_{o2}}{\partial out_{h1}} = 0.055399425 + -0.019049119 = 0.036350306$

Now that we have $\frac{\partial E_{total}}{\partial out_{h1}}$ , we need to figure out $\frac{\partial out_{h1}}{\partial net_{h1}}$ and then $\frac{\partial net_{h1}}{\partial w}$ for each weight:

$out_{h1} = \frac{1}{1+e^{-net_{h1}}}$

$\frac{\partial out_{h1}}{\partial net_{h1}} = out_{h1}(1 - out_{h1}) = 0.59326999(1 - 0.59326999 ) = 0.241300709$

We calculate the partial derivative of the total net input to $h_1$ with respect to $w_1$ the same as we did for the output neuron:

$net_{h1} = w_1 * i_1 + w_3 * i_2 + b_1 * 1$

$\frac{\partial net_{h1}}{\partial w_1} = i_1 = 0.05$

Putting it all together:

$\frac{\partial E_{total}}{\partial w_{1}} = \frac{\partial E_{total}}{\partial out_{h1}} * \frac{\partial out_{h1}}{\partial net_{h1}} * \frac{\partial net_{h1}}{\partial w_{1}}$

$\frac{\partial E_{total}}{\partial w_{1}} = 0.036350306 * 0.241300709 * 0.05 = 0.000438568$

You might also see this written as:

$\frac{\partial E_{total}}{\partial w_{1}} = (\sum\limits_{o}{\frac{\partial E_{total}}{\partial out_{o}} * \frac{\partial out_{o}}{\partial net_{o}} * \frac{\partial net_{o}}{\partial out_{h1}}}) * \frac{\partial out_{h1}}{\partial net_{h1}} * \frac{\partial net_{h1}}{\partial w_{1}}$

$\frac{\partial E_{total}}{\partial w_{1}} = (\sum\limits_{o}{\delta_{o} * w_{ho}}) * out_{h1}(1 - out_{h1}) * i_{1}$

$\frac{\partial E_{total}}{\partial w_{1}} = \delta_{h1}i_{1}$

We can now update $w_1$ :

$w_1^{+} = w_1 - \eta * \frac{\partial E_{total}}{\partial w_{1}} = 0.15 - 0.5 * 0.000438568 = 0.149780716$

Repeating this for $w_2$ , $w_3$ , and $w_4$

$w_2^{+} = 0.19956143$

$w_3^{+} = 0.24975114$

$w_4^{+} = 0.29950229$

Finally, we’ve updated all of our weights! When we fed forward the 0.05 and 0.1 inputs originally, the error on the network was 0.298371109. After this first round of backpropagation, the total error is now down to 0.291027924. It might not seem like much, but after repeating this process 10,000 times, for example, the error plummets to 0.0000351085. At this point, when we feed forward 0.05 and 0.1, the two outputs neurons generate 0.015912196 (vs 0.01 target) and 0.984065734 (vs 0.99 target).

If you’ve made it this far and found any errors in any of the above or can think of any ways to make it clearer for future readers, don’t hesitate to drop me a note. Thanks!

And while I have you…

Again, if you liked this tutorial, please check out Emergent Mind, a site I’m building with an end goal of explaining AI/ML concepts in a similar style as this post. Feedback very much welcome!

1,112 thoughts on “A Step by Step Backpropagation Example”

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katch says:

February 10, 2017 at 2:05 am

thanks!

Reply
Shekhar Prasad Rajak (@Shekharrajak) says:

February 13, 2017 at 11:17 pm

After this line :
“`
Here’s how we calculate the total net input for h_1:
“`

instead of w2 there should be w3. means w1*i1 + w3*i2 + b1*1

Reply
1. greenFr0g says:
  
  February 21, 2017 at 4:08 am
  
  yes, this part doesn’t make sense with the diagram, and I think should be corrected.
  
  Reply
2. greenFr0g says:
  
  February 21, 2017 at 4:14 am
  
  I stand corrected, w2 is the weight from i2 to h1, and w3 is the weight from i1 to h2
  
  Reply
Amit Kumar says:

February 14, 2017 at 3:27 am

You forgot to include how bias is updated.

Reply
Melo says:

February 18, 2017 at 2:58 pm

Great!

Reply
Edmund Gyasi says:

February 20, 2017 at 4:44 pm

Good job !, I like the way you explain stuff…..

Reply
DangHoang says:

February 22, 2017 at 11:14 pm

Thank you for your writing. It’s really interested.

Reply
Dimitris K says:

February 24, 2017 at 8:35 am

Dear Sir Mazur, I cant thank you enough for your exquisite tutorial. My only question is the following.
If we had more hidden layers, lets say 2, how can I calculate the derivative θΕ(tot)/θout(h-1), in order to correct the weights linking my input neurons with the first h.l. neurons? In the second hidden layer we wouldn’t have that kind of problem since we can link θΕ(tot)/θout(h1)=θΕ1/θout(h1)+θΕ2/θout(h1) as you wrote, but for a previews layer there is not “Error association”, as far as I can perceive. I apologise if my question is not well stated.

Thank you in advance

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Mohammed M. Alhessi says:

February 25, 2017 at 11:51 pm

Thank you so much. it is really interesting and very helpful.

Reply
Mark M says:

February 26, 2017 at 12:18 pm

Looks like you have an error in one of your formulas:
net_{h1} = w_{1} * i_{1} + w_{2} * i_{2} + b_{1} * 1

should be:
net_{h1} = w_{1} * i_{1} + w_{3} * i_{2} + b_{1} * 1

Reply
1. hallberg87 says:
  
  April 6, 2017 at 10:11 pm
  
  The diagram should have better spacing to eliminate the confusion, but,
  w2 is the weight from i2 to h1, and w3 is the weight from i1 to h2
  So his formula is good.
  
  Reply
Deepak Mandge says:

February 27, 2017 at 1:15 am

Excellent examples with simple explanations. Thank you

Reply
Shinny says:

March 1, 2017 at 2:30 am

Awesome!

Reply
josephkj20 says:

March 2, 2017 at 9:05 am

The Second formula under section: Hidden Layer

should be

\frac{\partial E_{total}}{\partial out_{h1}} = \frac{\partial E_{tot}}{\partial out_{h1}} + \frac{\partial E_{tot}}{\partial out_{h1}} rt?

instead of

\frac{\partial E_{total}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial out_{h1}} + \frac{\partial E_{o2}}{\partial out_{h1}}

Reply
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pete says:

March 9, 2017 at 9:25 pm

how it be used in DL,I mean calculated in image process(2 demension signal process),such as CNN.

Reply
Melodytch says:

March 11, 2017 at 12:46 am

Hi, I’m new to this topic so pardon me for asking simple questions or having wrong concepts. I understand that backpropagation algorithm uses gradient descent to find the minimized error, but you did not mention in your example that it stops when it reaches the minimum error value? Or rather, how do you know when you can stop the algorithm in this example?

Reply
1. BhanuKiran Chaluvadi says:
  
  April 18, 2017 at 11:20 pm
  
  Usually people set epoch. For example epoch = 10000, neural network will run 10000 times(forward and backward propagation).If they are not satisfied with the result they will tweak number of epoch.
  
  Reply
2. Mehrdad says:
  
  May 26, 2017 at 7:57 pm
  
  Hi Melodytch, lets I explain with an example. You have 1020 data input and you feed your network 1000 data as training and 20 as test data set. anytime you training data, you calculate error with this formula: MAPE=( |output-target| / target ) * 100 , and calculate this for test data separately. any iteration control the MAPE for the test if MAPE for test improves, it is mean your network get a better result to predict. When MAPE for train improved but MAPE for test stop and get worse its mean your network come to overfitting and you should stop training.
  We have other methods for stop training such as fix training epoch(iteration), those other friends mention about that.
  
  Reply
Murali Krx says:

March 13, 2017 at 12:50 pm

Great Explanation! Thanks a ton!!!

Reply
alpha says:

March 14, 2017 at 8:00 am

really very helpful

Reply
kris says:

March 14, 2017 at 4:01 pm

One of the best explanation for back-propagation that I have come across. Thank you!

Reply
Derrick says:

March 15, 2017 at 6:22 pm

Dear Matt,
I have been studying Machine Learning and Neural Networks intensely for the past 9 months and this IS the best explanation I have come across (including 2 online MOOC’s – Stanford and Johns Hopkins Universities). Brilliant!

Bet regards,

Derrick Attfield

Reply
Sur says:

March 18, 2017 at 4:38 pm

Thank you so much Matt ! You have explained the basic working backpropagation which is very helpful for the beginners of ML ! :)

Reply
endre says:

March 18, 2017 at 4:51 pm

Some of the best stuff I’ve read, extremely clear using a concrete example!

However, I would so much have appreciated that you did your calculations using Cross Entropy instead of (or in addition to!) the Squared Error as error function. Also, I’ll also point out that you do not calculate the Bias change (and the nodes should have separate biases..)

Also, while you’re at it (!), how does the derivation change using ReLU’s instead of Standard Logistc?

Reply
Sai Chaithanya Adapa says:

March 21, 2017 at 9:16 am

Great Explanation

Reply
Dimi says:

March 21, 2017 at 10:37 am

Great class! Thank you.

Reply
hgygty says:

March 22, 2017 at 1:01 pm

very good thing .

Reply
Sam says:

March 22, 2017 at 4:48 pm

This was very helpful for me. Thanks!

Reply
Pinto says:

March 28, 2017 at 9:57 pm

I would like to thank for this post and I would like to make one question about the partial derivative of the sigmoid function for out_o1 and out_h1.
the partial derivative of the function: out_o1=1/(1+exp(-net_o1) ) with respect to the net_o1 in this post is : out_1*(1-out_o1) why not :net_o1*(1-net_o1)?

Reply
Jaloliddin says:

March 29, 2017 at 9:46 pm

Hi there! Really need to clarify this point.
What steps will neural network run in case of a real data set? Let’s say, we run first record and found some error rate – e1. Will we pass forward our first record’s values again after finishing backpropagation with new weights or new weights will be applied to next record?
in other words, should we keep running backpropagation for one record until its weights are unsatisfactory?

Reply
Suresh says:

March 30, 2017 at 8:19 am

Great article indeed. I could not find any better resource than this one that explains backpropagation with an example.. Thanks for taking time to write this. This helped me a lot.

Reply
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pablo says:

March 30, 2017 at 9:30 pm

thank you this was very helpful!

Reply
CESAR says:

March 31, 2017 at 3:25 pm

Great, thanks

Reply
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young says:

April 4, 2017 at 9:49 pm

Thank you for your nice example.
I’m curious to know how weights of biases are updated in this example. This example includes only one starting weight for each bias.

Reply
Dominic Waithe (@dwaithe) says:

April 5, 2017 at 6:31 am

Hi, I really like this resource. Just wanted to reiterate other comments that I read which state that each node should have a separate bias weight (b1-b4) not just one bias per layer (e.g. b1,b2). Would be good to update the images to make that clear.

Reply
sunwaveapps says:

April 6, 2017 at 1:30 pm

https://s0.wp.com/latex.php?latex=%5Cfrac%7B%5Cpartial+net_%7Bo1%7D%7D%7B%5Cpartial+out_%7Bh1%7D%7D+%3D+w_5+%3D+0.40&bg=ffffff&fg=404040&s=0

But what if the network has two hidden layers? How am I supposed to get the weight connecting my hidden node to an output if there’s a whole other layer in between? somebody halp ;-;

Reply
Mohammed Adel says:

April 7, 2017 at 8:54 am

Great Topic with a smooth way of explaining. is there is any article for Convolutional neural network ?

Reply
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Pengshan Cai says:

April 9, 2017 at 10:59 pm

Very clear presentation and helps a lot. Thanks!

Reply
H Mann says:

April 12, 2017 at 8:20 pm

Great explanation. Good illustration and to the point.

Reply
Navy says:

April 13, 2017 at 10:21 am

Hi. Great article! I have a problem with computing net input for second hidden neuron in the forward pass.

Basically I put weights in a two-dimensional array, with input index and hidden neuron index. Then for every hidden neuron and for every input neuron, I calculate the net number: netNumber += weights[inputIndex][hiddenindex]*inputVectors[inputIndex];
Next, I add biases and use activation function. The result is wrong (out_h2 = 0.655318370173895), but the out_h1 gets computed correctly. Any hints?

Reply
yoba says:

April 16, 2017 at 5:05 am

Very comprehensive and useful example. Many thanks !

Reply
Himanshu says:

April 17, 2017 at 5:25 am

While calculating w1, shoul i use updated w5 or original w5?

Reply
1. Hitesh says:
  
  April 30, 2017 at 7:03 am
  
  original weight
  
  Reply
2. raghu says:
  
  May 5, 2017 at 4:21 am
  
  No, We perform the actual updates in the neural network after we have the new weights leading into the hidden layer neurons (ie, we use the original weights, not the updated weights, when we continue the backpropagation algorithm below).
  
  Reply
3. Nehi says:
  
  May 8, 2017 at 4:05 pm
  
  Original
  
  Reply
4. Sushovan Haldar says:
  
  May 17, 2017 at 6:29 am
  
  The forward pass and backward pass consists of one epoch.In this one epoch the weights are being updated using the just immediate previously calculated weights.
  Since in this example,the author has updated the weights only once,so the original value of w5 is to be used here.
  
  Reply
MaviccPRP says:

April 18, 2017 at 2:39 am

Really good tutorial Matt. I finally had something like a “aha” moment ;-)

With the inspiration of this tutorial I translated this real number example into matrix multiplication operations.

Reply
jerpint says:

April 18, 2017 at 1:13 pm

Hello, thanks for your explanation. Can you please clarify why the biases b1 and b2 are the same for their entire
respective layers? Shouldnt we have a bias vector b of the same length as the number of neurons per layer?

Reply
Paul Martin says:

April 18, 2017 at 2:39 pm

Like to print things out and read over. Seems like the print out starts omitting detail once it hits the hidden layers. Odd. All in all this is quite helpful to understand the back propagation.

Reply
Ab says:

April 24, 2017 at 5:35 pm

Really appreciate this, thanks!

Reply
KG says:

April 25, 2017 at 7:41 pm

thank you very much best explaining i have ever seen

Reply
mahdi says:

April 25, 2017 at 7:52 pm

that was very helpful < thanx from a fighter in
mahadi army

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