Backward Propagation Tutorial


 
Introduction
Backward Propagation is probably the most common method for
training forward-feed neural networks. A forward pass using an input pattern
propagates through the network and produces an actual output. The backward
pass uses the desired outputs corresponding to the input pattern and updates
the weights according to the error signal.
There are hundreds of papers covering the subject of backward
propagation. Unfortunately, many of them tend to exhibit a vast stockpile
of equations and complicated partial derivatives with undefined variables
to explain a concept that is really quite simple. Quite often, a pseudocode
algorithm or an example with pictures is the most effiecient method to convey
information.
Artificial Neuron Anatomy
請稍等，圖片正在下載中...
 
w --> weight

I --> input

O --> output

A --> activation value
Think of the activations as a pulses of electricity traveling
along the dendrite path. Each activation pulse is multiplied by the weight
of the dendrite it is traveling along. When all inputs to the neuron have
been "stimulated" with this pulse, the neuron adds up these products
and performs a thresholding or "squashing" operation on the resulting
sum. The result of the squashing function is then sent to each of its output
dendrites and the process is repeated for each neuron on the receiving end.
Thus, the net input of each neuron is:
(EQU 1)         請稍等，圖片正在下載中...
Read this equation as: The net input of a neuron is the sum
of (all weights connected as inputs to the neuron * all activation values
of the neurons at the end of those inputs)





i refers to a neuron that is an input to the neuron
we're looking at. (PREVIOUS)


j is the node we're currently looking at. (CURRENT)


k refers to a neuron that is an output to the neuron
we're looking at. (NEXT)


In the above diagram, the netin for the neuron would
be:
netin
= (w0 * A0) + (w1 * A1) + (w2 * A2);
The Activation Function
The most common squashing function is the sigmoidal function
defined as:
(EQU 2)         請稍等，圖片正在下載中...which
might look something like this if you graphed it on your TI-82 using F = 5:
請稍等，圖片正在下載中...
F is a factor that determines the pitch or "steepness"
of the function's graph. The sigmoid function will take an input and "squash"
it so that it's output is between 0 and 1.
 
So now we know how find netin for the neuron and use
it to give us an activation pulse A using Act(netin).
The Forward Pass
Consider the following network:
請稍等，圖片正在下載中...
For the forward pass, the only neuron variables we are interested
in are netin and A. Throughout the forward pass, all weights
(W) remain constant -- the only values that ever change are netin
and our activation pulse A.



Nodes 1 and 2 are input nodes.


Nodes 3 and 4 are hidden nodes.


Nodes 5 and 6 are output nodes.



The object of a forward pass is to give input nodes 1 and 2
your input pattern, and then compute netin and activation A
for each node. At the end of the pass, output nodes 5 and 6 will hold activation
values that essentially tell you what the network thinks the answer is.

At the start of a forward pass, nodes 1 and 2 are given input values.
Node1.A = some input
pattern bit

Node2.A = some input pattern bit

Then recalling equations 1 and 2, we can compute Node3.netin
and Node4.netin by the following:
(EQU 1)         請稍等，圖片正在下載中...
Node3.netin
= (Node1.A*
W13) + (Node2.A *
W23)

Node4.netin = (Node1.A*
W14) + (Node2.A *
W24)
Now that we know the netin values for
neurons 3 and 4, we can use the threshold function to find their activation
values.
(EQU 2)         請稍等，圖片正在下載中...
Node3.A
= 1 / (1 + e^(-F
* Node3.netin))

Node4.A = 1 / (1 + e^(-F * Node4.netin))

Now we can do the exact same things we just did above to give
us the activation values of neurons 5 and 6.
Node5.netin
= (Node3.A
* W35) + (Node4.A*
W45)

Node6.netin= (Node3.A *
W36) + (Node4.A*
W46)
Node5.A=
1 / (1 + e^(-F * Node5.netin))

Node6.A = 1 / (1 + e^(-F * Node6.netin))
Done! A5
and A6 are the outputs
of our network after a forward pass.
The Backward Pass
In the backward pass, we are given the desired outputs and propagate
backward through the network updating weight values for each dendrite. You
should perform a forward pass before running a backward pass. During a backward
pass, the variables A and netin remain constant and the values
that change are §, netout, D and W. A backward
pass takes one parameter -- the desired output (D) that you think the
previous input given in the forward pass should have produced as an output.



§ Is the error signal at each neuron. It emulates
a pulse much like the activation value A in forward propagation.


(I couldn't find a greek delta in html -- so pretend it's the same delta
in the network diagram above.)


netout is the sum of all output products for a neuron.
It is much like the netin in forward propagation.


Wjk is the weight of a particular
dendrite between neuron j and a neuron k that is an ouput
of neuron j.

D is the given desired value (used only on ouput nodes). (as the
"trainer", you provide this value)
A is the activation value that was calculated for each node during
the forward pass.
netin is the sum of all input units that was calculated for each
neuron during the forward pass.
F is a constant that specifies the slope/pitch of the sigmoidal function.
n (neta) is a constant that determines the learning rate of
the network.

For the output nodes, netout is not computed, (there are no outputs
on an output node). So for an output node, we set our netout to be the
difference between our desired output and the actual output that the network
gave when we did our last forward pass.
(EQU 3)         請稍等，圖片正在下載中...
Calculating netout in our example network above for nodes 5 and 6 is
as easy as:
Node5.netout = (Node5.D
- Node5.A)

Node6.netout = (Node6.D - Node6.A)
To find the error signal for any neuron, we calculate an error using:
(EQU 4)        請稍等，圖片正在下載中...
As you recall from our forward pass our sigmoidal activation function is:
(EQU 2)         請稍等，圖片正在下載中...
Taking the derivative of Act(netin) gives us:
(EQU 5)        請稍等，圖片正在下載中...
Substituting this into EQU 4 we get:
(EQU 4)        請稍等，圖片正在下載中...
This simplified version is the official equation we'll use for updating our
error signals (§). So, in the case
of our network above, we can find the error signals of the two output nodes
5 and 6 by:
Node5.§ = Node5.A
* (1 - Node5.A) * Node5.netout

Node6.§ = Node6.A * (1 - Node6.A) * Node6.netout
The error in a neuron is used to update all of its input dendrites' weights
using the following equation:
(EQU 5)        請稍等，圖片正在下載中...
Since we have the error signals Node5.§ and Node6.§
we can use this equation with neurons 3 and 4 in the diagram to update their
input weights.
W35
= W35 + n * Node5.§ * Node3.A

W36 = W36 + n *
Node6.§ * Node3.A

W45 = W45 + n *
Node5.§ * Node4.A

W46 = W46 + n *
Node6.§ * Node4.A
The method for calculating the general netout for all non-ouput nodes
is much like finding the general netin on a forward pass
(EQU 6)        請稍等，圖片正在下載中...
Implementing this with our example network diagram above, we calculate our
netout variables via:
Node3.netout
= (W35
* Node5.§) + (W36
* Node6.§)

Node4.netout = (W45
* Node5.§) + (W46
* Node6.§)
And use the netout variables to calculate our error signals
§.
Node3.§ = Node3.A
* (1 - Node3.A) * Node3.netout

Node4.§ = Node4.A * (1 - Node4.A) * Node4.netout
At this point, all equations have been introduced, and we continue back-propagating
by calculating each neuron's netout and §,
and then then updating its output weights until the entire network
is updated. Continuing our example, we move on to nodes 1 and 2 and update their
output weights as follows:
W13
= W13 + n * Node3.§ * Node1.A

W14 = W14 + n *
Node4.§ * Node1.A

W23 = W23 + n *
Node3.§ * Node2.A

W24 = W24 + n *
Node4.§ * Node2.A
If neurons 1 and 2 were hidden nodes, we would compute their error signals,
and keep going. However, since1 and 2 are input nodes and we have already updated
all weights in the network, there is no need to continue. That's all there is
to it!
Network Structure and Variations
Bi-connection: Most networks are bi-connected so that each layer is
fully connected with the layers immediately before and after it.
Biases: To help "normalize" a neural network, most networks
have bias nodes connected as an input to each non-input neuron. You initialize
their dendrite weights in the same way as any other neuron, and give them an
initial activation value of -1 or 1. Treat them as you would treat a neuron
in the input layer and update their dendrite weights during back-propagation
just as you would update the weights of any other dendrite.
Initialization: The simplest way to initialize a network's weights is
using small random values, say between +- 0.5. It doesn't matter how you initialize
the activation values, error signals, or any other parameters specific to a
neuron because they are over-written during the propagation processes. Two guys
named Nguyen and Widrow came up with an initialization scheme that seems to
help the network converge a bit faster.
F = ((0.7h^(1/i))/R)

BiasWeights = +-F

All Other Weights = F
F = the weight scale factor.

R is the range of input values

h = # of hidden neurons

i = # of input neurons

w = weight
The Squashing Function: Although the most popular thresholding technique
uses the sigmoidal method, there are several other functions that might yield
a better result in a particular situation. Some networks such as ANFIS use a
fuzzy logic lookup table to appropriately modify the type of sigmoidal function
as the network is trained. As far as I know, almost all activation functions
are continuous and differentiable to allow the backward propagation to calculate
the gradient error signal.
Here is a variation that allows you to adjust the asymptotes of the sigmoid function
so that it produces an output in any given desired range:
[(-L - H) / (1 + e^(-F * netin))] + L
L --> the low boundary

H --> the high boundary

F --> the steepness/pitch factor
Classification encoding schemes
A common use of neural networks is for classification of items. The object
is to take the data from a number of known samples, enter it as inputs to the
network and use backward propagation to train the network so that it later it
can correctly classify each sample. So once you have the back-prop algorithms
coded up, the next step is to figure out how to encode your information so that
the network outputs a meaningful value. There are many ways in which you can
encode your information, but there are two schemes that seem to be the most
popular for class separation.
Variables

Cc is the number of classes that you want to distinguish. (Class
Count)
Ci is the class index.
A is the output activation value. In this section it will always
refer to the activation value of the output neuron. (METHOD 1)
{A} is a set of activation values for all output neurons (METHOD
2)
D is the given desired value for the output node (as the "trainer",
you provide this value when training).

We will use an common example throughout this section. Lets say that we want
the network to classify each sample as either type 0, 1, 2, 3, or 4. (5 total
classes)
METHOD 1 -- One output neuron for all classes
The object here is to train the network so that the neuron's output activation
value A, (usually between 0 and 1), can be divided into a specified number
of classes.
請稍等，圖片正在下載中...
=====Identification====
To get the class index Ci of a given activation value A, you
would use the formula:
(EQU 7)       Ci = floor(A / (1.0
/ Cc))
Using our example, the following mapping shows how to convert an activation
value A into the corresponding class index Ci:
A Range         Class Index
(Ci)

0.00-0.20 --> 0

0.21-0.40 --> 1

0.41-0.60 --> 2

0.61-0.80 --> 3

0.81-1.00 --> 4
You can use this code to convert the output neuron's activation value A
into a class index Ci.
[blockquote]
// given the output neuron's activation output (dAct),
and the # of classes, return it's corresponding index

static int ClassGetIndex(double dAct, int nNumClasses) {

          return (int)(dAct
/ (1.0 / (double)nNumClasses));

}
[/blockquote]
So for example, if the network's output node's activation value A is
0.34, then to find the corresponding index we use (EQU 7):
In our example there are 5 total classes (0 to 4), so Cc is 5.

Ci = floor(0.43 / (1.0 / 5))

Ci = 1
**note that we can easily check using the above map to see that 0.21 < 0.34
< 0.40



int nClassIdx = GetClassIndex(0.43, 5); // puts type 1 into nClassIdx.
=====Training====
When training the network, you would use the value that is in the "middle"
of a class division to determine the desired value (D). When training,
you know Ci, and want to find out what value D should be to ensure optimal training
so that accurate identification is possible when the same inputs are used later.
To get D, you would use the formula:
(EQU 8)     D = (Ci * (1.0 / Cc))
+ (1.0 / Cc / 2.0)
Here is a mapping table for our example:
Ci       D

0 --> 0.1

1 --> 0.3

2 --> 0.5

3 --> 0.7

4 --> 0.9


You can use the following code to convert class index Ci to a desired
value D
[blockquote]
// given a class index and the # of classes, returns
the "desired" value

static double ClassGetDesired(int nClassIdx, int nClassCnt) {

           return (nClassIdx
* (1.0 / (double)nClassCnt)) + (1.0 / (double)nClassCnt / 2.0);

}
[/blockquote]
So in our example, if you were training and you happen to know that a certain
set of inputs are supposed to correspond to class index 3, you would use EQU
8 to find out what value should be used for D during backward propagation
training.
In our example there are 5 total classes (0 to 4), so Cc is 5.

D = (3 * (1.0 / 5)) + (1.0 / 5 / 2.0)

D = 0.7
double dDesired = ClassGetDesired(3, 5); // puts 0.7 into dDesired
METHOD 2 -- One output neuron for each class
The object is to use multiple output neurons, where each output neuron corresponds
to a specific class and the "winner" is the one with the largest value.
請稍等，圖片正在下載中...
=====Identification====
In our example, for identification, we would use the following mapping scheme:
LLLLH --> 0

LLLHL --> 1

LLHLL --> 2

LHLLL --> 3

HLLLL --> 4
Where H is the largest value, and the L's are low values.
To get the class index Ci of a given set of activation values A,
you would look at the activation value for each neuron and determine the index
of the largest one. The output neuron with the highest activation value is considered
the "winner".
For coding, lets put each output neuron's activation value into a slot in a
dynamically allocated array. Then we can find the largest value in the array
and get its index to convert an activation value A into a class index
Ci.
[blockquote]
// given the neural network's outputs, and # of classes,

// returns the class index that most closely matches the neural nets outputs

//

// ex: given [0.001][0.0012][0.987][0.0211], nClassCnt == 4, returns 2 since
the value of index 2 is the largest

static int ClassGetIndex(double* Outputs, int nClassCnt) {

          double dLargestVal
=DBL_MIN;

          int nLargestIdx
= 0;

          for (int I = 0;
I < nClassCnt; I++) {

                    if
(Outputs[I] > dLargestVal) { dLargestVal = Outputs[I]; nLargestIdx = I;
}

          }

          return nLargestIdx;

}
[/blockquote]
So for example if the 5 output neuron activation values {A} are:
    0       1       2       3       4

{0.01, 0.05, 0.98, 0.13, 0.02}
The corresponding class index Ci would be 2 since 0.98 is the largest
value and is in slot 2.
=====Training====
For training, you would simply give a desired value D of 1 for the neuron
corresponding to the index you want to activate with the given training pattern,
and 0 for all the other neurons.
0 --> 00001

1 --> 00010

2 --> 00100

3 --> 01000

4 --> 10000
Say you want the network to output class index Ci of type 3 for a given
input set. The output nodes should use the desired values of {0, 0, 0, 1, 0}
for

back-propagation.
I use the following C++ functions to translate between desired and activation
values when using METHOD 2:

**Note, strings are used instead of arrays here because most of the time they're
in string format being read from/written to a file. The strings could easily
be replaced with an array or a list.
[blockquote]
// given a class index and the # of classes,

// returns a string containing the desired outputs separated by strDelim

// (this is usually written to a training file)

// ex: given nClassIdx == 2, nClassCnt = 4, strDelim = "¦"
--> returns "0¦0¦1¦0"

static CString ClassGetOutputString(int nClassIdx, int nClassCnt, CString
strDelim) {

          CString strRet
= "";

          for (int I = 0;
I < nClassCnt - 1; I++) {

                    if
(nClassIdx == I) { strRet += "1" + strDelim; } else { strRet +=
"0" + strDelim; }

          }

          if (nClassIdx
== I) { strRet += "1"; } else { strRet += "0"; }

          return strRet;

}
// given a string such as "0¦0¦1¦0",
returns 2 because the 1 corresponds to index 2

// returns -1 on error

static int ClassGetOutputIndex(CString strOutputs) {

          int nNum = 0;
char ch;

          for (int I = 0;
I < strOutputs.GetLength(); I++) {

                    ch
= strOutputs[I];

                    if
(ch == '1') { return nNum; } else if (ch == '0') { nNum++; }

          }

          return -1;

}
[/blockquote]
Conclusion
I've found that METHOD 2 seems to be more accurate for most applications, but
it requires more nodes, which increases memory requirements. METHOD 2 is also
less susceptible to precision round-off error. As for speed, I'm not sure which
one trains faster. Each epoch takes longer using METHOD 2, but METHOD 2 doesn't
seem to require as many epochs to get a good separation.


Additional Introductory Information

http://home.cc.umanitoba.ca/~umcorbe9/perceptron.html
A perceptron java applet that lets you step through an example.
http://www.dontveter.com/bpr/public2.html
-- walks through an example of the XOR network. I found it extremely helpful.
Sometimes an example is worth more than 1000 greek variables.
http://www.ece.ubc.ca/~elec592/PDFfiles/7.0.pdf
-- gives some insight as to how the back-propagation method is derived, although
I think it's a bit of an overkill.

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症狀是用Control Panel的uninstall/repair，或者源安裝盤，都在load某個component時掛掉，不能進入正常安裝界面，然後彈出對話框，提示“A Problem has been encountered while loading the setup components. Canceling setup.”

而Log中顯示錯誤
MSITOSIT: [2] ERROR processed; exception was thrown for retail build
Developer Comment: : has no associated directory, verify the Feature table is correct
Build Time: Thu Dec 11 14:04:18 2003
HRESULT: -2147467259

網上查了一下，有的說是SP1的問題，有的說是Kaspersky的問題，我這裏都不管用。

最後的解決辦法是找到源安裝盤的vs_setup.msi，用Admin進入cmd，運行msiexec /x vs_setup.msi，它似乎就什麼都不檢查，就能卸載了。

之後重裝也沒問題。

To add additional information to domain users it is necessary to use the Profiles API. Since the release of this API, the Shared Contacts API should be use only for external contacts. For more information see:
http://code.google.com/apis/apps/profiles/developers_guide_protocol.html

錯誤789：L2TP 連接嘗試失敗,因為安全層在與遠程計算機進行初始協商時遇到處理錯誤

增加注冊表項(以下操作均在客戶端完成)

HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Services\RasMan\Parameters

"ProhibitIPSec"=dword:00000001

然後重啟機器