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Understanding of Backpropagation
Kang, Min-Guk
𝑋1
𝑋2
𝑧(1)
| 𝑎(1)
𝑧(2)
| 𝑎(2)
𝑧(3)
| 𝑎(3)
𝑧(4)
| 𝑎(4)
𝑧(5)
| 𝑎(5)
𝑧(6)
| 𝑎(6)
𝑀13 𝑀36
𝑀14
𝑀15
𝑀23
𝑀24
𝑀25
𝑀46
𝑀56
𝑀𝑖𝑛,1
𝑀𝑖𝑛,2
1
1. What is Backpropagation ?
1. Definition
Backpropagation is an algorithm for supervised learning of artificial neural networks using gradient descent.
2. History
Backpropagation algorism was developed in the 1970s, but in 1986, Rumelhart, Hinton and Williams
showed experimentally that this method can generate useful internal representations of incoming data in
Hidden layers of neural networks.
3. How to use Backpropagation?
Backpropagation consists of using simple chain rules. However, we often use non-linear functions
for activation functions, It is hard for us to use backpropagation.
(In this case, I will use sigmoid function for activation function.)
2
2. Preparations
1. Cost function(Loss function)
I will use below cost function(𝑊𝑎 is value of hypothesis, 𝑊𝑡 is value of true)
𝐶 =
1
2
(𝑊𝑎 − 𝑊𝑡)2
2. Derivative of sigmoid function
𝑑𝑆(𝑧)
𝑑𝑧
=
1
(1+𝑒−𝑧)2 × −1 × −𝑒−𝑧 = 𝑆(𝑧)(1 − 𝑆(𝑧))
∵ Sigmoid function = S(z) =
1
1+𝑒−𝑧 , 𝐹 𝑧 =(
1
𝑔 𝑧
) =
− 𝑔 𝑧
𝑔 𝑧 2
3. How to renew weights using Gradient descent
𝑊𝑖→𝑗,𝑛𝑒𝑀 = 𝑊𝑖→𝑗,𝑜𝑙𝑑 − η
𝜕𝐶
𝜕𝑊 𝑖→𝑗,𝑜𝑙𝑑
(η is learning rate)
3
3. Jump to the Backpropagation
1. Derivative Relationship between weights
1-1. The weight update is dependent on derivatives that reside previous layers.(The Word previous means it is located right side )
𝐶 =
1
2
(𝑊𝑎 − 𝑊𝑡)2
→
𝜕𝐶
𝜕𝑊2,3
= (𝑊𝑎 − 𝑊𝑡) ×
𝜕𝑊 𝑎
𝜕𝑊2,3
= (𝑊𝑎 − 𝑊𝑡) ×
𝜕
𝜕𝑊2,3
[σ{𝑧(3)
}] (σ is sigmoid function)
𝜕𝐶
𝜕𝑊2,3
= (𝑊𝑎 − 𝑊𝑡) ×σ{𝑧(3)
} × [1 − σ{𝑧(3)
}] ×
𝜕𝑍3
𝜕𝑊2,3
= (𝑊𝑎 − 𝑊𝑡) ×σ{𝑧(3)
} × [1 − σ{𝑧(3)
}] ×
𝜕
𝜕𝑊2,3
( 𝑎(2)
𝑀2,3)
∎
𝜕𝐶
𝜕𝑊2,3
= (𝑊𝑎 − 𝑊𝑡)σ{𝑧(3)
}[1 − σ{𝑧(3)
}] 𝑎(2)
1𝑋𝑖𝑛 2 3
𝑀1,2 𝑀2,3
𝑀𝑖𝑛,1
𝑧(2)
| 𝑎(2)
𝑧(1)
| 𝑎(1)
𝑧(3)
| 𝑎(3)
𝑎(3)
= 𝑊𝑎
Feed forward
Backpropagation
4
3. Jump to the Backpropagation
1. Derivative Relationship between weights
1-1. The weight update is dependent on derivatives that reside previous layers.(The Word previous means it is located right side )
𝐶 =
1
2
(𝑊𝑎 − 𝑊𝑡)2
→
𝜕𝐶
𝜕𝑊1,2
= (𝑊𝑎 − 𝑊𝑡) ×
𝜕𝑊 𝑎
𝜕𝑊1,2
= (𝑊𝑎 − 𝑊𝑡) ×
𝜕
𝜕𝑊1,2
[σ{𝑧(3)
}] (σ is sigmoid function)
𝜕𝐶
𝜕𝑊1,2
= (𝑊𝑎 − 𝑊𝑡) ×σ{𝑧(3)
} × [1 − σ{𝑧(3)
}] ×
𝜕𝑍3
𝜕𝑊1,2
= (𝑊𝑎 − 𝑊𝑡) ×σ{𝑧(3)
} × [1 − σ{𝑧(3)
}] ×
𝜕
𝜕𝑊1,2
( 𝑎(2)
𝑀2,3)
𝜕𝐶
𝜕𝑊1,2
= (𝑊𝑎 − 𝑊𝑡) ×σ{𝑧(3)
} × [1 − σ{𝑧(3)
}] × 𝑀2,3 ×
𝜕
𝜕𝑊1,2
𝑎(2)
= (𝑊𝑎 − 𝑊𝑡) ×σ{𝑧(3)
} × [1 − σ{𝑧(3)
}] × 𝑀2,3 ×
𝜕
𝜕𝑊1,2
σ{𝑧(2)
}
∎
𝜕𝐶
𝜕𝑊1,2
= (𝑊𝑎 − 𝑊𝑡)σ{𝑧(3)
}[1 − σ{𝑧(3)
}] 𝑀2,3 σ{𝑧(2)
}[1 − σ{𝑧(2)
} ] 𝑎(1)
1𝑋𝑖𝑛 2 3
𝑀1,2 𝑀2,3
𝑀𝑖𝑛,1
𝑧(2)
| 𝑎(2)
𝑧(1)
| 𝑎(1)
𝑧(3)
| 𝑎(3)
𝑎(3)
= 𝑊𝑎
Feed forward
Backpropagation
5
3. Jump to the Backpropagation
1. Derivative Relationship between weights
1-1. The weight update is dependent on derivatives that reside previous layers.(The Word previous means it is located right side )
𝐶 =
1
2
(𝑊𝑎 − 𝑊𝑡)2
→
𝜕𝐶
𝜕𝑊 𝑖𝑛,1
= (𝑊𝑎 − 𝑊𝑡) ×
𝜕𝑊 𝑎
𝜕𝑊 𝑖𝑛,1
= (𝑊𝑎 − 𝑊𝑡) ×
𝜕
𝜕𝑊 𝑖𝑛,1
[σ{𝑧(3)
}] (σ is sigmoid function)
Using same way, we will get below equation.
𝜕𝐶
𝜕𝑊 𝑖𝑛,1
= (𝑊𝑎 − 𝑊𝑡)σ{𝑧(3)
}[1 − σ{𝑧(3)
}] 𝑀2,3 σ{𝑧(2)
}[1 − σ{𝑧(2)
} ] 𝑀1,2 σ{𝑧(1)
}[1 − σ{𝑧(1)
} ] 𝑋𝑖𝑛
1𝑋𝑖𝑛 2 3
𝑀1,2 𝑀2,3
𝑀𝑖𝑛,1
𝑧(2)
| 𝑎(2)
𝑧(1)
| 𝑎(1)
𝑧(3)
| 𝑎(3)
𝑎(3)
= 𝑊𝑎
Feed forward
Backpropagation
6
3. Jump to the Backpropagation
1. Derivative Relationship between weights
1-2. The weight update is dependent on derivatives that reside on both paths.
To get the result, you have to do more tedious calculations than the previous one. So I now just write the result of it.
If you want to know the calculation process, look at the next slide!
𝜕𝐶
𝜕𝑊 𝑖𝑛,1
= (𝑊𝑎 − 𝑊𝑡) 𝑋𝑖𝑛[ σ{𝑧(2)
}[1 − σ{𝑧(2)
}]𝑀2,4 σ{𝑧(1)
}[1 − σ{𝑧(1)
}] 𝑀1,2 + σ{𝑧(3)
}[1 − σ{𝑧(3)
}]𝑀3,4 σ{𝑧(1)
}[1 − σ{𝑧(1)
}]𝑀1,3]
① ② ③ ④
2
3
𝑋𝑖𝑛 1
𝑀𝑖𝑛,1
4 𝑎(3) = 𝑊𝑎
Feed forward
𝑧(4)
| 𝑎(4)
𝑧(3)
| 𝑎(3)
𝑧(1)| 𝑎(1)
𝑀1,2 𝑀2,4
𝑀3,4𝑀1,3
𝑧(2)
| 𝑎(2)
①②
③④
7
2
3
𝑋𝑖𝑛 1
𝑀𝑖𝑛,1
4 𝑎(3)
= 𝑊𝑎
Feed forward
𝑧(4)
| 𝑎(4)
𝑧(3)
| 𝑎(3)
𝑧(1)
| 𝑎(1)
𝑀1,2 𝑀2,4
𝑀3,4𝑀1,3
𝑧(2)
| 𝑎(2)
𝜕𝐶
𝜕𝑊 𝑖𝑛,1
=
𝜕
𝜕𝑊 𝑖𝑛,1
1
2
(𝑊𝑎 − 𝑊𝑡)2
= 𝑊𝑎 − 𝑊𝑡 (
𝜕
𝜕𝑊 𝑖𝑛,1
(σ{𝑧 2
}𝑀2,4 + σ{𝑧 3
}𝑀3,4))
𝜕𝐶
𝜕𝑊 𝑖𝑛,1
= 𝑊𝑎 − 𝑊𝑡 [ 𝑀2,4
𝜕
𝜕𝑊 𝑖𝑛,1
σ{𝑧 2
} + 𝑀3,4
𝜕
𝜕𝑊 𝑖𝑛,1
σ{𝑧 3
} ] = 𝑊𝑎 − 𝑊𝑡 [ 𝑀2,4σ{𝑧 2
}
𝜕
𝜕𝑊 𝑖𝑛,1
(σ{𝑧 1
}𝑀1,2)+
𝑀3,4σ{𝑧 3
}
𝜕
𝜕𝑊 𝑖𝑛,1
(σ{𝑧 1
}𝑀1,3) ]
𝜕𝐶
𝜕𝑊 𝑖𝑛,1
= 𝑊𝑎 − 𝑊𝑡 [ 𝑀2,4σ{𝑧 2
}𝑀1,2σ{𝑧 1
}
𝜕
𝜕𝑊 𝑖𝑛,1
(𝑋𝑖𝑛 𝑀𝑖𝑛,1) + 𝑀3,4σ{𝑧 3
} 𝑀1,3 σ{𝑧 1
}
𝜕
𝜕𝑊 𝑖𝑛,1
(𝑋𝑖𝑛 𝑀𝑖𝑛,1) ]
𝜕𝐶
𝜕𝑊 𝑖𝑛,1
= 𝑊𝑎 − 𝑊𝑡 [ 𝑀2,4σ{𝑧 2
}𝑀1,2σ{𝑧 1
} 𝑋𝑖𝑛 + 𝑀3,4σ{𝑧 3
} 𝑀1,3 σ{𝑧 1
}𝑋𝑖𝑛 ]
= (𝑊𝑎 − 𝑊𝑡) 𝑋𝑖𝑛[ σ{𝑧(2)
}[1 − σ{𝑧(2)
}]𝑀2,4 σ{𝑧(1)
}[1 − σ{𝑧(1)
}] 𝑀1,2 + σ{𝑧(3)
}[1 − σ{𝑧(3)
}]𝑀3,4 σ{𝑧(1)
}[1 − σ{𝑧(1)
}]𝑀1,3]
𝑧(1)
𝑧(1)
8
3. Jump to the Backpropagation
1. Derivative Relationship between weights
1-3. The derivative for a weight is not dependent on the derivatives of any of the other weights in the same layer.
This is easy, so I will not explain it here.(homework )
𝑋1
𝑋2
1
2
3
4
5
6
𝑀13 𝑀36
𝑀14
𝑀15
𝑀23
𝑀24
𝑀25
𝑀46
𝑀56
𝑀(1)
𝑀(2)
Independant
9
3. Jump to the Backpropagation
2. Application of Gradient descent
𝑊𝑖→𝑗,𝑛𝑒𝑀 = 𝑊𝑖→𝑗,𝑜𝑙𝑑 − η
𝜕𝐶
𝜕𝑊 𝑖→𝑗,𝑜𝑙𝑑
(η is learning rate)
① At first, We initialize weights and biases with initializer  we know!
② we can control the learning rate  we know!
③ we can get this value through the equation  we know!
Then, we can renew the weights using above equation. But, is not it too difficult to apply?
So, We will define Error Signal for simple application.
① ② ③
10
3. Jump to the Backpropagation
3. Error Signals
1-1 Defintion: Ύ𝒋 =
𝝏𝑪
𝝏𝒁 𝒋
1-2 General Form of Signals
ÎŽj =
𝜕C
𝜕Zj
=
𝜕
𝜕Zj
1
2
(𝑊𝑎 − 𝑊𝑡)2
= (𝑊𝑎 − 𝑊𝑡)
𝜕𝑊 𝑎
𝜕𝑍 𝑗
------- ①
𝜕𝑊 𝑎
𝜕𝑍 𝑗
=
𝜕𝑊 𝑎
𝜕𝑎 𝑗
𝜕𝑎 𝑗
𝜕𝑍 𝑗
=
𝜕𝑊 𝑎
𝜕𝑎 𝑗
× σ(𝑧𝑗) (∵ 𝑎𝑗 = σ(𝑧𝑗))
Because neural network consists of Multiple units, we can think all of the units 𝑘 ∈ 𝑜𝑢𝑡𝑠 𝑗 .
So,
𝜕𝑊 𝑎
𝜕𝑍 𝑗
= σ(𝑧𝑗) 𝑘∈𝑜𝑢𝑡𝑠 𝑗
𝜕𝑊 𝑎
𝜕𝑧 𝑘
𝜕𝑧 𝑘
𝜕𝑎 𝑗
𝜕𝑊 𝑎
𝜕𝑍 𝑗
= σ(𝑧𝑗) 𝑘∈𝑜𝑢𝑡𝑠 𝑗
𝜕𝑊 𝑎
𝜕𝑧 𝑘
𝑀𝑗𝑘 (∵ 𝑧 𝑘 = 𝑀𝑗𝑘 𝑎𝑗)
By above equation ① and ÎŽk = (𝑊𝑎 − 𝑊𝑡)
𝜕𝑊 𝑎
𝜕𝑍 𝑘
ÎŽj = (𝑊𝑎 − 𝑊𝑡) σ(𝑧𝑗) 𝑘∈𝑜𝑢𝑡𝑠 𝑗
𝜕𝑊 𝑎
𝜕𝑧 𝑘
𝑀𝑗𝑘 = (𝑊𝑎 − 𝑊𝑡)σ(𝑧𝑗) 𝑘∈𝑜𝑢𝑡𝑠 𝑗
ÎŽk
(𝑊 𝑎−𝑊𝑡)
𝑀𝑗𝑘
∎ ÎŽj= σ(𝑧𝑗) 𝑘∈𝑜𝑢𝑡𝑠 𝑗 ÎŽk 𝑀𝑗𝑘 , and for starting, we define Ύ𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = (𝑊𝑎 − 𝑊𝑡)σ{𝑧(𝑖𝑛𝑖𝑡𝑖𝑎𝑙)
}[1 − σ{𝑧(𝑖𝑛𝑖𝑡𝑖𝑎𝑙)
}]
11
3. Jump to the Backpropagation
3. Error Signals
1-3 The General Form of weight variation
( ※ 𝑊3→6,𝑛𝑒𝑀= 𝑊3→6,𝑜𝑙𝑑 − η
𝜕𝐶
𝜕𝑊3→6,𝑜𝑙𝑑
)
( ※ ÎŽ6 = Ύ𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = (𝑊𝑎 − 𝑊𝑡)σ{𝑧(6)
}[1 − σ{𝑧(6)
}] )
∆𝑊3,6 = −η (𝑊𝑎 − 𝑊𝑡)σ{𝑧(6)
}[1 − σ{𝑧(6)
}] 𝑎(3)
= −ηΎ6 𝑎(3)
∆𝑊4,6 = −η (𝑊𝑎 − 𝑊𝑡)σ{𝑧(6)
}[1 − σ{𝑧(6)
}] 𝑎(4)
= −ηΎ6 𝑎(4)
∆𝑊5,6 = −η (𝑊𝑎 − 𝑊𝑡)σ{𝑧(6)
}[1 − σ{𝑧(6)
}] 𝑎(5)
= −ηΎ6 𝑎(5)
∆𝑊1,3 = −η (𝑊𝑎 − 𝑊𝑡)σ{𝑧(6)
}[1 − σ{𝑧(6)
}] 𝑊3,6 σ{𝑧(3)
}[1 − σ{𝑧(3)
}] 𝑎(1)
= −η ( 𝑘∈𝑜𝑢𝑡𝑠 𝑗 ÎŽ6) × 𝑀3,6 σ(𝑧3) 𝑎(1)
= −ηΎ3 𝑎(1)




∎ ∆𝑊𝑖,𝑗= −ηΎ𝑗 𝑎(𝑖)
 We can easily renew weights by using Error Signals ÎŽ and Equation ∆𝑟𝒊,𝒋= −𝜌𝜹𝒋 𝒂(𝒊)
𝑋1
𝑋2
𝑧(1)
| 𝑎(1)
𝑧(2)
| 𝑎(2)
𝑧(3)
| 𝑎(3)
𝑧(4)
| 𝑎(4)
𝑧(5)
| 𝑎(5)
𝑧(6)
| 𝑎(6)
𝑀13 𝑀36
𝑀14
𝑀15
𝑀23
𝑀24
𝑀25
𝑀46
𝑀56
𝑀𝑖𝑛,1
𝑀𝑖𝑛,2
12
4. Summarize
Although the picture below is a bit different from my description, Calculations will show you that this is exactly the same as my explanation.
(Picture Source: http://home.agh.edu.pl/~vlsi/AI/backp_t_en/backprop.html) 13
4. Summarize
Although the picture below is a bit different from my description, Calculations will show you that this is exactly the same as my explanation.
(Picture Source: http://home.agh.edu.pl/~vlsi/AI/backp_t_en/backprop.html) 14

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Backpropagation

  • 1. Understanding of Backpropagation Kang, Min-Guk 𝑋1 𝑋2 𝑧(1) | 𝑎(1) 𝑧(2) | 𝑎(2) 𝑧(3) | 𝑎(3) 𝑧(4) | 𝑎(4) 𝑧(5) | 𝑎(5) 𝑧(6) | 𝑎(6) 𝑀13 𝑀36 𝑀14 𝑀15 𝑀23 𝑀24 𝑀25 𝑀46 𝑀56 𝑀𝑖𝑛,1 𝑀𝑖𝑛,2 1
  • 2. 1. What is Backpropagation ? 1. Definition Backpropagation is an algorithm for supervised learning of artificial neural networks using gradient descent. 2. History Backpropagation algorism was developed in the 1970s, but in 1986, Rumelhart, Hinton and Williams showed experimentally that this method can generate useful internal representations of incoming data in Hidden layers of neural networks. 3. How to use Backpropagation? Backpropagation consists of using simple chain rules. However, we often use non-linear functions for activation functions, It is hard for us to use backpropagation. (In this case, I will use sigmoid function for activation function.) 2
  • 3. 2. Preparations 1. Cost function(Loss function) I will use below cost function(𝑊𝑎 is value of hypothesis, 𝑊𝑡 is value of true) 𝐶 = 1 2 (𝑊𝑎 − 𝑊𝑡)2 2. Derivative of sigmoid function 𝑑𝑆(𝑧) 𝑑𝑧 = 1 (1+𝑒−𝑧)2 × −1 × −𝑒−𝑧 = 𝑆(𝑧)(1 − 𝑆(𝑧)) ∵ Sigmoid function = S(z) = 1 1+𝑒−𝑧 , 𝐹 𝑧 =( 1 𝑔 𝑧 ) = − 𝑔 𝑧 𝑔 𝑧 2 3. How to renew weights using Gradient descent 𝑊𝑖→𝑗,𝑛𝑒𝑀 = 𝑊𝑖→𝑗,𝑜𝑙𝑑 − η 𝜕𝐶 𝜕𝑊 𝑖→𝑗,𝑜𝑙𝑑 (η is learning rate) 3
  • 4. 3. Jump to the Backpropagation 1. Derivative Relationship between weights 1-1. The weight update is dependent on derivatives that reside previous layers.(The Word previous means it is located right side ) 𝐶 = 1 2 (𝑊𝑎 − 𝑊𝑡)2 → 𝜕𝐶 𝜕𝑊2,3 = (𝑊𝑎 − 𝑊𝑡) × 𝜕𝑊 𝑎 𝜕𝑊2,3 = (𝑊𝑎 − 𝑊𝑡) × 𝜕 𝜕𝑊2,3 [σ{𝑧(3) }] (σ is sigmoid function) 𝜕𝐶 𝜕𝑊2,3 = (𝑊𝑎 − 𝑊𝑡) ×σ{𝑧(3) } × [1 − σ{𝑧(3) }] × 𝜕𝑍3 𝜕𝑊2,3 = (𝑊𝑎 − 𝑊𝑡) ×σ{𝑧(3) } × [1 − σ{𝑧(3) }] × 𝜕 𝜕𝑊2,3 ( 𝑎(2) 𝑀2,3) ∎ 𝜕𝐶 𝜕𝑊2,3 = (𝑊𝑎 − 𝑊𝑡)σ{𝑧(3) }[1 − σ{𝑧(3) }] 𝑎(2) 1𝑋𝑖𝑛 2 3 𝑀1,2 𝑀2,3 𝑀𝑖𝑛,1 𝑧(2) | 𝑎(2) 𝑧(1) | 𝑎(1) 𝑧(3) | 𝑎(3) 𝑎(3) = 𝑊𝑎 Feed forward Backpropagation 4
  • 5. 3. Jump to the Backpropagation 1. Derivative Relationship between weights 1-1. The weight update is dependent on derivatives that reside previous layers.(The Word previous means it is located right side ) 𝐶 = 1 2 (𝑊𝑎 − 𝑊𝑡)2 → 𝜕𝐶 𝜕𝑊1,2 = (𝑊𝑎 − 𝑊𝑡) × 𝜕𝑊 𝑎 𝜕𝑊1,2 = (𝑊𝑎 − 𝑊𝑡) × 𝜕 𝜕𝑊1,2 [σ{𝑧(3) }] (σ is sigmoid function) 𝜕𝐶 𝜕𝑊1,2 = (𝑊𝑎 − 𝑊𝑡) ×σ{𝑧(3) } × [1 − σ{𝑧(3) }] × 𝜕𝑍3 𝜕𝑊1,2 = (𝑊𝑎 − 𝑊𝑡) ×σ{𝑧(3) } × [1 − σ{𝑧(3) }] × 𝜕 𝜕𝑊1,2 ( 𝑎(2) 𝑀2,3) 𝜕𝐶 𝜕𝑊1,2 = (𝑊𝑎 − 𝑊𝑡) ×σ{𝑧(3) } × [1 − σ{𝑧(3) }] × 𝑀2,3 × 𝜕 𝜕𝑊1,2 𝑎(2) = (𝑊𝑎 − 𝑊𝑡) ×σ{𝑧(3) } × [1 − σ{𝑧(3) }] × 𝑀2,3 × 𝜕 𝜕𝑊1,2 σ{𝑧(2) } ∎ 𝜕𝐶 𝜕𝑊1,2 = (𝑊𝑎 − 𝑊𝑡)σ{𝑧(3) }[1 − σ{𝑧(3) }] 𝑀2,3 σ{𝑧(2) }[1 − σ{𝑧(2) } ] 𝑎(1) 1𝑋𝑖𝑛 2 3 𝑀1,2 𝑀2,3 𝑀𝑖𝑛,1 𝑧(2) | 𝑎(2) 𝑧(1) | 𝑎(1) 𝑧(3) | 𝑎(3) 𝑎(3) = 𝑊𝑎 Feed forward Backpropagation 5
  • 6. 3. Jump to the Backpropagation 1. Derivative Relationship between weights 1-1. The weight update is dependent on derivatives that reside previous layers.(The Word previous means it is located right side ) 𝐶 = 1 2 (𝑊𝑎 − 𝑊𝑡)2 → 𝜕𝐶 𝜕𝑊 𝑖𝑛,1 = (𝑊𝑎 − 𝑊𝑡) × 𝜕𝑊 𝑎 𝜕𝑊 𝑖𝑛,1 = (𝑊𝑎 − 𝑊𝑡) × 𝜕 𝜕𝑊 𝑖𝑛,1 [σ{𝑧(3) }] (σ is sigmoid function) Using same way, we will get below equation. 𝜕𝐶 𝜕𝑊 𝑖𝑛,1 = (𝑊𝑎 − 𝑊𝑡)σ{𝑧(3) }[1 − σ{𝑧(3) }] 𝑀2,3 σ{𝑧(2) }[1 − σ{𝑧(2) } ] 𝑀1,2 σ{𝑧(1) }[1 − σ{𝑧(1) } ] 𝑋𝑖𝑛 1𝑋𝑖𝑛 2 3 𝑀1,2 𝑀2,3 𝑀𝑖𝑛,1 𝑧(2) | 𝑎(2) 𝑧(1) | 𝑎(1) 𝑧(3) | 𝑎(3) 𝑎(3) = 𝑊𝑎 Feed forward Backpropagation 6
  • 7. 3. Jump to the Backpropagation 1. Derivative Relationship between weights 1-2. The weight update is dependent on derivatives that reside on both paths. To get the result, you have to do more tedious calculations than the previous one. So I now just write the result of it. If you want to know the calculation process, look at the next slide! 𝜕𝐶 𝜕𝑊 𝑖𝑛,1 = (𝑊𝑎 − 𝑊𝑡) 𝑋𝑖𝑛[ σ{𝑧(2) }[1 − σ{𝑧(2) }]𝑀2,4 σ{𝑧(1) }[1 − σ{𝑧(1) }] 𝑀1,2 + σ{𝑧(3) }[1 − σ{𝑧(3) }]𝑀3,4 σ{𝑧(1) }[1 − σ{𝑧(1) }]𝑀1,3] ① ② ③ ④ 2 3 𝑋𝑖𝑛 1 𝑀𝑖𝑛,1 4 𝑎(3) = 𝑊𝑎 Feed forward 𝑧(4) | 𝑎(4) 𝑧(3) | 𝑎(3) 𝑧(1)| 𝑎(1) 𝑀1,2 𝑀2,4 𝑀3,4𝑀1,3 𝑧(2) | 𝑎(2) ①② ③④ 7
  • 8. 2 3 𝑋𝑖𝑛 1 𝑀𝑖𝑛,1 4 𝑎(3) = 𝑊𝑎 Feed forward 𝑧(4) | 𝑎(4) 𝑧(3) | 𝑎(3) 𝑧(1) | 𝑎(1) 𝑀1,2 𝑀2,4 𝑀3,4𝑀1,3 𝑧(2) | 𝑎(2) 𝜕𝐶 𝜕𝑊 𝑖𝑛,1 = 𝜕 𝜕𝑊 𝑖𝑛,1 1 2 (𝑊𝑎 − 𝑊𝑡)2 = 𝑊𝑎 − 𝑊𝑡 ( 𝜕 𝜕𝑊 𝑖𝑛,1 (σ{𝑧 2 }𝑀2,4 + σ{𝑧 3 }𝑀3,4)) 𝜕𝐶 𝜕𝑊 𝑖𝑛,1 = 𝑊𝑎 − 𝑊𝑡 [ 𝑀2,4 𝜕 𝜕𝑊 𝑖𝑛,1 σ{𝑧 2 } + 𝑀3,4 𝜕 𝜕𝑊 𝑖𝑛,1 σ{𝑧 3 } ] = 𝑊𝑎 − 𝑊𝑡 [ 𝑀2,4σ{𝑧 2 } 𝜕 𝜕𝑊 𝑖𝑛,1 (σ{𝑧 1 }𝑀1,2)+ 𝑀3,4σ{𝑧 3 } 𝜕 𝜕𝑊 𝑖𝑛,1 (σ{𝑧 1 }𝑀1,3) ] 𝜕𝐶 𝜕𝑊 𝑖𝑛,1 = 𝑊𝑎 − 𝑊𝑡 [ 𝑀2,4σ{𝑧 2 }𝑀1,2σ{𝑧 1 } 𝜕 𝜕𝑊 𝑖𝑛,1 (𝑋𝑖𝑛 𝑀𝑖𝑛,1) + 𝑀3,4σ{𝑧 3 } 𝑀1,3 σ{𝑧 1 } 𝜕 𝜕𝑊 𝑖𝑛,1 (𝑋𝑖𝑛 𝑀𝑖𝑛,1) ] 𝜕𝐶 𝜕𝑊 𝑖𝑛,1 = 𝑊𝑎 − 𝑊𝑡 [ 𝑀2,4σ{𝑧 2 }𝑀1,2σ{𝑧 1 } 𝑋𝑖𝑛 + 𝑀3,4σ{𝑧 3 } 𝑀1,3 σ{𝑧 1 }𝑋𝑖𝑛 ] = (𝑊𝑎 − 𝑊𝑡) 𝑋𝑖𝑛[ σ{𝑧(2) }[1 − σ{𝑧(2) }]𝑀2,4 σ{𝑧(1) }[1 − σ{𝑧(1) }] 𝑀1,2 + σ{𝑧(3) }[1 − σ{𝑧(3) }]𝑀3,4 σ{𝑧(1) }[1 − σ{𝑧(1) }]𝑀1,3] 𝑧(1) 𝑧(1) 8
  • 9. 3. Jump to the Backpropagation 1. Derivative Relationship between weights 1-3. The derivative for a weight is not dependent on the derivatives of any of the other weights in the same layer. This is easy, so I will not explain it here.(homework ) 𝑋1 𝑋2 1 2 3 4 5 6 𝑀13 𝑀36 𝑀14 𝑀15 𝑀23 𝑀24 𝑀25 𝑀46 𝑀56 𝑀(1) 𝑀(2) Independant 9
  • 10. 3. Jump to the Backpropagation 2. Application of Gradient descent 𝑊𝑖→𝑗,𝑛𝑒𝑀 = 𝑊𝑖→𝑗,𝑜𝑙𝑑 − η 𝜕𝐶 𝜕𝑊 𝑖→𝑗,𝑜𝑙𝑑 (η is learning rate) ① At first, We initialize weights and biases with initializer  we know! ② we can control the learning rate  we know! ③ we can get this value through the equation  we know! Then, we can renew the weights using above equation. But, is not it too difficult to apply? So, We will define Error Signal for simple application. ① ② ③ 10
  • 11. 3. Jump to the Backpropagation 3. Error Signals 1-1 Defintion: Ύ𝒋 = 𝝏𝑪 𝝏𝒁 𝒋 1-2 General Form of Signals ÎŽj = 𝜕C 𝜕Zj = 𝜕 𝜕Zj 1 2 (𝑊𝑎 − 𝑊𝑡)2 = (𝑊𝑎 − 𝑊𝑡) 𝜕𝑊 𝑎 𝜕𝑍 𝑗 ------- ① 𝜕𝑊 𝑎 𝜕𝑍 𝑗 = 𝜕𝑊 𝑎 𝜕𝑎 𝑗 𝜕𝑎 𝑗 𝜕𝑍 𝑗 = 𝜕𝑊 𝑎 𝜕𝑎 𝑗 × σ(𝑧𝑗) (∵ 𝑎𝑗 = σ(𝑧𝑗)) Because neural network consists of Multiple units, we can think all of the units 𝑘 ∈ 𝑜𝑢𝑡𝑠 𝑗 . So, 𝜕𝑊 𝑎 𝜕𝑍 𝑗 = σ(𝑧𝑗) 𝑘∈𝑜𝑢𝑡𝑠 𝑗 𝜕𝑊 𝑎 𝜕𝑧 𝑘 𝜕𝑧 𝑘 𝜕𝑎 𝑗 𝜕𝑊 𝑎 𝜕𝑍 𝑗 = σ(𝑧𝑗) 𝑘∈𝑜𝑢𝑡𝑠 𝑗 𝜕𝑊 𝑎 𝜕𝑧 𝑘 𝑀𝑗𝑘 (∵ 𝑧 𝑘 = 𝑀𝑗𝑘 𝑎𝑗) By above equation ① and ÎŽk = (𝑊𝑎 − 𝑊𝑡) 𝜕𝑊 𝑎 𝜕𝑍 𝑘 ÎŽj = (𝑊𝑎 − 𝑊𝑡) σ(𝑧𝑗) 𝑘∈𝑜𝑢𝑡𝑠 𝑗 𝜕𝑊 𝑎 𝜕𝑧 𝑘 𝑀𝑗𝑘 = (𝑊𝑎 − 𝑊𝑡)σ(𝑧𝑗) 𝑘∈𝑜𝑢𝑡𝑠 𝑗 ÎŽk (𝑊 𝑎−𝑊𝑡) 𝑀𝑗𝑘 ∎ ÎŽj= σ(𝑧𝑗) 𝑘∈𝑜𝑢𝑡𝑠 𝑗 ÎŽk 𝑀𝑗𝑘 , and for starting, we define Ύ𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = (𝑊𝑎 − 𝑊𝑡)σ{𝑧(𝑖𝑛𝑖𝑡𝑖𝑎𝑙) }[1 − σ{𝑧(𝑖𝑛𝑖𝑡𝑖𝑎𝑙) }] 11
  • 12. 3. Jump to the Backpropagation 3. Error Signals 1-3 The General Form of weight variation ( ※ 𝑊3→6,𝑛𝑒𝑀= 𝑊3→6,𝑜𝑙𝑑 − η 𝜕𝐶 𝜕𝑊3→6,𝑜𝑙𝑑 ) ( ※ ÎŽ6 = Ύ𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = (𝑊𝑎 − 𝑊𝑡)σ{𝑧(6) }[1 − σ{𝑧(6) }] ) ∆𝑊3,6 = −η (𝑊𝑎 − 𝑊𝑡)σ{𝑧(6) }[1 − σ{𝑧(6) }] 𝑎(3) = −ηΎ6 𝑎(3) ∆𝑊4,6 = −η (𝑊𝑎 − 𝑊𝑡)σ{𝑧(6) }[1 − σ{𝑧(6) }] 𝑎(4) = −ηΎ6 𝑎(4) ∆𝑊5,6 = −η (𝑊𝑎 − 𝑊𝑡)σ{𝑧(6) }[1 − σ{𝑧(6) }] 𝑎(5) = −ηΎ6 𝑎(5) ∆𝑊1,3 = −η (𝑊𝑎 − 𝑊𝑡)σ{𝑧(6) }[1 − σ{𝑧(6) }] 𝑊3,6 σ{𝑧(3) }[1 − σ{𝑧(3) }] 𝑎(1) = −η ( 𝑘∈𝑜𝑢𝑡𝑠 𝑗 ÎŽ6) × 𝑀3,6 σ(𝑧3) 𝑎(1) = −ηΎ3 𝑎(1) 


 ∎ ∆𝑊𝑖,𝑗= −ηΎ𝑗 𝑎(𝑖)  We can easily renew weights by using Error Signals ÎŽ and Equation ∆𝑟𝒊,𝒋= −𝜌𝜹𝒋 𝒂(𝒊) 𝑋1 𝑋2 𝑧(1) | 𝑎(1) 𝑧(2) | 𝑎(2) 𝑧(3) | 𝑎(3) 𝑧(4) | 𝑎(4) 𝑧(5) | 𝑎(5) 𝑧(6) | 𝑎(6) 𝑀13 𝑀36 𝑀14 𝑀15 𝑀23 𝑀24 𝑀25 𝑀46 𝑀56 𝑀𝑖𝑛,1 𝑀𝑖𝑛,2 12
  • 13. 4. Summarize Although the picture below is a bit different from my description, Calculations will show you that this is exactly the same as my explanation. (Picture Source: http://home.agh.edu.pl/~vlsi/AI/backp_t_en/backprop.html) 13
  • 14. 4. Summarize Although the picture below is a bit different from my description, Calculations will show you that this is exactly the same as my explanation. (Picture Source: http://home.agh.edu.pl/~vlsi/AI/backp_t_en/backprop.html) 14