![]() |
#1
|
|||
|
|||
![]()
Counting weights. Just to clarify if I am counting right: suppose we have 2 input nodes and one hidden layer with 2 nodes and one output node. So then we have 4 weights, right? According to the problem statement the constant input nodes are counted in the "2". Similarly if every input or hidden layer has an even number of nodes then we must get an even number for the total number of weights. Do you agree? Or do I have an off-by-one error in this count somehow?
|
#2
|
|||
|
|||
![]()
Maybe I'm not right, but I think that in your example we have 6 weights. 4 to fully connect the input layer with the hidden layer, plus 2 for the hidden layer with the output layer
|
#3
|
|||
|
|||
![]()
4 nodes are not needed to connect the input layer with the hidden one, if he wants one of the 2 hidden layer nodes to be constant.
Last edited by rohanag; 05-14-2012 at 11:09 AM. Reason: grammar |
#4
|
|||
|
|||
![]()
so if you did count the constant nodes in your specification, I think 4 is correct for the final answer
|
#5
|
|||
|
|||
![]()
But even the constant is multiplied by a weight that connects it to the next layer, does not?
|
#6
|
|||
|
|||
![]()
Yes so, that's why there are 2 weights from the input to the hidden layer, and 2 nodes from the hidden layer to the final output node.
|
#7
|
|||
|
|||
![]()
Yes yes, now I understood what you meant. The constant unit does not receive connections from back layers.
|
#8
|
|||
|
|||
![]() Quote:
![]() |
#9
|
|||
|
|||
![]()
@mjbeeson, I am sorry for misled you in my first post.
|
#10
|
|||
|
|||
![]() Quote:
|
![]() |
Thread Tools | |
Display Modes | |
|
|