Homework 6 #10
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 offbyone error in this count somehow?

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so if you did count the constant nodes in your specification, I think 4 is correct for the final answer

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@mjbeeson, I am sorry for misled you in my first post.

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I don't think that's the way we are to read the homework. Notice problem 8, where it says,
"10 input units (including the constant value)...and 36 hidden units (include the constant inputs of each hidden layer). The "including" part important in the counting. 
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Has anybody got the correct answer? My answer was wrong and I can't see where is a mistake in my reasoning.
The maximum number of weights should be for the network looking like 1018181 (including constant nodes in the first three layers), right? But number of weights for it is not one in the answer key. 
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I got this one right. I used Excel (of all things) to explore all arrangements with one and two hidden layers, and also looked at portions of arrangements with more hidden layers. The spreadsheet does the arithmetic with simple copied formulas and gives the correct answer without too much work.

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Oh, I see. My stupid blunder.
Thnx! :) 
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Now I feel like an idiot! :(
You could also have used goal seek in Excel. 
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In case anyone is curious, there are 9227465 possible networks that satisfy the constraints of this problem.

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Professor, after the deadline, could you please post an explanation to this and Question 9, if there's a general case analytical solution, or if there's any other straightforward method?

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