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Timeline for A challenging question of ANN

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Feb 7, 2021 at 4:10 history edited user307505 CC BY-SA 4.0
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Jan 10, 2021 at 11:27 comment added Javier TG In the image added with the final update of the question, the data can be separated in different regions by using 3 lines. But note that this is not the case of the first image (the one associated with ($D$) solution). In this first image the data can be separated in different regions just by using 2 lines, and not 3 $\to$ In order to solve the first classification problem, we need at least 5 neurons if we are using bipolar/ step activation functions.
Jan 8, 2021 at 12:18 history edited user307505
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Jan 8, 2021 at 3:00 history tweeted twitter.com/StackStats/status/1347377636641792005
Jan 7, 2021 at 15:31 answer added Aksakal timeline score: 1
Jan 7, 2021 at 15:15 comment added Aksakal @kevin307505 I think D is correct. on first neuron layer all you need is to split into two linear areas, like a cross sign. no point in having more than two neurons.
Jan 7, 2021 at 15:13 comment added Aksakal @DaviedZuhraph i mean the first layer of neurons as 1st layer, not the inputs themselves.
Jan 7, 2021 at 14:54 comment added Aksakal the bipolar node allows you to split the space in three linear areas. so ideally each node in first layer can separate out the L pieces, then the second layer combines the two L areas. when the node is 0/1 then 1st layer splits up into halves that contain both L and not L pieces, and the second layer further splits halves into L and not L, finally the third layer picks up two L areas into one category. i would also edit your question title to make it more descriptive of the actual problem. "difficult" is subjective, I dont find this problem difficult
Jan 7, 2021 at 14:36 history edited user307505
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Jan 7, 2021 at 11:09 answer added Igor F. timeline score: 6
Jan 7, 2021 at 10:55 history edited user307505
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Jan 7, 2021 at 10:38 history edited user307505 CC BY-SA 4.0
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Jan 7, 2021 at 10:15 history edited user307505 CC BY-SA 4.0
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Jan 7, 2021 at 9:57 history edited user307505 CC BY-SA 4.0
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Jan 7, 2021 at 9:46 comment added Igor F. @kevin307505 Then I believe E is wrong. It could be right if the output neuron could multiply its inputs. Then you would have $1 \cdot 1 = 1$ and $-1 \cdot -1 = 1$ for one class, and $1 \cdot -1 = -1$ and $-1 \cdot 1 = -1$ for the other. But for additive neurons I don't see a solution.
Jan 7, 2021 at 9:26 comment added Thomas I would be interested to know where is the catch :D
Jan 7, 2021 at 9:20 comment added Thomas Maybe one should focus on why E does not work for the binary step function. If we call n_1,n_2 the hidden neurons, what the first layer is doing is dividing the space into 4, and assigning a different tuple (0,1),(1,0), (1,1), (0,0) to (n_1,n_2) in each of the 4 regions. The last output layer should just map these tuples to the desired output. It looks that this is impossible, but I do not see at the moment any difference between the bipolar neuron and the standard one since in (n1,n_2) space the points (0,1),(1,0), (1,1), (0,0) and (-1,1),(1,-1), (1,1), (-1,-1) are arranged similarly.
Jan 7, 2021 at 7:54 comment added Igor F. Are there any other assumptions we need to make about the neurons? By "the input" you simply mean the sum of all inputs? And each individual input is simply the output of the previous neuron multiplied by the weight? Do the neurons have a bias term?
Jan 7, 2021 at 7:51 history edited Igor F.
Removed python tag. The question is not software-specific.
Jan 7, 2021 at 7:39 history edited user307505
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Jan 7, 2021 at 6:36 history edited user307505
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S Jan 6, 2021 at 22:49 history suggested Davied Zuhraph CC BY-SA 4.0
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Jan 6, 2021 at 22:48 review Suggested edits
S Jan 6, 2021 at 22:49
Jan 6, 2021 at 22:39 history edited user307505 CC BY-SA 4.0
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Jan 6, 2021 at 22:28 comment added gunes What does Bipolar mean here?
Jan 6, 2021 at 22:27 review First posts
Jan 7, 2021 at 2:31
S Jan 6, 2021 at 22:25 history suggested Betty Andersson CC BY-SA 4.0
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Jan 6, 2021 at 22:22 history asked user307505 CC BY-SA 4.0