x'D Yep

no one cares

I care.

@@somedudeskilivinghislife3739 oof

@@sanchitverma2892 but no lie, Numero7 Mojeagering is a nerd.

thank you for taking your time

Thank you!

Thank you!

Nope, doesn't work, but don't know why

Will discuss more next stream!

The update equations for the weights will change if your function is a parabola. Taking the derivative with respect to m will now give you x^2, so maybe changing the update to error*(input^2) will work.

well a line is a line and not a graph

no the perceptron just found a function that satisfied the condition.... with more points closer to the line it would be a lot more accurate

Yeah thank you but I've understand that: as the second line is getting close to the "primary" line, the Perceptron is getting more accurate. Right. But... At the end, the spaces between those 2 lines - as it seems at the end of this video - could be a set of data that represents the error marging between Perceptron and the dataset, isn't it? I'm asking that because in Machine Mearning, there's also concepts about accuracy, confidence and error rate, to fine-tune algorithms...

Thanks for this important clarification!

;) u did gud

gufi7000 Hey David, Grüße aus Simbach 😀

S4N0I1 Moin 🙃

Yup!

Awesome

in the perceptron drawing you can see the inputs gets multiplied by weights and summed together then you compare that to a threshold"activation function " which makes it an inequality function wx+wy+wb > 0 , when you draw it you can just make it equal does not matter.

Sort of late, but I had trouble with this too. guessY() is supposed to return the y position of the classifier (the line of the perception). And since the range of the activation function is between -1 and 1, the absolute center or divider between whether to label the point green or red is when the activation function(sign in this case) outputs 0. Therefore, the perceptron's decision boundary (the line of the perception) is a line which has the perceptron's prediction set to 0 - the specific number at which the point is neither green nor red (although we label the point as green if the activation function outputs a value >=0). Hope that helps anyone coming here late.

Please suggest here! github.com/CodingTrain/Rainbow-Topics/issues

Why don't you take a look at the Processing reference page: processing.org/reference/

Isnt this typically the same concept as a discriminate function ? Similar to saving the "weights" as a discriminate function

@@DannyGriff97Oh my! I'm great at discrimination, do you think I could be of use here?

Lil_ZcrazyG not that kind of discrimination here ;)

I think it doesn't matter if its equal to zero or some other number. Hope someone can correct that for me. ax + by + c = 0 can also be represented as ax + by + c = d as you suggested. But now if you take d to LHS it becomes ax + by + (c - d) = 0. One could argue that (c-d) in and of itself is another constant. So we could call (c-d) as a k, so equation now becomes ax + by + k = 0. Which is similar to ax + by + c = 0. The value of constant(or the bias which we usually give as a 1) doesn't really matter as it is only there to make sure of that (0,0) thing that he explained in last video. Let's take an example. Lets say that the desired equation is x + y + 1 = 0. Now lets say that for our algorithm we fed inputs as (0,0,2) instead of say (0,0,1) meaning we are changing bias to be 2 instead of 1(coz we are crazy :P). Now the learning starts and we will end up with something like 2x + 2y + 2 = 0 (assuming that learning gives us the exact line implying that there is a lot of data that we don't end up with some other line that ALSO classifies our data). So 2x + 2y + 2 = 0 is same as x + y + 1 = 0; Meaning that the bias can be any number other than zero(why? coz of last video). So the bias value will not effect whether we will get final line or not. Bias effects other weights however. With a 0.5 bias in previous example we could end up with a line 0.5x + 0.5y + 0.5 = 0 or 0.25x + 0.25y + 0.25 = 0 which are all same as x + y + 1 = 0. So what I am trying to say is that the bias can be anything other than 0, so equating ax + by + c = 0 is pretty much the same as ax + by + c = d(any arbitrary d). Hope I am right and hope I helped. :D :P

I also keep thinking about this question. Any can kindly explain this?

The reason is "because math". With the equation "w0(x)+w1(y)+w2(b)=0" we can then make this "w0(x)+w2(b)=-w1(y)" which then becomes "y = -w0(x)/w1-w2(b)/w1"

The function x*w0+y*w1+bias has an output which is either >=0 or = 0, set2 contains all the points leading to an output

"w0(x) + w1(y) + w2(b) = 0" -> This is the equation of line. You can write it in this way -> " ax + by + c = 0" or "y = mx + c". The actual equation for the straight line in this example is "w0(x1) + w1(x2) + w3(b) = 0". Here 'y' is the output of the neural network. 'x1', 'x2' and 'b' are inputs of the neural network. And 'w0', 'w1' and 'w2' are the weights for the inputs. You may be confused in the 'y' notation, because he used 'y' to denote different things in different diagram. In the equation of straight line, he used it for Y-coordinates, and in the perceptron he used it for the output of the perceptron. Hopefully that makes sense.

Thanks for this feedback!

Fulgentius Willy Thanks! I will take that into consideration.

He is using a language called processing which is build on java. I haven't learn this language yet so i can't give you any advice but if you only want to learn processing just for this series then I think it is not necessary . If you know java you should be able to follow this tutorial. learning java would be more practical choice in that case, if you don't know that already. But if you find processing cool then go right ahead and fulfill your curiosity .

Charbel Sarkis a single perceptron can only solve linear seperation, so no. Dan explains this in the next video. Try changing the f(x) from 2*x + 1 to x * x + 1 and you will see that it doesn't find a solution

Great point!

I think i may have meant to comment on the first video. Sorry :(

x_ y_ was so ugly to me that i started looking for comment like this...

Hello! Love hearing from high school viewers! I would recommend 3Blue1Brown's calculus series and also maybe Khan academy videos?

The Coding Train Thank you, I checked out 3blue1brown's essence of calculus series and it's amazing.

Take a look here: github.com/shiffman/The-Nature-of-Code-Examples-p5.js/tree/master/chp10_nn

That helped. Thanks a lot!

fixed on the Point i was checking if x > lineY instead of y > lineY lol

eQuation

cure 44 processing is built on java and the code is basically java

no, java and javascript are 2 totally different languages

im confused this is java?

no, its javascript

nah im pretty sure its java

It works when y = x, but not other numbers

It was just a bug regarding the mapping of the points

You don't know how much I appreciate this, thanks!!!

Here you go: github.com/ShaunakSen/Data-Science-Updated/tree/master/Math%20of%20Intelligence/The%20Coding%20Train/Simple%20Perceptron/CC_SimplePerceptron2

xZen Combo will you go do that on your channel?