Watching it in 2024. In today's world we have brilliant tools like desmos to have better geometrical intuition of what he says in this series of tutorial. NPTEL lectures are generally very dry and Prof Khapra was not able to explain it clearly. Somehow I felt that the dots were not connected.
Take the expression on the right hand side, divide it by √k, and find the limit as k→∞. The division is how you would find the proportionality constant between two expressions; the limit will calculate the proportionality as k goes to infinity. It approaches −𝛿 as k grows. Another approach to show that "if k is unbounded then cos(β) is also unbounded" is to calculate the limit as k→∞ of just the right-hand side of the inequality in the slide (completely ignoring the proportionality to √k). Use L'Hôpital's rule and simplify to find the limit.
There is not assumption about w, It can be positive or negative. The only assumption about x is that its normalized and that it belongs to the "Positive class". If any x belongs to "Negetive class", the negative of that x is taken, which will inturn be in the positive class
In the entire proof, we haven't used the statement that the data is linearly separable. then even if the data is linearly inseparable, then wont this proof work? but it shouldnt. then how is the fact that data is linearly separable effecting this proof?
But if it is not linearly separable would there be w star? Please understand that w star in this case exists and is just unknown. But in your case, it does not exist. So how will you say anything about the Cos Beta between wt and w*?
This proof is done assuming w* exists, that is there could be a line completely separating these two sets of points. So the moment w* doesn't exist the proof will be invalid.
I think you are reffering to 1:44 A. wixi > Wo has summation from i=1 to n B. wixi > 0 has summation from i=0 to n If you look closely, B ----> w0x0 + w1x1 + w2x2... wnxn > 0 and A ----> w1x1 + w2x2... wnxn > W0 which is same as w0x0 + w1x1 + w2xn...wnxn > 0 with w0=-W0 and x0=1 and which is same as B. Here w0 is also called bias and W0 is called threshold. Hence proving Summation ( wixi > w0) with i=1 to n is same as proving Summation ( wixi > 0) with i=0 to n
If the delta represents minimum increase based on the set of points (p) then k could actually be more than 't' right? For example, imagine there is a value of p which is large and violates the angle requirement. Then, the increase in each iteration will be the minimum value of p which is represented by delta. If multiple times this iteration needs to run in order to overcome the violation for a specific point then overall k can exceed 't'. Am I missing something here?
I think you are confusing between 't' (the total number of time steps) and the total number of data points. In general, in one go, we iterate over all the data points, but there is no guarantee that it will converge if we do only that much. So, we have to repeat the process again. i.e, iterate over the data points again. Maybe after a few cycles of iterations, it will converge. The theorem only gurantess that it will converge after a finite set of cycles
You answer your own question and yet do not realize. If multiple iterations are needed for a particular point, then those multiple iterations are added to the 't'. Therefore, even in such a scenario, k
I Already said this in your BS degree lecture So I'm not going to repeat myself but No, the whole Proof is wrong because the whole statement is wrong. You can't prove that perceptron always reaches convergence when we clearly know it doesn't. it depends on the starting point... but to be honest this course is much more, far more better than that one😂 other than bad intro, bad intro music and bad outro🤣
Best explanation I have seen on Perceptron!
"Please, we are done!" - That's what she said.
That was an Awesome Explanation 👌👌👌
Watching it in 2024. In today's world we have brilliant tools like desmos to have better geometrical intuition of what he says in this series of tutorial. NPTEL lectures are generally very dry and Prof Khapra was not able to explain it clearly.
Somehow I felt that the dots were not connected.
Very clear explanation on proof of convergence.
Brilliant teacher
@ 14:14 + too good
😀
why is p(i) square = 1 when proving on the denominator ?
P is a unit norm
Kya hi horha hai
why || pi ||^2 = 1 at 12:01
@ravi gurnatham Why is 2*Wi*Pi taken negative?
because all the inputs are normalized before training the perceptron...refer setup @ 5:43
@@gauravlotey4263 Because pi was misclassified.
P is a unit norm
At 13:20, even though roughly speaking, we could say cos (beta) grows proportional to sqrt(k), how would one prove that mathematically?
Take the expression on the right hand side, divide it by √k, and find the limit as k→∞. The division is how you would find the proportionality constant between two expressions; the limit will calculate the proportionality as k goes to infinity. It approaches −𝛿 as k grows.
Another approach to show that "if k is unbounded then cos(β) is also unbounded" is to calculate the limit as k→∞ of just the right-hand side of the inequality in the slide (completely ignoring the proportionality to √k). Use L'Hôpital's rule and simplify to find the limit.
Shouldn't it be wi * xi >minus( w0)? because w0(bias) = minus(threshold) ?
Am I missing something here ?
Why is w and x considered positive throughout?
There is not assumption about w, It can be positive or negative. The only assumption about x is that its normalized and that it belongs to the "Positive class". If any x belongs to "Negetive class", the negative of that x is taken, which will inturn be in the positive class
In the entire proof, we haven't used the statement that the data is linearly separable. then even if the data is linearly inseparable, then wont this proof work?
but it shouldnt. then how is the fact that data is linearly separable effecting this proof?
But if it is not linearly separable would there be w star? Please understand that w star in this case exists and is just unknown. But in your case, it does not exist. So how will you say anything about the Cos Beta between wt and w*?
This proof is done assuming w* exists, that is there could be a line completely separating these two sets of points. So the moment w* doesn't exist the proof will be invalid.
This is suprisingly well explained, thank you.
For anyone watching this - don't be put off by his accent. The content is worth to be watched :)
Welcome to IIT
What's wrong with his accent?
@@ankitgupta8797 he might be foreigner bitch
why suprisingly?
no one was put off by his accent in the first place!
Sir why you take transpose of W
For matrix multiplication, dimensions have to be adjusted first.
Why do we Normalize the Inputs??
For sigmoidal activation functions the gradient descent can operate faster as the gradients are the large when the magnitudes are lower
@@SahanGamage99 I appreciate your replying but its already two years since I posted it and figured it out later. Thanks BTW
@@Abhisheism well others might have same doubt so its helpful for them.
*BTW i am others*
Why does summation ( wixi > w0) needs to be proved ? Shouldn't we prove that sumamtion(wixi > 0) instead of (wixi > w0) ?
Please somebody help
P is a unit norm
I think you are reffering to 1:44
A. wixi > Wo has summation from i=1 to n
B. wixi > 0 has summation from i=0 to n
If you look closely,
B ----> w0x0 + w1x1 + w2x2... wnxn > 0
and
A ----> w1x1 + w2x2... wnxn > W0
which is same as
w0x0 + w1x1 + w2xn...wnxn > 0 with w0=-W0 and x0=1
and which is same as B.
Here w0 is also called bias and W0 is called threshold.
Hence proving Summation ( wixi > w0) with i=1 to n
is same as proving Summation ( wixi > 0) with i=0 to n
p' means all possible positive points?
Technically it is the union of all the positive points and the NOT (negative) points.
Nicela
How is ||pi||² = 1?
It is the unit norm of pi
If the delta represents minimum increase based on the set of points (p) then k could actually be more than 't' right? For example, imagine there is a value of p which is large and violates the angle requirement. Then, the increase in each iteration will be the minimum value of p which is represented by delta. If multiple times this iteration needs to run in order to overcome the violation for a specific point then overall k can exceed 't'. Am I missing something here?
I think you are confusing between 't' (the total number of time steps) and the total number of data points. In general, in one go, we iterate over all the data points, but there is no guarantee that it will converge if we do only that much. So, we have to repeat the process again. i.e, iterate over the data points again. Maybe after a few cycles of iterations, it will converge. The theorem only gurantess that it will converge after a finite set of cycles
You answer your own question and yet do not realize. If multiple iterations are needed for a particular point, then those multiple iterations are added to the 't'. Therefore, even in such a scenario, k
Where can I get slides for these lectures?
Google CS7015 and download handouts.
Sir..it is somewhat confusing, whether the entities of P and N are in order (ascending/descending) or random.
Random.
Why would order matter if we randomly select points
I Already said this in your BS degree lecture So I'm not going to repeat myself but No, the whole Proof is wrong because the whole statement is wrong. You can't prove that perceptron always reaches convergence when we clearly know it doesn't. it depends on the starting point... but to be honest this course is much more, far more better than that one😂 other than bad intro, bad intro music and bad outro🤣