This is good up to a point. To calculate the next iteration, it isn't enough just to re-normalise the values. You have to re-calculate the authorities from the previous iteration's hubs; update the hubs from that; then normalise. For k=2, the numbers should look quite different. See en.wikipedia.org/wiki/HITS_algorithm#Pseudocode for pseudocode for doing the same thing.
for each iteration of k, we are simply taking unit vectors so it would naturally come the same in 2 iterations. So how is it useful? In some other cases such as page rank algorithm, i saw that the ranks converge to some values over several iterations, but here its not the case. Could you clear this doubt please?
Hello maam your videos really very helped to us .. thank you for making these videos :) i have little confusion at @4:02 where if in-degree is 4 than outdegree will be 0 but u were mention as 1 . for ( node 4 )
Thanks maam. I think there are some doubtful figures under K = 2. What I mean is that, during calculating authority score under K = 2 , N1 and N2 received different hub socres (respectively 0.445 from N3, 0.623 from N1). However you count both as same figures. This is why I am confused....
Madam,I am not getting what is wigen factor,why it takes and what is its importance.Then why you take damping factor as 0.8, .What are the restrictions on Matrix. Is any special type of matrices or not
mam your topic is HITS Algorithm for BIg data analytics but through out the video you have discuss about how to calculate hub and authority ranking. i dont find anything related to Big data analytics. HITS algorithm for ranking would be better.
How can you normalize already normalized vectors, I guess each time we got to multiply by matrix but you are simply normalizing again so no matter what answer won't change. sorry to say but I think wrong concept
Step count in algorithm implementation. A k-step application of the Hub-Authority algorithm entails applying for k times first the Authority Update Rule and then the Hub Update Rule.
i want to ask some other question regarding the information retrieval techniques kindly advise me from which forum i can join and able to ask the questions? w8ng for reply
is it necessary to have Adjacency matrix to solve this problem? but ma'm the matrix that you have considered do not have diag elements "0" so I am confused here. please clear my doubt
Thank you maam! We still miss you from my diploma days
This is good up to a point. To calculate the next iteration, it isn't enough just to re-normalise the values. You have to re-calculate the authorities from the previous iteration's hubs; update the hubs from that; then normalise. For k=2, the numbers should look quite different. See en.wikipedia.org/wiki/HITS_algorithm#Pseudocode for pseudocode for doing the same thing.
Yeah, I was wondering the same thing. Looks like what she's done here isn't right.
for each iteration of k, we are simply taking unit vectors so it would naturally come the same in 2 iterations. So how is it useful? In some other cases such as page rank algorithm, i saw that the ranks converge to some values over several iterations, but here its not the case. Could you clear this doubt please?
Its probably the wrong way.. for k=2, we get updated hub vectors by multiplying Ateanspose with previous Hub vector
Program implementing the above algorithm:
import numpy as np
NAMING_ARRAY = []
k = int(input())
nodes = int(input())
adjacency = []
strength_adjacency = nodes*nodes
for i in range(strength_adjacency):
if i < nodes:
NAMING_ARRAY.append(f"N{i+1}")
adjacency.append(int(input()))
A = np.array(adjacency)
A.shape = (nodes, nodes)
A_T = np.transpose(A.copy())
u = np.ones((nodes, 1))
v = np.matmul(A_T, u)
u = np.matmul(A, v)
n = 1
while k >= n:
print()
print("ITERATION NO: ", n)
hub = u.copy().flatten()
hub_normalizer = 1/sum(np.square(hub))**0.5
authority = v.copy().flatten()
authority_normalizer = 1/sum(np.square(authority))**0.5
print("Node\tHub Scores\tAuthority Scores")
for i in range(0, nodes):
print(f"{NAMING_ARRAY[i]}\t{round(hub[i], 3)}\t\t{round(authority[i], 3)}")
sorted_hubs = sorted(zip(NAMING_ARRAY, hub), key=lambda x : x[1], reverse=True)
sorted_authorities = sorted(zip(NAMING_ARRAY, authority), key=lambda x : x[1], reverse=True)
print()
print("HUB:", end=" ")
is_a_tie = False
for node, _ in sorted_hubs:
if np.count_nonzero(hub == _) > 1:
is_a_tie = True
print(node, end=" ")
if is_a_tie:
print("{Tie}")
else:
print()
print("AUTHORITY:", end=" ")
is_a_tie = False
for node, _ in sorted_authorities:
if np.count_nonzero(authority == _) > 1:
is_a_tie = True
print(node, end=" ")
if is_a_tie:
print("{Tie}")
v = v*authority_normalizer
u = u*hub_normalizer
n += 1
print()
# 3
You explain every topic very easy and perfect 💯🤝thanks
thanx a lot mam., your teaching was so useful....
Hello maam your videos really very helped to us .. thank you for making these videos :) i have little confusion at @4:02 where if in-degree is 4 than outdegree will be 0 but u were mention as 1 . for ( node 4 )
Hi,
The out degree is 1, as the node points to itself.
Thank you :)
Best of Luck.
Thank you :) ppr was good :) maam plz upload DWM videos :)
Good.
Sure will do.
Thanks maam.
I think there are some doubtful figures under K = 2.
What I mean is that, during calculating authority score under K = 2 , N1 and N2 received different hub socres (respectively 0.445 from N3, 0.623 from N1). However you count both as same figures. This is why I am confused....
What about spider traps
we are dividing all the ranks by the same value so won't they remain the same, no matter how many iterations it takes?
we have to multiply A Transpose with new hub weights to get new authority vector
This is a very helpful video, thank you
Wow, explanation made so simple and easy to understand
Ganesh Vadcar Thank you
Madam,I am not getting what is wigen factor,why it takes and what is its importance.Then why you take damping factor as 0.8, .What are the restrictions on Matrix. Is any special type of matrices or not
Hello mam, if you can upload a video on Latent semantic Indexing it will be great.. Looking forward.
Awesome tutorial thanks ma'am
In hubs and authorities
In 2nd iteration are we suppose to use normalised value or normal values of h & a ?
Normal
mam your topic is HITS Algorithm for BIg data analytics but through out the video you have discuss about how to calculate hub and authority ranking. i dont find anything related to Big data analytics. HITS algorithm for ranking would be better.
If k is not mention in the question then can we assume as k=1?
Wow aapnay bohat acchay say samjhaya hai ayr aik hamaray sir jinko sirf bhaagnay ki jaldi rehti hai 😡😡😡 thanks a lot
mam was the intial hub vector 'u' is same for all the given graphs?
Divya Sampathirao yes
@@AnuradhaBhatia mam tq so much ....this video helped me a lot
waaaoo mam u r great teacher in world u make it very simple for me.
How can you normalize already normalized vectors, I guess each time we got to multiply by matrix but you are simply normalizing again so no matter what answer won't change. sorry to say but I think wrong concept
Thank You, Ma'am!
thank you sooo much. It helps me a lot
Would have been a great video. if question and use of variables were shown in a better way
Mam in exams they given a diagram to find hubs and authority where there are 7 edges so how can we consider 7*7 matrix
yes
So answer will be much more lengthy than this na?
yes.. can you send me the diagram.
what is 'k'
Step count in algorithm implementation. A k-step application of the Hub-Authority algorithm entails applying for k times first the Authority Update Rule and then the Hub Update Rule.
Nice Explanation mam..Please can u upload video on k-mediods with example
Rashi Jain ok..will put
i want to ask some other question regarding the information retrieval techniques kindly advise me from which forum i can join and able to ask the questions? w8ng for reply
Mehtab Ahmed Hello,
What queries you have regarding Information Retrieval, kindly let me know.
Regards
Hello Madam
I have many quetion but this time i want to ask
Is page ranking query dependent?
What is Faign Algorithm
is it necessary to have Adjacency matrix to solve this problem? but ma'm the matrix that you have considered do not have diag elements "0" so I am confused here. please clear my doubt
Ganesh Vadcar kindly mail ur doubt..go on to the website and post a query please
yes, I have done that, thank you
Extremely helpful, thank you!
Superb mam …thanks alot
the video is wrong after k=2
what is k in adjacency matrix?
very nice explanation mam , thank you
Hi Can you please make a video with an example for "Dynamic time warping" for Time series . I just could find any good resources on RUclips. Thanks.
what is the hadoop architecture?,is it the same HDFS ?,plz reply fast.thanks in advance
Rahul Bharsadiya yes...it is Hadoop Distributed File System..
Rahul Bharsadiya Refer to my notes on my site for Big Data
Rahul Bharsadiya all the best
Thanks for your videos, support and for sure the notes ,thanks for good wishes
You are always welcome. You are from which college? Any help please feel free to contact.
Mam can u teach friends of friends algorithm
Thank You Mam. Best explanation.
Please share the ppt madam
youre a great teacher :)
Korkor Kotey-Afutu Thanks
No fake Accent. Crisp to the point explanation. Thanks a lot!!
She still has to learn a lot from you to be where you are right now.
Really helpful
Mansingrao More Thanks
yes it was because of you my papar was damn easy maam greatwork..
thank u maam....ur videos of BDA n CSM helped us a lot in our exams
Thank you so very much.
Thank you so much Ma'am much appreciated !
Thank you
Mam can u explain fuzzy system
gayu ravi Sure..by when do you need it
Thank u mam😍
Thank u a ton....maam
Kunal Borge
Thank you..
thx alot madam
6 star video
BDA k liye kon aaya hai ??😂
Thanks a lot!!
You are the savior. :)