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Congrats!

can you make video on hardware implementation from scratch ?

why "rejection" and "CBD" used ? can we use only one of them to generate A and s,e ? (Also, since rejection's output that is used for A is assumed to be NTT domain. why is so ? )

@1:23:30 how to give the seed in keccak ? will it be randomly generated or it will secret ?

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0:48 0:49

Thats very interesting! Can you maybe upload your FPGA Code to realize the not algorithm?

سلام مجتبی. خوشحال شدم یک ایرانی تبار در رابطه با موضوع رمزنگاری تولید محتوا کرده. اگر امکان مکالمه بیشتری بین ما وجود داره خوشحال میشم مکالمه ای داشته باشیم یکی از داشنجوی های دانشگاه فردوسی مشهد هستم. من حسین هستم

sir can i get the pdf of this presentation , as im a second year undergrade student so this pdf will help me alot . so it will be great if you can provide me the pdf and one thing more how can i use or implement kyber . im having some difficulty because im trying it in windows and the cmd are for linux and mac os . Your help would be great. thank you .

There is an error at slide 11. The system of equations yields result x≈0.46109, y≈-0.096502.

Sir very good presentation can u share code for kyber for this

The slide at 29:43, I was figuring out the storage of M-LWE, and I think it should be O(kn) instead. Considering that the whole matrix is divided into k^2 matrix, and for each divided matrix we only need to store the first column. The column length is divided by k everytime it is divided, so each divided matrix instead of needing to store n column vector, they only need to store n/k column vector. So the storage big O should be O((n/k)*k^2), which should be O(kn). I am not sure how it will affect the computation big-O.

@Mojtaba In slide 28 (52:00) you write m*((q+1)>>1). This will fix 0 to zro and map 1 to 7. In slide 29 also, you write if (f2[0]>(q/4)) & (f2[0]<(3*q/4)): f2[0]=1; For q=13, q/4 is 3 and 3q/4 is 9. These are boundary values for making decision. This also indicates that message is not encoded with 3 and 9 but 0 and 7. When f2[0]>3 and f2[0]<9, it contains 7 and hence message is 7. Otherwise f2[0]<=3 and f2[0]>=-4, the interval contains zero, it means the message will be zero. Please check it.

A nice presentation, would you please share the PDF file?

Thank you for this video.

why in 22:00 say as=t is easy to find s, and 22:10, say its hard to find e in ae=t? whats relationship with ae=t and as+e=t?

@Mojtaba Great presentation! Why in 47:29 in the t's formula you didn't consider to compute A*s + e ? You only compute t = A*s in your example. Thanks

Very Nice presentation Naisar ji...can we get your Presentation if possible for educational usages

Hello!! really nice explanation!! I have one question!! In @5:21, presentation shows "Bob computes u as follows". I think that "Alice" computes ~ is correct!! Therefore, could you explain what is correct sentence?? Thank you for your outstanding presentation!!:)

51:28 The student posed an excellent question. Nonetheless, the answer appears to be incorrect! We do not raise the size of the k to accommodate a larger message! In fact, the size of the message is 32Bytes without having a 32Bytes k in Crystal Kyber!

I'm wondering if you are going to give a presentation about Attacks on Kyber

thank you so much it's very beneficial

Actually, actually, actually, 😵‍💫

very good job .I work in this subject in my memory .I want this document please

Can you provide me the Kyber 512 Python code

Could you to send slide

I need code and presentation plz

Please, could you share the code?

I really enjoyed the presentation and I would like to replicate the examples. Where did the modular_mult() and modular_add functions come from?

Very nice introduction, thank you very much!

This lecture should be the first stop for anyone new to the topic. Nicely done.

First time listening to Lattice cryptography. Very informative and thanks for explaining with simple examples.

The only lecture which actually takes a simple example to help you understand the concept of lattices. A bundle of thanks Sir. 👍

Is there a link to the paper itself?

Interesting talk...

good job could you share the slide please?

very informative sir...

Could you share the slide?