Martin Hell
Martin Hell
  • Видео 95
  • Просмотров 141 475
Communicating Cybersecurity Vulnerability Information: A Producer-Acquirer Case Study
Presentation of paper on Profes 2021.
Authors: Martin Hell (Lund University) and Martin Höst (Lund University)
Title: Communicating Cybersecurity Vulnerability Information: A Producer-Acquirer Case Study
Presentation by Martin Hell
Просмотров: 113

Видео

Using Program Analysis to Identify the Use of Vulnerable Functions - SECRYPT 2021
Просмотров 1443 года назад
Conference presentation by Rasmus Hagberg.
The SMARTY Research Project
Просмотров 1483 года назад
A short overview of a research project conducted at Lund University. The project is entitled SMARTY: Secure Software Update Deployment for the Smart City. It is a collaboration between the Electrical and Information Technology department and the Computer Science department. It is supported by the Swedish Foundation for Strategic Research.
Bitcoin technology explained in a simple way
Просмотров 1864 года назад
An introductory overview of the technology that enables Bitcoin. The underlying technical problems and challenges are given and I show how Bitcoin solves these problems. The description is nontechnical, yet explains how the technical challenges are handled by the transactions and the blockchain.
5.5 Binary, Octal and Hexadecimal Representation of Numbers
Просмотров 9144 года назад
In digital circuits, numbers are treated in their binary form. Here we introduce the binary, octal, and hexadecimal representation of numbers. We also show how one can be derived from the other.
14.6 Observer Canonical Form for LFSRs
Просмотров 1,2 тыс.4 года назад
We define the observer canonical form for an LFSR and show that when clocking an LFSR of this form, it is the same as performing the division P(D)/C(D).
14.7 Period of Connection Polynomial
Просмотров 1,3 тыс.4 года назад
We define the period of the connection polynomial and link this period to the period of the sequence generated by the LFSR.
14.5 Derive P(D) from Starting State
Просмотров 1,3 тыс.4 года назад
With the connection polynomial and the starting state, we show how we can derive the polynomial P(D).
14.3 LFSR Theorem and Euclid's Algorithm for LFSRs
Просмотров 2,1 тыс.4 года назад
The LFSR theorem connects the sequence generated by the LFSR with the connection polynomial. We state the theorem and show how we can use Euclid's algorithm to find the shortest LFSR that generates a given sequence.
14.4 Derive Starting State From P(D)
Просмотров 1,4 тыс.4 года назад
With the connection polynomial and P(D), we show how we can derive the starting state of the LFSR.
14.1 LFSR - Definition and State Transitions
Просмотров 2 тыс.4 года назад
We define the linear feedback shift register and show how it implements transitions between states.
13.4 Rank, Diagnostic Matrix and Reduced Form
Просмотров 1,5 тыс.4 года назад
To understand how we can reduce the linear circuit, we need to define the rank of a matrix and define the diagnostic matrix. This will allow us to define the reduced form of the matrices A, B, C, and H.
14.2 LFSR - Connection Polynomial and Recursion
Просмотров 3,9 тыс.4 года назад
The connection polynomial defines the LFSR. We define this polynomial and give a few examples of LFSRs and their connection polynomial. We also prove the shift register recursion.
13.7 D-transform of Periodic Sequences
Просмотров 1,7 тыс.4 года назад
We introduce periodic sequences and show how such infinite sequences can be written on the D-transform.
13.5 Example of a Linear Sequential Circuit, Part 2 - Reducing the Circuit
Просмотров 1,5 тыс.4 года назад
We continue the example by showing how the reduced form of our matrices can be computed and finally give the reduced circuit.
13.6 The D-transform
Просмотров 1,7 тыс.4 года назад
13.6 The D-transform
13.3 Example of a Linear Sequential Circuit, Part 1 - Defining the Circuit
Просмотров 1,3 тыс.4 года назад
13.3 Example of a Linear Sequential Circuit, Part 1 - Defining the Circuit
13.2 Linear Sequential Circuits
Просмотров 1,2 тыс.4 года назад
13.2 Linear Sequential Circuits
13.1 Linear Boolean Functions
Просмотров 2,5 тыс.4 года назад
13.1 Linear Boolean Functions
12.6 Two CMOS Gates
Просмотров 8434 года назад
12.6 Two CMOS Gates
12.4 The Counter 74LS162A
Просмотров 1,3 тыс.4 года назад
12.4 The Counter 74LS162A
12.5 Introduction to CMOS
Просмотров 1,1 тыс.4 года назад
12.5 Introduction to CMOS
11.8 The D-element - Cascading Two Latches
Просмотров 1,5 тыс.4 года назад
11.8 The D-element - Cascading Two Latches
12.1 Programmable Logic Array (PLA)
Просмотров 1,3 тыс.4 года назад
12.1 Programmable Logic Array (PLA)
12.2 Multiplexer and Demultiplexer
Просмотров 9404 года назад
12.2 Multiplexer and Demultiplexer
12.3 Tri-state Buffers
Просмотров 1,3 тыс.4 года назад
12.3 Tri-state Buffers
11.7 The Latch - A Simple Memory Circuit
Просмотров 1,5 тыс.4 года назад
11.7 The Latch - A Simple Memory Circuit
11.5 Hazard-Free Realisation
Просмотров 1,2 тыс.4 года назад
11.5 Hazard-Free Realisation
11.6 Asynchronous Sequential Circuit
Просмотров 1,1 тыс.4 года назад
11.6 Asynchronous Sequential Circuit
11.3 Asynchronously Realisable Graphs
Просмотров 1,3 тыс.4 года назад
11.3 Asynchronously Realisable Graphs

Комментарии

  • @MoukiBouagal
    @MoukiBouagal 12 дней назад

    Thanks a lot

  • @BitRepairs
    @BitRepairs Месяц назад

    Awesome, really appreciate this, helped a lot, thanks :)

  • @Pepe2708
    @Pepe2708 Год назад

    Thanks a lot for this series, it has been very helpful!

  • @efi3825
    @efi3825 Год назад

    Now I am starting to wonder if all the state transition graphs except for 00...0 have the same cylce length. But I feel that at least, any cycle length would have to divide (2^N - 1) with N as the number of slots in the register.

    • @martinhell4596
      @martinhell4596 Год назад

      If they have the same cycle length, then this length will divide 2^n - 1, but it will depend on the polynomial that defines the LFSR. There are three cases, primitive, irreducible and reducible polynomials. If the polynomial is primitive then there is a maximum length sequence of length 2^n - 1. If it is irreducible (it can not be factored into the product of polynomials of smaller degree), then all sequences are the same length, which of course must divide 2^n - 1. If they are reducible (it can be factored), then the cycle lengths are more complex and will depend on the cycle lengths of the irreducible polynomials that it is factored into. There is old and well established theory on this. The first I know of is from 1958 by Birdsall and Ristenbatt (introduction to linear and shift-register generated sequences). It is actually quite readable and you will find the PDF on Google scholar.

    • @efi3825
      @efi3825 Год назад

      @@martinhell4596 Awesome, thank you so much!

  • @mansleifsson8277
    @mansleifsson8277 Год назад

    CNF var ju inte så svårt ju!

  • @chaos9790
    @chaos9790 Год назад

    where are you martin i miss your videos

  • @TalksWithGift
    @TalksWithGift Год назад

    Great Job!

  • @baronvonbeandip
    @baronvonbeandip Год назад

    Interesting. How did you develop your definition for addition? EDIT: Nvm, it seems you explain it in the video about Boolean Algebra <-> Boolean Rings

  • @vickymronga8998
    @vickymronga8998 Год назад

    Very nice video, clear explanation! You deserve more attention.

  • @sarsax123
    @sarsax123 Год назад

    great video

  • @stefanro95
    @stefanro95 2 года назад

    Thank you very much for this video, really helped my on my final exam! 😁

  • @mathematicalexpert208
    @mathematicalexpert208 2 года назад

    weldon

  • @mathematicalexpert208
    @mathematicalexpert208 2 года назад

    eeldon

  • @mathematicalexpert208
    @mathematicalexpert208 2 года назад

    nice

  • @levetbyck
    @levetbyck 2 года назад

    2:12 “3 times 3 is idempotent because it equals 3”? maybe wish the multiplication was explained (at the “idempotency”-table)

    • @martinhell4596
      @martinhell4596 2 года назад

      It is a ring operation, so it is modulo 6. 3x3 = 3 mod 6.

    • @levetbyck
      @levetbyck 2 года назад

      mhm

  • @kapeusw
    @kapeusw 2 года назад

    I think you made a careless mistake at 1:25 . I think it is not "x1y1" but "x1x2".

    • @martinhell4596
      @martinhell4596 2 года назад

      Yes. Thanks for pointing it out. It should of course be x1x2.

  • @asdhanya3798
    @asdhanya3798 2 года назад

    Hi Martin.. your presentation was very nice.. i have one doubt. In a Boolean table a+b equation, is it solve for all the elements in that table? . If you solve only 4+1. In the same way I will try to solve 3+4 but it is not working.can you help me to solve that.

    • @martinhell4596
      @martinhell4596 2 года назад

      Hi. It will work. You have 3 + 4 = 3 (+) 4 (+) 4 (x) (3 (x) 4) mod 6 = 7 (+) 48 mod 6 = 55 mod 6 = 1 mod 6. I hope the notation makes sense.

  • @hakan6379
    @hakan6379 2 года назад

    Thank you sir. You explained better than my teacher :)

  • @amirghorban2044
    @amirghorban2044 3 года назад

    you are awsome

  • @cydrus2856
    @cydrus2856 4 года назад

    Ska inte C vara x3x5 och inte x2x5?

  • @victorschack4537
    @victorschack4537 4 года назад

    This is pretty cool.

  • @CritZzHD
    @CritZzHD 4 года назад

    Just a note; In the last part of the video its written : "rj = sjn1 + tjn3 for all j", but should be ""rj = sjn1 + tjn2 for all j"

  • @michalpomorski4394
    @michalpomorski4394 4 года назад

    2:28 slip of tongue ("when all inputs are zero")

    • @martinhell4596
      @martinhell4596 4 года назад

      Yes, you are right. It should be "when all inputs are one" .