Computer Science Theory Explained
Computer Science Theory Explained
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Binary vs. Unary Number Encodings and Strong NP-completeness
Textbooks:
Computational Complexity: A Modern Approach by S. Arora and B. Barak.
Algorithm Design by J. Kleinberg and E. Tardos.
Lecture slides by K. Wayne accompanying the latter textbook:
www.cs.princeton.edu/~wayne/kleinberg-tardos/
Переглядів: 1 464

Відео

The Lemke-Howson Algorithm - Complementary Pivoting
Переглядів 3,8 тис.3 роки тому
The Lemke-Howson Algorithm - Complementary Pivoting
The Lemke-Howson Algorithm - Best Response Polytopes
Переглядів 3,4 тис.3 роки тому
The Lemke-Howson Algorithm - Best Response Polytopes
The Lemke-Howson Algorithm - Best Response Diagrams
Переглядів 4,9 тис.3 роки тому
The Lemke-Howson Algorithm - Best Response Diagrams
Colorability of Planar Graphs
Переглядів 1,7 тис.3 роки тому
Textbooks: Computational Complexity: A Modern Approach by S. Arora and B. Barak. Algorithm Design by J. Kleinberg and E. Tardos. Lecture slides by K. Wayne accompanying the latter textbook: www.cs.princeton.edu/~wayne/kleinberg-tardos/
The Complexity Class PPAD
Переглядів 1,5 тис.3 роки тому
The Complexity Class PPAD
Scarf's Theorem
Переглядів 6643 роки тому
Scarf's Theorem
Computing a Nash Equilibrium
Переглядів 1,3 тис.3 роки тому
Computing a Nash Equilibrium
Nash's Theorem
Переглядів 2,7 тис.3 роки тому
John Nash's PhD Thesis: library.princeton.edu/special-collections/sites/default/files/Non-Cooperative_Games_Nash.pdf
Brouwer's Fixed Point Theorem
Переглядів 5 тис.3 роки тому
Brouwer's Fixed Point Theorem
Sperner's Lemma
Переглядів 5 тис.3 роки тому
Sperner's Lemma
Two-Player Zero-Sum - a Second Example
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Two-Player Zero-Sum - a Second Example
Solving Rock-Paper-Scissors
Переглядів 6 тис.3 роки тому
Solving Rock-Paper-Scissors
Existence and Computation of Nash Equilibria in Two-Player Zero-Sum Games
Переглядів 2 тис.3 роки тому
Existence and Computation of Nash Equilibria in Two-Player Zero-Sum Games
A Brief Linear Programming Refresher
Переглядів 1,2 тис.3 роки тому
A Brief Linear Programming Refresher
Two-Player Zero-Sum Games
Переглядів 3,7 тис.3 роки тому
Two-Player Zero-Sum Games
The Poisened Drink and the Mixed Nash Equilibrium
Переглядів 1,2 тис.3 роки тому
The Poisened Drink and the Mixed Nash Equilibrium
The Battle of the Sexes and Burning Money
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The Battle of the Sexes and Burning Money
Pure Nash Equilibrium - a Further Example
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Pure Nash Equilibrium - a Further Example
The Battle of the Sexes
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The Battle of the Sexes
Weak Dominance
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Weak Dominance
The Iterated Elimination of Dominated Strategies
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The Iterated Elimination of Dominated Strategies
Dominating Strategies in the Prisoner's Dilemma
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Dominating Strategies in the Prisoner's Dilemma
The Prisoner's Dilemma
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The Prisoner's Dilemma
Games in Strategic Form
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Games in Strategic Form
The "Tragedy" in the Tragedy of the Commons
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The "Tragedy" in the Tragedy of the Commons
The Tragedy of the Commons
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The Tragedy of the Commons
Algorithmic Game Theory - Introduction
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Algorithmic Game Theory - Introduction
An FPTAS for the Knapsack Problem
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An FPTAS for the Knapsack Problem
Another Dynamic Program for the Knapsack Problem
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Another Dynamic Program for the Knapsack Problem

КОМЕНТАРІ

  • @queenpost
    @queenpost 2 дні тому

    Now, if |x| = n, how can Alice send x to Bob using log(n) bits??

    • @queenpost
      @queenpost День тому

      O.K., I confused sth.

  • @arhamkhxn
    @arhamkhxn 18 днів тому

    what if we have 3 literals in each of the 3 clause, do we still assign a single buffer too?

  • @Liad138
    @Liad138 20 днів тому

    Where is the algorithm soloution?

  • @seeblu
    @seeblu 22 дні тому

    Thank you very much.

  • @akshatgupta7471
    @akshatgupta7471 Місяць тому

    10:51 Can you please tell why did you take a triangle as the mixed strategy simplex of the first player? Like can't we take a 1x1x1 cube where the strategy probabilities at any point within it would be its coordinates? The triangle approach doesn't seem so obvious

    • @akhilkammila6910
      @akhilkammila6910 23 дні тому

      Sum of probabilities must be 1. The triangle is effectively the region in the 1x1x1 cube where the sum of points is 1.(try graphing x1 + x2 + x3 = 1)

  • @acliooeye2881
    @acliooeye2881 Місяць тому

    Should it not be strict subset at 0:40?

  • @蒲昊-s6z
    @蒲昊-s6z Місяць тому

    After normalization of V, why didn't the equation y4+y5=1/V exit?

  • @mahirrahman3363
    @mahirrahman3363 2 місяці тому

    Why does evaluating the formula require "some space" c. Why can't we say c = n?

  • @MegaSávioMiranda
    @MegaSávioMiranda 2 місяці тому

    that means that points (2, 1) and (4, 5) are also Nash Equilibrium because 3 is played with probability 0, right?

  • @A-_AkramahFaizi
    @A-_AkramahFaizi 2 місяці тому

    Thanks for these great lectures. You videos are very easy to understand and follow. Unfortunately in this video I got lost as to why we need n/3 bits of zeros in middle. We could just work with 1 bit of zero. And also, O(k) bits includes all the k times that the tape head leaves the cell in the entire running of TM for palindromes. Then why are we multiplying it with n/3 in the end?

    • @queenpost
      @queenpost День тому

      Well, we need n/3 zeros in the middle to do the multiplication at the end, and this only shows, that we need OMEGA(n^2) moves.

  • @jibbmng
    @jibbmng 2 місяці тому

    I definitely prefer proof 1, it's a nice elegant counting argument, love a good bit of combinatorics. While also valid, the construction in proof 2 feels more contrived, the altering of the original graph obscures the argument a little.

  • @A-_AkramahFaizi
    @A-_AkramahFaizi 2 місяці тому

    Thank You for the Great videos. I have a doubt in this video. You said that after the construction of φ, we can put value of x and leave out t, and if the SAT gives a satisfying assignment, that means a t exist which satisfies φ. Connecting it to one of your starting videos where you mentioned that when constructing TM for taking sum of two numbers, we can represent each bit of numbers with duplicates and the plus sign with 10. So, what if, the satisfying assignment that t gives doesn't correspond to any useful info of the certificate? like in case of sum of two nos. example, the t turns out to be 100110101.... or something, which the TM won't recognize cuz it was expecting duplicates and a 10 in between for sum. I hope I am clear enough in my question. It would be great if you could clear my doubt soon.

  • @mohamedamineaydoun6364
    @mohamedamineaydoun6364 3 місяці тому

    this Problem is truly a fascinating product of the human imagination, i'm obsessed with it, and yes your channel is amazing

  • @ZhengyuanZhou-l4q
    @ZhengyuanZhou-l4q 3 місяці тому

    "There's a subsequence of such triangles such that the three corners all converge to the same point z" may not be true and you don't need it, right? You only need there's a subsequence of red points converging to z, a subsequence of blue points converging to z and a subsequence of green points converging to z. But these three subsequences may come from different triangles.

    • @沈骁瑜
      @沈骁瑜 2 місяці тому

      Without the triangle things, how would you argue there’s a subsequence of red, green, blue points that converge to the same point z?

  • @brendawilliams8062
    @brendawilliams8062 3 місяці тому

    If you didn’t consider 324/3244 then it’s all in a mess

  • @brendawilliams8062
    @brendawilliams8062 3 місяці тому

    It’s hard after simple

  • @brendawilliams8062
    @brendawilliams8062 3 місяці тому

    Thankyou.

  • @brendawilliams8062
    @brendawilliams8062 3 місяці тому

    Thankyou. I’ve been desiring information on this. Greatly appreciated

  • @JikeWimblik
    @JikeWimblik 4 місяці тому

    From partial reductions of the napsack problem to combinations of 2 items to sniff out more item variants of the problem with blocks in some parts of the bag as to find clues for your particular napsack problem

  • @taha7shaikh
    @taha7shaikh 4 місяці тому

    Thank you !!!!!

  • @shantanukaushikmathoholic
    @shantanukaushikmathoholic 4 місяці тому

    Lf is basically how to define a set, for which the output is "yes".

  • @shantanukaushikmathoholic
    @shantanukaushikmathoholic 4 місяці тому

    Sire, any book recommendation to along this playlist ?

  • @jasoncotton9804
    @jasoncotton9804 5 місяців тому

    Wouldn't a set of non-result altering transformations which result in True be a verifiable certificate for the Tautology problem? Thanks for the well articulated video.

  • @cikechukwujohn
    @cikechukwujohn 5 місяців тому

    Who is here in August, 2024. I am.

  • @akshatgupta7471
    @akshatgupta7471 5 місяців тому

    can you pls proved the pdfs of your notes? they would be really helpful for revising the concepts

  • @akshatgupta7471
    @akshatgupta7471 5 місяців тому

    thank you very much... this was really helpful :) would be great if you could provide the pdfs of your handwritten notes

  • @adrianho7165
    @adrianho7165 5 місяців тому

    The first proof seems to be easier to come up with

  • @ProfJigsaw
    @ProfJigsaw 5 місяців тому

    Hello @compscilessons Great work you are doing on this channel. I have a favour to ask of you. I have a paper on computational complexity, more specifically on SAT solving and would like a peer review. Unfortunately, I am unaffiliated with any institutions. Can you give me an endorsement on the preprint server arXiv so that I can make a submission or provide an email address where I can share my paper with you? Any assistance on this is appreciated. Thanks.

  • @dan7nm
    @dan7nm 6 місяців тому

    Really helpful thanks!

  • @MohdYaseenmas
    @MohdYaseenmas 6 місяців тому

    Thank you

  • @michaelperry886
    @michaelperry886 6 місяців тому

    Why is the g in the definition of a reduction taking in both of x and y?

  • @queenpost
    @queenpost 7 місяців тому

    Very clear explanation - Thanks! One Question: What would we do, if we insist, that the numbers given as instance for the SUBST SUM problem are different to each other?

    • @queenpost
      @queenpost 7 місяців тому

      O.K., the solution is given at the end!

  • @nicolasgioanni606
    @nicolasgioanni606 7 місяців тому

    No other video explains the pricing method on weighted vertex covers as well as yours, thank you so much, great video

  • @FabiolaMariaVillatoroGarcia
    @FabiolaMariaVillatoroGarcia 8 місяців тому

    You are genuinely awesome.

  • @halchen1439
    @halchen1439 8 місяців тому

    you are single handedly saving my ass. may your soul be blessed

  • @olxxa4967
    @olxxa4967 8 місяців тому

    Anyway, thank you. This video helped me a lot 🎉

  • @olxxa4967
    @olxxa4967 8 місяців тому

    Can you please do the colonel Blotto game

  • @nothingspecial2018
    @nothingspecial2018 8 місяців тому

    you did good job but my stoopid self could not understand it.

  • @ElIrracional
    @ElIrracional 9 місяців тому

    Why is that true? What about a clause of the form: x and ~y? It is not satistied when x and y are true and not satisfied when x and y are false.

  • @liamtolkkinen5025
    @liamtolkkinen5025 9 місяців тому

    you really don't explain things well

  • @welfarewagonrepairs
    @welfarewagonrepairs 9 місяців тому

    This was a very helpful explanation. The part after 13:00 was especially helpful explaining the purpose of the additional rows

  • @laautaro11
    @laautaro11 9 місяців тому

    Your channel is the best theoretical CS channel I've seen, fantastic job! thank you!

  • @lucasarif4387
    @lucasarif4387 9 місяців тому

    when you wrote down that proof idea at 2:30 my mouth started watering lets see where this goes

  • @lucasarif4387
    @lucasarif4387 9 місяців тому

    this module is so so good

  • @gate_dose
    @gate_dose 9 місяців тому

    Thank You!!

  • @ShakrinJahanMozumder
    @ShakrinJahanMozumder 9 місяців тому

    Thanks!

  • @AamirMohamedThahirAli-x5z
    @AamirMohamedThahirAli-x5z 10 місяців тому

    thank you so so so so much!!! you explain everything so clearly!!!!

  • @manawa3832
    @manawa3832 10 місяців тому

    i am going to now define the concept of mebeingfkingright. this foundational principle, now called mebeingfkingright, will be proven in the following way. i propose a machine, called the RightMachine that lays out a sequence of adjacent symbols. in each symbol is encoded the final position of the symbol to the left of itself or non at all. the machine beingfkingright at r0 and ends at r-finished. any sequence that this machine can lay down is said to be mebeingfkingright. therefore mebeingfkingright is any seqquence of symbols of mebeingfkingright that the RightMachine can process i.e. can go from r0-r-finish. tada the same level of proof as computability in this turing complete nonsense. the entire field of computer science folks.. like seriously without information theory and linguistics, computer science would just be a joke based on the inane crackpot nonsense of two guys during ww2.

  • @alieser7770
    @alieser7770 10 місяців тому

    Sir, you are doing God's work

  • @abdurrahmanyakar3040
    @abdurrahmanyakar3040 11 місяців тому

    Teacher, time 1:01, you write a f:{0,1}*: What does asterisk (*) mean?

    • @compscilessons
      @compscilessons 11 місяців тому

      The * is referred to as the Kleene star. {0,1}* is the set of all bitstrings of arbitrary (but finite) length. This includes the empty string, i.e., the string that has length 0.