SackVideo
SackVideo
  • Видео 27
  • Просмотров 4 945 186
An impossible game at the heart of math
Strategy stealing, the axiom of determinacy, and why it's incompatible with the axiom of choice. #SoME3
Resources to learn more and other interesting notes:
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Chomp:
Play online: www.math.ucla.edu/~tom/Games/chomp.html
Wikipedia article: en.wikipedia.org/wiki/Chomp
List of known best first moves: sites.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/chompc.html
Also check out chapter 18 in the book "Winning Ways for Your Mathematical Plays"
Zermelo's Theorem: en.wikipedia.org/wiki/Zermelo%27s_theorem_(game_theory)
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The Axiom of Determinacy:
Wikipedia article: en.wikipedia.org/wiki/Axiom_of_determinacy
Borel Determinacy Theorem: en.wikipedia.org/wiki/Borel_de...
Просмотров: 92 220

Видео

ChatGPT can't multiply, but can AI do math?
Просмотров 7 тыс.Год назад
A discussion about AI and math research. Resources to learn more and other interesting notes: SAT solvers: en.wikipedia.org/wiki/SAT_solver Pythagorean Triples paper: arxiv.org/pdf/1605.00723.pdf A nice writeup on the use of SAT solvers on another recent problem: bsubercaseaux.github.io/blog/2023/packingchromatic/ Neural Networks in pure math: arxiv.org/pdf/2104.14516.pdf arxiv.org/pdf/2304.126...
Cover's Paradox is very cool
Просмотров 2,8 тыс.Год назад
For more about Cover's paradox, check out these links: fermatslibrary.com/s/pick-the-largest-number math.stackexchange.com/questions/655972/help-rules-of-a-game-whose-details-i-dont-remember/656426#656426 Corrections:
Why cutting a Möbius strip is so weird
Просмотров 10 тыс.Год назад
An explanation for why cutting a Möbius strip down the middle keeps it connected. Resources to learn more and other interesting notes: en.wikipedia.org/wiki/Möbius_strip The boundary of a Möbius strip is a circle. If you make 3 half-twists, you still get a Möbius strip, but the boundary is knotted into what's called a "trefoil knot." We call a surface that bounds a knot a Seifert surface. More ...
Why care about infinite dimensions?
Просмотров 12 тыс.Год назад
Infinite dimensions and problem solving Resources to learn more: Banach fixed-point theorem: en.wikipedia.org/wiki/Banach_fixed-point_theorem ruclips.net/video/9jL8iHw0ans/видео.html Complete metric space: en.wikipedia.org/wiki/Complete_metric_space ruclips.net/video/kdKYV0B145k/видео.html math.stackexchange.com/questions/97356/is-the-space-c0-1-complete Function spaces: en.wikipedia.org/wiki/F...
What are the odds of always winning?
Просмотров 7 тыс.Год назад
What are the odds of always winning?
What is 2^π?
Просмотров 466 тыс.Год назад
What is 2^π?
The Most Important Sequence: The Catalan Numbers
Просмотров 44 тыс.Год назад
The Most Important Sequence: The Catalan Numbers
If it probably exists, then it does
Просмотров 380 тыс.Год назад
If it probably exists, then it does
Counting with Calculus: The Magic of Generating Functions
Просмотров 39 тыс.3 года назад
Counting with Calculus: The Magic of Generating Functions

Комментарии

  • @MysteriousObjectsOfficial
    @MysteriousObjectsOfficial 6 дней назад

    me when im rating something to 1 from 10 and i choose 4: (the middle finger)

  • @Bird1502
    @Bird1502 8 дней назад

    Thank you! You explained this very well

  • @be1tube
    @be1tube 14 дней назад

    Another reason "real" numbers are anything but.

  • @user-cu9ww9tj4i
    @user-cu9ww9tj4i 15 дней назад

    모든게 무한하다면 무한히 겹치는 것도 무한하지 않을까요?

  • @cf6755
    @cf6755 15 дней назад

    if you have infinite money then there is no purpose in making a profit.

  • @jrgamma
    @jrgamma 18 дней назад

    Awesome youtuber

  • @Alanalan12297
    @Alanalan12297 18 дней назад

    Pick the first on the left!

  • @drewduncan5774
    @drewduncan5774 23 дня назад

    What you're describing here isn't the axiom of choice. That you can pick an arbitrary element from any one set is a principle of first order logic. The axiom of choice is only needed when you need to pick an element from each of an infinite collection of sets.

    • @EdwardCurrent
      @EdwardCurrent 18 дней назад

      Wrong.

    • @ASackVideo
      @ASackVideo 18 дней назад

      By asking for a rule that can pick an element out of any set of reals, we’re asking for a choice function on the power set of the reals! That’s the infinite collection we’re dealing with. A choice function on the reals would let us construct non-measurable sets, but these require choice!

    • @drewduncan5774
      @drewduncan5774 17 дней назад

      @@ASackVideo True, but that's not what you say in the video: "If you want to say, just reach your hand in and pick out an element, this requires what's called the axiom of choice." This is the problem. In this and all of your examples you're taking just one set. "Reach your hand in and pick out an element" is a perfectly good "rule" if you're only working with one set (or any finite number of sets) at a time, no AC needed.

  • @ryanmarck7901
    @ryanmarck7901 24 дня назад

    Mathematically this is correct!

  • @co9681
    @co9681 Месяц назад

    Basically, for any nonempty set, there exists a function that can choose an element from the set to make a new set. I’ve also heard it as a choice function for the power set of a set to make a set from one element of each set in the power set. Must exist*

  • @rektator
    @rektator Месяц назад

    "Choosing" an element from a non-empty set doesn't require the axiom of choice. Having a "uniform" choice or a "simultaneous" choice over infinitely many sets does require. One can prove the finite axiom of choice. If you have n-many non-empty sets, then there exists a uniform choice of elements from the sets. Equivalently, any surjective function between finite sets has a section.

    • @ASackVideo
      @ASackVideo Месяц назад

      This is why I focus on a “rule” for choosing, i.e. a choice function.

  • @BeenYT
    @BeenYT Месяц назад

    if A is good A^c isnt always bad, imagine the rule is A is good if A^c = {n + 1 : n exists in A, 1 ... m -1 : m is the lowest number in A}, then if A is the set of odd numbers, both A and A^c are good sets, unless ive typed the definition of A^c incorrectly this will disprove that (1 ... m means all the numbers 1 up to m, im not good with set theory notation )

    • @ASackVideo
      @ASackVideo Месяц назад

      You’re correct, but we aren’t claiming that for all possible games that A is good if and only if A^c is bad. Instead we’re doing the work to construct a specific game where that would be the case.

  • @Paul-A01
    @Paul-A01 Месяц назад

    Just pick your favorite one, how is this hard?

  • @mubarakvodel5763
    @mubarakvodel5763 Месяц назад

    Good job

  • @armsport9324
    @armsport9324 Месяц назад

    3 persons you , Alice & Bob ,So there have to be atleast 2 off the same kind to her but she doesnt know you ,so she tells not the truth but is kinda guessing therefore a knave bob is the Knight

  • @MintCloudyta
    @MintCloudyta Месяц назад

    Thank you so much. This video helps me so muh 🫶

  • @pedroribeiro1536
    @pedroribeiro1536 Месяц назад

    Absolutely wonderful explanation. Thank you, brother.

  • @Fictionarious
    @Fictionarious Месяц назад

    Rule: Label each continuous sub-interval in the evidently discontinuous set of real numbers you have been asked to consider with a pair of labels I and S, corresponding to the infimum and supremum of that sub-interval. Take the mean of the smallest such infimum label and its associated supremum label, then choose that number. Job done. . . . The axiom of choice is bullshit, by the way.

    • @stefanwrobel8042
      @stefanwrobel8042 Месяц назад

      What if there is an infinite amount of sub-intervals? It seems to me that it may not have the smallest element of the set of infimi, no?

    • @tenebrae711
      @tenebrae711 Месяц назад

      Counterexample: see cantor set

    • @tenebrae711
      @tenebrae711 Месяц назад

      Or the set of irrational numbers (not reals), or, heck, even rational numbers

  • @niloymondal
    @niloymondal Месяц назад

    Axiom of Choice is a scam.

  • @user-cu9ww9tj4i
    @user-cu9ww9tj4i Месяц назад

    간단함.우리의 뇌는 이미 양자역학적으로 돌아가고 인간의 뇌는 양자컴퓨터임.풍부한 감정 옳은 의지 좋은 환경 고난 좋은 사람들 사이에 있으면 필연적 우연으로 성취할 수 있음.스메일 그로텐딕 허준이씨처럼

  • @samueldeandrade8535
    @samueldeandrade8535 Месяц назад

    I really didn't want to know about McNugget Numbers ...

  • @OL9245
    @OL9245 Месяц назад

    Infinite money isn't possible. So the expectancy is still zero, but not rvenly distributed..you win a dollar most of the time and you tlose a lot with very small probably. The thing is they balance each other

  • @tiripoulain
    @tiripoulain Месяц назад

    “Just pick the smallest element” And this is why the AoC is equivalent to the Well-ordering principle!

  • @markwrede8878
    @markwrede8878 Месяц назад

    The Choice Theorem is born of the inconsistency (not incompleteness) of Arithmetic. The theorem poses the undefined "plus one" of Addition againat the definable prospects of Multiplication.

  • @yomo_13
    @yomo_13 Месяц назад

    Wow I love it

  • @abiyyupanggalih854
    @abiyyupanggalih854 Месяц назад

    jojo refrence

  • @GlacialScion
    @GlacialScion Месяц назад

    The animations showing internal/external rotation force when using a barbell were the best I've ever seen in a fitness video.

  • @rosiefay7283
    @rosiefay7283 Месяц назад

    Hang on. Wikipedia says that the axiom of choice says that, given some indexing set I and a family of non-empty sets, {S_i for each i in I}, it's possible to construct an indexed set {x_i for each i in I}, where x_i\in S_i. No restriction on the cardinality of I. But you say that the axiom of choice needs to be invoked in order to pick one element out of just *one* non-empty set of reals? That's certainly not true --- if I is countable, then the axiom of countable choice is enough.

    • @ASackVideo
      @ASackVideo Месяц назад

      I'm asking for a rule that given any set of real numbers picks out an element of that set. That is, an explicit choice function on the powerset of the reals. Our indexing set here is the powerset of the reals, which has cardinality 2^𝔠

  • @shubhankgautam935
    @shubhankgautam935 2 месяца назад

    At 2:39 , where you got F(z) = -z/(z²+z-1), what if we input z = 1...? Since F(z) = 0z⁰ + 1z¹ + 1z² + 2z³...., F(1) should be equal to the sum of first n ( as n-->inf) fibonacci numbers, And if we compute F(1) from the formula you deduced, we get -1/(1²+1-1), which equals -1, But how is this possible, F(1) by definition should be the summation of the fibonacci series, how can it be negative 1...? Ik this video was posted 3 years ago and i have no hopes of getting a reply, but by chance if anybody sees this, plz reply so that i can be reminded of this video sometime in the future, I am still a student so maybe by then i would have gotten the mathematical maturity to be able to solve this by myself... Also, if i made a mistake plz feel free to correct me in the comments..!!

    • @ASackVideo
      @ASackVideo 2 месяца назад

      This is a good question! The radius of convergence of this series is 1/phi which is about .618. So when plugging in a value outside of the radius of convergence (such as 1), we shouldn't expect the power series to be equal to the function in question! In a similar vein, we know that the geometric series 1+r+r^2 + r^3 + ... = 1/(1-r). But we only know this when |r| < 1! If we try plugging in a value outside of this, such as for r = 2, it would look like saying 1+2+4+8 + ... = -1. I did unfortunately make a mistake a little later in the video at 3:16, where there should be a minus sign between the sums and between the two phis.

  • @ILoveMaths07
    @ILoveMaths07 2 месяца назад

    This is the best explanation of the axiom of choice.

  • @tcoren1
    @tcoren1 2 месяца назад

    6:39 alice wins. Since the harmonic series diverges she can always color enough squares to equal (exceed) 1 at every step, no matter how much bob colors before. 1+1+1+1+1...=infinity

  • @piotr780
    @piotr780 2 месяца назад

    ok - Alice and Bob colors tape on which some TM with infinite tape is run - if TM stops then Alice wins... it is uncomptuable :)

  • @piotr780
    @piotr780 2 месяца назад

    can we encode turing machine number this why and say that first person which TM halts wins ? this is clearly uncomputable...

  • @CS_n00b
    @CS_n00b 2 месяца назад

    I found the proof challenging, what’s the easiest intuitive proof for the Catalan numbers formula?

    • @ASackVideo
      @ASackVideo 2 месяца назад

      In the video, there was a slight error in the proof. At 6:43 the formula should only go up to 2n-1 instead of 2n+1. However I believe the easiest way to get the formula is using Bertrand's ballot theorem. I have a video titled "What are the odds of always winning?" that explains this theorem. For Catalan numbers, you want to consider a contest with n+1 votes for Alice and n votes for Bob. Hopefully it's not too hard to see why this should be in bijection with Dyck paths. With a little algebra you can get the formula.

  • @BooleanDisorder
    @BooleanDisorder 2 месяца назад

    Pretty sure the problem is the tokenization, not neural networks per say.

    • @Takyodor2
      @Takyodor2 2 месяца назад

      Neural networks don't understand multiplication, getting the correct result every time would mean training on enough samples to "remember" every solution. I don't think it would be very difficult to train a neural network to recognize an arithmetic problem, and hard-code the behavior "put the numbers and operator into a calculator instead of giving an answer directly, return the result given by the calculator app".

  • @FireyDeath4
    @FireyDeath4 2 месяца назад

    It also intakes text tokenistically. Since patterns like "123" can be single tokens, it creates a lot of confusion when it tries to process data with unique numbers in it. It's much harder to train it and make it predict digital interactions properly when there are just random variations introduced like that

  • @gandalfthegrey9116
    @gandalfthegrey9116 2 месяца назад

    So the Dyck Paths were very funny but this was also a pretty helpful video so thanks I guess but I still am pretty sure the Dyck Paths were the best part idk

  • @skarluna2976
    @skarluna2976 3 месяца назад

    Alice is a knave and Bob is a knight

  • @SunShine-xc6dh
    @SunShine-xc6dh 3 месяца назад

    Can you win a game without an end? Going to infinity is always impossible. The game has to have an end to have a winner at the end

  • @cecilarthur3579
    @cecilarthur3579 3 месяца назад

    If Alice is a knave why did she say she was indeed a knave?🤔

    • @ASackVideo
      @ASackVideo 3 месяца назад

      She didn’t say “I am a knave.” She said “Both of us are knaves.”

  • @kaminoeugene
    @kaminoeugene 3 месяца назад

    It's often told that math with axiom of choice is weird because it produces weird things like the famous Banarch Tarski paradoxon (which isn't a mathematical paradox afaik), but there are a lot more weird consequences in maths without the axiom of choice

  • @heethanthen
    @heethanthen 3 месяца назад

    I know nothing about math. Why would anyone pick the smallest element?

    • @xinpingdonohoe3978
      @xinpingdonohoe3978 Месяц назад

      They just want to demonstrate that they can pick *an* element, and definitively state what that element is. Picking the smallest would make that job simple. If it existed.

  • @Emiyuna
    @Emiyuna 3 месяца назад

    the 🤘 means 18 when you use the finger counting technique, 18 is 6x3, means 666

  • @codywohlers2059
    @codywohlers2059 4 месяца назад

    2:26 I have a counter example to the first player could have always made a move to make the same game state as after the second player's first move. If instead of the top right I go one to the left of the top right and then the second player plays one below the top right, the resulting game shape is an inverted L removed from the top right. There is no way to make an inverted L shape with one move...?

    • @ASackVideo
      @ASackVideo 3 месяца назад

      The claim is that player 1 has at least one first move that can guarantee a win, not that every first move player 1 makes can guarantee a win.

  • @blucat4
    @blucat4 4 месяца назад

    Thank you for this great explanation. I wish I had seen it when I was a kid. I used to be smart, now I'm 57 and forget my own phone number!! I have a question for you, similar, maybe you could do the video on it. If the formula for the surface area of a circle is πr2, why isn't the formula for the volume of a sphere πr3? Doing the visuals would require some CGI though, still, it would be good. Cheers. 🙂

  • @tictactoe101
    @tictactoe101 4 месяца назад

    I like your magic words Keep going!

  • @user-lw7ss7to8l
    @user-lw7ss7to8l 4 месяца назад

    my head hurts

  • @martinhazard5982
    @martinhazard5982 4 месяца назад

    Wtf

  • @pra.
    @pra. 4 месяца назад

    super cool

  • @shalinado
    @shalinado 4 месяца назад

    bababa