The Rubik's Cube is a Calculator

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  • เผยแพร่เมื่อ 21 ม.ค. 2025

ความคิดเห็น • 457

  • @EvilSandwich
    @EvilSandwich 2 หลายเดือนก่อน +962

    Man. Abstract Algebra is a hell of a drug.

    • @mekaindo
      @mekaindo 2 หลายเดือนก่อน +10

      wait till calculus

    • @minerscale
      @minerscale 2 หลายเดือนก่อน +96

      @@mekaindo who's taking abstract algebra before calculus?

    • @PFnove
      @PFnove 2 หลายเดือนก่อน +11

      ​@@minerscaleme later today cuz this sounds interesting

    • @Lilly-Lilac
      @Lilly-Lilac 2 หลายเดือนก่อน +2

      @@minerscale I mean... you could, I suppose, but it is just a bit unorthodox

    • @mekaindo
      @mekaindo 2 หลายเดือนก่อน +3

      @@minerscale i dont know i tried to be funny

  • @1.4142
    @1.4142 2 หลายเดือนก่อน +960

    when you can't bring a calculator to a test

    • @petersusilo9588
      @petersusilo9588 2 หลายเดือนก่อน +23

      However it looks really complicated and probably took a long time.

    • @petersusilo9588
      @petersusilo9588 2 หลายเดือนก่อน +11

      Also, for one who aren't able to play rubik, that would be extremely hard. I think it is just better to use the usual method.

    • @lollol-tt3fx
      @lollol-tt3fx 2 หลายเดือนก่อน

      he didnt say that its a good method he said its a method​@@petersusilo9588

    • @PFnove
      @PFnove 2 หลายเดือนก่อน +21

      ​@@petersusilo9588no shit Sherlock

    • @ynycu
      @ynycu 2 หลายเดือนก่อน

      🍑🧮🧮🧮⚖️😞💀🌝🌝🌚🌚

  • @excelmaster2496
    @excelmaster2496 2 หลายเดือนก่อน +684

    0×0
    "Remove the stickers"
    *removes the stickers*
    "Remove the stickers"
    *starts panicking*

    • @mr.duckie._.
      @mr.duckie._. 2 หลายเดือนก่อน +56

      then just remove the cube
      wait what do you do in case of 0x0x0???

    • @wj11jam78
      @wj11jam78 2 หลายเดือนก่อน +59

      ​@@mr.duckie._.
      Remove stickers
      Stickers existed on cube, so remove cube
      cube existed in your hands, so...

    • @olegtarasovrodionov
      @olegtarasovrodionov 2 หลายเดือนก่อน +25

      just fix the rule like this: "Remove the stickers if they are not removed yet"

    • @katie-ampersand
      @katie-ampersand 2 หลายเดือนก่อน +8

      ​@@mr.duckie._. hammer

    • @cheeseburgermonkey7104
      @cheeseburgermonkey7104 2 หลายเดือนก่อน +7

      @@katie-ampersand 0^4: remove the hammer

  • @danielleidulvstadpereda5481
    @danielleidulvstadpereda5481 2 หลายเดือนก่อน +219

    I've been speedcubing for quite a few years, and this is by far the coolest thing involving Rubik's cubes I've come across!

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน +27

      Thank you, I'm glad you enjoyed the video!

    • @tymikaseawood2596
      @tymikaseawood2596 2 หลายเดือนก่อน +3

      I like it too! 10:51

    • @ajiprnk4821
      @ajiprnk4821 หลายเดือนก่อน

      @@TheGrayCuber i have a question can a 4d rubiks cube do mod 1000 ?

  • @thegnugod2108
    @thegnugod2108 2 หลายเดือนก่อน +117

    My gosh doing 3*67 and watching the cube turn back to its starting position was amazingly satisfying

    • @cubingbox
      @cubingbox 2 หลายเดือนก่อน +5

      Isn't 67 then the inverse of 3?

    • @Boxytablet
      @Boxytablet 2 หลายเดือนก่อน +1

      @@cubingboxSo then he would be using mod 70

    • @PMA_ReginaldBoscoG
      @PMA_ReginaldBoscoG หลายเดือนก่อน +1

      ​@@Boxytablet He is using modular multiplication, not addition. You are right about 67 being the inverse of 3. However the underlying operation and the number we use to divide the sum or product differs.

    • @Boxytablet
      @Boxytablet หลายเดือนก่อน +1

      @@PMA_ReginaldBoscoG bro I’m not talking about addition, it’s because 67 is 3 less than 70, so 67 in mod 70 could be -3

    • @PMA_ReginaldBoscoG
      @PMA_ReginaldBoscoG หลายเดือนก่อน

      @@Boxytablet you really are talking about addition. In the case of the set of all integers, -3 is the additive inverse of 3, whereas in the case of the set of integers modulo 70, 67 is the additive inverse of 3.

  • @DarkAlgae
    @DarkAlgae 2 หลายเดือนก่อน +146

    by tricking me into being entertained by modular arithmatic, you earn my subscription.

  • @memetech-
    @memetech- 2 หลายเดือนก่อน +130

    1:05 hey, I *AM* thirsty, I should drink water.

    • @error_6o6
      @error_6o6 2 หลายเดือนก่อน +7

      Fr bro caught me

    • @obsidianflight8065
      @obsidianflight8065 2 หลายเดือนก่อน +3

      not only once, he made me think about it 3 times!
      I was thirsty sure, but now I'm REALLY thirsty and can't contain myself anymore, to the water I go!

    • @PMA_ReginaldBoscoG
      @PMA_ReginaldBoscoG หลายเดือนก่อน +1

      Hi Thirsty! I'm your subscriber.

  • @rarebeeph1783
    @rarebeeph1783 2 หลายเดือนก่อน +384

    is this just a sneaky introduction to group homomorphisms?

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน +163

      It sure is!

    • @mathgeniuszach
      @mathgeniuszach 2 หลายเดือนก่อน +52

      But without any of the complex math terms that obscure away the beauty of math to the layman viewer.

    • @wymarsane7305
      @wymarsane7305 2 หลายเดือนก่อน +12

      I KNEW IT

    • @spicetea4060
      @spicetea4060 2 หลายเดือนก่อน +8

      I haven't been in a group homomorphism since college

  • @clementdato6328
    @clementdato6328 2 หลายเดือนก่อน +101

    “Can we doNO” 😂

    • @msolec2000
      @msolec2000 2 หลายเดือนก่อน +7

      "We can make a religion out of"NO, don't.

  • @romywilliamson4981
    @romywilliamson4981 หลายเดือนก่อน +6

    I love that you are teaching the essenece of group theory and elementary number theory to cube-enthusiasts, without ever saying that's what you are doing :)

  • @art-of-imagination
    @art-of-imagination 2 หลายเดือนก่อน +16

    I'm a cuber as well as a math student but I've never thought anything like this. I appreciate you for giving this kinda mind-boggling aspect to see or use the mod. ❤

    • @philia08
      @philia08 หลายเดือนก่อน +1

      Another speedcuber here!
      Hope You aren't fan of Max Park or somebody random from new gen speedcubers ☺️

  • @ProactiveYellow
    @ProactiveYellow 2 หลายเดือนก่อน +25

    Working through this is an interesting exploration of the normal subgroups of the rubik's group. It must be a challenge to make sure the algorithms for the larger cyclic group elements end up commutative.

  • @SamuelLiJ
    @SamuelLiJ 2 หลายเดือนก่อน +65

    Interesting video. You can actually do all four elementary operations (addition, subtraction, multiplication, division) mod 10 on the 3x3, where defined and invertible, for the dumb reason that the symmetric group S10 embeds into the Rubik's group. So you just perform the action corresponding to how the elements 0 through 9 permute. (Actually S12 fits as well.)
    Note that you can add or subtract any number, but only multiply and (modular) divide by units.

    • @alejrandom6592
      @alejrandom6592 2 หลายเดือนก่อน

      Nice, can you elaborate?

    • @louisrustenholz7642
      @louisrustenholz7642 2 หลายเดือนก่อน +1

      @@alejrandom6592 For S12, you can simply use the 12 edge cubes, ignore both edge orientation and all about corner cubes, and notice that you can perform swaps between any two edges. (These swaps also swap corner cubes, but you choose to ignore it.)

    • @louisrustenholz7642
      @louisrustenholz7642 2 หลายเดือนก่อน +3

      Then, for the encodings, assign the names '0', '1', ..., '11' to each edge (arbitrarily). For each operation of the form '+4', '-3', etc., encode it as the corresponding permutation (e.g. for +4, you get 0->4, 1->5, ...), which can always be built out of simple swaps.
      For multiplication/division, do the same game, restricting yourself to invertibles mod 12.

    • @louisrustenholz7642
      @louisrustenholz7642 2 หลายเดือนก่อน +1

      For mod 10, play the same game and just ignore two edges.

    • @beansprugget2505
      @beansprugget2505 2 หลายเดือนก่อน

      I'm not a cuber so maybe I'm missing something obvious, but I don't understand your addition. Are you saying that to add, eg 3 + 4, you first do 3, and then you make the 7 permutation? Where does the 4 come in? Or is it that one set of moves goes from 1, 2 , 3 etc each iteration (like how repeating U adds one in mod 4).

  • @arisweedler4703
    @arisweedler4703 2 หลายเดือนก่อน +13

    If your stickers all had arrows on them, that would introduce additional state for your cube.
    I once had a Rubik’s cube (or… a derivative of one) that had directioned stickers. The 9 stickers formed a picture with the right orientation. With the wrong orientation it looked scrambled. It was harder for this reason.

    • @error_6o6
      @error_6o6 2 หลายเดือนก่อน +6

      Actually, only the center stickers’ direction matter, but I think that should be enough to improve the highest number it can multiply by.

    • @arisweedler4703
      @arisweedler4703 2 หลายเดือนก่อน +1

      @@error_6o6 that’s super cool! I never noticed that with my cube. From what you’re saying, there are only 6*4 additional legal states that are added by taking this type of “sticker rotation” into account? Well maybe it’s not 6*4… but it’s a number small enough to be entirely represented by the orientation of all the middle stickers. I can understand that. I gotta think abt it a bit 😁

    • @error_6o6
      @error_6o6 2 หลายเดือนก่อน +2

      @@arisweedler4703 I’m pretty sure the amount of states are multiplied by 4^6 divided by 2 because of weird parity stuff, but that should total to a multiplier of 2048, or, in simpler terms, a lot.

  • @HyperCubes
    @HyperCubes 2 หลายเดือนก่อน +4

    one of the most impressive and informative videos I've seen, and bonus points for using a rubiks cube (I'm a speedcuber so it makes me so happy to see videos like these) great video!!

  • @Higgsinophysics
    @Higgsinophysics 2 หลายเดือนก่อน +8

    A math video that also reminds you to drinks water. This is just the summit of youtube. Loved the video

  • @PaulFisher
    @PaulFisher 2 หลายเดือนก่อน +79

    I’m not very skilled with the Rubik’s Cube so I couldn’t handle the complex algorithms for 10, but I have found one for modulo 2:
    1: do nothing
    and the decoding process:
    • cube has been destroyed: 0
    • cube exists: 1 (orientation unimportant)

    • @catstone
      @catstone 2 หลายเดือนก่อน +19

      I have found one for modulo 1:
      0: destroy cube
      and the decoding process:
      • cube has been destroyed: 0

    • @R6nken
      @R6nken 2 หลายเดือนก่อน +9

      ​@@catstone i think it's more like:
      0: cube
      Decoding:
      * cube: 0

    • @rayzhao491
      @rayzhao491 2 หลายเดือนก่อน +12

      @@R6nken idea for mod 0: you don't need the cube. the cube does not exist. the cube has never existed. multiplication does not exist. the universe does not exist. there is only an eternal void.

  • @itzmetanjim
    @itzmetanjim 2 หลายเดือนก่อน +34

    0:15 setting it to its default position is not as easy as the other steps (depending on who you are)

    • @Rb_Perm
      @Rb_Perm 2 หลายเดือนก่อน +1

      No its pretty Easy actually

  • @MooImABunny
    @MooImABunny 2 หลายเดือนก่อน +15

    I'm relearning group theory right now and I'm delighted that this video came out right now 😁
    Also, embedding an Abelian group within the Rubik's cube in order to multiply numbers mod n is stick is a wild idea. I know these (non-trivial) subgroups exist, I know how group isomorphism works, but there's a whole other set of steps I'd need to do to come to the idea "I'll use the Rubik's cube to compute ab mod n"

  • @wymarsane7305
    @wymarsane7305 2 หลายเดือนก่อน +12

    9:58 The reason for this is, of course, that cube algorithms aren't abelian. The irony here is that commutators are an extremely useful concept for solving twisty puzzles precisely because the piece movements of one algorithm messes with the pieces of another algorithm.

  • @kyanilcauli9002
    @kyanilcauli9002 หลายเดือนก่อน +4

    When the obscure theorems in Dummit and Foote finally appear to you in a visual form. Bless you. You just made my day.

    • @PMA_ReginaldBoscoG
      @PMA_ReginaldBoscoG หลายเดือนก่อน +1

      It's a pain in the ass because our professor beat the shit outta us!😂

  • @vaughnp3913
    @vaughnp3913 2 หลายเดือนก่อน +14

    I've spent far too much of my life watching cubing videos on youtube, and this has to be one of my all-time favorites! Thank you for making this - it's excellently done :)

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน +2

      Thank you for watching, I'm glad you enjoyed it!

  • @escthedark3709
    @escthedark3709 2 หลายเดือนก่อน +10

    This was supremely disappointing when it turned out that you couldn't do 5x8, but then supremely interesting when it turned out that you could do all sorts of other stuff.

  • @tl_dragonstars2877
    @tl_dragonstars2877 2 หลายเดือนก่อน +5

    I was brushing teeths while watching, this :
    => 1) I have to watch the video a second time
    => 2) Iwon't be able to sleep because of too intense curiosity

  • @voliol8070
    @voliol8070 2 หลายเดือนก่อน +16

    Ah, the cliffhanger. Looking forwards to the next video!

  • @have-bear
    @have-bear 2 หลายเดือนก่อน +12

    If you need to construct a n cycle algorithm, it doesn't have to find n positions for stickers. For example, one can construct a 45 cycle algorithm that move an edge piece between 9 positions and move a corner piece between 5 positions. Unforntunately, a 25 cycle algorithm still can't be constructed with this technique.

    • @BlueDog15391
      @BlueDog15391 2 หลายเดือนก่อน +3

      That's what was bugging me while I was watching. Thanks for answering my question before I asked it =)

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน +4

      Yes, this is a great point! 25 and 23 are problematic because they are only divisible by one prime, and therefore can't be constructed from smaller cycles.

    • @aloi4
      @aloi4 2 หลายเดือนก่อน +2

      Because Z45 = Z9 × Z5

    • @Anonymous-df8it
      @Anonymous-df8it 2 หลายเดือนก่อน +3

      @@TheGrayCuber Couldn't you have two simultaneous five cycles for 25?

    • @ajiprnk4821
      @ajiprnk4821 หลายเดือนก่อน

      woudnt make you have to use multiple algorithms for one number?

  • @GhostShadow.0316
    @GhostShadow.0316 2 หลายเดือนก่อน +5

    this is literally what I had looking for for the past months!
    this is so smart, thank you so much to make this video

  • @RowanFortier
    @RowanFortier 2 หลายเดือนก่อน +3

    You could use a 1x2xN cuboid to do N/2-digit binary multiplication

    • @qwerty_qwerty
      @qwerty_qwerty 2 หลายเดือนก่อน

      rowanfortier?!?! 0 likes 0 replies??!?!!?

  • @applimu7992
    @applimu7992 2 หลายเดือนก่อน +8

    The multiplicative group of units are one of my favorite constructions in ring theory!!!

  • @jesusvera7941
    @jesusvera7941 หลายเดือนก่อน +1

    oh man i was having a terrible insomnia, but this video had me falling sleep like 7 times, thanks you for your help! now... i have to take advantage of this feeling.

  • @Danitux11
    @Danitux11 2 หลายเดือนก่อน +4

    I know you won't read this, but this actually just made my day better. This is such a cool concept and I'm very thankful this appeared in my page. Subbed.

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน +2

      I did read this! Thanks for the positive comment, I'm very glad that you enjoyed the video

  • @JoannBaron
    @JoannBaron 19 วันที่ผ่านมา +1

    Your videos are so good, I genuinly hope youll manage to become more popular

  • @studypurposeonly69
    @studypurposeonly69 20 วันที่ผ่านมา +2

    Don't know how useful it is going to be in industries but definitely a very cool topic of study to make a video!

  • @Fur0rem
    @Fur0rem 2 หลายเดือนก่อน +4

    Wow that was such a great video, didn't know rubix cubes could go that deep!
    Now it makes me wonder if this, or a similar concept, could somehow be used by computers, with arithmetic modulo 256, 65536, (all the powers of 2), since right now they still do multiplication the old fashioned way

  • @kotmooson6459
    @kotmooson6459 หลายเดือนก่อน +1

    This video shows that pretty much anything with memory and operations over it can be computer of some sort if really want to

  • @gwenhidwy
    @gwenhidwy 28 วันที่ผ่านมา +1

    As a lover of group theory and Rubick's cube I thoroughly enjoyed this! You should explore comutators in the same light

  • @willlagergaming8089
    @willlagergaming8089 2 หลายเดือนก่อน +113

    Then my chess board is a graphing calculator. Also most sane mathematician.

    • @mekaindo
      @mekaindo 2 หลายเดือนก่อน +4

      if the chess is a graphin calculator, what is checkers???

    • @Bangaudaala
      @Bangaudaala 2 หลายเดือนก่อน +9

      ​@@mekaindo binary?

    • @mekaindo
      @mekaindo 2 หลายเดือนก่อน +3

      @@Bangaudaala thats a good idea

    • @willlagergaming8089
      @willlagergaming8089 2 หลายเดือนก่อน +3

      The Chinese calculator thing. Can't remember it's name

    • @brenatevi
      @brenatevi 2 หลายเดือนก่อน +2

      @@willlagergaming8089 Abacus?

  • @alejrandom6592
    @alejrandom6592 2 หลายเดือนก่อน +8

    U crazy bro. This math so good it seems forbidden.

  • @Phylaetra
    @Phylaetra 2 หลายเดือนก่อน +5

    I love your encoding schema! That is a great way to map modular arithmetic onto a non-abelian group! Although I am very disappointed that R2D2 was not an alg...

    • @artemisSystem
      @artemisSystem 2 หลายเดือนก่อน +3

      The reason it works is that the subgroup is abelian. But if you want R2 and D2 to be valid algs, you can't do that, because they don't commute, and your subgroup is then not abelian. Though i suppose R2D2 could be a base alg in itself, perhaps. It has a period of 6 though, so not sure it can be used for this? It's not clear to me what determines what cycles you need for a given mod, how that's determined, and if you can have multiple different cycle sets. I guess i'll have to wait for the next video.

    • @mr.vladislav5746
      @mr.vladislav5746 2 หลายเดือนก่อน +1

      ​@@artemisSystem I guess you could just do ℤ/6ℤ or (ℤ/7ℤ)× with R2D2, as both groups have one cycle of length 6.
      To answer your question, first and foremost, yes, you can have multiple different cycle sets by the Chinese Remainder Theorem. So one 6-cycle is isomorphic to a 2-cycle and a 3-cycle because these two numbers are coprime. However, a 4-cycle is NOT the same as two 2-cycles. So essentially, if we break our cycles into "elementary cycles", they all have a length that is a power of a prime. These are sometimes called something along the lines of "elementary divisors."
      For a given n, to find which cycles you need (the elementary ones, i.e. powers of primes), you need to analyze the multiplicative group (ℤ/nℤ)× (which has φ(n) elements where φ is the Euler Totient function; in the video he calls these the units, e.g. φ(10) = 4 because the four units mod 10 are 1, 3, 7, 9).
      This, in turn, is easily Googlable, i.e. to find what product of cyclic groups (ℤ/nℤ)× is isomorphic to. But if you want to find it yourself, there is something called the Structure Theorem of Finitely Generated Abelian Groups (SToFGAG), which states that any finitely generated (thus also any finite) abelian group is a direct product of cyclic groups, i.e. that is what allows this entire exercise.
      If we take the example of mod 15, there are 8 elements in (ℤ/15ℤ)× (specifically, 1, 2, 4, 7, 8, 11, 13, 14). Then, SToFGAG clearly says it must be isomorphic to one of the following:
      ➡ ℤ/8ℤ
      ➡ ℤ/4ℤ × ℤ/2ℤ
      ➡ ℤ/2ℤ × ℤ/2ℤ × ℤ/2ℤ
      simply because if an abelian group (we know modular multiplication is abelian) with 8 = 2³ elements is a direct product of cyclic groups, there simply are no other ways. In other words, every abelian groups with 8 elements is isomorphic to one of the three above.
      Furthermore, since (ℤ/15ℤ)× has no element of order 8 (easily checkable) but it has an element of order 4 (for example 2*2*2*2 = 16 ≡ 1 mod 15), it must be the middle case, i.e.
      (ℤ/15ℤ)× ≅ ℤ/4ℤ × ℤ/2ℤ
      so we conclude that the "structure" of the multiplicative group mod 15 is a 4-cycle and a 2-cycle, which can be encoded using the ways discussed in the video (e.g. by setting 2 to be R, 11 to be L2, and then everything else is generated by 2 and 11).
      However, it's another question whether the Rubik's cube is "big enough" to contain so and so many different cycles.

  • @amogus4868
    @amogus4868 2 หลายเดือนก่อน +25

    This is why Rubik's cubes are loved by many. They are more than some toys.

    • @alansun70
      @alansun70 2 หลายเดือนก่อน +3

      I had one in Louisiana. I never thought of it this way.

  • @yeokonma
    @yeokonma 2 หลายเดือนก่อน +4

    2 of my favorite things in one video. thank you

  • @BaranCemCesme
    @BaranCemCesme 2 หลายเดือนก่อน +3

    Great video. My idea for multiplying by 0 was exploding the cube but removing the stickers is way better.

  • @katherinek6166
    @katherinek6166 หลายเดือนก่อน +5

    "Can I interest you in a video about group theory?" - "No." - "How about a Rubik's Cube one?"

  • @MeepMu
    @MeepMu 2 หลายเดือนก่อน +6

    Removing the stickers was really funny to me for some reason

  • @wyattstevens8574
    @wyattstevens8574 2 หลายเดือนก่อน +11

    "Can we do mod one thou-"
    "NO."

    • @error_6o6
      @error_6o6 2 หลายเดือนก่อน +5

      *multiple angry mathematicians staring at you*
      Edit: btw this reply was made before watching this video so I thought the comment said “can we do mod 1 though” (oh well too late to change it now)

  • @Rhys_1000
    @Rhys_1000 2 หลายเดือนก่อน +5

    It is actually possible to input even numbers if you use 6 * 5 = 0 in modular arithmetic:
    Since 6 * 6 = 6, 6 would be peeling the stickers partially and not doing anything else.
    And the other numbers:
    2 = 6 * 7
    4 = 6 * 9
    8 = 6 * 3
    And any face (ignoring the colors) that has peeled the other stickers is 5
    Hope this helps!

    • @vytah
      @vytah 2 หลายเดือนก่อน +2

      You still need to figure out which stickers to remove so that a cube with partially removed stickers can still be unambiguously interpreted as the correct result.

    • @Rhys_1000
      @Rhys_1000 2 หลายเดือนก่อน

      5:27 6 and 5 can be these two

    • @aloi4
      @aloi4 2 หลายเดือนก่อน

      ​@@Rhys_1000 No, because 5×3=5×7=5×9=5
      5 need to remove all stickers from the up layer (except the center)

    • @Rhys_1000
      @Rhys_1000 2 หลายเดือนก่อน

      ​@@aloi4Actually, it still only matters if that specific kind of stickers are peeled to be 5

  • @roxashikari3725
    @roxashikari3725 2 หลายเดือนก่อน +2

    This was an immediate like and subscribe for me. I love it.

  • @jonathanshuman5859
    @jonathanshuman5859 2 หลายเดือนก่อน +2

    This is an amazing video, loved it!

  • @cubingbox
    @cubingbox 2 หลายเดือนก่อน

    You could maybe also use corner twists as moves, but I don't know if you would use that

  • @szlanty
    @szlanty 2 หลายเดือนก่อน +10

    the Gray Cuber doing a video with Cubes mentioned?
    its more likely than you think!

  • @Knighttwister
    @Knighttwister 2 หลายเดือนก่อน +5

    when you said "I'm a little thirsty" i was literally grabing for my water bottle

  • @skmgeek
    @skmgeek 2 หลายเดือนก่อน +3

    this is a really well-made video!

  • @lyvindy0527
    @lyvindy0527 9 วันที่ผ่านมา +1

    "remove the stickers"
    People with stickerless cubes:

  • @MrConverse
    @MrConverse 2 หลายเดือนก่อน +5

    12:56, small error: the audio says 106 but the graphic shows 107. I’m fairly certain that 107 is correct. Hope it helps. Great video!

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน +3

      yes, thank you. 107 is correct

  • @cosmicScience-p5m
    @cosmicScience-p5m หลายเดือนก่อน +2

    Just a a nuclear fussion reaction started in my brain. 🤣🤣🤣
    I can power whole city by this video.

  • @abhijeetghodgaonkar
    @abhijeetghodgaonkar 2 หลายเดือนก่อน +1

    Insane yo, good explanation!

  • @genandnic
    @genandnic 2 หลายเดือนก่อน +6

    it goes in the square hole

  • @whitefwr
    @whitefwr หลายเดือนก่อน +1

    So you made a calculator from the Rubik's Cube? Congrats!

  • @Leonardo-cw1dd
    @Leonardo-cw1dd 26 วันที่ผ่านมา +1

    holy shit you earned a sub from me. the math behind this is insane, it broken my brain lmfao

  • @adityakhanna113
    @adityakhanna113 2 หลายเดือนก่อน +3

    Oh my gosh, this is brilliant. I'm definitely very jealous to not have thought of it , considering all of my years of cube experience and "it's a group" propaganda. Couldn't put 2 and 2 together to make a 4 xD
    Also, i believe you can only use units because the cube's moves are reversible (i.e. a group). I like your idea of 2*5 ≈ 0 mod 10, but just to hammer in the point that this means 2 doesn't have an inverse, which every move on the cube does.

  • @aviralsood8141
    @aviralsood8141 2 หลายเดือนก่อน +2

    This is beautiful

  • @reyuki-i
    @reyuki-i 2 หลายเดือนก่อน

    12:27 what subject I must learn to understand the reason behind the technicalities?

    • @reyuki-i
      @reyuki-i 2 หลายเดือนก่อน

      Ah, it seems about abstract algebra (just read the video description)

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน +1

      depends on which technicalities. Abstract algebra is needed to determine which cycles, and then familiarity with the Rubik's cube to determine what can fit on the cube. or just wait bit because the next few months I'll be posting videos about both topics!

  • @lionelinx7
    @lionelinx7 หลายเดือนก่อน +1

    Beautiful

  • @Zufalligeule
    @Zufalligeule 2 หลายเดือนก่อน +1

    Really cool. makes me wonder, whether there is a largest prime modulus that can be represented on a cube.

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน

      This is a really interesting problem!

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน +1

      991 is the highest prime modulus possbile

    • @Zufalligeule
      @Zufalligeule 2 หลายเดือนก่อน

      @TheGrayCuber wow, it's surprisingly large! I've expected it to be around 100-200.

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน +2

      @@Zufalligeule 1260 is the maximum cycle possible on a cube, but 1261 is not prime. 990 is the second highest cycle that fits on a 3x3, and then adding 1 we get 991 which is prime!

  • @Salsmachev
    @Salsmachev 2 หลายเดือนก่อน +16

    Wow it's like is a slide rule took 10 times as many moves to use and gave you the least significant digits instead of the most significant digits. How... useful?

    • @HzyMkwii
      @HzyMkwii 2 หลายเดือนก่อน +1

      BUT ITS A RUBIX CUBE so it’s cool

    • @joeshmoe4207
      @joeshmoe4207 6 วันที่ผ่านมา

      It’s a nice applied group theory for non mathematics people.

  • @ConradoPeter-hl5ij
    @ConradoPeter-hl5ij 2 หลายเดือนก่อน +1

    really creative thinking like this

  • @emmettdja
    @emmettdja 14 วันที่ผ่านมา +1

    Finally I can put my modern algebra class to use.

  • @Sjoerd-gk3wr
    @Sjoerd-gk3wr 2 หลายเดือนก่อน +1

    Great video can’t wait for the next one

  • @אביבשקד-נ2ד
    @אביבשקד-נ2ד 2 หลายเดือนก่อน

    Try using a prime number as the mod so no numbers will multiply to that, meaining you can use all numbers

  • @cheeseburgermonkey7104
    @cheeseburgermonkey7104 2 หลายเดือนก่อน +1

    12:26 Does someone know how to explain why this doesn't work

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน +2

      If there were a 23-cycle, then 23 of the stickers would move around to each other's locations. But then that 24th sticker must not move - otherwise it would hit one of the 23 and interfere with the 23-cycle. But if that 24th sticker can't move, then neither can the other stickers on the same piece, which also interferes with the 23-cycle

    • @Anonymous-df8it
      @Anonymous-df8it 2 หลายเดือนก่อน +2

      @@TheGrayCuber What's the largest cycle that *_does_* work?

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน

      1260 is the largest possible

    • @Anonymous-df8it
      @Anonymous-df8it 2 หลายเดือนก่อน +1

      @@TheGrayCuber Why?

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน +1

      i dont know how to answer the 'Why?' other than 'nothing bigger fits on the cube'. A sort of satisfying answer is to look at the prime factorization. 1260 = 4*9*5*7. This is about the max that even fits on the cube. You can push the power of 2 up to 16 but that would require dropping either 5 or 7, resulting in a lower cycle. You can also use 11, but again that would require dropping some combination of other cycles that would result in something lower. And then any other prime factorization is impossible. Any possible cycle must divide 16*9*5*7*11

  • @reyuki-i
    @reyuki-i 2 หลายเดือนก่อน

    9:10 are you just guessing each possibility and ended up with 7^4, or is there a systematic way to arrive at that expression?

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน +1

      This is largely just guessing

  • @adityakhanna113
    @adityakhanna113 2 หลายเดือนก่อน +3

    It might be possible to do for larger numbers by using multiple cubes and exploiting chinese remainder theorem right?
    To do modulo 1000, you could do 125 (if possible) and 8

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน +3

      You've got the right idea, the CRT does help breakdown the structure, but the units mod 125 still need a 4 cycle and a 25 cycle. There isn't really a way to do that 25 cycle on an nxn

    • @adityakhanna113
      @adityakhanna113 2 หลายเดือนก่อน

      ​@@TheGrayCuber oh that's so true. The cycles are given by factors and CRT requires the same factors, so they possibly inherit the impossiblities

  • @oincapaz
    @oincapaz 22 วันที่ผ่านมา

    6:02
    i could could chose: 1, 2, 3, 7, 9 and no multiplication of these would make a 0.
    why aren't they units?

    • @TheGrayCuber
      @TheGrayCuber  22 วันที่ผ่านมา

      Using this system of numbers causes the problem that 2*3 is not defined, since you didn't include 6. But if you did also include 6, then 1*2 = 6*2, meaning that 1 alg = 6 alg so you're actually just using mod 5

  • @rodrigoqteixeira
    @rodrigoqteixeira 2 หลายเดือนก่อน +1

    Possible definition for units those whose gcd with b is 1, those that are co-prime with b

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน

      This is a very good definition for the units!

  • @KLR-3
    @KLR-3 2 หลายเดือนก่อน +1

    I wonder if anyone has adapted this as an alternative to the current popular blindfold method.

  • @Sw3d15h_F1s4
    @Sw3d15h_F1s4 2 หลายเดือนก่อน +1

    dumb/random idea:
    since the center squares never change with respect to eachother, can you use the orientation of the cube itself to get larger cycles? mix in pitch, yaw, and roll of the cube, and say define white up with some color facing you as the default starting position?

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน

      Yes this would allow at least an additional 4-cycle! But then I think you'd also need to 'fix' the algs, like saying that U must always be the white face instead whatever is on top

  • @crumblinggolem6327
    @crumblinggolem6327 2 หลายเดือนก่อน

    Could you bypass the 24 cycle limit by tying two stickers together? like rather than consider just the edge or just the corner, 1 'position' would be (edge1 at pos1 x corner 1 at pos 1), then two would be (edge1 at pos1 x corner 1 at pos 2), etc... up to (edge1 at pos1 x corner 1 at pos 24) then (edge1 at pos2 x corner 1 at pos 1) which would allow for cycles up to 24².

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน

      This is a great point! You can get a cycle higher than 24, it's just that 25 and 29 don't work specifically because they're prime powers over 24.

  • @vinesthemonkey
    @vinesthemonkey 2 หลายเดือนก่อน

    I didn't watch it, but it's an application of the Chinese Remainder Theorem. For finite Abelian groups, there's a unique factorization analogous to the fundamental theorem of algebra. The Rubik's group isn't abelian (for example R U is not equivalent to U R) but the cyclic subgroup generated by a sequence of moves as one element is (for example )

  • @miners_haven
    @miners_haven 2 หลายเดือนก่อน

    I wonder what could be done on a Rubik's Tesseract

  • @d1tnhauxa0rau
    @d1tnhauxa0rau หลายเดือนก่อน +1

    my act of drinking water predates the thirst reminder in the video by 3 seconds

  • @reyuki-i
    @reyuki-i 2 หลายเดือนก่อน

    How the hell does he come up with this brilliant idea!? awesome.

  • @durza4297
    @durza4297 2 หลายเดือนก่อน

    5:49 Does that means we can't multiply 2 by 7 mod 10 ? Can someone explain it to me, because it feels quite unsatisfying...

  • @titimathrosgui5109
    @titimathrosgui5109 2 หลายเดือนก่อน

    can you give a link or a list of every modular multiplication below a number, like 100 ?

  • @onur-karakus
    @onur-karakus หลายเดือนก่อน

    you should use de esser for your audio records

  • @ben_adel3437
    @ben_adel3437 2 หลายเดือนก่อน

    I love this because this year i was feeling so desperate that i wanted to cheat using more 5x5 i didnt because like memorazing a cheating method is harder than actually learning the topic i needed for the exam but it's cool knowing i could've done it

  • @tepan
    @tepan 2 หลายเดือนก่อน

    Can I caculate the number by which I can multiply the cube into its starting position?

  • @rossthebesiegebuilder3563
    @rossthebesiegebuilder3563 2 หลายเดือนก่อน

    What do you mean that you solve 4x4 parity by ignoring it? You still have to deal with it at some point...?

  • @YATAQi
    @YATAQi 2 หลายเดือนก่อน

    Great video! I've always known this trick was a thing, but I've never fully understood it great detail. This could be a great video in an advanced abstract algebra series - you thinking of diving more into it?
    Also, do you mind if I ask how you did the 3D Rubik's cube animation? I'm getting my hands dirty with Manim right now, but I haven't dealt with any 3D components yet so I'm just wondering if that's what you used or if it was something entirely different. I want to try to create a 2x2 chess cube if you're curious :)

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน

      There is a link in the description to that Rubik's cube I made on OpenProcessing. You can view the source code and even make a fork to make your own version!

  • @SumiOfficial-uj4uz
    @SumiOfficial-uj4uz 4 วันที่ผ่านมา

    Bro, how are you so genius ?

  • @StewartStewart
    @StewartStewart 2 หลายเดือนก่อน

    I'll watch this later, but my big question going in is what abelian subgroups you used to make it commutative.

  • @creepinator4587
    @creepinator4587 2 หลายเดือนก่อน +2

    Thoughts:
    The prime numbers seem really important for this, since they avoid the "multiply to 0" problem, and seem to be related to the unit cycles
    Would each prime number just have 1 cycle?
    If they do each have 1 cycle, than would 23 be the largest prime you can fit on a cube? And therefore are you able to fit any modulus that only has prime factors less than 23?
    I eagerly await the next videos in this series, since they seem like they'll answer some of these questions

    • @creepinator4587
      @creepinator4587 2 หลายเดือนก่อน +1

      Scratch that, the "prime numbers have 1 cycle" conjecture is easily disproved by 7 having 3 cycles
      2-4-6-1, 3-6-1, and 5-3-1

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน +2

      Yes, primes are important for this and they do each only have 1 cycle. Good observations!
      It does turn out that you can get higher primes than 23 onto the cube though. 29 needs a 28 cycle, but you can achieve a 28 cycle by mixing a 4 cycle and a 7 cycle.
      So therefore the problematic primes are ones like 83, where p-1 is divisible by a prime > 24. 82 = 2*41

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน +1

      7 is really interesting! It can be represented as just one 6 cycle, or a 2 cycle and a 3 cycle. It's the smallest number that offers such a choice

    • @vytah
      @vytah 2 หลายเดือนก่อน +1

      @@creepinator4587 If you have two cycles A and B, you can always combine them into a single cycle lcm(A,B) by doing them both at the same time. If A and B are coprime, this just means A×B.

  • @artemis_furrson
    @artemis_furrson 2 หลายเดือนก่อน +1

    Finally a use for the multiplicative group of integers modulo n

  • @πτΩαπσ
    @πτΩαπσ 2 หลายเดือนก่อน +1

    What is the largest modulus possible on 3x3?

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน +2

      533,520 is the largest that I have found

  • @TheBookDoctor
    @TheBookDoctor 2 หลายเดือนก่อน

    I hope you're doing this as part of a paper for some math journal.

  • @lanevalhalla1225
    @lanevalhalla1225 หลายเดือนก่อน

    i actually derived a formula for optimizing soludtions using a similar concept to that of yours

  • @lazerwastaken
    @lazerwastaken 2 หลายเดือนก่อน

    Help! All of my cubes is stickerless and I can't peel the colors!

  • @Candlest1112-hb8tj
    @Candlest1112-hb8tj 2 หลายเดือนก่อน +1

    Pretty nice video!!!!!!!

  • @CosmicHase
    @CosmicHase 2 หลายเดือนก่อน

    How about larger cubes/smaller cubes? Would they be more accurate/less accurate,?

  • @RandomBurfness
    @RandomBurfness 2 หลายเดือนก่อน

    If you throw in a gigaminx or similar but bigger, can you do stuff a megaminx can't? Probably.

    • @TheGrayCuber
      @TheGrayCuber  2 หลายเดือนก่อน

      Yes! The gigaminx has way more pieces to work with. It's still constrained by the 60 sticker limit just like the megaminx, so it can't fit any cycles over 60. But it can fit more sub-60 cycles than a megaminx

  • @pradeepmalar327
    @pradeepmalar327 หลายเดือนก่อน

    "I'm a little thirsty" and "this answer is wrong" are the 2 things I thought.