Q: "Why do I need to study math?" A: "Well, if you ever want to maximize points of failure..."

@MrSonny6155

4 жыл бұрын

Some people just like to watch the world burn, so that's why we invented maths.

@fiveoneecho

4 жыл бұрын

But by understanding something new, which you would not otherwise, you are reducing points of failure! D: This effort is hopeless!

@just_a_rock

4 жыл бұрын

This comment proves that engineering & mathematics are the exact opposite of eachother.

@trickytreyperfected1482

2 жыл бұрын

@@MrSonny6155 math is definitely one of the greatest inventions (or maybe "discoveries" is the right word) of all time! Allowed for so much stuff we would never be able to have otherwise. Like that phone you carry around in your pocket ;)

@diynevala

2 жыл бұрын

@@MrSonny6155 It takes science to make an atom bomb. :)

Not only does this generalise to _n_ hooks, it also answers the question "How long is a piece of string?" Turns out, very long if you use more than 2 hooks.

@abhiyadav9633

4 жыл бұрын

In this solution, if you increase number of hooks by 1, the string length needed is more than double of what was being used. String length grows exponentially with number of hooks. But there could be a better solution for which string length required is polynomial in n. Harder problem would be to find such a solution or to prove that no such solution exists.

@nicholasgunter3838

4 жыл бұрын

It takes much less string if your hooks are infinitely close together and your string is thinner than the gaps

@WhattheHectogon

4 жыл бұрын

@@nicholasgunter3838 that would just be multiplying 2^x by a tiny tiny scalar, but it still blows up exponentially :P

@kevinorr54

4 жыл бұрын

Interestingly, the multiple isn't super tiny. It's an interesting problem to find the closed form solution of f(1) = 1, f(n+1) = 2*f(n) + 2, which produces the sequence {1, 4, 10, 22, 46, 94, ...}. Answer: f(n) = 3/2 * 2^n - 2

@filipsperl

4 жыл бұрын

hi again :)

"We should thank Tom Scott eventhough he has done literally nothing" well. Thanks for nothing also is a phrase that exists.

@sinom

4 жыл бұрын

@@monikafiori that's the joke.

@rosebuster

4 жыл бұрын

Aw, I was looking for a comment from the real Tom Scott somewhere under this video.

@anandsuralkar2947

4 жыл бұрын

Lol i was literally reading ur comment while he was saying the exact same words cool

@Henrix1998

4 жыл бұрын

First take I guess

@AaronTheGerman

4 жыл бұрын

Everyone go on random Tom Scott videos and comment "Thanks for nothing"

4:17 me when doing a group project but don’t want to do actual work

@skippyflatrock

4 жыл бұрын

lol

@hrishikeshkulkarni9450

3 жыл бұрын

lol

@krzyche86

3 жыл бұрын

hahaha

Only when the standupmaths theme started playing I realized which channel I was watching.

@T3sl4

4 жыл бұрын

Steve's apartment and/or pipe is very distinctive, it's true

@KaliTakumi

4 жыл бұрын

Same lol

@gorikbienstman6240

4 жыл бұрын

wie we daar hebben

: "I have generalized this problem mathematically " : *Draws a minion.*

@bensmith9253

4 жыл бұрын

Lol

@gorillaau

4 жыл бұрын

Can we let the minion work the rest of it out?

@aditya95sriram

4 жыл бұрын

Lol, I think Steve's n=1 solution can be counted as Stuart

@aceman0000099

2 жыл бұрын

Amogus

The optimal solution for 4 hooks: ABA̅B̅CDC̅D̅BAB̅A̅DCD̅C̅ (16 loops). Not hard to work out and can be generalised. The generalised formula is: solution = x₁x₂⁻x₁⁻x₂ such that ⁻xᵤ cancels out xᵤ when concatenated (xᵤ⁻xᵤ = null). That is solved by writing xᵤ in reverse and inverting each letter. 1 hook: A 2 hooks: ABA̅B̅ (straightforward substitution for x₁x₂⁻x₁⁻x₂ where x₁=A, x₂=B) 3 hooks: ABA̅B̅CBAB̅A̅C̅ (x₁ = ABA̅B̅ and x₂ = C. ⁻x₁ must cancel out x₁, so is = BAB̅A̅ which is x₁ written backwards with each letter negated). 4 hooks: ABA̅B̅CDC̅D̅BAB̅A̅DCD̅C̅ (x₁ = ABA̅B̅ and x₂ = CDC̅D̅. This is the shortest solution another solution is x₁ = ABA̅B̅CBAB̅A̅C̅, x₂ = D) 5 hooks: ABA̅B̅CDC̅D̅EDCD̅C̅E̅BAB̅A̅ECDC̅D̅E̅DCD̅C̅ (x₁ = ABA̅B̅, x₂ = CDC̅D̅EDCD̅C̅E̅. Again this is the shortest solution, another solution is x₁ = ABA̅B̅CDC̅D̅BAB̅A̅DCD̅C̅, x₂ = E) And so on.

@ArupRoy_fromPlanetEarth

3 жыл бұрын

your comment deserves to be pinned (or up there)

@qxtr5853

3 жыл бұрын

Legend

@sofakartoffel1392

3 жыл бұрын

That's an awesome generalisation of their generalisation =D

@AractusPuphlicus

3 жыл бұрын

@@sofakartoffel1392 I honestly don't understand why Matt didn't find the optimal solution, typical Parker solution it's half-baked! Matt's solution for 4 hooks was as I mentioned x₁ = ABA̅B̅CBAB̅A̅C̅ (or the equivalent thereof), x₂ = D but with the generalised solution of x₁x₂⁻x₁⁻x₂ it's easy to see that there's a more optimal substitution to solve the equation using the solution found for 2 hooks as the the substitution. You can use the 5 hook solution to generate a 112 loop solution for 10 hooks, you can also substitute the solutions for 4 hooks and 6 hooks into x₁x₂⁻x₁⁻x₂ to get a 112 loop solution as well and either is optimal.

@tubebrocoli

3 жыл бұрын

so the smallest solution in number of loops with 10 hooks would be made of 2 solutions of 5 hooks, meaning 4 lots of 28 loops. This naively looks like the first step in order to optimize for smallest rope length, but I suppose there's probably at least some extra step in how we order the loops themselves, they don't intuitively seem to all use the same amount of rope due to potential distant jumps... the way we position the hooks among themselves probably also makes a difference.

0:00 Now you got me curious so i went on a spree to see if Steve ever makes contact to those pipes. and HAH I found it! On the video named "Why November is the most dangerous month for trains" at 2:32 he touches the red pipe, so it does exists!

@irenezs5992

4 жыл бұрын

No, it was expected

@soupisfornoobs4081

4 жыл бұрын

The power of determination

@douglasjackson295

3 жыл бұрын

I checked the video and he does

@ShaunakDe

3 жыл бұрын

I needed to know this in my life. Thank you for this.

@DanielSultana

3 жыл бұрын

Why didn't you link the video?

Now ask James Grime if he can come up with a 4th different method!

@nanamacapagal8342

4 жыл бұрын

13:49 maybe a 6th method. (gold plated goof did a 5th method (same principle as matt, different generalization)) He did A B A' B' C D C' D' B A B' A' D C D' C' (16 moves compared to Matt and Steve's 22) Where A' = A inverse

@aceman0000099

2 жыл бұрын

I thought the name James Grime was a play on Steve Mould at first hahahaha Now we need a guy called Adam Filth or something

@mrgabgob4675

2 жыл бұрын

@@aceman0000099 😆

Abstract: a commutator-based approach reduces the cost of the n-hook problem to quadratic, instead of exponential. Also, numerical investigation rules out non-commutator-based solutions for 3 hooks. This is further evidence against the possibility of non-commutator-based approaches. In this post, a "sequence" is a sequence of elements of the form "loop around a specific hook, clockwise (or counter-)". Letters (a, b, ...) indicate such elements. The order of a sequence is the number of hooks in the problem. The length of a sequence is the number of its elements. The inverse of an element "loop around hook x, with this orientation" is "loop around the same hook x, with opposite orientation". The inverse of a sequence is the sequence of the inverses of its elements, in inverse order. For example: "a b c" has inverse " (inverse c) (inverse b) (inverse a)". Notice that in this way a sequence cancels out with its inverse. A commutator, indicated as [a, b], is the sequence "a b (inverse a) (inverse b)". Notice that a commutator's length is 2*(length of a + length of b). Notice that erasing any element from a commutator makes everything else cancel out. This way it is possible to prove that any sequence constructed entirely of commutators will solve the hanging picture problem. Further notice: [[[a, b], c], d] has cost 22, while [[a, b], [c, d]] has cost 16. Nesting commutators is costly: a better result is achieved "flattening" the commutators. This improves enormously over Parker's method. For sequences of order 6: [[[[[a, b], c], d], e], f] costs 94. In general, for n elements, this costs 2^n + 2^(n-1) - 2. [ [[a, b], c] , [[d, e], f] ] costs 40. For 10 elements, as they want to do: Parker's method costs 1534, while [ [ [[a, b], c], [d, e] ] , [ [[f, g], h], [i, j] ] ] costs only 112. Nesting commutators this way is easy of you have a power of 2 amount of hooks. In this case, the cost for n=2^k hooks satisfies: Cost(k) = 4*Cost(k-1) This is because if n hooks produce a pattern, 2n hooks produce [pattern, pattern] i.e. the commutator of two distinct copies of the original pattern, with cost 4 times the original cost. This recurrence relation reduces to: Cost(k) = 4^k, and since n=2^k, k=log2(n): Cost = 4^(log2(n)) = n^2. This applies only for power-of-2 integers, but non-power-of-2 integers are easily bounded above by their next bigger power of 2. Credits to user mstmar for the idea of nesting commutators. For three hooks, the 10-long sequence found in the video is of minimal length - I checked all combinations via a script. (I can provide the ugly unoptimized python code upon request). This leads me to suspect that the commutator-based approach cannot be improved upon, except by rearranging the commutators. While there are many optimizations to be made, I think doing a search for n=4 is unfeasible without industrial-size compute, and unlikely to yield significant improvement. The size of any pattern of order power-of-2 is directly proportional to the size of the pattern of order 4 that underlies it, but the current order 4 pattern has length 16 and no significant improvement appears likely. Finally, for the real-life version of the problem that takes into account the real-world distance between hooks, further work may be required. Thank you for your attention.

@kevinhart4real

4 жыл бұрын

🤔

@tavern.keeper

4 жыл бұрын

You're basically solving the 2-hook problem many times and nesting the solutions in a binary tree. And I think the strategy is the most efficient if all you've got it the same 2-hook solution. I wonder though if there are other ways. Eg, are there any 3-hook solutions that don't look like [[a,b],c]?

@ElchiKing

4 жыл бұрын

On the other hand, optimizing the commutator based approach is quite easy: If (N) is the minimal cost for a commutator based approach on N hooks, then (N)=min{2*(k)+2*(N-k)|1=2*(k)+2*(N-k) (since A and B must in total contain at least N letters). On the other hand, for A and B as optimal commutators for k and N-k, we get equality here, hence 2*(k)+2*(N-k) is a minimum for A containing the fixed number of k letters. Finding k such that this term gets minimal thus yields the minimum cost for N. Using dynamic programming we can find these values efficiently.

@ElchiKing

4 жыл бұрын

Ok, I implemented it and it looks like this: If the biggest power of two smaller or equal to N is P, then the/an optimal split is P/2, N-(P/2). e.g. N=10: [10]=[4,6]=[[2,2],[2,4]]=[[[1,1],[1,1]],[[1,1],[2,2]]]=[[[1,1],[1,1]],[[1,1],[[1,1],[1,1]]]] has cost 112. written down: a^{-1}=A [4]=abABcdCDbaBAdcDC [6]=efEFghGHijIJhgHGjiJIfeFEijIJghGHjiJIhgHG [10]=abABcdCDbaBAdcDCefEFghGHijIJhgHGjiJIfeFEijIJghGHjiJIhgHGcdCDabABdcDCbaBAghGHijIJhgHGjiJIefEFijIJghGHjiJIhgHGfeFE

@ElchiKing

4 жыл бұрын

@Neel Shukla Actually, as I read some other comments, I saw that any split in the range (n/2-p/2,n/2+p/2) is optimal, my cut just happened to be the smallest optimal cut.

You just need to arrange hooks from the middle to sides, like 9, 7, 5, 3, 1, 2, 4, 6, 8, 10 to have final string shorter. Just because first hooks are being meet in algorithm more frequently. Also I think that the most suitable thread for this purpose is one of those used for fishing. They have some cords which flex not much and they designed to reduce friction

@anandsuralkar2947

4 жыл бұрын

Cool

@jacksonpercy8044

4 жыл бұрын

Seems pretty obvious in hind-sight.

@xaytana

4 жыл бұрын

Fishing line would be terrible for visual representation.

@NoobLord98

4 жыл бұрын

@@xaytana Glow in the dark fishing line. turn of the lighs and you might see something.

@joshandrews8913

4 жыл бұрын

@@NoobLord98 Glow in the dark fishing line? What's the point of that? Definitely not fishing. The point of fishing is that the fish DOESN'T see anything suspicious.

Steve was incredibly whelmed by that gift. It was hilarious. Thanks guys. You're always great together.

@filipsperl

4 жыл бұрын

He's got the humble tau poster already

@zeikjt

4 жыл бұрын

@@filipsperl Pretty sure the Tau is just this gift but modded since it has the same signature and note from Matt.

@yousorooo

3 жыл бұрын

Just whelmed. Not overwhelmed or underwhelmed. Just whelmed.

@BenKonosky

2 жыл бұрын

So, you can just be whelmed in Europe.

0:15 Steve: It's good to have you. Matt: I KNOW!!!

I love it when KZreadrs I watch reference other KZreadrs I watch guest starring on a KZread channel I watch

You don't actually have a face? No I add it in post every time *Face disappears* "It's very convincing"

"great to have you." "I know!"

@Talaxianer

3 жыл бұрын

0:14

I just love the fact that these two are such great friends and can geek out together about some maths problem and enjoy every minute of it. 😊

16:49 "So the challenge is for some large number of hooks, what's a really short way of doing it." Challenge accepted: My suggestion to extend from n to n+1 hooks would be that: 0) Use solution for n hooks 1) search first least used character x (y is the successor of x in natural order == A 2) Shift every bigger character (in natural order) by 1 3) Replace first letter used least with solution for two hooks x and y From 2 to 3 hooks: 0) use: A B A̅ B̅ 1) x = A, y = B 2) shift => A C A̅ C̅ 3) replace A with A B A̅ B̅ and A̅ with B A B̅ A̅ => A B A̅ B̅ C B A B̅ A̅ C̅ length: 10 cord length: 10 From 3 to 4 hooks: 0) use: A B A̅ B̅ C B A B̅ A̅ C̅ 1) x = C, y = D 2) shift => A C A̅ C̅ 3) replace C with C D C̅ D̅ and C̅ with D C D̅ C̅ => A B A̅ B̅ C D C̅ D̅ B A B̅ A̅ D C D̅ C̅ length: 16 cord length: 18 From 4 to 5 hooks: 0) use: A B A̅ B̅ C D C̅ D̅ B A B̅ A̅ D C D̅ C̅ 1) x = A, y = B 2) shift => A C A̅ C̅ D E D̅ E̅ C A C̅ A̅ E D E̅ D̅ 3) replace A with A B A̅ B̅ and A̅ with B A B̅ A̅ => A B A̅ B̅ C B A B̅ A̅ C̅ D E D̅ E̅ C A B A̅ B̅ C̅ B A B̅ A̅ E D E̅ D̅ length: 28 cord length: 33 Similar from 5 to 10 hooks (use solution for 5 hooks and insert solution for 2 hooks; imlicit shifting using appropriate letters) use: A B A̅ B̅ C B A B̅ A̅ C̅ D E D̅ E̅ C A B A̅ B̅ C̅ B A B̅ A̅ E D E̅ D̅ replace E with I J I̅ J̅ and E̅ with J I J̅ I̅ => A B A̅ B̅ C B A B̅ A̅ C̅ D I J I̅ J̅ D̅ J I J̅ I̅ C A B A̅ B̅ C̅ B A B̅ A̅ I J I̅ J̅ D J I J̅ I̅ D̅ replace D with G H G̅ H̅ and D̅ with H G H̅ G̅ => A B A̅ B̅ C B A B̅ A̅ C̅ G H G̅ H̅ I J I̅ J̅ H G H̅ G̅ J I J̅ I̅ C A B A̅ B̅ C̅ B A B̅ A̅ I J I̅ J̅ G H G̅ H̅ J I J̅ I̅ H G H̅ G̅ replace C with E F E̅ F̅ and C̅ with F E F̅ E̅ => A B A̅ B̅ E F E̅ F̅ B A B̅ A̅ F E F̅ E̅ G H G̅ H̅ I J I̅ J̅ H G H̅ G̅ J I J̅ I̅ E F E̅ F̅ A B A̅ B̅ F E F̅ E̅ B A B̅ A̅ I J I̅ J̅ G H G̅ H̅ J I J̅ I̅ H G H̅ G̅ replace B with C D C̅ D̅ and B̅ with D C D̅ C̅ => A C D C̅ D̅ A̅ D C D̅ C̅ E F E̅ F̅ C D C̅ D̅ A D C D̅ C̅ A̅ F E F̅ E̅ G H G̅ H̅ I J I̅ J̅ H G H̅ G̅ J I J̅ I̅ E F E̅ F̅ A C D C̅ D̅ A̅ D C D̅ C̅ F E F̅ E̅ C D C̅ D̅ A D C D̅ C̅ A̅ I J I̅ J̅ G H G̅ H̅ J I J̅ I̅ H G H̅ G̅ replace A with A B A̅ B̅ and A̅ with B A B̅ A̅ => A B A̅ B̅ C D C̅ D̅ B A B̅ A̅ D C D̅ C̅ E F E̅ F̅ C D C̅ D̅ A B A̅ B̅ D C D̅ C̅ B A B̅ A̅ F E F̅ E̅ G H G̅ H̅ I J I̅ J̅ H G H̅ G̅ J I J̅ I̅ E F E̅ F̅ A B A̅ B̅ C D C̅ D̅ B A B̅ A̅ D C D̅ C̅ F E F̅ E̅ C D C̅ D̅ A B A̅ B̅ D C D̅ C̅ B A B̅ A̅ I J I̅ J̅ G H G̅ H̅ J I J̅ I̅ H G H̅ G̅ length: 112 (= 28 * 4) cord length: 154 I hope, there are no flaws... Edit: just noticed, i posted the length of the rule (in letters) instead that of the cord, so i added the cord length too. Hopefully that's not irritating.

@nosignal5804

4 жыл бұрын

Neat! There are no flaws that I can find. But what is step 2 "Shifting", I didn't understand what do you do in it.

I found a way to get shorter sequences. I'm sure it's not optimal, but at least better than the one presented. My approach does xyXY where x and y are solutions for smaller problems and capitals are inverses. your approach is a special case of mine where x contains all hooks but one and y is the last hook. my approach checks all combinations of x and y to find the best partition of hooks that gives the lowest length. an example for n = 4 can split its hooks 1/3 or 2/2. The 1/3 case gives your solution to n = 4 and the 2/2 case is abAB cdCD baBA dcDC, length 16 vs 22, so 2/2 split is chosen. I'm pretty sure that having balanced hooks (or 1 off for odd n's) gives the smallest length (for this approach anyways). however, for n=6 you can get a length of 40 in 2 ways (3/3 split or 2/4 split). so for big n's, there might be a better split than even. an other advantage of this approach is that you stay localized the same half of hooks for each 1/4, most likely saving even more rope (in addition to needing less loops), compared to jumping to the end and back every 4 loops. this approach lowers the n=10 that you wanted to attempt from 1534 loops to 112, a much more manageable feat. oddly, n = 10 also has 2 ways of getting to 112 loops: 5/5 or 4/6. It also would have saved the 5 people on the stage a little trouble too, as it would only have taken 28 loops instead of the 46. Could have even added a 6th person with a similar tangle (40 loops).

@jordanlinus6178

4 жыл бұрын

After studying it for some time, I can prove by induction that taking balanced splits (1 off for odd n) is always best. It is also relatively easy to show by induction that if we call f(n) the number of steps needed (with this method) is always at least n^2, with equality if and only if n is a power of 2. What is much more surprising is that we can always spread a split until we reach a power of 2, for example f(12) = 2(f(6) + f(6)) = 2(f(5) + f(7)) = 2(f(4) + f(8)) This explains why we often have several solutions, especially for large n (but remember, we can't be better than an even split). Still, as it only grows as n^2, this is much more efficient than their solution, which grows as 2^n. You can easily show f(n) < 4n^2 (take the smallest k such that 2^k ≥ n, and check that f is strictly growing, then we have f(n) < f(2^k) = (2^k)^2 < (2n)^2 = 4n^2), and with some work f(n) < 5/2n^2 (first prove that f(3*2^k) = 10/9(3*2^k)^2, then use the same argument as before). It even seems like f(n) < 1.3n^2, which would be really close to n^2.

@rikwisselink-bijker

4 жыл бұрын

This reasoned approach is much better than my first thought: brute force. With a solution length that is at most approximately 2^n we get a search space of only about a 1000 long for n=10, so that should be feasible, right? Nope. That ignores the fact that with a pure brute force method (n hooks and k loops for your shortest solution) you actually have a search space of (2n)^k (both directions for each hook, repeat pick k times). For n=10, k=112, and t_check=50ms, a pure brute force would finish after about 8e135 years, which is well after we expect supermassive black holes to have evaporated into Hawking radiation (1e100 years). Turns out a pencil can still beat a CPU.

@cubethesquid3919

4 жыл бұрын

Following this thread

@swerasnym

4 жыл бұрын

This is also the solution given in this paper: arxiv.org/pdf/1203.3602.pdf

@Devilogic

4 жыл бұрын

@@swerasnym That paper gives the algorithm but does not establish optimality, it only conjectures it. This paper also establishes its optimality and counts the number of different minimal-length solutions: eudml.org/doc/282667 . The provenly optimal lengths of solutions for n nails are also given by the OEIS sequence A073121. Namely, for n = 2^m + k with k Intriguingly, all of this applies only to the case when the string is only allowed to loop aroud the nails, but not around itself (giving a "simple Brunnian link" for solutions of the puzzle, in the parlance of the paper). If loops of the string around itself are allowed (i.e. "knotting"/linking the string with itself) then only a linear number (O(n)) of crossings of the string are required (see Fig. 8 in your cited arXiv paper), not the quadratic number (O(n^2)) if these aren't allowed. Your cited arXiv paper gives 8*n as a candidate solution in this case but does not claim it is optimal. I don't know if it is an open question or not as to what the lowest number of string crossings required is, but it does seem interesting to consider. :)

There's a nice topological way to view this. You're looking at the fundamental group of the plane with n-many holes and discussing what elements of the group, when projected onto the plane with (n-1)-many holes, return the trivial element of the group. To be more precise, the fundamental group of the plane (or, homotopically equivalent, the n-petal graph), is the free group generated by n-many variables. Consider then the projection map, say, called P_i, from F(x_1, x_2, ..., x_n) to F(x_1, x_2, ..., x_(i-1), x_(i+1), ..., x_n). This is a group homomorphism, and its kernel is the subgroup generated by g^m * x_i^n * g^(-m), for g in F(x_1, x_2, ..., x_(i-1), x_(i+1), ..., x_n). The problem you're asking about in the video, or just the general problem, is finding elements of the intersection of the kernels of all the project maps P_i. I gave this a shot and came up with the same answer found in the video, i.e. a middle element serving as a barrier between a string of mirrored inverses on each side.

I've got a solution to the scaling problem which leads to fewer symbols (e.g., less rope) for longer sequences. Conceptually, rather than focusing on adding new pegs at the end, I imagine that the pegs we already have are getting duplicated: so the rope that used to go around one peg is now going around two pegs together. The question then becomes how to ensure that the picture falls if either of those two pegs is removed. We can consider going from the trivial one-peg case to the two-peg case to be exactly this problem, and so can adapt the two-peg solution. To be specific, any time we have a clockwise turn which we want to turn into turns around two pegs, we replace it with AbaB (where A is the first peg, B is the second, and capital letters mean clockwise turns while lower-case mean counterclockwise). Likewise, we replace any counterclockwise turns with bABa. This will quadruple the total number of symbols, but in the long run, quadrupling the number of symbols whenever we double the number of pegs is better than (more than) doubling the number of symbols when we add a single peg. To give an example of how effective this is, here's my four-peg solution, as before with capital letters being clockwise turns and lower-case being counterclockwise: AbaBcDCdbABaDcdC If you remove any of the pegs (e.g., remove any of Aa, Bb, Cc, or Dd from that sequence), the rest will collapse. This 16-symbol sequence is shorter than the 22-symbol sequence presented in the video, already demonstrating the effectiveness of this approach. The savings would be even more evident if going to higher numbers of pegs.

Now Humble Tau in Steve's light video makes sense

@VaradMahashabde

4 жыл бұрын

It's humble pi btw 😂😂😂

@Maninawig

4 жыл бұрын

@@VaradMahashabde if you check Steve Mould's newest video about levitating water, the picture is featured in the background. At 6:37, he edited it to say "Humble Tau" for a bit, being a refference to a Numberphile video where Matt and Steve debated for which is better. (Steve stating Tau is better abd Matt stating it depends on 2Pi or half Tau)

Funny, that solution looks a lot like commutators in rubik's cube The notation for them is [A; B] and that means ABA'B' So the solutions with 3 hooks can be noted [[A; B] ; C] I wonder how you could add conjugates, another tool in rubik's cube ([A: B] = ABA') in the picture frame story...

@bobthegiraffemonkey

4 жыл бұрын

That's because they are commutators! I'm not sure how to add conjugates to the story either.

@JNCressey

4 жыл бұрын

I wanna look at [[[A;B];C];D] vs [[A;B];[C;D]]. [[[A;B];C];D] = ABA'B'CBAB'A'C'DCABA'B'C'BAB'A' [[A;B];[C;D]] = ABA'B'CDC'D'BAB'A'DCD'C' hmmmmmm... dat one's shorter...

@MyrciaAshlan

4 жыл бұрын

There is another video that explains that this is exactly the case. kzread.info/dash/bejne/oGGqmNKmit3efJM.html

@HagenvonEitzen

4 жыл бұрын

@@JNCressey We want a nested commutator involving every variable (i.e., every hook) and that is as "shallow" as possible (to minimize the expanded length). That's how [[[A;B];[C;D]];[[[E;F];[G;H]];[I;J]]] gives us a solution with only 112 loops for 10 hooks

@AlgyCuber

4 жыл бұрын

fellow cubers

The red pipe is going to keep me in suspense for a long time

@anandsuralkar2947

4 жыл бұрын

Same

@KartheekTammana123

4 жыл бұрын

When the pipe vanishes at the beginning, you can see a little bit of the pipe that he wasn't able to crop out

@Jkirek_

4 жыл бұрын

@@KartheekTammana123 Ah, there he got you: he wasn't able to crop out a piece of pipe that he added in before. The pipe still isn't there.

@KartheekTammana123

4 жыл бұрын

@@Jkirek_ Touché

@barongerhardt

4 жыл бұрын

What about the framed Humble Tau by Patt Marker?

Now, I occasionally make a habit of constructing a fractal called a dragon curve. It's made with a series of clockwise and anticlockwise 90 degree turns. Each increasing step adds a clockwise turn, then mirrors every previous step in reversing order, inversed the same way as done in this video when demonstrating the addition of D. That's fascinating to me.

@inigo8740

4 жыл бұрын

Everyone talking about commutators, yet here you are bringing in fractals to the show. I love it.

8:35 In Parker notation, Steve is actually putting a C at the beginning, not the end here because he grabbed the other string.

@justanormalyoutubeuser3868

3 жыл бұрын

It's not meant to be exact, it's a Parker notation

@debblez

3 жыл бұрын

fake fan... that’s actually an anti-C... smh

You both are brilliant, thanks for showing all the details of your thought process. learned how to think in terms of maths

"Shout out to Tom Scott for doing nothing."

Watching a new Standupmaths video just makes my day every time!

When you step up to 3 all you are doing is treating the A and B hooks as 1 hook and treat it like 2 hooks again: AB and C Is this true?

@walterkipferl6729

4 жыл бұрын

omegadan Should be, and the „reading it in reverse and negativ“ is simply the result of winding counterclockwise around a combined hook.

@PhilBoswell

4 жыл бұрын

That's interesting. So is the solution for ABC·D the same as AB·CD? In other words, if you make a four-hook solution by combining a three-hook solution with an extra hook, do you get the same as combining two two-hook solutions together?

@Gihntemos

4 жыл бұрын

That's my thought as well, by grouping like that, you should be able to get any number of hooks by treating A•B•C•D as A•B hook and C•D hook .

@vitomarchino

4 жыл бұрын

@@PhilBoswell YES! THAT WORKS!!

@cealvan8941

4 жыл бұрын

The AB•BC solution only has 16 rather than Matt's 22

Interestingly, there's a lovely acronym that explains exactly Matt's algorithm. DRII: Duplicate, Reflect / Reverse (makes abAB -> BAba), Inverse (makes abAB -> ABab), and Inject (makes abAB + baBA -> abABcbaBAC). DRII can be expanded to the nth integer term, which I think is really interesting.

when you ask a man of physics and a man of mathematics to solve a question...you find different answers...then things comlicate when you add a compurer programmer to the mix...SCOTT SOLVE THIS QUESTION!(by scott i mean tom scott)

@satibel

4 жыл бұрын

Tom solves the question using linguistics.

I deeply hope that Steve's follow up video features your lovely framed picture. But with Pi scribbled out and replaced with Tau.

@qwertyasdf66

4 жыл бұрын

He already did that in the background if his most recent video.

Now we have a new pattern for completing multiple-choice tests!

because you guys seemed to be coming across stuff in real time, this watches almost like a visual podcast - was a really cool video style! hope to see more similar stuff in the future, keep it up guys!

i love the way you guys understand each other yet have completely different approaches.

"good to have you" ,"i know" loool

@theunpopularcuber9554

4 жыл бұрын

What does loool stand for?

@uwootmviii8695

4 жыл бұрын

@@theunpopularcuber9554 laughing out oridinary loud

7:40 "How to draw angry robots with Matt and Steve"

@JNCressey

4 жыл бұрын

Or how to draw bikini tops

@sixdfx

4 жыл бұрын

I was thinking of Dr. Zoidberg

@Jamie_kemp

4 жыл бұрын

I thought it looked like squidward with glasses

The frame we spotted in last video haha!

@ToMeK3001pro

4 жыл бұрын

?

I love when the time lapses start cause I get to listen to a great song everytime

honestly seeing Matt’s method and being able to compare it with what Steve and Jade have done, I’m quite amazed by how a flexible thing this naughty knotty problems actually are. I mean steve’s method is something I would never think off knowing myself but what Matt has done is like a „method to madness” kind of approach. God i love how with those carts you completely forgot about the physical situation by just knowing those two clowiseness-hook axioms God i just had to write this down somehow cause it is kind of a illumination type a moment for me

It's the magicians knot yall.

I'm surprised that you guys didn't point out the nice expandable.. almost fractal-like pattern that you're doing; You're even wearing a recursive T-Shirt! ( AB A'B' ) Just becomes the new A, and B is the next pin, and you can expand that out forever. It's the same for ( AB' A'B )

@georgesmith4768

4 жыл бұрын

L system!

@Spacepixel1

4 жыл бұрын

@@georgesmith4768 I'll be honest, I didn't know there was a name for that. Thank you!

Your solution(s) is very elegant! It's a really nice problem, love it.

It's not worse than doubling, because as the number of hooks goes to oo, it _basically_ becomes doubling. So it's still O(2^n). Big O Notation

@josephcunningham5882

4 жыл бұрын

Asymptotically it may be no worse than doubling. Here we are talking about no more than 10 hooks

@Geevs80

4 жыл бұрын

I have no idea what you are saying and I believe everything you said

@justanormalyoutubeuser3868

3 жыл бұрын

It's doubling and adding 2

I love that these two and Tom Scott are all aware of each other. That would be the most epic collab ever, right?

Gotta admit, that looks like a superpermutation with how horrific the sequence is. That being said, the superpermutation for n is still much, much more horrifying...

God, I love Matt's theme song. One of my absolute favorites and it lives in my brain permanently. I would say "rent-free", but it's probably a drag on my productivity to be humming it so often.

The ending bit reminded me a LOT of a solitaire card game called "Accordion", where similar sandwich/removal occurs. I'm wondering if some of the thought process of playing that game would come into play here.

WARNING: long comment ahead, but I believe there is an efficient algorithm for *doubling* the number of hooks at each step, rather than adding 1 at a time. I've tried to explain it as coherently as possible over text, and I'd greatly appreciate anyone who could verify this method or point out any mistakes. *ALGORITHM EXPLANATION* Steps of the algorithm: Step 1: Start with the 2 hook solution: A B A' B', where A stands for a clockwise string turn around hook A, and A' is anticlockwise (counterclockwise for us 'muricans) Step 2: Consider an entirely separate 2 hook solution, with hooks C and D. of course, the solution is still the same: C D C' D' Step 3: Consider both separate two hook systems as two individual hooks, let's call them X and Y. Step 4: Naturally, the solution for the XY system is: X Y X' Y' Step 5: Substitute our original hooks, ABCD, into the new solution. BUT WAIT: we know X and Y, but what does X' and Y' look like? that's simple: X' is just everything in X, but in reverse order. So if X = A B A' B', then X' = B' A' B A. this ensures that X and X' will fully cancel each other out when they "collide" Final result: X Y X' Y' becomes: A B A' B' C D C' D' B' A' B A D' C' D C You can test out yourself: remove any one of the hooks, and the others will collapse and cancel each other out. *COMPARING THIS ALGORITHM WITH MATT AND STEVE'S ALGORITHM* Now, the interesting thing to note is how many "turns" of string we have: the final solution ends up with ONLY 16 turns, whereas Matt and Steve's algorithm resulted in 22 turns for 4 hooks. The equations for # of turns vs. number of hooks for each algorithm are as follows: Matt/Steve algorithm: t(n) = 2 * t(n-1) + 2 (where: n > 1) (in layman's terms: adding 1 hook causes the number of turns to double, plus 2) My doubling algorithm: t(n) = 4 * t(n/2) (where n >= 4, and MUST be a power of 2) (in layman's terms: doubling the number of hooks quadruples the number of turns) *TACKLING THE CHALLENGE OF 10 HOOKS* Of course, you can use this algorithm to continue doubling the number of hooks as much as you want. For example: if you want 8 hooks, simply take the solution for 4 hooks above (X Y X' Y') and name it as a single hook, lets say W. Then, let's also define V, which is another system of 4 hooks. Then, a system of 8 hooks has the solution W V W' V', with a total of 16 * 4 = 64 turns. Regarding Matt and Steve's challenge of 10 hooks: Using my algorithm, you can solve an 8 hook system with 64 turns of string, but unfortunately, the efficient doubling method can't add the last 2 hooks. For those, we need to use the algorithm in the video. So, using the formula t(n) = 2 * t(n-1) + 2, we can see that: t(9) = 2 * t(8) + 2 = 2 * 64 + 2 = 130 t(10) = 2 * t(9) + 2 = 2 * 130 + 2 = 262 So I predict that a 10 hook system can be solved with at least 262 turns of string, but there may be a more efficient way to add those last 2 hooks to the system. I'll leave that to someone smarter to figure out ;)

@zaheenahmed304

4 жыл бұрын

After some thinking, I realized this can be generalized to not just doubling the # of hooks, but adding any amount of hooks (m) to an existing system of n hooks. Let X = existing system of n hooks, represented as a single hook. Let Y = new system of m hooks being added, represented as a single hook. Solution for X + Y is X Y X' Y' The new number of "turns" of string needed for this solution is: t(n+m) = 2 * t(n) + 2 * t(m) *(Generalized solution)* We can show that both Matt and Steve's algorithm, and my algorithm above, are special cases of this general solution: Matt and Steve's algorithm is where m = 1, so: t(n+m) = 2 * t(n) + 2 * t(m) = 2 * t(n) + 2 * t(1) = 2 * t(n) + 2 (since we know that the solution for n = 1 is trivial: just 1 turn of string) The doubling algorithm: t(n+m) = 2 * t(n) + 2 * t(m) = 2 * t(n) + 2 * t(n) = 4 * t(n) (since m = n) Given this, we can compute the minimum (theoretically) # of turns for 10 hooks: t(10) = 2 * t(8) + 2 * t(2) = 2 * 64 + 2 * 4 = 128 + 8 = *136* So we should expect 10 hooks to be solved with at least 136 turns of string

@mstmar

4 жыл бұрын

Small mistake in step 5. X' isn't just X reversed, but also each term inverted. So if X = A B A' B', then X' = B A B' A' so that in XX' you get B'B, A'A... which cancel and not B'B', A'A'... which don't. Also as an addition to what Zaheen Ahmed said, t(10) can also be split other ways than 8/2. if we split it 5/5, we get t(10) = 4*t(5) = 4*(2*t(3)+2*t(2)) = 8*(10+4) = 112. Even more savings!

@verfmeer

4 жыл бұрын

@@zaheenahmed304 I was thinking about the same strategy, and wrote the sequences down. You are right. This is the 4 hook sequence (length 16), broken up in its four components: ABA'B' CDC'D' BAB'A' DCD'C'. And this is the 8 hook sequence (length 64), split up in fours as well: ABA'B'CDC'D'BAB'A'DCD'C' EFE'F'GHG'H'FEF'E'HGH'G' CDC'D'ABA'B'DCD'C'BAB'A' GHG'H'EFE'F'HGH'G'FEF'E'. We can use this to create the 10 hook sequence (length 136): ABA'B'CDC'D'BAB'A'DCD'C'EFE'F'GHG'H'FEF'E'HGH'G'CDC'D'ABA'B'DCD'C'BAB'A'GHG'H'EFE'F'HGH'G'FEF'E' KLK'L' EFE'F'GHG'H'FEF'E'HGH'G'ABA'B'CDC'D'BAB'A'DCD'C'GHG'H'EFE'F'HGH'G'FEF'E'CDC'D'ABA'B'DCD'C'BAB'A' LKL'K' You can clearly see the structure. First comes the 8-hook sequence, followed the 2-hook sequence, followed by the reverse 8-hook sequence, and the reverse 2-hook sequence seals the deal.

@zaheenahmed304

4 жыл бұрын

@@mstmar Thanks for catching that!

@zaheenahmed304

4 жыл бұрын

@@verfmeer Yep, that should be the sequence. I salute you for actually writing it out, I didn't have the patience for it haha (Now, we patiently await Matt and Steve attempting to do this with actual hooks and string...)

That switch is on in the background and it's killing me

I love the intuitive mathematical solution with the cards. that was very well done!

As you were laying out the cards I was also thinking of the Towers of Hanoi. 🤣

Still the only channel I'll turn notifications on for.

I love these two together

You can make them shorter by substituting your last expression into both a and b, not just one of them (i.e. doubling the amount of nails each time). This way you get around n^2 turns for n nails, not 2^n+2^(n-1)-2 like you get.

Man you’re really paying that steve mould premium

When I watched Jane's original I tried to correlate it to graph theory and got close to your cube solution using the chinese mailman problem, you guys should look into it for a possible smaller solution.

I love when you two talk about maths and you are always amazed at what the other just said :)

Very great explanations - all three of them.

11:54 could someone please make a GIF out of this?!

@restinpeace1916

3 жыл бұрын

Sorry about green dots (have no idea what's up with that) and late response.

Here is a paper formalizing this kind of things: erikdemaine.org/papers/PictureHanging_TOCS/paper.pdf I've had good use for it in a programming contest once. :)

@PhilBoswell

4 жыл бұрын

Erik Demaine is awesome, I watched his MIT Origami class and was blown away.

Brilliant to show how a problem can be aproached in many ways.

This is amazing. Thank you both. I love how concepts can be intertwined like this, and how one puzzle can manifest itself in different ways (ie, rope vs braiding vs towers of hanoi. I feel like it wouldn't be difficult to come up with (following either of these algorithms) a python/perl/c/intercal/whatever program to generate these anti-knotting sequences.... Thinking more about this, it's also very similar to the way you do commutators in puzzle cubes. (ie, do a sequence.. R F U, then do a change (B) then back out the sequence... U' F' R'

@yorgle

4 жыл бұрын

And now I'm thinking this is also related to charlieplexing...

I found simpler solutions than what your method gives for a bigger number of hooks (>3). You just have to encapsulate a group of hook into a single abstract hook with an algorithm that represent the clockwise an anti-clockwise turn. If one of the hook in the group is removed, the whole abstract hook is removed. And if a clockwise turn is followed by an anti-clockwise turn (or vice versa), everything is undone just like how a real hook would behave. Here an example with 4 hooks : - Your method in 22 steps : ABabCBAbacD CABabcBAbad - This method in 16 steps : ABab CDcd BAba DCdc I grouped the hooks A and B (respectively C and D) into an abstract hook with the abstract clockwise turn ABab (respectively CDcd). Here's the best results I get with the 10 first hooks : Note: the x operator means I combine two previous solutions and the +1 operator means I use your method to use the previous solution Hooks: Steps -> Combinations of Hooks 1: 1 2: 4 3: 10 4: 16 = 4*4 -> 2x2 5: 34 = 2*16+2 -> 2x2+1 6: 40 = 4*10 -> 2x3 7: 82 = 2*40+2 -> 2x3+1 8: 64 = 16*4 -> 2x4 9: 100 = 10*10 -> 3x3 10: 136 = 4*34 -> 2x5 We might add this sequence to the OEIS eventually ...

@DS-xh9fd

4 жыл бұрын

You can do 5 in 28, and therefore 10 in 112: ABab CDcdEDCdce BAba ECDcdeDCdc

@DS-xh9fd

4 жыл бұрын

Check out sequence A073121 in OEIS

@Cookie_Wookie_7

4 жыл бұрын

And you can do 9 in 88 because 4+5=9 and 2(16+28)=88

@clem494949

4 жыл бұрын

@@DS-xh9fd Yeah, it"s a solved problem A073121, A254575

Yo dawg, I heard you like commutators! What ever happened to GoldPlatedGoof?

@PhilBoswell

4 жыл бұрын

No idea, the latest thing I can see is a tweet from October 2018 :-(

I really love this kind of thing where it's a real solution, multiple methods, all the same results. In algebra and all the maths up from there, I have always struggled with that kind of thing, (i know rearranging a problem doesn't fundamentally change it, but messy hand writing and a tendency to misread characters really interferes with that) so seeing the same solution, solved with three different trains of thought is just awesome. It's why I really like Those videos where Pi is hiding everywhere, and why it would be. Great content! Just hit the part of the video with the N dimensional cubes and I am losing it.

Matt and Steve, such an excellent pairing!

You could group them, correct? If you need 10 hooks, simplify to 5 groups of 2, because it's effectively the same. Treat each group as a single hook, since removing one part of the group would remove the group as a whole. Since you've shown it possible to be done with 3 as well, you should be able to do any number, I would think.

@JBergmansson

4 жыл бұрын

Wow, cool, I think you are right! But although this makes the number of logical steps smaller, the actual dependencies of which hooks the string pass over do not get simpler.

@youtubeuniversity3638

4 жыл бұрын

So you'd only have to prove prime numbers!

@pseudotaco

4 жыл бұрын

Well yes, but since there is an algorithm, it should be possible for any number anyways, shouldn't it?

@JBergmansson

4 жыл бұрын

@@youtubeuniversity3638 Not really, that would only be true if we had to divide them evenly. Take 7 for example, it can be done by making three groups, with 3, 2 and 2.

@youtubeuniversity3638

4 жыл бұрын

@@JBergmansson Good point! Guess I looked too hard at the example given!

Here's a question: is there a way to do this so the weight of an object hung from the string is as evenly distributed as possible across the N points? My thinking is that this could have some sort of practical application where something is being suspended evenly and could be quick released. In the examples, the Nth point would have a significantly less amount of string around it than the earlier points. Maybe the points could even be extended into a second dimension to suspend 3-dimensional objects (or just orthogonal 2-dimensional objects).That bit would be wholly impractical but it's fun to ponder.

I've seen the puzzle a bunch of times so I wasn't that excited, but when Matt then showed it was analogous to pathing around n-cubes it suddenly became extremely exciting.

I liked how you showed it falling at the end with the cards

ABABCBABAC is like a Parker Thue-Morse sequence.

@thankyouforthismanysubscri4521

4 жыл бұрын

@@iykury So it's even worse then. It's like a Parker Parker Thue-Morse sequence. A Parker squared Thue-Morse sequence.

@JNCressey

4 жыл бұрын

You mean the Thue-Morse-Morse-Thue sequence.

Hi everyone! I hope you all are having a wonderful and blessed day. 🖤

I also loved how up and atom did that demonstration.

Maths at its finest! When there's multiple different methods of approaching a problem, using diverse areas of mathematics, and they all provide equally valid solutions! Love it!

isn't that book Humble Tau? was on Steve's recent video

@hammerth1421

4 жыл бұрын

Go away Tauist! No one wants you here!

@fanq_

4 жыл бұрын

@@hammerth1421 it's a joke? Steve had it in his video and made it say Humble Tau for a second or two

@KaliTakumi

4 жыл бұрын

@@fanq_ Their comment was also a joke

@krissp8712

4 жыл бұрын

☯

@pseudotaco

4 жыл бұрын

@@krissp8712 Oh come on, we don't need a Humble Tao here

I like how Steve Mold does all sorts of collabs. Not to mention his solo videos are peng.

I think there's a Harter-Heighway dragon curve turn map in here. Take what you've got, copy it, reverse it, invert it, and glue the mangled copy to what you started with, with a new right turn in the middle. It would take testing, but I think taking a right turn as a clockwise wrap would result in yet another solution to this problem.

Most interesting problem in a while! Thanks for sharing!

I'll dare to post a non-mathematical comment: Is this really the largest table in Steve's house?

Am I the only one that thinks Matt’s solution looks an awful lot like a commutator?

A few things: first, I was thinking permutations just before they said it (mostly because the cards reminded me of the permutation cards in the other video). Second, I am watching this on a laptop that is perched on a folding table exactly like the one in the video (my chair is merely similar to Steve's). Third, it was very satisfying to watch the cards annihilate each other (via card:anticard reactions) at the end.

You Don't need to double each time, for four you can go aba/b/cdc/d/bab/a/dcd/c/ You put in the new letter as a commutator with a previous letter. for replace a clockwise it goes in the form aba/b/ where a is the old letter and b is the new letter, and when replacing anticlockwise you go bab/a/. So starting from 1 hook 1 hook solution: a 2 hook solution: aba/b/ 3 hook solution: aca/c/bcac/a/b/ 4 hook solution: aca/c/bdb/d/cac/a/dbd/b/ 5 hook solution: aea/e/ceae/a/c/bdb/d/caea/e/c/eae/a/dbd/b/ however when moving from 2 hooks to three hooks it is more efficient to use your doubling method since adding the commutator adds more than double the letters. this changes the solutions to 1 hook: a 2 hook: aba/b/ 3 hook: aba/b/cbab/a/c/ 4 hook: aba/b/cdc/d/bab/a/dcd/c/ 5 hook: aea/e/beae/a/b/cdc/d/baea/e/b/eae/a/dcd/c/ doubling each odd number of hooks is also less efficient due to the fact that when using this commutator extension method you add three per letter you are commutating with whereas with the doubling method you multiply the terms by 2 and add 2.

How to do this from an engineer's perspective: make the strings weak enough to fail under any stress such as removing a hook.

@pfeilspitze

3 жыл бұрын

Or make the hooks weak enough that all N are needed to support the load, so that removing one causes a cascading failure of the other N-1 hooks.

16:08 My brother walked in with a gerbil and asks "why are you watching a video on how to hang a picture, why is it 18 minutes long and why are you 16 minutes in?" What am I supposed to say?

@aarondurst2151

4 жыл бұрын

Let him know that it relates to hanging the picture in space, and that the Gerbil Space Program is conditioning the gerbils to understand that hooks and loops will allow zero-G environments to sustain a hanging picture, on as many or as few hooks as needed, provided it is not hung with any of the methods mentioned in the video. Also, explain that 18 minutes is roughly the total attention span in minutes of a gerbil, and that you are 16 minutes in, but only by appearances, because it was paused a few times and you would have been done by now if you didn't have to explain the previous information.

Humble Parker a comedy of Matt’s errors

You never fail to completely change my outlook on life - thanks once again Matt!!!

Haha funny, I saw Tom Scotts Video of this topic 👌🏼

Should be able to make an arbitrarily large number of hooks solution, right? I.E. 3 hooks = AB'A'BCAB'ABA'C'. LET X = AB'A'B and X' = B'ABA'. We then substitute to: XCX'C' which is in the form of the 2 hook solution Using the symbols to make a 3 hook solution: XC'X'CDC'XCX'D'. Substituting the Xs back should give a 4 hook solution: AB'A'BC'B'ABA'CDC'AB'A'BCB'ABA'D'. I think this could be done in a script fairly efficiently, in case you want to make one that's 200 long or whatever.

@Qwerasd

4 жыл бұрын

itchykami yes, this is the algorithm (more or less) that they presented in this video. Their question is if there are solutions that produce shorter patterns than this algorithm does.

@sebastianmestre8971

4 жыл бұрын

It can be done really efficiently in the sense that writing a program that runs in time linear to the size of the solution is fairly easy. It cannot be done really efficiently in the sense that the final solution is exponential on the amount of hooks. Maybe a different family with a linear or quadratic (or, generally, polynomial) amount of turns exists, but we don't know it. Now THAT would be a lot more efficient. By the way, generating the solution for 200 hooks using this family of solutions would give something like 2^201 = 2*2^200 = 2*(2^20)^10 turns, which is about 2*(10^6)^10 or 2*10^60. Assuming you can process a billion (10^9) turns per second (reasonable estimate for a modern x64 CPU), and that your algorithms instantly finds the sequence and all it has to do is put it in memory, finding the solution for 200 hooks takes at least 2*10^60/10^9 s = 2*10^51s =~ 10^46 years (very rough approximation on this last step, i did all calculations in my head)

I'm so excited that they're so excited!

Expanding on Matt Parker pointing out that the solutions to the puzzle can be represented as a path through vertices on the cube, we can generalize to N dimensions by considering an n-dimensional cube with volume of 1 and a corner fixed at the origin and the axes defined by all the edges at that vertex. We have now the following observations: a) Any solution must pass through all the vertices that are distance 1 from the origin. b) No solutions can contain path segments that backtrack (ie, in the 3-dimensional case, (0,0,1) -> (0,1,1) -> (0,0,1) and the like would be backtracking). c) The distance between each vertex in the path must be 1. Axiom a) ensures all hooks are utilised, axiom b) removes isomorphic solutions, and axiom c) comes from the clockwise/anticlockwise turns around a hook translating to positive/negative direction of travel along the axis defined by the coordinate of the hook. The minimal solutions (shortest amount of string), are paths such that the magnitude of the vector from the origin is minimized at each step.* There is an algorithm to do compute this efficiently (O(n) in both space and time), but the youtube comments section are not large enough to contain it.

Matt saying "Maths!" with his hands in the air after noting that three people came up with the same solution different ways made me laugh so hard. 11:49 - 11:55

Well i think i have a great solution that works for any number of hooks but it requires one extra item sadly, a pair of scissors, now you might call that cheating but i tend to think its just a creative way of solving the problem :P

@TheLuuuuuc

4 жыл бұрын

How will pulling a hook make the picture fall down just because you have a pair of scissors lying around?

"But you can't have knots in higher dimensions" "You can, if you consider them as sheets."

I love the facts that this collaboration video is inspired by another colaboration video

That switch left on on the wall is bugging me...

Only in Matths can you talk about the same thing as if its 2 different things and convince all other Matthsers to go along with you~

My brain just freaked out when you pulled out the cards. So cool

I'm so sorry I have to bring this up. In the last seconds when you connected D and anti-D the order was reversed and therefore incorrect

@onemadscientist7305

4 жыл бұрын

No that's the point you need to reverse it