How to mathematically hang a picture (badly).
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Channel: Stand-up Maths
Views: 363,521
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Keywords: maths, math, mathematics, comedy, stand-up, picture, hang, matt parker, steve mould, up and atom, puzzle, knots, challenge
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Length: 18min 26sec (1106 seconds)
Published: Fri Aug 02 2019
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Use ' instead of bar to denote the inverse (for a single clockwise turn A this will be the usual counter-clockwise turn A'), Matt's general procedure to go from n to n+1 hooks is:
(n hook solution) A_(n+1) (n hook solution)' A_(n+1)'
Where when you take the inverse of a multi-step sequence you reverse the order and take in inverse of each item (like with matrices). E.g., (ABA'B')' = (B')'(A')'(B)'(A)' = BAB'A'
So, stepping up from 1 hook to 4 we get (using bold to indicate the solution from the previous line, or the full inverse of that solution):
I'll stop there, because there's another construction we can make easily for 4 hooks that only requires 16 steps: take two of the 2-hook solutions and put them together as if each pair of hooks were a single hook. We get (bolding just one of the 2-hook solutions for visual clarity):
(ABA'B')(CDC'D')(ABA'B')'(CDC'D')' = ABA'B'CDC'D'BAB'A'DCD'C'
I have a feeling this is a minimal-length solution for 4 hooks, but even if it isn't it definitely demonstrates another family of solutions that can have fewer steps than Matt's algorithm. Though it still uses the heart of Matt's algorithm.
I also think even this type of construction won't always produce minimal-length solutions. As the number of hooks increases, I would predict sporadic solutions would be likely to take some of the minimal-length solutions. Prime numbers of hooks also don't work well with my construction based on earlier solutions.
Edit: Fixed a copy-paste error.
Notice the cover says Humble Pi, not Humble Tau
I may have discovered a solution that's better (at least when the number of hooks gets larger) than Matt's solution.
Matt's solution takes ~2^n turns, whereas mine takes 2*(n^2). The only downside of my solution is that it only works for an odd number of hooks.
The solution I came up with was inspired by (but does not require) a property of this system that neither Jade on Tom Scott's channel nor Matt used: any solution sequence can be 'rotated' all you want. Since it's a loop of string, a CW loop at the beginning of the sequence can be canceled out by a CCW loop at the end of the sequence. So this means that the loop sequence abcbca = aabcbc = bcbc (I'm using bold for inverses since they're easier to read).
The general idea of this solution is that for n hooks labeled 0 through (n-1), we have n Segments X_0 through X_(n-1). Each segment X_i, when hook i is removed, collapses by itself. Then adjacent segments cancel quite neatly. I originally meant for segments equal distance from X_i to cancel with each other, but I ended up with half of each segment canceling with its neighbor in the next half.
I think my solution may be the most efficient one. The fact that it 'looks the same' from the perspective of every hook makes it more beautiful, at least to me. I can't figure out a neat way to write down the formula for this thing, so I'll just write out the patterns I got for 3, 5, and 7 and I think you'll get the picture.
(I'm using letters instead of numbered terms here, for readability)
3: cbabca acbcab bacabc
5: edcbabcdea aedcbcdeab baedcdeabc cbaedeabcd dcbaeabcde
7: gfedcbabcdefga agfedcbcdefgab bagfedcdefgabc cbagfedefgabcd dcbagfefgabcde edcbagffFRICK this simplifies to the trivial solution I'm a stupidddddd
EDIT: wait, nevermind, I not only got turned around, but also the algorithm I was using works for 3 but not for more.
3: bcacba cabacb abcbac
without a: bccb cbcb bcbc
without b: caca caac acac
without c: baba abab abba
Sorta pulled a parker square on that one.