Here's a mathematical sort of question that occurred to me last year. I've been pondering it off and on ever since, but not reached any useful conclusions; and I just came across my file of notes on it, and thought perhaps I should post it somewhere for other people to ponder as well.
Let m and n be positive integers. Suppose you have m sticks of length n, and you want to turn them into n sticks of length m, by cutting them into pieces and gluing the pieces back together. And the tricky bit is that you want to do this in a way that maximises the length of the shortest fragment involved in the dissection.
One obvious approach is to glue all your sticks end-to-end into one long stick of length mn, then cut that into n equal pieces. If you do that, the shortest fragment will have a length equal to gcd(m,n). So that's a lower bound on the solution in general, and in some cases (e.g. m=3, n=2) it really will be the best you can do.
But gcd(m,n) is not always the best you can do. For example, consider m=6 and n=5. The obvious gcd solution gives a shortest fragment of 1, but actually this case has a solution with shortest fragment 2: cut each of the six 5-sticks into pieces of length 2 and 3, then reassemble as three lots of 3+3 and two lots of 2+2+2. (In fact this was the case that first brought the problem to my attention.)
I also know that the best solution will not necessarily involve fragments of integer length. For example, consider m=5 and n=7. If you constrain yourself to only use integer-length fragments, then you can't make the shortest fragment any longer than 1. (If you try to make it 2, then you have no option but to cut up each 7-stick as 2+2+3, but then you have ten 2s and five 3s, out of which you can't make the seven 5s you need without bisecting a 2.) But with fractional fragment lengths you can improve on a shortest fragment of 1, by cutting up three 7-sticks as (8/3 + 8/3 + 5/3) and the other two as (7/3 + 7/3 + 7/3), then reassembling as six lots of (8/3 + 7/3) and one (5/3 + 5/3 + 5/3). So now your shortest fragment is one and two-thirds units long.
So, does anyone have thoughts? Other than the above examples, I haven't managed very many myself. I haven't even come up with a sensible computer search algorithm to reliably determine the best answer for a given m,n pair – so, in particular, I don't even know that 5/3 is the best answer for the example above, only that it's the best answer I know of. I have a vague idea that the best answer generally seems to involve at least one stick (on either the source or the destination side) being cut into equal-length fragments, but I wouldn't be at all surprised to find that wasn't true in all cases.
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As I was saying to someone in another forum:
One would like to think that there's some nice proof involving local perturbations – pick the smallest fragment size and adjust it by ε, then adjust some other fragment of the same stick by −ε to make up for it, which means in turn that a fragment of the appropriate stick at the other end of the dissection goes by +ε, and ... you hope to end up with a closed loop of adjustments, or several such loops superimposed, and hence an adjustment to your original dissection which is legal for some sufficiently small ε > 0 and increases the length of every currently minimal fragment. If you can do that, then the dissection you started with was not a local optimum, and hence not a global optimum either.
The point of doing this would be that it gives you a reason why the smallest fragment keeps turning out to be achieved by cutting a single stick into equal pieces – because then clearly no such perturbation can increase the length of all those pieces at once, or else the result wouldn't fit in that stick any more.
But I don't actually have a proof along those lines, only a sketch of the sort of thing I'd like the proof to achieve. And even that wouldn't prove that an optimum dissection had to have all fragments rational, only that the smallest fragment had to be rational.
In fact I'd guess it's quite likely that optimal dissections need not have the longer fragments all rational (there's probably some situation in which you can find a closed loop of those local perturbations that never touch a minimal fragment, in which case you can apply that change with irrational ε to get a partially irrational and equally good answer), but I'd at least hope that there exists a rational optimal dissection in every case...
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In fact, yes, the 7/5 example in my original post admits just such a transformation. Where we previously dissected three 7-sticks as (8/3 + 8/3 + 5/3) and the other two as (7/3 + 7/3 + 7/3), we now do:
- two lots of (8/3+ε) + (8/3−ε) + (5/3)
- one unmodified (8/3) + (8/3) + (5/3)
- two lots of (7/3+ε) + (7/3−ε) + (7/3)
which reassembles as:- two lots of (8/3+ε) + (7/3−ε)
- two lots of (8/3−ε) + (7/3+ε)
- two lots of (8/3 + 7/3)
- one lot of (5/3) + (5/3) + (5/3) as before.
And then you can set ε = π/1000 or some such and you have some irrational pieces – but not interestingly irrational.At a guess, you can probably rule this out by introducing a tie-breaking rule, in which solutions with the same shortest fragment length are now compared by their second shortest, and so on. That'd probably put a stop to frivolous irrationality :-)
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I've written a thing that looks for integer solutions; but it's slow and crap and obviously "I searched exhaustively and found this" is much less interesting than "I thought really hard about it and realised that all solutions look like this". Also it can't handle irrationals and so far can do fractions only where all stick-fragments share a denominator....
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(In fact,
I've written a thing that looks for integer solutions
Ooh, I'd like to see whatever data it's generated.
Also it can't handle irrationals and so far can do fractions only where all stick-fragments share a denominator
In any rational dissection there must be some denominator shared by all fragments (just take the lcm of all denominators), so the latter isn't a problem. And I'm still convinced that irrationals can't appear in any solution unless there's an equally good or better one without them (I have a half-thought-out proof idea involving treating R as a vector space over Q), so I'm not worried about the former either.
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5 x 8 sticks in half integer steps is best with 4 sticks cut into 2/3 and 4 uncut assembled into 4 lots of 3/5 and 1 of 2/2/2/2 which gives more than 3 stick-parts.
With m and n <= 10 and m<n in half integer steps 12 combinations have a smallest-stick-fragment larger than gcd(m,n); 4 of which are non-integer (5 x7 is the smallest). I have no idea how to prove whether any given answer could be bested by taking smaller steps other than by trying progressively smaller steps and seeing.
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Borrowing
So this dissection for 5 into 8 cannot be beaten even if you were to increase the denominator, and hence
(I do wonder how far that proof technique can be automated. It might give rise to a better search algorithm!)
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Choose a basis B for R as a vector space over Q, and choose it so that it includes an actual rational q. Write every fragment length in the dissection as its unique weighted sum of elements of B. Rational lengths will therefore be represented as straightforward multiples of q.
The shortest irrational fragment length, being irrational, must have a nonzero coefficient for at least one element of B that isn't q; let b be such an element. Look at the coefficients of b across all fragments of the dissection. They must sum to zero in every input and output stick (because those sticks' lengths are all rational, hence have zero coefficient of everything other than q). So if we pick any real ε and adjust every fragment length by ε times its coefficient of b, we must get a system of adjustments that preserves every input and output stick length, and which always adjusts equally long sticks by the same amount. So in particular we can choose ε so that it has the correct sign to slightly increase the shortest irrational length (and has absolute value small enough not to make any two lengths cross over each other), and then applying those adjustments yields a strictly better solution.
Hence, no dissection with any irrational fragment can be optimal (under the extended criterion including tie-breaking), and in particular no dissection with an irrational shortest fragment can be optimal even under the original criterion. []
(Slightly icky because it uses the Axiom of Choice, almost certainly unnecessarily. But AC is obviously true, right? So no problem really. :-)
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What confused me in the pub is, you use the axiom of choice to get a basis for R over Q. But don't you just need a basis over Q for the vector space generated by the fragment lengths you actually have, not over any possible irrationals? Isn't that just some subset of the fragment lengths, throwing away any non-linearly-independent ones?
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(I leave the exact dissection of 7 into 4 as an exercise, but it shouldn't be hard given the minimum length :-)
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I'm flattered you call my comment a conjecture, when it was only really saying I wasn't convinced by your conjecture. But I'm glad we found an answer deciding it :)
Stupid question, if the search program finds a best answer, how does it show that's the best? Is there an obvious limit on something? Previously we seemed to have no way of clearly showing an answer _is_ the best, except by special-case proofs like writing-hawk's.
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Not a stupid question at all, since nobody else had thought of an answer to that problem!
Ian's basic idea is to start by (WLOG) assuming each input stick contributes at most one piece to each output stick (clearly if that's not true then you can merge fragments until it is without making your score worse). So now you have an n × m matrix of fragment lengths. Next, we want to treat the problem as a linear-programming optimisation exercise, trying to maximise one variable ("smallest nonzero matrix entry") under a collection of linear constraints (matrix entries all positive, rows and columns sum to the right thing). The trouble with that is that it's not actually a linear problem, since the objective criterion is nonlinear (taking min of all the matrix entries isn't a linear function). But if you knew the adjacency matrix (that is, you knew which input sticks contribute at all to which output sticks) it would be, because you could introduce the minimum frag length itself as an extra variable, so that some matrix elements were constrained to zero and others were of the form (min_frag + some positive extra value). So Ian simply iterates over all possible adjacency matrices and appeals to a linear-programming library to solve that optimisation problem for each one, and once he's gone through all possible matrices then he knows he's considered every answer!
Of course, then you optimise as hard as you can (by e.g. iterating over adjacency matrices in a sensible order that means you see the best ones first, stopping as soon as your matrix gets too dense because then some stick has to be cut into enough pieces that the smallest won't beat your existing best answer, parallelising over all the CPUs you can find, etc). But that's the basic idea.
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Has any pattern emerged in the results found?
I wonder if it's possible to make any theoretical progress following the same approach? (But I'm just speculating I haven't even had time to start thinking about it yet.)
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Oops! I hope it comes into shape.
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Or rather, what happens if you try to increase all the shortest lengths by epsilon? That obviously fails if several of them compose a single stick. That's what you were trying to explain to me before. But if you follow through adjusting all other stick lengths epsilon as you described, it has to fail *somewhere* or you didn't start with a local optimum. Another way it could fail is if the *complements* of the shortest fragments are equal divisions of a stick. But there may be other ways as well, if you end up with a stick made of fragments that all have to get longer or all have to get shorter, but came from a different number of steps from the original perturbation.
But if the optimal solution for some division is 5/3, I can't help but wonder if that means *something* has to be divided into three equal pieces.
I wonder if this might be easier to imagine looking at perturbing a matrix as you described, instead of sticks.
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I don't think that's worded quite right, but I think something like that. If so, maybe a perturbations argument could show that if at some step you divide into the same number of fragments of slightly different lengths, you can follow through the remainder of the steps in the same way. And if so, the non-equal division would be sub-optimal (because one stick of that length gets shorter).
But I'm not sure if that's actually right, or just another plausible guess that will turn out to have exceptions.
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