From: Simon Tatham Date: Fri, 28 Mar 2014 18:51:55 +0000 (+0000) Subject: Filled in some gaps in the writeup. X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ianmdlvl/git?a=commitdiff_plain;h=a165e3e81d16cae015a37fa952d6c355270fe6d4;p=matchsticks-search.git Filled in some gaps in the writeup. --- diff --git a/template.html b/template.html index 6d2d755..f8e82d1 100644 --- a/template.html +++ b/template.html @@ -24,6 +24,11 @@ table, td, th { .range { background-color: #ff8888; } +sub { + vertical-align: -0.3em; + font-size: 70%; + line-height: 0; +} @@ -111,17 +116,206 @@ case m = n requiring no cuts at all is so uninteresting that we may as well ignore it.)

+

No irrationals

+ +

+We've established that the minimum fragment size need not be an +integer, so an obvious next question is, need it even +be rational? +

+ +

+The answer is yes, and we can prove it as follows. Consider an +arbitrary dissection of m sticks into n containing at +least one irrational fragment length. The lengths of all the fragments +generate a finite-dimensional vector space over ℚ; let +{ e1, â€¦, ek } +be a basis for that space in which e1 is rational. +Then the shortest irrational fragment length must have a nonzero +coefficient of some basis element other than e1, +say ei. +

+ +

+Now we construct a new dissection based on the old one, in which we +adjust every fragment length by adding ε times its original +value's coefficient of ei, for some +constant ε. Doing this must preserve the property that the +fragments making up each input and output stick sum to the right total +length (proof: the original ei coefficients +must have summed to zero in order to generate an integer stick length +previously, so adding a constant multiple of all +those ei coefficients doesn't change the +overall sum). So now we choose |ε| small enough that no two +original fragment lengths change their ordering and no fragment length +becomes negative or more than the entire stick size. (We can do this, +since those constraints give us finitely many upper bounds all of +which are strictly positive, so we can certainly find an ε +smaller than all of them and still nonzero. The only interesting case +is when a fragment was the whole of its stick – in which case it has a +zero coefficient of ei anyway, so we aren't +changing it at all and no bound is placed on ε by the +requirement for it not to overflow.) Finally, we choose the sign +of ε so that it lengthens rather than shortening the smallest +irrational fragment. +

+ +

+So now we've constructed a new dissection of basically the same shape +as the old one, in which every fragment length is adjusted by at most +a small amount, with the properties that: +

+This means that no dissection with the shortest fragment irrational +can be optimal, since the above construction would improve it. +

+ +

+Also, if we were to strengthen our optimisation criterion by imposing +a tie-breaking condition in which dissections with the same shortest +fragment were compared by looking at the next shortest +fragment and so on, then any dissection optimal under that stronger +condition would have to have every fragment length rational +(because, again, we could always lengthen the shortest irrational one, +without changing any of the rational fragment lengths shorter than +it). This implies that not only must the minimum fragment size always +be rational, but also there must always be an all-rational dissection +which attains it. So we can rule irrationals completely out of +consideration, which is a relief! +

+ +

Upper bounds

+ +

+A problem with searching for dissections is, once you've found one, +how do you prove there isn't a completely different one that does +better? +

+ +

+One way to solve that problem is to find a genuinely exhaustive search +strategy, one of which is described in the next section. But +exhaustive searches are expensive, so it would be nice to have some +other tools in our toolbox as well, such as strategies for proving +that no solution for a given n,m can be better than a +given value. Then, if you can find an upper bound proof and a +dissection attaining that bound, you know you don't have to look any +further. +

+ +

+One such upper bound proof, largely due +to writinghawk, +works by counting the pieces in the dissection. For this proof, we +start by assuming that every length-m stick (i.e. every shorter +stick) is cut into at least two pieces; this doesn't lose generality, +because if any stick remains whole in a dissection, we can just cut it +in half and put both the pieces into the same n-stick. (And +this does not decrease the minimum fragment length +unless all m-sticks were whole, i.e. m +divides n, which is a special case with such an obvious answer +that we don't need this proof anyway.) +

+ +

+So, let's suppose that every piece is at least length s. Also, +every piece must be at most +length m âˆ’ s (because it must have been part +of some m-stick, which had a piece of at least s cut off +it). +

+ +

+That means that each stick of length n has to be cut into at +least n/(m−s) and at most n/s +pieces (otherwise some piece would have to be too long or too short). +Because the number of pieces also has to be an integer, that really +means it's at least ⌈n/(m−s)⌉ and at most +⌊n/s⌋. Summing those expressions over +all n-sticks tells us that the total number of pieces P +in the whole dissection must satisfy the +inequality m⌈n/(m−s)⌉ â‰¤ P â‰¤ m⌊n/s⌋. +

+ +

+By similarly considering the number of pieces that a stick of +length m is cut into, we derive a second +inequality n⌈m/(m−s)⌉ â‰¤ P â‰¤ n⌊m/s⌋. +(This is identical to the first inequality, but with n +and m swapped everywhere they appear except in the +divisor m−s). +

+ +

+So when s becomes big enough that either of those inequalities +is self-inconsistent (with its LHS strictly larger than its RHS), or +when the two are incompatible (because they define ranges for P +that do not meet), there can be no dissection with s that +large. This allows us to establish an upper bound on s for a +given instance of the problem, which is often (but, results below +indicate, not always) actually achievable. +

+

-FIXME: min frag is rational. +A second upper-bound proof, due to Tom Womack, applies in a restricted +set of cases, some of which are not solved optimally by the above +proof. Specifically: if m/3 is an integer and is also a +multiple of n−m, then no solution can be better +than m/3.

-FIXME: empirical observation suggests conjecture that denominator is -bounded by n, but no proof of this. +Proof: suppose the shortest fragment is m/3 + Îµ for +some ε > 0. Then if any segment of stick less +than twice that length arises while you're dissecting, it +must remain whole. In particular, any fragment shorter than +2m/3 + 2ε must be monolithic in this way.

-FIXME: state the known upper bound proofs. +So consider a stick of length m from which we cut a piece of +length m/3+ε. The other piece of that stick has length 2m/3−ε and +hence is monolithic; so it must have been cut as a whole from a longer +stick of length n. Let n = m+k, because we'll want to talk about k a +lot. Now we reason as follows: +

+... and so on, so that we derive the existence of pieces of lengths +m/3+ε, m/3+k+ε, +m/3+2k+ε, …, 2m/3+ε (which we must +reach since m/3 is an exact multiple of k). That +2m/3+ε is still monolithic (recall that only a piece of +at least 2m/3+2ε can avoid being so) and so it must +share a short stick with a piece of length m/3−ε, which +is shorter than our presumed minimum fragment. This is a +contradiction, and hence we reject the initial assumption that a +minimum fragment greater than +m/3 was possible in the first place.

Computer search

@@ -131,7 +325,7 @@ FIXME: describe our two search strategies.

-FIXME: stress that one is exhaustive but the other depends on the +FIXME: stress that one is exhaustive but the other depends on a conjectured denominator bound.