





 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
In fact, yes, the 7/5 example in my original post admits just such a transformation. Where we previously dissected three 7sticks 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 tiebreaking 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 :) 
   


naath  Fri 20140221 13:08 
 Because you have to match up the + and  irrationals I don't think you get a longer smallest stick doing this; although obviously you can have the samelength smallest stick.
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 stickfragments share a denominator.... 
   


 Indeed, you certainly can't get a longer smallest stick by small perturbations of that answer, because one of the 5sticks is entirely made up of smallestfragments and so if you lengthen them all by any ε > 0 then that stick overflows. What I was hoping to show was that the converse is also true – that the only way in which a dissection can be locally optimal is if some stick is made up entirely of smallestfragments, by showing that in all other cases you can find a system of εadjustments that lengthens every smallestfragment. No luck yet, though!
(In fact, writinghawk has proved over on LJ that my dissection for 5into7 is globally as well as locally optimal, which is more than I had previously known.)
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 stickfragments 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 halfthoughtout proof idea involving treating R as a vector space over Q), so I'm not worried about the former either. 
   


naath  Fri 20140221 14:17 
 So far it is tediously slow and hasn't got anything harder than 5,7... 
   


naath  Fri 20140221 16:50 
 Apparently I'm not exactly doing this the optimal way... because 7 x 10 in 1/3 steps is, er, ENOMEMORY. Fantastic.
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 stickparts.
With m and n <= 10 and m<n in half integer steps 12 combinations have a smalleststickfragment larger than gcd(m,n); 4 of which are noninteger (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.

   


 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 stickparts.
Borrowing writinghawk's proof technique: in a better solution, each of the five 8sticks would have to be cut into at most three pieces (if you cut one in four then a piece must be <=2), which gives 15 pieces overall, and so at least one of the eight 5sticks must end up as a single piece (15 isn't enough pieces for two each). The 8stick including that whole 5stick then has 3 units of length left over; if you divide that in two then you have a piece <=1.5 (no good) and OTOH if you leave it whole then that leaves 2 units on the 5stick you cut it off.
So this dissection for 5 into 8 cannot be beaten even if you were to increase the denominator, and hence gerald_duck's question is indeed answered.
(I do wonder how far that proof technique can be automated. It might give rise to a better search algorithm!) 
   
 