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[personal profile] simont Fri 2014-02-21 17:18
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.

Borrowing [livejournal.com profile] writinghawk's proof technique: in a better solution, each of the five 8-sticks 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 5-sticks must end up as a single piece (15 isn't enough pieces for two each). The 8-stick including that whole 5-stick 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 5-stick you cut it off.

So this dissection for 5 into 8 cannot be beaten even if you were to increase the denominator, and hence [personal profile] 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!)
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