...and block of 9... You could have it take as inputs the original layout (mostly blank cells) as well as your proposed answer. The interpersonal problem still arises, though, I agree--why should your friend believe that you've entered it all accurately/honestly. I'm not sure what you're after...
Perhaps part of the difficulty of persuading doubters is related to the fact that the proposed argument does not seem to depend on bullets being fired at one second intervals, and in fact doesn't appear to depend on the probability distribution of velocities?
I'm with Gavin--we are not talking about sorting algorithms--we already know a vector of final index numbers (or two vectors, one for forward sorted and one for backward). I think we would like a logistical algorithm--one that requires the fewest Steps as defined clearly by Pickett: remove "i", slide some down, insert "i" in the "i-th" slot, slide some up The OP doesn't require this, but the OP DOES request to know the maximum fewest steps for N DVDs. I did exhaustively verify that every permutation of 12345 can be changed to either 12345 or 54321 in two Steps or less. I'm struck at how hard even this seemingly easy problem is: I can't straighten out BAKFHJCIDGE in under 7 moves. Nice problem, BMAD.
Part A ) I can identify weights of all bells in 6 weighings or less. I don't know what B ) means. In what follows, "L" means "Low", "H" means "High". The results of a weighing are always Equal (all three pans equal weight), LL (two Low pans are equal, third is lighter), L (one Low pan is lowest, other two pans are lighter and indeterminate)
Even more mature reflection suggests that relaxing ( 3 ) is not realistic--you would NOT have left off such a key rule. So, I propose relaxing my interpretation of ( 2 ). Please tell us if this is acceptable:
Gavin, mature reflection suggests to me that my assumptions ( 1 ) and ( 2 ) above are good assumptions, that make an interesting, challenging puzzle, but that ( 3 ) makes it impossible. ( 1 ) makes it possible to keep some information about each flag; ( 2 ) if you had infinite precision arithmetic, then one number could act as all of memory, large enough to contain a map of the entire maze, so restricting to finite sized numbers makes the puzzle interesting; ( 3 ) If the robot could leave breadcrumbs (Tremaux's algorithm, noted by DejMar), then you could imagine a strategy in which the robot uses the maze itself as the unbounded memory, tallying flags that have no breadcrumbs on them, and figuring out a way to determine that it has seen all of the boundary and all of the insides. So, did you somehow accidentally leave out one of the rules (I doubt it, as you a careful guy, but I can hope...), and the robot IS permitted to leave more than just a flag in a maze cell?