CaptainEd

a blastomere characterized by:

dgreening and I have noticed the same thing:

It's been a while, and I have no more clue, but...I'm guessing that when we want to make an 18th number, there must already be an 18th that contains two numbers. I can't prove that, but I'm wondering whether the overall problem is experienced in this version: ( a ) pick 9 numbers, so that each 9th is populated ( b ) pick 7 more numbers, so that each 16th is populated ( c ) pick 1 more number, so that each 17th is populated ( d ) pick 1 more number, so that each 18th is populated--can we experience the same apparent impossibility as before?

An observation without an explanation

It took me a while to follow Guest's reasoning, like that Far Side cartoon where a person is writing a solution on the blackboard, and step 2 is in a cloud, and the other person is saying, "We need more detail in step 2". How did Guest get to "the partner of the person who knows 9 can only know one person (their partner)"?

I'll try

I feel SO lame: The hour hand points halfway between 1 and 2 at 1:30 and at no other time, I would have thought. What am I missing? (Light bulb goes off), it crows if the hour hand touches halfway between ANY 1 and ANY 2, either way around. So...

how bout...

Rephrasing my addition to Jasen's answer in response to Harey's hint:

Better bound, but not certain yet...

...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...

one approach

Pickett (or anyone), Try this sequence: AGDICFEBKJH

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?

Wrong again! N/2 is not ruled out!

Now I doubt we can order them in N/2 steps

My answer is

A method and a couple of questions that might lead to a theorem:

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.

A trivial strategy:

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) Case 1b: Case 1c:

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:

I'll go for 9 calls-18 minutes.

(bows long and deeply) that IS a pretty solution!