gavinksong

Perhaps... but sometimes there is more than meets the eye, my friend. I shall put forth another guess.

I'm sorry if I wasn't clear. I was asking for an answer in terms of N (number of weighings allowed). After all, the maximum number of coins from which you can pick out the fake is not constant and in fact increases with N. On the other hand, the method you outlined is a reasonable first attempt. You do have one blind spot... It is possible to calculate the maximum number of coins your method could work for (in terms of N). There is a more efficient method that works for larger numbers of coins. P. S. It is common to hide any revealing content inside spoilers (like I did above). The "eye" icon on the very left of the toolbar allows you to insert them into your responses.

Here's what it looks like when I solve for the error. Unless I've made a mistake?

Oh, I see. That makes more sense. In that case, there's one equation that pops into my head. Seems a bit messy.

I thought we needed to calculate how far we needed to move the corner that was off? Is it unnecessary to use the fact that one of the diagonals is 6 cm longer than the other?

Hmm. Since distances aren't bounded, we can't store them in memory. You could mean... Also, On further reflection, it appears that you are right.

Actually, I just did the math...

I will confirm that it is possible even when it is not known whether the fake is lighter or heavier. In your first guess, you had a power of two, which makes me think you were inspired by binary search. Remember that a balance scale has three possible outcomes. Even then, the answer is not as simple as 3^n/2.

OH, haha. Is one of the answers...

Once again, I'm not sure I understood the question correctly (sorry). I think you're asking for three numbers, but I only have one. It's in the spoiler.

Assuming all the men are highly intelligent,

True, but do you see where I am confused? If there are multiple possible locations for the fourth vertex, which one do we use when we calculate how far we have to move it?

I think I see the intuition behind that guess, but alas that is not it. Counterexample in the spoiler.

So only one of the corners is off. The other three corners form a right triangle with legs of length L and W. Here's where I'm confused: This is where I have been confused. Please help if I misunderstood.

A reversal of a classic problem. You have a bag of gold coins. They are all of the exact same weight except one, which is a fake. You have a balance scale, which you can only use N times. What is the maximum number of coins from which you can pick out the fake?

That's actually a pretty interesting point. Perhaps this is reading too much into the problem, but if that's the case, why wouldn't A simply ask how many pieces B got?

No, BMAD's 5-step solution is incorrect for the reason that CaptainEd brought up. Hole 4 hasn't been eliminated for Day 4, only for Day 3. On Day 4, it could still be in either hole 2 or 4.

I've always wanted to play one of these! This would be my first time. Can I sign up? Is that okay?

This makes a lot of sense. The fewest number of moves to sort a sequence must be at least the length of the sequence minus the length of the largest monotone subsequence. At the same time, if you have a monotone subsequence of a certain length, then you can simply move the remaining DVDs into place. In other words, the fewest number of moves to sort a sequence is exactly the length of the sequence minus the length of the largest monotone subsequence. The captain is once again correct! When I revisited this thread, I immediately thought of the Erdos-Szekeres theorem as well. Each of the shelves containing BMAD's DVD collection must have a monotonic subsequence of at least length 4, which means the upper bound for minimum moves is 7. If we can find a sequence with no length 5 monotonic subsequence, then it should prove definitively that the maximum number of minimum moves to sort BMAD's collection is exactly 7. Of course, CaptainEd's already beat us to the punch. I think the case is finally closed. Wouldn't you say? I've been awake for far too long. I need reassurance.

I think the example in the OP has a mistake in it? Anyways, here's a length 120 sequence.

Hmm? I'm a little confused. :/ Do the values L and W tell us something about the defective rectangle? Is only one of the corners off?

That seems to be a reasonable assumption, CaptainEd.

If we assume that the men are perfectly intelligent, I think this puzzle may be trickier than it seems. In fact, it appears impossible? Like Pickett, I will adopt a positive world view and assume that each man received at least one piece of beef jerky.

I assume you can't make any calls yourself.