Sixteen bottles of wine riddle

SPOILERS FOLLOW as I will be discussing the answer.

Looking at the table, device 3 obviously tells you if the bottle is from the "high" group (8-15) or the "low" group (0-7). So you line up the bottles and start using device 3 on them, and move them into two groups, 0-7 on the left and 8-15 on the right, as you get the results of each test.

Also, once you've found all eight bottles of one group, you can stop testing because all the remaining bottles will be in the other group. If you're lucky this might happen as soon as test 8, but worst case you must test 15 bottles, then you'll know which group the 16th belongs to without needing to check it.

Worst case: 15 tests done so far.

Now look at what device 2 does. For each group, 0-7 and 8-15, it tells you whether that bottle belongs to the "low" half of the group (0-3 or 8-11) or the "high" half of the group (4-7 or 12-15). Furthermore, in each group of eight, once you've identified four "highs" or four "lows" you can skip testing the rest. Worst case, you have to test 7 bottles of each group before you find four of a kind, and can skip at most 1 bottle per group. 2 skips total, 14 tests.

Worst case: 15+14 = 29 tests done so far.

Now you have four groups, 0-3, 4-7, 8-11, 12-15. You use device 2 which will tell you whether each bottle is in the "high" or "low" pair for each group (0-1 or 2-3, 4-5 or 6-7, and so on). Worst case you have to test three bottles from each group before you are guaranteed to find a pair and be able to skip the fourth bottle. So worst case here is 12 tests.

Worst case: 15+14+12 = 41 tests done so far.

Now you have eight pairs that are 0-1, 2-3, 4-5 and so on. The final device, device 0, will tell you whether the bottle you tested is the "low" or "high" bottle of that pair, so you can arrange each pair in correctly-sorted order after testing one bottle. Guaranteed to need 8 tests, with no possibility of luck of the draw changing that number.

Worst case: 15+14+12+8 = 49 tests done and you've arranged the bottles in order from 0 to 15, so you now know the year of every bottle.

Bonus points if you convince her to leave, and rig it up right above the door for when she comes back, home alone style.

Without looking at the answer I came up with a geometric interpretation of it (explained below as this is a spoiler to the Poison and Rat puzzle).

But to move the spoiler down a little, my initial solution was to pour all the bottles out into a barrel, mix them together, and refill the bottles. Nothing explodes since you never measure anything, but you know the exact set of years in each bottle.

SPOILER:

T(n) is the number of measurements required for n bits.

If you measure the first bit (of n) of all the bottles, you'll never need to measure the last bottle. At that point, you have grouped the bottles into 2 groups, and need to figure out the remaining n-1 bits. So T(n) = (2^n)-1 + 2T(n-1).

    T(1) = 1 (if you have 2 bottles, you only need to measure one.)
    T(2) = 2^2-1 + 2*1 = 5
    T(3) = 2^3-1 + 2*5 = 17
    T(4) = 2^4-1 + 2*17 = 49

I've no idea how to do better than 49 in the worst case, though. Hm... is it necessarily the case that 45 is possible? (log2(16!))? It seems like that is only true if each measurement can divide the total set of possibilities in half, and I'm not sure why that would be the case.

Just go left to right on each bottle, and keep track of how often each prefix has appeared (i.e. on the first bottle, if you get 1, 0, 0, 1), we'd keep track of: {"1": 1, "10": 1, "100": 1}. Now, if a prefix of length 1 appears 7 times, or a prefix of length 2 appears 3 times, we stop measuring (because there's only 1 left).

In all cases, for 8 bottles you will need 4 measurements, for 4 bottles you will need 3 measurements, 2 bottles will require 2 measurements, and 2 bottles will require 1 measurement. (4 * 8) + (4 * 3) + (2 * 2) + (2 * 1) = 32 + 12 + 4 + 2 = 50. But for the very last bottle, you can just do 0 measurements, by way of process of elimination. so 50 - 1 = 49.

Sorry about the poor formatting of the algorithm but I'm typing on my phone and don't want to submit something AI generated

I was trying to figure out the runtime of this…it captures your best case scenario, and I think the worst as well, but what about the average?

"What's the best average value with worst result being no worse than" seems like the perfect question, tho

Has anyone found any good collections of these? Whenever I try to search for riddles online, I end up with mostly results containing wordplay riddles like "what has a mouth but doesn't eat, ..."

Lol

   The wines are from different years
   If you apply an arbitrary order to the bottles, the number of possible year-arrangements of the bottles is 16!
   Each test gives you one bit of information
   Since 2^50 is only greater than 16! by about 50x < 2^6, you only have about 5 tests to spare.
   There's probably some clever way to express the solution beyond just the brute force the above implies, but I haven't thought about it past this point

Optimal solutions are known for B <= 13 only, and some asymptotic bounds are known; the rest is conjecture. It is essentially modifying the “standard” wine riddle to allow two poisoned bottles as a possibility.

(This is OEIS A286874 / A303977; https://oeis.org/A286874 https://oeis.org/A303977)