[jlblatt]

I feel like this is solvable using binary to assign which bottles to which prisoners, but only if there were 16 bottles. For example, bottle 11 dec = 1011 bin. So bottle 11 would be tasted by prisoners 1, 3, and 4. Then if 1, 3, and 4 die, we know bottle 11 is poisoned. Every combination of prisoner deaths would point to a unique bottle.

Except 17. Fencepost error on my part maybe?

[sleepychu]

No, you're correct I typo'd.

[benibela]

Actually it seems you can have a 17th bottle.

Use binary search to find the poisoned bottle among bottle 1 to 16. If you fail to find a poisoned bottle, it is the 17th

[Kapow]

No, that's already covered by 0000, nobody drinks that one. Number the bottles 0 to 15 and it's clearer.