Number all bottles in binary, starting from 0000000000. Then the Nth prisoner drinks all wines where the Nth digit is “1”. have each prisoner drinking the wines where a certain digit is “1”.
So for example. If you had 8 bottles and 3 prisoners (exact same logic):
number your wines 000, 001, 010, 011, 100, 101, 110, 111
Prisoner 1 drinks wines 100, 101, 110, 111; if he dies the leftmost digit of the poisoned wine is 1, if he lives the poisoned wine starts with 0
Prisoner 2 drinks wines 010, 011, 110, 111; if he dies the mid digit is 1, else it’s 0
Prisoner 3 drinks wines 001, 011, 101, 111; if he dies the right digit is 1, else it’s 0
If nobody dies the poisoned wine is numbered 000. And if all die it’s the 111.
Initially I simplified the problem to one prisoner. The best way to reduce uncertainty was to split the bottles into two sets with 500 bottles each; the prisoner drinks from one, if he dies the poisonous wine is there, otherwise it’s in one of the leftover 500 bottles.
Then I added in a second prisoner. The problem doesn’t give me enough time to wait for the first prisoner to die, to know which set had the poisonous wine; so I had to have the second prisoner drinking at the same time as the first, from a partially overlapping set. This means splitting the bottles into four sets instead - “both drink”, “#1 drinks it”, “#2 drinks it”, “neither drinks it”.
Extending this reasoning further to 10 prisoners, I’d have 2¹⁰=1024 sets. That’s enough to uniquely identify which bottle has poison. Then the binary part is just about keeping track of who drinks what.
solution
Number all bottles in binary, starting from 0000000000. Then the Nth prisoner drinks all wines where the Nth digit is “1”. have each prisoner drinking the wines where a certain digit is “1”.
So for example. If you had 8 bottles and 3 prisoners (exact same logic):
If nobody dies the poisoned wine is numbered 000. And if all die it’s the 111.
Did ya need the hint?
No, I only saw it after I solved the problem.
my reasoning / thought process
Initially I simplified the problem to one prisoner. The best way to reduce uncertainty was to split the bottles into two sets with 500 bottles each; the prisoner drinks from one, if he dies the poisonous wine is there, otherwise it’s in one of the leftover 500 bottles.
Then I added in a second prisoner. The problem doesn’t give me enough time to wait for the first prisoner to die, to know which set had the poisonous wine; so I had to have the second prisoner drinking at the same time as the first, from a partially overlapping set. This means splitting the bottles into four sets instead - “both drink”, “#1 drinks it”, “#2 drinks it”, “neither drinks it”.
Extending this reasoning further to 10 prisoners, I’d have 2¹⁰=1024 sets. That’s enough to uniquely identify which bottle has poison. Then the binary part is just about keeping track of who drinks what.