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'Everything I Say Leaks,' Zuckerberg Says in Leaked Meeting Audio

www.404media.co 'Everything I Say Leaks,' Zuckerberg Says in Leaked Meeting Audio

"There are a bunch of things that I think are value-destroying for me to talk about, so I’m not going to talk about those."

'Everything I Say Leaks,' Zuckerberg Says in Leaked Meeting Audio
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  • Surly he is smart enough to do some combinatorics to figure out who is leaking the meetings.

    It's the classic riddle of u have 1000 bottles of wine 1 is poisoned and u have 10 prisoners to test the wine. U must test all the bottles in 4 days, however it takes 3 days for the poison to take effect.

    hint

    With 10 prisoners u can test up to 1024 bottles

    • 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):

      • 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.

      • 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.

26 comments