3/19/2023 0 Comments Rocks pirates![]() ![]() So, at some point we'll probably find out what the other Titan Captains have done and who they've pissed off. We already know that Blackbeard, Catarina Devon and Vasco Shot attacked Hancock.įrom the chapter 1046 cover page, it seems that Kuzan (left) and Van Augur (right) attacked Chocolat Town.Īlso, in chapter 1048 Brulee is alerting Cracker that she believes the Vinsmoke's are in Chocolat Town, we then don't see Cracker helping Katakuri and Oven fight the Vinsmoke's, meaning that Cracker is probably dead, since he doesn't stand a chance against Kuzan, no one in Totto Land does. Now that Blackbeard has targeted both Hancock and Katakuri, it seems apparent that he's building up antagonism towards Blackbeard from these characters. Hint 3: Try proving the proposition by induction.I made this theory a few months ago about Blackbeard reviving the Rocks pirates with Moriah's powers and how Luffy would get allies like Katakuri and Hancock that would help him fight the new Rocks crew. Hint 2: Formulate a proposition with the pattern found. You may find it helpful to add more rows. Hint 1: Write out the above solutions in a table and try to look for a pattern. The final outcome is therefore:Īs an extension problem, can you say something about the general case of n pirates? And if so, can you prove it by induction? If we allocate a single coin to each of them, they will vote in favour of our proposal (since the alternative sees them receive zero coins without the ability to change anything). We can see from the above scenario with four pirates that pirates 3 and 5 can be most easily bribed. We have finally worked our way back up to the original problem. ![]() Therefore, pirate 2 only needs to give 1 coin to pirate 4 to ensure their vote and therefore a winning proposal. Since they are intelligent and rational, they know that if this proposal fails and only three pirates remain they will end up with nothing (and most importantly can do nothing to stop this). With four pirates remaining, the key vote lies with pirate 4. Therefore, pirate 3 needs to only sacrifice 1 coin to pirate 5 to ensure a winning vote and does not need to give anything to pirate 3. Pirate 4 knows that they will receive 100 coins in the next round of voting and so will vote against any proposal made here. If there are three pirates remaining, pirate 3 needs only give pirate 5 a single coin and they will already be better off. Outcome for two pirates (4, 5) = (100, 0) If pirates 4 and 5 are the only one remaining, pirate 4 can simply propose that they have all 100 coins and given only one vote is required for the plan to triumph, they can vote in favour and receive all of the treasure. We start by numbering the pirates from 1 (tallest) to 5 (shortest) and work backwards through the problem. Since pirates are generally known to be bloodthirsty, if a pirate will get the same number of coins if voting for or against a proposal, they will choose to vote against so that the proposer is thrown overboard.Īssuming that all 5 pirates are intelligent, rational, greedy, and do not wish to die, what will happen? Otherwise, the pirate proposing the scheme is thrown overboard, and the process is repeated with the pirates that remain. If 50% or more of the pirates vote for it, then the coins will be shared that way. ![]() ![]() The tallest pirate proposes how to share the coins, and ALL pirates (including the tallest) vote for or against it. They decide to split the coins using the following scheme: Five pirates of different heights have a treasure of 100 gold coins, which they plan to share amongst themselves. ![]()
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