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L _ R R Y Clue
THE JAILOR NEEDS 10 PRISONERS. BINARY MATHS IS NEEDED TO SOLVE THIS PUZZLE. BELOW IS AN EXAMPLE ON HOW YOU USE BINARY LOGIC TO FIND THE NUMBER OF PRISONERS FOR 8 BOTTLES OF WINE. ASSUME THE WINE BOTTLES ARE NAMED W1, W2, W3...W8. THE PRISONERS ARE NAMED P1, P2 AND P3. THE ABOVE CHART SUMMARISES WHICH PRISONER HAS TO DRINK FROM WHICH WINE BOTTLE. '1' INDICATES THAT THE PRISONER HAS TO DRINK FROM THAT BOTTLE. BOTTLE W1 IS NOT FED TO ANY PRISONER. BOTTLE W2 IS FED TO PRISONER P3. BOTTLE W3 IS FED TO PRISONER P2 AND SO ON. IF NO ONE DIES, THEN WINE BOTTLE W1 IS POISONED. IF ONLY PRISONER P3 DIES, BOTTLE W2 IS POISONED. IF ONLY PRISONER P2 DIES, BOTTLE W3 IS POISONED. IF BOTH PRISONERS P2 AND P3 DIE, BOTTLE W4 IS POISONED. IF ONLY PRISONER P1 DIES, BOTTLE W5 IS POISONED. IF BOTH PRISONERS P1 AND P3 DIE, BOTTLE W6 IS POISONED. IF BOTH PRISONERS P1 AND P2 DIE, BOTTLE W7 IS POISONED. IF ALL 3 PRISONERS DIE, BOTTLE W8 IS POISONED. SO TO TEST 1000 BOTTLES OF WINE, 10 PRISONERS ARE SUFFICIENT AS THAT WILL ALLOW (2^10) 1024 UNIQUE COMBINATIONS THE GREEN ONE IS TELLING THE TRUTH. LETS ASSUME THAT ONE OF THEM IS TELLING THE TRUTH AND THEN TRY TO PROVE THAT. SINCE ALL FOUR SERVANTS ARE DISAGREEING THEN 3 OF THEM MUST BE LYING. THE SERVANT TELLING THE TRUTH WILL HAVE EITHER 6 OR 8 LEGS. THE OTHER 3 SERVANTS WILL HAVE 7 LEGS SINCE THEY LIE. SO THE TOTAL NUMBER OF LEGS SHOULD BE EITHER 27 (6 + 7 + 7 + 7) LEGS OR 29 (8 + 7 + 7 +7) LEGS. ONLY GREEN SERVANT COULD BE TELLING THE TRUTH AS IT SAID 27 LEGS. ALTERNATIVELY, LET'S SAY BLUE IS TELLING THE TRUTH: SO THE BLUE ONE HAS EITHER 6 OR 8 LEGS. AND EACH OF THE OTHER OCTOPUSES ARE LYING HENCE HAVE 7 LEGS EACH. SO OUR TOTAL NUMBER OF LEGS: 6 + 7 + 7 + 7 = 27 LEGS OR 8 + 7 + 7 + 7 = 29 LEGS. BUT SINCE BLUE SAID THAT ALTOGETHER THEY HAVE 28 LEGS, WE KNOW HE IS LYING. IF YOU FOLLOW THIS SAME LOGIC FOR ALL OF THEM, YOU REALIZE THAT ONLY THE GREEN OCTOPUS CAN BE TELLING THE TRUTH AS THE NUMBER OF LEGS ADDS UP LARRY IS THE PRESIDENT. SUPPOSE MATT'S FIRST STATEMENT IS TRUE. THIS WOULD MEAN THAT HE IS NOT A TYPE B PERSON (WHO ALWAYS LIE), AND LARRY'S SECOND STATEMENT IS TRUE. THIS WOULD MEAN THAT LARRY IS NOT A TYPE B PERSON, AND MATT'S SECOND STATEMENT IS TRUE. SINCE NEITHER LARRY NOR MATT IS THE PRESIDENT, NICK MUST BE PRESIDENT, MAKING BOTH HIS STATEMENTS FALSE. SINCE NICK'S FIRST STATEMENT IS FALSE, AND SINCE HE'S A TYPE B PERSON WHILE MATT IS A TYPE A PERSON, LARRY MUST BE A TYPE C PERSON, MAKING HIS FIRST STATEMENT FALSE. BUT SINCE THE PRESIDENT BELONGS TO A DIFFERENT RACE FROM EACH OF THE OTHER TWO, LARRY'S FIRST STATEMENT MUST BE TRUE. THIS MEANS THAT MATT'S FIRST STATEMENT IS FALSE. SUPPOSE MATT IS PRESIDENT. THIS WOULD MEAN THAT LARRY'S SECOND STATEMENT IS FALSE AND MATT'S AND NICK'S SECOND STATEMENTS ARE TRUE. THEN, SINCE MATT IS BOTH THE PRESIDENT AND A TYPE C PERSON, AND SINCE LARRY MUST ALSO BE A TYPE C PERSON IF HIS FIRST STATEMENT IS TRUE (WHICH WOULD MAKE HIS FIRST STATEMENT FALSE), LARRY'S FIRST STATEMENT MUST BE FALSE, MAKING HIM A TYPE B PERSON. THEN IF NICK'S FIRST STATEMENT IS TRUE, HE IS A TYPE A PERSON, WHICH MEANS THAT ALL THREE ARE DIFFERENT RACES AND MAKES NICK'S FIRST STATEMENT FALSE, BUT IF HIS FIRST STATEMENT IS FALSE, HE IS A TYPE C PERSON, WHICH MEANS THAT HE AND MATT BOTH BELONG TO A RACE THAT LARRY DOES NOT AND MAKES NICK'S FIRST STATEMENT TRUE. SO MATT CANNOT BE PRESIDENT. SUPPOSE NICK IS PRESIDENT. THIS WOULD MEAN THAT LARRY'S AND MATT'S SECOND STATEMENTS ARE BOTH TRUE, BUT NICK'S IS FALSE. THEN, SINCE MATT'S FIRST STATEMENT AND NICK'S SECOND STATEMENT ARE BOTH FALSE, NICK'S FIRST STATEMENT MUST BE TRUE. THEN, SINCE MATT AND NICK ARE BOTH TYPE C PEOPLE, LARRY'S FIRST STATEMENT MUST BE FALSE AND ALL THREE MUST BE TYPE C PEOPLE, MAKING NICK'S FIRST STATEMENT FALSE. SO NICK CANNOT BE THE PRESIDENT EITHER, LEAVING ONLY LARRY IF THERE WAS ONLY ONE BLUE-EYED PERSON ON THE ISLAND, THEN THAT PERSON WOULD LOOK AROUND AND SEE THAT THERE IS NO OTHER BLUE-EYED PERSON. SO HE REALIZES THAT HE IS THE ONLY PERSON WITH BLUE EYES ON THE ISLAND AND LEAVES ON THE DAY OF THE ANNOUNCEMENT. IF THERE ARE 2 BLUE-EYED PEOPLE, THEN THEY LOOK AT EACH OTHER. EACH ONE EXPECTS THE OTHER TO LEAVE ON THE DAY OF THE ANNOUNCEMENT. HOWEVER, ON THE NEXT DAY, WHEN THEY REALIZE THAT NEITHER OF THEM LEFT THE ISLAND, THEY WOULD BE ABLE TO DEDUCE THAT BOTH OF THEM HAVE BLUE EYES. THEY BOTH LEAVE THE ISLAND ON THE SECOND DAY. THROUGH MATHEMATICAL INDUCTION, THIS LOGIC CAN BE APPLIED TO THE 100 BLUE-EYED PEOPLE ON THE ISLAND. SO ON THE 100TH DAY, ALL THE 100 BLUE-EYED PEOPLE LEAVE THE ISLAND