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_ S L _ N D Clue
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 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 THE KING KEEPS POISON 12 SO THAT HE CAN NEUTRALIZE ANY POISON THAT THE CLOWN GIVES HIM. HE POURS POISON 10 (HIS SECOND STRONGEST POISON) AND GIVES IT TO THE CLOWN TO DRINK. CLOWN CAN DEDUCE THIS SO HE KEEPS POISON 11 FOR HIMSELF TO NEUTRALIZE POISON 10. THAT'S HOW HE SURVIVES. HE ALSO KNOWS THAT THE KING, AFTER DRINKING THE CUP HE GAVE HIM, WILL DRINK POISON 12 TO NEUTRALIZE THE POISON. SO THE CLOWN POURS PLAIN WATER INSTEAD OF POISON INTO THE CUP FOR THE KING. THE KING DRINKS THE WATER (WATER IS NOT POISON) AND RIGHT AFTER THAT HE DRINKS THE STRONGEST POISON POSSIBLE (12) AND DIES THEIR HATS ARE ALL BLUE IN COLOR. THERE ARE 3 POSSIBLE HAT COLOR COMBINATIONS: [A] 1 BLUE, 2 WHITE [B] 2 BLUE, 1 WHITE [C] 3 BLUE THE COLOR COMBINATION OF 3 WHITE HATS IS NOT POSSIBLE SINCE THE KING HAS ALREADY SAID THAT AT LEAST ONE OF THE WISE MEN HAS A BLUE HAT. SO, LET'S START OUR ANALYSIS. WHAT IF THERE WERE ONE BLUE HAT AND TWO WHITE HATS? THEN THE WISE MAN WITH THE BLUE HAT WOULD HAVE SEEN TWO WHITE HATS AND IMMEDIATELY CALLED OUT THAT HIS OWN HAT WAS BLUE, SINCE HE KNEW THERE IS AT LEAST ONE BLUE HAT. THIS DIDN'T HAPPEN, AND THUS THE HAT COLOR COMBINATION [A] IS RULED OUT. NOW, THE WISE MEN KNEW THAT ONLY 2 HAT COLOR COMBINATIONS ARE POSSIBLE - COMBINATION [B] OR [C]. WHAT IF THERE WERE TWO BLUE HATS AND ONE WHITE HAT? THE MEN WITH THE BLUE HATS WILL SEE ONE WHITE HAT AND ONE BLUE HAT. THEY WILL CONCLUDE THAT COLOR COMBINATION [B] IS THE CASE AND WOULD CALL OUT BLUE AS THEIR HAT COLOR. THIS ALSO DID NOT HAPPEN, AND THUS THE HAT COLOR COMBINATION [B] IS ALSO RULED OUT. AFTER SOME TIME, WHEN NONE OF THE WISE MEN ARE ABLE TO IDENTIFY THE COLOR OF THEIR OWN HATS, COMBINATION [C] (OF 3 BLUE HATS) BECOMES THE ONLY POSSIBLE OPTION