Devedander Posted February 23, 2007 Author Report Share Posted February 23, 2007 Well if you followed the answers of the people behind you, you should know that at this point there should either be more blacks or more whites, if you see 6 of each and lets say there are supposed to be more whites than blacks at this point then you would know that your hat should be white. Sorry jdkno I edited that, see above. Link to comment Share on other sites More sharing options...
Devedander Posted February 23, 2007 Author Report Share Posted February 23, 2007 (edited) Well, it seems simple. Were I a Harvard-educated pirate, when I leaned over to whisper the color of my hat to my ninja captor, I'd look him in the eye, and see the my reflection, thus revealing the color of the hat, which is exactly what I'd whisper to him. Ninjas dude... Ninjas... masters of invisibility... they keep their eyes closed lest they reflect light and give away their position like a sniper with his scope. There would be no reflection option. Even with their eyes closed their hearing and finely tuned senses is enough to strike fear in the heart of the most hardened pirate. Edited February 23, 2007 by Devedander Link to comment Share on other sites More sharing options...
tenacious_b Posted February 23, 2007 Report Share Posted February 23, 2007 Ninjas dude... Ninjas... masters of invisibility... they keep their eyes closed lest they reflect light and give away their position like a sniper with his scope. There would be no reflection option. I have never seen a ninja movie where Ninja's covered their eyes, as a matter of fact, that is usually the only part that is not covered. Since the stipluation of no eye reflection was not stipulated, and since it involves no pirate to pirate funny business, I feel that you should be compelled to concede that my solution is valid, and more than likely much simpler than whatever solution was originally intended. Link to comment Share on other sites More sharing options...
Devedander Posted February 23, 2007 Author Report Share Posted February 23, 2007 (edited) I have never seen a ninja movie where Ninja's covered their eyes, as a matter of fact, that is usually the only part that is not covered. Since the stipluation of no eye reflection was not stipulated, and since it involves no pirate to pirate funny business, I feel that you should be compelled to concede that my solution is valid, and more than likely much simpler than whatever solution was originally intended. Do you suppose they have real ninjas in movies? They are actors, actors are not illusive masters of invisibilty, they just play one on the big screen. If you ever saw a real ninja you would notice that only in very rare instances (for instance scoping out a total hottie) are their eyes open. But then if you ever saw a real ninja you would be dead. BTW I do not think they teach eyeball reflection technique at Harvard. And even should the ninja guard choose to open his eyes (say to scope out a hottie) he might not be looking at said pirate during the answer phase, screwing him and all the other pirates (assuming it was the second pirate to get it wrong). I see you pug I haven't seen any input on the other pirate post yet... are you just holding your tongue? Edited February 23, 2007 by Devedander Link to comment Share on other sites More sharing options...
Pugwash Posted February 23, 2007 Report Share Posted February 23, 2007 ps. I've solved this. Devedander can confirm this. It's a pure logic problem. I'll go look at the other. Link to comment Share on other sites More sharing options...
Devedander Posted February 23, 2007 Author Report Share Posted February 23, 2007 (edited) Confirmed that pug does know the answer, but is playing nice as he was ripping through my previous logic puzzles like a ninja through something easy to rip through As of yet no one else has provided a fully correct answer althought it's definitely getting close. Edited February 23, 2007 by Devedander Link to comment Share on other sites More sharing options...
eddhead Posted February 23, 2007 Report Share Posted February 23, 2007 Third guy now sees 2 blacks and 1 white ahead... but what can he do with this info? But you are on the right track. Nice hint!!! just got it. the last guy sees the 99 hats in front of him. 99 is an odd number so that means there has to be an odd number of black or white hats but NOT BOTH!! in other words there has to be one odd and one even number of hats. He identifies the odd number of hats.. say black. The guy in front of him now knows there are an odd number of black hats in the bunch not includint th guy behind him.. he sees 98 hats in front of him... he cannot see his own. these 98 hats can be odd black, odd white even black even white he now knows there are an odd number of black hats all told including his, but not including the one behind him. - if he sees an even number of black hats ahead of him, he knows his hat is black - any odd number of colored hats cannot be his the third poerson now has 97 hats in front of him. he knows that of the 99 (excluding the last guy) there are an odd number of black hats, and the buy behind him (second gusser) has a black hat too. he can go through the same calculations as the second guy including what he learned from the second guy's guess. I want you to know i have not worked this hard all day! Link to comment Share on other sites More sharing options...
Devedander Posted February 23, 2007 Author Report Share Posted February 23, 2007 Very good!!! Now here's the real kicker... I actually posted the problem altered from it's original state... the original question did not state the number of pirates... So at this point it should be easy, but can you adjust your answer to work with 101 pirates (ie the last guy could look at head and indeed see 50 white and 50 black hates)? How about any number of pirates? Link to comment Share on other sites More sharing options...
eddhead Posted February 23, 2007 Report Share Posted February 23, 2007 Very good!!! Now here's the real kicker... I actually posted the problem altered from it's original state... the original question did not state the number of pirates... So at this point it should be easy, but can you adjust your answer to work with 101 pirates (ie the last guy could look at head and indeed see 50 white and 50 black hates)? How about any number of pirates? i could but i would have to kill you. seriously, where there are an even number of pirates ahead of you the combinations have to be both odd nos both even nos but the second guessor sees odd/even so the last guy calls out black if there are odd nos of both, and white if there are even nos of both. The second to last guy can see the odd color, knows the total are either odd or even and can deduce from there. eg: 100 hats 50 B 50W Last buy calls White to signify even nos of both colors second to last guy sees 50 B 49W he now knows his hat is white. the guy in front of him can deduce the color of his hat by knowing the color of the guy behind his and the number of hats of each color in front. Link to comment Share on other sites More sharing options...
Devedander Posted February 23, 2007 Author Report Share Posted February 23, 2007 i could but i would have to kill you. I am a ninja... I am not worried Link to comment Share on other sites More sharing options...
Pugwash Posted February 23, 2007 Report Share Posted February 23, 2007 How about any number of pirates? Doesn't matter. It's an XOR parity problem. The only guy that could get it wrong is the first. Link to comment Share on other sites More sharing options...
Devedander Posted February 23, 2007 Author Report Share Posted February 23, 2007 Doesn't matter. It's an XOR parity problem. The only guy that could get it wrong is the first. Actually while very close, eddheads solution depends on there being an odd number of pirate in front of the last guy. Link to comment Share on other sites More sharing options...
Pugwash Posted February 23, 2007 Report Share Posted February 23, 2007 Actually while very close, eddheads solution depends on there being an odd number of pirate in front of the last guy. Doesn't matter. Link to comment Share on other sites More sharing options...
eddhead Posted February 23, 2007 Report Share Posted February 23, 2007 Actually while very close, eddheads solution depends on there being an odd number of pirate in front of the last guy. re-read the more detailed edited post.. i think you will find it is correct. and if it is not it is friday night and happy hour in NYC anyway so it is all good!! Link to comment Share on other sites More sharing options...
Devedander Posted February 24, 2007 Author Report Share Posted February 24, 2007 Unless I am missing something, eddheads solution for even number of pirates requires using the first guys guess to tell which group is odd (white or black), but if there are an odd number of pirates, the first guys guess tells whether the groups are both even or both odd. So the solution for X Pirates must be able to account for both scenarious with the same rule set (ie first guys "white" or "black" tells all the other pirates what they need to know regardless of how many pirates there are total). Link to comment Share on other sites More sharing options...
eddhead Posted February 24, 2007 Report Share Posted February 24, 2007 Unless I am missing something, eddheads solution for even number of pirates requires using the first guys guess to tell which group is odd (white or black), but if there are an odd number of pirates, the first guys guess tells whether the groups are both even or both odd. So the solution for X Pirates must be able to account for both scenarious with the same rule set (ie first guys "white" or "black" tells all the other pirates what they need to know regardless of how many pirates there are total). but you can always get back to the odd number by taking into account the guy behind you.. or at least i think so... actually, i am still recovering from a long nite.. and i doubt i am up for this now... Link to comment Share on other sites More sharing options...
Devedander Posted February 24, 2007 Author Report Share Posted February 24, 2007 (edited) but you can always get back to the odd number by taking into account the guy behind you.. or at least i think so... actually, i am still recovering from a long nite.. and i doubt i am up for this now... Well rather than beat this horse (since you obviously get the idea of the solution) the generally accepted answer is they decide that if the last guy in line sees and odd number of black hats, he calls black, and if he sees an even number he calls white. Everyone working their way up can then deduce what their own hat is depending on whether you see an odd or even number of black hats and how many hats were called behind you of what color. This works for any number of pirates Well done! Pug for somehow seeing the solution in minutes (this one took me literally hours) and edd for sticking through and getting it! Edited February 24, 2007 by Devedander Link to comment Share on other sites More sharing options...
tenacious_b Posted February 24, 2007 Report Share Posted February 24, 2007 Devedander, Even though you attempted to invalidate my response, based on stipulations that were not included before the guessing began, I want to show you there are no hard feelings by sharing my love of Ninja's with you. Check out my favorite site on the net (Besides RWG): http://www.realultimatepower.net/index4.htm Regards, B Link to comment Share on other sites More sharing options...
Devedander Posted February 25, 2007 Author Report Share Posted February 25, 2007 Devedander, Even though you attempted to invalidate my response, based on stipulations that were not included before the guessing began, I want to show you there are no hard feelings by sharing my love of Ninja's with you. Check out my favorite site on the net (Besides RWG): http://www.realultimatepower.net/index4.htm Regards, B TB I give you honorary mention for your outside the box solution! Yes, technically it was not stipulated that eyeball reflection reading was not allowed Link to comment Share on other sites More sharing options...
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