Guest Posted December 1, 2007 Report Share Posted December 1, 2007 Josephine - Back to the Logic Puzzles It is easy to tidy up the induction to show that the n UHs will all be shot just on the n'th midnight. It's impossible to "tidy up the induction..." because your logic is absolutly WRONG!!!!!! If anyone think i'm wrong i would like to see him do it for only 5 unfaithfull husbands!! btw. i dont know the real answer and if it really exists Link to comment Share on other sites More sharing options...
rookie1ja Posted December 1, 2007 Author Report Share Posted December 1, 2007 Josephine - Back to the Logic Puzzles It is easy to tidy up the induction to show that the n UHs will all be shot just on the n'th midnight. It's impossible to "tidy up the induction..." because your logic is absolutly WRONG!!!!!! If anyone think i'm wrong i would like to see him do it for only 5 unfaithfull husbands!! btw. i dont know the real answer and if it really exists the logic is absolutely correct ... below is explanation for 1 and 2 unfaithful husbands ... apply the same logic and you will see that if there were 5 unfaithful husbands then all 5 would be shot the 5th night Josephine - solution The two questions for scroll #1 were: 1. How many husbands were shot on that fateful night? 2. Why is Queen Henrietta I revered in Mamajorca? The answers are: If there are n unfaithful husbands (UHs), every wife of an UH knows of n-1 UH's while every wife of a faithful husband knows of n UHs. [this because everyone has perfect information about everything except the fidelity of their own husband]. Now we do a simple induction: Assume that there is only one UH. Then all the wives but one know that there is just one UH, but the wife of the UH thinks that everyone is faithful. Upon hearing that "there is at least one UH", the wife realizes that the only husband it can be is her own, and so shoots him. Now, imagine that there are just two UH's. Each wife of an UH assumes that the situation is "only one UH in town" and so waits to hear the other wife (she knows who it is, of course) shoot her husband on the first night. When no one is shot, that can only be because her OWN husband was a second UH. The wife of the second UH makes the same deduction when no shot is fired the first night (she was waiting, and expecting the other to shoot, too). So they both figure it out after the first night, and shoot their husbands the second night. It is easy to tidy up the induction to show that the n UHs will all be shot just on the n'th midnight. Link to comment Share on other sites More sharing options...
Guest Posted December 3, 2007 Report Share Posted December 3, 2007 This is not a proper answer, you just repeated the solution given in the first place! Yes, the explanation goes well for 1 and 2, unfaithfull husbands, but I'd like to see anyone doing it for only 5 like, I said before! IT IS IMPOSSSIBLE to assume that if the truth is that there are 40 UHs anyone would think that there were 2 or 3 UHs! Link to comment Share on other sites More sharing options...
rookie1ja Posted December 3, 2007 Author Report Share Posted December 3, 2007 This is not a proper answer, you just repeated the solution given in the first place! Yes, the explanation goes well for 1 and 2, unfaithfull husbands, but I'd like to see anyone doing it for only 5 like, I said before! IT IS IMPOSSSIBLE to assume that if the truth is that there are 40 UHs anyone would think that there were 2 or 3 UHs! yes - I repeated the solution, since I assumed that it is clear ... I will try to take it a step further ... so let's assume that I am a wife and I know that there are 3 UH's 1st night Let's see if all 3 UH's will be shot ... but nothing happens ... I could have known that ... I guess that each of the 3 wives expected 2 shots ... hmmm, maybe the next night 2nd night let's wait what happens ... I guess that each of the 3 wives is anxiously awaiting the 2 shots now ... I imagine that one of them might think as follows: "I know there are 2 UH's. They were not shot the first night, since both of the 2 wives thought that there is just 1 UH. So this 2nd night both of them must kill their UH's. What the ...? I hear no shots. And now I know why - not only those 2, but also I do have an UH. We'll shoot them the 3rd night" ... I think that is the way one of the 3 wives might have thought 3rd night I'll open the window ... time is ticking away and then I hear something ... nope, was just some cat outside ... what is going on here? no shots? it is morning already ... the wives must have thought the way I imagined it ... unless ... wait a minute ... each of them does not know 2 UH's but 3 of them ... so there are 4 UH's? ... but who is the 4th one??? oh noooooo ... where is my gun ... Link to comment Share on other sites More sharing options...
Guest Posted December 3, 2007 Report Share Posted December 3, 2007 i think u hit it right on the money there. It was hard to follow but i see where your coming from. Link to comment Share on other sites More sharing options...
Guest Posted February 1, 2008 Report Share Posted February 1, 2008 I agree this could be a flawed riddle. Each woman could ask one or more other women simply how many unfaithful husbands they know of. The instant any other woman says a number one higher than the number she personally knows, then she immediately knows her own husband is unfaithful. She has not directly asked at all about her own husband this way. This could be determined on the very first day, and the entire situation resolved by *everyone* at midnight. I saw a couple of people pose this explanation, so I wanted to point out why it won't work. Only "The queens of Mamajorca are truthful", so the other women could lie about their answer. Link to comment Share on other sites More sharing options...
Guest Posted February 12, 2008 Report Share Posted February 12, 2008 I guess it's hard to grasp this answer unless you understand the simple fact that each wife knows every other husband's status but their own. Or, phrased another way. If I am a wife of Mamajorca, every other woman in the town knows whether my husband is unfaithful or faithful. So, from my perspective, if I know that 20 husbands are unfaithful SOBs, then I will anxiously await the 20th night to hear the gunshots. If my husband is faithful, then there really are only 20 unfaithful men in the town. This means that 20 wives in the town will be believing that there are only 19 unfaithful husbands and would be waiting for the 19th night instead. When the 19th night has come and gone with no gunshots fired, then they will know that their husbands are unfaithful and will wait for midnight on the 20th night to shoot them. I will know that my husband is indeed faithful and probably breath a sigh of relief. If my husband is unfaithful, then there are really 21!!! unfaithful husbands, the 20 that I know of plus my own. So, again, I still wait for the 20th night, hoping that gunshots will be heard (from the women who would have believed 19 unfaithful husbands exist), but none would come, because instead there are 21 women waiting for the 20th night (believeing there were 20 unfaithful husbands) and did not shoot their husbands. All of those 21 women will now know their husband is the unfaithful one, because they know all the other women in town are thinking of 21 unfaithful husbands (the 20 that I know about plus my own husband). So, on midnight of the 21st night, those 21 women (including me) will shoot their husbands, knowing, full out, that those husbands are unfaithful. Long winded! Wow. Link to comment Share on other sites More sharing options...
Guest Posted March 7, 2008 Report Share Posted March 7, 2008 When is midnight of today? If I find out something today...I cant do something at midnight today, because midnight occured in the past for today. or Since midnight is neither AM or PM, then its neither today or tomorrow. Today ends after 11:59:59 Link to comment Share on other sites More sharing options...
Guest Posted March 12, 2008 Report Share Posted March 12, 2008 I'd say 40 husbands were shot, because "at least one husband was cheating" and in 39 days no one was shot, leaving the "at least one" as your own. Henrietta eliminated all the men in the town, thereby creating an effecient, organized, smart society! hehehe Link to comment Share on other sites More sharing options...
Guest Posted March 14, 2008 Report Share Posted March 14, 2008 Surely the shots cannot be counted if everyone fired at midnight they all shoot at the same time, so individual shots cannot be heard. If the Queen tells the truth and she knows there is at least one UH, then she must be being unfaithful with at least one of the others' husbands. The other women might lie, but if asked, she would tell the truth. Maybe the Queen has a husband. I dunno the answer, but this is the logic I came up with... Link to comment Share on other sites More sharing options...
Brandonb Posted April 10, 2008 Report Share Posted April 10, 2008 I HAVE ANOTHER SOLUTION! THE QUEEN slept with a new man every night, the next day the women knew that a new husband had been unfaithful but does not know if it is her own husband. On the 40th night the queen ran out of men to sleep with, so the women know that there is no man left to sleep with the queen, and ALL 39 of the men are shot. The queen is revered for sleeping with ALL the men in the kingdom AND having them ALL killed. The black widow queen. How do ya like dem apples? Link to comment Share on other sites More sharing options...
Guest Posted July 2, 2008 Report Share Posted July 2, 2008 I have been struggling with the solution for a while and I was about to post my opinion on the fact that critical information was missing, but then I re-read and realised... Take an Arbitrary number of couples (singles are irrelevant). Let's choose 250 Married Couples. We know there are 50 UH (Unfaitful Husbands). Each W (wife) only knows of the total number of OTHER UH's, not the ACTUAL number of UH So... 200 W believe there are 50 UH (but they don't know if there are 51) 50 W believe there are 49 UH (but they don't know if there are 50) 0 W believe there are less than 49H The reason for the (nested) inductive logic is as follows... The 200 W believe there are 50 UH and believe that each of those 50 W believe that... There are 49 UH and will shoot their husbands on night 50 and that those 49 W believe that... There are 48 UH and will shoot their husbands on night 49 and that those 48 W believe that... There are 47 UH and will shoot their husbands on night 48 and that those 47 W believe that... There are 46 UH and will shoot their husbands on night 47 and that those 46 W believe that... .... There are 2 UH and will shoot their husbands on night 3 and that those 2 W believe that... There are 1 UH and will shoot their husbands on night 2 and that those 1 W believe that... THERE MUST BE AT LEAST 1 UH So as the nights increase the woman see that logic unravel. When it does reach night 50 and there is no shots, ALL 50 W realise that there was no-one out there that believed there were 48 UH. If someone believed there were 48 UH they would have shot their husband on night 49 because the logic dictates that if my husband is Faithful, then 49 W must have believed there were 48 UH, and if no-one believed there were 48 UH, then all 50 W realise their Husbands are cheaters and shot them the next night, night 51. It works because all woman are perfectly equal in their logic. Although at night 50, the 200 W might have started sweating and been extremely overjoyed by the mass murder on night 51. Link to comment Share on other sites More sharing options...
Guest Posted July 26, 2008 Report Share Posted July 26, 2008 The math is still very slightly wrong. There should be 39 unfaithful husbands instead of 40, if gun shots were in fact heard at midnight of the 40th night. As everyone has said many times before, each wife will be waiting for the numbered night of the number of UHs they know of. For a wife who has an unfaithful husband, this would be 38. On the 38th night, when no gunshots are heard, the wife of a UH will think, "The other wives must have only just found out their husbands are unfaithful, but cannot shoot until midnight rolls around again." But when the 39th midnight rolls around and still no gunshots are heard, the wife of a UH will discover that her husband is unfaithful, and will shoot him the next midnight; the 40th. Link to comment Share on other sites More sharing options...
Guest Posted July 28, 2008 Report Share Posted July 28, 2008 (edited) Maybe there are only 39 women in the city? and each one tells another about ANOTHER person's husbands being unfaithful, and so they all concur and kill once they each decide to kill another person's husband??? EDIT: Wow. To the person two posts above me. Bravo. xD seems much logical than my answer Edited July 28, 2008 by FreudianComplex Link to comment Share on other sites More sharing options...
Guest Posted October 21, 2008 Report Share Posted October 21, 2008 (edited) Interesting puzzle - had fun with that one! I know this is old, but i just registered, as this logic puzzle just came to my attention via igoogle... anyway, i get it, but have another bit of logic to point out. The women obviously were not allowed to talk amongst themselves about either their husbands fidelity, or how many husbands were cheating - otherwise it all would have been taken care of the first night... So the only way for the women to know how many husbands were unfaithful would be for each of the unfaithful husbands to have slept with every wife in the city - including their own. Wives of unfaithful husbands knew of n-1 unfaithful husbands because they slept with them, but didn't know of the nth because sleeping with their own cheating husband was not unscrupulous. You might say, well, those were new rules applied after the cheating, however, it still applies as true because any of the wives discussing it prior to the proclamation by the queen would know of any disparity, and would then know that their own husband was guilty prior to the proclamation because they knew of one less than everyone else except other wives of UHs. Therefore, while only 40 of the men were unfaithful, EVERY wife was unfaithful, and slept with 40 different men while married! Edited October 21, 2008 by cyphen Link to comment Share on other sites More sharing options...
Guest Posted November 2, 2008 Report Share Posted November 2, 2008 I think they should be able to figure out how many unfaithful men there are. They can just ask eachother how many unfaithful men the other woman knows. If they ever get an answer that higher than their number then their husband is a cheater! They're not allowed to discuss it. Link to comment Share on other sites More sharing options...
k-man Posted October 6, 2010 Report Share Posted October 6, 2010 This is a pretty old riddle and I saw it in a few varieties, but I believe the logic is wrong here. It works only for one and two unfaithful husbands (UH), but cannot be inductively applied for any number >2. Case 1: When there is only one UH in the kingdom his wife doesn't know of any cheaters, so the queen's announcement is a revelation to her. She kills her husband the next midnight. All good! Case 2: When there are only two UHs in the kingdom their respective wives know of only one cheater and incorrectly think that they are in the case 1 and the queen's announcement will be a revelation to the other cheater's wife. When nothing happens on the first night they now know that their husband is a cheater too and kill their husbands the next midnight. All good! Case 3: When there are three or more UHs in the kingdom there is nobody who thinks they are in the case 1. In other words, no wife thinks that there is one UH. No wife will expect anything to happen on the first night or any other night because the queen's announcement is not news to anybody and they all know that it's not news to anybody. Here are the contradictions in the explanations given in this thread: yes - I repeated the solution, since I assumed that it is clear ... I will try to take it a step further ... so let's assume that I am a wife and I know that there are 3 UH's 1st night Let's see if all 3 UH's will be shot ... but nothing happens ... I could have known that ... I guess that each of the 3 wives expected 2 shots ... hmmm, maybe the next night 2nd night let's wait what happens ... I guess that each of the 3 wives is anxiously awaiting the 2 shots now ... I imagine that one of them might think as follows: "I know there are 2 UH's. They were not shot the first night, since both of the 2 wives thought that there is just 1 UH. So this 2nd night both of them must kill their UH's. What the ...? I hear no shots. And now I know why - not only those 2, but also I do have an UH. We'll shoot them the 3rd night" ... I think that is the way one of the 3 wives might have thought 3rd night I'll open the window ... time is ticking away and then I hear something ... nope, was just some cat outside ... what is going on here? no shots? it is morning already ... the wives must have thought the way I imagined it ... unless ... wait a minute ... each of them does not know 2 UH's but 3 of them ... so there are 4 UH's? ... but who is the 4th one??? oh noooooo ... where is my gun ... The highlighted part of the logic is faulty because everybody knows at least 2 UHs and everybody knows that everybody knows that there are at least 2 UHs. I guess it's hard to grasp this answer unless you understand the simple fact that each wife knows every other husband's status but their own. Or, phrased another way. If I am a wife of Mamajorca, every other woman in the town knows whether my husband is unfaithful or faithful. So, from my perspective, if I know that 20 husbands are unfaithful SOBs, then I will anxiously await the 20th night to hear the gunshots. If my husband is faithful, then there really are only 20 unfaithful men in the town. This means that 20 wives in the town will be believing that there are only 19 unfaithful husbands and would be waiting for the 19th night instead. When the 19th night has come and gone with no gunshots fired, then they will know that their husbands are unfaithful and will wait for midnight on the 20th night to shoot them. I will know that my husband is indeed faithful and probably breath a sigh of relief. If my husband is unfaithful, then there are really 21!!! unfaithful husbands, the 20 that I know of plus my own. So, again, I still wait for the 20th night, hoping that gunshots will be heard (from the women who would have believed 19 unfaithful husbands exist), but none would come, because instead there are 21 women waiting for the 20th night (believeing there were 20 unfaithful husbands) and did not shoot their husbands. All of those 21 women will now know their husband is the unfaithful one, because they know all the other women in town are thinking of 21 unfaithful husbands (the 20 that I know about plus my own husband). So, on midnight of the 21st night, those 21 women (including me) will shoot their husbands, knowing, full out, that those husbands are unfaithful. Long winded! Wow. This logic recursively assumes that the other women will be waiting and counting the nights. Why would they do that? They haven't learned anything new from the queen's announcement. This logic relies on the assumption that this logic is correct to prove that it's correct. I have been struggling with the solution for a while and I was about to post my opinion on the fact that critical information was missing, but then I re-read and realised... Take an Arbitrary number of couples (singles are irrelevant). Let's choose 250 Married Couples. We know there are 50 UH (Unfaitful Husbands). Each W (wife) only knows of the total number of OTHER UH's, not the ACTUAL number of UH So... 200 W believe there are 50 UH (but they don't know if there are 51) 50 W believe there are 49 UH (but they don't know if there are 50) 0 W believe there are less than 49H The reason for the (nested) inductive logic is as follows... The 200 W believe there are 50 UH and believe that each of those 50 W believe that... There are 49 UH and will shoot their husbands on night 50 and that those 49 W believe that... There are 48 UH and will shoot their husbands on night 49 and that those 48 W believe that... There are 47 UH and will shoot their husbands on night 48 and that those 47 W believe that... There are 46 UH and will shoot their husbands on night 47 and that those 46 W believe that... .... There are 2 UH and will shoot their husbands on night 3 and that those 2 W believe that... There are 1 UH and will shoot their husbands on night 2 and that those 1 W believe that... THERE MUST BE AT LEAST 1 UH So as the nights increase the woman see that logic unravel. When it does reach night 50 and there is no shots, ALL 50 W realise that there was no-one out there that believed there were 48 UH. If someone believed there were 48 UH they would have shot their husband on night 49 because the logic dictates that if my husband is Faithful, then 49 W must have believed there were 48 UH, and if no-one believed there were 48 UH, then all 50 W realise their Husbands are cheaters and shot them the next night, night 51. It works because all woman are perfectly equal in their logic. Although at night 50, the 200 W might have started sweating and been extremely overjoyed by the mass murder on night 51. The highlighted statement is true, but forgotten in the explanation that follows. If all know that no wife believes that there are less than 49 UH then the logical chain breaks. Link to comment Share on other sites More sharing options...
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