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I found this cool problem in Santos' number theory book, page 12, but he gives the credit to Halmos.Supposedly it must be solved by mathematical induction.

Every man in a village knows instantly when another’s wife is unfaithful, but never when his own is. Each man is completely intelligent and knows that every other man is. The law of the village demands that when a man can PROVE that his wife has been unfaithful, he must shoot her before sundown the same day. Every man is completely law-abiding. One day the mayor announces that there is at least one unfaithful wife in the village. The mayor always tells the truth, and every man believes him. If in fact there are exactly forty unfaithful wives in the village (but that fact is not known to the men,) what will happen after the mayor’s announcement?

I defined $P(n)$ to be the number of unfaithful wives in the village. Then since everyone believes the mayor and he always tells the truth, $P(1)$ must be true. This is why I defined $P(n)$ to be the number of unfaithful wives in the village, because we've been told that $P(1)$ is true. Now I have to prove that if $P(k)$ is true, i.e. $k$ women are unfaithful, then $P(k+1)$ is true. I guess this must be somehow related to men knowing the unfaithful women in the village, but I can't proceed. I have made some different guesses in my mind, but I can't logically entail anything out of them yet.

Any hints or ideas will be appreciated.

user66733
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  • Are you sure that the $P(n)$ you defined is useful for the problem? What are you trying to prove (:? Perhaps it would be more useful to have $P(n)$ represent what might happen upon knowing that there are $n$ unfaithful wives. So, for $P(1)$, the statement might be: $X$ happens when there is 1 unfaithful wife, for some statement $X$. Fill in $X$! – Raymond Cheng Aug 26 '13 at 16:05
  • the same puzzle is also seen here with a more thorough discussion – Ben Grossmann Aug 27 '13 at 00:27
  • See also here. – Andrés E. Caicedo Jan 15 '14 at 07:15

1 Answers1

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Suppose that there is only one unfaithful wife. Then all men except for her husband perceive the wife to be unfaithful. However, he knows by the mayor's announcement that there is at least one unfaithful wife. Since he is unable to otherwise account for the unfaithful wife, he deduces that the unfaithful wife must be his own.

Having deduced that his wife is unfaithful, he shoots her before sundown the same day.

Now, suppose that there are $2$ unfaithful wives. All men except for the two husbands perceive two wives to be unfaithful. The two husbands, however, perceive only one unfaithful wife. Following the above logic, they know that if there is exactly one unfaithful wife...

Perceiving no deaths by the end of the first day, both men may now deduce that their wives are unfaithful, and shoot them by sundown the second day.

The pattern continues.

A "more mathematically rigorous" phrasing, as requested:

Let $P(n)$ be the statement: "if $n$ wives are unfaithful, then all $n$ wives will be shot before sunset on the $n^{th}$ day, starting the day of the announcement, with no deaths preceding this day". Our inductive proof is as follows:

By the above argument, we know that $P(1)$ is true. For our inductive step, we must show that $P(k)\implies P(k+1)$. We may do so as follows:

Suppose that there are $k+1$ unfaithful wives (UWs for short). The $k+1$ corresponding husbands (CHs for short) each perceive $k$ UWs, and each deduce that either there are $k$ total UWs or there are $k+1$ UWs total, which means that their wives are unfaithful.

By their perfect knowledge and intelligence, each CH is aware of $P(k)$. On the $(k+1)^{th}$ day, the CHs perceive no deaths, and deduce via $P(k)$ that there cannot be only $k$ total UWs. They thereby deduce that their wives are unfaithful after sundown of the $k^{th}$ day.

$P(k+1)$ follows.

Thus, $P(k)\implies P(k+1)$. Our inductive proof is complete.

If you are still confused about why all wives are not shot, try to trace the reasoning of a man whose wife is faithful. He knows $P(n)$ to be true. When the $n^{th}$ day comes, all $n$ of the unfaithful wives he perceived are shot. Having witnessed this event, he knows that his wife is faithful.

Ben Grossmann
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  • +1. So, the fact there are exactly 40 unfaithful ones is completely ignored by mathematical induction and all wives will finally be shot, including the innocent faithful ones. Why does this happen? – user66733 Aug 26 '13 at 16:13
  • You've gone wrong with your induction somewhere. What is the general pattern? – Ben Grossmann Aug 26 '13 at 16:13
  • To be more clear: not all wives will be shot, supposing there are innocent wives. Also, an important aspect of the question is not only if the wives will be shot, but when. – Ben Grossmann Aug 26 '13 at 16:19
  • Also, note that the death of any innocent wives supposes that some man broke the law or came to an incorrect conclusion. This contradicts the supposition that each man is law-abiding and completely intelligent. – Ben Grossmann Aug 26 '13 at 17:23
  • How does the mayor's announcement make any difference or start a "count-down to death" clock? According to the conditions of the problem, every man knows of 39 unfaithful wives; all but 40 know of 40 unfaithful wives, before the mayor does anything... – DJohnM Aug 26 '13 at 21:57
  • @User58220: With the announcement, each man knows that each man knows that... that each man knows of an unfaithful wife, and each man knows that each man knows that... that each man will act accordingly. It is this knowledge of mind-states that starts the counter, not the fact on its own. Consider what happens with $2$ unfaithful wives. At the beginning of the problem, each man was aware of at least one other faithful wife. However the announcement sets off the timer nevertheless. – Ben Grossmann Aug 26 '13 at 22:34
  • @User58220: if you'd like to find a flaw with this solution (and I am all ears, because this classical solution has never sit well with me), you must either find a reason that this line of deduction won't work for $n=2$, or find the lowest $n$ such that this line of reasoning fails. If there is no lowest such $n$, then the pattern must continue. – Ben Grossmann Aug 26 '13 at 22:37
  • @User58220 Here's another way to think about it: if you are one of the 40 men with a UW (unfaithful wife), you see 39 men with UWs. From your perspective, it could be that one of the men with UWs sees only 38 men with UWs. From the perspective of one of those 38 hypothetical men, there could be 37 men with UWs. And so on. – Ben Grossmann Aug 26 '13 at 22:46
  • @Omnomnomnom: I seriously doubt that I can find a hole in such a widely discussed proof, beyond the doubt I expressed: How does the mayor's announcement change anything. How is any Supremely Intelligent Man's knowledge base changed? Suppose there was a public announcement: The mayor will make an "Infidelity in the Village: Y/N?" report in 11 days. All the SIMs knows what the report will be, so does that start the clock? – DJohnM Aug 26 '13 at 23:16
  • @Omnomnomnom: I just don't see how this pattern terminates at the end when all the unfaithful 40 wives have been shot. Maybe you need to make it a bit more mathematically rigorous? For example you precisely define a P(n) and make your argument a bit more detailed? If you do that I'd appreciate it. – user66733 Aug 26 '13 at 23:26
  • @some1.new4u I've updated the argument accordingly. Interestingly, if the mayor's statement is a lie (and no citizens realize that this is the case), then all wives are innocent, and every wife is shot on the first day. – Ben Grossmann Aug 27 '13 at 00:03
  • Perhaps the problem as stated has a fatal flaw. I would be happier if, when the mayor announced the infidelity in the village, he also revealed the previously secret "Shoot With Proof" law... – DJohnM Aug 27 '13 at 00:15
  • @User58220 I think that fix avoids the heart of the problem. – Ben Grossmann Aug 27 '13 at 00:22