Puzzle: Intelligent Women

Puzzle: There are some couples in a town and some of the husbands are cheating. Once a wife deduces her husband is cheating, she throws the husband out of town. Also, there is an informer who knows all who are cheating and answers if there is any cheater in the town. Each woman also knows if any of the other men are cheating (except her own husband) but they cannot inform each other.

The informer informs that there are cheats in the town. But for the first 9 days after this announcement, nothing happens. But on the 10th day, all the cheaters are kicked out. How many cheaters were in the town? 

Solution: 

Case 1: Only one cheater:

Suppose a couple is Mr. X and Mrs. X and the cheater is Mr. X. Every woman in the town except Mrs. X knows who he is. Also, Mrs. X has not observed any man cheating in the town. But since there has to be a cheater in the town Mrs. X deduces it is none other than Mr. X and throws him out of the house the next morning. 

One cheater in town

Case 2: Two cheaters in town:

Let the two couples be  

• Mr. X and his wife Mrs. X
• Mr. Y and his wife Mrs. Y

and the cheaters are  Mr. X and Mr. Y. Mrs. X knows Mr. Y has cheated and Mrs. Y knows Mr. X has cheated. 

Each thinks there is only one cheater in the town. So when on the first morning no one is thrown out of the house both are shocked. Now Mrs. X thinks, Mrs. Y is not so foolish that she cannot deduce that her husband is the only cheater unless there is one more cheater in the town who Mrs. Y thought would be thrown out. Who can Mrs. Y see cheating, when the only person seen cheating by Mrs. X was Mr. Y? The only possible answer that she can deduce is Mr. X. The same above argument holds for Mrs. Y.

So on day 2 both Mr. X and Mr. Y are thrown out.

Two cheaters in town

Case 3: Three cheaters in town:

Let the three couples be (Mr. X, Mrs. X), (Mr. Y, Mrs. Y), and (Mr. Z, Mrs. Z). and the cheaters are Mr. X, Mr. Y, and Mr. Z. 

In this case, Mrs. X, Y, and Z all think there are exactly 2 cheaters. However, in 2^nd morning, they find no one is kicked out. Now, they can deduce, had there been exactly 2 cheaters then they would have been thrown out of the house in 2^nd morning. But since this was not done, it means that there are more than 2 cheaters. Since each of Mrs. X, Y, and Z has seen 2 cheaters only, they deduce that their own husband is also a cheater.

So on day 3 both Mr. X, Y, and Z are thrown out. 

Generalization:

Let us generalize the above deductions using mathematical induction. 

Assume there are n cheaters, they will be kicked out on n^th day. If the above statement holds true for n cheaters, and a wife sees n cheaters but finds no one on the street on n^th morning, she has to conclude that her husband is a cheater. 

That is, if there are exactly n+1 cheats,  they are thrown out on (n+1)^th morning. That is if our assumption is true for n cheaters it is also true for n+1 cheaters. 

Also we have proved it for n = 1 and n = 2. So by mathematical induction we can say that if there are exactly n cheaters, they will thrown out on n^th morning. 

As they are kicked on the 10^th day after the information, there will be exactly 10 cheaters on the streets.