#cherylsbirthday

This past week the hash tags were all chirping about how challenging this logic puzzle was. It seems that Singaporean middle school students are posed with a question that challenges their deductive reasoning as well as their ability to remain objective even when the posing of the question invites them to assume a subjective position within the problem. Technically, it is a contest question for the Singaporean and Asian School Math Olympiad.

You the solver are shown that one of two characters in the problem possesses enough information that they can deduce that the other does not in fact possess enough information to deduce the solution. This fact when announced provides the second character with enough information that he claims to now be able to deduce the answer. That proclamation of deducibility incites the first character to then declare that he too has deduced the answer.

The problem goes something like this:

Cheryl gives a list of dates that could be her birthday to Albert and Bernard. She tells Albert the month and Bernard the day. Albert declares that he does not know the date but that he does know that Bernard does not. Bernard declares that he didn’t know the date but now does. Albert then declare that he too knows. So what is Cheryl’s birthday? Since SQL is my hammer let’s see if this is a nail.

For starters, we have a short list of dates so let’s put them into a table variable.

Next, we can select those dates not in months with days that are unique to a month. We do this because if it were a day unique to the month it would be possible for Bernard to determine which. Albert has decided is impossible (we are assuming these characters are not wrong in their deductions).

After that we select those dates not sharing a day. This is because if it were one, it would be impossible to have deduced the answer knowing only the day and now that it is not in a month with a unique day. It was this second piece of information that enabled Bertrand to deduce the date already knowing the first.

What remains are three dates only one of which is unique to the month, which is a requirement to deduce it knowing only the month and that solution can be deduced knowing the day and that it is not in a month with a unique day. Gosh, Albert and Bernard are smart.

DECLARE @candidates TABLE (rowid INT IDENTITY(1,1), candiDATE DATE);

INSERT INTO @candidates VALUES

(‘May 15 2015’),(‘May 16 2015’),(‘May 19 2015’),

(‘June 17 2015’),(‘June 18 2015’),

(‘July 14 2015’),(‘July 16 2015’),

(‘August 14 2015’),(‘August 15 2015’),(‘August 17 2015’);

WITH COUNTS AS (

SELECT candiDATE, DATEPART(DAY, candiDATE) [Day], DATEPART(MONTH, candiDATE) [Month],

COUNT(DATEPART(DAY, candiDATE))

OVER (PARTITION BY DATEPART(DAY, candiDATE)) [IfItWereAMonthWithAUnqiueDayBernardCouldKnow]

FROM @candidates

) , NEXTCOUNT AS (SELECT candiDATE,

DATEPART(DAY, candiDATE) [Day],

DATEPART(MONTH, candiDATE) [Month],

COUNT(DATEPART(DAY, candiDATE))

OVER (PARTITION BY DATEPART(DAY, candiDATE)) [IfItWasNotAUniqueDayRemainingBernardWouldNotKnow]

FROM @candidates WHERE DATEPART(MONTH, candiDATE) NOT IN (SELECT [Month] FROM COUNTS WHERE [IfItWereAMonthWithAUnqiueDayBernardCouldKnow]=1)

)

SELECT candiDATE, COUNT(DATEPART(DAY, candiDATE))

OVER (PARTITION BY DATEPART(MONTH, candiDATE)) [IfBernardDidnotKnowItWouldnotBeUniqueTotheMonth]

FROM @candidates WHERE candiDATE IN

(SELECT candiDATE FROM NEXTCOUNT n WHERE n.[IfItWasNotAUniqueDayRemainingBernardWouldNotKnow]=1)

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