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Riddle

ttwtt78640 said:
CanadaJohn came up with a one drop scenario. We were given that the phone broke at floor 100 so if it did not break when dropped at floor 99 then only one drop would be required.
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The problem with this riddle is precisely that you are NOT told at what height it will NOT break as a minimum.

There is therefore insufficient data given.

You cannot answer the question without making a major assumption.
 
CanadaJohn said:
Comparing me to Trump is laughable, and yet you've still, not once, shown why my answer is wrong.
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Yes I have, the answer has to work for every possible floor not just the one you decide to pick.

Its a math riddle not a trick question.
 
Riveroaks said:
I really do not believe enough engineering data are given to solve this query.

We need to know the minimum height at which it will NOT break.

Everyone is assuming that it won't break if dropped. But nobody knows how high it is guaranteed NOT to break.

14 my ass.
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Thats what you are supposed to find out with your answer
 
Crovax said:
Thats what you are supposed to find out with your answer
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Not enough engineering data are given however.

I would say in the interview that you not only need a maximum, you need a mini-max. This is a mathematical concept from theoretical math.

It is a formal version of the maxim "never ass-u-me."
 
Riveroaks said:
The problem with this riddle is precisely that you are NOT told at what height it will NOT break as a minimum.

There is therefore insufficient data given.

You cannot answer the question without making a major assumption.
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Not so, you are simply not given enough phones to use a binary search method which assures the answer in a maximum of 7 drops. But you were asked what a theoretical minimum numbers of drops would be.

I also pointed out that an un-dropped phone is most likely structurally superior to a perviously dropped phone, making any method that relies on dropping a phone more than once is suspect; for example, a phone already dropped from floors 1 through 7 may break at floor 8 yet a previously un-dropped phone may be found not to break until floor 12. ;)
 
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ttwtt78640 said:
Not so, you are simply not given enough phones to use a binary search method which assures the answer in a maximum of 7 drops. But you were asked what a theoretical minimum numbers of drops would be.

I also pointed out that an un-dropped phone is most likely structurally superior to a perviously dropped phone, making any method that relies on dropping a phone more than once is suspect; for example, a phone already dropped from floors 1 through 7 may break at floor 8 yet an un-dropped phone may be found not to break until floor 12. ;)
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The theoretical minimum is two. And that would depend on a lucky calculation and a corresponding drop right on the money.
 
Riveroaks said:
The theoretical minimum is two. And that would depend on a lucky calculation and a corresponding drop right on the money.
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Unless you picked floor 99 and the phone did not break then you know the answer with only one drop. ;)
 
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