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Riddle

Crovax

Crovax

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From 538 What’s The Best Way To Drop A Smartphone? | FiveThirtyEight

You work for a tech firm developing the newest smartphone that supposedly can survive falls from great heights. Your firm wants to advertise the maximum height from which the phone can be dropped without breaking.

You are given two of the smartphones and access to a 100-story tower from which you can drop either phone from whatever story you want. If it doesn’t break when it falls, you can retrieve it and use it for future drops. But if it breaks, you don’t get a replacement phone.

Using the two phones, what is the minimum number of drops you need to ensure that you can determine exactly the highest story from which a dropped phone does not break? (Assume you know that it breaks when dropped from the very top.)
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Crovax said:
From 538 What’s The Best Way To Drop A Smartphone? | FiveThirtyEight
Click to expand...

If you take one of your phones and drop it from the highest floor, or the second highest if you are assured that at the highest floor it breaks, and it doesn't break, you've solved the puzzle with just one drop. If it breaks, then you can use your fallback phone and go down to the first floor and work your way up, one floor at a time. Therefore, the minimum number of drops would be one.
 
ttwtt78640 said:
2 drops would be the minimum. That assumes that you picked floor X and the phone broke and then tried floor X-1 and the other phone did not break or that you picked floor X and the phone did not break and then tried floor X+1 and the other phone broke.
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CanadaJohn said:
If you take one of your phones and drop it from the highest floor, or the second highest if you are assured that at the highest floor it breaks, and it doesn't break, you've solved the puzzle with just one drop. If it breaks, then you can use your fallback phone and go down to the first floor and work your way up, one floor at a time. Therefore, the minimum number of drops would be one.
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The minimum they are referring to is the minimum drops to a guarantee that you find out the right floor no matter what the exact floor would be.

So while you might find the answer with 3 drops that wouldnt work for every possibility, what is the minimum amount of drops to cover every possibility?
 
Crovax said:
The minimum they are referring to is the minimum drops to a guarantee that you find out the right floor no matter what the exact floor would be.

So while you might find the answer with 3 drops that wouldnt work for every possibility, what is the minimum amount of drops to cover every possibility?
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I stand by my answer, based on the question posed in the OP. The minimum number is one if you start at the top and the phone doesn't break - that becomes the highest story from which the phone doesn't break.
 
CanadaJohn said:
I stand by my answer, based on the question posed in the OP. The minimum number is one if you start at the top and the phone doesn't break - that becomes the highest story from which the phone doesn't break.
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Read the riddle again, "Assume you know that it breaks when dropped from the very top"
 
Crovax said:
Read the riddle again, "Assume you know that it breaks when dropped from the very top"
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Yes, I did. This is why I said in my first answer that if it's a given that it breaks from the highest floor, the next highest floor is where you start and if with that first drop it doesn't break, you have your answer in a minimum of one drop.
 
Crovax said:
The minimum they are referring to is the minimum drops to a guarantee that you find out the right floor no matter what the exact floor would be.

So while you might find the answer with 3 drops that wouldnt work for every possibility, what is the minimum amount of drops to cover every possibility?
Click to expand...

Unless you consider a previously dropped phone as structurally identical to a non-previously dropped phone then there is no number other than 2 drops, one for each phone - one pass (did not break) and one fail (did break). I suppose you that could use a binary search technique but that would likely require more than two phones for more than one failed drop. Thus the minimum number of drops remains two but is totally dependent upon luck.
 
CanadaJohn said:
Yes, I did. This is why I said in my first answer that if it's a given that it breaks from the highest floor, the next highest floor is where you start and if with that first drop it doesn't break, you have your answer in a minimum of one drop.
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That only covers one possibility you have to cover them all, if you use your first drop at the top floor and it breaks then you are going to have a tough time finding out what floor it is. The object of the riddle is to find out the numbers of drops that would cover if it broke on every floor in separate scenarios.
 
ttwtt78640 said:
Unless you consider a previously dropped phone as structurally identical to a non-previously dropped phone then there is no number other than 2 drops, one for each phone - one pass and one fail. I suppose you that could use a binary search technique but that would likely require more than two phones for more than one failed drop. Thus the minimum number of drops remains two but is totally dependent upon luck.
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Its not the minimum possible but the minimum to ensure that you find the highest floor it breaks no matter what that floor is.
 
Crovax said:
Its not the minimum possible but the minimum to ensure that you find the highest floor it breaks no matter what that floor is.
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The highest floor was given, it was stated that it broke at floor 100, so then the answer would be ZERO more drops are required. ;)
 
ttwtt78640 said:
The highest floor was given, it was stated that it broke at floor 100, so then the answer would be ZERO more drops are required. ;)
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Whoops, you know what I meant. The highest floor it doesnt break at
 
Crovax said:
That only covers one possibility you have to cover them all, if you use your first drop at the top floor and it breaks then you are going to have a tough time finding out what floor it is. The object of the riddle is to find out the numbers of drops that would cover if it broke on every floor in separate scenarios.
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Sorry, but your description of the riddle is illogical. Logic dictates that finding the highest floor where a phone could be dropped without breaking with the least number of drops would be to use one of your phones and drop it from the highest floor eligible. If you succeed in not breaking the phone, you've answered the question with one drop. It's that simple. I grant you that if it breaks and you have to go to the bottom floor and work your way up with the one remaining phone, you've got a harder task, but that wasn't the question you asked or the problem to be solved.

Anyway, you have my answer and it's the right one to the question asked so nothing is gained by my further involvement.
 
Crovax said:
Whoops, you know what I meant. The highest floor it doesnt break at
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That is what I initially gave you - 2 more drops, one floor apart, with one resulting in broken phone and one resulting in an intact phone. I fully admit that this minimum requires luck, yet that is still a valid possibility even if not a probability.

EDIT: Another method, is a binary search starting at floor 49. If the phone breaks then you know that all floors above that need not be tested so you would then take the only remaining phone and start at floor 1 working your way up until it too was broken. If the phone did not break, at floor 49, then you may try floor 74 and so forth.
 
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CanadaJohn said:
Sorry, but your description of the riddle is illogical. Logic dictates that finding the highest floor where a phone could be dropped without breaking with the least number of drops would be to use one of your phones and drop it from the highest floor eligible. If you succeed in not breaking the phone, you've answered the question with one drop. It's that simple. I grant you that if it breaks and you have to go to the bottom floor and work your way up with the one remaining phone, you've got a harder task, but that wasn't the question you asked or the problem to be solved.

Anyway, you have my answer and it's the right one to the question asked so nothing is gained by my further involvement.
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Yes it was.

Here is the exact question from the riddle, "what is the minimum number of drops you need to ensure that you can determine exactly the highest story from which a dropped phone does not break?"

Your method has to be able to work to find any floor. Using your method it would take 100 drops to ensure that you find out what floor it is.
 
ttwtt78640 said:
That is what I initially gave you - 2 more drops, one floor apart, with one resulting in broken phone and one resulting in an intact phone. I fully admit that this minimum requires luck, yet that is still a valid possibility even if not a probability.
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You have to find a method that could find every possibility not just one possibility.
 
Crovax said:
Yes it was.

Here is the exact question from the riddle, "what is the minimum number of drops you need to ensure that you can determine exactly the highest story from which a dropped phone does not break?"

Your method has to be able to work to find any floor. Using your method it would take 100 drops to ensure that you find out what floor it is.
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Nonsense.

My method was simple. To find out the highest floor from which a phone can be dropped without breaking in the fewest number of drops you drop it from the highest possible floor first to see if it breaks or not - if it doesn't break, you have your answer in one try. The answer is one, regardless how convoluted you want to make the question.
 
CanadaJohn said:
Nonsense.

My method was simple. To find out the highest floor from which a phone can be dropped without breaking in the fewest number of drops you drop it from the highest possible floor first to see if it breaks or not - if it doesn't break, you have your answer in one try. The answer is one, regardless how convoluted you want to make the question.
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The answer is not one because if it doesnt break you will have to drop it more than once. Your choice of method has to work for all floors to me the minimum to ensure that you find out what floor it is. Your method of one drop does not ensure that you find out with one drop. Yes you can get lucky and find it with one drop but you have to ensure that you will find out the floor and there is no way to do that with one drop.
 
Crovax said:
The answer is not one because if it doesnt break you will have to drop it more than once. Your choice of method has to work for all floors to me the minimum to ensure that you find out what floor it is. Your method of one drop does not ensure that you find out with one drop. Yes you can get lucky and find it with one drop but you have to ensure that you will find out the floor and there is no way to do that with one drop.
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Did you actually read what you posted here? Seriously?

Here's your question from the OP to refresh your memory:

Using the two phones, what is the minimum number of drops you need to ensure that you can determine exactly the highest story from which a dropped phone does not break? (Assume you know that it breaks when dropped from the very top.)

So, would you agree there are two parts to the question? First part, what's the highest story from which a dropped phone doesn't break and second part, what's the minimum number of drops you'd have to make in order to get the first answer? Correct?

So, I dropped a phone from the highest eligible story and it didn't break. Ipso, facto, the highest story is the answer to part one and the answer to part two is one because that's all it took for me to get the answer to part one.

That's the last time I'm going to explain it. Have fun and enjoy your day.
 
ttwtt78640 said:
That is what I initially gave you - 2 more drops, one floor apart, with one resulting in broken phone and one resulting in an intact phone. I fully admit that this minimum requires luck, yet that is still a valid possibility even if not a probability.

EDIT: Another method, is a binary search starting at floor 49. If the phone breaks then you know that all floors above that need not be tested so you would then take the only remaining phone and start at floor 1 working your way up until it too was broken. If the phone did not break, at floor 49, then you may try floor 74 and so forth.
Click to expand...

Your problem is made much more difficult because a failed result (broken phone) elminates a test possibility for each such failure. Once you get the first failure then you must abandon the binary search technique.


A binary search may be the coolest search going because it is so simple. Don't let the name binary fool you - it just means we're cutting the search space in half each time we look. The elements have to be in some sorted order. so you start in the middle of the array. if this case element 100. so now you test that element is it greater than your search parameter or less. if greater you only search the lower half of the array if its less then only the upper half.

So now you just repeat the procedure on what's left. Suppose the search parameter is less than the element we just looked at then we now only search from 100 to 200 in the array. We do the same thing we did before start in the middle in this case it will be element 150. again we compare the element against the search parameter.

So we keep cutting the search space in half with each look up. We can cut an array in half only so many times, Say S is the maximum number of cuts. In the case of any number of elements, say N. We can only cut it in half 2^S times before we reach 0 cuts - our number. so N < 2^S or log2(N) = S which we round up to the next whole integer. log2(200)= 7.6 ... rounding to 8.

of course you can get lucky and cut and compare right at some earlier half way point. then you will be less the maximum cuts.
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https://answers.yahoo.com/question/index?qid=20110111135643AAvVGZs

Wooops! I hit the reply instead of the edit button.
 
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CanadaJohn said:
Did you actually read what you posted here? Seriously?

Here's your question from the OP to refresh your memory:

Using the two phones, what is the minimum number of drops you need to ensure that you can determine exactly the highest story from which a dropped phone does not break? (Assume you know that it breaks when dropped from the very top.)

So, would you agree there are two parts to the question? First part, what's the highest story from which a dropped phone doesn't break and second part, what's the minimum number of drops you'd have to make in order to get the first answer? Correct?

So, I dropped a phone from the highest eligible story and it didn't break. Ipso, facto, the highest story is the answer to part one and the answer to part two is one because that's all it took for me to get the answer to part one.

That's the last time I'm going to explain it. Have fun and enjoy your day.
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But that doesnt ensure that you found the highest story if it breaks. You arent able to claim that one drop will ensure you find the highest story when your method only works if the floor is the 100th floor.
 
Crovax said:
Now youve got it. Whats the answer you came up with?
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It totally depends on luck, as I had stated from the very beginning. The problem is that a binary search is impossible once you get the first failure (broken phone) - at that point you must revert to a linear (sequential?) search. The answer is then that the higher the floor at which the phone breaks (which is an unknown) then the fewer number of tries which are required too find that value. CanadaJohn made that same point as well.
 
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