Riddle me this: Let's say you have 2 phones and a 100 story building. What's the minimum # of drops you can do?

Can I take the elevator?

This is impossible, because even if the phones are identical, they are not assured to impact the ground at the exact same angle, thus two drops from the exact same floor could yield a broken phone or not, depending upon the angle at which they land.

BUT...If you could get them to land the exact same way, the minimum drops would be 2: the first drop would be from the highest floor it doesn't break and the next drop would be from the floor above, at which it breaks.

Martian wrote:

HKop wrote:

Your fancy math still doesn't explain to me what happens when the first phone breaks after the initial drop from the 14th floor.

You go floor by floor from the first at that point. If it's an iPhone it breaks on floor 1 and you're done.

If you read the manual, you'd find the shock values and calculate accordingly.

Floor 100 meets the criteria for minimal number of drops.

One phone to test, and one to verify.

i = 1

j = 0

tries = 0

breaksFrom = 0

for j = 0 to 98 step 2

tries = tries + 1

If phoneBreaksFromFloor(i+j) = True Then:

breaksFrom = i + j

#Phone 1 breaks, then test floor below

If j >0 then:

tries = tries + 1

If phoneBreaksFromFloor(i+j -1) = True Then:

breaksFrom = i + j - 1

End if

End if

Exit For

End if

Next j

This is a 5 minute answer and probably wrong...

Provided it doesn't break at floor 1 and does at 100, the minimum number of drops you'd need is 2. Drop at floor two. If phone 1 breaks, drop from floor one.

If you want the most efficient way for a non extreme case, my quick calculation for what I think is the ideal way to drop it is:

I think it's a matter of picking a big increment for phone 1 (between 2 and 50 stories). Then dropping phone 2 story by story higher from the last story that phone 1 survived.

If you pick an increment for phone 1 of i, you drop it 100/i times at most.

Then you're going to have to drop phone two a maximum of i-1 times.

Total drops (T) is then:

T=i-1+100/i

dT/di = 1-100*i^(-2) = 0 (at extreme) --> i=10

(d2T/di2=200i^(-3) -> curve (+) given i>0 so extreme point is a minimum.)

Easy wrote:Since you're lucky, you drop it from the right floor on the first try. However, you aren't certain unless you test the one above it. So you trudge up the stairs to the next floor and drop that sucker. It breaks.

Post Hole Digger wrote:

Easy wrote:Since you're lucky, you drop it from the right floor on the first try. However, you aren't certain unless you test the one above it. So you trudge up the stairs to the next floor and drop that sucker. It breaks.

This is the answer because the question was the "minimum", not which algorithm will give the lowest maximum.

How many dozen times does this unclever interpretation of the question need to be repeated? It's not funny anymore.

Real life questions:

Are there phones out there that can survive a 10 foot fall 100% of the time?

What is the terminal velocity of the average phone?

How high must it be dropped from to reach terminal velocity?

thank you

stop already wrote:

Post Hole Digger wrote:

This is the answer because the question was the "minimum", not which algorithm will give the lowest maximum.

How many dozen times does this unclever interpretation of the question need to be repeated? It's not funny anymore.

It wasn't meant to be funny. It's the real answer. I suspect that the question was meant to get people to come up with the algorithms for the lowest maximum trials and then reveal that two is the real answer to the actual question for the reasons above.

My daughter was always bringing me these. The question sounds like something that requires a complicated mathematical algorithm and I'd get out pencil and paper, only for her to reveal the very simple true answer that I should have gotten had I read the question more exactly.

btw, I admittedly didn't read every post in the thread. Can you count for me how many dozen times this answer was given?

Post Hole Digger wrote:

stop already wrote:

How many dozen times does this unclever interpretation of the question need to be repeated? It's not funny anymore.

It wasn't meant to be funny. It's the real answer. I suspect that the question was meant to get people to come up with the algorithms for the lowest maximum trials and then reveal that two is the real answer to the actual question for the reasons above.

My daughter was always bringing me these. The question sounds like something that requires a complicated mathematical algorithm and I'd get out pencil and paper, only for her to reveal the very simple true answer that I should have gotten had I read the question more exactly.

btw, I admittedly didn't read every post in the thread. Can you count for me how many dozen times this answer was given?

Maybe you should read the thread then. Although it was indeed poorly worded, obvious what the question meant and that it wasn't a "gotcha" question. The unclever misinterpretation has been posted 5-10 times and the correct answer has also been posted 3 or 4 times. Reading before writing (as well as listening before speaking) is always a good thing.

stop already wrote:

Maybe you should read the thread then. Although it was indeed poorly worded, obvious what the question meant and that it wasn't a "gotcha" question. The unclever misinterpretation has been posted 5-10 times and the correct answer has also been posted 3 or 4 times. Reading before writing (as well as listening before speaking) is always a good thing.[/quote]

OK, I read the thread. Mine was the 3rd to offer the solution of 2, one of which I quoted, so there was one answer of 2 that I didn't read. There were a whole bunch of attempts to be funny and clever without offering a real answer.

I'm don't know why you insist it is not a "gotcha". Reading the question again, the answer is clearly 2 and it is very much like the many "gotcha" riddles like this that I have seen before. I actually like this one. I'm a engineering professor and I might use this with students. They'll start laying out the optimization problem but hopefully it will dawn on one of them that the minimum is when you guess right on the first try so the answer is two.

I clicked on it because I thought it was going to be a variation on the classic freshman engineering question: "how do you use a barometer to measure the height of a building?". The obvious answer is to calculate the elevation difference from the pressure difference, but other answers are drop the barometer off the top and time how long it takes to hit the ground, tie it to the end of a string that you hang off the top, find the building supervisor and say "I'll give you this cool barometer if you tell me how tall the building is" and others. The point is to get them to "think outside the box" (that phrase come from the famous 9-dot exercise given to freshman engineers) and look for other solutions.

This riddle is along the same lines of teaching students to not automatically assume the obvious solution to a problem but instead examine the problem more carefully and see if there are simpler solutions.

I did a quick google, and others believe, like I do, that the answer is 2. e.g.,

In fact, based on "Assume you know that it breaks when dropped from the very top", you could say that the minimum is 1 if you first try the 99th (or 100th if "top" means roof) and it doesn't break. So you could make a good case that our answer of 2 is wrong and 1 is the real answer. If I give it to students, I might leave off the break from the top assumption. That leaves open the option that it doesn't break even from the top, so it'd have to be rephrased to find the highest floor it doesn't break at and confirm that it does break at the next floor, or something like that.

Either the 2 or 1 answer requires luck, but that's the minimum. What's the minimum number of trials required to make a full-court basketball shot? The answer is 1. You are unlikely to make it on the first try, but that is the minimum.