Or like this.
merthsoft wrote:
That is right, so what's the maximum tries it will take in that case, and what's the general solution for n bulbs and m floors?
Look up the manufacturer's documentation on the height at which 95% of bulbs break?
If you knew the height from which a lightbulb will reach terminal velocity, you only need to go that high at most. Anything higher should return identical results.
Nice outside-the-box thinking for both of you. I think those would be good interview question answers.
DShiznit wrote:
If you knew the height from which a lightbulb will reach terminal velocity, you only need to go that high at most. Anything higher should return identical results.
I doubt a standard lightbulb shape is going to reach terminal velocity so close to the ground. Forms approximating a teardrop should do very well aerodynamically speaking.
it's going to rapidly flip around and trail behind so that the bulbous portion is in front. To be aerodynamically stable the center of mass should be in front of the center of drag.
That's true, but the flat end would create a small vacuum as the air rushes past it, pulling on the bulb. This would only matter of course if you're using super heavy duty industrial encased lights that could survive a fall at terminal velocity.
KermMartian wrote:
But does being a CS major have to do with being physically fit? I'm fit and a CS major. I find this question offensive.
My apologies - didn't intend for it to be offensive
KermMartian wrote:
Also, do we have to take into account retrieving the unbroken lightbulb each time?
No. If you drop a bulb and it does not break, it will be replaced such that you do not need to walk all the way down.
@Qwerty.55 Having an expected value _could_ help out a lot. The "correct" answer assumes that the probability of a a certain level being the point at which a lightbulb breaks is evenly distributed. If you knew that the expected threshold floor was 5, then instead of dropping the first bulb from floor 50, you could start from floor 4.
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