Interesting maths problem - Page 2

yeah, 512

(don't forget you have just two eggs, so you can't apply binary searching; suppose you take your SUPPOSEDLY unbreakable egg and drop it from the 25th floor, now you have one egg left and no clue what the critical floor is)

drop at floor f and it breaks, so you drop at f-2; if it breaks at f-2, you're done. if it doesn't break, you have to drop at f-1
drop at floor f and it doesn't break, drop at f+3
so you start from floor 3

so if they were unbreakable you would drop all 16 times to get to the 48th floor but because 3 goes into 50 with a remainder , you have to drop again at the 50th floor (if it were to break at the 50th, you would have to use the 18th drop on the 49th floor, but otherwise, it would take 17 drops)

if they are unbreakable, you have to drop 17 times to SAFELY verify that they are unbreakable; if you drop more than this, you're fine, because you would just be doing a linear evaluation like start at floor 1 and go to 50 (if the eggs' integrity isn't affected by the impacts). but if you drop fewer times than this, you get fucked up if they DO break like the people who want to do binary searching.

after you drop at 48, you don't have to drop at 49 before 50 which is why it doesn't take 50/3 + remainder drops (that would be 18)

it takes 17 drops because we don't KNOW that they're unbreakable, if we knew they were unbreakable it would only take 1 drop (at floor 50, but if your egg broke at floor 50, you would have to do a linear search and drop a max of 49 times to safely figure it out). we have to test it as though we didn't know, which is the above method. 17 drops.

The correct answer for the first problem should be 6. You start on the 25th floor and go up half the floors left if the egg doesn't break. At best case scenario, where the egg breaks only at the 50 floor or doesn't break at all, you would test 6 times.

25, 38, 44, 47, 49, 50

If the egg breaks at any point before the 50th floor, then you would have to start testing from the floor above the last floor that it didn't break from. But yeah, the least number of tests to find the floor which the egg breaks is 6.
Nevermind. Was only thinking about best case scenario.

People already got this one. It's 8^3, which is 512.

stop trying to use binary searching. it's tantamount to saying "I know this egg is unbreakable so I will just drop it on the 50th floor so it only takes me 1 drop"

My 17 drop method is in fact not the most efficient (I retract any implications I made to that effect) - although it's miles better than the binary searching bullshit which isn't even valid. I see the 10 drop method now. That's really elegant, and I'm going to remember that.

10,19,27,34,40,45,48,50,49

or, if it breaks on 10
10,1,2,3,4,5,6,7,8,9




actually, i guess the best way to go would be
10,18,26,33,39,44,47,48,49,50

The answer is 2 drops. That is the minimum number you would need to find out at which floor it would break.

Example: Let us say the egg will break if you drop it off the 30th floor or higher. For your first attempt you just happen to drop it off the 30th floor. For your 2nd attempt, you go down one floor and drop it off the 29th floor. You now know that it will break on the 30th floor but no lower. Problem solved, 2 drops.

Alternatively, if the eggs really are unbreakable, you could figure that out with one drop off the 50th floor. In that case, the answer is 1 drop.

512, everybody already posted that one. 8x8x8 == 512

if you had infinite number of eggs you'd just binary search, starting at 25 (wiki can explain that better then i can). but obviously with only two eggs thats out.

so you have to pick a large increment, drop the egg, if it doesn't break, you move up by one more increment and try again until it does. once you find the unsafe large increment floor, you then work your way up to that in one's from the last highest "safe" known floor, one at a time.

so

d=number of drops with egg2
(1+d)+(1+(d-1))+(1+(d-2))+.....+1+0 = no floors
let 1+d=x=total number of drops
x+(x-1)+(x-2).....= floors
x(x+1)/2 = floors
x(x+1) = 2(floors)
x(x+1) - 2(floors) = 0
x^2 +x - 2(floors) = 0

hope you see where i'm coming from now.

i never realized typing formulae was such a pain in the ass