Fri Dec 07, 2007 6:14 pm by tartle 


you have two identical crystal orbs. you need to figure out how high an orb can fall from a 100 story building before it breaks. you know nothing about the toughness of the orbs: they may be very fragile and break when dropped from the first floor, or they may be so tough that dropping them from the 100th floor doesn't even harm them.
what is the largest number of orbdrops you would ever have to do in order to find the right floor? (i.e. what's the most efficient way you could drop the orbs to find your answer?)
you are allowed to break both orbs, provided that in doing so you uniquely identify the correct floor. 




Tue Dec 18, 2007 3:24 am by Rick 


It may be less but I think I could do it in at most 14 drops. 




Tue Jan 08, 2008 5:02 am by aatifahmed 


The answer is 7 drops.
The solution I am submitting below is encrypted so as not to take the fun out of your efforts! To decrypt use: www.rotn.spam
The cipher used is ROT18ROT5 & ROT13.
Guvax bs vg nf n ovanel ceboyrz, naq pbaireg vg vagb ovanel. Fgneg sebz bayl 65 fgbel uvtu ohvyqvat. Guhf lbh unir 6655 sbe 65. Rnpu cbfvgvba bs gur qvtvgf va 6655 qrfpevorf gur fgbel sebz juvpu gur beo vf gb or qebccrq, va gur qbjajneq snfuvba. Fb sbe rknzcyr bs 65, vg fubhyq or qebccrq sebz fgbel ahzore =, 9, 7 naq 6 erfcrpgviryl.
Abgr  O zrnaf Oernx, AO zrnaf Ab Oernx.
Guhf sbe 65:
Q=AOQ65AO65 vf gur znk sybbe
Q=AOQ65OQ>AO> vf gur znk
Q=AOQ65OQ>O= vf gur znk sybbe
Q=OQ9AOQ;AOQ<O; vf gur znk sybbe
Q=OQ9AOQ;AOQ<AO< vf gur znk sybbe
Q=OQ9AOQ;OQ:AO: vf gur znk sybbe
Q=OQ9AOQ;OQ:O9 vf gur znk sybbe
Q=OQ9OQ7AOQ8AO8 vf gur znk sybbe
Q=OQ9OQ7AOQ8O7 vf gur znk sybbe
Q=OQ9OQ7OQ6AO6 vf gur znk sybbe
Q=OQ9OQ7OQ6O5 vf gur znk sybbe
Nf h pna frr sebz nobir, abar bs gurz gnxr zber guna 9 qebcf! Fvzvyneyl vg pna or jbexrq bhg sbe 655, juvpu va ovanel vf 6655655, univat < qvtvgf naq urapr < qebcf!! 




Tue Jan 08, 2008 6:52 am by aatifahmed 


My solution holds true if infinite number of Orbs are available, for minimum number of drops.
The correct answer to the problem is 14. 




Mon Dec 22, 2008 10:24 pm by alexonfyre 


yeah, radically different math for the two solutions as well ;) 




Mon Jun 28, 2010 3:26 pm by shrek619 


i think the anwser is 7 (minimum no.). 




Thu Jun 02, 2011 5:26 pm by Unni 


Correct answer is 10 




Thu Jun 23, 2011 9:44 am by titu 


I guess only [b]18[/b] drops would be sufficient to test the orb being dropped from 1 through 100 floors. Please let us know the answer. 




Thu Feb 14, 2013 8:34 am by asheeshg 


largest number of drop = 18.
suppose you start from 10 and then 20 and then 30........ 100
now consider the situation...
first orb cracked at 10: then you have to come from 1 to 9 as well = 10
first orb cracked at 20: then you have to come from 11 to 19 as well = 11
first orb cracked at 30: then you have to come from 21 to 29 as well = 12
...
...
first orb cracked at 90: then you have to come from 81 to 89 as well = 18
but if orb is not cracked at 90, here you can change the strategy to come in 13 drop.
10,20,30,40,50,60,70,80,90,94,97,99, 100
if 94: 91,92,93 = 13
if 97: 95,96 = 13
if 99: 98 = 13 




Thu May 30, 2013 3:29 am by chrisfortytwo 


Sorry, asheeshg, but your 13 solution misses the mark at 80. If the first one breaks at 80, you might need 9 more drops to get to 90.
The key is to balance the number of drops left with the number gained.
So, first you drop at X, and if it breaks, you need X1 drops to fill from 1
Then at X + (X1), you have X2 drops to fill from X.
Then at X + (X1) + (X2), you have X3 drops to fill from 2X1.
Then at X + (X1) + (X2) + (X3), etc.
Until you top out at X + (X1) + (X2) + ... + (X(X1)) + (XX), which, when reversed is Sum( 0: X ), or X/2*(X+1).
Then we just need to find out the smallest integral value of X where X/2*(X+1) > 100.
It turns out that 13/2*14 = 91, and 14/2*15 = 105, so we need at least 14 drops in the worst case.
First orb drops are at 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99 and 100.
Whenever the first bulb breaks, you start at the next lower number and count up by one. It is impossible to do with a worst case of 13 drops. 




Tue Sep 24, 2013 12:39 pm by hasz921 


i think the answer is 14. Those claiming 7 shall justify. 




Thu Oct 17, 2013 9:26 am by daleman21 


The answer is 7.
Drop from 50
If it breaks drop from 25, if it doesn't drop from 75
If it breaks from 25 drop from 13 if it doesn't drop from 38
(Similarly, if it breaks from 75 drop from 63, if it doesn't drop from 88)
Assuming each time we cut in half how many floors it could be (rounding up due to no halflevels) we get:
after 1 drop  50 possibilities
after 2 drops  25 possibilities
after 3 drops  13 possibilities
after 4 drops  7 possibilities
after 5 drops  4 possibilities
after 6 drops  2 possibilities
after 7 drops  1 possibility AKA our correct floor 




Fri Nov 01, 2013 12:22 am by BaggyT 


Daleman21,
Sorry, but your solution doesn't work:
You only have 2 orbs. If the first one breaks at floor 50, and the second at floor 25, then you can't progress.
I have a definite method for an answer of 14.
:) 




Mon Nov 11, 2013 11:51 pm by Balognadiddler 


I'm thinking 19. You have two orbs, so you have to divide the floors into two subsets. If you graphed the ideal way to break it up, the square root of the total number of floors would be at the apex of a quadratic curve. That means we group it into ten sets of ten.
First we go up ten floors at a time, and the most tests we need to get the first break will be 10 if it breaks at floor 100. This means that the break point is anywhere from 91100. Next we test floors 9199. If the breaking point is floor 100, then the second orb will not break on any of these tests, meaning we have to test nine times.
So in a worstcase scenario, the first orb gets tested ten times, the second gets tested nine times, for a total of 19.
I must respectfully disagree with those who have suggested the halving method, or whatever it's called. Say you break orbs on the first two drops. Now you're stuck at 25 and only know that the correct floor is 25 or lower.
Edit: My method also gives a lowest number of trials of two. That happens if the first breaks at 10 and the second breaks at 1. 




Tue Dec 10, 2013 5:56 pm by hakan 


49 






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