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Interesting maths problem - Page 2

Blogs > MakkurtE
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Prev 1 2 3 Next All
oBlade
Profile Blog Joined December 2008
United States5496 Posts
November 10 2009 19:33 GMT
#21
#2

+ Show Spoiler +
yeah, 512


#1

+ Show Spoiler +

(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.

"I read it. You know how to read, you ignorant fuck?" - Andy Dufresne
AeTheReal
Profile Joined June 2009
United States108 Posts
Last Edited: 2009-11-10 21:44:49
November 10 2009 20:04 GMT
#22
+ Show Spoiler [Question #1] +
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.

+ Show Spoiler [Question #2] +
People already got this one. It's 8^3, which is 512.
oBlade
Profile Blog Joined December 2008
United States5496 Posts
November 10 2009 20:15 GMT
#23
Question #1 revisited


+ Show Spoiler +
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.

"I read it. You know how to read, you ignorant fuck?" - Andy Dufresne
Deleted User 3420
Profile Blog Joined May 2003
24492 Posts
Last Edited: 2009-11-10 22:31:44
November 10 2009 20:31 GMT
#24
@LiAIH4

i think ur rightbut ur order for dropping can be improved on. here is what i put

+ Show Spoiler +

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


AeTheReal
Profile Joined June 2009
United States108 Posts
Last Edited: 2009-11-10 21:31:38
November 10 2009 21:23 GMT
#25
Edit: Oops, hit back too much and resubmitted old stuff.
Edit2: Looks like LiAIH4 has the right answer.
gyth
Profile Blog Joined September 2009
657 Posts
November 10 2009 21:44 GMT
#26
For all integers 4x^2 + x = 3y^2 + y


Are there any non zero solutions to that equation?
The plural of anecdote is not data.
Bearigator
Profile Blog Joined July 2009
United States233 Posts
Last Edited: 2009-11-10 21:58:18
November 10 2009 21:55 GMT
#27
Question 1
I think everybody is over complicating this.
+ Show Spoiler +
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.

It won't work without fail obviously, but it accurately answers the question as given. Maybe I am that jerk who looks at the wording too hard though, idk.

Question 2
+ Show Spoiler +
512, everybody already posted that one. 8x8x8 == 512
MakkurtE
Profile Blog Joined July 2009
United States46 Posts
November 10 2009 22:04 GMT
#28
yes, you're reading it badly. so badly.

It won't work without fail obviously
Opinions in the above post are less informed then they appear
Bearigator
Profile Blog Joined July 2009
United States233 Posts
Last Edited: 2009-11-10 22:32:52
November 10 2009 22:30 GMT
#29
On November 11 2009 07:04 MakkurtE wrote:
yes, you're reading it badly. so badly.

Show nested quote +
It won't work without fail obviously

The question never says it has to work 100% of the time, just the minimum number. Hence my statement that I might just be reading it too literally.
MakkurtE
Profile Blog Joined July 2009
United States46 Posts
November 10 2009 22:39 GMT
#30
it also never said "if the first floor you randomly choose is the one the above breaking point" but don't let that stop ya

shine on you pedantic diamond
Opinions in the above post are less informed then they appear
conCentrate9
Profile Blog Joined December 2007
United States438 Posts
November 10 2009 22:50 GMT
#31
On November 11 2009 02:39 MakkurtE wrote:
solution:

+ Show Spoiler +

x^2 + x - 2(floors)=0 will solve for 2 eggs and any number of floors

in this case optimal is 9.xxxx, but you can't drop an egg 9 and a half times - so 10.

for the answer to be a perfectly round number the number of floors has to be a triangular number

i love the elegance of the method to get to the equation more then anything else really


Could you explain how you arrived at the equation? I'm not seeing the method.
JeeJee
Profile Blog Joined July 2003
Canada5652 Posts
November 10 2009 22:55 GMT
#32
On November 11 2009 07:39 MakkurtE wrote:
it also never said "if the first floor you randomly choose is the one the above breaking point" but don't let that stop ya

shine on you pedantic diamond


no need to get uppity, makkurte, he's not wrong. the question should have been phrased better.
(\o/)  If you want it, you find a way. Otherwise you find excuses. No exceptions.
 /_\   aka Shinbi (requesting a name change since 27/05/09 ☺)
Bearigator
Profile Blog Joined July 2009
United States233 Posts
Last Edited: 2009-11-10 23:16:32
November 10 2009 23:11 GMT
#33
On November 11 2009 07:39 MakkurtE wrote:
it also never said "if the first floor you randomly choose is the one the above breaking point" but don't let that stop ya

shine on you pedantic diamond

I pointed out why my answer was probably flawed. I was very clear that my answer was just taking advantage of the wording. No need to try to use ridiculous extremes and sarcasm to point out what I already did.

Edit: Also I think your formula will not solve for any number of floors. Edit again after I actually do the math.
Impervious
Profile Blog Joined March 2009
Canada4198 Posts
Last Edited: 2009-11-10 23:15:28
November 10 2009 23:14 GMT
#34
Tried to quote someone and their post was deleted.....
~ \(ˌ)im-ˈpər-vē-əs\ : not capable of being damaged or harmed.
MakkurtE
Profile Blog Joined July 2009
United States46 Posts
November 10 2009 23:16 GMT
#35
jeejee/bear - thats i know it's awkwardly worded, thats why i clarified it properly, several times. going back and picking at the original statement is pointless and tiresome


and lol, are you kidding me?

x^2 + x - 2(floors) = 0
x^2 + x - 2(10) = 0
x^2 + x - 20 = 0
(x+5)(x-4) = 0
x = -5 x = 4
x = -5 inadmissable

x=4 in the answer
Opinions in the above post are less informed then they appear
Bearigator
Profile Blog Joined July 2009
United States233 Posts
Last Edited: 2009-11-10 23:21:34
November 10 2009 23:18 GMT
#36
On November 11 2009 08:16 MakkurtE wrote:
jeejee/bear - thats i know it's awkwardly worded, thats why i clarified it properly, several times. going back and picking at the original statement is pointless and tiresome


and lol, are you kidding me?

x^2 + x - 2(floors) = 0
x^2 + x - 2(10) = 0
x^2 + x - 20 = 0
(x+5)(x-4) = 0
x = -5 x = 4
x = -5 inadmissable

x=4 in the answer

How would you go about dropping an egg 4 times to solve it 100% of the time? Seriously, not being an annoying ass, I'm doing different combinations of floor drops and I can't find one that works every time in 4.
MakkurtE
Profile Blog Joined July 2009
United States46 Posts
November 10 2009 23:20 GMT
#37
conCentrate9:

+ Show Spoiler +
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
Opinions in the above post are less informed then they appear
Impervious
Profile Blog Joined March 2009
Canada4198 Posts
November 10 2009 23:21 GMT
#38
4 - 7 - 9 - 10

If it breaks on 4, start at 1 and go to 3. Max of 4 drops.
If it breaks on 7, start at 5, and go to 6. Max of 4 drops.
If it breaks on 9, start at 8. Max of 4 drops.
If it breaks on 10, you know where it broke.

Easy.
~ \(ˌ)im-ˈpər-vē-əs\ : not capable of being damaged or harmed.
MakkurtE
Profile Blog Joined July 2009
United States46 Posts
November 10 2009 23:22 GMT
#39
yup
Opinions in the above post are less informed then they appear
Bearigator
Profile Blog Joined July 2009
United States233 Posts
November 10 2009 23:22 GMT
#40
On November 11 2009 08:21 lMPERVlOUS wrote:
4 - 7 - 9 - 10

If it breaks on 4, start at 1 and go to 3. Max of 4 drops.
If it breaks on 7, start at 5, and go to 6. Max of 4 drops.
If it breaks on 9, start at 8. Max of 4 drops.
If it breaks on 10, you know where it broke.

Easy.

Ah, my bad. I even did that combination of drops. I must have counted wrong or something. My bad.
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