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cushioncrawler
09-11-2012, 04:44 AM
Dropping Balls Puzzle
The Puzzle:
You have to do an experiment to determine the highest floor on a 100-floor building from which a manufactured snooker ball may be dropped without breaking.

You are given two identical snooker balls, which you can drop from various floors of the building, to carry out your experiment.

If a ball doesn't break after being dropped, it may be reused without suffering any loss of quality. But if both balls break before you have determined the highest floor, then you are an incompetent bungler and your boss is ultimately going to fire you.

What is the least number of times you must drop the snooker balls in order to determine the highest floor?

Jal
09-21-2012, 12:32 AM
Mac, if I may, I think the question should read "what is the most number of times you would need to drop the balls in the worst case," otherwise you could get lucky with just two.

If each floor is 10 feet high and the terminal velocity of the balls is 120 mph, then you would only need to go up to the 49'th floor in the worst case. The balls will not pick up any more speed beyond that.

Dividing the floors into groups and testing the lowest floor of each group, when you've found the group in which one of the balls breaks, you then test the individual floors of the group below that one with the remaining ball. But what would be the most efficient way of dividing the floors up into groups for the worst case?

Let's say there are x floors in each group. The maximum number of drops, Nmax, would then be:

Nmax = 49/x + x - 1

The x-1 comes in because, working upward, you've already tested the first floor of the group just below the one in which a ball breaks. From calculus, Nmax is a minimum when:

0 = -49/x^2 + 1

which gives x = sqrt(49) = 7. Nmax is then equal to:

49/7 + 7 - 1 = 13

It's the same number if the first ball never breaks.

Jim

cushioncrawler
09-21-2012, 02:08 AM
On the other hand by uzing the word "must" it more or less negates the trick answer of "2".

I dunno about that terminal velocity bizness. My own tests show that Cd = 0.49 rather than 0.47. Hmmmm, no, 0.49 would make your assertion even truerer.

I think i worked out that 11 drops iz the least u kood get away with uzing a minimal attack, ie allowing for the best case.
But the answer iz more than that, ie allowing for the worst case, ie the worst floor. And it iz more than 13.
mac.

Jal
09-21-2012, 02:12 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: cushioncrawler</div><div class="ubbcode-body">On the other hand by uzing the word "must" it more or less negates the trick answer of "2".

I dunno about that terminal velocity bizness. My own tests show that Cd = 0.49 rather than 0.47. Hmmmm, no, 0.49 would make your assertion even truerer.

I think i worked out that 11 drops iz the least u kood get away with uzing a minimal attack, ie allowing for the best case.
But the answer iz more than that, ie allowing for the worst case, ie the worst floor. And it iz more than 13.
mac. </div></div>
So, ignoring the terminal velocity stuff such that you would have to consider all 100 floors, the optimal number of floors in a group, using the math in my first post, would be sqrt(100) = 10 to cover the worst case. Then, for the worst case, the most number of drops that would have to be performed is 10 + 9 = 19.

By worst case, I mean that which would require the most number of drops, which would, of course, be the best case in terms of ball durability. (I'm not sure if we're using "best" and "worse" in the same sense, or whether your "11" includes the terminal velocity reduction of the maximum height?)

Jim

cushioncrawler
09-21-2012, 03:36 PM
The answer will of course kover all possibilitys, and the best possibility uzing this worst possibility iz i think (not sure) when the kritikal floor iz 100, which will require 11 drops.

But by trial (and error), the only way (i think), i eventually (after an embarrassinnggllyy long time) got the korrekt (i think) answer which iz greater than 13 but less than 19.
mac.

Jal
09-21-2012, 05:44 PM
I'll await your solution.

Jim

cushioncrawler
09-21-2012, 06:27 PM
The drop count shalt be 14, no more, no less. 14 shall be the number thou shalt count, and the number of the counting shall be 14.
15 shalt thou not count, neither count thou 13, excepting that thou then proceed to 14.
15 is right out. Once the number 14, being the fourteenth number, be reached, then lobbest thou thy Holy Krappamith no more.

First drop iz at 14, if no breakage then next drop iz at 27, if no breakage then next drop iz at 39, if no breakage then next drop iz at 50, if no breakage then next drop iz at 60, if no breakage then next drop iz at 69, if no breakage then next drop iz at 77, if no breakage then next drop iz at 84, if no breakage then next drop iz at 90, if no breakage then next drop iz at 95, if no breakage then next drop iz at 99, if no breakage then answer iz 100. Here u hav had 11 drops and zero breakages.

If u get breakage at 14 then next drop iz at 1 then 2 then 3 etc, ie up to 13, ie total of 14 drops.

If u get breakage at 27 then next drop iz at 15, then 16 then 17 etc up to 26, ie total of 14 drops.
etc etc etc etc.

So, the answer iz 14 -- but the minimum possible number of drops iz 11, but this only applys if the balls break at floor 100.
mac.

Jal
09-22-2012, 10:04 AM
Very good, Mac. I would have never gotten it as I was too happy with the fixed grouping size.

Jim

cushioncrawler
09-22-2012, 04:42 PM
Jim -- U will like the 8 camels problem. This iz billiards related koz the two teams are travelling to their pool venues.
mac.

1hit1der
09-23-2012, 12:14 AM
Nice solution. But there is an assumption that the ball will break at all. I started down the fixed grouping size method myself, but realized there had to be a better way since the number of drops fluctuated so much depending on which floor within the grouping the ball would break on.

So mathematically, you could write for a building with N floors, the maximum number of drops can be minimized if the sum from 1 to i of x is &gt;= N-1. If it's not guaranteed to break at all, then the sum from 1 to i of x is &gt;= N. And the maximum number of drops is then i.

The camel puzzle isn't too hard. Took me a couple tries though.

cushioncrawler
09-23-2012, 12:56 AM
Yes that seems to work. So 14 drops would accommodate up to 105 floors. And the minimum least number of drops iz allways 2.
mac.

Some peeple (uzually senior mangagers) take weeks to find the camel solution. Would never get a job in arabia.
mac.

1hit1der
09-23-2012, 11:05 PM
I'm not sure how you derive the mathematical solution from scratch though.

Jal
09-24-2012, 10:25 AM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Jal</div><div class="ubbcode-body">...If each floor is 10 feet high and the terminal velocity of the balls is 120 mph, then you would only need to go up to the 49'th floor in the worst case. The balls will not pick up any more speed beyond that....</div></div>
I unthinkingly derived this using a constant acceleration of g up to the time when it reaches terminal velocity, which is of course, nonsense. The acceleration of the ball is ever diminishing and therefore, in the mathematical sense, it never actually reaches terminal velocity.

Jim

cushioncrawler
09-24-2012, 03:52 PM
Plus gravitational attraktion, if indeed that iz what it iz, iz greatest at floor 1.
I suppoze that some purists would say gravitational acceleration g, not attraktion.
And some purists would say it woz neither attraktion nor acceleration but merely a bending of space-time.
And us aetheristians would say that it iz drag, ie a sort of gravitational push.

Unless of course the building haz been built so far underground that the greatest attraktion g bending drag iz at the top floor.
mac.

Duzz air density get greater quicker than g gets greater?????
If it duzz then V might dekreec (a little) az u get lower.
But this wouldnt change the original question and answer.

No, it would/might change the answer. Koz if the final V woz lesserer az u dropped from higherer then the whole logik etc goze out the window.

Jal
09-24-2012, 11:37 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: 1hit1der</div><div class="ubbcode-body">I'm not sure how you derive the mathematical solution from scratch though. </div></div>
I can't derive the fact that it should be an arithmetic series from pure math. But with hindsight (provided by Mac), I think you could argue for it, and one in which each successive term is <u>one</u> less than the previous term. (Mac would be the best source for this.)

At any rate, given that it is an arithmetic series, the first term "a" (size of the first group, 14 in this case), can be obtained from the fact that the last term must be greater than zero. It (a) is then given by the odd looking formula:

a = (1/2)[√(8N+9)-1]

which for N = 100, yields a &gt; 13.72. Use either the next higher or lower integer, whichever works.

Jim

cushioncrawler
09-25-2012, 12:30 AM
........But with hindsight (provided by Mac).........

Mac missed grade 1 and grade 2 but went straight to grade 3 but then spent 2 yrs in grade 3, whence he changed skools and went up to grade 4 and promptly got zero out of 25 for a spelling test. After that things went downhill. Anyhow u would probly find mac up in that skyskraper dropping krapps, to save time.
mac.

Jal
09-25-2012, 10:02 AM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Jal</div><div class="ubbcode-body"><div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: 1hit1der</div><div class="ubbcode-body">I'm not sure how you derive the mathematical solution from scratch though. </div></div>
I can't derive the fact that it should be an arithmetic series from pure math. But with hindsight (provided by Mac), I think you could argue for it, and one in which each successive term is <u>one</u> less than the previous term. (Mac would be the best source for this.)

At any rate, given that it is an arithmetic series, the first term "a" (size of the first group, 14 in this case), can be obtained from the fact that the last term must be greater than greater than zero. It (a) is then given by the odd looking formula:

...garbage...

Jim </div></div>
I've been having a problem with basic math, but finally got it right. /forums/images/%%GRAEMLIN_URL%%/blush.gif The formula should read:

a ≥ (1/2)[√(8N+1)-1]

which for N = 100, yields a ≥ 13.65. Round up to the next integer when a fractional part exists.

Jim

Jal
09-25-2012, 10:48 AM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: cushioncrawler</div><div class="ubbcode-body">...
Duzz air density get greater quicker than g gets greater?????
If it duzz then V might dekreec (a little) az u get lower.</div></div>
Mac, I should think the drag force increases faster than mg, since g has the whole radius of the earth behind it, so to speak...and despite the fact that it varies with R^2. As far as V decreasing, I'd have to do the math, which at this point isn't such a good idea. My feeling is that it still wouldn't reach a steady velocity in the strict mathematical sense, much less decrease, since air density is continuously increasing as opposed to a sudden jump. But maybe there is a rate of increase, a threshold gradient, in which that can happen?

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: cushioncrawler</div><div class="ubbcode-body">But this wouldnt change the original question and answer.

No, it would/might change the answer. Koz if the final V woz lesserer az u dropped from higherer then the whole logik etc goze out the window. </div></div>
I thought the terminal velocity thing might be a trick you had in store for us, but it really isn't in the spirit of your puzzle (and probably irrelevant according to the above).

By the way, I counted and got the camel puzzle in 1.97 thoughts, so I'll have to lay claim to the world record.

Jim...on the other hand, I probably counted wrong

cushioncrawler
09-25-2012, 04:43 PM
I am thinking that everyone gets camel gridlock with their first attempt. Then by not rushing most get it on their second go, ie by putting their toe in the water to check.

But in fakt i think that it shood be ok to rush, ie to make 10 mistakes along the way, koz in most cases the rusher would get the korrekt solution before the timid scaredycat, allbeit with 10 mistakes instead of 1.1111111111 mistakes.

Bearing in mind that the way the camel puzzle iz worded, all who hav made even 1 mistake are skrewed, and (all or some of) the camels will die. So tiptoeing so az not to make any more errors aint gonna help after the gate iz klozed.
mac.

I do well in IQ tests, except that i am so slow, so i take my hat off to speedsters.

Jal
09-25-2012, 05:42 PM
I didn't get it until I realized it's the death knell to have two consecutive camels from the same group with the space behind them. Aside from which group makes the first move, I think every move then follows "naturally" by just avoiding that situation.

(I was kidding about getting it immediately.)

Jim