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# Classic eggs problem

6 replies to this topic

### #1 ujjagrawal

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Posted 09 July 2012 - 11:47 AM

You have been given three eggs and your job is to figure out how high an egg can fall from a 120 story building before it breaks. The eggs might break from the first floor, or might even survive a drop from the 120th floor, you have no prior information about it. Except all three eggs are know to be of exactly same strength.

What's the most efficient way to drop the eggs i.e. reducing the number of times you need to drop eggs and still able to determine the answer? You are allowed to break all three eggs, as long as you identify the correct floor afterwards.

After you've solved the above problem, generalize. Define the "break floor" as the lowest floor in a building from which an egg would break if dropped. given an n story building and a supply of m eggs, find the strategy which minimizes (in the worst case) the number of experimental drops required to determine the break floor.

Edited by ujjagrawal, 09 July 2012 - 11:49 AM.

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### #2 phil1882

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Posted 09 July 2012 - 03:06 PM

Spoiler for 3 eggs

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### #3 jim

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Posted 09 July 2012 - 03:50 PM

Reverse the problem and let m be the number of drops, and then n is the tallest building that can be done using this number of drops. If you only have one egg n=m. With two eggs n = 1 + 2 +...+m = m(m+1)/2---call this b(m). With three eggs we get n = b(1) + ...+b(n) which we can call c(n) and so on. Now c(m) is half the sum of 1+...+m and 1+...+m^2 or half m(m+1)/2 and m(m+1)(2m+1)/6 so we get c(m) =(1/6)m^3 + (1/2)m^2 + (1/3)m or c(m) = (m^3 + 3m^2 +2m) / 6. For three eggs we get c(8) = 120 so the answer to the basic problem is 8.
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### #4 vinay.singh84

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Posted 09 July 2012 - 06:49 PM

Spoiler for My guess

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### #5 jim

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Posted 09 July 2012 - 09:07 PM

Correction. Let h(k,m) be the height that can be handled with k eggs and up to m drops. Above I used a recursion of h(k,n) = h(k,n-1) + h(k-1,m) when I should have used a slightly different formula. Let the base at any time be the highest known safe floor. With k eggs and m drops left we can afford to skip h(k-1,m-1) floors and do our next drop at base + h(k-1,m-1) + 1. When an egg breaks the base remains the same and both k and m decrease by 1. If the egg doesn't break, increase the base by h(k-1,m-1) + 1 and then decrease just m by 1.

h(1,m) = m where start dropping at the lowest floor and work upwards one at a time.

h(2,m) = (m-1)+1 + (m-2)+1 + ... + (m-m)+1 = SUM of m, m-1, m-2, ...1 = m(m+1) / 2

h(3,m) = (m-1)m/2 +1 + (m-2)(m-1)/2+1 + ... = SUM (1/2)j^2 - (1/2)j +1 for j from 1 to m and we get
(1/12)m(m+1)(2m+1) - (1/4)m(m+1) + m = (1/6)(m^3 -m) + m = (1/6)(m-1)m(m+1) + m. so 8 is not enough bu 9 drops would let us go as high as 120 + 9 which is enough.
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### #6 ujjagrawal

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Posted 10 July 2012 - 05:30 AM

Good work dude... the same answer I have got but with different working...
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### #7 ~andy~

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Posted 26 October 2012 - 03:04 PM

Spoiler for It's simple

Can't believe you guys didn't get that one!
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