Three egg proble​m

If we throw from floor 14 again, and it breaks, then we recurse. n(n+1)/2=k where k is now 13... but that gives us 4.815, if we ceil and that and add our previous drop we get 6, which is lower than the actual solution, so something here is wrong...

What if it doesn't break? Then you have a three-egg problem with 86 floors, which takes maybe one drop less to solve than the 100-floor problem.

Say you drop the first egg from the 50th floor. If it breaks, you have a two-egg problem with 49 floors, which takes up to 10 additional drops. So that would give you a worst-case of 11 drops (since if it doesn't break, the 50-floor three-egg problem takes at most 7 additional drops).

If you choose the 37th floor for the first drop, if it breaks, you have a 36-floor two-egg problem, needing up to 8 additional drops. If it doesn't break, you have a 63-floor three-egg problem left. You want to solve that problem with at most 8 drops, so if the next drop breaks the egg, the remaining two-egg problem should be solvable in at most 7 drops, thus the highest floor you can choose for the second drop is 37 + 28 + 1 = 66, since 28 floors is the highest you can solve with at most 7 drops and two eggs. If the egg doesn't break, you have a 34-floor three-egg problem with 7 drops left. The highest you can certainly solve with the remaining 6 drops if the egg breaks is 21 (6*7/2), so you can choose floor 66 + 21 + 1 = 88. If the egg doesn't break, you have 12 floors left with 6 drops, which is already doable with only two eggs.

Systematically, the highest number of floors you can certainly solve with d drops and e eggs is

          / 1, if d == 1
F(e,d) = |  d, if e == 1
          \ F(e-1,d-1) + 1 + F(e,d-1), if e > 1 and d > 1

If you have only one drop, you have no choice but to choose the lowest floor of which you do not yet know that the egg doesn't break. If that breaks it, and you tried a higher floor, you don't know the first floor to break the egg.

If you have only one egg, you have to check every floor in order until the egg breaks or you run out of drops.

Otherwise, if the first drop is from a floor higher than F(e-1,d-1) + 1, you possibly can't find the first breaking floor if the egg breaks. If the first drop is from a lower floor, you can't reach as high with d-1 drops if the egg doesn't break, so the first drop should be from floor F(e-1,d-1) + 1. If it breaks, you can solve with the remaining e-1 eggs and d-1 drops by assumption. If not, you can solve for the next F(e,d-1) floors with the remaining drops and eggs.

Conversely, to find how many drops you may need for f floors with e eggs, you have to find

D(e,f) = min { d | F(e,d) >= f }

You can find that by calculating the F(e,d) matrix, or you can use dynamic programming:

If you choose floor s for the first drop, if the egg breaks, you need up to D(e-1,s-1) drops to determine the floor. If the egg doesn't break, you need up to D(e,f-s) drops to determine the floor. So the worst case for choosing floor s for the first drop is

WC(s,e,f) = 1 + max { D(e-1,s-1), D(e,f-s) }

and the best of the worst case is

D(e,f) = minimum { WC(s,e,f) | 1 <= s <= f }

(where of course D(e,0) = 0).


This problem can be solved with following 3 approaches (that I know) :

  1. Dynamic Programming
  2. Solution using Binary Search Tree
  3. Solution by obtaining the direct mathematical formula for maximum number of floors that can be tested or covered with given number of eggs and given number of drops

Let me first define some symbols as follows: 

e = number of eggs
f = number of floors in building
n = number of egg drops 
Fmax(e, n) = maximum number of floors that can be tested with e eggs and n drops

The crux for dynamic programming approach lies in following recursive formula for Fmax:

Fmax(e, n) = 1 + Fmax(e-1, n-1) + fmax(e, n-1)

And the crux for obtaining the direct mathematical formula for Fmax lies in following recursive formula for Fmax:

Fmax(e, n) = { ∑Fmax(e-1,i) for i = 1 to n } - Fmax(e-1, n) + n

Alternative solution using Binary Search Tree (BST) is also possible for this problem. In order to facilitate our analysis, let us draw BST with slight modifications as follows:

1.    If egg breaks then child node is drawn on left down side
2.    If egg does not break then child node is drawn straight down side

If we draw BST with above kind of representation then width of the BST (i.e. number of vertical columns in BST) represents the number of eggs.

Any BST with f number of nodes, drawn with above kind of representation and subjected to the constraint width of BST <= e (number of eggs) is a solution but it may not be the optimal solution.

Hence obtaining the optimal solution is equivalent to obtaining the arrangement of nodes in BST with minimum height subjected to the constraint: width of BST <= e

For more details about all the above 3 approaches, check my blog at: 3 approaches for solving generalized egg drop problem

Tags:

Algorithm

Math

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