Thu Jul 21, 2016 9:57 am by bathfilms |
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Hi Dzallen,
Yes, I see your point..
After reconsidering the problem, I have worked out that the optimum pseudo-binary splitting would yield a new answer of 21 drops:
We start on floor 5 and then move to floor 10 or floor 1 (depending on ball survival or breakage respectively)... see the flow chart below:
[Floor numbers]
..............................25..etc...
..........................20
........................./...16...etc...
......................../
.....................15....
..................../....\...12
.................10......11
................/....\.......10
............../.......\..7
............5......4..6
..............\...3......5
...............\./..2
................1
..................0
This allows a maximum of 2 breakages.
If the flow chart is extended, it shows that 21 drops are required to determine 1 to 100 floors inclusively.
However, to my surprise this pseudo-binary splitting has not produced the minimum.
Therefore I agree with the answer of 14 as described by others. |
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