OpenEuphoria

Re: Puzzle challenge:

• Posted by "Mike Nelson" <MichaelANelson at worldnet.att.net> Aug 10, 2004
• 442 views

Mr Trick wrote <snip>
> So, 19 tries. I can do it in 14 though....14 or less tries, no matter
> what floor it breaks on... (Guess how)
> Haven't figured out a general case algorithm yet though....


First drop is on 14: if it breaks, then dropping on 1-13 (at worst 13 drops)
will find the right floor.
If it doesn't break on 14, second drop is on 27: if it breaks, then dropping
15-26 (at worst 12 drops) will find the right floor.
If it doesn't braek on 27, the third drop is on 39 . . .

The sequence of drops until the first ball breaks is 14, 27, 39, 50, 60, 69,
77, 84, 90, 95, 99

For b=2, the number of drops (d) for a given f  is the smallest d such that
d*(d+1)/2 is greater than or equal to f.

Later I will try to generalize this.

-- Mike Nelson