Posted: 8/14/2007 7:06 PM
Post subject: Eternity II (real)puzzle
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maTO
Skilled player (1514)
Joined: 4/28/2005
Posts: 240
Location: Finland
Sega TASer of 2009
How come there's no topic on this puzzle? I saw a topic on another forum and had to show it to you :) Homepage First one to solve this puzzle wins $2M. Eternity1 was solved in 7 months [picture]. This second, harder version, was released over two weeks ago. Maybe someone has already bought this?
Projects: Warlock, Ren & Stimpy (impr) / Generations Lost (impr.)
Posted: 8/14/2007 7:13 PM
Post subject: Re: Eternity II (real)puzzle
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Bisqwit
Editor, Active player (297)
Joined: 3/8/2004
Posts: 7469
Location: Arzareth
Ah, I've read about it. They kind of overlooked some mathematical similarity for algorithmical solving in the previous version. Guess they didn't do the same mistake this time, so it's really up to luck this time.
Posted: 8/14/2007 8:14 PM
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Tub
Joined: 6/25/2005
Posts: 1377
Good news! That's only 4! * 56! * 196! * 4^196 possible combinations! Less than 10^560! seriously, does anyone know the amount of colors they use, and the general distribution of the tiles? edit: Wikipedia mentions 22 colors (+ gray edges) that leaves us with around 60K possible pieces, so I'll just assume that every piece is unique. I'll also assume that colors and combinations are evenly distributed. (probably wrong, but who cares) If two adjacent fields are already placed, you'd expect around two of the 256 pieces to match. With a brute force strategy starting at the upper left corner and filling in tile after tile, row after row, you'd need to search less than 2^256 ~ 10^77 possible combinations. Way lower, although still way too much.
m00
Posted: 8/16/2007 6:59 AM
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Ramzi
Joined: 3/25/2004
Posts: 459
Well, I thought about it and maybe I'm about to give the winning solution to the Internet. But, realistically, I'm posting this to get your opinions. If we use intelligent backtracking, it would see that the order you place the pieces matter. For example, suppose you place: a b c d Where you pick a randomly, then choose b sharing the right edge of a, choose c sharing the right edge of b, and choose d choosing the right edge of c. Well, had you done: a b c d Then when you choose d, you have to satisfy two edges, namely, that of b and c. Because there are fewer possibilities for d, this recursive solution should be faster than the aforementioned one.