Optimal algorithm to lose game 2048
By now, especially after this post and other similar internet resources, I guess most people have figured out how to easily win at Gabriele Cirulli's game 2048: even manually, by observing simple rules, reaching 2048 is not a problem.
However, losing at this game seems far more challenging than winning! As much as I try, the minimum tile I got so far was 16. It seems to me that losing depends on chance much more than winning. Is there any strategy that can guarantee to lose with no tile more than 8?
(Of course, some of the hints suggested here might help, such as calculating all possible moves for n steps and choosing the combination of moves that maximises the probability to get the tiles stuck and end the game. But is there a more logical principle to obtain that?)
In the luckiest case, you would alternate 2s and 4s; alternating 2s, 4s and 8s should be easier. Actually I have just made it with five 8s, seven 4s and four 2s.
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