[I'm expecting this to fail as a post to the mailing list. Fred, Dan, could one of you forward it if necessary?] On 11 April 2010 16:20, Dan Asimov <dasimov@earthlink.net> wrote:
Alex Fink wrote (forwarded):
<< Sticky Towers of Hanoi is played with the disks of an n-disk ToH set (the pegs can be disregarded). In the initial position all of the disks are separate. In a general position there will be stacks of fused disks, with top radius and bottom radius differing. A move is to pick up a stack S and set it on another stack whose top radius exceeds the bottom radius of S; this causes the two disks that come in contact to fuse.
I'm at least as perplexed as Fred here. Where it says "In a general position there will be stacks of fused disks" does that mean each stack is entirely fused, or just that it consists of a stack of fused stacks?
Each stack is entirely fused. I should've said that every stack automatically becomes fused: there are only separate fused stacks in the game, there's no circumstance in which a disk or stack gets put on top of another without fusing.
I don't see how the description of play can lead to anything but each stack being entirely fused, yet as Fred points out, that makes "solving" the puzzle all too obvious. (There is nothing about what constitutes a solution, so maybe that's where the confusion lies?) Perhaps Alex can clarify this?
Right. Like most of the games we were studying at the 2005 Banff workshop (BIRS is in Banff), this is a two-player game, not a single-player puzzle. The players take turns making fusing moves. Eventually someone will be left without a move, since no stack fits on another anymore. That player loses (this is the "normal play convention"; alternatively you could use the misère convention, by which that player wins). Thus, for example, if your move unifies all the disks into one (fused) stack, you win -- but most games don't end this way, since nonadjacent disks get stuck together. Alex