Hi all. I'm new to this list. I just got my pieces yesterday and I'm quite fond of playing with them :) On a whim I decided to analyze Treehouse. Here are some highlights: Each trio can be in one of 204 positions (that's 2 * 2 * 3 * 17). Six of those positions are one big stack; thirty-six of them are a small stack and a lone piece (there are 3! permutations of sizes, 3 orientations of the lone piece and 2 positions of it, relative to the stack). The remaining 162 are the positions with no stacks (3**3 orientations, 3! permutations). This means that a two-player game can be in one of 8,489,664 states (204**3), one for each of the trios. I imagine they'll be clustered around the 41616 (204**2) ones where the house hasn't been changed. Some of the moves one can make have more than one "type", i.e. they're both a tip and an aim, or both a dig and a hop, etc.. As it turns out, it's possible to arrange the types (except WILD) in a sequence, such that only neighbouring types have moves in common. Furthermore, that ordering is unique, modulo mirroring. The ordering is this: TIP - AIM - DIG - HOP - SWAP The number of moves (summed up for all positions) for the individual types, or shared between two neighbouring types, are: TIP only: 84 TIP or AIM: 570 AIM only: 162 AIM or DIG: 312 DIG only: 594 DIG or HOP: 78 HOP only: 588 HOP or SWAP: 84 SWAP only: 528 That's 3000 moves in total, 1044 of which are multi-typed. If we sum up the above numbers per type (including "TIP or AIM", "AIM only" and "AIM or DIG" under the "AIM" label), include WILD, and sort this, we get: WILD: 3000 AIM: 1044 DIG: 984 HOP: 750 TIP: 654 SWAP: 612 If we add this up, we see that if we're in a random position, we (expectedly) have to choose between one out of (1044+984+750+654+612+3000) / 204 / 6 options. That's 5.7549019607843137. One might want to use the above list as a measure of how flexible each kind of move is. However, it weighs all positions equally; I'd guess that you'll rarely see one big stack once your trio has left such a position. There are 84 combinations of positions and die rolls which force you to alter the house (if possible), or 14 per die outcome. If you're in a random position, then you have approximately 7% (0.068627450980392163) of having to alter the house. A thing that might be interesting to compute: given any two positions, the expected number of turns it'll take to transform one into the other. And of course, an optimal strategy :) Here's to numbers! Have fun playing :) -- Jonas Kölker <jonaskoelker@xxxxxxx> <URL:http://jonaskoelker.ignorelist.com>
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