Could we consider a smaller board and tile set, then generalize that to a 225?
--- On Thu, 3/5/09, Chris Lipe <[email protected]> wrote: From: Chris Lipe <[email protected]> Subject: Re: [quackle] Re: Scrabble Complexity To: [email protected] Date: Thursday, March 5, 2009, 12:55 PM It would be 225C99 if all the tiles were blank. Given our distribution, the ways to put tiles uniquely on the board would be 225C9 * 216C2 * 214C2 ... * 130C2 * 128C1 * 127C2 (for the two blanks) This (I think) works out to about 3.46*10^181. This of course doesn't necessarily cover the center star, or ensure that all the tiles are connected to each other. Chris Lipe On Thu, Mar 5, 2009 at 12:45 PM, Steven Gordon <sgordon...@gmail. com> wrote: > On Thu, Mar 5, 2009 at 10:25 AM, Lone Locust of the Apocalypse > <zo...@ninthbit. com> wrote: >> On Thu, 5 Mar 2009 zax_...@yahoo. com wrote: >>> What would the space be if we didn't care about the validity of the >>> words? How many possible ways can 100 tiles fit on a 225 square grid? >>> That would seem like the upper bound. >> >> With Steven's correction about 99, wouldn't the answer just be 225C99? >> That comes out to: >> >> 5688531038830326202 2843177354307569 3828810926090255 948388293937600 >> >> But even that wouldn't be an upper bound, because there are multiple >> ways to reach the same position of tiles on the board depending which >> "words" are formed when, and by which hooks/extensions etc. >> >> -- S. Spencer Sun >> > > Interestingly, this still comes out to be 10 > > Other complications: > - the squares on the board would have to be connected (no islands), > - the center square would have to be occupied (224C98 ?) > - the number of unplayed tiles could be any number between 1 and 7 > (assuming no resignations or 6 consecutive passes). > >
