--- In fc-solve-discuss_at_yahoogroups.com, "Gary Campbell" <gary_at_...> wrote:
>
I'm not sure what was different between your two posts, but I'll respond to this one.
*** I originally wrote 2C instead of 2H at one place in my message, so I had to delete and repost the message FOR ONE STINKIN' CHARACTER.
My definition of a pruning opportunity is (1) all moves can be ignored if a layout has been reached before, or (2) all moves but one can be ignored from a layout if it can be proven that the one move cannot possibly prevent a solution, or (3) a particular move can be ignored if it can be proven (with little or no look ahead) that it cannot possibly lead to a solution. Have I missed anything?
*** I'll need to review your comments further.
In your scenario, we have 2 choices at move 10: (10. 61) and (10. 15). While I would always try (10. 15) first, both moves are irreversible, and I don't see how you could prove that (10. 61) could not possibly be necessary to a solution (without some kind of look ahead). Neither subsequent sequence of moves leads to a repeated layout (until you come to the last move, at which point there are no more moves possible).
Somehow, I get the feeling you don't see it as being this simple, so I must be missing something.
*** I've reviewed the five moves -- 15 65 a6 5a 5h : 3C 2H 9S 2H 3C -- many times while looking for an idea on how to prune it. It occurs to me that there is a pruning option.
*** After making the final move for 3C, I find the previous move for 3C and see if not performing it would still allow the intermediate moves -- 2H-3C 9S-TH 2H-FC 3C-HC -- to occur. If the first 3C move doesn't prevent the subsequent moves, then it can be pruned. Looking back at the two move sequences that I originally posted, this deduction now seems obvious. (see below)
*** I suspect that this falls under one of Gary's (1-3) above.
>
> From: dannyjones183
> Sent: Monday, October 08, 2012 12:26 PM
> To: fc-solve-discuss_at_yahoogroups.com
> Subject: Missed Pruning Opportunity ?
>
> #88445005 Attempt: 1 NumFcs=4 (Hrn Super)
>
> 7a 7b 7c 7d b7 5b d7 17 6d 15 65 a6 5a 5h
> 9S 9H 5S 8C 9H 7H 8C 7D 5D 3C 2H 9S 2H 3C
> | | | | | | | | |
> 7a 7b 7c 7d b7 5b d7 17 6d 61 a6 1a 1h
> 9S 9H 5S 8C 9H 7H 8C 7D 5D 2H 9S 2H 3C
>
Received on Mon Oct 08 2012 - 16:50:59 IST