dailysudoku.com

PoppaPoppa · Joined: 06 Nov 2005 Posts: 21 Location: Arkansas USA

Nice look and feel to your flash solver, Bram but dude it blew up on Monday's Nov 7 "very hard" puzzle! Even after I fed additional numbers it couldn't do it.

PoppaPoppa · Joined: 06 Nov 2005 Posts: 21 Location: Arkansas USA

I apologize for my last posting. Brad's solver worked fine on the Nov 7 puzzle. I had input it incorrectly! Embarassed

Sorry! Nice program!

David Bryant · Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

AZ Matt · Posted: Tue Nov 15, 2005 8:32 pm Post subject:

someone_somewhere posted a puzzle on Oct 30 (above) that I just got around to seeing. He said he broke his teeth on it, so I know it’s tough.

But do we know this puzzle is “solvable”? (Using the very broadest definition.)

I too saw how to eliminate the 5 as a candidate from six cells -- though I didn’t see it as a swordfish. Interestingly, because of the placement of the given 5s, the six-cell elimination gives you no really helpful information. It was fun, though.

David Bryant (or anyone else, but he raised the issue) – How is this a swordfish? I got to here with 5s as candidates.

000 005 500
000 055 505
000 000 000

000 000 000
050 000 500
550 000 505

550 055 000
000 000 000
050 005 000

I circled each 5 that could be coupled with only one other 5 in each row and column. From there it was fairly easy to see that each un-circled 5 resulted in a contradiction and could be eliminated as a candidate (r2c6, r2c7, r6c2, r6c7, r7c2, r7c7).

I suppose I see now that it is a sort-of double swordfish – I think.

In any event, has anyone made any progress…?

PS/FYI: “card203” appears to be a fairy sprinkling poker-link dust…

David Bryant · Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Hi, AZ Matt! I'm glad you came out from under your desk at work long enough to ask about the "Bix X" puzzle! Smile

A guest named Mars Plastic was looking for a very hard puzzle, so I popped this one on him a couple of weeks ago, and he came up with a (very tough, but impressively elegant) way to solve it. You can read that discussion starting right here.

The swordfish in the "5"s is in row 1, row 5, and row 9 -- specifically, in r1c6 & r1c7, in r5c2 & r5c7, and in r9c2 & r9c6. These are six points of a "complete" swordfish that could lie in r1c2, r1c6, & r1c7 + r5c2, r5c6, & r5c7 + r9c2, r9c6, & r9c7. But these six points are enough to define the pattern -- we know the "5" in column 2 must lie in r5c2 or r9c2, that the "5" in column 6 must lie in r1c6 or r9c6, and that the "5" in column 7 must lie in r1c7 or r5c7. So we can eliminate "5" from the six cells you mentioned.

Have you read the description of a swordfish at Sad Man Software? That's a pretty good introduction. dcb