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prakash · Joined: 02 Jan 2008 Posts: 67 Location: New Jersey, USA

My way involved a straight forward XY wing on 389 which eliminated the 8 in R9C7. The rest was smooth sailing.

Marty R. · Posted: Thu Jul 24, 2008 12:34 am Post subject:

I saw at least three solutions. Highlight inside the quote if you want to see them.

nataraj · Posted: Thu Jul 24, 2008 6:01 am Post subject:

Basics really take you almost all the way...

After basics:

andras · Joined: 31 Oct 2007 Posts: 56 Location: Mid Wales

A very nice puzzle, I thought; basics solve most of it, and I saw the xy-wing without tearing out too much of my hair or losing too much of my crossword time later!

John

LloydB · Joined: 09 Jul 2007 Posts: 21 Location: Skagit Co., WA USA

The 389 xy-wing solved it for me.

The 378 xy-wing in box 5 Nataraj mentioned doesn't seem to do any thing for me.

tlanglet · Posted: Thu Jul 24, 2008 1:47 pm Post subject:

Part way through basics I felt that the puzzle would be solved by a BUG, and sure enough, the bivalues keep popping up like daffodils in the spring. However I could not solve the BUG+2 that I reached, and finally looked for another solution that readily appeared as the <389> xy-wing.

If anyone solved the BUG+2 condition, I would appreciate the details.

Ted Crying or Very sad

cgordon · Joined: 04 May 2007 Posts: 769 Location: ontario, canada

The Bug+2 references are interesting because I could clearly see after basics that there were only two cells with tri-values. I'd heard of Bug+2's but after a lot of searching I could only find one explanation. This suggested that unless the two Bug+1 eliminations were stongly linked, one had to use "forcing chains" (that's how I read it anyway). I didn't see strong links here - and I see that doing two independent Bug+1s doesn't work either. (But then I guess it's not supposed to).
So my interpretation of Bug+2s is that unless the Bug+1 eliminations offer a solution by "seeing" each other - one has to guess. Right???

Marty R. · Posted: Thu Jul 24, 2008 3:43 pm Post subject:

As to the BUG+2, either r5c5 must be 7 or r3c7 must be 8. The former leads to an invalid solution, thus reducing it to a BUG+1.

cgordon · Joined: 04 May 2007 Posts: 769 Location: ontario, canada

Marty R. · Posted: Thu Jul 24, 2008 5:06 pm Post subject:

tlanglet · Posted: Thu Jul 24, 2008 5:37 pm Post subject:

Marty,

Thanks for the explanation. In fact, I had performed just what you indicated and found the same conclusion: that r5c5=7 resulted in an invalid puzzle, but I did not then eliminate the 7 from that cell to derive a BUG+1 situation.

I understand your logic in eliminating a candidate that creates an invalid solution and will keep it in mind for the future.

Thanks again...

Ted

nataraj · Posted: Thu Jul 24, 2008 6:09 pm Post subject:

Clement · Posted: Thu Jul 24, 2008 6:50 pm Post subject: Daily Sudoku: Thu 24- Jul-2008 VH

XY-Wing 389 Pivot {3,8}r6c7 with Pincers {3,9}r5c8 and {8,9}r9c7 eliminating 9 in r9c8 solving the puzzle.

nataraj · Posted: Thu Jul 24, 2008 8:27 pm Post subject:

I think I found a way to exploit that BUG+2 without resorting to forcing chains:

The BUG+2 establishes a strong link between r5c5 and r3c7 (either r5c5=7 or r3c7=8).
That means, if r5c5 is NOT 7 then r3c7 must be 7 (and vice versa).

One way to make r5c5<>7 is if r4c6=7 which is the case if r4c7=3.

Now we have a nice elimination:
If r6c7 is not 8 then (r6c7=3,r6c4=7,r5c5<>7) then r4c7=8.
That means r9c7 cannot be 8

Written as an AIC:
-(8=3)r6c7-(3=7)r6c4-[BUG+2:(7)r5c5=(8)r4c7]- ;r9c7<>8

keith · Posted: Fri Jul 25, 2008 5:05 am Post subject:

Or, a finned XY-wing!

nataraj · Posted: Fri Jul 25, 2008 5:50 am Post subject:

Thinking about that BUG+2 elimination some more, I come to view it as something very similar to what we do in multi coloring, like a kite for example.
I once called such a pattern (where there are two strong rods connected by a weak chain) a nunchuck Wink

Only this time the strong links are not in one single candidate, but in two different candidates (that is how the strong link can "jump" from r5c5 to r3c7).

Look at this drawing - I marked the strong link induced by the BUG+2 and the strong link in row 6.

Both links connect a "7" candidate in one cell with an "8" candidate in another cell. And box 5 provides the necessary (weak) link between the two strong links.

Finally, the poor "8" that sees both ends of the nunchuck gets killed

Steve R · Posted: Fri Jul 25, 2008 11:06 am Post subject:

As an alternative to the BUG+2 I think you can simply place 3 in r5c5 using the deadly pattern marked by + signs:

DennyOR · Joined: 12 Sep 2007 Posts: 33 Location: Portland, Oregon

The 38-84-47-78-83 xy-chain solved it for me, eliminating the 3 in r5c8.