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blackwood
Joined: 12 Sep 2006 Posts: 4 Location: Lehi, Utah, USA
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Posted: Fri Sep 29, 2006 5:34 pm Post subject: Sep 29 - Logic behind an elimination? |
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Basic moves got me to the point shown below. While looking at the 7's I noticed that no matter where the 7 was in block 5, (r5c5 or r6c6) the 7 in block 6 would end up in r4c9. (The highlighted square). Not knowing why this was the case, I looked at the hint and it showed 5 for r6c8. I can't seem to see the reason for that either, although I had already figured it out since I knew the 7. What I don't understand is the logic behind solving either of these. Any ideas?
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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado
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Posted: Fri Sep 29, 2006 5:52 pm Post subject: Have you learned the "X-Wing"? |
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Hello, Blackwood!
The short answer is that you have an "X-Wing" formation in columns 1 and 6, on digit 7. In other words, since there are only two places to fit a "7" in column 1 (at r6c1 or r9c1) and also only two places to fit a "7" in column 6 (at r6c6 or r9c6) you can be certain that the "7" in row 6 must appear either at r6c1 or r6c6 (and, similarly, the "7" in row 9 must appear either at r9c1 or r9c6). So you can eliminate the possibility of a "7" at r6c8, and also at r9c2. dcb |
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blackwood
Joined: 12 Sep 2006 Posts: 4 Location: Lehi, Utah, USA
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Posted: Fri Sep 29, 2006 6:11 pm Post subject: |
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Ok, I hate it when these puzzles humble us. I looked all over for x-wings, swordfish, unique rectangles, and remote pairs. I've been staring at it too long... :shock:
I just missed that x-wing, even after figuring out the 7. (I even wrote down all the 7's on a blank puzzle and STILL missed it!) :oops:
Blackwood |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Sat Sep 30, 2006 1:38 pm Post subject: Another way to solve it |
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Note that there is almost an X-wing on <4> in C25, except for the possibility <4> in R9C2. This is a "finned" X-wing: Either R9C2 is <4>, or the X-wing will place a <4> in R7C2 or R7C5. Either way, R7C1 cannot be <4>, it must be <1>.
Keith |
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blackwood
Joined: 12 Sep 2006 Posts: 4 Location: Lehi, Utah, USA
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Posted: Mon Oct 02, 2006 2:32 pm Post subject: |
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I'm not familiar with "finned" x-wings. does the logic you expained work because the "finned" part of the x-wing is in the same box as part of the x-wing or would it still apply even if they where in different boxes?
So does the finned x-wing only remove the <4> from R7C1 or does it also remove the <4> from R7C6? The former seems correct to me, but just trying to understand the logic.
Blackwood |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Mon Oct 02, 2006 4:48 pm Post subject: |
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blackwood wrote: | I'm not familiar with "finned" x-wings. does the logic you expained work because the "finned" part of the x-wing is in the same box as part of the x-wing or would it still apply even if they where in different boxes?
So does the finned x-wing only remove the <4> from R7C1 or does it also remove the <4> from R7C6? The former seems correct to me, but just trying to understand the logic.
Blackwood |
I'm not Keith, but I'll try and answer.
Yes, the fin, or extra possibility or possibilities, must be in the same box as one corner of the wing. The actual finned X-Wing technique eliminates only the "4" from r7c1 because we either do or do not have an X-Wing and "4" can't be present in r7c1 under either possibility.
Or to state the same logic slightly differently: r9c2 must be equal or unequal to "4." If equal, then all other "4s" in the box must go. If unequal, there would be a true X-Wing which zaps the "4" from r7c1.
By the way, there is also a Finned X-Wing in rows 6 and 9 which accomplishes the same thing of eliminating the "4" from r7c1.
Removal of the "4" from any other cell must be done with other techniques. R7c6 has the "4" removed if it were a true X-Wing, but the finned technique, per se, can't remove it. |
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