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storm_norm
Joined: 18 Oct 2007 Posts: 1741
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Posted: Mon Nov 23, 2009 7:20 am Post subject: |
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Mogulmeister wrote: | I believe in some quarters it is referred to as an AIC Type 2. |
you are right.
Code: | (A) = A - (A=B) - B = (B)
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----see each other--- |
in this particular pattern, the cell containing the first A has to see the Cell of the last B. this is how the elimination is made. hence why its called a type 2, especially on the scanraid site.
if the first A and the last B reside in the same cell it would be a loop.
If you think about it, explaining this technique would be a bit of a shift in reasoning for the solvers who like to form chains with "pincers".
the W-wing, M-wing, xy-wing, even the skyscraper, two-string kite and turbot-fish have pincers on both ends of the chain that see a victim. the victim in these techniques never resides in the same cell as the pincers.
but!!
this type 2 AIC pattern would require that the victim resides in a pincer cell (one of the end cells of the chain). think about that one. it seems the sudoku community likes chains with CLAWS
fyi, some other named techniques that include the victim candidates in the same cells of the technique pattern: hidden pair, hidden triple, just to name two. |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Mon Nov 23, 2009 8:37 am Post subject: |
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Mogulmeister wrote: | Be gentle with me - I haven't been on this board for 2 years but....
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I'm guessing you are tired. There is only one <6> in box 1 [b1] ... and you delete it by setting r4c3=6.
Please check to see if your solver has the ability to produce pencilmark (PM) output.
Code: | +--------------------------------------------------------------+
| 157 15 3 | 4 16 2 | 567 8 9 |
| 2 9 16 | 8 5 7 | 4 3 16 |
| 157 4 8 | 69 3 19 | 2 167 1567 |
|--------------------+--------------------+--------------------|
| 3 167 16 | 2 8 5 | 9 167 4 |
| 4 8 2 | 679 16 19 | 67 5 3 |
| 156 1567 9 | 67 4 3 | 8 1267 1267 |
|--------------------+--------------------+--------------------|
| 8 26 5 | 1 9 4 | 3 267 267 |
| 16 126 4 | 3 7 8 | 56 9 256 |
| 9 3 7 | 5 2 6 | 1 4 8 |
+--------------------------------------------------------------+
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Regards, Danny
Last edited by daj95376 on Mon Nov 23, 2009 9:09 am; edited 2 times in total |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Mon Nov 23, 2009 8:52 am Post subject: |
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storm_norm wrote: | Code: | (A) = A - (A=B) - B = (B)
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----see each other--- |
In this particular pattern, the cell containing the first A has to see the cell of the last B. |
I don't know an AIC Type 2 (by name) from a hole in the wall. What I do know is that this particular chain has what I call a gotcha. It can perform eliminations on two different candidate values in two different cells!
Here's my original chain and PM.
Code: | (6)r1c5 = r5c5 - (6=7)r5c7 - r1c7 = (7)r1c1 => r1c1<>6
+--------------------------------------------------------------+
| 157-6 15 3 | 4 16 2 | 567 8 9 |
| 2 9 16 | 8 5 7 | 4 3 16 |
| 157 4 8 | 69 3 19 | 2 167 156 |
|--------------------+--------------------+--------------------|
| 3 167 16 | 2 8 5 | 9 167 4 |
| 4 8 2 | 679 16 19 | 67 5 3 |
| 156 1567 9 | 67 4 3 | 8 126 126 |
|--------------------+--------------------+--------------------|
| 8 26 5 | 1 9 4 | 3 26 7 |
| 16 126 4 | 3 7 8 | 56 9 256 |
| 9 3 7 | 5 2 6 | 1 4 8 |
+--------------------------------------------------------------+
# 45 eliminations remain
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Now, imagine there's a <7> in cell r1c5. The chain would now be written:
Code: | (6)r1c5 = r5c5 - (6=7)r5c7 - r1c7 = (7)r1c1 => r1c1<>6, r1c5<>7
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Mogulmeister
Joined: 03 May 2007 Posts: 1151
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Posted: Mon Nov 23, 2009 9:42 am Post subject: |
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Danny,
Indeed
As in eliminating the only 6 in b1 also produces an absurdity meaning r4c3 <> 6.
Is it mandatory that we do pencilmarks ? (mine I think unfortunately doesn't) I tend to prefer a more pictorial output so I can use colours to make a point.
Thanks M |
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Mogulmeister
Joined: 03 May 2007 Posts: 1151
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Posted: Mon Nov 23, 2009 10:03 am Post subject: |
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Quote: | Now, imagine there's a <7> in cell r1c5. The chain would now be written: |
Yes thats the type 2.
Last edited by Mogulmeister on Mon Nov 23, 2009 10:23 am; edited 3 times in total |
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Mogulmeister
Joined: 03 May 2007 Posts: 1151
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Posted: Mon Nov 23, 2009 10:05 am Post subject: |
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Quote: | If you think about it, explaining this technique would be a bit of a shift in reasoning for the solvers who like to form chains with "pincers".
the W-wing, M-wing, xy-wing, even the skyscraper, two-string kite and turbot-fish have pincers on both ends of the chain that see a victim. the victim in these techniques never resides in the same cell as the pincers.
but!!
this type 2 AIC pattern would require that the victim resides in a pincer cell (one of the end cells of the chain). think about that one. it seems the sudoku community likes chains with CLAWS Cool |
Indeed Norm. Totally agree. It was one ot the things I had to get my mind around when I was learning AICs having been a pincer man. |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Mon Nov 23, 2009 2:28 pm Post subject: |
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Mogulmeister wrote: | Be gentle with me - I haven't been on this board for 2 years but....
We got to here
Now consider the <16> pair at r24c3. There is an absurdity created if you let r4c3 become 6. Why ?
r2c3 is forced to be 1 which leaves a locked pair <57> at r13c1 thus removing both 1 and 5 from r1c2 - an impossibility.
(Edit - and as Danny pointed out a 6 in r4c3 removes all 6's from block 1!)
So r4c3 is not 6.
After that it is just singles and the puzzle is over. |
Yes, once the 6 is removed from R1C1, the puzzle is solved.
In the position you post, there is a 157 triplet in B1, so R2C3 must be 6.
Keith |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Mon Nov 23, 2009 5:49 pm Post subject: |
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Mogulmeister wrote: | Is it mandatory that we do pencilmarks ? (mine I think unfortunately doesn't) I tend to prefer a more pictorial output so I can use colours to make a point.
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It's not mandatory that you use pencilmarks. However, image files are an annoyance for me and I tend to ignore them.
Regards, Danny |
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