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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Fri Mar 11, 2011 10:32 pm Post subject: Puzzle 11/03/11: ~ Moderate/Difficult |
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| Code: | +-----------------------+
| 1 . . | . . . | . . . |
| . 4 . | . . 3 | . . . |
| . . 3 | 1 . . | 8 2 . |
|-------+-------+-------|
| . . 2 | 6 . . | . . . |
| . . . | . 8 . | 3 6 . |
| . 3 . | . . . | 1 . 2 |
|-------+-------+-------|
| . . 9 | . 6 5 | . 3 8 |
| . . 8 | . 9 . | 2 4 1 |
| . . . | . . 8 | 6 9 . |
+-----------------------+
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sat Mar 12, 2011 9:14 am Post subject: |
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A 2-SIS hidden pair move - does some damage
| Quote: | hp(49)r1c67=(4)r1c5 - (4=6)r3c6 ; r1c6<>6
w-wing(57) ; (7=5)r2c3 - r3c2=r5c2 - (5=7)r5c4 ; r2c4<>7 |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Mar 13, 2011 1:50 am Post subject: |
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I'm stuck here after one unhelpful move. I have no idea of what Peter's 2-SIS hidden pair means.
| Code: |
+--------------+-------------+------------+
| 1 2 567 | 8 457 469 | 459 57 3 |
| 8 4 567 | 579 2 3 | 59 1 679 |
| 5679 569 3 | 1 457 46 | 8 2 467 |
+--------------+-------------+------------+
| 479 8 2 | 6 3 1 | 45 57 479 |
| 4579 59 1 | 579 8 2 | 3 6 479 |
| 5679 3 567 | 579 457 49 | 1 8 2 |
+--------------+-------------+------------+
| 2 1 9 | 4 6 5 | 7 3 8 |
| 56 56 8 | 3 9 7 | 2 4 1 |
| 3 7 4 | 2 1 8 | 6 9 5 |
+--------------+-------------+------------+
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Mar 13, 2011 2:40 am Post subject: |
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| Marty R. wrote: | I'm stuck here after one unhelpful move. I have no idea of what Peter's 2-SIS hidden pair means.
| Code: | +--------------+-------------+------------+
| 1 2 567 | 8 457 469 | 459 57 3 |
| 8 4 567 | 579 2 3 | 59 1 679 |
| 5679 569 3 | 1 457 46 | 8 2 467 |
+--------------+-------------+------------+
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Marty,
<9> has to be in one of the cells r1c67. IF there wasn't a <4> in r1c5, would you agree that a hidden pair must exist for <49> in r1c67?
This represents strong link/inference logic:
| Code: | (4)r1c5 = (49)r1c67 -or-
(94)r1c67 = (4)r1c5
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Another way to visualize it is to ask yourself where in [r1] can <4> exist if it's not part of a hidden pair <49> in r1c67.
This is the first of two Strong Inferences that peter needed to get an elimination. (I forgot what the last "S" stands for.)
===== ===== ===== ===== ===== ===== ===== =====
As for the bottleneck, look at [r3] & [c3] and think <5>. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Mar 13, 2011 4:14 am Post subject: |
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| daj95376 wrote: |
Marty,
<9> has to be in one of the cells r1c67. IF there wasn't a <4> in r1c5, would you agree that a hidden pair must exist for <49> in r1c67?
This represents strong link/inference logic:
| Code: | (4)r1c5 = (49)r1c67 -or-
(94)r1c67 = (4)r1c5
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Another way to visualize it is to ask yourself where in [r1] can <4> exist if it's not part of a hidden pair <49> in r1c67.
This is the first of two Strong Inferences that peter needed to get an elimination. (I forgot what the last "S" stands for.)
===== ===== ===== ===== ===== ===== ===== =====
As for the bottleneck, look at [r3] & [c3] and think <5>. |
I believe that SIS stands for "Strong Inference Set"
Ted |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Mar 13, 2011 4:15 am Post subject: |
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I used several steps for this one; no idea if all were useful.
ANP(46=7)r3c69-(7=5)r1c8-(5=9)r2c7-r2c4=(9-6)r1c6=(6)r3c6; r3c12<>6
Flightless w-wing(57)r1c8|r5c4 with grouped SL(5)r136c5; r1c3,r3c9<>7
AXY-wing (5-67) vertex (57)r1c8 with fin (9)r2c9; r2c3<>6
If fin is true: (9)r2c9-r2c4=)9-6)r1c6=(6)r1c3
Flightless xy-wing (5-79) vertex (59)r5c2 plus transport: (7=9)r4c1-(9=5)r5c2-(5=7)r5c3-r2c4=(7)r3c5; r3c1<>7
Ted |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Mar 13, 2011 5:19 am Post subject: |
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| Quote: | | <9> has to be in one of the cells r1c67. IF there wasn't a <4> in r1c5, would you agree that a hidden pair must exist for <49> in r1c67? |
Certainly.
| Quote: | | Another way to visualize it is to ask yourself where in [r1] can <4> exist if it's not part of a hidden pair <49> in r1c67. |
Obviously in r1c5. However obvious the answers to these questions are, I can't carry it further to arrive at Peter's conclusion that r1c6<>6. However, that conclusion can be drawn if one wants to look at the whole thing as an "almost" naked triple of 567. I have no idea how much this differs from the 2-SIS thing.
| Quote: | | As for the bottleneck, look at [r3] & [c3] and think <5>. |
Sorry, I have no idea what thinking 5 does here other than being an ERI of a non-existent ER.
I appreciate the help, but my brain is just not wired to think things through like most of the others here.  |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Mar 13, 2011 6:07 am Post subject: |
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Marty,
Forget about SIS for a second and just look a the pattern of the three cells r1c567.
First, I believe that you agree if r1c5<>4, then we have a hidden pair (49)r1c67 which makes r1c6<>6 and r1c7<>5.
Second, what if r1c5=4? To find that answer we simply check the implications: If r1c5=4 then r3c6<>4=6, then r1c6<>6.
So, if either the hidden pair (49)r1c67 is true or r1c5=4 is true, r1c6<>6.
I view either/or conditions as in this puzzle just like a possible UR or a BUG+n where at least one of the free digits must be true and we check all those free digits to determine if we can derive a common result. All "almost" patterns, including the finned and sashimi single digit patterns, use this technique.
Ted |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sun Mar 13, 2011 9:14 am Post subject: |
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Sorry to bring up the whole SIS thing! I don't really know what it means either - I just meant "short chain with only two strong links".
I have started to look for these hidden-pair moves more - they are actually easier to find than the corresponding, often large, ALS. Find a strong link in a house, see if another digit is common to both those cells plus one other. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Mar 13, 2011 2:15 pm Post subject: |
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| tlanglet wrote: | Marty,
Forget about SIS for a second and just look a the pattern of the three cells r1c567.
First, I believe that you agree if r1c5<>4, then we have a hidden pair (49)r1c67 which makes r1c6<>6 and r1c7<>5.
Second, what if r1c5=4? To find that answer we simply check the implications: If r1c5=4 then r3c6<>4=6, then r1c6<>6.
So, if either the hidden pair (49)r1c67 is true or r1c5=4 is true, r1c6<>6.
I view either/or conditions as in this puzzle just like a possible UR or a BUG+n where at least one of the free digits must be true and we check all those free digits to determine if we can derive a common result. All "almost" patterns, including the finned and sashimi single digit patterns, use this technique.
Ted |
Ted,
Thank you. If I understand you correctly, you are agreeing that this is an "almost" triple of 567, since saying that there's a hidden pair of 49 if r1c5<> 4 is the same as saying there's a triple of 567 if r1c5<>4. I'm not criticizing Peter, but had he said "almost triple" rather than 2-SIS, I would have never needed to pose my initial question. The whole thing boils down to terminology, not the actual technique. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Mar 13, 2011 3:20 pm Post subject: |
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| Marty R. wrote: | | tlanglet wrote: | Marty,
Forget about SIS for a second and just look a the pattern of the three cells r1c567.
First, I believe that you agree if r1c5<>4, then we have a hidden pair (49)r1c67 which makes r1c6<>6 and r1c7<>5.
Second, what if r1c5=4? To find that answer we simply check the implications: If r1c5=4 then r3c6<>4=6, then r1c6<>6.
So, if either the hidden pair (49)r1c67 is true or r1c5=4 is true, r1c6<>6.
I view either/or conditions as in this puzzle just like a possible UR or a BUG+n where at least one of the free digits must be true and we check all those free digits to determine if we can derive a common result. All "almost" patterns, including the finned and sashimi single digit patterns, use this technique.
Ted |
Ted,
Thank you. If I understand you correctly, you are agreeing that this is an "almost" triple of 567, since saying that there's a hidden pair of 49 if r1c5<> 4 is the same as saying there's a triple of 567 if r1c5<>4. I'm not criticizing Peter, but had he said "almost triple" rather than 2-SIS, I would have never needed to pose my initial question. The whole thing boils down to terminology, not the actual technique. |
Marty,
When I look at row1, I see a couple of patterns.
First is the hidden pair posted by Peter: AHP(49=4)r1c567.
I also see an almost naked pair; ANP(57=4)r1c58-(4=6)r3c6-(6)r1c6=(6-57)r1c3; r1c3<>57 which does the same damage as the almost hidden pair.
However, I do not see an almost triple(567). What happens if r1c5<>4?
Initially the code is
| Code: | *-----------------------------------------------------------*
| 1 2 567 | 8 457 469 | 459 57 3 |
| 8 4 567 | 579 2 3 | 59 1 679 |
| 5679 569 3 | 1 457 46 | 8 2 467 |
*-----------------------------------------------------------* |
if r1c5<>4 then we get
| Code: | *-----------------------------------------------------------*
| 1 2 567 | 8 57 469 | 459 57 3 |
| 8 4 567 | 579 2 3 | 59 1 679 |
| 5679 569 3 | 1 457 46 | 8 2 467 |
*-----------------------------------------------------------* |
So, we do have a (sort of) triple (567)r1c358 but I think it is really viewed as a naked pair (57)r1c58.
How do you detect a triple 567 in row 1 if r1c5<>4
Ted |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Mar 13, 2011 4:38 pm Post subject: |
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| Quote: | | How do you detect a triple 567 in row 1 if r1c5<>4 |
In my haste I didn't notice that it was a 57 pair as opposed to a 567 triple. But the pair makes r1c3=6, making r1c6<>6, having the same effect as the triple.
Sorry. |
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