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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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 Posted: Mon Jun 22, 2009 1:52 pm Post subject: Free Press 12 June, 2009 |
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I neglected to post this one last week. | Code: | Puzzle: FP061209
+-------+-------+-------+
| . . . | 7 . 8 | . . 2 |
| . 1 . | . . 4 | 5 . . |
| . . . | . 9 . | . . . |
+-------+-------+-------+
| . 6 8 | . . . | . 3 . |
| . . 7 | . . 2 | 6 . . |
| 9 3 . | . . . | 8 1 . |
+-------+-------+-------+
| 7 . . | . 8 . | . . . |
| . . . | 5 . . | . 6 . |
| 4 . . | 9 . 6 | . . . |
+-------+-------+-------+ |
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Jun 23, 2009 4:48 am Post subject: |
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I'm having a mental block with this thing, it's a horror story. A useless W-Wing and Hidden UR leaves me here, stuck, and refusing to waste any more time on it. 
| Code: |
+-------------+---------------+----------------+
| 6 59 3459 | 7 15 8 | 1349 49 2 |
| 28 1 39 | 236 26 4 | 5 7 3689 |
| 28 7 345 | 1236 9 135 | 134 48 1368 |
+-------------+---------------+----------------+
| 5 6 8 | 14 147 9 | 2 3 47 |
| 1 4 7 | 8 3 2 | 6 59 59 |
| 9 3 2 | 46 567 57 | 8 1 47 |
+-------------+---------------+----------------+
| 7 259 6 | 1234 8 13 | 149 2459 159 |
| 3 289 19 | 5 1247 17 | 1479 6 189 |
| 4 258 15 | 9 27 6 | 137 258 1358 |
+-------------+---------------+----------------+
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Play this puzzle online at the Daily Sudoku site |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Tue Jun 23, 2009 5:05 am Post subject: |
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| Quote: | | and refusing to waste any more time on it |
then you probably don't want to see this solution
| Code: | .------------------.------------------.------------------.
| 6 *59 3459 | 7 *15 8 | 1349 49 2 |
| 28 1 3-9 | 236 26 4 | 5 7 3689 |
| 28 7 345 | 1236 9 135 | 134 48 1368 |
:------------------+------------------+------------------:
| 5 6 8 | 14 147 9 | 2 3 47 |
| 1 4 7 | 8 3 2 | 6 59 59 |
| 9 3 2 | 46 *4567 *57 | 8 1 47 |
:------------------+------------------+------------------:
| 7 259 6 | 1234 8 13 | 149 2459 159 |
| 3 289 *19 | 5 1247 *17 | 1479 6 189 |
| 4 258 15 | 9 127 6 | 137 258 1358 |
'------------------'------------------'------------------' |
(9=5)r1c2 - (5)r1c5 = (5)r6c5 - (5=7)r6c6 - (7=1)r8c6 - (1=9)r8c3; r2c3 <> 9 |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Tue Jun 23, 2009 6:06 am Post subject: |
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After basics: | Code: | +----------------+----------------+----------------+
| 6 +59 3+459 | 7 15 8 | 1349 4+9 2 |
| 28 1 39 | 236 26 4 | 5 7 3689 |
| 28 7 34-5 | 1236 9 135 | 134 48 1368 |
+----------------+----------------+----------------+
| 5 6 8 | 14 147 9 | 2 3 47 |
| 1 4 7 | 8 3 2 | 6 59 59 |
| 9 3 2 | 46 4567 57 | 8 1 47 |
+----------------+----------------+----------------+
| 7 259 6 | 1234 8 13 | 149 2459 159 |
| 3 289 19 | 5 1247 17 | 1479 6 189 |
| 4 258 15 | 9 127 6 | 137 258 1358 |
+----------------+----------------+----------------+ |
I do not remember how I solved this one last week.
Sudoku Susser shows the "comprehensive chain" above.
How to read it: If R3C3 is <5>, the cell values "+" must be true, so R3C3 is not <5>.
Seems to me, a simple, visual way to explain it.
I don't want to enter the notation wars, but there must be a simpler way! I am looking for a common way to annotate the 9 x 9 grid.
Best wishes,
Keith |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Tue Jun 23, 2009 6:24 am Post subject: |
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| storm_norm wrote: | | Quote: | | and refusing to waste any more time on it |
then you probably don't want to see this solution
| Code: | .------------------.------------------.------------------.
| 6 *59 3459 | 7 *15 8 | 1349 49 2 |
| 28 1 3-9 | 236 26 4 | 5 7 3689 |
| 28 7 345 | 1236 9 135 | 134 48 1368 |
:------------------+------------------+------------------:
| 5 6 8 | 14 147 9 | 2 3 47 |
| 1 4 7 | 8 3 2 | 6 59 59 |
| 9 3 2 | 46 *4567 *57 | 8 1 47 |
:------------------+------------------+------------------:
| 7 259 6 | 1234 8 13 | 149 2459 159 |
| 3 289 *19 | 5 1247 *17 | 1479 6 189 |
| 4 258 15 | 9 127 6 | 137 258 1358 |
'------------------'------------------'------------------' |
(9=5)r1c2 - (5)r1c5 = (5)r6c5 - (5=7)r6c6 - (7=1)r8c6 - (1=9)r8c3; r2c3 <> 9 |
Norm's chain might be:
| Code: | +----------------+----------------+----------------+
| 6 +59 3459 | 7 +15 8 | 1349 49 2 |
| 28 1 3-9 | 236 26 4 | 5 7 3689 |
| 28 7 345 | 1236 9 135 | 134 48 1368 |
+----------------+----------------+----------------+
| 5 6 8 | 14 147 9 | 2 3 47 |
| 1 4 7 | 8 3 2 | 6 59 59 |
| 9 3 2 | 46 4+567 5+7 | 8 1 47 |
+----------------+----------------+----------------+
| 7 259 6 | 1234 8 13 | 149 2459 159 |
| 3 289 1+9 | 5 1247 +17 | 1479 6 189 |
| 4 258 15 | 9 127 6 | 137 258 1358 |
+----------------+----------------+----------------+ |
Keith |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Tue Jun 23, 2009 6:36 am Post subject: |
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There is a short chain that cracks the puzzle. It's essentially Keith's solution, but as an AIC.
The strong link on <4> in [c3] lets everything unfold.
| Code: | (4)r3c3 = (4)r1c3 - (4=9)r1c8 - (9=5)r1c2 => [r3c3]<>5
+--------------------------------------------------------------+
| 6 *59 *3459 | 7 15 8 | 1349 *49 2 |
| 28 1 39 | 236 26 4 | 5 7 3689 |
| 28 7 *345 | 1236 9 135 | 134 48 1368 |
|--------------------+--------------------+--------------------|
| 5 6 8 | 14 147 9 | 2 3 47 |
| 1 4 7 | 8 3 2 | 6 59 59 |
| 9 3 2 | 46 4567 57 | 8 1 47 |
|--------------------+--------------------+--------------------|
| 7 259 6 | 1234 8 13 | 149 2459 159 |
| 3 289 19 | 5 1247 17 | 1479 6 189 |
| 4 258 15 | 9 127 6 | 137 258 1358 |
+--------------------------------------------------------------+
# 80 eliminations remain
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There are a couple of UR eliminations that I couldn't get to go anywhere.
===== ===== ===== AIC Notation as Forcing Chain on <4>
| Code: | [r3c3]= 4 forces [r3c3]<>5 -- or else
[r3c3]<>4 forces [r1c3]=4 forces [r1c8]=9 forces [r1c2]=5 forces [r3c3]<>5
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Therefore, logic on the two possibilities for <4> in [r3c3] force [r3c3]<>5. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Tue Jun 23, 2009 7:04 am Post subject: |
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| daj95376 wrote: | There is a short chain that cracks the puzzle. It's essentially Keith's solution, but as an AIC.
The strong link on <4> in [c3] lets everything unfold.
| Code: | (4)r3c3 = (4)r1c3 - (4=9)r1c8 - (9=5)r1c2 => [r3c3]<>5
+--------------------------------------------------------------+
| 6 *59 *3459 | 7 15 8 | 1349 *49 2 |
| 28 1 39 | 236 26 4 | 5 7 3689 |
| 28 7 *345 | 1236 9 135 | 134 48 1368 |
|--------------------+--------------------+--------------------|
| 5 6 8 | 14 147 9 | 2 3 47 |
| 1 4 7 | 8 3 2 | 6 59 59 |
| 9 3 2 | 46 4567 57 | 8 1 47 |
|--------------------+--------------------+--------------------|
| 7 259 6 | 1234 8 13 | 149 2459 159 |
| 3 289 19 | 5 1247 17 | 1479 6 189 |
| 4 258 15 | 9 127 6 | 137 258 1358 |
+--------------------------------------------------------------+
# 80 eliminations remain
|
There are a couple of UR eliminations that I couldn't get to go anywhere.
===== ===== ===== AIC Notation as Forcing Chain on <4>
| Code: | [r3c3]= 4 forces [r3c3]<>5 -- or else
[r3c3]<>4 forces [r1c3]= 4
[r1c3]= 4 forces [r1c8]= 9
[r1c8]= 9 forces [r1c2]= 5
[r1c2]= 5 forces [r3c3]<>5
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Therefore, logic on the two possibilities for <4> in [r3c3] force [r3c3]<>5. |
Danny,
So, rather than "*" on the relevant cells, is there a better notation? I have suggested "+" or "-" before the relevant candidate.
Best wishes,
Keith |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Tue Jun 23, 2009 7:44 am Post subject: |
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| keith wrote: | Danny,
So, rather than "*" on the relevant cells, is there a better notation? I have suggested "+" or "-" before the relevant candidate.
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Hello Keith, long time no chat!
It's late here, so I may need to update this reply in the morning when I'm fresher. Here goes for a first cut.
I'm just falling back onto old habits where the (basic) relevent cells are marked with an asterisk (*) -- or the alphabet (a..z) -- and other symbols are used to mark special cells that may be part of the process. Otherwise, the chain/UR/Fish/etc contains the details of what's happening and why.
I saw your use of +/-, but don't feel comfortable with it. Without notation, I had to keep checking the linkage from one cell to the next. I also wondered how it would handle my AIC where it starts with one candidate assumed negative in a cell ... and ends with another candidate shown to be negative in that cell.
Explaining what's happening is very important (to me), but I have no idea how it can be presented to a mixed audience. I worked hard learning notation, but I'm still a beginner compared to the notation that I've seen others use!
Regards, Danny |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Tue Jun 23, 2009 12:45 pm Post subject: |
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It is the <5> elimination in r3c3 that I found interesting about this puzzle. I saw it as a Finned XY-Wing in r1c28|r3c3 with the <3> in r3c3 as the fin. It depends upon the conjugate <4>s in c3. One way to see it is: if the XY-Wing is true then r3c3 is <4>. If the fin is true, it is <3>. Either way, it's not <5>.
The Finned XY-Wing provides the strong inference (3)r3c3=(4)r1c8|r3c3. I then saw it as the (very short!) continuous AIC loop:
Loop: (3)r3c3 = XYWr1c28|r3c3[(4)r1c8|r3c3] - (4)r1c3=(4-3)r3c3; r3c3<>5
While this may be interesting, Danny's 2-node XY Chain plus conjugate <4>s is certainly simpler to follow. But, I miss the Finned XY-Wing idea that had a brief flurry some time back. And, continuous AIC loops are always a treat and I can't recall seeing one this short before. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Tue Jun 23, 2009 12:55 pm Post subject: |
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P.S.:
I don't like that +/- stuff either. It looks like forcing to me.
Regarding Danny's AIC, I prefer to notate the weak inference discontinuity explicitly:
(5-4)r3c3=(4)r1c3 - (4=9)r1c8 - (9=5)r1c2 - (5)r3c3 => [r3c3]<>5
It is now clear that the <5> in r3c3 occurs at a weak inference discontinuity and therefor must be false. No assumptions about anything are necessary.
(Similarly, in my continuous loop, all the inferences become conjugate links, which means that r3c3 becomes a 34 bivalue cell, eliminating <5>. No assumptions are required to read the AIC notation.) |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Jun 23, 2009 3:59 pm Post subject: |
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| keith wrote: | After basics: | Code: | +----------------+----------------+----------------+
| 6 +59 3+459 | 7 15 8 | 1349 4+9 2 |
| 28 1 39 | 236 26 4 | 5 7 3689 |
| 28 7 34-5 | 1236 9 135 | 134 48 1368 |
+----------------+----------------+----------------+
| 5 6 8 | 14 147 9 | 2 3 47 |
| 1 4 7 | 8 3 2 | 6 59 59 |
| 9 3 2 | 46 4567 57 | 8 1 47 |
+----------------+----------------+----------------+
| 7 259 6 | 1234 8 13 | 149 2459 159 |
| 3 289 19 | 5 1247 17 | 1479 6 189 |
| 4 258 15 | 9 127 6 | 137 258 1358 |
+----------------+----------------+----------------+ |
I do not remember how I solved this one last week.
Sudoku Susser shows the "comprehensive chain" above.
How to read it: If R3C3 is <5>, the cell values "+" must be true, so R3C3 is not <5>.
Seems to me, a simple, visual way to explain it.
I don't want to enter the notation wars, but there must be a simpler way! I am looking for a common way to annotate the 9 x 9 grid.
Best wishes,
Keith |
This might be an example of why I have such difficulties following some of this stuff. It certainly looks to me that if r3c3=5, then the "+" values would be false.  |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Tue Jun 23, 2009 9:10 pm Post subject: |
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Clearly, the consensus is that my suggestion for notation within the puzzle will not fly.
Too bad. I was just reading how Richard Feynman revolutionized particle physics with his Feynman diagrams.
Keith |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Wed Jun 24, 2009 9:11 am Post subject: |
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| Keith wrote: | Clearly, the consensus is that my suggestion for notation within the puzzle will not fly.
Too bad. I was just reading how Richard Feynman revolutionized particle physics with his Feynman diagrams. |
Well... I can't blame you for trying. And, I say that as someone who actually met Richard Feynman. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Tue Sep 01, 2009 12:51 pm Post subject: |
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| Asellus wrote: | Regarding Danny's AIC, I prefer to notate the weak inference discontinuity explicitly:
(5-4)r3c3=(4)r1c3 - (4=9)r1c8 - (9=5)r1c2 - (5)r3c3 => [r3c3]<>5
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I'm in a messy discussion elsewhere and was reviewing posts here when I ran across Asellus' comment. This strikes at the heart of another messy discussion a long time back.
Back then, I used WI-based AICs -- aka a discontinuous nice loop ... aka (by me) a contradiction chain. They're nothing more than an SI-based AIC with an incorrect premise added to the front and a contradiction derived at the end. It is definitely more informative ... and doesn't even need the final deduction IMO because it's self-implied.
However, when I used a WI-based AIC a long time back, I was summarily informed that it qualified as T&E because I had no logical basis for assuming that a specific candidate was true in a specific cell. That comment still smarts to this day, and it's why I only list WI-based AICs as alternatives to my SI-based AICs.
Ironically, WI-based AICs are now popular. My wrong then is now considered right. 
Regards, Danny
BTW: I haven't determined how to list the following SI-based AIC (and deduction) other than as two, separate WI-based AICs.
(2)r2c2 = ... = (5)r7c2; => [r2c2]<>5, [r7c2]<>2
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun Sep 27, 2009 3:41 am Post subject: |
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| daj95376 wrote: | BTW: I haven't determined how to list the following SI-based AIC (and deduction) other than as two, separate WI-based AICs.
(2)r2c2 = ... = (5)r7c2; => [r2c2]<>5, [r7c2]<>2 |
How about:
(5)r2c2|(2)r7c2 - (2)r2c2= ... =(5)r7c2 - (5)r2c2|(2)r7c2
It is not unusual to show multiple instances of a particular digit at the weak inference discontinuity when the "pincer" digits match, so why not show multiple different digits at the discontinuity when they don't match? It's really the same thing.
In other words, if an AIC with <2> pincers eliminates 4 separate <2>s we don't consider it to be 4 separate AICs. So why do so if the pincers digits are dissimilar?
I find the "SI-based" and "WI-based" distinction to be artificial. The AIC loop is the same regardless of ones chosen notation convention. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Mon Sep 28, 2009 7:21 pm Post subject: |
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| Asellus wrote: | | daj95376 wrote: | BTW: I haven't determined how to list the following SI-based AIC (and deduction) other than as two, separate WI-based AICs.
(2)r2c2 = ... = (5)r7c2; => [r2c2]<>5, [r7c2]<>2 |
How about:
(5)r2c2|(2)r7c2 - (2)r2c2= ... =(5)r7c2 - (5)r2c2|(2)r7c2
It is not unusual to show multiple instances of a particular digit at the weak inference discontinuity when the "pincer" digits match, so why not show multiple different digits at the discontinuity when they don't match? It's really the same thing.
In other words, if an AIC with <2> pincers eliminates 4 separate <2>s we don't consider it to be 4 separate AICs. So why do so if the pincers digits are dissimilar?
I find the "SI-based" and "WI-based" distinction to be artificial. The AIC loop is the same regardless of ones chosen notation convention. |
Good suggestion. Thanks!
Another contact suggested:
| Quote: | Why not just terminate both ends with weak links like this?
| Code: | |<------- loop 1 ------>|
(2)r7c2 - (2)r2c2= ... =(5)r7c2 - (5)r2c2 ==> r7c2<>2, r2c2<>5
|<------- loop 2 ------>|
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The "loop1" and "loop2" are just there to "help beginners". I see the Eureka people extend chains like this quite often, especially David P Bird. |
Another good suggestion ... and I'm glad the loop1 and loop2 were included for this beginner. |
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