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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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 Posted: Fri Sep 30, 2011 9:00 pm Post subject: Free Press Sep 30, 2011 |
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May be many steps ...
| Code: | Puzzle: FP093011
+-------+-------+-------+
| 6 . . | 2 3 . | . . 8 |
| . . . | 5 . . | . . . |
| 3 5 . | 8 . 7 | . . 2 |
+-------+-------+-------+
| . . 8 | . . . | . 3 . |
| 5 . 3 | . . . | 2 . 6 |
| . 9 . | . . . | 8 . . |
+-------+-------+-------+
| . . . | 1 . 9 | . 7 5 |
| . . . | . . 4 | . . . |
| . . . | . 7 5 | . . 3 |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Sep 30, 2011 11:26 pm Post subject: |
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OK, I hope I didn't screw this up. ABCDEFGH are a potential DP on 28. R7c1=4 or r2c1=149, which combines with other cells in that box to form a 1479 quad. Several cells are a common outcome, solving the puzzle.
| Code: | +-------------------+-----------+---------------+
| 6 47 479 |2 3 1 | 57 459 8 |
| 12489A 28B 12479 |5 49 6 | 3 149 1479 |
| 3 5 149 |8 49 7 | 19 6 2 |
+-------------------+-----------+---------------+
| 14 16 8 |479 56 2 | 57 3 1479 |
| 5 47 3 |479 1 8 | 2 49 6 |
| 124 9 12467 |47 56 3 | 8 145 147 |
+-------------------+-----------+---------------+
| 248C 3 246 |1 28D 9 | 46 7 5 |
| 7 16 5 |3 28E 4 | 169 28F 19 |
| 149 28G 149 |6 7 5 | 14 28H 3 |
+-------------------+-----------+---------------+ |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Oct 01, 2011 1:10 am Post subject: |
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Marty,
Your DP is correct, but I do not see how it solves the puzzle.
| Code: | +-------------------+-------------------+-------------------+
| 6 47 479 | 2 3 1 | 57 459 8 |
|28+149 28 12479 | 5 49 6 | 3 149 1479 |
| 3 5 149 | 8 49 7 | 19 6 2 |
+-------------------+-------------------+-------------------+
| 14a 16 8 | 479 56 2 | 57 3 1479 |
| 5 47 3 | 479 1 8 | 2 49 6 |
|2-14 9 12467 | 47 56 3 | 8 145 147 |
+-------------------+-------------------+-------------------+
|28+4 3 246 | 1 28 9 | 46 7 5 |
| 7 16 5 | 3 28 4 | 169 28 19 |
| 149b 28 149 | 6 7 5 | 14 28 3 |
+-------------------+-------------------+-------------------+ |
Type 3: To break the DP, there is a pseudocell 149 in R27C1. It makes a triple with ab, eliminating 14 in R6C1.
Type 4: One of R27C1 must be 8. Neither can be 2.
Leaving me here:
| Code: | +-------------------+-------------------+-------------------+
| 6 47 479 | 2 3 1 | 57 459 8 |
| 1489 28 12479 | 5 49 6 | 3 149 1479 |
| 3 5 149 | 8 49 7 | 19 6 2 |
+-------------------+-------------------+-------------------+
| 14 16 8 | 479 56 2 | 57 3 1479 |
| 5 47 3 | 479 1 8 | 2 49 6 |
| 2 9 1467 | 47 56 3 | 8 145 147 |
+-------------------+-------------------+-------------------+
| 48 3 246 | 1 28 9 | 46 7 5 |
| 7 16 5 | 3 28 4 | 169 28 19 |
| 149 28 149 | 6 7 5 | 14 28 3 |
+-------------------+-------------------+-------------------+ |
Keith
Go, Tigers! |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Oct 01, 2011 1:46 am Post subject: |
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| Keith, I'll look further later. However, I looked at the 149 in r2c1 as combining with other cells in that box to form a 1479 quad, setting r2c3=2. Carrying that further and comparing the results to what I got when I tried r7c1=4, yielded several common outcomes which solved it. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Oct 01, 2011 3:37 am Post subject: |
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Keith, I still get the same result. Common outcomes solve the puzzle.
I do appreciate the fact that you post these. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Oct 01, 2011 9:38 am Post subject: |
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| Marty R. wrote: | | Keith, I still get the same result. Common outcomes solve the puzzle. |
Marty, I'll agree, for example, that 4 in R7C1 or 2 in R2C3 both drive 6 in R4C2, which solves the puzzle.
IMHO, something of a stretch.
Keith |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Oct 01, 2011 10:41 am Post subject: |
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After basics, an XYZ-wing takes out 9 in R1B8. That sets up a W-wing -14 in B14 that takes out 1 in R2C1 and solves the puzzle.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Oct 01, 2011 3:56 pm Post subject: |
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| Quote: | Marty, I'll agree, for example, that 4 in R7C1 or 2 in R2C3 both drive 6 in R4C2, which solves the puzzle.
IMHO, something of a stretch. |
I guess by a stretch, you mean that the chains get extended longer than you'd prefer before a common outcome is seen.
As to the XYZ- and W-Wings, I never got that far, being fascinated by an eight-cell DP. |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Sat Oct 01, 2011 4:46 pm Post subject: |
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| Marty R. wrote: | | Quote: | Marty, I'll agree, for example, that 4 in R7C1 or 2 in R2C3 both drive 6 in R4C2, which solves the puzzle.
IMHO, something of a stretch. |
I guess by a stretch, you mean that the chains get extended longer than you'd prefer before a common outcome is seen. |
For that "stretch", did anyone have anything shorter than this?
(1=4)r4c1 - BUG-Lite:[(4)r7c1 = (2-7)r2c3] = (7)r2c9 - (7)r1c7 = (7-5)r4c7 = (5-6)r4c7 = (6)r4c2 ==> r4c2<>1 |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Sat Oct 01, 2011 6:24 pm Post subject: |
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| ronk wrote: | | (1=4)r4c1 - BUG-Lite:[(4)r7c1 = (2-7)r2c3] = (7)r2c9 - (7)r1c7 = (7-5)r4c7 = (5-6)r4c7 = (6)r4c2 ==> r4c2<>1 |
Beautiful! 
shouldn't the (5-6)r4c7 be (5-6)r4c5?
Nice find. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Oct 01, 2011 6:40 pm Post subject: |
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| ronk wrote: | | Marty R. wrote: | | Quote: | Marty, I'll agree, for example, that 4 in R7C1 or 2 in R2C3 both drive 6 in R4C2, which solves the puzzle.
IMHO, something of a stretch. |
I guess by a stretch, you mean that the chains get extended longer than you'd prefer before a common outcome is seen. |
For that "stretch", did anyone have anything shorter than this?
(1=4)r4c1 - BUG-Lite:[(4)r7c1 = (2-7)r2c3] = (7)r2c9 - (7)r1c7 = (7-5)r4c7 = (5-6)r4c7 = (6)r4c2 ==> r4c2<>1 |
Ron, I had:
a) R7C1=4; R4C1=1; R4C2=6.
b) R3C3=2; R2C9=7; R4C5=5; R4C2=6.
I think it's the same as yours.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Oct 01, 2011 7:00 pm Post subject: |
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Being notationally challenged, I have no idea of what Ron's chain is.
With the 1479 quad in box 1, r2c3=2-->r2c9=7-->r1c7=5
R7c1=4-->r4c1=1-->r4c2=6-->r4c5=5-->r4c7=7-->r1c7=5
R1c7=5 solves it. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sat Oct 01, 2011 10:04 pm Post subject: |
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Aaaaagh!!! I just burned out half of my brain cells! Now there's only one left. _  _
| Code: | after basics
+------------------------------------------------------------------------+
| 6 47 479 | 2 3 1 | 57 459 8 |
| *28+149 *28 12479 | 5 49 6 | 3 149 1479 |
| 3 5 149 | 8 49 7 | 19 6 2 |
|------------------------+-----------------------+-----------------------|
| 14 16 8 | 479 56 2 | 57 3 1479 |
| 5 47 3 | 479 1 8 | 2 49 6 |
| 124 9 12467 | 47 56 3 | 8 145 147 |
|------------------------+-----------------------+-----------------------|
| *28+4 3 246 | 1 *28 9 | 46 7 5 |
| 7 16 5 | 3 *28 4 | 169 *28 19 |
| 149 *28 149 | 6 7 5 | 14 *28 3 |
+------------------------------------------------------------------------+
# 69 eliminations remain
(194)r2c1=DP=r7c1-(4=91)r9c13-r8c2=r4c2-(1=4)r4c1-r27c1=DP=(19)r2c1 => r2c1<>28
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Yes, it would be simpler/smarter to derive r7c1<>4 first, and then derive r2c1<>28. But, no one seems to post simpler/smarter solutions anymore.
Marty: Thanks for spotting the DP!!! |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Sun Oct 02, 2011 2:03 pm Post subject: |
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| daj95376 wrote: | (194)r2c1=DP=r7c1- (4=91)r9c13-r8c2=r4c2-(1=4)r4c1 -r27c1=DP=(19)r2c1 => r2c1<>28
...
Yes, it would be simpler/smarter to derive r7c1<>4 first, and then derive r2c1<>28. But, no one seems to post simpler/smarter solutions anymore. |
r7c1<>4 is what you did (in blue).  |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Oct 02, 2011 3:36 pm Post subject: |
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| ronk wrote: | r7c1<>4 is what you did.
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Almost. r7c1<>4 is what I could/should have done for a two-stepper.
| Code: | (4=91)r9c13-(1=6)r8c2-(6=1)r4c2-(1=4)r4c1 => r7c1<>4
DP on <28> => r2c1=149 => r2c1<>28
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Instead, I created a messy single-stepper. _  _ |
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