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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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 Posted: Thu Nov 20, 2008 4:15 pm Post subject: Vanhagen Fiendish 11/20 |
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Here is another Vanhegan Fiendish puzzle that involved many varied VH steps for me. | Code: |
+-------+-------+-------+
| 5 1 . | . 4 . | . . 9 |
| . 6 . | 1 8 . | . 7 5 |
| . . . | . . 5 | . . . |
+-------+-------+-------+
| . . 7 | 9 . 4 | . 2 . |
| 2 3 . | . . . | . 9 4 |
| . 5 . | 3 . 8 | 6 . . |
+-------+-------+-------+
| . . . | 8 . . | . . . |
| 8 7 . | . 5 9 | . 4 . |
| 1 . . | . 3 . | . 5 8 |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site
Ted
Edited to correct puzzle. Thanks Danny and sorry to all......
Last edited by tlanglet on Thu Nov 20, 2008 4:52 pm; edited 1 time in total |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Nov 20, 2008 7:20 pm Post subject: |
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Here's where I found myself after a UR and a couple of ERs:
| Code: |
+-------------+-----------+-------------+
| 5 1 238 | 7 4 236 | 23 368 9 |
| 39 6 239 | 1 8 23 | 4 7 5 |
| 7 24 2348 | 26 9 5 | 123 368 16 |
+-------------+-----------+-------------+
| 6 8 7 | 9 1 4 | 5 2 3 |
| 2 3 1 | 5 67 67 | 8 9 4 |
| 49 5 49 | 3 2 8 | 6 1 7 |
+-------------+-----------+-------------+
| 34 249 5 | 8 67 1 | 79 36 26 |
| 8 7 36 | 26 5 9 | 13 4 126 |
| 1 29 26 | 4 3 267 | 79 5 8 |
+-------------+-----------+-------------+
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Play this puzzle online at the Daily Sudoku site
I used the potential 38 DP in boxes 13. A 6 in either of the cells in box 3 breaks it up, as does a 2 in either cell in box 1, as well as a 4 in r3c3. The latter leads to an invalidity and the others both force the same solution for the cells in c1.
This looks like it would be easily solvable with Medusa as well, although I'd have been happier with other methods. I wonder if there was a productive W-Wing in there that I missed. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Thu Nov 20, 2008 7:35 pm Post subject: |
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I see that you've already used the type 4 UR (49).
There is one w-wing right here: (6) r3c4=r7c9 via SL (2) row 8
But the next step for me was coloring (3) to remove 3 from r1c7:
3:-r1c6=r2c6-r2c1=r7c1-r7c8=r8c7-
After that, there's another w-wing
(3) r1c6=r8c3 via SL (6) col 4 |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Thu Nov 20, 2008 7:37 pm Post subject: |
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| Code: | after basics
+--------------------------------------------------------------------+
| 5 1 238 | 7 4 236 | 23 368 9 |
| 349 6 2349 | 1 8 23 | 234 7 5 |
| 7 24 2348 | 26 9 5 | 1234 368 126 |
|----------------------+----------------------+----------------------|
| 6 8 7 | 9 1 4 | 5 2 3 |
| 2 3 1 | 5 67 67 | 8 9 4 |
| 49 5 49 | 3 2 8 | 6 1 7 |
|----------------------+----------------------+----------------------|
| 34 249 5 | 8 67 1 | 79 36 26 |
| 8 7 236 | 26 5 9 | 123 4 126 |
| 1 29 26 | 4 3 267 | 79 5 8 |
+--------------------------------------------------------------------+
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| Code: | finned X-Wing r38\c37
+-----------------------------------+
| . . 3 | . . 3 | -3 3 . |
| 3 . 3 | . . 3 | -3 . . |
| . . *3 | . . . | *3 #3 . |
|-----------+-----------+-----------|
| . . . | . . . | . . 3 |
| . 3 . | . . . | . . . |
| . . . | 3 . . | . . . |
|-----------+-----------+-----------|
| 3 . . | . . . | . 3 . |
| . . *3 | . . . | *3 . . |
| . . . | . 3 . | . . . |
+-----------------------------------+
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Marty: I'd be willing to bet a gold coin against a stale donut that you'd have found both eliminations if only one of [r3c78] had a (3) in it.
Leaving [r8c3]<>2 and a W-Wing to complete the puzzle.
Last edited by daj95376 on Thu Nov 20, 2008 10:12 pm; edited 1 time in total |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Thu Nov 20, 2008 9:00 pm Post subject: |
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| Marty R. wrote: | Here's where I found myself after a UR and a couple of ERs:
| Code: |
+-------------+-----------+-------------+
| 5 1 238 | 7 4 236 | 23 368 9 |
| 39 6 239 | 1 8 23 | 4 7 5 |
| 7 24 2348 | 26 9 5 | 123 368 16 |
+-------------+-----------+-------------+
| 6 8 7 | 9 1 4 | 5 2 3 |
| 2 3 1 | 5 67 67 | 8 9 4 |
| 49 5 49 | 3 2 8 | 6 1 7 |
+-------------+-----------+-------------+
| 34 249 5 | 8 67 1 | 79 36 26 |
| 8 7 36 | 26 5 9 | 13 4 126 |
| 1 29 26 | 4 3 267 | 79 5 8 |
+-------------+-----------+-------------+
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I wonder if there was a productive W-Wing in there that I missed. |
Marty, I used multi-coloring to eliminate <3> at r1c7, which makes that cell equal to <2> thereby making r1c6 = <36>. Then a w-wing36 between r1c6 & r8c3 with strong link <6> in col4 removes the <3> in r1c8.
Ted |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Nov 20, 2008 10:21 pm Post subject: |
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| Quote: | | I wonder if there was a productive W-Wing in there that I missed. |
Nataraj,
I had a hunch there was one hiding in there somewhere. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Fri Nov 21, 2008 12:55 am Post subject: |
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| Code: | .------------------.------------------.------------------.
| 5 1 238 | 7 4 236 | 23 C368 9 |
|H3-49 6 2349 | 1 8 23 | 234 7 5 |
| 7 F24 2348 |E26 9 5 | 1234 C368 D126 |
:------------------+------------------+------------------:
| 6 8 7 | 9 1 4 | 5 2 3 |
| 2 3 1 | 5 67 67 | 8 9 4 |
| 49 5 49 | 3 2 8 | 6 1 7 |
:------------------+------------------+------------------:
|A34 G2-49 5 | 8 67 1 | 79 B36 26 |
| 8 7 236 | 26 5 9 | 123 4 126 |
| 1 29 26 | 4 3 267 | 79 5 8 |
'------------------'------------------'------------------' |
care for a one stepper?
(4=3)A - (3=6)B - (6)CC = (6)D - (6=2)E - (2=4)F; G|H <> 4 |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Fri Nov 21, 2008 1:51 am Post subject: |
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I know its past halloween, but if you really wanna get spooky with this puzzle, then please enter.
| Code: | .------------------.------------------.------------------.
| 5 1 C238 | 7 4 236 | 23 B368 9 |
|D349 6 D2349 | 1 8 E23 | 234 7 5 |
| 7 24 C2348 |F26 9 5 | 1234 B368 G126 |
:------------------+------------------+------------------:
| 6 8 7 | 9 1 4 | 5 2 3 |
| 2 3 1 | 5 67 67 | 8 9 4 |
| 49 5 49 | 3 2 8 | 6 1 7 |
:------------------+------------------+------------------:
| 34 249 5 | 8 67 1 | 79 A36 H26 |
| 8 7 236 | 26 5 9 | 123 4 H126 |
| 1 29 26 | 4 3 267 | 79 5 8 |
'------------------'------------------'------------------' |
consider the cells marked above... cell A is either 3, or if not 3 then...
A=6 => BB<>6
UR{3,8}, CC<>3
DD=3
E<>3, E=2
F<>2, F=6
G<>6
HH=6
A<>6
A has to be 3. done.
whew!
the AIC would start with a group and end with a group.
(6)BB=UR{3,8}BBCC - (3)CC = (3)DD - (3=2)E - (2=6)F - (6)G = (6)HH; A is not 6
Asellus, one more time with the way you would write it?
Danny, can't wait for the analysis.
the more I think about the structure of the UR, from the point that was made that a UR is always false, the UR is a good way to start a forcing chain. its probably not really a node for a AIC.
the forcing chains are initiated by the UR structure. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Fri Nov 21, 2008 5:42 am Post subject: |
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| storm_norm wrote: | cell A = 3|6 ...
A=6 => BB<>6
UR{3,8}, CC<>3
DD=3
E<>3, E=2
F<>2, F=6
G<>6
HH=6
A<>6
A has to be 3. done.
Danny, can't wait for the analysis.
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You had fun with this one ... didn't you Norm! 
Technically, the red statements are unnecessary because the blue statements eliminate all of the 6s in [b3]. Once this happens, your conclusion automatically follows. Of course, you will need the red statements if you want it to be a chain instead of a SIN -- Single Inference Network.
Regards!!! |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Fri Nov 21, 2008 8:07 am Post subject: |
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| storm_norm wrote: | | Asellus, one more time with the way you would write it? |
First, I would make it clear that the 38UR in r13c38 induces a strong inference between the grouped <6>s in r13c8 and the grouped <3>s in r2c13 since if both are false the DP results. This I'd write as:
38URr13c38[(6)r13c8=(3)r2c13]
Now, for Norm's AIC:
(6)r7c8 - 38URr13c38[(6)r13c8=(3)r2c13] - (3=2)r2c6 - (2=6)r3c4 - (6)r3c9=(6)r78c9 - (6)r7c8; r7c8<>6
But, as Danny points out, this can be shortened:
(6)r3c9 - 38URr13c38[(6)r13c8=(3)r2c13] - (3=2)r2c6 - (2=6)r3c4 - (6)r3c9; r3c9<>6
The end result is the same. And, we are still in the realm of AICs. (I'm not convinced that SINs are necessary ... other than, maybe, in the enjoyment of life outside sudoku  .) |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Fri Nov 21, 2008 12:25 pm Post subject: |
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Norm, it's the middle of the night and I woke with thoughts of your (38) UR running through my head. When I was first introduced to URs, the concept that first jelled is that one of the non-UR candidates must be true. So, anything common from assuming that each (in turn) was true must also be true. How might this work with this PM.
| Code: | +--------------------------------------------------------------+
| 5 1 38+2 | 7 4 236 | 23 38+6 9 |
| 349 6 2349 | 1 8 23 | 234 7 5 |
| 7 24 38+24 | 26 9 5 | 1234 38+6 126 |
|--------------------+--------------------+--------------------|
| 6 8 7 | 9 1 4 | 5 2 3 |
| 2 3 1 | 5 67 67 | 8 9 4 |
| 49 5 49 | 3 2 8 | 6 1 7 |
|--------------------+--------------------+--------------------|
| 34 249 5 | 8 67 1 | 79 36 26 |
| 8 7 236 | 26 5 9 | 123 4 126 |
| 1 29 26 | 4 3 267 | 79 5 8 |
+--------------------------------------------------------------+
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| Code: | [r1|3c 8]=6 [r7c8]<>6 [r7c8]=3
[r1|3c3 ]=2 [r3c2]=4 [r7c2]<>4 [r7c1]=4 [r7c8]=3
[r 3c3 ]=4 [r3c2]=2 [r3c4]=6 [r3c89]<>6 [r1c8]=6 [r7c8]<>6 [r7c8]=3
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And we have a common assignment [r7c8]=3. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Fri Nov 21, 2008 4:50 pm Post subject: |
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After some URing and a lot of ERing, I was left with this grid for <3>.
So looking at the strong link in C1. If it’s the bottom <3> there’s an ER in Box 9 that removes <3> in R1C7. If it’s the top <3> there’s a transported pincer(?) in Box 2 that also removes R1C7.
I don’t know what you call this technique – other than brilliant.
It didn’t get me very far – but that’s not the point.
| Code: |
+-------+-------+-------+
| . . 3 | . . 3 | 3 3 . |
| 3 . 3 | . . 3 | . . . |
| . . 3 | . . . | 3 3 . |
+-------+-------+-------+
| . . . | . . . | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
+-------+-------+-------+
| 3 . . | . . . | 3 3 . |
| . . 3 | . . . | 3 . . |
| . . . | . . . | . . . |
+-------+-------+-------+
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Fri Nov 21, 2008 6:08 pm Post subject: |
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"brilliant" would be a good name.
Congratulations, Craig!
You've discovered a very powerful technique: "grouped" (multi-) coloring.
Forget the "coloring" part, that's how the method was invented. No actual colors involved any more...
What you described is the relationship of some of the cells in a "house" (row, column or box) with regard to a candidate. There are basically two such relationships:
a) "can't be both" x. Like in row 2: it is not possible that both r2c1=3 and r2c6=3.
This relationship is called "weak", the connection between the two cells, a "weak link". I usually draw weak links as dotted lines.
b) "if one is not x then the other must be x" (or: "at least one of the two must be x") Like in column 1: at least one of the two cells r2c1 and r7c1 must be "3". This is called a "strong link". I usually draw these as a solid line.
The trick is that one can "daisy chain" weak and strong links as long as there is always a strong link followed by a weak link followed by a strong link and so on...
The "grouped" part comes in when you apply the relationship not to a single cell but to a group of cells. Like you did in row 7: "if it's the bottom ..." translates into: "r7c1 and r7c78 cannot be BOTH 3" (weak link) and the ER (which is actually a strong link because of:) "if it's NOT r7c78 then it must be r8c7.
Your relationships look like this, when drawn as a chain of weak and strong links:
Pretty, innit?
And useful. Because every such chain that begins and ends with a strong link has the potential to destroy one or more candidates. Like in this case: 3 in r1c7.
The daisy chain is like a mantra: a sentence repeted again and again. The sentence is: "if cell ... is not x, then cell ...must be x. And then cell ... cannot be x"
repeat.
"if cell ... is not x, then cell ...must be x. And then cell ... cannot be x"
Let's do it for real:
"if cell r1c6 is not 3, then cell r2c6 must be 3. And then cell r2c1 cannot be 3"
repeat
"if cell r2c1 is not 3, then cell r7c1 must be 3. And then cells r7c78 cannot be 3"
repeat
"if cells r7c78 are not 3, then cell r8c7 must be 3. And then ... "
STOP!
What did we just say?
"if r1c6 is not 3 then r8c7 must be 3" But then we can remove 3 from r1c7!!!!
If one is comfortable with the other phrasing, one could say: "at least one of r1c6 and r8c7 must be 3". Same result.
Simple yet very powerful. Takes some experience to quickly see the daisy chain. But some smart people can find the elimination even without drawing the chain...  Now you've got the name for it. Grouped Multi Coloring. Stupid name, really. But the method is "brilliant", indeed.
Happy hunting!
Last edited by nataraj on Fri Nov 21, 2008 9:01 pm; edited 1 time in total |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Fri Nov 21, 2008 6:28 pm Post subject: |
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I usually find the "simple", i.e. "non grouped" chains before I find the grouped ones. In the same, grid, the same elimination can be made by this chain:
But that was not the point...
Neither is "better", it's just what one finds first
Last edited by nataraj on Fri Nov 21, 2008 6:30 pm; edited 1 time in total |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Fri Nov 21, 2008 6:28 pm Post subject: |
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| Quote: | But, as Danny points out, this can be shortened:
(6)r3c9 - 38URr13c38[(6)r13c8=(3)r2c13] - (3=2)r2c6 - (2=6)r3c4 - (6)r3c9; r3c9<>6
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thank you, both.
| Quote: | | First, I would make it clear that the 38UR in r13c38 induces a strong inference between the grouped <6>s in r13c8 and the grouped <3>s in r2c13 since if both are false the DP results |
the inference isn't between the 6's and the 3's in r13c3, instead the inference is with the 3's in r2c13, got it!! |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Fri Nov 21, 2008 9:14 pm Post subject: |
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| cgordon wrote: | After some URing and a lot of ERing, I was left with this grid for <3>.
So looking at the strong link in C1. If it’s the bottom <3> there’s an ER in Box 9 that removes <3> in R1C7. If it’s the top <3> there’s a transported pincer(?) in Box 2 that also removes R1C7.
I don’t know what you call this technique – other than brilliant.
It didn’t get me very far – but that’s not the point.
| Code: | +-------+-------+-------+
| . . 3 | . . 3 | 3 3 . |
| 3 . 3 | . . 3 | . . . |
| . . 3 | . . . | 3 3 . |
+-------+-------+-------+
| . . . | . . . | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
+-------+-------+-------+
| 3 . . | . . . | 3 3 . |
| . . 3 | . . . | 3 . . |
| . . . | . . . | . . . |
+-------+-------+-------+
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Nice observation, but you could have also started in [c6] and derived this X-Chain.
| Code: | (3) r1c6 = r2c6 - r2c1 = r7c1 - r7c78 = r8c7 => r1c7<>3
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It's also a ...
| Code: | finned Franken Swordfish c16b9\r127 w/fin cell [r8c7] => r1c7<>3
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Do the base units of the Swordfish -- [c1], [c6] and [b9] -- look familiar to what you used?
FWIW: An ER pattern (in a box) often leads to a Franken fish when working with single-digit patterns!
===== ===== ===== =====
Now, look at [c1], [c6], and [c8] ... and think about this X-Chain.
| Code: | (3) r1c6 = r2c6 - r2c1 = r7c1 - r7c8 = r13c8 => r1c7<>3
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It's also a ...
| Code: | finned Swordfish c168\r127 w/fin cell [r3c8] => r1c7<>3
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