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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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 Posted: Thu May 08, 2008 6:08 am Post subject: May 8 vh |
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Basics were a real challenge this time ...
(hint: naked quad in box 6, then naked quint/hidden pair in row 5)
The usual xy-wings (look for candidates greater than 2 and less than 6 in boxes 4,5 and 6) finished it off ... |
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andras
Joined: 31 Oct 2007
Posts: 56
Location: Mid Wales
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| Posted: Thu May 08, 2008 6:15 am Post subject: |
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Actually I thought that it solved very quickly right up to the point where something more advanced was needed; indeed, I thought it might work out without any 'dodges' at all.
In the end it turned out to be a one-trick puzzle, with a simple xy-wing (pivot in C9, not to be too much of a spoiler!), but a very satisfying puzzle for all that. 
John |
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George Woods
Joined: 28 Mar 2006
Posts: 304
Location: Dorset UK
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| Posted: Thu May 08, 2008 8:31 am Post subject: w_wing alos works! |
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| a w-wing 34 in boxes 5 and 6 operating on row 4 solves the problem! |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Thu May 08, 2008 11:43 am Post subject: |
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After finding the quad in Box 6, it seemed like a medium with a basic UR (269).
Last edited by cgordon on Thu May 08, 2008 5:31 pm; edited 1 time in total |
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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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| Posted: Thu May 08, 2008 12:32 pm Post subject: |
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George: | Quote: | | a w-wing 34 in boxes 5 and 6 |
Yep, or a 45 w-wing in rows 4 & 5.
Or a type 1 UR in r56 in 29s.
Or XY-wing - the one I see is pivoted on r4c1.
Many choices in a fairly straightforward puzzle. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Thu May 08, 2008 12:46 pm Post subject: |
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After basics, I found three moves to solve the puzzle, all in box 4. An xy-wing on {345} with pivot at r4c1, an xy-wing on {345} with pivot at r5c2, and a Type 1 UR on {269} which makes r6c1 = <9>.
It was fun just finding all the choices to complete the puzzle.
Ted |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu May 08, 2008 3:35 pm Post subject: |
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| The Type 4 UR on 26 did it for me. I happened to notice that before searching for other techniques. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Thu May 08, 2008 4:34 pm Post subject: |
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| put me down for the UR 26 |
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stevieboy
Joined: 25 Jan 2008
Posts: 31
Location: Michigan
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| Posted: Thu May 08, 2008 5:35 pm Post subject: |
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After all of your help a couple of weeks ago, I've been completely successful in solving the Very Hards here, as well as those Extreme puzzles on the 'playr' website!! So, I believe that I've finally got it!
For this one (after the basics), I looked for a pattern of "pair links" (I'm still lousy with the terminology  ) , and found the following:
35 35 (in R3C13)
259 59 25 (in R8C123)
...I "linked" the 2's in R8C13; the 5's from R8C3 to R3C3; and, the 3's in R3C31 -> this enabled me to eliminate the 5's in R4C1 and R5C1, leaving me a lone 5 in R5C2.....puzzle solved!
Now, apparently, this is all legitimate, or, I've been VERY lucky in the last 6 puzzles I've solved this way!
What is this method called? It IS legitimate, right?
Thanks. |
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crunched
Joined: 05 Feb 2008
Posts: 168
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| Posted: Thu May 08, 2008 6:24 pm Post subject: |
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| stevieboy wrote: | After all of your help a couple of weeks ago, I've been completely successful in solving the Very Hards here, as well as those Extreme puzzles on the 'playr' website!! So, I believe that I've finally got it!
For this one (after the basics), I looked for a pattern of "pair links" (I'm still lousy with the terminology ) , and found the following:
35 35 (in R3C13)
259 59 25 (in R8C123)
...I "linked" the 2's in R8C13; the 5's from R8C3 to R3C3; and, the 3's in R3C31 -> this enabled me to eliminate the 5's in R4C1 and R5C1, leaving me a lone 5 in R5C2.....puzzle solved!
Now, apparently, this is all legitimate, or, I've been VERY lucky in the last 6 puzzles I've solved this way!
What is this method called? It IS legitimate, right?
Thanks. |
I hope you get an answer. Is this a legitimate trick? |
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stevieboy
Joined: 25 Jan 2008
Posts: 31
Location: Michigan
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| Posted: Thu May 08, 2008 6:46 pm Post subject: |
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I mean, you can use DIFFERENT numbers for these trail of links...as long as the first and last link have number(s) that can be eliminated in the common-sharing cells, right? I've seen that done before here.
Isn't there something about an ODD or EVEN number of these links affecting the accuracy of the solution (i.e., the puzzle being incorrectly solved)?
Again, if this is legitimate, then I didn't have to use ONE single XY-Wing, XYZ-Wing, etc....the only other thing I did use here, was a 3-4 UR in R8C45 and R5C5; this eliminated the 3-4 in R5C4...after that was the "trail of links" I've been talking about. |
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Dougs
Joined: 24 Apr 2008
Posts: 2
Location: London. UK
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| Posted: Thu May 08, 2008 7:30 pm Post subject: |
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My first post - have been doing VHs for the last few of weeks and used the Forum for huge amounts of guidance. Finally cracked it on this one. Thanks to you all for the help  - used the UR type 1 ( I think) on 269 ... am now to find the other types of UR .. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Thu May 08, 2008 7:47 pm Post subject: |
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| Quote: | | ...I "linked" the 2's in R8C13; the 5's from R8C3 to R3C3; and, the 3's in R3C31 -> this enabled me to eliminate the 5's in R4C1 and R5C1, leaving me a lone 5 in R5C2.....puzzle solved! |
I lost you here - though perhaps someone more proficient than me can identify this as colouring - which I'm not into. It does look to me like you are guessing - ie making a random assumption and seeing if it works.
The UR that has been repeatedly mentioned above is probably the most straightforward solution. The 269 has to be a <9> -otherwise the puzzle doesn't have a unique solution.
| Code: |
+-------+-------+-------+
| . . . | . . . | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
+-------+-------+-------+
| . . . | . . . | . . . |
| 26 . .| . . . | . . 26|
| 269 . | . . . | . . 26|
+-------+-------+-------+
| . . . | . . . | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
+-------+-------+-------+
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stevieboy
Joined: 25 Jan 2008
Posts: 31
Location: Michigan
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| Posted: Thu May 08, 2008 8:05 pm Post subject: |
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Nope, it wasn't guessing.
Every pair-link that I described was just that, a pair. There were no other
'2', '5', and '3' along those columns or rows - just the pair-links indicated.
The trail basically led from R8C1 to R3C1 - both of THOSE cells (the "pincers", if you will), have a '5' as the common number, meaning, one of those 2 cells has to be a '5'....so, you can eliminate both of the '5's in Box4C1 (the common cells), leaving the only 5 in Box 4 in R5C2.
I was only asking if this is a legitimate solving method - I 'believe' it is, based on the fact that I've correctly solved a bunch of Sudoku's in the last few days using that method!
I've gone from, pretty much, gathering all of the "expert" solvers' advice in our discussions, to actually (I believe) using a (seemingly) sure-fire method of solving these babies! I just need verification.
Thanks. |
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Clement
Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania
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| Posted: Thu May 08, 2008 8:06 pm Post subject: Daily Sudoku: Thu8-May-2008 VH |
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Two UR's Solved The Puzzle for me: UR (3,4) r58c45
UR (2,6) r56c19. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Thu May 08, 2008 8:27 pm Post subject: |
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| stevieboy wrote: | Nope, it wasn't guessing.
Every pair-link that I described was just that, a pair. There were no other
'2', '5', and '3' along those columns or rows - just the pair-links indicated.
The trail basically led from R8C1 to R3C1 - both of THOSE cells (the "pincers", if you will), have a '5' as the common number, meaning, one of those 2 cells has to be a '5'....so, you can eliminate both of the '5's in Box4C1 (the common cells), leaving the only 5 in Box 4 in R5C2.
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Hi stevieboy,
I am also a recent addition to the Very Hard puzzles and still have a lot to learn including how to post "code" info.
In any case, I definitely have a different solution after basics and the Type 1 UR on {349}. Specifically, I eliminated the <5> from r8c1 because of the naked double on<35> in r34c1. How did you get rid of the <35> in r4c1?
Ted |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Thu May 08, 2008 8:57 pm Post subject: |
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| Stevieboy: I'll defer further comments to the Masters - but I would note that after basics there are a pair of <35>s in R34C1 which eliminate all other 3's and 5's in that column. But as far as I can see there is no functional link that can eliminate the remaing two 5's in each of col. 2 and 3. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Thu May 08, 2008 9:01 pm Post subject: |
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Good news and bad news.
Bad news: no, it is not a method. It does not have a name. it is not legitimate. Do not pass "GO". Do not collect 200 $.
Good news: stevieboy, you are halfway there. More than halfway, in fact. The difficult part is to recognize patterns. You got that part right.
There are, in fact, "links", and you found them and desribed them correctly.
There is only a tiny bit missing - the connection between the links.
Or, if you will, the "meaning" of these links. And I will go into the meaning of the links to show why the deduction is vot valid here:
| stevieboy wrote: | | both of THOSE cells (the "pincers", if you will), have a '5' as the common number, meaning, one of those 2 cells has to be a '5' |
First, the grid after basics (plus the additional "5" in r8c1, which can of course be removed by the naked pair 35 in c1. But the nice thing with sudoku is, you dont have to find all the possible eliminations. I am concentrating on the logical soundness of the "trail" argument, here ):
| Code: |
+--------------------------+--------------------------+--------------------------+
| 8 6 4 | 2 7 9 | 3 15 15 |
| 1 2 9 | 6 5 3 | 4 7 8 |
| 35# 7 35# | 1 8 4 | 6 2 9 |
+--------------------------+--------------------------+--------------------------+
| 35 1 7 | 34 6 2 | 9 8 45 |
| 26 45 8 | 9 34 7 | 1 345 26 |
| 269 49 23 | 5 1 8 | 7 34 26 |
+--------------------------+--------------------------+--------------------------+
| 7 3 6 | 8 9 5 | 2 14 14 |
| 259# 59 25# | 34 34 1 | 8 6 7 |
| 4 8 1 | 7 2 6 | 5 9 3 |
+--------------------------+--------------------------+--------------------------+
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I've marked "the trail".
Let us look at the links and what the term "link" means.
There are only 2 "2"s in row 8. That means, if one is "2" the other cannot be "2" (that would be the definition af a "weak" link).
Or, if one cell is NOT "2", then the other must be "2" (that is the definition of a "strong" link).
Similarly, there are only 2 "5"s in col 3, and only 2 "3"s and two "5"s in row 3.
This means, yes there are strong links in those houses.
The question is, now, to verify if any eliminations can be made.
It helps to try and use the links as signposts and read aloud the associated meaning. The goal is to construct a logical chain that - taken together - leads to a valid elimination.
An easy to remember rule of thumb: if you would like to make an elimination on X, always start with the sentence "if ... is NOT X" (in other words, start with a strong link).
Start with row 3. There is a potentially useful strong link: "if r3c1 is not 5 then r3c3 is 5"
Continue with the link in col 3: "if r3c3 is 5 then r8c3 is not 5" (it is "2")
and the final link in row 8: "if r8c3 is 2, then r8c1 is not 2"
This is what the trail means.
Unfortunately the logical meaning of the links do not allow an elimination. Why?
Taken together, they read:
"if r3c1 is not 5 then (r3c3 is 5 then r8c3 is 2 then) r8c1 is not 2"
So there is NO usable connection between the "5"s in r3c1 and r4c1.
Therefore no valid elimination. Sorry.
IF we would have been able to find links between r3c1 and r8c1 that amounted to the sentence "if r3c1 is not 5 then r8c1 is 5", then we could have eliminated 5 from all cells that see r3c1 and r8c1.
As an exercise in strong and weak links, follow the "trail" # (xy-wing, eliminates 4) and o (w-wing, eliminates 3) in this grid:
| Code: | +--------------------------+--------------------------+--------------------------+
| 8 6 4 | 2 7 9 | 3 15 15 |
| 1 2 9 | 6 5 3 | 4 7 8 |
| 35 7 35 | 1 8 4 | 6 2 9 |
+--------------------------+--------------------------+--------------------------+
| 35# 1 7 | 34# 6 2 | 9 8 45 |
| 26 45#o 8 | 9 34o 7 | 1 345 26 |
| 269 49o 23 | 5 1 8 | 7 34o 26 |
+--------------------------+--------------------------+--------------------------+
| 7 3 6 | 8 9 5 | 2 14 14 |
| 29 59 25 | 34 34 1 | 8 6 7 |
| 4 8 1 | 7 2 6 | 5 9 3 |
+--------------------------+--------------------------+--------------------------+
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Thu May 08, 2008 10:11 pm Post subject: |
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| nataraj wrote: | An easy to remember rule of thumb: if you would like to make an elimination on X, always start with the sentence "if ... is NOT X" (in other words, start with a strong link).
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I think I better explain this one.
In order to make a sound logical elimination, you have to cover all your bases. Just saying "if r3c1 is 5 then r4c1 cannot be 5" is kind of trivial. Of course it can't be because we supposed that there is already another 5 in c1.
But if we start the sentence with "if r3c1 is NOT 5, then .." and end it with "r4c1 is 5", then we've covered all possibilities and no matter what happens in r3c1 or r4c1, all other cells in col 1 could not be 5. This is the reasoning behind all "pincer" arguments: even if cell 1 is not X, then cell 2 is X and all cells seeing cell 1 and cell 2 cannot be X. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Fri May 09, 2008 1:16 am Post subject: |
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| nataraj wrote: | | nataraj wrote: | An easy to remember rule of thumb: if you would like to make an elimination on X, always start with the sentence "if ... is NOT X" (in other words, start with a strong link).
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I think I better explain this one.
In order to make a sound logical elimination, you have to cover all your bases. Just saying "if r3c1 is 5 then r4c1 cannot be 5" is kind of trivial. Of course it can't be because we supposed that there is already another 5 in c1.
But if we start the sentence with "if r3c1 is NOT 5, then .." and end it with "r4c1 is 5", then we've covered all possibilities and no matter what happens in r3c1 or r4c1, all other cells in col 1 could not be 5. This is the reasoning behind all "pincer" arguments: even if cell 1 is not X, then cell 2 is X and all cells seeing cell 1 and cell 2 cannot be X. |
Nataraj,
THANK YOU, THANK YOU 
I have been struggling with chains for some time without real success, and I usually start with "if <CellCandidate> is <SomeValue> then" . Your simple explanation makes it clear why that is wrong.
Another day, another sudoku trick.
Ted |
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