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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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 Posted: Mon Apr 21, 2008 6:06 am Post subject: April 21 VH |
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nice and long, just the way I like 'em !
Even getting to the "advanced" point took longer than usual, some noteable points on the way: naked pair (NP) 1,6 in r3 (=> r2c4=9), 4 in c5 confined to box 2 (r4c6=4, r4c4=7), 4 in r3 confined to bax 3 (r7c1=4), 6 in c1 confined to box 7 (r4c1=3)
In the end, it was a skyscraper that solved the puzzle (8 in rows 2 and 7, solves r1c4=3 and r9c5=5) |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Mon Apr 21, 2008 6:19 am Post subject: |
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on the contrary, nataraj...
I used the sweep function on the site and ran into some naked pairs and some pointing pairs which opened up the {4,8} UR, and was quite visible before all basics were done. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Mon Apr 21, 2008 6:57 am Post subject: |
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| storm_norm wrote: | on the contrary, nataraj...
I used the sweep function on the site ... |
Now, that's just too bad - over before you know it...
As I said, I like to spend time and effort. No sweep with pencil and paper 
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"The 48 UR" ??... don't see a useable one right away at this point, which is what I believe is end of basics:
| Code: |
+--------------------------+--------------------------+--------------------------+
| 1 57 458 | 38 348 2 | 6 9 78 |
| 27 6 48 | 9 48 5 | 3 1 27 |
| 28 3 9 | 16 7 16 | 48 45 258 |
+--------------------------+--------------------------+--------------------------+
| 3 8 6 | 7 1 4 | 5 2 9 |
| 5 2 1 | 368 9 368 | 47 47 68 |
| 9 4 7 | 5 2 68 | 18 3 168 |
+--------------------------+--------------------------+--------------------------+
| 4 157 58 | 138 6 9 | 2 57 135 |
| 6 9 2 | 4 35 137 | 17 8 135 |
| 78 157 3 | 2 58 178 | 9 6 4 |
+--------------------------+--------------------------+--------------------------+
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play online
oh, and BTW, there is an xy-wing, too (78-85-57) |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Mon Apr 21, 2008 7:33 am Post subject: Re: April 21 VH |
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| nataraj wrote: | | In the end, it was a skyscraper that solved the puzzle |
Why is it that in sudoku all skyscrapers look like the Citicorp building ? |
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andras
Joined: 31 Oct 2007
Posts: 56
Location: Mid Wales
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| Posted: Mon Apr 21, 2008 8:40 am Post subject: |
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The 4,8 UR broke it for me - a very quick solution once I'd got there, but the build-up to it took a little time. 
John |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Mon Apr 21, 2008 12:08 pm Post subject: |
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| After struggling badly with the last VH, I redeemed myself here and brilliantly used an ER in Box 7 and the two 8's in R2 to remove the 8 from R9C5. I ain't doing it again, but I suspect my ER is related to Nataraj's skyscraper. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Mon Apr 21, 2008 12:39 pm Post subject: |
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Another variation on the theme is to use multi-coloring to eliminate the <8> at r1c4 thereby solving the puzzle. This is one of the two eliminations provided by the skyscraper.
Ted |
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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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| Posted: Mon Apr 21, 2008 1:34 pm Post subject: VH |
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A 578 xy-wing eliminated <7> from R7C2 and solved the puzzle for me.
Earl |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Mon Apr 21, 2008 2:02 pm Post subject: |
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| Please, John or Norm or any of you UR gurus: please explain the 4,8 UR in more detail. All I can see is a possible DP in r12c35, and with 4 being locked into c35 we can remove 8 from r1c35. Then what? Clearly, I am missing something here. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Apr 21, 2008 3:34 pm Post subject: |
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| Quote: | "The 48 UR" ??... don't see a useable one right away at this point, which is what I believe is end of basics:
| Code: |
+--------------------------+--------------------------+--------------------------+
| 1 57 458 | 38 348 2 | 6 9 78 |
| 27 6 48 | 9 48 5 | 3 1 27 |
| 28 3 9 | 16 7 16 | 48 45 258 |
+--------------------------+--------------------------+--------------------------+
| 3 8 6 | 7 1 4 | 5 2 9 |
| 5 2 1 | 368 9 368 | 47 47 68 |
| 9 4 7 | 5 2 68 | 18 3 168 |
+--------------------------+--------------------------+--------------------------+
| 4 157 58 | 138 6 9 | 2 57 135 |
| 6 9 2 | 4 35 137 | 17 8 135 |
| 78 157 3 | 2 58 178 | 9 6 4 |
+--------------------------+--------------------------+--------------------------+ |
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Nataraj, there are only two 4s in row 1, making it a Type 4.
I had an early W-Wing on 57, I used the UR and an XY-Wing (34-35-45) with pincer coloring finished it off. |
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sheryl
Joined: 24 Sep 2007
Posts: 64
Location: New York
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| Posted: Mon Apr 21, 2008 4:36 pm Post subject: |
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| I had a UR on the 168s in r57, c69 making r5c6 the 3. don't know if this is kosher but it did work, the solve for me was a skyskraper on 8s making a 3 in r1c4 solving the puzzle. Did i do this right??? |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Mon Apr 21, 2008 5:24 pm Post subject: |
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| Marty R. wrote: | Nataraj, there are only two 4s in row 1, making it a Type 4.
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Thanks, Marty! I thought so (it was the only UR with 4,8 I could see) - sorry, I am notoriously bad with taxonomy so never can remember the types correctly. I did try and make the elimination of 8 from r1c3 and r1c5 but that did not get me anywhere. Still need either the skyscraper or the xy-wing.
But John said:"The 4,8 UR broke it for me", and Norm said something to the same effect... Hm, still puzzled. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Mon Apr 21, 2008 5:29 pm Post subject: |
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| sheryl wrote: | | I had a UR on the 168s in r57, c69 making r5c6 the 3. don't know if this is kosher but it did work, the solve for me was a skyskraper on 8s making a 3 in r1c4 solving the puzzle. Did i do this right??? |
Sheryl, I cannot comment on the UR (if you read my other posts here you'll see why - URs are not my speciality) but you are absolutely right with the skyscraper. And indeed the elimination of 8 in r1c4 (making r1c4=3) is enough to solve the puzzle (as an optional step, the same skyscraper removes 8 from r9c5). |
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zx
Joined: 21 Apr 2008
Posts: 6
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| Posted: Mon Apr 21, 2008 6:12 pm Post subject: |
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| Stupid question: how does the 4 in r3 confined to box 3 then unlock the 4 in r7c1? I still have possibilities for 4s in r7c3 and r2c1. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Mon Apr 21, 2008 6:31 pm Post subject: |
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zx, welcome to this forum!
There is a naked pair {4,8} in row 2 (cols 3 and 5) that should take care of the 4 in r2c1. Another 4 in r3c1 is eliminated by the box/line interaction I mentioned: in box 3, one of r3c78 must be 4. That leaves a hidden single 4 in column 1: r7c1=4.
Enjoy! |
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Clement
Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania
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| Posted: Mon Apr 21, 2008 6:37 pm Post subject: Daily Sudoku: Mon 21-Apr-2008 VH |
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Assume r1c2 to be 5, this creates a pair of (1,7) in r79c2 in BOX7, leaving 6 in r8c1 and 8 in r9c1. Therefore, r9c5 is 5 and r8c5 must be 3. With 5 in r1c2 and 3 in r8c5 we get a UR of 4,8 in BOX1 and BOX2 i.e r12c35. Therefore,r1c2 cannot be 5 as assumed, it must be 7.This solves the puzzle.
How do your find this Sudoku experts? |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Mon Apr 21, 2008 6:52 pm Post subject: Re: Daily Sudoku: Mon 21-Apr-2008 VH |
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| Clement wrote: | Assume r1c2 to be 5, ...
How do your find this... |
Exactly, how do you find this?
Why r1c2 ?
Why 5?
That's the problem with forcing chains: never a good reason for the starting point other than "because it works". There are 23 bi-value cells in the "grid after basics", with two alternatives each, that makes 46 possible assumptions. Some of them will lead to a contradiction, some of them to a solution. For me: None of them lead to satisfaction.
If other methods work, I use them. If not I throw the puzzle into the garbage.
But that's a personal choice, of course |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Mon Apr 21, 2008 7:48 pm Post subject: |
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| nataraj wrote: | | Marty R. wrote: | Nataraj, there are only two 4s in row 1, making it a Type 4.
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Thanks, Marty! I thought so (it was the only UR with 4,8 I could see) - sorry, I am notoriously bad with taxonomy so never can remember the types correctly. I did try and make the elimination of 8 from r1c3 and r1c5 but that did not get me anywhere. Still need either the skyscraper or the xy-wing.
But John said:"The 4,8 UR broke it for me", and Norm said something to the same effect... Hm, still puzzled. |
John must have found something more than I did or I made a mistake somewhere because I finished the puzzle after the UR on{4,8}... I found the ur on 4,8 and mistakenly thought it was the only advanced move needed. 
since the topic had been brought up, I thought it was kind of cool that the UR was there only after you find the naked pair {4,8} in row 2. this, of course, after the discussion on whether you need to write in candidates and if you can spot advanced moves before basics are done.
I was kind of curious to see how many puzzles would "give up" advanced moves before you needed to eliminate all the basics and have been using the sweep function ever since. its fascinating to watch the puzzle unfold in this way. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Mon Apr 21, 2008 9:43 pm Post subject: |
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I'm not sure if sheryl reasoned her UR correctly. However, it is interesting. There is a 68 UR in r56c69. And, it is an example of what I believe is called Type "6a". There is a diagonal pair of 68 bivalues and a strong link on <6> in r6. So, <6> can be eliminated from the trivalue corner in r5 (r5c6). (This is not immediately helpful.)
However, the strong inference induced by the UR between the <3> in r5c6 and the <1> in r6c9 can be combined with the <3> strong link in c6 and the {135} ALS in r8 to provide a short AIC:
(3)r5c6=(1)r6c9-(1={35})r8c59-(3)r8c6=(3)r5c6; r5c6=3
In ordinary language: If r5c6 is not 3, then r6c9 must be <1> (to kill the DP), resulting in a {35} pair in r8, removing <3> from r8c6, and so r5c6 must be <3> (due to the strong link in c6). But this is a contradiction. So the assumption (in italics) is false and r5c6 must be <3>.
Perhaps this is what sheryl saw. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Mon Apr 21, 2008 9:47 pm Post subject: |
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| cgordon wrote: | | I suspect my ER is related to Nataraj's skyscraper. |
I believe it is more correctly related to the Kite in r2 and c1 that pivots in b1. |
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