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sdq_pete
Joined: 30 Apr 2007
Posts: 119
Location: Rotterdam, NL
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 Posted: Sun Jun 10, 2007 11:46 am Post subject: June 10 - wrongly graded? |
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Did I do something wrong and get lucky or was this easy in the extreme?
Peter |
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sdq_pete
Joined: 30 Apr 2007
Posts: 119
Location: Rotterdam, NL
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| Posted: Sun Jun 10, 2007 1:50 pm Post subject: |
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I guess the answer is "yes" - I did make a lucky mistake. Second time through was rather harder. I came to an interesting move however - did others find another way? The situation was this:
| Code: |
+-----------+-----------+------------+
| 7 5 2 | 68 689 1 | 389 4 39 |
| 189 4 189 | 3 89 2 | 5 6 7 |
| 389 6 39 | 4 7 5 | 89 2 1 |
+-----------+-----------+------------+
| 19 3 19 | 5 2 6 | 4 7 8 |
| 6 7 4 | 9 3 8 | 2 1 5 |
| 2 8 5 | 1 4 7 | 369 39 369 |
+-----------+-----------+------------+
| 58 2 68 | 7 1 3 | 69 59 4 |
| 4 1 368 | 268 568 9 | 7 35 236 |
| 35 9 7 | 26 56 4 | 1 8 236 |
+-----------+-----------+------------+
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Play this puzzle online at the Daily Sudoku site
What I noticed was that the 89 in blocks 2 and 3 have the same state - if one is 8 the other is 8 and... Now the XY-wing with pivot at R7C7 eliminates not only the 8 at R3C3 but also the 8 at R2C3.
Peter |
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jLo
Joined: 30 Apr 2007
Posts: 55
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| Posted: Sun Jun 10, 2007 2:30 pm Post subject: |
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| W-wing is isles 8 and 9. |
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sdq_pete
Joined: 30 Apr 2007
Posts: 119
Location: Rotterdam, NL
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| Posted: Sun Jun 10, 2007 3:28 pm Post subject: |
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The only W-wing I see is on the 39's in blocks 3 and 6 - is that what you mean? The 3 in column 7 must be in block 3 or 6 so one of those two 39's must be 9.
Peter |
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George Woods
Joined: 28 Mar 2006
Posts: 304
Location: Dorset UK
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| Posted: Sun Jun 10, 2007 3:36 pm Post subject: Right answer but only via an unpopular route! |
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I must admit I used a forcing chain If r7c8 is 5 then r8c7 is 3 and naked pair 26 in row 9 means r9c5 is 5 - now there is nowhere for a 5 in BOX 7 so r7c8 is 9
Well done Texcat! below. I am kicking myself I was looking for W wings but my entry for r9c1 was foolishly 358 [/img]
Last edited by George Woods on Sun Jun 10, 2007 5:06 pm; edited 1 time in total |
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TexCat
Joined: 07 Jul 2006
Posts: 32
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| Posted: Sun Jun 10, 2007 3:49 pm Post subject: |
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| My new friend the W-wing in boxes 7 8 9! The 35 pair in boxes 7 and 9, plus that there is no 5 in row 7 of box 8, allowed me to eliminate the 3's from box 7 row 8 and from box 9 row 9. |
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George Woods
Joined: 28 Mar 2006
Posts: 304
Location: Dorset UK
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| Posted: Sun Jun 10, 2007 3:49 pm Post subject: |
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| Quote: | | What I noticed was that the 89 in blocks 2 and 3 have the same state - if one is 8 the other is 8 and... Now the XY-wing with pivot at R7C7 eliminates not only the 8 at R3C3 but also the 8 at R2C3. |
I can't see this. they look like a "conjugate pair"? if one is 8 other is 9 and vice versa |
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sdq_pete
Joined: 30 Apr 2007
Posts: 119
Location: Rotterdam, NL
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| Posted: Sun Jun 10, 2007 4:07 pm Post subject: |
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| George Woods wrote: | | they look like a "conjugate pair"? if one is 8 other is 9 and vice versa |
Hmm... I do believe you are right. I must have been lucky a second time - I'm not usually lucky!!
Peter |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Jun 10, 2007 4:41 pm Post subject: |
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| sdq_pete wrote: | | George Woods wrote: | | they look like a "conjugate pair"? if one is 8 other is 9 and vice versa |
Hmm... I do believe you are right. I must have been lucky a second time - I'm not usually lucky!!
Peter |
There's no one or the other; r2c5 must be=9. In boxes 1 and 4 there is a Type 2 rectangle on 19. Either r2c1 or r2c3 must be=8 to avoid the "deadly pattern", therefore, no other cells in row 2 can be=8. |
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George Woods
Joined: 28 Mar 2006
Posts: 304
Location: Dorset UK
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| Posted: Sun Jun 10, 2007 7:57 pm Post subject: |
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| sdq_pete wrote: | The only W-wing I see is on the 39's in blocks 3 and 6 - is that what you mean? The 3 in column 7 must be in block 3 or 6 so one of those two 39's must be 9.
Peter |
The w wing in 39s works by recognising that there is no 3 in col 7 of box9 so to allow a 3 in col7 one of the 39s must be 9 hence r6c9 cannot be 9. Hence r1c9 is 9 ... getting to the same position as was achieved by recognizing the locked 8s in Box 1 (to avoid the DR).
Unfortunately this set of moves does not solve the sudoku. So has anyone found a decent total solution other than the W wing in 35 in the the bottom 3 rows? |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Sun Jun 10, 2007 9:21 pm Post subject: |
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Looking at Sdq Pete's grid, the 8 comes out of R8C3 cos there's naked 8s in box 8. Then by my reckoning, the 6s come out of R8C4 and R8C5 cos of an xy wing on R9C1, R8C3 and R9C5 (alternately just basic x wings on Rows 7 and  . Then the 6 comes out R9C9 cos of naked 6s in box 8. Then there's an xyz wing in R8C9, R8C3 and R9C9 that takes out the 3 in R8C8. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun Jun 10, 2007 10:10 pm Post subject: George's Challenge |
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| Quote: | | So has anyone found a decent total solution other than the W wing in 35 in the the bottom 3 rows? |
In Peter's grid, the <8> in R8C3 can be eliminated due to Box 8. This creates an XY-Wing (R9C1 Pivot) to eliminate <6> in R8C5. I also applied Marty's UR. This leaves:
| Code: | +-----------+-----------+------------+
| 7 5 2 | 68 68 1 | 39 4 39 |
| 18 4 18 | 3 9 2 | 5 6 7 |
| 39 6 39 | 4 7 5 | 8 2 1 |
+-----------+-----------+------------+
| 19 3 19 | 5 2 6 | 4 7 8 |
| 6 7 4 | 9 3 8 | 2 1 5 |
| 2 8 5 | 1 4 7 | 369 39 369 |
+-----------+-----------+------------+
| 58 2 68 | 7 1 3 | 69 59 4 |
| 4 1 36 | 268 58 9 | 7 35 236 |
| 35 9 7 | 26 56 4 | 1 8 236 |
+-----------+-----------+------------+
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Now here, I believe I see a sort of UR (though I don't use UR solutions much!) on 68 in R18C45. The 58 in R8C5 acts as a sort of pivot to eliminate <6> from R8C4. This in turn eliminates <6> from R9C9.
Now, there's an XYZ-Wing that removes <3> from R8C8. And the solution follows immediately.
I didn't use the W-Wing! |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun Jun 10, 2007 10:14 pm Post subject: Scooped! |
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| cgordon posted while I was held up with interuptions. He's right, I don't need the UR to get rid of the second <6>. Shoulda seen that! |
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George Woods
Joined: 28 Mar 2006
Posts: 304
Location: Dorset UK
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| Posted: Sun Jun 10, 2007 11:21 pm Post subject: |
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| Code: |
+----------+---------+-----------+
| 7 5 2 | 68 68 1 | 3 4 9 |
| 18 4 18 | 3 9 2 | 5 6 7 |
| 39 6 39 | 4 7 5 | 8 2 1 |
+----------+---------+-----------+
| 19 3 19 | 5 2 6 | 4 7 8 |
| 6 7 4 | 9 3 8 | 2 1 5 |
| 2 8 5 | 1 4 7 | 69 39 36 |
+----------+---------+-----------+
| 58 2 68 | 7 1 3 | 69 59 4 |
| 4 1 36 | 28 58 9 | 7 35 236 |
| 35 9 7 | 26 56 4 | 1 8 23 |
+----------+---------+-----------+
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[url=http://www.dailysudoku.com/sudoku/play.shtml?p=7:5:2:68:68:1:3:4:9:18:4:18:3:9:2:5:6:7:39:6:39:4:7:5:8:2:1:19:3:19:5:2:6:4:7:8:6:7:4:9:3:8:2:1:5:2:8:5:1:4:7:69:39:36:58:2:68:7:1:3:69:59:4:4:1: 36:28:58:9:7:35:236:35:9:7:26:56:4:1:8: 23:]Play this puzzle online[/url] at the Daily Sudoku site
This is the position after using the w-wing in 39 in boxes 3 and 6 and the XY wing 63-35-56 to eliminate the 6s in row8 of box8.
There are at least 4 ways of finishing it off
a) the XYZ wing to make r8c8 5
b) forcing chain that shows that r7c8 cannot be 5 so is 9
c) W-Wing in 35 to make r9c1 3
d) Forgotten what it is called - when all cells bar 1 are filled with doublets(and no obvious solution) the triplet has the value that turns up too many times in the "set"
e) I guess at this stage almost any trial of a wrong value to a cell would lead to a quick failure i.e. there are probably many forcing chains! |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Jun 11, 2007 4:06 am Post subject: |
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| Quote: | | d) Forgotten what it is called - when all cells bar 1 are filled with doublets(and no obvious solution) the triplet has the value that turns up too many times in the "set" |
BUG+1 |
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alain
Joined: 11 Jun 2007
Posts: 1
Location: Mons, Belgium
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| Posted: Mon Jun 11, 2007 2:08 pm Post subject: Back to step1 |
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| I don't see how you eliminate 8 from r3c3 |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Jun 11, 2007 4:48 pm Post subject: Re: Back to step1 |
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| alain wrote: | | I don't see how you eliminate 8 from r3c3 |
I did the puzzle yesterday with pencil and paper, but I just tried it online and got here with basic moves:
| Code: |
+-------------+------------+--------------+
| 7 5 2 | 689 689 1 | 389 4 139 |
| 189 4 189 | 3 89 2 | 5 6 7 |
| 1389 6 1389 | 4 7 5 | 389 239 1239 |
+-------------+------------+--------------+
| 19 3 19 | 5 2 6 | 4 7 8 |
| 368 7 3468 | 89 3489 8 | 2 1 5 |
| 2 8 5 | 1 4 7 | 369 39 369 |
+-------------+------------+--------------+
| 568 2 68 | 7 1 3 | 69 59 4 |
| 4 1 368 | 268 2568 9 | 7 235 236 |
| 356 9 7 | 26 2456 4 | 1 8 236 |
+-------------+------------+--------------+
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Play this puzzle online at the Daily Sudoku site
Note in boxes 1 and 4 there is a rectangle on 19. R2c1 or r2c3 must be an 8 to avoid the deadly pattern, so other 8s in row two are eliminated. R2c5 has to be 9 and that creates a 18 pair in box 1 which gets rid of other 8s in that box. |
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MarieHelene
Joined: 11 Jun 2007
Posts: 1
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| Posted: Mon Jun 11, 2007 5:39 pm Post subject: |
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[code]
+----------+---------+-----------+
| 7 5 2 | 68 68 1 | 3 4 9 |
| 18 4 18 | 3 9 2 | 5 6 7 |
| 39 6 39 | 4 7 5 | 8 2 1 |
+----------+---------+-----------+
| 19 3 19 | 5 2 6 | 4 7 8 |
| 6 7 4 | 9 3 8 | 2 1 5 |
| 2 8 5 | 1 4 7 | 69 39 36 |
+----------+---------+-----------+
| 58 2 68 | 7 1 3 | 69 59 4 |
| 4 1 36 | 28 58 9 | 7 35 236 |
| 35 9 7 | 26 56 4 | 1 8 23 |
+----------+---------+-----------+
[quote]
There are at least 4 ways of finishing it off
a) the XYZ wing to make r8c8 5
b) forcing chain that shows that r7c8 cannot be 5 so is 9
c) W-Wing in 35 to make r9c1 3
d) Forgotten what it is called - when all cells bar 1 are filled with doublets(and no obvious solution) the triplet has the value that turns up too many times in the "set"
e) I guess at this stage almost any trial of a wrong value to a cell would lead to a quick failure i.e. there are probably many forcing chains!
Either R9C1 or R9C9 must be 3 which is forcing 5 in R8C8. This simply solves the puzzle, isn't it? |
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sdq_pete
Joined: 30 Apr 2007
Posts: 119
Location: Rotterdam, NL
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| Posted: Thu Jun 14, 2007 3:48 pm Post subject: |
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| Code: | +----------+---------+-----------+
| 7 5 2 | 68 68 1 | 3 4 9 |
| 18 4 18 | 3 9 2 | 5 6 7 |
| 39 6 39 | 4 7 5 | 8 2 1 |
+----------+---------+-----------+
| 19 3 19 | 5 2 6 | 4 7 8 |
| 6 7 4 | 9 3 8 | 2 1 5 |
| 2 8 5 | 1 4 7 | 69 39 36 |
+----------+---------+-----------+
| 58 2 68 | 7 1 3 | 69 59 4 |
| 4 1 36 | 28 58 9 | 7 35 236 |
| 35 9 7 | 26 56 4 | 1 8 23 |
+----------+---------+-----------+ |
MarieHelene - I thought perhaps someone else might have answered by now... For what it's worth, I see that a 3 at R9C1 leads to a 5 at R8C8, but I don't see (by any reasonably short chain of logic) how 3 at R9C9 leads to 5 at R8C8.
Peter |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Jun 14, 2007 5:11 pm Post subject: |
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| sdq_pete wrote: | | Code: | +----------+---------+-----------+
| 7 5 2 | 68 68 1 | 3 4 9 |
| 18 4 18 | 3 9 2 | 5 6 7 |
| 39 6 39 | 4 7 5 | 8 2 1 |
+----------+---------+-----------+
| 19 3 19 | 5 2 6 | 4 7 8 |
| 6 7 4 | 9 3 8 | 2 1 5 |
| 2 8 5 | 1 4 7 | 69 39 36 |
+----------+---------+-----------+
| 58 2 68 | 7 1 3 | 69 59 4 |
| 4 1 36 | 28 58 9 | 7 35 236 |
| 35 9 7 | 26 56 4 | 1 8 23 |
+----------+---------+-----------+ |
MarieHelene - I thought perhaps someone else might have answered by now... For what it's worth, I see that a 3 at R9C1 leads to a 5 at R8C8, but I don't see (by any reasonably short chain of logic) how 3 at R9C9 leads to 5 at R8C8.
Peter |
Peter,
I'm not sure what you're looking at, but I'd suggest re-reading your post and looking at the grid. If r9c9=3, r8c8 is not a 3. |
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