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Kdelle
Joined: 20 Mar 2008
Posts: 59
Location: Hudson, NH
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 Posted: Thu Mar 20, 2008 1:30 am Post subject: Tough Puzzle from Sudoku.com.au |
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| Code: |
+-----------------+------------------+----------------+
| 2467 2468 247 | 678 1 9 | 348 5 38 |
| 9 4568 457 | 5678 3567 3567 | 2 48 1 |
| 1 58 3 | 58 2 4 | 6 9 7 |
+-----------------+------------------+----------------+
| 3 24569 2459 | 4567 45679 567 | 5789 1 5689 |
| 46 4569 1 | 2 345679 8 | 3579 367 3569 |
| 8 7 59 | 156 3569 1356 | 359 2 4 |
+-----------------+------------------+----------------+
| 5 234 8 | 9 467 267 | 1 3467 36 |
| 47 349 6 | 1457 457 157 | 389 38 2 |
| 247 1 2479 | 3 8 267 | 4579 467 569 |
+-----------------+------------------+----------------+
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Play this puzzle online at the Daily Sudoku site
This is as far as I can get. I read through the comments on the site and cannot follow their proof. I was hoping someone here could help. I feel like I must be missing something as several people found it quite easy.
Thanks.
Kathy |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Thu Mar 20, 2008 4:09 am Post subject: |
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hello,
welcome...
yikes, from where did this monster arise? it reminds me of the extremes from sudokulist weekly competition.
your grid as it stands can still be broken down a little bit more using basics, all be it not the easist basics.
in row 8 is a naked quad on 1457
but it might be easier to see it as a hidden triple on 389, the cells with {3,8,9} already in them are the only cells that {3,8,9} can go so you can eliminate anything else in those cells.
also, there are some box/line interactions that can be taken care of in your grid.
if you will notice box 5 and row 6... the number 6 can only go in row 6 in that box. so you can remove all the 6's in box 5 that are not in row 6.
this is your grid now...
| Code: | 2467 2468 247 | 678 1 9 | 348 5 38
9 4568 457 | 5678 3567 3567 | 2 48 1
1 58 3 | 58 2 4 | 6 9 7
------------------------------------------------------------------
3 24569 2459 | 457 4579 57 | 5789 1 5689
46 4569 1 | 2 34579 8 | 3579 367 3569
8 7 59 | 156 3569 1356 | 359 2 4
------------------------------------------------------------------
5 234 8 | 9 467 267 | 1 3467 36
47 39 6 | 1457 457 157 | 389 38 2
247 1 2479 | 3 8 267 | 4579 467 569 |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Thu Mar 20, 2008 12:08 pm Post subject: |
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| Code: | +-----------------+------------------+----------------+
| 2467 2468 247 | 678 1 9 | 348 5 38 |
| 9 4568 457 | 5678 3567 3567 | 2 48 1 |
| 1 58 3 | 58 2 4 | 6 9 7 |
+-----------------+------------------+----------------+
| 3 24569 2459 | 4567 45679 567 | 5789 1 5689 |
| 46 4569 1 | 2 345679 8 | 3579 367 3569 |
| 8 7 59 | 156 3569 1356 | 359 2 4 |
+-----------------+------------------+----------------+
| 5 234 8 | 9 467 267 | 1 3467 36 |
| 47 349 6 | 1457 457 157 | 389 38 2 |
| 247 1 2479 | 3 8 267 | 4579 467 569 |
+-----------------+------------------+----------------+
|
Cant find anything nice for this puzzle so far.
But there is a brute force solution by testing 3 in r1c7:
[edit: formulated it better now (hopefully)]
r1c7=3 => r9c7=4 => r9c9=5
Then with r9c7<>9 there are 2 strong links for 9, in box 9 and row 9
r9c9=5 => r8c7=9
r9c9=5 => r9c3=9 => r6c3=5 => r6c7=9 (because r6c7<>3)
i.e. we have two 9's in column 7, r1c7 cannot be 3.
Solves with x-wing and coloring then.
Can you provide a link to the "comments on the site" ? |
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Kdelle
Joined: 20 Mar 2008
Posts: 59
Location: Hudson, NH
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| Posted: Thu Mar 20, 2008 1:57 pm Post subject: |
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| storm_norm wrote: | hello,
welcome...
yikes, from where did this monster arise? it reminds me of the extremes from sudokulist weekly competition.
your grid as it stands can still be broken down a little bit more using basics, all be it not the easist basics.
in row 8 is a naked quad on 1457
but it might be easier to see it as a hidden triple on 389, the cells with {3,8,9} already in them are the only cells that {3,8,9} can go so you can eliminate anything else in those cells.
also, there are some box/line interactions that can be taken care of in your grid.
if you will notice box 5 and row 6... the number 6 can only go in row 6 in that box. so you can remove all the 6's in box 5 that are not in row 6.
this is your grid now...
| Code: | 2467 2468 247 | 678 1 9 | 348 5 38
9 4568 457 | 5678 3567 3567 | 2 48 1
1 58 3 | 58 2 4 | 6 9 7
------------------------------------------------------------------
3 24569 2459 | 457 4579 57 | 5789 1 5689
46 4569 1 | 2 34579 8 | 3579 367 3569
8 7 59 | 156 3569 1356 | 359 2 4
------------------------------------------------------------------
5 234 8 | 9 467 267 | 1 3467 36
47 39 6 | 1457 457 157 | 389 38 2
247 1 2479 | 3 8 267 | 4579 467 569 |
|
Norm Thanks. I had the Naked Quad...just inadvertently left that 4 in r8c2 in my haste to get this posted. Should have seen the sixes..But after that I'm still stuck, other than "Brute Force" as suggested in the previous post. |
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Kdelle
Joined: 20 Mar 2008
Posts: 59
Location: Hudson, NH
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| Posted: Thu Mar 20, 2008 2:05 pm Post subject: |
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| ravel wrote: | | Code: | +-----------------+------------------+----------------+
| 2467 2468 247 | 678 1 9 | 348 5 38 |
| 9 4568 457 | 5678 3567 3567 | 2 48 1 |
| 1 58 3 | 58 2 4 | 6 9 7 |
+-----------------+------------------+----------------+
| 3 24569 2459 | 457 4579 57 | 5789 1 5689 |
| 46 4569 1 | 2 34579 8 | 3579 367 3569 |
| 8 7 59 | 156 3569 1356 | 359 2 4 |
+-----------------+------------------+----------------+
| 5 234 8 | 9 467 267 | 1 3467 36 |
| 47 39 6 | 1457 457 157 | 389 38 2 |
| 247 1 2479 | 3 8 267 | 4579 467 569 |
+-----------------+------------------+----------------+
|
Cant find anything nice for this puzzle so far.
But there is a brute force solution by testing 3 in r1c7:
[edit: formulated it better now (hopefully)]
r1c7=3 => r9c7=4 => r9c9=5
Then with r9c7<9> r8c7=9
r9c9=5 => r9c3=9 => r6c3=5 => r6c7=9 (because r6c7<>3)
i.e. we have two 9's in column 7, r1c7 cannot be 3.
Solves with x-wing and coloring then.
Can you provide a link to the "comments on the site" ? |
Ravel,
Using Norm's adjusted grid, I can follow your "Brute Force" solution.....Should we have to resort to that? Seems like a guess to me.
Here's the ling to the "comments"....ttt from Vietnam loses me at step 3. They're using different conventions (rows 1 to 9 from the bottom up and colums a to i from left to right).
http://sudoku.com.au/1V18-3-2008-sudoku.aspx#Comments
Can you see it?
Kathy |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Thu Mar 20, 2008 4:18 pm Post subject: |
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| Kdelle wrote: | | Using Norm's adjusted grid, I can follow your "Brute Force" solution.....Should we have to resort to that? Seems like a guess to me. |
Yes, you somewhat have to guess, where to start. Of course i would not start with the first candidate and work through to the last (if i have no success), but first study the grid to check out, where i have good chances. So its a kind of educated guessing.
You will have to do that with the method of this site also in some way (starting points might bring no success at all).
Thanks for the link. I only looked at the first chain (point 3) and try to explain it here.
| Code: | a b c d e f g h i
*-------------------------------------------------------------*
9| 2467 2468 247 | 678 1 9 | 348 5 38 |
8| 9 4568 457 | 5678 3567 3567 | 2 48 1 |
7| 1 58 3 | 58 2 4 | 6 9 7 |
|--------------------+--------------------+-------------------|
6| 3 24569 2459 | 457 4579 57 | 5789 1 5689 |
5| 46 4569 1 | 2 34579 8 | 3579 367 3569 |
4| 8 7 59 | 156 3569 1356 | 359 2 4 |
|--------------------+--------------------+-------------------|
3| 5 234 8 | 9 467 267 | 1 3467 36 |
2| 47 39 6 | 1457 457 157 | 389 38 2 |
1| 247 1 2479 | 3 8 267 | 4579 467 569 |
*-------------------------------------------------------------*
|
3) c4=5==c4=9 - c1=9==b2=9 - b2=3==b3=3 - i3=3==i3=6 - i6=6==b6=6 - b6=2==c6=2 => bc6<>5, c6<>9
[Edit:] My first interpretation of this notation as Medusa coloring was not correct. The eliminations are valid anyway:
c1<>9 => b2=9 => b3=3 => i3=6 => i6<>6 => b6=6 => c6=2
So bc6 cant be 5 (=> c4=9 => c1<>9) and c6 cant be 9.
Written as AIC:
(5=9)c4-(9)c1=(9-3)b2=(3)b3-(3=6)i3-(6)i6=(6)b6-(2)b6=(2)c6, i.e. c4=5 or c6=2
Last edited by ravel on Thu Mar 20, 2008 5:54 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 Mar 20, 2008 5:54 pm Post subject: |
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ttt's (3) is what we call an AIC here. The 'official' blog of the site (which is very good) uses different notation from that of ttt. S/he would write 5c4 = 9c4 - 9c1 = ..... = 2c6, summarise it as 5c4 = 2c6, and call it a forbidding chain - a series of alternating strong and weak links, starting and finishing with strong.
The AIC / forbidding chain essentially says that if c4 <> 5 then c6 =2 and if c6 <> 2 then c4 = 5, from which you can work out the eliminations. But finding the chain in the first place is no easy matter - certainly beyond me. The current experts writing in this forum would probably use some sort of extended Medusa (3-D) colouring to find it. Too hard for me!
Curious to know whether you've done other Toughs. I've tried a few, and found them of very mixed standard - the first couple I tried happened to be quite easy, but after that they were out of my range & I gave up on them.
(It doesn't help much, but I did find a pair of ALS that eliminated 3 from g2 and 6 from i1. Couldn't find anything else of my own.) |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Mar 20, 2008 9:10 pm Post subject: |
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| Quote: | | Curious to know whether you've done other Toughs. I've tried a few, and found them of very mixed standard - the first couple I tried happened to be quite easy, but after that they were out of my range & I gave up on them. |
I know you weren't asking me, but I've done them on and off for awhile. I just started on them again a few weeks ago. They seem to be typified by a small number of clues and a large number of cells with 4–5 candidates. But some of them work out fairly easily and others are more difficult. The one in this thread is the hardest one I've seen. |
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Kdelle
Joined: 20 Mar 2008
Posts: 59
Location: Hudson, NH
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| Posted: Fri Mar 21, 2008 9:54 pm Post subject: |
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| Victor wrote: | ttt's (3) is what we call an AIC here. The 'official' blog of the site (which is very good) uses different notation from that of ttt. S/he would write 5c4 = 9c4 - 9c1 = ..... = 2c6, summarise it as 5c4 = 2c6, and call it a forbidding chain - a series of alternating strong and weak links, starting and finishing with strong.
The AIC / forbidding chain essentially says that if c4 <> 5 then c6 =2 and if c6 <> 2 then c4 = 5, from which you can work out the eliminations. But finding the chain in the first place is no easy matter - certainly beyond me. The current experts writing in this forum would probably use some sort of extended Medusa (3-D) colouring to find it. Too hard for me!
Curious to know whether you've done other Toughs. I've tried a few, and found them of very mixed standard - the first couple I tried happened to be quite easy, but after that they were out of my range & I gave up on them.
(It doesn't help much, but I did find a pair of ALS that eliminated 3 from g2 and 6 from i1. Couldn't find anything else of my own.) |
Actually I look at other sites if the day's puzzle is easy or medium....I'm only just starting to understand some of the more advanced solving techniques. The reason I posted here is that I thought that this particular puzzle seemed very difficult for me, but many of the comments on the other site indicated that it was relatively easy for quite a few people and I thought I must be missing something obvious. Needless to say, I was at least reassured that the puzzle was difficult. I'm still having trouble "seeing" the proof, but I will keep it on my desk and continue to look at it...I really hate to give up without understanding the solution.
Thanks for everyone's input. |
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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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| Posted: Thu Mar 27, 2008 11:36 pm Post subject: |
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Marty: forgot to say thanks. I ran off three more, and though 2 are in my 'can't get any further at present' pile I did manage to solve one. As I recall it was March 21st (this year), and quite a nice puzzle.
Kdelle: good luck with your sudoku learning stuff. I realise, glancing through what I've written, that I may have given the impression that Ravel is not one of the current experts. That's not at all true! If you can tap into the mindset you'll be doing well. (I haven't a clue how he figured out how to do it - wish I did!) |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Mon Mar 31, 2008 6:46 am Post subject: |
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I've been out of commission for a while... traveling and dealing with urgent stuff. So, I was catching up on board activity and looking for something with which to grapple. This puzzle is certainly a challenge.
First, in the grid posted, there is a Sue de Coq that no one has mentioned:
| Code: | +--------------------+--------------------+--------------------+
| 2467 2468 247 | 678 1 9 | 348 5 38 |
| 9 4568 457 | 5678 3567 3567 | 2 @48 1 |
| 1 58 3 | 58 2 4 | 6 9 7 |
+--------------------+--------------------+--------------------+
| 3 24569 2459 | 457 4579 57 | 5789 1 5689 |
| 46 4569 1 | 2 34579 8 | 3579 #36-7 3569 |
| 8 7 59 | 156 3569 1356 | 359 2 4 |
+--------------------+--------------------+--------------------+
| 5 234 8 | 9 467 267 | 1 @3467 @36 |
| 47 39 6 | 1457 457 157 |#-389 @38 2 |
| 247 1 2479 | 3 8 267 |#45-79 @467 #5-69 |
+--------------------+--------------------+--------------------+ |
The 5 digits {34678} are confined to the 5 cells marked @ in a way that forms overlapping locked sets, plus a conjugate pair of <7>s. There can be no {478} outside the @ cells in c8 (removing <7> from r5c8), no {36} outside the @ cells in b9 (removing <3> from r8c7 and <6> from r9c9) and no {7} outside the @ cells in b9 (removing <7> from r9c7).
This Sue de Coq doesn't get us very far, but is nice nonetheless.
After this, I turned to Medusa multi-coloring. It was a bit of a slog. Starting with 3 clusters:
| Code: | +--------------------+--------------------+-------------------+
| 2467 2468 247 | 678 1 9 | 3B4A8 5 3b8B |
| 9 4568 457 | 5678 3567 3567 | 2 4a8A 1 |
| 1 58 3 | 58 2 4 | 6 9 7 |
+--------------------+--------------------+-------------------+
| 3 24569 2459 | 457 4579 57 | 578B9 1 -568b9 |
| 46 4569 1 | 2 34579 8 | 3579 36 3569 |
| 8 7 59 | 156 3569 1356 | 359 2 4 |
+--------------------+--------------------+-------------------+
| 5 23A4 8 | 9 467 267 | 1 3467 36 |
| 47 3a9A 6 | 1457 457 157 | 8A9a 3A8a 2 |
| 247 1 2479a | 3 8 267 | 4a5c9 467 5C9c |
+--------------------+--------------------+-------------------+
Strong pairs: ab AC
bC |
<5> is removed from r4c9 by a bC trap. Next, I added 4 more clusters:
| Code: | +----------------------+--------------------+-------------------+
| 2467 2468 247 | 678 1 9 | 3B4A8 5 3b8B |
| 9 4568 457 | 5678 3567 3567 | 2 4a8A 1 |
| 1 58 3 | 58 2 4 | 6 9 7 |
+----------------------+--------------------+-------------------+
| 3 2E456G9 2e45-9 | 457 4579 57 | 578B9 1 6g8b9 |
| 46 4569 1 | 2 34579 8 | 3579 36 35c69 |
| 8 7 59 | 156 3569 1356 | 359 2 4 |
+----------------------+--------------------+-------------------+
| 5 2D3A4 8 | 9 467 2d67 | 1 3467 3F6f |
| 47 3a9A 6 | 1457 457 157 | 8A9a 3A8a 2 |
| 247 1 -2479a | 3 8 2D67 | 4a5c9 467 5C9c |
+----------------------+--------------------+-------------------+
Strong pairs: ab AC ad de af Bf BG eg
bC Cd Cf CG Be
Ce aG ae |
<9> is removed from r4c3 and <2> is removed from r9c3, both by ae traps. This allows some cluster expansion and another ae trap that removes <5> from r4c3:
| Code: | +----------------------+--------------------+-------------------+
| 2D467 2468 2E47 | 678 1 9 | 3B4A8 5 3b8B |
| 9 4568 457 | 5678 3567 3567 | 2 4a8A 1 |
| 1 58 3 | 58 2 4 | 6 9 7 |
+----------------------+--------------------+-------------------+
| 3 2E456G9 2e4-5 | 457 4579 57 | 578B9 1 6g8b9 |
| 46 4569 1 | 2 34579 8 | 3579 36 35c69 |
| 8 7 5a9A | 156 3569 1356 | 35-9 2 4 |
+----------------------+--------------------+-------------------+
| 5 2D3A4 8 | 9 467 2d67 | 1 3467 3F6f |
| 47 3a9A 6 | 1457 457 157 | 8A9a 3A8a 2 |
| 2d47 1 479a | 3 8 2D67 | 4a5c9 467 5C9c |
+----------------------+--------------------+-------------------+ |
The Aa coloring is expanded and leads to an Aa trap of <4> in r2c3:
| Code: | +----------------------+--------------------+-------------------+
| 2D467 2468 2E47 | 678 1 9 | 3B4A8 5 3b8B |
| 9 4568 -45A7 | 5678 3567 3567 | 2 4a8A 1 |
| 1 58 3 | 58 2 4 | 6 9 7 |
+----------------------+--------------------+-------------------+
| 3 2E456G9 2e4 | 457 4579 57 | 578B9 1 6g8b9 |
| 46 4569 1 | 2 34579 8 | 3579 36 35c69 |
| 8 7 5a9A | 156 3569a 1356 | 35 2 4 |
+----------------------+--------------------+-------------------+
| 5 2D3A4 8 | 9 467 2d67 | 1 3467 3F6f |
| 47 3a9A 6 | 1457 457 157 | 8A9a 3A8a 2 |
| 2d47 1 479a | 3 8 2D67 | 4a5c9 467 5C9c |
+----------------------+--------------------+-------------------+ |
Next, I add a tiny Hh cluster:
| Code: | +----------------------+--------------------+-------------------+
| 2D467 2468 2E47 | 678 1 9 | 3B4A8 5 3b8B |
| 9 4A568 5A7a | 5678 3567 3567 | 2 4a8A 1 |
| 1 58 3 | 58 2 4 | 6 9 7 |
+----------------------+--------------------+-------------------+
| 3 2E456G9 2e4 | 457 4579 57 | 578B9 1 6g8b9 |
| 46 4569 1 | 2 34579 8 | 3579 -36 35c69 |
| 8 7 5a9A | 156 3569 1356 | 3H5h 2 4 |
+----------------------+--------------------+-------------------+
| 5 2D3A4 8 | 9 467 2d67 | 1 3467 3F6f |
| 47 3a9A 6 | 1457 457 157 | 8A9a 3A8a 2 |
| 2d47 1 479a | 3 8 2D67 | 4a5c9 467 5C9c |
+----------------------+--------------------+-------------------+
New strong pair: AH |
This creates an AH trap of <3> in r5c8 and there is some significant simplification. Also, Hh become bB:
| Code: | +----------------+--------------------+-------------------+
| 6 2 4a7A | 7a8A 1 9 | 3B4A8 5 3b8B |
| 9 4A58 5A7a | 5678 356 357 | 2 4a8A 1 |
| 1 58 3 | 58 2 4 | 6 9 7 |
+----------------+--------------------+-------------------+
| 3 6 2 | 457 459 57 | 58B9 1 8b9B |
| 4 5A9a 1 | 2 3b59 8 | 7 6 3B5c9 |
| 8 7 5a9A | 156 3569a 135 | 3b5B 2 4 |
+----------------+--------------------+-------------------+
| 5 3A4a 8 | 9 7 2 | 1 3a4A 6 |
| 7 3a9A 6 | 145 45 15 | 8A9a 3A8a 2 |
| 2 1 4A9a | 3 8 6 | 4a5c9 7 5C9c |
+----------------+--------------------+-------------------+
Strong pairs: AC bC ab Ab => b is true and B is false. |
This is the Medusa Multi-Coloring form of a "color wrap." The conjugate color pair A and a each have a strong inference with b. Since one of A or a must be false, b MUST be true (and its conjugate, B, must be false).
The grid is now:
| Code: | +------------------+------------------+------------------+
| 6 2 47 | 78 1 9 | 48 5 3 |
| 9 458 57 | 5678 56 3 | 2 48 1 |
| 1 58 3 | 58 2 4 | 6 9 7 |
+------------------+------------------+------------------+
| 3 6 2 | 45 459 7 | 59 1 8 |
| 4 59 1 | 2 3 8 | 7 6 59 |
| 8 7 59 | 156 569 15 | 3 2 4 |
+------------------+------------------+------------------+
| 5 34 8 | 9 7 2 | 1 34 6 |
| 7 39 6 | 145 45 15 | 89 38 2 |
| 2 1 49 | 3 8 6 | 459 7 59 |
+------------------+------------------+------------------+ |
There are various URs one could have fun with. However, an XY chain does the trick:
{59}r9c9-{95}r5c9-{59}r5c2-{93}r8c2-{38}r8c8-{89}r8c7-{95}r4c7
This removes <5> from r9c7 and solves the puzzle. |
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