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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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 Posted: Wed Oct 31, 2007 7:18 am Post subject: Oct 31 vh - surprisingly nice (at the end) |
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Nothing but (hidden, naked) singles/pairs up to here:
| Code: |
+--------------------------+--------------------------+--------------------------+
| 4 5 3 | 2 8 1 | 7 9 6 |
| 8 16 16 | 4 9 7 | 2 5 3 |
| 7 9 2 | 6 5 3 | 8 1 4 |
+--------------------------+--------------------------+--------------------------+
| 2 3 5 | 179 17 8 | 4 6 79 |
| 9 67 4 | 3 67 2 | 1 8 5 |
| 16 8 167 | 5 4 69 | 3 2 79 |
+--------------------------+--------------------------+--------------------------+
| 5 1267 167 | 17 1267 4 | 9 3 8 |
| 16 1267 8 | 179 3 69 | 5 4 12 |
| 3 4 9 | 8 12 5 | 6 7 12 |
+--------------------------+--------------------------+--------------------------+
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Looked like one of those very easy very hard ones.
Turned out to have swordfish, color wing, xy-wing, xyz-wing, you name it ...
or is there the "silver bullet"/"golden shot" that does it all at once?
happy hunting/fishing/winging !
nataraj |
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andras
Joined: 31 Oct 2007
Posts: 56
Location: Mid Wales
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| Posted: Wed Oct 31, 2007 10:34 am Post subject: |
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A bit round the houses, I found - an xyz, an x-wing, and a couple of xy wings.
Not too hard in the end though!
John |
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re'born
Joined: 28 Oct 2007
Posts: 80
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| Posted: Wed Oct 31, 2007 12:24 pm Post subject: Re: Oct 31 vh - surprisingly nice (at the end) |
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| nataraj wrote: |
or is there the "silver bullet"/"golden shot" that does it all at once? |
A little 3d-coloring does the job in one shot (note that I've suppressed superfluous coloring).
| Code: | .------------------.------------------.------------------.
| 4 5 3 | 2 8 1 | 7 9 6 |
| 8 16 16 | 4 9 7 | 2 5 3 |
| 7 9 2 | 6 5 3 | 8 1 4 |
:------------------+------------------+------------------:
| 2 3 5 | 179 17 8 | 4 6 79 |
| 9 67B 4 | 3 6B7b 2 | 1 8 5 |
| 16 8 167 | 5 4 6b9 | 3 2 79 |
:------------------+------------------+------------------:
| 5 1267 167 | 17 1267 4 | 9 3 8 |
| 16 1267a 8 | 17A9B 3 6B9b | 5 4 12 |
| 3 4 9 | 8 12 5 | 6 7 12 |
'------------------'------------------'------------------' |
As B sees both a and A, B must be false.
Here is the corresponding nice loop:
r8c4 =7= {r8c2 -7- r5c2 =7= r5c5 =6= r6c6 -6- r8c6} -9- r8c4, => r8c4 <> 9, solving the puzzle. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Wed Oct 31, 2007 2:18 pm Post subject: |
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I used a plethora of talents here. Two skyscrapers on 7s in R58. An X-wing on 6s in C16. An xy wing with target R2C3. Trouble is, I then noticed I'd missed a naked single (I have a problem with these) - so it may have been a lot easier.
Nonetheless the coup-de-grace was a BUG+1 that put a 1 in R8C4. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Wed Oct 31, 2007 3:43 pm Post subject: |
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| Quote: | ... the corresponding nice loop:
r8c4 =7= {r8c2 -7- r5c2 =7= r5c5 =6= r6c6 -6- r8c6} -9- r8c4, => r8c4 <> 9 |
is indeed very nice. One shot - boom.
May I take this opportunity to advance my meager knowledge of Eureka and try to re-write the loop? Somehow I think the weak and strong links should alternate. Let me start with the summary of the implication chain in (almost) natural language (basically I think this is what the curly brackets prove):
if r8c4<>7 then ... r8c6=9 (both ends of an AIC), either way r8c4<>9.
(9-7)r8c4=(7)r8c2-(7=6)r5c2-(6)r5c5=(6)r6c6-(6=9)r8c6-(9)r8c4
I've never used Eureka notation before so I would appreciate all comments. Thx. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed Oct 31, 2007 3:50 pm Post subject: |
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| Code: | .------------------.------------------.------------------.
| 4 5 3 | 2 8 1 | 7 9 6 |
| 8 16 16 | 4 9 7 | 2 5 3 |
| 7 9 2 | 6 5 3 | 8 1 4 |
:------------------+------------------+------------------:
| 2 3 5 | 179 17 8 | 4 6 79 |
| 9 67B 4 | 3 6B7b 2 | 1 8 5 |
| 16 8 167 | 5 4 6b9 | 3 2 79 |
:------------------+------------------+------------------:
| 5 1267 167 | 17 1267 4 | 9 3 8 |
| 16 1267a 8 | 17A9B 3 6B9b | 5 4 12 |
| 3 4 9 | 8 12 5 | 6 7 12 |
'------------------'------------------'------------------' |
I don't recall if my grid looked exactly like this. But the W-Wing in boxes 1 and 7 zaps the 6 in r7c3, creating a 17 pair in row 7. That exposed an XY-Wing on 16-26-12 which finished things off. |
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duffy
Joined: 13 Sep 2007
Posts: 26
Location: Toronto Canada
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| Posted: Wed Oct 31, 2007 5:47 pm Post subject: |
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Having reached the following BUG+1 position that cgordon notes, using that x-wing on 6's and an only slightly different xy-wing, I still cannot understand a proof I read about a BUG having no solution or at least two. (In this case the BUG is there if we remove the solution's 1 as a candidate for r8c4.)
| Code: |
+----------+-----------+--------+
| 4 5 3 | 2 8 1 | 7 9 6 |
| 8 1 6 | 4 9 7 | 2 5 3 |
| 7 9 2 | 6 5 3 | 8 1 4 |
+----------+-----------+--------+
| 2 3 5 | 19 17 8 | 4 6 79 |
| 9 67 4 | 3 67 2 | 1 8 5 |
| 16 8 17 | 5 4 69 | 3 2 79 |
+----------+-----------+--------+
| 5 26 17 | 17 26 4 | 9 3 8 |
| 16 27 8 | 179 3 69 | 5 4 12 |
| 3 4 9 | 8 12 5 | 6 7 12 |
+----------+-----------+--------+
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Play this puzzle online at the Daily Sudoku site
Marty's xy-wing does solve the puzzle from here, however. (I noted instead that a 6 in r7c2 would force in impossible condition with 1 and 2 in the next row, but I didn't like that either as a clean solution.)
I have two questions:
1) Has anyone seen a clear proof for the BUG that I could accept?
2) For a p&p guy like me, colouring does not seem like a good idea. Or am I misunderstanding the concept?
Don D. |
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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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| Posted: Wed Oct 31, 2007 6:04 pm Post subject: Oct31 VH |
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Trying to use the most basic techniques:
an x-wing in 6's removes the 6 from R6C3 and R8C2, then an xy-wing removes the 1 from R2C3, and finally another xy-wing removes the 2 from R8C2, leaving only singles.
Earl |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Wed Oct 31, 2007 7:07 pm Post subject: |
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| Quote: | | I still cannot understand a proof I read about a BUG having no solution or at least two. (In this case the BUG is there if we remove the solution's 1 as a candidate for r8c4.) |
Duffy: I just do techniques by rote without getting into the rationale - all those strong links, weak links and bivalues give me a headache. So my understanding of BUGs+1 is that near the end of the puzzle when there are only pairs left plus ONE triple - look for the other two pairs that contain the same numbers as the triple. In your grid there's a 179, 17 and 19 in col 4. LOOK FOR THE NUMBER THAT OCCURS THREE TIMES(here it's a 1) then SHOVE IT IN THE TRIPLE CELL. (You don't remove the 1 from the triple cell as you stated because then you can't solve the puzzle and Sudoku.com promised us we could).
| Quote: | | For a p&p guy like me, colouring does not seem like a good idea. Or am I misunderstanding the concept? |
I agree - I think colouring is just a fancy way of trail and error. Though I'll probably be told to shove it also. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Wed Oct 31, 2007 8:54 pm Post subject: |
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Don, I'll try an answer to your questions:
| Quote: |
1) Has anyone seen a clear proof for the BUG that I could accept?
2) For a p&p guy like me, colouring does not seem like a good idea. Or am I misunderstanding the concept?
Question
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ad 1) "proof" is an elusive term. Any proof really depends on the frame of reference (the set of axioms plus the whole theory up to the point of the proof for a new theorem). And "I could accept" depends so much on the "I" that probably there is no answer. But - taking the heuristic approach - let me say it's worked for me and many others every time. If you are looking for a formal approach I think there are probably some people on this forum more familiar with the math literature on the subject of sudoku.
ad 2) I am solving all sudokus with p&p and (again with the cautious remark that "like me" is a bit ambiguous) I really like the idea of visualizing dependencies between possible candidates throughout the grid. True, I don't use real colors (I make little line drawings on the side of the sheet of paper) but the effect is the same: it becomes so much easier to spot those x-wings, swordfish, turbot fish, skyscrapers and longer inference chains. The terms "coloring" or "color wing" are really often a shorthand way of saying 'I used alternating inference chains of weak and strong links in "7"'. But I have read in one of the other sudoku forums that supposedly there are two types of solving strategies: pattern A and pattern B (coloring belonging to "pattern b" in that scheme) and they appeal to different people, maybe use different parts of the brain. That would explain why some methods feel "more natural" than others. |
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eddieg
Joined: 12 Jan 2006
Posts: 47
Location: San Diego, CA USA
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| Posted: Wed Oct 31, 2007 9:14 pm Post subject: For less advanced players |
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Hope this helps someone who is stuck, and doesn't want to ask for help.
| Code: |
+------------+-------------+--------+
| 4 5 3 | 2 8 1 | 7 9 6 |
| 8 16 16 | 4 9 7 | 2 5 3 |
| 7 9 2 | 6 5 3 | 8 1 4 |
+------------+-------------+--------+
| 2 3 5 | 179 17 8 | 4 6 79 |
| 9 67 4 | 3 67 2 | 1 8 5 |
| 16 8 17 | 5 4 69 | 3 2 79 |
+------------+-------------+--------+
| 5 267 167 | 17 1267 4 | 9 3 8 |
| 16 127 8 | 179 3 69 | 5 4 12 |
| 3 4 9 | 8 12 5 | 6 7 12 |
+------------+-------------+--------+
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Play this puzzle online at the Daily Sudoku site
I could not figure out this puzzle for some time, Was basically a little past where nataraj's sample puzzle above was. I persisted, and based on the moves below, finished the puzzle.
The moves from where nataraj was:
I removed the 6 from R6C3 using the W-Wing in columns 1 and 6
I removed the 1 from R7C2 using the XYZ-Wing involving 16-167-17
I removed the 6 from R2C2 using XY-Wing involving 67-17-16
R2C3 became 6
Used Naked pair in row 7 for elimination
Eliminated 7 from R4C4
Eliminate 1 in R8C4 because of BUG+1 |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Wed Oct 31, 2007 9:25 pm Post subject: |
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| Quote: | | a fancy way of trial and error |
Isn't the whole solving process one of trial and error?
I am sure most people are familiar with the "squeeze" method: I always start by looking at the first row. Try each given clue in turn and look for a second occurrence of that number in rows two and three. VERY effective method to quickly find hidden singles. But of course: pure "trial and error"
"hidden pairs", "naked pairs": very basic techniques. But surely we all scan the grid over and over and try diffent combinations of cells in a row:
"12",... "13" ah might be a triple.
"56" - no.
next: "13" - ah we found a naked pair ...
and so on. trial and error.
I have not yet seen a sudoku with a big x or an arrow saying "the single is hidden HERE".
Who can say they spot an xy wing immediately? I certainly cannot.
I try all bivalue (there's the bad word. - but those bi-values are our friends!) cells in turn
ah "23" - ok
where's the next one in the same row?
"59" no, useless,
"56" no, not good,
"39" hm, might be on to something, is there a "29" somewhere ?
and so on and so on.
So - yes. Coloring is a fancy way of trial and error. It saves a lot of trial and avoids a lot of error. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed Oct 31, 2007 9:31 pm Post subject: |
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| Quote: | | For a p&p guy like me, colouring does not seem like a good idea. Or am I misunderstanding the concept? |
I don't know what your understanding is, but I'm a paper-and-pencil guy and I've used coloring often, although probably not often enough.
| Quote: | | I agree - I think colouring is just a fancy way of trail and error. Though I'll probably be told to shove it also. |
OK, shove it!! 
I'm happy with coloring and don't view it as trial and error. It's another technique based on pattern recognition. To me, T&E is like a Forcing Chain where you plug a value into a cell and see what happens. But I'm sure different folks have their own idea of what constitutes T&E. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Wed Oct 31, 2007 10:34 pm Post subject: |
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Don,
I'll take a stab at giving a BUG explanation you might accept.
Let's take the grid above with the single trivalue cell {179} at R8C4. Per BUG+1, we must place the <1>. So, instead, let's remove it so that we have all bivalue cells with two instances of each candidate in each house. Then, let's do some coloring:
| Code: | +------------+--------------+--------+
| 4 5 3 | 2 8 1 | 7 9 6 |
| 8 1 6 | 4 9 7 | 2 5 3 |
| 7 9 2 | 6 5 3 | 8 1 4 |
+------------+--------------+--------+
| 2 3 5 | 1g9r 1r7g 8 | 4 6 79 |
| 9 67g 4 | 3 67r 2 | 1 8 5 |
| 16 8 17g | 5 4 69 | 3 2 79 |
+------------+--------------+--------+
| 5 26 17r | 17g 26 4 | 9 3 8 |
| 16 27 8 | 7r9g 3 69 | 5 4 12 |
| 3 4 9 | 8 12 5 | 6 7 12 |
+------------+--------------+--------+ |
In this case, we quickly find two "green" <7>s in B4. This means we either have 2 <7>s in B4 or zero <7>s in B4. Neither is acceptable, so the puzzle is invalid. That means R8C4 had to be <1>, since we can assume the puzzle has a valid solution.
Now, sometimes, we would be able to color in all the remaining candidates without finding such an invalid conflict. But, what would that mean?
Well, since all the remaining candidates would be colored and there are no invalid conflicts, we have two valid solutions: a red solution and a green solution. By uniqueness, this isn't valid either. So, again, R8C4 would have to be <1> to get a unique solution.
I hope this works for you. |
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re'born
Joined: 28 Oct 2007
Posts: 80
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| Posted: Wed Oct 31, 2007 10:34 pm Post subject: |
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| cgordon wrote: |
I agree - I think colouring is just a fancy way of trail and error. |
Coloring is no more trial and error than a naked pair. Both can be applied without ever assuming the truth or falsehood of any value in the pattern. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Thu Nov 01, 2007 12:07 am Post subject: |
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| Quote: | | Isn't the whole solving process one of trial and error? |
| Quote: | | Coloring is no more trial and error than a naked pair. |
| Quote: | | But I'm sure different folks have their own idea of what constitutes T&E. |
Sooo many philosophical arguments. I wish Harry Stotle or Des Cartes were still around. I'm sure they'd find truth from naked pairs and T&A. (Sorry T&E) |
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duffy
Joined: 13 Sep 2007
Posts: 26
Location: Toronto Canada
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| Posted: Thu Nov 01, 2007 1:59 pm Post subject: |
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Thanks for all the feedback! Clearly people have varying feelings about certain things that affect how we proceed with suduko--and with life.
As for colouring, eh (Canadian spelling), I must do some further reading, because solutions do depend on interdependencies that can be sketched or coloured. 
The BUG issue is a different matter for me, because i was trying to understand a generalized theorem proof presented by user "nick67" in the Suduko Players Forums in Nov 2005 under the curious heading, "Is there a simpler way to solve this?" 
Working by p&p, one may reach an apparent BUG+k situation (where k=1, 2,....), but because of human fallibility one needs to do careful checking to be sure that the pattern of candidates is legitimate--and for me this can prove to be more work than I am prepared to do, and other methods should be available to solve more cells and reduce the value of k. In the simplest situation one gets BUG+1, with a small number of unsolved cells, and then the solution is much more apparent. 
Don |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Thu Nov 01, 2007 9:07 pm Post subject: |
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| Quote: | | I wish Harry Stotle or Des Cartes... |
Good one, cgordon ! I'll ask Mo Dzar'd to jam with us ... |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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