Solving without pencilmarks

Advanced methods and approaches for solving Sudoku puzzles

Postby tarek » Tue Apr 25, 2006 11:14 pm

here is the next puzzle (from Ruud's #062 in the superior 100 list):
Code: Select all
 . . 6 | . 9 . | 3 . . 
 9 5 . | . 3 . | . 6 . 
 2 3 . | 5 6 1 | . 4 9 
-------+-------+------
 . . 9 | . . . | 2 . 3 
 7 . . | 3 5 . | . . 6 
 . . 3 | . . . | 4 . . 
-------+-------+------
 6 . . | 7 4 3 | . . 1 
 1 8 . | . 2 . | . 3 4 
 3 . 4 | 1 8 5 | 6 . . 


What is your next step following the x-wing(s)?

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Postby RW » Wed Apr 26, 2006 4:39 pm

Nice puzzle, maybe I should start working my way through the superior list!:)

Saw four immediate uniqueness reductions, but unfortunately none of them helped me, so I had to start looking for the x-wings you asked for. Found one that gave me a naked pair (18) in r2c37 => r1c4 or r1c6=8 => r46c1=58 => r1c1=4. So my last non-single move would have been the hidden pair, I usually spot them before naked singles.

Definitely harder than your first example, the x-wing(s) were not very hard to find, but it took me quite a while to realise that it revealed a naked pair. Once again proved: naked subsets (and singles) are hard in a non pencilmarkgrid.

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Postby tarek » Wed Apr 26, 2006 5:03 pm

Thanx RW for your great help.......

Indeed spotting Hidden doubles prior to spotting naked doubles is easier on the non PM grid......

But that wasn't what I was looking for.......

There was a hidden triple in the same row as the naked double, (247) in r2c469......

You saw the naked double before that hidden triple, what do you think of that !!!

Would you say that spotting naked doubles is easier than hidden triples on the non PM grid?

on the PM grid the hidden triple is a nightmare, but I was under the impression that it would be easier to spot on the non PM grid than naked subsets (I'm sure it is easier than a naked triple, but I wanted to see if it was easier to spot than the naked double)......

I don't have that many STABLE hidden triples with a naked double counterpart [& no uniqueness reductions] to see if that happens again.[& no uniqueness reductions]

Thanx RW for your help again.......

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Postby RW » Wed Apr 26, 2006 7:53 pm

tarek wrote:There was a hidden triple in the same row as the naked double, (247) in r2c469......

You saw the naked double before that hidden triple, what do you think of that !!!

Would you say that spotting naked doubles is easier than hidden triples on the non PM grid?


It very much depends. Inside a box I spot the hidden triplets very easily, definately before a naked double as I can do it by "drawing lines":

Code: Select all
...|...|.12
.5.|.46|7..
...|...|..3

or

...|..2|...  ...|..2|...
...|...|...  ...|...|...
...|...|...  ...|...|...
---+---+---  ---+---+---
...|..7|...  ...|...|...
1..|5.6|.23  1..|...|.23
...|4..|...  ...|4..|...
---+---+---  ---+---+---
...|...|...  ...|...|...
...|..1|...  ...|..1|...
...|..3|...  ...|..3|...


[edit: note that the hidden triplet is actually easier to see when it's got a naked quintuplet counterpart, not any extra information "blocking your vision"]

In rows and columns I search in a different way, even for hidden singles. I mentioned in the original post how I count through the row and find the missing numbers, then I compare each cell to this set of numbers. In row 2 of your example I found that the missing numbers were 12478, then I compared each cell, c3 could see 247, c4-17, c6-1, c8-247. Then I didn't need to check the last cell, I knew that if two cells can see three out of five missing numbers, there must be a naked pair. Then I rechecked the remaining cells (c469) against the remaining three numbers only (to find possible singles). When I do it like this I think there isn't any difference if the set is naked or hidden, I'll find it anyway .:)

This is of course only how I do it, other people may experience it in another way, but the order I spot things is something like:

-Hidden singles within boxes
-Locked candidates
-Hidden pairs/triplets within boxes
-UR
-naked singles
-subsets in rows/columns

I usually don't look for x-wings before I scanned all of the above, I think the puzzle you posted would have required a lot more time if you hadn't mentioned the x-wing.

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Postby tarek » Wed Apr 26, 2006 11:09 pm

Interesting RW, I have been ranking the superior puzzles with a bias towards the hidden triple, the naked triple would receive 1/3 of what the hidden triple receives. From what your saying, the naked triple in the superior is the ultimate challenge........

As the superior list contains too many high level superiors, it is difficult to have a pure consistan naked triple with no counterpart.....

The alternative would be some mix & match with hidden triples & doubles:

I would recommend vidarino's #007 for a nice treat.....

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Postby Heuresement » Thu Apr 27, 2006 9:38 pm

You may be interested to know that this last puzzle (Ruud's #062) posed me a real problem, and I couldn't make any headway for a long time. I couldn't spot any URs, but did find two x-wings (1s and 7s) which finally opened up the puzzle. I sometimes find spotting x-wings quite difficult with or without pencilmarks, but these two were quite straightforward to spot considering the cells already completed.

In my experience, in general, spotting triples using locked candidates is often easier than spotting doubles, as there a less intermediate facts that you need to remember along the way.

Unfortunately this week, I have not had much time to spend on Sudoku, but I will try your next suggestion (vidarino's #007) when I have the opportunity, and add some comments if necessary. (I ration myself to 1 or 2 fiendish puzzles per day)
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Postby RW » Fri Apr 28, 2006 9:16 am

Heuresement wrote:I couldn't spot any URs, ...


They are not the easiest ones, but there is a few:

Code: Select all
 . . 6 | . 9 . | 3 . . 
 9 5 . | . 3 . | . 6 . 
 2 3 . | 5 6 1 | . 4 9 
-------+-------+------
 . . 9 | . . . | 2 . 3 
 7 . . | 3 5 . | . . 6 
 . . 3 | . . . | 4 . . 
-------+-------+------
 6 . . | 7 4 3 | . . 1 
 1 8 . | . 2 . | . 3 4 
 3 . 4 | 1 8 5 | 6 . . 


-UR in r46c58 => r6c8<>1
-UR in r68c46 => r6c6<>9
-UR in r79c28 => r7c8<>2

[edit: typo, 2nd should be r6c6<>6]

I also removed two candidates like this, not a direct UR:
If r6c2=6 => r6c2<>2 => r5c23=2 => r5c6<>2
If r4c2= 6 => (AUR in r68c46) => r5c6=9 => r5c6<>2 => r5c23=2=> r6c2<>2

That was a short course in UR:s!:) Lots of fun to find these!

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Last edited by RW on Wed May 03, 2006 3:35 am, edited 1 time in total.
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Postby Heuresement » Tue May 02, 2006 7:25 am

RW Thanks for the tips!:D
I can see that I haven't fully got to grips with this the partial URs. When I have been able to find the time, I have been looking at your examples, and following the logic of the UR deductions. Unfortunately, the logic is not always immediately apparent, but as you have previously said, this is perhaps an advantage because it is thus more likely that the puzzle creator (man or machine) has also not seen the URs, and thus could, if spotted, lead to shortcuts in the solving process. I am slightly unclear about one of the deductions but hopefully I will understand it when I am able to take another look.:)
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Postby RW » Tue May 02, 2006 12:57 pm

Heuresement wrote:I have been looking at your examples, and following the logic of the UR deductions. Unfortunately, the logic is not always immediately apparent


I know these can be difficult to find, but you'll get used to it. Earlier I had to think these over several times before making any conclusions, but I've searched for them so much by now that they become more and more obvious all the time. In all patterns that I refer to as UR, the reduction can be found by only considering the involved cells. The implication would let us solve all other corners as singles (example 1: if r6c8=1 => r6c5=7 => r4c5=1 => r4c8=7). I believe uniqueness rectangles like these have been discussed here. I'm not sure if these particular variations have been mentioned there, I haven't really studied that thread. I think learning which strong links let us make reductions isn't very relevant when solving by hand, if you can see the effect your implication has on the three other cornercells you can cover all of the examples from that thread + any possibilities they might have missed. Only computers need to be told exactly what to do in specific situations.:)

Heuresement wrote:I am slightly unclear about one of the deductions but hopefully I will understand it when I am able to take another look.


In my way of thinking there's three different reasons that may let us make uniqueness reductions:

-At least one of the cells in the possible deadly pattern must hold a number that is not included in the pattern.

-At least one of the numbers involved in the pattern must be placed somewhere outside the pattern.

-If any implication causes a deadly pattern to form, that implication is false.

My first three deductions was made because of the third reason. The last one (AUR => r5c6=9) can be derived from the second reason. Hope this helped.

regards
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Postby ronk » Tue May 02, 2006 1:55 pm

RW wrote:example 1: if r6c8=1 => r6c5=7 => r4c5=1 => r4c8=7
How did you eliminate the possiblity of r4c9=7?

RW wrote:-UR in r46c58 => r6c8<>1
-UR in r68c46 => r6c6<>9
-UR in r79c28 => r7c8<>2
The 2nd should be r6c6<>6.
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Postby RW » Wed May 03, 2006 7:34 am

ronk wrote:How did you eliminate the possiblity of r4c9=7?


In the posted grid r4c9=3...

ronk wrote:The 2nd should be r6c6<>6


Thanks, I'll fix the typo.

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Postby ronk » Wed May 03, 2006 11:10 am

RW wrote:
ronk wrote:How did you eliminate the possiblity of r4c9=7?
In the posted grid r4c9=3...
Oops. I meant to write r6c9=7 ... but I started the potential "contradiction elimination" at the wrong corner (r4c8) anyway.
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Postby Heuresement » Wed May 03, 2006 12:50 pm

Thanks RW. I believe I understand now completely the logic.

Though I had no problem in identifying regular URs, in these partial UR situations, I had wrongly assumed that I needed to be sure that the UR would still exist if I swapped around the deadly pair. I now realise that it is much simpler than that. All I need to do is that by picking one of the deadly pair on one corner, I can force another two corners, and then I cannot have the original number in the last corner. Really simple after all.

Thank you for taking the time on these explanations.

Onwards, upwards, more logic and more solving.:)
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Keep it going

Postby Jumper » Sun May 07, 2006 6:03 pm

Thanks, RW, for your contributions. I have been steadfastly solving without marking since I began, and your incisiveness has assisted me, most especially this morning as I worked out a puzzle from several days ago. I will incorporate more of your ideas in my noteless methods.

I propose we submit to others not only our most difficult but solvable puzzles, but also some that are challenging, yes, but also fun.

Having said that, I am chagrined that I have not a challenging but fun puzzle, to be solved without markings, to submit to you all today!

I will promise to post the next one I find that strikes me as quite enjoyable. Anyone else have one now?
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Postby RW » Mon May 08, 2006 7:44 pm

Thanks Jumper, I'm glad if I've been able to help.

Jumper wrote:I propose we submit to others not only our most difficult but solvable puzzles, but also some that are challenging, yes, but also fun.


Fun puzzles are always welcome. I haven't really had time to solve that many puzzles lately, so I don't have any up my sleeve right now. When there is more time I'll probably go through the superior thread, I believe there's lots of fun and challenging puzzles there.

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