The New Sudoku Players' Forum

[Yogi]

Post.png (24.04 KiB) Viewed 1014 times

....1.5..2.86597..15.....297.5...2.6.2.8.5.4...1.6.9.5.1293..5...9...4.2....2..9.

The box analysis of this puzzle gave some possibilities for single-digit eliminations but nothing came up until I tried candidate 7 which gave conjugate pairs in rows 5 & 7, which formed a skyscraper that eliminated 7 from r6c6 and actually solved r8c5 = 8. But wait . . . There's more!
Now, there are two new 7CPs in row6 and column5, and they form a kite. Maybe that could produce some more eliminations? No. In fact, I don’t recall ever finding further single-digit eliminations flowing on from such newly-discovered conjugate pairs. Is there a principle at work here which means that this will in fact never happen?
I think SpAce mentioned something about this once but I can’t find the post the comment was made in.
Anyway, I’d be interested in any further comments on this issue.

[SpAce]

Hi Yogi,

First I want to thank you again for introducing me to Keith's box analysis technique. It has been very helpful in quickly determining which digits have single-digit potential and which can be ruled out immediately. Sorry I ever doubted that! With a bit of practice the lack of potential can be seen with a glance if you have the digit's candidates filtered. It has become a standard procedure in my solving because it avoids many unnecessary checks.

The box analysis of this puzzle gave some possibilities for single-digit eliminations

Which digits did your box analysis flag for single-digit potential? I could rule out 1,2,5,9 for good, which left 3,4,6,7,8 as potentials. Does your procedure agree with that? Out of those 3,6,7 actually have immediate single-digit eliminations available, though it's not very easy to see for the 3s. 4,8 don't, but could have later if certain eliminations were made.

but nothing came up until I tried candidate 7 which gave conjugate pairs in rows 5 & 7, which formed a skyscraper that eliminated 7 from r6c6 and actually solved r8c5 = 8.

Yes, 7 has the simplest single-digit eliminations. Your Skyscraper gets them both. You could also get them with two Kites or a Kite and an Empty Rectangle, but the Skyscraper is obviously better.

Note that there's also a Finned X-Wing (aka Grouped Skyscraper) with digit 6, but that's not quite as easy to spot.

The most difficult single-digit elimination is with the 3s which have several Finned Franken or Mutant Jellyfishes available (all producing the same elimination, though). It's probably not realistic to spot them as fishes for most manual players (certainly not for me), but it's not that hard as a grouped chain. At least it's a much more natural approach for me. Either way, it solves the puzzle.

solved with the 3s: Show

Code: Select all

.---------------------.------------------.-----------------------.
| 3469   34679  d3467 | 2347  1    23478 |  5       6-3     348  |
| 2     e34      8    | 6     5    9     |  7     f(3)1   f(3)14 |
| 1      5      d3467 | 347   478  3478  |  368     2       9    |
:---------------------+------------------+-----------------------:
| 7      3489    5    | 134   49   134   |  2     a[3]8     6    |
| 369    2      c36   | 8     79   5     | b13      4      b137  |
| 348    348     1    | 2347  6    2347  |  9     a[3]78    5    |
:---------------------+------------------+-----------------------:
| 468    1       2    | 9     3    4678  |  68      5       78   |
| 3568   3678    9    | 157   78   1678  |  4     a[3]167   2    |
| 34568  34678  c3467 | 1457  2    14678 | b1368    9      b1378 |
'---------------------'------------------'-----------------------'

(3)r468c8 = r5c79&r9c79 - r59c3 = r13c3 - r2c2 = (3)r2c89 => -3 r1c8; stte

...or using a couple of almost Franken X-Wings to help:

(3)r468c8 = B69\r59 - r59c3 = (3)B12\r13 => -3 r1c8; stte

...or as a single Finned Franken Jellyfish:

(3)C3B269\r1359c8 => -3 r1c8; stte

But wait . . . There's more!
Now, there are two new 7CPs in row6 and column5, and they form a kite. Maybe that could produce some more eliminations? No. In fact, I don’t recall ever finding further single-digit eliminations flowing on from such newly-discovered conjugate pairs. Is there a principle at work here which means that this will in fact never happen?

I can't say anything definite about that, but you're probably on the right track. It seems logical that executing single-digit eliminations doesn't make new ones immediately available (I guess it could simplify some, though, but not sure about that either). Template analysis would tell exactly what single-digit eliminations are possible for a certain digit at a certain puzzle state. Once you get them all, there shouldn't be anything left (for the moment). Therefore, if you can execute multiple single-digit eliminations in a row for the same digit, it probably means they were all available when you started instead of new ones being generated on the fly.

However, new single-digit eliminations can still become available later (unless all potential for that digit has been exhausted), but that requires eliminations through other techniques first. Here the 7s haven't lost their overall single-digit potential, even though they have no further immediate eliminations available. Box analysis would agree with that assessment.

I think SpAce mentioned something about this once but I can’t find the post the comment was made in.

Unfortunately I can't remember what discussion you might be referring to.

[pjb]

Hi Yogi

Have you ever looked at patterns? Certainly a single digit method. Not easy for hand solving, but then how many advanced methods are? In this case no valid pattern for 3 includes the 3 at r1c8!

Otherwise, avoiding complex one steppers, I like the continuous chain: (6)r5c3 - r3c3 = r3c7 - (6=8)r7c7 - (8=7)r7c9 - r5c9 = (7-9)r5c5 = (9-6)r5c1 = r5c3 => -6 r1c3, -6 r9c3, -6 r9c7, -8 r7c16, -8 r9c79, -7 r9c9, -3 r5c1, followed by the grouped continuous chain: (4=7)r9c3 - r13c3 = (7-6)r1c2 = r89c2 - (6=4)r7c1 - (4=7)r9c3 => -6 r89c1, -4 r9c1, -349 r1c2; stte

Phil

[SpAce]

Hi Phil,

Have you ever looked at patterns?

I guess you mean POM aka Templates? Everything in sudoku is a "pattern", so that's a bit ambiguous name for a specific solving technique

Certainly a single digit method.

It's the ultimate single-digit method, of course. It reveals even eliminations and placements that the most complex fishes fail to find. That being said, I seriously doubt a manual solver like Yogi (or myself) would find it enjoyable. Feel free to convince us otherwise!

Not easy for hand solving,

That's probably an understatement. Or actually, the procedure is very easy to understand, but the amount of hard labor seems ridiculous for anyone to do manually. I've never even tried, nor probably will. Seems even less fun than manual full tagging (which I've tried exactly once). Both are powerful methods for software solvers, but I see almost zero applicability for manual solving, except for some very simple scenarios which are probably solvable otherwise more easily.

but then how many advanced methods are?

By definition, advanced methods aren't supposed to be easy. However, almost all of them can be learned to do relatively efficiently, even on paper. I don't see how that's possible with POM. At this point in my sudoku "career" I'm probably aware of almost all of the advanced methods that exist, and have also applied most of them manually. I'm no longer aware of any method or pattern that I wouldn't understand (partly thanks to your awesome solver!). Some may be hard to learn, but few remain that hard or laborious once you know what you're doing. Most of them are fun to apply, especially once you figure out some ways that make it easier and more efficient (figuring them out is part of the fun too).

Unfortunately I can't imagine any fun in applying POM manually, because I can't really see how it could be done efficiently without software help. Besides, it kind of stinks of brute force. Maybe I just lack imagination or have misunderstood something. If you have any viable suggestions, I'm all ears. Seriously. It's one of the last well-known methods that I've never tried, for the above-mentioned reasons. I'd like to, but only if I'm convinced that there's a smart way to do it manually.

In this case no valid pattern for 3 includes the 3 at r1c8!

Yes, but that elimination is not that hard to find using normal, manually applicable, solving methods. Did you look at my hidden solution? I don't think that chain is awfully difficult at all. It's not for beginners, obviously, but it's certainly findable by many manual solvers here.

Some might even be able to spot it through a fish, though not I. However, the shape of the 3s pattern in boxes 6 and 9 does give a hint that some complex fishes might be present. I just rather use that information to look for a grouped chain which often -- but not always -- accompanies a complex fish and is typically much simpler to find for a manual player.

Can you show us how you'd use POM manually to find that elimination?

[Yogi]

BA1.png (5.46 KiB) Viewed 953 times

Whether the Box Analysis process is useful to the manual Sudoku solver depends (I think) on a factor that I call Critical Mass. It’s a nominal threshold which means that you are more likely to find single-digit eliminations if you are starting the process with at least 35 solved cells. Here we are starting with 34. That forecasts a lot of work for possibly little result, but as I could not find any more singles, off I went:
This diagram might have been drawn by someone new to this technique. The more experienced with it would not have written 8 in Box4 or 1 in Box8, because by looking ahead you can see that these candidates will not count in those boxes as there are no more of those candidates to work with them in the same chutes. So, as Space said, we get possibilities for candidates 3467 & 8. This is a relatively large number. You tend to get fewer possibilites and generally better results if you are starting with more solved cells.
Usually having no reason to favour any one candidate, I tend to work through them numerically. I found nothing that I could find a way to work with in 34 & 6, but when 7 solved r8c5 = 8 that was a breakthrough, although it still did not reduce the puzzle to solving by singles.

[SpAce]

Whether the Box Analysis process is useful to the manual Sudoku solver depends on...

...nothing. It's always useful, whether it produces immediate results or not. My process of using it is so light that it's automatic. I even do it with basic puzzles because it doesn't cost me anything. (Who says you can't solve a basic puzzle with an X-Wing or a Skyscraper? They're often easier to see than subsets, especially if solving without pencil marks. URs are even easier.)

The process is obviously even more useful if you know other single-digit techniques besides the simplest Turbot Fishes. Here the puzzle was solvable with a single-digit chain (or a fish) on 3s. An experienced player can look for those kinds of non-trivial openings for the digits that are known to have single-digit potential. It's not limited to "kite-hunting".

This diagram might have been drawn by someone new to this technique. The more experienced with it would not...

...need any diagrams. You're making the whole process seem much more complicated than it is (or needs to be). It's great that you've found ways to make it work for you, and I'm not criticizing your way at all, but the essence of the technique is much simpler. Otherwise I never would have adopted it. If solving on paper, all I have is a tiny bookkeeping row written in some empty space near the puzzle:

Code: Select all

xx  x   x
123456789

I mark an 'x' on top of a digit once I deem it incapable of single-digit eliminations. I do that while solving basics instead of a separate process, because Keith's simple rules make it very easy to see once a digit loses its potential. I don't even have to think about it. There's no need to know anything about the strong links for that, and even pencil marks are unnecessary.

I also mark another 'x' below a digit once it's completely solved. This simple bookkeeping tells me which digits are still in play, and which ones of those have potential for single-digit eliminations. Once basics are exhausted (or sometimes before that), the single-digit check is usually the second non-basic thing I do (after URs). Only then do I need to look at the strong links, and just for the relevant digits. That's a real time-saver, because it's the heaviest part of the process in paper solving. If solving with Hodoku, filters make it easy, and there's no need for the bookkeeping either.